Solution of the System 10x+12y1 25x-8y+8

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Copyright © by The McGraw-Hill Companies, Inc. All rights reserved. Except as permitted under the United States Copyright Act, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without prior permission of the publisher. Send all inquiries to: Glencoe/McGraw-Hill 8787 Orion Place Columbus, OH 43240-4027 ISBN: 0-02-834177-5 Printed in the United States of America. 4 5 6 7 8 9 10 024 08 07 06 05 04  Chapter 1 Linear Relations and Functions 8.  Relations and Functions  1-1  Pages 8–9  Check for Understanding  1. x 
4 6 0 8 2 
4  y  y 2 1 5 
4 2 0  4 2 
8 
4 O 
2  4  8x  y  x  3. Determine whether a vertical line can be drawn through the graph so that it passes through more than one point on the graph. Since it does, the graph does not represent a function. 4. Keisha is correct. Since a function can be expressed as a set of ordered pairs, a function is always a relation. However, in a function, there is exactly one y-value for each x-value. Not all relations have this constraint. 5. Table: Graph: x 1 2 3 4 5 6 7  y 
3 
2 
1 0 1 2 3  x 1 2 3 4 5 6 7 8  y 
5 
5 
5 
5 
5 
5 
5 
5  y 12 8 4 
4  O  2  4x  
4  y  O  x  10. {
3, 0, 1, 2}; {
6, 0, 2, 4}; yes; Each member of the domain is matched with exactly one member of the range. 11. {
3, 3, 6}; {
6, 
2, 0, 4}; no; 6 is matched with two members of the range. 12a. domain: all reals; range: all reals 12b. Yes; the graph passes vertical line test. 13. f(
3)  4(
3)3  (
3)2 
 5(
3)  
108  9  15 or 
84 14. g(m  1)  2(m  1)2 
 4(m  1)  2  2(m2  2m  1) 
 4m 
 4  2  2m2  4m  2 
 4m 
 4  2  2m2 15. x  1  0 x  
1 The domain excludes numbers less than 
1. The domain is {xx  
1}. 16a. {(83, 240), (81, 220), (82, 245), (78, 200), (83, 255), (73, 200), (80, 215), (77, 210), (78, 190), (73, 180), (86, 300), (77, 220), (82, 260)}; {73, 77, 78, 80, 81, 82, 83, 86}; {180, 190, 200, 210, 215, 220, 240, 245, 255, 260, 300}  y  O  y 
7 
4 
1 2 5 8 11 14 17  9.  
4  2. Sample answer:  O  x 
4 
3 
2 
1 0 1 2 3 4  x  Equation: y  x 
4 6. {(
3, 4), (0, 0), (3,
4), (6, 
8)}; D  {
3, 0, 3, 6}; R  {
8, 
4, 0, 4} 7. {(
6, 1), (
4, 0), (
2, 
4), (1, 3), (4, 3)}; D  {
6, 
4, 
2, 1, 4}; R  {
4, 0, 1, 3}  1  Chapter 1  16b.  22. {(4, 0), (5, 1), (8, 0), (13, 1)}; D  {4, 5, 8, 13}; R  {0, 1} 23. {(
3, 
2), (
1, 1), (0, 0), (1, 1)}; D  {
3, 
1, 0, 1}; R  {
2, 0, 1} 24. {(
5, 5), (
3, 3), (
1, 1), (2, 
2), (4, 
4)}; D  {
5, 
3, 
1, 2, 4}; R  {
4, 
2, 1, 3, 5} 25. {(3, 
4), (3, 
2), (3, 0), (3, 1), (3, 3)}; D  {3}; R  {
4, 
2, 0, 1, 3} y 26. O x y  300 280 260 Weight 240 (lb) 220 200 180  O  70 72 74 76 78 80 82 84 86 Height (in.)  16c. No; a vertical line at x  77, x  78, x  82, or x  83 would pass through two points.  Pages 10–12  Exercises  17. Table x 1 2 3 4 5 6 7 8 9  y 24 18 12 6  O  2  4  6  10 x  8  x 
4 
3 
2 
1 0 1 2 3 4  y O  y 
11 
10 
9 
8 
7 
6  Equation: y  x 
 5 19. Table: y 4 5 6 7 8 9 10 11 12  y 
1 
2 
3 
4 
5 
6  x 
5 
4 
3 
2 
1 0 1  x  y x  O  y  y 5 4 3 2 1 0 1  x  O  29. x 1 2 3 4 5  Graph:  y  y  y 0 3 6 9 12  x  O 30.  O  x 11 11  x  Equation: y  8  x 20. {(
5, 
5), (
3, 
3), (
1, 
1), (1, 1)}; D  {
5, 
3, 
1, 1}; R  {
5, 
3, 
1, 1} 21. {(
10, 0), (
5, 0), {0, 0), (5, 0)}; D  {
10, 
5, 0, 5}; R  {0}  Chapter 1  x 1 2 3 4 5 6 28.  Equation: y  3x 18. Table: x 
6 
5 
4 
3 
2 
1  
9 
8 
7 
6 
5 
4  27.  Graph: y 3 6 9 12 15 18 21 24 27  
4 
3 
2 
1 0 1  x  y 3 
3  4 2  O –2 –4  2  y  4  8  12 x  31.  x 4 4  y  y 2 
2  O  3
  
5  51c.  x  
2, 2  Number 5 Attending (thousands) 4 3 2  O 12 16 20 24 28 Number Applied (thousands)  52c. Yes; no member of the domain is paired with more than one member of the range.   1)  
1  x  6  2 44. j(2a)  1 
 4(2a)3  1 
 4(8a3)  1 
 32a3 45. f(n 
 1)  2(n 
 1)2 
 (n 
 1)  9  2(n2 
 2n  1) 
 n  1  9  2n2 
 4n  2 
 n  1  9  2n2 
 5n  12 b2  51b.  7  1  3
  1  52a. {(13,264, 4184), (27,954, 4412), (21,484, 6366), (23,117, 3912), (16,849, 2415), (19,563, 5982), (17,284, 6949)}; {13,264, 16,849, 17,284, 19,563, 21,484, 23,117, 27,954}; {2415, 3912, 4184, 4412, 5982, 6366, 6949} 52b.   43. h(0.5)   0.5   46. g(b2  1)   5  (b2  1)  x  x  32. {4, 5, 6}; {4}; yes; Each x-value is paired with exactly one y-value. 33. {1}; {
6, 
2, 0, 4}; no; The x-value 1 is paired with more than one y-value. 34. {0, 1, 4); {
2, 
1, 0, 1, 2}; no; The x-values 1 and 4 are paired with more than one y-value. 35. {0, 2, 5}; {
8, 
2, 0, 2, 8}; no; The x-values 2 and 5 are paired with more than one y-value. 36. {
1.1, 
0.4, 
0.1}; {
2, 
1}; yes; Each x-value is paired with exactly one y-value. 37. {
9, 2, 8, 9}; {
3, 0, 8}; yes; Each x-value is paired with exactly one y-value. 38. domain: all reals; range: all reals; Not a function because it fails the vertical line test. 39. domain: {
3, 
2, 
1, 1, 2, 3}; range: {
1, 1, 2, 3}; A function because each x-value is paired with exactly one y-value. 40. domain: {x
8  x  8}; range: {y
8  y  8}; Not a function because it fails the vertical line test. 41. f(3)  2(3)  3  6  3 or 9 42. g(
2)  5(
2)2  3(
2) 
 2  20 
 6 
 2 or 12  (b2  51a.  x
1  53. x  2m  1, so 2  m.  2 
 b2     6  b2 or 6  b2  47. f(5m)  (5m)2 
 13  25m2 
 13 2 48. x 
 5  0 x2  5 x  
5 ; x 
5  49. x2 
 9  0 x2  9 
3  x  3; x  
3 or x  3 50. x2 
 7  0 x2  7 

7   x  
7 ; x  

7  or x  x
1 Substitute 2 for m 24m3  36m2  26m x
1 3  in f(2m  1) to solve for f(x),  x
1 2  x
1   242  362  262 x3 
 3x2  3x 
 1  x2 
 2x  1  x
1   24   364  262 8   3x3 
 9x2  9x 
 3  9x2 
 18x  9  13x 
 13  3x3  4x 
 7 54a. t(500)  95 
 0.005(500)  92.5°F 54b. t(750)  95 
 0.005(750)  91.25°F 54c. t(1000)  95 
 0.005(1000)  90°F 54d. t(5000)  95 
 0.005(5000)  70°F  
7   3  Chapter 1  54e. t(30,000)  95 
 0.005(30,000)  
55°F 55a. d(0.05)  299,792,458(0.05)  14,989,622.9 m d(0.02)  299,792,458(0.2)  59,958,491.6 m d(1.4)  299,792,458(1.4)  419,709,441.2 m d(5.9)  299,792,458(5.9)  1,768,775,502 m 55b. d(0.008)  299,792,458(0.08)  23,983,396.64 m  3.  (1)(2)  1  56. P(4)  3  1 (2)(3)  1  P(5)  1  7 (3)(1)  1  4. Sample answer: The (sum/difference/product/ quotient) of the function values is the function values of the (sum/difference/product/quotient) of the functions. 5. Sample answer: For functions f(x) and g(x), (f  g)(x)  f(x)  g(x); (f 
 g)(x)  f(x) 
 g(x);  4  P(6)  7  7 57. 72 
 (32  42)  49 
 (9  16)  49 
 25 or 24 The correct choice is B.  f  1-2 Page 13  Composition of Functions  Page 17  Graphing Calculator Exploration  0  Check for Understanding  1. Sample answer: f(x)  2x 
 1 and g(x)  x  6; Sample explanation: Factor 2x2  11x 
 6. 2. Iteration is composing a function on itself by evaluating the function for a value and then evaluating the function on that function value. 3. No; [f 
 g](x) is the function f(x) performed on g(x) and [g 
 f ](x) is the function g(x) performed on f(x). See students' counter examples. 4. Sample answer: Composition of functions is performing one function after another. An everyday example is putting on socks and then putting shoes on top of the socks. Buying an item on sale is an example of when a composition of functions is used in a real-world situation. 5. f(x)  g(x)  3x2  4x 
 5  2x  9  3x2  6x  4 f(x) 
 g(x)  3x2  4x 
 5 
 (2x  9)  3x2  2x 
 14 f(x) g(x)  (3x2  4x 
 5)(2x  9)  6x3  35x2  26x 
 45  1.  2.  f  f(x)  g(x)  g(x)  3x2  4x 
 5  , x  2x  9  6. [f 
 g](x)  f(g(x))  f(3  x)  2(3  x)  5  2x  11 [g 
 f ](x)  g(f(x))  g(2x  5)  3  (2x  5)  2x  8  Chapter 1  f(x)   (f g)(x)  f(x) g(x); and g(x)   g(x) , g(x)  4  9  
2  7. [f 
 g](x)  f(g(x))  f(x2 
 2x)  2(x2 
 2x) 
 3  2x2 
 4x 
 3 [g 
 f ](x)  g(f(x))  g(2x 
 3)  (2x 
 3)2 
 2(2x 
 3)  (4x2 
 12x  9) 
 4x  6  4x2 
 16x  15 8. Domain of f(x): x 1 Domain of g(x): all reals g(x)  1 x31 x  
2 Domain of [f 
 g](x) is x 
2. 9. x1  f(x0)  f(2)  2(2)  1 or 5 x2  f(x1)  f(5)  2(5)  1 or 11 x3  f(x2)  f(11)  2(11)  1 or 23 5, 11, 23 10a. [K 
 C](F)  K(C(F)) 5  K9(F 
 32) 5  9(F 
 32)  273.15 5 10b. K(
40)  9(
40 
 32)  273.15  
40  273.15 or 233.15  x  2  f(x) 
 g(x)   x  1 
 (x 
 1)  f(x)  g(x)    13.   
17.78  273.15 or 255.37 5   0  273.15 or 273.15  14.  5  K(212)  9(212 
 32)  273.15  100  273.15 or 373.15  Pages 17–19  Exercises  11. f(x)  g(x)  x2 
 2x  x  9  x2 
 x  9 f(x) 
 g(x)  x2 
 2x 
 (x  9)  x2 
 3x 
 9 f(x) g(x)  (x2 
 2x)(x  9)  x3  7x2 
 18x x2 
 2x   g(x)   x9 , x f  9  x  2  12. f(x)  g(x)   x1  x 
 1      
x3 
 x2  2x  1  , x1 x  (x2 
 1)  x1 x(x  1)(x 
 1)   x1 2  x 
 x, x 
1  x  
1  (x)   x2 
 1 x 1    x  1 x2 
 1 x   
1 or 1 x3  x2 
 x 
 1 , x 3 2  f(x)  g(x)   x 
 7  x  5x 3 (x2  5x)(x 
 7)    x
7  x
7 x3 
 7x2  5x2 
 35x 3    x
7  x
7 x3 
 2x2 
 35x  3   ,x 7 x
7 3 2  f(x) 
 g(x)   x 
 7 
 (x  5x) (x2  5x)(x 
 7) 3    x
7 x
7 
 3 x3 
 7x2  5x2 
 35x    x
7 
 x
7 x3 
 2x2 
 35x 
 3  
  ,x 7 x
7 3 2  f(x) g(x)   x 
 7 (x  5x) 2 3x  15x  7  x
7 , x 3  x
7 f  (x)   g x2  5x 3 1    x 
 7 x2  5x 3 , x  
5, 0, 7 x3 
 2x2 
 35x 2x  f(x)  g(x)  x  3   x
5 (x  3)(x 
 5) 2x     x
5 x
5 x2 
 2x 
 15 2x     x
5 x
5 x2 
 15   5 x
5 , x 2x  f(x) 
 g(x)  x  3 
  x
5 (x  3)(x 
 5) 2x  
   x
5 x
5 x2 
 2x 
 15 2x  
   x
5 x
5 x2 
 4x 
 15 , x  5 x
5 2x  f(x) g(x)  (x  3) 
  x
5 2x2  6x   5 x
5 , x    5   9(32 
 32)  273.15  x3  x2 
 x 
 1  x  x1  f  g   
24.44  273.15 or 248.71  K(32)  x    5   9(0 
 32)  273.15  (x2 
 1)(x  1)     x1 
 x1  K(
12)  9(
12 
 32)  273.15 K(0)  x     x1 
 x1          x3  g(x)   2x  f  (x2 
 1)(x  1) x    x1 x1 x x3  x2 
 x 
 1    x1 x1 x3  x2 
 1 , x 
1 x1  x
5  x
5  2x x2 
 2x 
 15 , x 2x  x3   5  0 or 5  Chapter 1  15. [f 
 g](x)  f(g(x))  f(x  4)  (x  4)2 
 9  x2  8x  16 
 9  x2  8x  7 [g 
 f ](x)  g(f(x))  g(x2 
 9)  x2 
 9  4  x2 
 5 16. [f 
 g](x)  f(g(x))  f(x  6)  21. [f 
 g](x)  f (g(x))    f  x 
 1 1  1    x
1  1  x    x 
 1, x  1  [g 
 f ](x)  g(f(x))  g(x  1) 1    x1
1 1   x, x  1   2(x  6) 
 7  0  22. Domain of f(x): all reals Domain of g(x): all reals Domain of [f 
 g](x): all reals 23. Domain of f(x): x 0 Domain of g(x): all reals g(x)  0 7
x0 7x Domain of [f 
 g](x) is x 7. 24. Domain of f(x): x  2 Domain of g(x): x 0 g(x)  2  1   2x  3 
 7 1   2x 
 4 [g 
 f ](x)  g(f(x)) 1   g(2x 
 7) 1   2x 
 7  6 1   2x 
 1 17. [f 
 g](x)  f(g(x))  f(3x2)  3x2 
 4 [g 
 f ](x)  g(f(x))  g(x 
 4)  3(x 
 4)2  3(x2 
 8x  16)  3x2 
 24x  48 18. [f 
 g](x)  f(g(x))  f(5x2)  (5x2)2 
 1  25x4 
 1 [g 
 f ](x)  g(f(x))  g(x2 
 1)  5(x2 
 1)2  5(x4 
 2x2  1)  5x4 
 10x2  5 19. [f 
 g](x)  f(g(x))  f(x3  x2  1)  2(x3  x2  1)  2x3  2x2  2 [g 
 f ](x)  g(f(x))  g(2x)  (2x)3  (2x)2  1  8x3  4x2  1 20. [f 
 g](x)  f(g(x))  f(x2  5x  6)  1 x2  5x  6  x2  5x  7 [g 
 f ](x)  g(f(x))  g(1  x)  (x  1)2  5(x  1)  6  x2  2x  1  5x  5  6  x2  7x  12  Chapter 1  x
1  1     x
1  x
1  1 x 4  2  1  8x 1  8  x 1  Domain of [f 
 g](x) is x  8, x 25. x1  f(x0)  f(2)  9 
 2 or 7 x2  f(x1)  f(7)  9 
 7 or 2 x3  f(x2)  f(2)  9 
 2 or 7 7, 2, 7 26. x1  f(x0)  f(1)  (1)2  1 or 2 x2  f(x1)  f(2)  (2)2  1 or 5 x3  f(x2)  f(5)  (5)2  1 or 26 2, 5, 26 27. x1  f(x0)  f(1)  1(3 
 1) or 2 x2  f(x1)  f(2)  2(3 
 2) or 2 x3  f(x2)  f(2)  2(3 
 2) or 2 2, 2, 2  6  0.  34a. I  prt  5000(0.08)(1)  400 I  prt  5400(0.08)(1)  432 I  prt  5832(0.08)(1)  466.56 I  prt  6298.56(0.08)(1)  503.88 I  prt  6802.44(0.08)(1)  544.20 (year, interest): (1, $400), (2,$432), (3, $466.56), (4, $503.88), (5, $544.20) 34b. {1, 2, 3, 4, 5}; {$400, $432, $466.56, $503.88, $544.20} 34c. Yes; for each element of the domain there is exactly one corresponding element of the range. 35. {(
1, 8), (0, 4), (2, 
6), (5, 
9)}; D  {
1, 0, 2, 5}; R  {
9, 
6, 4, 8} 36. D  {1, 2, 3, 4}; R  {5, 6, 7, 8}; Yes, every element in the domain is paired with exactly one element of the range.  28. $43.98  $38.59  $31.99  $114.56 Let x  the original price of the clothes, or $114.56. Let T(x)  1.0825x. (The cost with 8.25% tax rate) Let S(x)  0.75x. (The cost with 25% discount) The cost of clothes is [T 
 S](x). [T 
 S](x)  T(S(x))  T(0.75x)  T(0.75(114.56))  T(85.92)  1.0825(85.92)  93.0084 Yes; the total with the discount and tax is $93.01. 29. Yes; If f(x) and g(x) are both lines, they can be represented as f(x)  m1x  b1 and g(x)  m2x  b2. Then [f 
 g](x)  m1(m2x  b2)  b1  m1m2x  m1b2  b1 Since m1 and m2 are constants, m1m2 is a constant. Similarly, m1, b2, and b1 are constants, so m1b2  b1 is a constant. Thus, [f 
 g](x) is a linear function if f(x) and g(x) are both linear. 30a. Wn  Wp 
 Wf  Fpd 
 Ff d  d(Fp 
 Ff) 30b. Wn  d(Fp 
 Ff)  50(95 
 55)  2000 J 31a. h[f(x)], because you must subtract before figuring the bonus 31b. h[f(x)]  h[f(400,000)]  h(400,000 
 275,000)  h(125,000)  0.03(125,000)  $3750 32. (f 
 g)(x)  f(g(x))  f(1 
 x2)  (
4)3  5   37. g(
4)   4(
4) 
64  5    
16 
59  38.  x2(x2  1)    1  x2   x2  
(1 
 x2)  1 7p  1  1  33b. r(v)  0.84v  33a. v(p)  47  x 
2 
1 0 1 2 3  y 6 3 0 3 6 9  y  O  x  39. f(n 
 1)  2(n 
 1)2 
 (n 
 1)  9  2(n2 
 2n  1) 
 n  1  9  2n2 
 5n  12 The correct choice is C.  So, f(x)  
x  1 and f 2  
2  1  2. 1  11     
16 or 3 16  33c. r(p)  r(v(p)) 7p   r47  7p   0.8447 5.88p  1-3  14 7p    4 7 or 1175  Graphing Linear Equations  147(4 23.18)   33d. r(423.18)   1175  Page 23   $52.94  Check for Understanding  1. m represents the slope of the graph and b represents the y-intercept 2. 7; the line intercepts the x-axis at (7, 0) 3. Sample answer: Graph the y-intercept at (0, 2). Then move down 4 units and right 1 unit to graph a second point. Draw a line to connect the points.  147(2 25.64)   r(225.64)   1175   $28.23 147(7 97.05)   r(797.05)   1175   $99.72  7  Chapter 1  4. Sample answer: Both graphs are lines. Both lines have a y-intercept of 8. The graph of y  5x  8 slopes upward as you move from left to right on the graph and the graph of y  
5y  8 slopes downward as you move from left to right on the graph. 5. 3x 
 4(0)  2  0 3(0) 
 4y  2  0 3x  2  0 
4y  2  0 3x  
2 
4y  
2 2  1  9. 2x  6  0 1 x 2  y 12 8 4  1  x  
3  y  2  
12 
8 
4 O 
4  y  (
 23 ,0)   
6  x  
12  x  10. Since m  0 and b  19, this function has no x-intercept, and therefore no zeros.  (0, 12 )  y  x  O  24 16  6. x  2(0) 
 5  0 x
50 x5  0  2y 
 5  0 2y 
 5  0 2y  5  8 
4 
2 O  5  y  2  2  4x  11a. (38.500, 173), (44.125, 188)  y  188 
 173   11b. m   44.125 
 38.500  (0, 52 )  15    5.625 or about 2.667  11c. For each 1 centimeter increase in the length of a man's tibia, there is an 2.667-centimeter increase in the man's height.  x O  (5, 0)  7. The y-intercept is 7. Graph (0, 7). The slope is 1.  y  Pages 24–25  Exercises  12. The y-intercept is 
9. The slope is 4.  y  (1, 8) (0, 7)  x  O  y  4x 
 9  O  x  8. The y-intercept is 5. Graph (0, 5). The slope is 0.  13. The y-intercept is 3. The slope is 0.  y  y  y3  (0, 5)  O O  Chapter 1  x  8  x  19. 2x  y  0 y  
2x The y-intercept is 0. The slope is 
2.  14. 2x 
 3y  15  0 
3y  
2x 
 15 2  y  3x  5  y  2  The y-intercept is 5. The slope is 3.  y O  2x 
 3y  15  0  x  2x  y  0  2  20. The y-intercept is 
4. The slope is 3.  O x  y  15. x 
 4  0 x4 There is no slope. The x-intercept is 4.  O  x  y y  23 x 
 4  x
40  x  O  21. The y-intercept is 150. The slope is 25.  y y  25x  150 100  16. The y-intercept is 
1. The slope is 6.  50  y  
6 
4 
2 O 50  y  6x 
 1  O  2x  22. 2x  5y  8 5y  
2x  8  x  2  8  y  
5x  5 8  2  The y-intercept is 5. The slope is 
5.  17. The y-intercept is 5. The slope is 
2.  y  y  2x  5y  8  y  5 
 2x  x O O  x 23. 3x 
 y  7 
y  
3x  7 y  3x 
 7 The y-intercept is 
7. The slope is 3.  18. y  8  0 y  
8 The y-intercept is 
8. The slope is 0.  y O  y  x  x  O 3x 
 y  7  y80  9  Chapter 1  24. 9x  5  0 9x  
5  33a. (1.0, 12.0), (10.0, 8.4)  f (x )  8.4 
 12.0   m 10.0 
 1.0  f (x )  9x  5  5  x  
9  
3.6   9 or 
0.4  The y-intercept is 5.  
(
0.4)  0.4 ohms  (
 59 , 0)  12 
 v  25. 4x 
 12  0 4x  12 x3 The y-intercept is 
12.   33b. 
0.4   1.0 
 25.0  x  O  12 
 v   
0.4   
24  f (x )  9.6  12 
 v v  2.4 volts  f (x )  4x 
 12  9
7  (3, 0)  2  26. 3x  1  0 3x  
1    
7  x  O  (
 13 ,0)  The y-intercept is 1.  m  x    f (x )  1  14 (0, 0)  there is a 4-pascal increase in the pressure.  f (x )  14x  35c. 100 P  x  80 60  28. None; since m  0 and b  12, this function has no x-intercepts, and therefore no zeros.  40  f (x )  20  f (x )  12 8  O  4  f (x ) f (x )  5x 
 8 2  8  5  The y-intercept is 
8.  
2 O 
4 
6 
8  ( 85 , 0) x  2  30. 5x 
 2  0 5x  2 2  x  5 3  10,000  31. The y-intercept is 3. The slope is 
2. 3  
2x  3  0  20  y   
3 x2  60  80 T  (0, 10,440)  8000  y  3 
2x  40  36. No; the product of two positives is positive, so for the product of the slopes to be 
1, one of the slopes must be positive and the other must be negative. 37a. 10,440 
 290t  0 
290t  
10,440 t  36 The software has no monetary value after 36 months. 37b. 
290; For every 1-month change in the number of months, there is a $290 decrease in the value of the software. v (t ) 37c.  x  29. 5x 
 8  0 5x  8 x  100 – 90  126.85 – 86.85 10 1  or  40 4  35b. For each 1 degree increase in the temperature,  28  O  O  
 32 x  3  6000 4000  (2, 0)  O  2000  x  (36, 0)  O  32. Sample answer: f(x)  5; f(x)  0  Chapter 1  
8  35a. (86.85, 90), (126.85, 100)  O 27. 14x  0 x0 The slope is 14.  2  
2(a  4)  
56 
2a 
 8  
56 
2a  
48 a  24  f (x )  3x  1  1  1
9   
7   a4  f (x )  x  
3  2   
7   a 
 (
4)   34. m   
4 
 3  10  8  16  24  32  t  38. A function with a slope of 0 has no zeros if its y-intercept is not 0; a function with a slope of 0 has an infinite number of zeros if its y-intercept is 0; a function with any slope other than 0 has exactly 1 zero. 39a. (56, 50), (76, 67.2)  1-3B Graphing Calculator Exploration:  Analyzing Families of Linear Graphs  Page 26  67.2 
 50   m 76 
 56  1. See students' graphs. All of the graphs are lines with y-intercept at (0, 
2). Each line has a different slope. 2. A line parallel to the ones graphed in the Example and passing through (0, 
2). 3. See students' sketches. Sample answer: The graphs of lines with the same value of m are parallel. The graphs of lines with the same value for b have the same y-intercept.  17.2   2 0 or 0.86 39b. 1805(0.86)  $1552.30 39c. 1 
 MPC  1 
 0.86  0.14 39d. 1805(0.14)  $252.70 40. (f  g)(x)  2x  x2 
 4 or x2  2x 
 4 (f 
 g)(x)  2x 
 (x2 
 4)  
x2  2x  4 41a. 1 
 0.12  0.88 d(p)  0.88p 41b. r(d)  d 
 100 41c. r(d(p))  r(0.88p)  0.88p 
 100 41d. r(799.99)  0.88(799.99) 
 100  603.9912 or about $603.99 r(999.99)  0.88(999.99) 
 100  779.9912 or about $779.99 r(1499.99)  0.88(1499.99) 
 100  1219.9912 or about $1219.99 42. [f 
 g](
3)  f(g(
3))  f(
3 
 2)  f(
5)  (
5)2 
 4(
5)  5  25  20  5 or 50 [g 
 f ](
3)  g( f(
3))  g((
3)2 
 4(
3)  5)  g(9  12  5)  g(26)  26 
 2 or 24 43. f(9)  4  6(9) 
 (9)3  4  54 
 729 or 
671 44. No; the graph fails the vertical line test. 45. x y 
3 14 
2 13 
1 12 {(
3, 14), (
2, 13), (
1, 12), (0, 11)}, yes 0 11  1-4  Page 29 Check for Understanding 1. slope and y-intercept; slope and any point; two points 2. Sample answer: Use point-slope Use slope-intercept form: form. y 
 y1  m(x 
 x1) y  mx  b 1  y 
 (
4)  4(x 
 3) 1  3  y  4  4x 
 4 x 
 4y 
 19  0  1  
4  4(3)  b 19  
4  b Substitute the slope and intercept into the general form. 1  19  y  4x 
 4 Write in standard form. x 
 4y 
 19  0 3. 55 represents the hourly rate and 49 represents the fee for coming to the house. 
3 
 0  1   4. m   0
6 
3  y  2x 
 3 1     
6 or 2  5. Sample answer: When given the slope and the y-intercept, use slope-intercept form. When given the slope and a point, use point-slope form. When given two points, find the slope, then use pointslope form.  46. Let s  sum. s  4  Writing Linear Equations  1  6. y  mx  b → y  
4x 
 10   15  7. y 
 2  4(x 
 3) y 
 2  4x 
 12 y  4x 
 10  s  60 The correct choice is D.  9
2   8. m   7
5 7   2 7  y 
 2  2(x 
 5) 7  35  7  31  y 
 2  2x 
 2 y  2x 
 2 9. y 
 2  0(x 
 (
9)) y
20 y2  11  10a. y  5.9x  2  Chapter 1  10b. y  5.9(7)  2  41.3  2 or 43.3 in. 10c. Sample answer: No; the grass could not support it own weight if it grew that tall.  Pages 30–31  27b. Using sample answer from part a, 9  171  3  13. y  mx  b → y  
4x  3
9  14. y  mx  b → y  
12x   9
5  4  y
  20. x  
4 22. y 
 0  0.25(x 
 24) y  0.25x 
 6 23.  y 
 (
4)  y4 1 x 2  4  y 
 5  
9x  9  18. (
8, 0), (0, 5)  4 49 y  
9x  9 5 0  8(x 
 (
8)) 5 y  8x  5  3  3x 
 2y 
 5  0  5  y  2x 
 2  O  y 
 1  0(x 
 8) y
10 y1 21. x  0  82,805 
 70,583   m 1997–1999 12,222   2 or 6111; $6111 billion 31b. The rate is the slope. 32. g[f(
2)]  g(f(
2))  g((
2)3)  g(
8)  3(
8) or 
24 33. (f g)(x)  f(x) g(x)  x3(x2 
 3x  7)  x5 
 3x4  7x3  1 
2(x 
 (
2)) 1 
2x 
 1  y50  
3 
 0  x , where g(x) g(x)   x 
 3x  7 3  f   24. m   1 
 (
2) 
3  3  34.   
1 y 
 0  
1(x  2) y  
x 
 2 x  y  
2  x 
4 
3 
2  y 16 9 4  {(
4, 16), (
3, 9), (
2, 4)}, yes  x 
 7000   25a. t  2   2000  35. x   14,494 
 7000   25b. t  2   2000  1  y  y1  y  C  y2  1   y  A  y2  1   2 y  27a. Sample answer: using (20, 28) and (27, 37),  The correct choice is A.  9   m 27 
 20  y 
 28  7(x 
 20)  9  y 
 28  7x 
 7  9  180  9  16  y  7x  7 Chapter 1  1   y   2   2  y  
Bx 
 B; m  
B   7  y2  y  y2  1  y   2  3.747 or 5.747; about 5.7 weeks 26. Ax  By  C  0 By  
Ax 
 C  37 
 28  0  2  A  x  31a. (1995, 70,583), (1997, 82,805)  x  2y  10  0 x  2y  
10   3  lines would have the same slope and would share a point, their equations would be the same. Thus, they are the same line and all three points are collinear. y 30. 3x 
 2y 
 5  0 
2y  
3x  5  4  4  19.  6
3    (
3, 3), and (1, 6) is  1 
 (
3) or 4 . Since these two  y 
 5  
9(x 
 1)    
9  3    is  
3 
 5 or 4 . The slope of the line through  1  2  16. x  12   17. m   
8 
 1  187  27c. Sample answer: The estimate is close but not exact since only two points were used to write the equation. 28a. See students' work. 28b. Sample answer: Only two points were used to make the prediction equation, so many points lie off of the line. 29. Yes; the slope of the line through (5, 9) and (
3, 3)  Exercises  5
0  m 0 
 (
8) 5  8 1
1  m 
3 
 8 0   
11 or 0  16   7  7 or 7 or about 26.7 mpg  11. y  mx  b → y  5x 
 2 12. y 
 5  8(x 
 (
7)) y 
 5  8x  56 y  8x  61  15. y 
 5  6(x 
 4) y 
 5  6x 
 24 y  6x 
 19  16  y  7(19)  7  12  1  2  Page 31  10b. y – 6,478,216  170,823.7(x – 1990) y – 6,478,216  170,823.7x – 339,939,163 y  170,823.7x – 333,460,947  Mid-Chapter Quiz  1. {
2, 2, 4}, {
8, 
3, 3, 7}; No, 
2 in the domain is paired with more than one element of the range. 3  3. g(n  2)   2. f(4)  7 
 42 n2
1  7 
 16 or 9 3   n1 4. Let x  original price of jacket Let T(x)  1.055x. (The cost with 5.5% tax rate) Let S(x)  0.67x. (The cost with 33% discount) The cost of the jacket is [T 
 S](x). [T 
 S](x)  T(S(x))  T(0.67x)  1.055(0.67x) The amount paid was $49.95. 45.95  1.055(0.67x) 43.55  0.67x 65  x; $65 5. [f 
 g)(x)  f(g(x)  f(x  1)  Pages 35–36    4  4 3  
3; 4 4. All vertical lines have undefined slope and only horizontal lines are perpendicular to them. The slope of a horizontal line is 0. 5. none of these 6. perpendicular 7. y  x 
 6 x
y80 yx8 parallel 8. y  2x 
 8 4x 
 2y 
 16  0 y  2x 
 8 coinciding 9. y 
 9  5(x 
 5) y 
 9  5x 
 25 5x 
 y 
 16  0 10. 6x 
 5y  24  1    g( x 
 1) 1    x
1  1  x
1     x
1  x
1 x    x 
 1, x  1  y  6. 2x 
 4y  8 
4y  
2x  8 1  y  2x 
 2  x  O  2x 
 4y  8  6  7. 3x  2y 3 x 2  19  y  
3x 
 3  [g 
 f ](x)  g(f(x))  1  Check for Understanding  1. If A, B, and C are the same or the ratios of the As and the Bs and the Cs are proportional, then the lines are coinciding. If A and B are the same and C is different, or the ratios of the As and the Bs are proportional, but the ratio of the Cs is not, then the lines are parallel. 2. They have no slope. 3. 4x  3y  19  0  1  x1
1 1 , x 0 x    Writing Equations of Parallel and Perpendicular Lines  1-5  24  y  5x 
 5  y  5  y 
 (
5)  
6(x 
 (
10))  y  5  25  y  5  
6x 
 3 6y  30  
5x 
 50 5x  6y  80  0  O x  3x  2y  8
4   11. m of EF: m   4
3 4   1 or 4  8. 5x 
 3  0 5x  3  2
6   m of GH: m   6
7 
4    
1 or 4  3  x  5 8
5  y 
 5  5(x 
 2)  3  3  2   
3  6  3  Pages 36–37  19  
5x  y 
 5  0  12. y  5x 
18  3x 
 5y  19  0 10a. (1990, 6,478,216), (2000, 8,186,453)    6
8   m of FG: m   7
4  y 
 5  5x 
 5   5  m  2   
3  parallelogram  3   9. m   7
2  2
4   m of EH: m   6
3  2x  10y  10  0 1  y  
5x  1  8,186,453 
 6,478,216  2000 – 1990 1,708,237  10  Exercises  slopes are opposite reciprocals; perpendicular  or about 170,823.7  13  Chapter 1  13. y 
 7x  5  0 y 
 7x 
 9  0 y  7x 
 5 y  7x  9 same slopes, different y-intercepts; parallel 14. different slopes, not reciprocals; none of these 15. horizontal line, vertical line; perpendicular 16. y  4x 
 3 4.8x 
 1.2y  3.6 y  4x 
 3 same slopes, same y-intercepts; coinciding 17. 4x 
 6y  11 3x  2y  9 y  2 x 3  
  11  6  y  3 
2x    4 4  y 
 8  5x  12 5y 
 40  4x  60 4x 
 5y  100  0 5  28b. perpendicular slope: 
4 5  y 
 8  
4(x 
 (
15))  9  2  5  y  y  5 
9x    4y 
 32  
5x 
 75 5x  4y  43  0 29a. 8x 
 14y  3  0 kx 
 7y  10  0 8  4  7  29b.  same slopes, same y-intercepts; coinciding 20. y  4x 
 2  0  k   7 → k  4  7   
4 49  k  
4 30a. Sample answer: y 
 1  0, x 
 1  0 30b. Sample answer: x 
 7  0, x 
 9  0 31. altitude from A to BC: 
5 
 (
5)   m of BC   10 
 4 0   6 or 0 m of altitude is undefined; x  7 altitude from B to AC:  4    23. m  
 (
9) or 9  
5 
 10  4  y 
 (
15)  9(x 
 12) 4  k  7   m of AC   4
7 
15  y  15  9x 
 3  16    
3 or 5  9y  135  4x 
 48 4x 
 9y 
 183  0 24. y 
 (
11)  0(x 
 4) y  11  0  m of altitude  
5  25.  1  1  y 
 (
5)  
5(x 
 10) 1  y  5  
5x  2 5y  25  
x  10 x  5y  15  0 altitude from C to AB:  1  y 
 (
3)  
5(x 
 0) y3  1 
5x  5y  15  
x x  5y  15  0  
5 
 10   m of AB   10 
 7  6  
15   3 or 
5  1    26. m  
 (
1) or 6; perpendicular slope is 
 6  1  m of altitude  5  1  y 
 (
2)  
6(x 
 7) 1  1  y 
 (
5)  5(x 
 4)  7  y  2  
6x  6 6y  12  
x  7 x  6y  5  0 27. x  12 is a vertical line; perpendicular slope is 0. y 
 (
13)  0(x 
 6) y  13  0  Chapter 1  k    m  
 (
7) or 7  4k  
49  y  4x  1  0  y  
4x  2 y  
4x 
 1 same slopes, different y-intercepts; parallel 21. None of these; the slopes are not the same nor opposite reciprocals. 22. y 
 (
8)  2(x 
 0) y  8  2x 2x 
 y 
 8  0 4  k  4    m  
 (
14) or 7  14  9  14  9    75  y 
 8  
4x 
 4  y  
3x  2 different slopes, not reciprocals; none of these 19. 5x  9y  14  4    m  
 (
5) or 5  y 
 8  5(x 
 (
15))  3x  y  2  5 
9x  4  4x 
 5y  10  0  Slopes are opposite reciprocals; perpendicular. 18. y  3x  2  5y 
 4x  10  28a.  1  4  y  5  5x 
 5 5y  25  x 
 4 x 
 5y 
 29  0 32. We are given y  m1x  b1 and y  m2x  b2 with m1  m2 and b1 b2. Assume that the lines intersect at point (x1, y1). Then y1  m1x1  b1 and y1  m2x1  b2. Substitute m1x1  b1 for y1 in y1  m2x1  b2. Then m1x1  b1  m2x1  b2. Since m1  m2, substitute m1 for m2. The result is m1x1  b1  m1x1  b2. Subtract m1x1 from each side to find b1  b2. However, this contradicts the given information that b1 b2. Thus, the  14  36a. (40, 295), (80, 565)  assumption is incorrect and the lines do not share any points. 33a. Let x  regular espressos. Let y  large espressos. 216x  162y  783 248x  186y  914 y  4 
3x    29  6  y  4 
3x    565 
 295   m 80 
 40 270   4 0 or 6.75 y 
 565  6.75(x 
 80) y 
 565  6.75x 
 540 y  6.75x  25 36b. $6.75 36c. $25 37. 3x 
 2y 
 6  0  457  93  No; the lines that represent the situation do not coincide. 33b. Let x  regular espressos. Let y  large espressos. 216x  162y  783 344x  258y  1247 4  29  y  
3x  6  4  y  y  3x 
 2y 
 6  0  29  O  x  38. [g 
 h](x)  g(h(x))  g(x2)  x2 
 1 39. Sample answer: {(2, 4), (2, 
4), (1, 2), (1, 
2), (0, 0)}; because the x-values 1 and 2 are paired with more than one y-value. 40. 2x  y  12 x  2y  
6 y  
2x  12 x  2(
2x  12)  
6 x 
 4x  24  
6 
3x  
30 x  10 2(10)  y  12 y  
8 2x  2y  2(10)  2(
8)  20  (
16) or 4  1943.09 – 1939.20  m   16 – 15 3.89   1 or 3.89 y 
 1943.09  3.89(x 
 16) y 
 1943.09  3.89x 
 62.24 y  3.89x  1880.85 (16, 1943.09), (17, 1976.76) 1976.76 – 1943.09  m  17 – 16 33.67   1 or 33.67 y 
 1976.76  33.67(x 
 17) y 
 1976.76  33.67x 
 572.39 y  33.67x  1404.37 (17, 1976.76), (18, 1962.44) 1962.44 
 1976.76  18 
 17 
14.32   or –14.32 1 y 
 1962.44  –14.32(x 
 18) y 
 1962.44  –14.32x  257.76 y  –14.32  2220.2 (18, 1962.44), (19, 1940.47) m  
3  y  
3x  6  Yes; the lines that represent the situation coincide. 34a. (15, 1939.20), (16, 1943.09)  m  3 x 2  1-6  Modeling Real-World Data with Linear Functions  Pages 41–42  Check for Understanding  1. the rate of change 2. Choose two ordered pairs of data and find the equation of the line that contains their graphs. Find a median-fit line by separating the data into three sets and using the medians to locate a line. Use a graphing calculator to find a regression equation. 3. Sample answer: age of a car and its value 4a. Personal Consumption on Durable Goods  1940.47 
 1962.44  19 
 18 –21.97   1 or –21.97 y 
 1940.47  
21.97(x 
 19) y 
 1940.47  
21.97x  417.43 y  
21.97x  2357.9 34b. parallel lines or the same line; no 34c. y  3.89(22)  1880.85  1966.43 y  33.67(22)  1404.37  2145.11 y  
14.32(22)  2220.2  1905.16 y  
21.97(22)  2357.9  1874.56 No; the equations take only one pair of days into account. 35. y 
 5  
2(x 
 1) y 
 5  
2x  2 y  
2x  7  3500 3000 2500 2000 Dollars 1500 1000 500 0 1995 1997 1999 2001 2003 Year  15  Chapter 1  4b. Sample answer: using (1995, 2294) and (2002, 3158) m  7a.  3158 – 2294  2002 – 1995 864   7 or 123.4 y 
 3158  123.4(x 
 2002) y  123.4x 
 243,888.8 4c. y  132.8x 
 262,621.2; r  0.98 4d. y  132.8(2010) 
 262,621.2  4306.8 $4306.80; yes, the correlation coefficient shows a strong correlation. Students per Computer 5a.  0 1995 1997 1999 2001 2003 Year  7b. Sample answer: using (1995, 23,255) and (2002, 30,832) m   25 15 10 5 0 '89  '91  '93  '95 Year  '97  '99  '01  5b. Sample answer: using (1997, 6.1) and (2001, 4.9) m   30,832 – 23,255  2002 – 1995 7577  or 1082.43 7  y 
 23,255  1082.43(x 
 1995) y  1082.43x 
 2,136,192.85 7c. y  1164.11x 
 2,299,128.75; r  0.99 7d. y  1164.11(2010) – 2,299,128.75  40,732.35 $40,732.35; yes, r shows a strong relationship. Car Weight and Mileage 8a.  20 Average  Personal Income 35 30 25 Dollars (thousands) 20 15  80  4.9 – 6.1  2001 – 1997 –1.2  or 
0.3 4  60 Average Mileage 40  y 
 6.1  
0.3(x 
 1997) y  
0.3x  605.2 5c. y  
1.61x  3231.43; r  
0.97 5d. 1  
1.61x  3231.43 
3230.43  
1.61x 2006.47  x 2006; yes, the number of students per computer is decreasing steadily.  20 0 0  15 20 25 30 35 Weight (hundreds of pounds)  8b. Sample answer: using (17.5, 65.4) and (35.0, 27.7) 27.7 
 65.4   m 35.0 
 17.5 
37.7    17.5 or 
2.15  Pages 42–44  y 
 65.4  
2.15(x 
 17.5) y  
2.15x  103.0 8c. y  
1.72x  87.59; r  
0.77 8d. y  
1.72(45.0)  87.59 y  10.19 10.19; no, r doesn't show a particularly strong relationship. 9a. Acorn Size and Range  Exercises  6a.  All-Time NFL Coaching Victories 400 Wins 200 0 0  10  20  30  40  Years  6b. Sample answer: using (18, 170) and (40, 324) m   30,000  324 – 170  40 – 18 154  or 7 22  20,000 Range (hundreds of km2) 10,000  y 
 324  7(x 
 40) y  7x  44 6c. y  7.57x  33.38; r  0.88 6d. y  7.57(16)  33.38  154.5 155; yes, r is fairly close to 1. (Actual data is 159.)  Chapter 1  0 0  16  2 4 6 8 10 12 Acorn Size (cm3)  9b. Sample answer: using (0.3, 233) and (3.4, 7900) m   12b. Sample answer: a medication that proves to help delay the progress of a disease; because any positive correlation is better than none or a negative correlation. 12c. Sample answer: comparing a dosage of medicine to the growth factor of cancer cells; because the greater the dosage the fewer cells that are produced. 13. Men's Median Salary Women's Median Salary LinReg LinReg y  ax  b y  ax  b a  885.2867133 a  625.041958 b  –1,742,768.136 b  –1,234,368.061 r  .9716662959 r  .9869509009 The rate of growth, which is the slope of the graphs of the regression equations, for the women is less than that of the men's rate of growth. If that trend continues, the men's median salary will always be more than the women's. 14a. Let x  computers. Let y  printers. 24x  40y  38,736 y  
0.6x  968.4 30x  50y  51,470 y  
0.6x  1029.4 No; the lines do not coincide. 14b. Let x  computers. Let y  printers. 24x  40y  38,736 y  
0.6x  968.4 30x  50y  48,420 y  
0.6x  968.4 Yes; the lines coincide. 15. y 
 (
4)  
6(x 
(
3)) y  4  
6x 
 18 6x  y  22  0 y 16a.  7900 
 233  3.4 
 0.3 7667  or 2473.23 3.1  y 
 7900  2473.23(x 
 3.4) y  2473.23x 
 508.97 9c. y  885.82  6973.14; r  0.38 9d. The correlation value does not show a strong or moderate relationship. Working Women 10a. 40 35 30 25 Percent in 20 Management 15 10 5 0 '86 '88 '90 '92 '94 '96 '98 '00 '02 '04 Year  10b. Sample answer: using (1990, 26.2) and (2003, 37.6) 37.6 – 26.2  m   2003 – 1990 11.4   1 3 or 0.88 y 
 26.2  0.88(x 
 1990) y  0.88x 
 1725 10c. y  0.84x 
 1648.27; r  0.984 10d. y  0.84(2008) 
 1648.27 or 38.45 38.45%; yes, r is very close to 1. 11a. World Population 7000 6000 5000 Millions 4000 of People 3000 2000 1000 0  25 20 y  0.82x  24 15 10 5 0  1000 Year  2000  O  11b. Sample answer: using (1, 200) and (2000, 6050) m   1 2 3 4 5x  16b. $24 billion 16c. If the nation had no disposable income, personal consumption expenditures would be $24 billion. For each 1 billion increase in disposable income, there is a 0.82 billion dollar increase in personal consumption expenditures. 17. [f 
 g](x)  f(g(x))  f(x  1)  (x  1)3  x3  3x2  3x  1 [g 
 f ](x)  g(f(x))  g(x3)  x3  1 18. Yes; each domain value is paired with exactly one range value.  6050 – 200  2000 – 1 5850  or 2.93 1999  y 
 200  2.93(x 
 1) y  2.93x  197.07 11c. y  1.65x 
 289.00; r  0.56 11d. y  1.65(2010) 
 289.00  3027.5 3028 million; no, the correlation value is not showing a very strong relationship. 12a. Sample answer: the space shuttle; because anything less than perfect could endanger the lives of the astronauts.  17  Chapter 1  19. The y-intercept is 1. The slope is 
3 (move down 3 and right 1). The correct choice is C.  Piecewise Functions  1-7  Pages 48–49 1. 2. 3. 4. 5.  8.  Check for Understanding    x 
1 0 1 2 3 4  y  f(x) 4 3 2 1 0 1  x  O  9. greatest integer function; h is hours, c(h) is the 50h if [[h]]  h cost, c(h)  50[[h  1]] if [[h]]  h  x if x  0 f(x)  
x if x  0 reals, even integers x  2 if x  0 f(x)  x if 0  x  4 x 
 2 if x 4 Alex is correct because he is applying the definition of a function. y      x 0x1 1x2 2x3 3x4  y  f(x) 50 100 150 200  400 300 200 100  x O  O y  Pages 49–51 11.  x 
3  x  
2 
2  x  
1 
1  x  0 0x1 1x2 2x3 3x4 4x5  10  y  x  12.  x 1 3 5 7 9  y  f(x) 4 2 0 2 4  x  O  Chapter 1  8  Exercises  O  f(x) 3 2 1 0 
1 
2 
3 
4  y  O  6  x  O 7.  4  10. long term lot: 2(6)  3(1)  12  3 or 15 shuttle facility: 4(3)  12 shuttle facility  x  6.  2  13.  x  18  x 
2  x  
1 
1  x  0 0x1 1x2 2x3  f(x) 0 1 2 3 4  y  O  x  14.  x 
5 
3 
1.5 0 2  x 
2  x  
1 
1  x  0 0x1 1x2 2x3  x  x  f(x) 
3 
2 
1 0 1  f(x)  
1  x   2 
3  3  2 
3  1 
3  2  x  1 
3  O 15.  18.  y  f(x) 7 3 0 3 7  x0  1  0x  1  3  0  1  3  x  2  3  
1  2  3  x1  
2  y 4 2  y O  –1  1  x  –2 –4  O  x 19.  20.  y  y 16. ART y  O x  O 21.  x  O 17.  x 0 1 2 3 4  f(x) 6 4 2 0 2  y  x  x  f(x)  
5  x  
4  
5  
4  x  
3  
2  
3  x  
2  
3  
2  x  
1 
1  x  0 1x2 2x3  
1 
2 2 1  3x4  2  3  4x5  1  2  5x6  2  5  2 1 2  y 2  O  x  1  x  O 
1 
2  19  Chapter 1  22.  x 
5 
3 
1 0 1 3 5  26c.  y  f(x) 6 0 6 9 6 0 6  Shipping (dollars)  O x  O  23. Step; t is the time in hours, c(t) is the cost in dollars,  6 if t  1 2  1    t  1 10 if c(t)   2  16 if 1  t  2   24 if 2  t  24  27. If n is any integer, then all ordered pairs (x, y) where x and y are both in the interval [n, n  1) are solutions. 28a. absolute value 28b. d(t)  65 
 t 28c. d (t ) 80  d(t) 24  60  16  40  8  20  O  2  4  6  8  10 12 14 16 18 20 22 24 t  2  37  2  c(w)  60  80  t   19.5 heating degree days  5% if x  $10,000  29b. t(x)  7.5% if $10,000  x  $30,000  9.3% if x $30,000  0.8 0.6  29c.  0.4  y  0.2  10 O  2  4 w  Tax Rate (percent)  5  d(w)  f(x) 1 0 1 2  0  O  10  20 30 Income (thousands of dollars)  x  29d. 9.5% 30. No; the functions are the same if x is positive. However, if x is negative, the functions yield different values. For example, [g 
 f ](1.5)  1 and [f 
 g](1.5)  1; [g 
 f ](
1.5)  2 and [f 
 g](
1.5)  1.  w  26a. step 26b. v is the value of the order, s(v) is the shipping,  3.50 if 0.00  v  25.00   5.95 if 25.01  v  75.00 s(v)    7.95 if 75.01  v  125.00   9.95 if 125.01  v  Chapter 1  40  29a. step  1.0  25. Absolute value; w is the weight in pounds, d(w) is the discrepancy, d(w)  1 
 w x 0 1 2 3  20  28d. d(63)  65 
 63 or 2 d(28)  65 
 28 or 37    f(x) 0.37 0.37 0.60 0.60 0.83 0.83  d (t )  |65 
 t |  O  24. Greatest integer; w is the weight in ounces, c(w) is the cost in dollars, 0.37  0.23(w 
 1) if [[w]]  w c(w)  0.37  0.23[[w]] if [[w]]  w x 0x1 1 1x2 2 2x3 3  25 50 75 100 125 150 Value of Order (dollars)  20  31a.  Public Transport  Graphing Linear Inequalities  1-8  60 50  Pages 54–55  Percent who use 40 public transportation 30 20 10 0 0  5  10  15  20  25  30  35  Working Population (hundreds of thousands)  31b. Sample answer: using (3,183,088, 53.4) and (362,777, 3.3) m   Check for Understanding  1. y  2x 
 6 2. Graph the lines 
3  2x  y and 2x  y  7. The graph of 
3  2x  y is solid and the graph of 2x  y  7 is dashed. Test points to determine which region(s) should be shaded. Then shade the correct region(s). 3. Sample answer: The boundaries separate the plane into regions. Each point in a region either does or does not satisfy the inequality. Using a test point allows you to determine whether all of the points in a region satisfy the inequality. 4. y  3.3 
 53.4  362,777 
 3,183,088 
50.1  or 0.0000178 
2,820,311  y 
 53.4  0.0000178(x 
 3,183,088) y  0.0000178x 
 3.26 31c. y  0.0000136x  4.55, r  0.68 31d. y  0.0000136(307,679)  4.55 y  8.73 8.73%; No, the actual value is 22%. 32. y 
 2  2(x 
 4) y 
 2  2x 
 8 2x 
 y 
 6  0 15 
 29  33a. (39, 29), (32, 15) 33b. m   32 
 39  x  O xy!4  5. ART  
14  y x  O    
7 or 2 33c. The average number of points scored each minute. 34. p(x)  (r 
 c)(x)  (400x 
 0.2x2) 
 (0.1x  200)  399.9x 
 0.2x2 
 200 35. Let x  the original price, or $59.99. Let T(x)  1.065x. (The cost with 6.5% tax rate) Let S(x)  0.75x. (The cost with 25% discount) [T 
 S](x)  (T(S(x))  (T(0.75x))  (T(0.75(59.99))  (T(44.9925)  1.065(44.9925)  $47.92 36. {
7, 
2, 0, 4, 9}; {
2, 0, 2, 3, 11}; Yes; no element of the domain is paired with more than one element of the range. 37. 5 612  10,883,911,680 5  612  2,176,782,341 So, 5  612 is not greater than 5 612. The correct choice is A.  3x 
 y ^ 6  6.  y  7!xy^9  x  O 7.  y  y ! |x  3|  21  O  x  Chapter 1  8a. c(m)  45  0.4m 8b. c(m)  y  14.  70 60  c (m) ^ 45  0.4m  x  O  50  –4 ^ x 
 y ^ 5  40 30  15.  20  y  16.  10  O  20  40  60  x  100 m  80  O  8c. Sample answer: (0, 45), (10, 49), (20, 50)  Pages 55–56  O  x  Exercises  9.  17.  y  y  y 6 |x |  4  y<3  x  O  x  O 10.  18.  y  y  y ! |2x  3|  x  O x–y>–5  11.  x  O 19.  y  y 
8 ^ 2x  y ! 6  2x  4y 6 7  x  O  x  O 12.  
y ! 2x 
 1  y |x  3| ! y 
 1  y 6 2 x  19 5  O  20.  y  13.  y  5  x O O  Chapter 1  y 
2 ^ x  2y ^ 4  y 6 |x |  x  22  x  21.  y  y  26b.  4.252  x60 y60  4.25 4.248  O  O  x  27a. 0.6(220 
 a)  r  0.9(220 
 a) 27b. r  y  22.  4 3 2 1 O  200  2 4 6 8 10x  
10
8
6
4
2 
2 
3 
4  0.6 (220 
 a) ^ r ^ 0.9 (220 
 a)  100  23a. 8x  10y  8(60) 8x  10y  480 23b. y  O  60 40  60  20  30  40  50  60  70 a  275 220 165 110 55  20 40  10  28a. step 28b. Let c(h) represent the cost for h hours. 55h if [[h]]  h Then c(h)  55[[h  1]] if [[h]]  h 28c. y  80  O 20  x  7.998 8 8.002  80  x  23c. Sample answer: (0, 48) (60, 0), (45, 6) 23d. Sample answer: using complex computer programs and systems of inequalities. 24. y  O  2 3 4 Hours  5  x  29a. 3x 
 y  10 y  3x 
 10 y 
 (
2)  3(x 
 0) 3x 
 y 
 2  0  |y| 6 x  x  O  1  1  29b. perpendicular slope: 
3 1  y 
 (
2)  
3(x 
 0) 1  y  2  
3x 3y  6  
x x  3y  6  0  25a. points in the first and third quadrants 25b. If x and y satisfy the inequality, then either x  0 and y  0 or x  0 and y  0. If x  0 and y  0, then x  x and y  y. Thus, x  y  x  y. Since x  y is positive, x  y  x  y. If x  0 and y  0, then x  
x and y  
y. Then x  y  
x  (
y) or 
(x  y). Since both x and y are negative, (x  y) is negative, and x  y  
(x  y). 26a. 8 
 x   1 ; 500  1 44  
 y   7
4   30. m   5
1  3  y 
 7  4(x 
 5)  3  3  1  y  4x  34   4 31a. (0, 23), (16, 48); 48 
 23   m 16 
 0 25   16 or 1.5625 31b. the average change in the temperature per hour  1  500  32.  95 
 94  8  59,049 
 6561   8 52,488   8 or 6561 or 94 The correct choice is E.  23  Chapter 1  20. (f  g)(x)  f(x)  g(x)  4 
 x2  3x or 4  3x 
 x2 (f 
 g)(x)  f(x) 
 g(x)  4 
 x2 
 (3x)  4 
 3x 
 x2 (f g)(x)  f(x) g(x)  (4 
 x2)(3x)  12x 
 3x3  Chapter 1 Study Guide and Assessment Page 57 1. 3. 5. 7. 9.  Understanding the Vocabulary 2. 4. 6. 8. 10.  c d i h e  f g a j b   (x)  f  g    Pages 58–60  15. g   (x)  f  g  2  1 5 3       2  5  3  or    6  5    f  g     (x)  f  g       4     x3 
 8x2  16x  4  x
4 4  x2 
 4x   x 
 4,  (f 
 g)(x)  f(x) 
 g(x)  x2 
 4x 
     (x2 
 4x) x  2   (x)  f  g       4x, x 
 4 f(x)  g(x) x2 
  x
4    4  x
4  x3 
 8x2  16x 
 4  x
4 4 2  x 
 4x 
  x 
 4,  (f g)(x)  f(x) g(x)  f(x)    f(x)  g(x) x2 
 1  x1 (x 
 1)(x  1)  x1    x2 
 4x   x
4   (x)   g(x ) x2  4x , x
2  f(x)  g(x) x2  7x  12  x4 (x  4)(x  3)  x4   x  1, x 
 
1 23. (f  g)(x)  f(x)  g(x)  f(x)  g(x) 6x 
 4  2   3x 
 2 19. (f  g)(x)  f(x)  g(x)  x2  4x  x 
 2  x2  5x 
 2 (f 
 g)(x)  f(x) 
 g(x)  x2  4x 
 (x 
 2)  x2  3x  2 (f g)(x)  f(x) g(x)  (x2  4x)(x 
 2)  x3  2x2 
 8x f  g  0   x  3, x 
 
4 22. (f  g)(x)  f(x)  g(x)  x2 
 1  x  1  x2  x (f 
 g)(x)  f(x) 
 g(x)  x2 
 1 
 (x  1)  x2 
 x 
 2 (f g)(x)  f(x) g(x)  (x2 
 1)(x  1)  x 3  x2 
 x 
 1  16. k(4c)  (4c)2  2(4c) 
 4  16c2  8c 
 4 17. f(m  1)  (m  1)2  3(m  1)  m2  2m  1  3m  3  m2  5m  4 18. (f  g)(x)  f(x)  g(x)  6x 
 4  2  6x 
 2 (f 
 g)(x)  f(x) 
 g(x)  6x 
 4 
 (2)  6x 
 6 (f g)(x)  f(x) g(x)  (6x 
 4)(2)  12x 
 8   (x)   x  21. (f  g)(x)  f(x)  g(x)  x2  7x  12  x  4  x2  8x  16 (f 
 g)(x)  f(x) 
 g(x)  x2  7x  12 
 (x  4)  x2  6x  8 (f g)(x)  f(x) g(x)  (x2  7x  12)(x  4)  x3  11x2  40x  48  Skills and Concepts  11. f(4)  5(4) 
 10  20 
 10 or 10 12. g(2)  7 
 (2)2  7 
 4 or 3 13. f(
3)  4(
3)2 
 4(
3)  9  36  12  9 or 57 14. h(0.2)  6 
 2(0.2)3  6 
 0.016 or 5.984 1  3  f(x)  g(x) 4 
 x2 , 3x  x  4    4  x
4  x3 
 8x2  16x 4x   or  , x 
 4 4 4  x
4  Chapter 1  24  24. [f 
 g](x)  f(g(x))  f(2x)  (2x)2 
 4  4x2 
 4 [g 
 f ](x)  g(f(x))  g(x2 
 4)  2(x2 
 4)  2x2 
 8 25. [f 
 g](x)  f(g(x))  f(3x2)  0.5(3x2)  5  1.5x2  5 [g 
 f ](x)  g(f(x))  g(0.5x  5)  3(0.5x  5)2  0.75x2  15x  75 26. [f 
 g](x)  f(g(x))  f(3x)  2(3x)2  6  18x2  6 [g 
 f ](x)  g( f(x))  g(2x2  6)  3(2x2  6)  6x2  18 27. [f 
 g](x)  f(g(x))  f(x2 
 x  1)  6  (x2 
 x  1)  x2 
 x  7 [g 
 f ](x)  g(f(x))  g(6  x)  (6  x)2 
 (6  x)  1  x2  11x  31 28. [f 
 g](x)  f(g(x))  f(x  1)  (x  1)2 
 5  x2  2x 
 4 [g 
 f ](x)  g( f(x))  g(x2 
 5)  x2 
 5  1  x2 
 4 29. [f 
 g](x)  f(g(x))  f(2x2  10)  3 
 (2x2  10)  
2x2 
 7 [g 
 f ](x)  g(f(x))  g(3 
 x)  2(3 
 x)2  10  2x2 
 12x  28 30. Domain of f(x): x  16 Domain of g(x): all reals g(x)  16 5 
 x  16 x  
11 Domain of [f 
 g](x) is x  
11.  31. The y-intercept is 6. The slope is 3. 18  y  12 6  y  3x  6  O 6  
6  12 x  32. The y-intercept is 8. The slope is 
5. y 8  y = 8 – 5x  4 –4 O  x  4  33. y 
 15  0 y  15 The y-intercept is 15. The slope is 0.  y 20 y 
 15  0 10 
10 O  10  x  34. 0  2x 
 y 
 7 y  2x 
 7 The y-intercept is 
7. The slope is 2.  y O  x  y  2x 
 7  35. The y-intercept is 0. The slope is 2.  y y  2x  x  O  36. The y-intercept is 
2. The slope is 
8.  y O  x  y  
8x 
 2  25  Chapter 1  37. 7x  2y  
5 y  1  7 
2x  
  The y-intercept is  50. y 
 (
8)  2(x 
 4)  5  2  1  5 
2.  The slope is  y  8  2x 
 2  7 
2.  2y  16  x 
 4 x 
 2y 
 20  0  y  1  4    51. m  
 (
2) or 2, perpendicular slope is 
 2  7x  2y  
5  1  y 
 4  
2(x 
 1)  x  O  1  1  y 
 4  
2x  2 38. The y-intercept is 
6. The slope is y 
4 O 2 4 x  2y 
 8  
x  1 x  2y 
 9  0 52. x  
8 is a vertical line; perpendicular slope is 0. y 
 (
6)  0(x 
 4) y60 Overseas Visitors 53a.  1 . 4  y  1x 
 6  28  4  21  39. y  2x 
 3  40. y  
x  1  Visitors 14 (millions)  1  41. y 
 2  2(x 
 (
5)) 1  5  y 
 2  2x  2 42.  7  1 9 y  2x  2 5
2  m 2 
 (
4) 3 1  6 or 2 1 y 
 5  2(x 
 2) 1 y 
 5  2x 
 1 1 y  2x  4  0 1994 1996 1998 2000 Year  53b. Sample answer: using (1994, 18,458) and (2000, 25,975) m   43. (1, 0), (0, 
4) m   
4 
 0  0
1 
4  or 4 
1  y 
 (
4)  4(x 
 0) y  4  4x y  4x 
 4 44. y  
1 46. y 
 0  0.1(x 
 1) y  0.1x 
 0.1 47. y 
 1  1(x 
 1) y
1x
1 x
y0  25,975 – 18,458  2000 – 1994 7517  or 1252.8 6  y 
 25,975  1252.8(x 
 2000) y  1252.8x 
 2,479,625 53c. y  1115.9x 
 2,205,568; r  0.9441275744 53d. y  1115.9(2008) 
 2,205,568  35,159.2 35,159,200 visitors; Sample answer: This is a good prediction, because the r value indicates a strong relationship. 54. f (x )  45. y  0  1  y 
 6  3(x 
 (
1))  48.  1  O  1  y 
 6  3x  3 3y 
 18  x  1 x 
 3y  19  0 2  49. m  
1 or 
2 y 
 2  
2(x 
 (
3)) y 
 2  
2x 
 6 2x  y  4  0  Chapter 1  26  x  55.  y  60. ART  h (x )  x^5  x  O  O  x  61.  56.  x 
2  x  
1 
1  x  0 0x1 1x2 2x3  f(x) 
1 0 1 2 3  y  x  O xy^1  y  62.  f (x )  2y 
 x ^ 4  x  O  f (x )  x   1  x  O  63.  64.  y  y y 
 3x 1 2  O 57.  x 
2 
1 0 1 2  O  y ^ |x |  g (x )  f(x) 8 4 0 4 8  x  65.  x  y y ! |x| 
 2  g (x )  |4x |  O  x  O x  58.  x 
2 
1 0 1 2  k (x)  f(x) 6 4 2 4 6  66.  y  O O  y ! |x 
 2|  x  x  y  59. y14  O  x  27  Chapter 1  Page 61 67a. d   Applications and Problem Solving  2. Since this is a multiple-choice question, you can try each choice. Choice A, 16, is not divisible by 12, so eliminate it. Choice B, 24, is divisible by both 8 and 12. Choice C, 48, is also divisible by both 8 and 12. Choice D, 96, is also divisible by both 8 and 12. It cannot be determined from the information given. The correct choice is E. 3. Write the mixed numbers as fractions.  1 (20)(1)2 2   10 1  d  2(20)(2)2  40 1  d  2(20)(3)2  90 d  1  1  d  2(20)(5)2  1  43  3  25   250 10 m, 40 m, 90 m, 160 m, 250 m 67b. Yes; each element of the domain is paired with exactly one element of the range. 68a. (1999, 500) and (2004, 636)   636 – 500  2004 – 1999 136  or 27.2; 5  about $27.2 billion  Open-Ended Assessment  1. Possible answer: f(x)  4x 
 4, g(x)  x2; [f 
 g](x)  f(g(x))  4(x2) 
 4  4x2 
 4 2a. No; Possible explanation: If the lines have the same x-intercept, then they either intersect on the x-axis or they are the same line. In either case, they cannot be parallel. 2b. Yes; Possible explanation: If the lines have the same x-intercept, they can intersect on the x-axis. If they have slopes that are negative reciprocals, then they are perpendicular.  4 if x  4 3a. y   2x 
 4 if x  4 x 
 1 if x  
1 3b. y   3x  1 if x  
1  13   3  1  5  13  5   3  1  1  3  13  13     16  1 6  4  16  16  16 16  The correct choice is A. 7. Use your calculator. First find the total amount per year by adding. $12.90  $16.00  $18.00  $21.90  $68.80 Then find one half of this, which is the amount paid in equal payments. $68.80 2  $34.40 Then divide this amount by 4 to get each of 4 monthly payments. $34.40 4  $8.60. The correct choice is A.  SAT and ACT Practice  1. Prime factorization of a number is a product of prime numbers that equals the number. Choices A, B, and E contain numbers that are not prime. Choice C does not equal 54. Choice D, 3 3 3 2, is the prime factorization of 54. The correct choice is D.  Chapter 1  13  3  13  5  (2)
4  24  1 6  Chapter 1 SAT & ACT Preparation Page 65  13  The correct choice is B. 4. Since this is a multiple-choice question, try each choice to see if it answers the question. Start with 10, because it is easy to calculate with tens. If 10 adult tickets are sold, then 20 student tickets must be sold. Check to see if the total sales exceeds $90. Students sales  Adult sales $90 20($2.00)  10($5.00)  40  50  $90 So 10 is too low a number for adult tickets. This eliminates answer choices A, B, C, and D. Check choice E. Eleven is the minimum number of adult tickets. 19($2.00)  11($5.00)  38  55  $93 The correct choice is E. 5. Recall the definition of absolute value: the number of units a number is from zero on the number line. Simplify the expression by writing it without the absolute value symbols. 
7  7 

7  
7 

7

5
3
4  
7 
 5 
 12  
24 The correct choice is A. 6. Write each part of the expression without exponents. (
4)2  16  68b. y 
 500  27.2(x 
 1999) y  27.2x 
 53,872.8 69. y  
0.284x  12.964; The correlation is moderately negative, so the regression line is somewhat representative of the data.  Page 61  3  25  5  Remember that dividing by a fraction is equivalent to multiplying by its reciprocal   160  m  13  43  3  1 (20)(4)2 2  28  10. First find the number of fish that are not tetras.  8. First combine the numbers inside the square root symbol. Then find the square root of the result. 
64    
100 36   10 The correct choice is A. 9. 60  2 2 3 5  22 3 5 The number of distinct prime factors of 60 is 3. The correct choice is C.  18(24) or 3 are tetras. 24 
 3 or 21 are not 2  tetras. Then 3 of these are guppies.  23(21)  14  The answer is 14.  29  Chapter 1  Chapter 2 Systems of Linear Equations and Inequalities y  6.  Solving Systems of Equations in Two Variables  2-1  no solution  y
x2  O  Page 69  Graphing Calculator Exploration  y
x5  1. (1, 480) 3  2. 3x 
 4y  320  →  5x  2y  340  y  4x 
 80 5  y  
2x  170  7. 7x  y  9 5x 
 y  15 12x  24 x2 8. 3x  4y  
1 2(6x 
 2y)  2(3)  (76.923077, 
22.30769) 3. accurate to a maximum of 8 digits 5  4. 5x 
 7y  70  →  10x 
 14y  120  y  7x 
 10 5  60  y  7x 
 7  Inconsistent; error message occurs. 5. See students' systems and graphs; any point in TRACE mode will be the intersection point since the two lines intersect everywhere.  7x  y  9 7(2)  y  9 y  
5 (2, 
5) 3x  4y  
1 12x 
 4y  6 15x  5  →  x  1  3  3x  4y  
1 33  4y  
1 1  Page 70  4y  
2  Check for Understanding  →  4y  16 
 6x    1  3    9. 30 3x 
 2y  30(
4) 
2(5x 
 4y)  
2(14)  →  1  10x 
 45y  
120 
10x  8y  
28 
37y  
148 y4  5x 
4y  14 5x 
 4(4)  14 5x  30 x6 (6, 4) 10. Let b represent the number of baseball racks and k represent the number of karate-belt racks. b  6k b  6k 3b  5k  46,000  6(2000) 3(6k)  5k  46,000  12,000 23k  46,000 k  2000 12,000 baseball, 2000 karate  y  
2x  3 3  y  
2x  4  Pages 71–72  Inconsistent; Sample answer: The graphs of the equations are lines with slope  3 
2,  Exercises  11. x  3y  18  but each  
x  2y  7  equation has a different y-intercept. Therefore, the graphs of the two equations do not intersect and the system has no solution. 5. (1, 3) y  1  y  
3x  6  →  1  7  y  2x  2  consistent and independent 12. y  0.5x y  0.5x → 1 2y  x  4 y  2x  2 inconsistent 13. 
35x  40y  55  y 
 5x  2  (1, 3)  7x  8y 
 11  y 
 2x  5  Chapter 2  1  3  4. 2y  3x  6  O  3, 
2  1  y  
2  1. Sample answer: 4x  7x  21 y  2x 
 1 The substitution method is usually easier to use whenever one or both of the equations are already solved for one variable in terms of the other. 2. Sample answer: Madison might consider whether the large down-payment would strap her financially; if she wants to buy the car at the end of the lease, then she might also consider which lease would offer the best buyout. 3. Sample answer: consistent systems of equations have at least one solution. A consistent, independent system has exactly one solution; a consistent, dependent system has an infinite number of solutions. An inconsistent system has no solution. See students' work for examples and solutions.  →  consistent and dependent  x  30  7  11  7  11  y  8x  8 y  8x  8  14.  y  15.  y x
5  O  y 
 3  y  16.  O y 
 x  2  (4, 3)  (4, 
3) 17. y  (5, 0)  x  y 
 13 x  56  y 
 3x  10  x  O  (2, 4)  y
  →  
6x  20y  10 6x  21y  72 41y  82 y2  2x  7y  24 2x  7(2)  24 2x  10 x5 (5, 2) 26. 2x 
 y  
7 x  2y 
 8 2(2y 
 8) 
 y  
7 x  2(3) 
 8 3y  9 x  
2 y3 (
2, 3) 27. 3(2x  5y)  3(4) 6x  15y  12 
2(3x  6y)  
2(5) → 
6x 
 12y  
10 3y  2 2   2x  5y  4 y 3  x  O  x  (5, 0)  25. 2(
3x  10y  5) 3(2x  7y  24)  x
4  y 
45 x  4  13 x  2x  53  4 2  (2, 
4)  y  18.  19. y
x2  no solution y (0, 3)  O  2  2x  3 1  28. 55x 
 6y  5(1) 3  x  y 
 12 x  2  O  (0, 2)  y 
 32 x  3  1 x 5  x  1  3, 3 1 2  x  3  y 
  14 x  3  5   6y  1  →  5  3x 
 6y  5 1 5 x  y 5 6 16 x 5   11  16  x5 (0, 
2) 20. 3x 
 8y  10  →  16x 
 32y  75  21.  22.  23.  24.  1 x 5  5  6y  11 1 5 (5)  y  11 5 6 5 y  10 6  (0, 3) 3  5  1  75  y  8x 
 2 y  2x 
 32  Consistent and independent; if each equation is written in slope-intercept form, they have different slopes, which means they will intersect at some point. 3(5x 
 y)  3(16) 15x 
 3y  48 → 2x  3y  3 2x  3y  3 17x  51 x3 5x 
 y  16 5(3) 
 y  16 
y  1 y  
1 (3, 
1) 3x 
 5y  
8 3x 
 5y  
8 
3(x  2y)  
3(1) → 
3x 
 6y  
3 
11y  
11 x  2y  1 y1 x  2(1)  1 x  
1 (
1, 1) x  4.5  y y6
x x  4.5  6 
 x y  6 
 5.25 2x  10.5 y  0.75 x  5.25 (5.25, 0.75) 5(2x  3y)  5(3) 10x  15y  15 → 12x 
 15y  
4 12x 
 15y  
4 22x  11 1 2x  3y  3 x  2  y  12 (5, 12) 29. 7(4x  5y)  7(
8) 28x  35y  
56 5(3x 
 7y)  5(10) → 15x 
 35y  50 43x  
6 6 4x  5y  
8 x  
4 3    320 64  y  
4 3  
463 , 
6443   30. 
2(3x 
 y)  
2(
9) 
6x  2y  18 → 4x 
 2y  
8 4x 
 2y  
8 
2x  10 3x 
 y  
9 x  
5 3(
5) 
 y  
9 
y  6 y  
6 (
5, 
6) 31. Sample answer: Elimination could be considered easiest since the first equation multiplied by 2 added to the second equation eliminates b; substitution could also be considered easiest since the first equation can be written as a  b, making substitution very easy. a
b0 3a  2b  
15 ab 3(b)  2b  
15 5b  
15 b  
3 a
b0 a 
 (
3)  0 a  
3 (
3, 
3) 32a. B  1  3y  2 2    5y  
4 3  22  3y  3 y  3  6  4 
4 3  5y  
8  2, 3 1 2  31  Chapter 2  32b. S 
 4V  0 S  4V  S  V  30,000 4V  V  30,000 5V  30,000 V  6000  38. Let x represent the number of people in line behind you. 200  x represents the number in front of you. Let 
 represent the whole line. 200  x  1  x  
 
  3x 200  x  1  x  3x 
  3x 201  x 
  3(201) 603 people  603 y 39.  S 
 4V  0 S 
 4(6000)  0 S 
 24,000  0 S  24,000 Spartans: 24,000; visitors: 6000 33a. Let b represent the base and 
 represent the leg. Perimeter of first triangle: b  2
  20 Perimeter of second triangle: 6  b  
  20 b  2
  20 b  20 
 2
 6  b  
  20 b  2
  20 6  20 
 2
  
  20 b  2(6)  20 

  
6 b8 
6 6, 6, 8; 6, 6, 8 33b. isosceles 34. y 
 (
3)  4(x 
 4) y  4x 
 19  2x  7  y  40.  x 
2 
1 0 1 2  1  y 
 (
3)  
4(x 
 4) 1  y  
4x 
 2 35a. Let x represent the number of refills. Then, x  1  number of drinks purchased. C  2.95  0.50x C  0.85  0.85x C  2.95  0.50x 0.85  0.85x  2.95  0.50x 0.35x  2.1 x6 x17 C  2.95  0.50x C  2.95  0.50(7) or 5.95 (7, 5.95) 35b. If you drink 7 servings of soft drink, the price for each option is the same. If you drink fewer than 7 servings of soft drink during that week, the disposable cup price is better. If you drink more than 7 servings of soft drink, the refillable mug price is better. See students' choices. 35c. Over a year's time, the refillable mug would be more economical. 36a. 36b. 36c.  a d   b e a d   , c b e a d   , c b e  45.  f (x )  O f (x) 
 2|x|  3  
25   5    5  5   
1  1 The correct choice is A.  2-2  Solving Systems of Equations in Three Variables  f f  Page 76  3.5 x 4 3.5 (516 4  Check for Understanding  1. Solving a system of three equations involves eliminating one variable to form two systems of two equations. Then solving is the same.   264  y  264  y   y)  264  y  451.5  0.875y  264  y y  $1500  Chapter 2  f(x) 1 
1 
3 
1 1  41. y 
 6  2(x 
 0) y  2x  6 42. $12,500 43. [f 
 g](x)  f(g(x))  f(x  2)  3(x  2) 
 5  3x  1 44. {18}, {
3, 3}; no, because there are two range values paired with a single domain value.  37. Let x represent the full incentive. Let y represent the value of the computer. x  516  y 3.5 x 4  x  O  32  x  1  2. The solution would be an equation in two variables. Sample example: the system 2x  4y  6z  12, x  2y  3z  6, and 3x 
 5y 
 6z  27 has a solution of all values of x and y that satisfy 5x 
 y  39. 2x  4y  6z  12 
2(x  2y  3z)  
2(6) 3x 
 5y 
 6z  27 
2x 
 4y 
 6z  
12 5x 
 y  39 ↓ 
2x 
 4y 
 6z  
12 2x  4y  6z  12 0 0 all reals 3. Sample answer: Use one equation to eliminate one of the variables from the other two equations. Then eliminate one of the remaining variables from the resulting equations. Solve for a variable and substitute to find the values of the other variables. 4. 4(4x  2y  z)  4(7) 2(4x  2y  z)  2(7) 2x  2y 
 4z  
4 x  3y 
 2z  
8 ↓ ↓ 16x  8y  4z  28 8x  4y  2z  14 2x  2y 
 4z  
4 x  3y 
 2z  
8 18x  10y  24 9x  7y  6 18x  10y  24 
2(9x  7y)  
2(6) →  
2.5(75)  
2.5(0.5a  v0  s0) 75  3.125a  2.5v0  s0 ↓ 
187.5  
1.25a 
 2.5v0 
 2.5s0 75  3.125a  2.5v0  s0 
112.5  1.875a 
 1.5s0 
4(75)  
4(0.5a  v0  s0) 3  8a  4v0  s0 ↓ 
300  
2a 
 4v0 
 4s0 3  8a  4v0  s0 
297  6a 
 3s0 
2(
112.5)  
2(1.875a 
 1.5s0) 
297  6a 
 3s0  18x  10y  24 
18x 
 14y  
12 
4y  12 y  
3 4x  2y  z  7 4(3)  2(
3)  z  7 z1  9x  7y  6 9x  7(
3)  6 x3 (3, 
3, 1) 5. 2(x 
 y 
 z)  2(7) 
x  2y 
 3z  
12 ↓ 2x 
 2y 
 2z  14 
x  2y 
 3z  
12 x 
5z  2 
2(x 
 5z)  
2(2) 2x  4z  18  7. 75  2a(1)2  v0(1)  s0  0.5a  v0  s0 1 75  2a(2.5)2  v0(2.5) s0  3.125a  2.5v0  s0 1 3  2a(4)2  v0(4)  s0  8a  4v0  s0  ↓ 225  
3.75a  3s0 
297  6a 
 3s0 
72  2.25a 
32  a 
297  6a 
 3s0 3  8a  4v0  s0 
297  6(
32) 
 3s0 3  8(
32)  4v0  s0 35  s0 56  v0 acceleration 
32 ft/s2, initial velocity: 56 ft/s, initial height: 35 ft  
x  2y 
 3z  
12 3x 
 2y  7z  30 2x  4z  18  Pages 76–77  
2x  10z  
4 2x  14z  18 14z  14 z1 x 
 5z  2 x
y
z7 x 
 5(1)  2 7
y
17 x7 y  
1 (7, 
1, 1) 6. 
2(2x 
 2y  3z) 
2(2x 
 3y  7z)  
2(6)  
2(
1) 4x 
 3y  2z  0 4x 
 3y  2z  0 ↓ ↓ 
4x  4y 
 6z  
12 
4x  6y 
 14z  2 4x 
 3y  2z  0 4x 
 3y  2z  0 y 
 4z  
12 3y 
 12z  2 
3(y 
 4z)  
3(
12) 3y 
 12z  2  →  →  Exercises  8. 3(5x  3y 
 z)  3(
11) x  2y  3z  5 ↓ 15x  9y 
 3z  
33 x  2y  3z  5 16x  11y  
28 
2(5x  3y 
 z)  
2(
11) 3x  2y 
 2z  
13 ↓ 
10x 
 6y  2z  22 3x  2y 
 2z  
13 
7x 
 4y  9 4(16x  11y)  4(
28) 64x  44y  
112 → 11(
7x 
 4y)  11(9) 
77x 
 44y  99 
13x  
 13 x1 16x  11y  
28 x  2y  3z  5 16(1)  11y  
28 1  2(
4)  3z  5 y  
4 z4 (1, 
4, 4)  
3y  12z  36 3y 
 12z  2 0  38  no solution  33  Chapter 2  9. 7(
x  3y  2z)  7(16) 7x  5y  z  0 ↓ 
7x  21y  14z  112 7x  5y  z  0 26y  15z  112 
15(
3y  z)  
15(
2) 26y  15z  112 
3y  z  
2 
3(2)  z  
2 z4 (
2, 2, 4) 10. 2(2x  y 
 2z)  2(11) 
x 
 2y  9z  13 ↓ 4x  2y 
 4z  22 
x 
 2y  9z  13 3x  5z  35 
3(x 
 3z)  
3(7) 3x  5z  35  13. 3(x  y 
 z)  3(3) 2(x  y 
 z)  2(3) 4x 
 3y  2z  12 
2x 
 2y  2z  5 ↓ ↓ 3x  3y 
 3z  9 2x  2y 
 2z  6 4x 
 3y  2z  12 
2x 
 2y  2z  5 7x  z  21 0  11 no solution 14. 3(36x 
 15y  50z)  3(
10) 
5(54x 
 5y  30z)  
5(
160) ↓ 108x 
 45y  150z  
30 
270x  25y 
 150z  800 
162x 
 20y  770  
x  3y  2z  16 x 
 6y 
 2z  
18 
3y  2z  
 2  45y 
 15z  30 26y  15z  112 71y  142 y2 x 
 6y 
 z  
18 x 
 6(2) 
 4  
18 x  
2 →  81(2x  25y)  81(40) 
162x 
 20y  770  162x  2025y  3240 → 
162x 
 20y  770 2005y  4010 y2 36x 
 15y  50z  
10 36(
5) 
 15(2)  50z  
10 z4  →  2x  25y  40 2x  25(2)  40 x  
5 (
5, 2, 4) 15. 4(
x 
 3y  z)  4(54) 4x  2y 
 3z  
32 ↓ 
4x 
 12y  4z  216 4x  2y 
 3z  
32 
10y  3z  184  
3x  9z  
21 3x  5z  35 14z  14 z1 2x  y 
 2z  11 2(10)  y 
 2(1)  11 y  
7  x 
 3z  7 x 
 3(1)  7 x  10 (10, 
7, 1) 11. 
3(x 
 3y 
 2z)  
3(
8) 
5(x 
 3y 
 2z)  
5(
8) 3x 
 5y  z  9 5x 
 6y  3z  15 ↓ ↓ 
3x  9y  6z  24 
5x  15y  10z  40 3x 
 5y  z  9 5x 
 6y  3z  15 4y  7z  33 9y  13z  55 9(4y  7z)  9(33) 36y  63z  297 → 
4(9y  13z)  
4(55) 
36y 
 52z  
220 11z  77 z7 4y  7z  33 x 
 3y 
 2z  
8 4y  7(7)  33 x 
 3(
4) 
 2(7)  
8 y  
4 x  
6 (
6, 
4, 7) 12. 8x  y 
 z  4 yz5 8x  y 9 
1(8x  y)  
1(9) 11x  y  15 8x 
 z  4 8(2) 
 z  4 z  12 (2, 
7, 12)  Chapter 2  →  5(2y 8z)  5(78) 
10y  z  184  →  2y  8z  78 2y  8(14)  78 y  
17 (11, 
17, 14) 16. 1.8x  1.2y 
 z  0.7 1.2y  z  0.7 1.8x  1.2y  0  10y  40z  390 
10y  z  184 41z  574 z  14 
x 
3y  z  54 
x 
 3(
17)  14  54 x  11  3(1.8x  1.2y)  3(0) 5.4x  3.6y  0 1.2(1.5x 
 3y)  1.2(3) → 1.8x 
 3.6y  3.6 7.2x  3.6 x  0.5 1.5x 
 3y  3 1.8x 
 z  0.7 1.5(0.5) 
 3y  3 1.8(0.5) 
 z  0.7 y  
0.75 z  0.2 (0.5, 
0.75, 0.2) 17. y  x  2z z  
1 
 2x y  (y 
 14)  2z 7  
1 
 2x 7z 
4  x x  y 
 14 
4  y 
 14 10  y (
4, 10, 7)  
8x 
 y  
9 11x  y  15 3x  6 x2 yz5 y  12  5 y  
7  34  18.      52(
12)  5 3 1 1  x  y 
 z 2 4 6 3 1 2 5 x 
 y  z 8 3 6  20c. Sample answer: x  y  z  6; 2x 
 y 
 2z  8; x 
 2y 
 3z  2   
8  1  21. 124  2a(1)2  v0(1)  s0  ↓ 15 x 8 1 x 8  5  1  5  272  2a(3)2  v0(3)  s0  5  82  2a(8)2  v0(8)  s0    1 2 y 
 6 z  
30 2 y 3 1 y 4  
  2x 
  1   6z  
8  
38      7 3 1 1 
4 4x  6y 
 3z 3 7 5 x 
 y 
 z 16 12 8  124  272   7 
4(
12)  7  
 
  5 y 8 11 y 12    1 2x 
 4y 11 9 
8x 
 12y  
1(124)  
12a  v0  s0  21  9  272  2a  3v0  s0  7  
 1 2 z  
25  ↓ 1 
124  
2a 
 v0 
 s0   
4    9  16  9 (
38) 16  →   
4  9 x 8 9 
8x  9 
 6 4y 11 
 12y 203  
 192 y    272  148   171 
8   
4   2x 
 1  3 x 4   
38  2x 
 4(24)  
38  3 (
16) 4   1  1 y 6  9 a 2   3v0  s0 4a  2v0  
1(124)  
12a  v0  s0 82  32a  8v0  s0 ↓ 1 
124  
2a 
 v0 
 s0 82  32a  8v0  s0 
42  31.5a  7v0 1  203 
8  y  24 1 y 4   3v0  s0  1    
1 6 x 
 24 y  12 z  3 x 16 9 
8x  ↓  v0  s 0  82  32a  8v0  s0   
25  ↓ 7  21  1 a 2 9 a 2  1  
 3z  
12 1   6(24) 
 3z  
12  x  
16 z  12 (
16, 24, 12) 19. Let x represent amount in International Fund, y represent amount in Fixed Assets Fund, and z represent amount in company stock.  7(148)  7(4a  2v0) 
2(
42)  
2(31.5a  7v0) ↓ 1036  28a  14v0 84  
63a 
 14v0 1120  
35a 
32  a 1 124  2a  v0  s0 148  4a  2v0  x  y  z  2000 x  2z 0.045x  0.026y 
 0.002z  58  148  4(
32)  2v0  x  y  z  2000 2z  y  z  2000 y  3z  2000 y  2000 
 3z  1  124  2(
32)  138  s0  138  v0 2  s0 (
32, 138, 2) 22a. Sample answer: A system has no solution when you reach a contradiction, such as 1  0, as you solve. 22b. Sample answer: A system has an infinite number of solutions when you reduce the system to two equivalent equations such as x  y  1 and 2x  2y  2.  0.045x  0.026y 
 0.002z  58 0.045(2z)  0.026y 
 0.002z  58 0.026y  0.088z  58 0.026y  0.088z  58 0.026(2000 
 3z)  0.088z  58 z  600 x  2z x  y  z  2000 x  2(600) 1200  y  600  2000 x  1200 y  200 International Fund  $1200; Fixed Assets Fund  $200; company stock  $600 20a. Sample answer: x  y  z  15; 2x  z  1; 2y 
 z  7 20b. Sample answer: 4x  y  z  12; 
4x 
 y 
 z  
10; 5y 
 z  9  35  Chapter 2  23. x  yz  2 y  xz  2 z  xy  2 x  yz  2 y  xz  2  y  26.  D (3, 6) (1, 3) A  x yz  2 
y 
 xz  
2 (x 
 y)  (yz 
 xz)  0 (x 
 y) 
 z(x 
 y)  0 (1 
 z)(x 
 y)  0 1 
 z  0 or x 
 y  0 z1 xy y  xz  2 y  xz  2 → z  xy  2 
z 
 xy  
2 ( y 
 z) 
 (xz 
 xy)  0 ( y 
 z) 
 x(y 
 z)  0 (1 
 x)(y 
 z)  0 1 
 x  0 or y 
 z  0 x1 yz If z  1 and x  1, 1  1y  2 and y  1. If z  1 and y  z, x  1 1  2 and x  1. If x  y and x  1, 1  1z  2 and z  1. If x  y and y  z, x  y  z. xx x2 x  x2  2 2 x x
20 (x  2)(x 
 1)  0 x  2  0 or x 
 1  0 x2  
2 x1 If x  
2, y  
2 and z  
2. The answers are (1, 1, 1) and (
2, 
2, 
2). 24. 3x  4y  375 → 3x  4y  375 
2(5x  2y)  
2(345) 
10x 
 4y  
690 
7x  
315 x  45 5x  2y  345 5(45)  2y  345 y  60 (45, 60) y 25. →  C (6, 2)  O  B (2, 1)  x  AB: d  
 (
1 
 3)2  (2 
  (
1))2  5 2 (2 
 ( 
1))   (6 
 2)2  5 BC: d  
 CD: d  
 (6 
 2 )2  (3 
 6)2  5 2 AD: d  
 (6 
 3 )  (3 
 (
 1))2  5 
1 
 3   AB: m   2 
 (
1)  2 
 (
1)   B C: m    6
2  4  3   
3   4  AB  BC  CD  AD  5 units; ABCD is 4  3  rhombus. Slope of A B   
3 and slope of B C   4, so  AB ⊥B C . A rhombus with a right angle is a square. 27a. (20, 3000), (60, 5000) 5000 
 3000   m 60 
 20 2000   40 or 50 y 
 5000  50(x 
 60) y  50x  2000 C(x)  50x  2000 27b. $2000; $50 27c. c (x ) 4  3 Cost 2 ($1000) 1  c(x) 
 50x  2000  O 1 2 3 4 5 6 7 8 9x Televisions Produced  x  28.  O y  13 x  2  A A  r2 2  s2  (
2  )2 
2 s  2 The correct choice is C.  2-3  s2  Modeling Real-World Data with Matrices  Pages 82–83  Check for Understanding  1. Sample answer: film pain reliever blow dryer (24 exp.) (100 ct) $4.03 $6.78 $18.98  $4.21 $7.41 $20.49  $3.97 $7.43 $32.25   $7.08 $36.57 $63.71          Atlanta Los Angeles Mexico City Tokyo 2. 2 4 3. The sum of two matrices exists if the matrices have the same dimensions. Chapter 2  36  4. Anthony is correct. A third order matrix has 3 rows and 3 columns. This matrix has 4 rows and 3 columns. 5. 2y  x 
 3 2y  y  5 
 3 xy5 y2 xy5 x25 x7 (7, 2) 6. 18  4x 
 y 24  12y 24  12y 2y 18  4x 
 y 18  4x 
 2 5x (5, 2) 7. 16  4x 16  4x 0y 4x 2x  8 
 y (4, 0) 13 8. X  Z  4  (
1) 
2  0 6  (
2) 3 4  
2 4 9. impossible 
1 
 4 3
1 10. Z 
 X  0 
 (
2) 
2 
 6 2  
5 2 
8 4(4) 4(1) 4(
2) 4(6) 16 4  
8 24 12. impossible 4 1 13. YX  [0 
3] 
2 6  [0(4)  (
3)(
2) 0(1)  (
3)(6)] or [6 
18] 14. Budget ($ million) soft-drink 40.1 package delivery 22.9 telecommunications 154.9  x  2y x  2(2) x4 (4, 2) 19. 27  3y 8  5x 
 3y  27  3y 9y  4x 
 3y  11 3(x  y)  1 xy1 2y1 y  
1 21. 2x  
10 y  3x 
y  15 (
5, 
15) 22. 
12  6x 2y1 12y  10 
 x 23. x  y  0 y  y2 3  2y 
 x 6  4 
 2x 6  4 
 2x 
1  x  Viewers (million) 78.6  21.9  88.9   24. x2  1  2 xy5 5
yx y
42 y
42 y6       Exercises  15. y  2x 
 1 xy
5  25.  y  2(y 
 5) 
 1 y  11  xy
5 x  11 
 5 x6 (6, 11) 16. 9  x  2y 13  4x  1 9  x  2y 9  3  2y 3y (3, 3) 17. 4x  15  x 5  2y 5  2y 2.5  y  y  2(2y) 
 6 y2  8  5x 
 3y 8  5x 
 3(9) 7x (7, 9) 20. 4x 
 3y  11 xy1  11. 4X   Pages 83–86  18. x  2y y  2x 
 6  4x 
 3y  11 3x  3y  3 7x  14 x2  (2, 
1) 2x  
10 x  
5  
12  6x 
2  x  y  3x y  3(
5) y  
15 2y1 1y (
2, 1)  xy0 
1  y  0 y1  (
1, 1)  xy5 x65 x  
1  (
1, 6)  6 3 x y 
 1  15 4 3z 6z 3x  y 3x 3y 
 3 15 6  12 9z 6z 3x  y 3x  15 12  6z 3y 
 3  6 9z  3x  y 3x  15 x5 (5, 3, 2)  13  4x  1 3x  →  12  6z 2z  3y 
 3  6 y3  4x  15  x x5  (5, 2.5)  37  Chapter 2  26.  
4 
2 w  5 x 
 z  
16 3y 8 6 2x  8z 
2w 
 10 
2x  2z 
16 
4  
6y 
16 6 2x  8z 
2w 
 10  
16 
2w 
 10  
16 
6y  6 
6y  6 w3 y  
1 
2x  2z  
4 
16  2x  8z 
2x  2z  
4 2x  8z  
16 10z  
20 z  
2  5(5) 5(7) 5(
6) 5(1) 25 35  
30 5 3 5 5 7 38. BA  
1 8 
6 1 3(5)  5(
6) 3(7)  5(1)  
1(5)  8(
6) 
1(7)  8(1) 
15 26 or 
53 1 39. impossible 4 
2 3 40. FC  
6 
1 0 5 0 
1 1 4 0 9 0 1 
6(4)  (
1)(5)  0(9)  1(4)  4(5)  0(9) 37. 5A   
2x  2z  
4 
2x  2(
2)  
4 x0 (3, 0, 
1, 
2)  53 75 
6  (
1) 1  8 8 12  
7 9 28. impossible 29. impossible 0  4 1  (2) 23 30. D  C  
2  5 3  0 0  (
1) 49 40 
2  1 4 
1 5  3 3 
1 13 4 
1 27. A  B   
6(
2)  (
1)(0)  0(0) 1(
2)  4(0)  0(0) 
6(3)  (
1)(
1)  0(1) 1(3)  4(
1)  0(1) 
29 12 
17  24 
2 
1 0 1 2 8 
4 2 
2 3 0 41. ED  3 1 
5 4 4 
2  8(0)  (
4)(
2)  2(4) 3(0)  1(
2)  (
5)(4)  31. B 
 A  B  (
A) 3 5  
5 
7  
1 8 6 
1 3  (
5) 5  (
7) 
2 
2  or 
1  6 8  (
1) 5 7 32. C 
 D  C  (
D) 4 
2 3 0 
1 
2 0 
1   5 2 
3 0 9 0 1 
4 
4 2 4  0 
2  (
1) 3 (
2) 52 0  (
3) 
1  0  9  (
4) 0  (
4) 12 4 
3 1 or 7 
3 
1 5 
4 3 33. 4(0) 4(1) 4(2) 4D  4(
2) 4(3) 4(0) 4(4) 4(4) 4(
2) 0 4 8  
8 12 0 16 16 
8 34. 
2F  
2(
6) 
2(
1) 
2(1) 
2(4) 12 2 0  
2 
8 0 35. F 
 E  F  (
E) 
6 
1 0   1 4 0 
14 3 
2  
2 3 5 36. E 
 F  E  (
F) 8 
4 2   3 1 
5 14 
3 2  2 
3 
5  Chapter 2   42. AA    43. FD    8(1)  (
4)(3)  2(4) 3(1)  1(3)  (
5)(4) 8(2)  (
4)(0)  2(
2) 3(2)  1(0)  (
5)(
2) 16 4 12 
22 
14 16 5 7 5 7 
6 1 
6 1 5(5)  7(
6) 5(7)  7(1) 
6(5)  1(
6) 
6(7)  1(1) 42 or 
17 
36 
41 0 1 2 
6 
1 0 
2 3 0 1 4 0 4 4 
2 
6(0)  (
1)(
2)  0(4) 1(0)  4(
2)  0(4)  
6(1)  (
1)(3)  0(4) 1(1)  4(3)  0(4) 
6(2)  (
1)(0)  0(
2) 1(2)  4(0)  0(
2) 2 
9 
12  
8 13 2 8  2 
4  (
9) 2  (
12) E  FD  3  (
8) 1  13 
5  2 10 
13 
10  
5 14 
3  
2(0) 
2(0)  
8 4 
2 
3 
1 5  6 1 0 
1 
4 0  38  5 7 3 5 
6 1 
1 8 (
3)(5) 
3(7) 3 5  (
3)(
6) 
3(1) 
1 8 3 5  
15 
21 18 
3 
1 8  
15(3)  (
21)(
1) 18(3)  (
3)(
1) 
15(5)  (
21)(8) 18(5)  (
3)(8) 
24 
243  57 66 3 5 5 7 8 
4 2 45. (BA)E  
1 8 
6 1 3 1 
5 3(5)  5(
6) 3(7)  5(1)  
1(5)  8(
6) 
1(7)  8(1) 8 
4 2 3 1 
5 8 
4 2  
15 26 
53 1 3 1 
5  
15(8)  26(3) 
15(
4)  26(1) 
53(8)  1(3) 
53(
4)  1(1) 
15(2)  26(
5) 
53(2)  1(
5) 
42 86 
160  
421 213 
111 46. F 
 2EC  F  (
2EC) 4 
2 3 2 
2EC  
2 8 
4 5 0 
1 3 1 
5 9 0 1 
2(8) 
2(
4) 
2(2)  
2(3) 
2(1) 
2(
5) 4 
2 3 5 0 
1 9 0 1 4 
2 3 
16 8 
4  5 0 
1 
6 
2 10 9 0 1 
16(4)  8(5)  (
4)(9)  
6(4)  (
2)(5)  10(9)  44. 
3AB  
3      2 4 3 
3 6 8 
4 5 4 
2 
2 6 3(2) 3(4) 3 
3 6 3(8) 3(
4) 5 4 
2 3(
2) 3(6) 6 12 3 
3 6 24 
12 5 4 
2 
6 18 6(3)  12(5) 6(
3)  12(4) 24(3)  (
12)(5) 24(
3)  (
12)(4) 
6(3)  18(5) 
6(
3)  18(4) 6(6)  12(
2) 24(6)  (
12)(
2) 
6(6)  18(
2) 78 30 12 12 
120 168 72 90 
72  47. 3XY  3        5 48. 2K 
 3J  2 1 
7  (
3) 
4 3 2 1 
1 
3(5)  2(1) 2(
7)  (
3)(
4) 2(3) 2(2) (
3)(1) 
3(
1)  2 
14  12 
15 6 4 
3 3 2  12 
14  (
15)  6  (
3) 43 14 
29  3 7 49. Sample answer: 1996 2000 2006 18 to 24 8485 8526 8695    10,102 9316 9078  25 to 34  35 to 44 8766 9036 8433    45 to 54 6045 6921 7900    55 to 64 2444 2741 3521   65 and older  2381 2440 2572  18 24 19 26 31 24 16 24 17 22 28 21 50a. 
 6 6 6 12 9 7 12 2 4 17 4 6 18 24 19 
26 
31 
24  26 24 17  
22 
28 
21 6 6 6 
12 
9 
7 12 2 4 
17 
4 
6 18  (
26) 24  (
31) 19  (
24)  16  (
22) 24  (
28) 17  (
21) 6  (
12) 6  (
9) 6  (
7) 12  (
17) 2  (
4) 4  (
6) 
8 
7 
5 
6 
4 
4 or 
6 
3 
1 
5 
2 
2  
16(
2)  8(0)  (
4)(0) 
6(
2)  (
2)(0)  10(0) 
16(3)  8(
1)  (
4)(1) 
6(3)  (
2)(
1)  10(1)  
60 32 
60 56 12 
6 
6  (
60) 
1  32 0  (
60) F  (
2EC)  1  56 4  12 0  (
6) 
66 31 
60  57 16 
6  Classical Jazz Opera Musicals  TV 
8 
6 
6 
5  Radio 
7 
4 
3 
2  Recording 
5 
4 
1 
2        50b. classical performances on TV  39         Chapter 2  51a.  
2 3 a b 
2 3  4 
5 c d 4 
5 
2(a)  3(c) 
2(b)  3(d) 
2 3  4(a)  (
5)(c) 4(b)  (
5)(d) 4 
5 
2a  3c  
2 
2b  3d  3 4a 
 5c  4 4b 
 5d  
5  55.  2(
2a  3c)  2(
2) → 
4a  6c  
4 4a 
 5c  4 4a 
 5c  4 c 0 4a 
 5c  4 4a 
 5(0)  4 a1 2(
2b  3d)  2(3) → 
4b  6d  6 4b 
 5d  
5 4b 
 5d  
5 d 1 4b 
 5d  
5 4b 
 5(1)  
5 b0 51b. a matrix equal to the original matrix 52a. [42 59 21 18] 52b. 33.81 30.94 27.25    15.06 13.25 8.75   [42 59 21 18]  54 54 46.44   52.06 44.69 34.38     153  20z  10 2x  63  84  5 1 1 z  4 x  2 12, 13, 14 7 56. 4x 
 2y  7 y  2x 
 2 → 7 
12x  6y  
21 y  2x 
 2 consistent and dependent y 57. 1  1  1  6  3x  y  12  x  O   4233.81  5915.06  21(54)  1852.06  58.  4230.94  5913.25  21(54)  1844.69 4227.25  598.75  2146.44  1834.38   [4379.64 4019.65 3254.83] July, $4379.64; Aug, $4019.65; Sep, $3254.83 53. The numbers in the first row are the triangular numbers. If you look at the diagonals in the matrix, the triangular numbers are the end numbers. To find the diagonal that contains 2001, find the smallest triangular number that is greater than or equal to 2001. The formula for the n(n  1) n(n  1) nth triangular number is 2. Solve 2  2001. The solution is 63. So the 63rd entry in the 63(63  1) first row is 2  2016. Since 2016 
 2001  15, we must count 15 places backward along the diagonal to locate 2001 in the matrix. This movement takes us from the position (row, column)  (1, 63) to (1  15, 63 
 15)  (16, 48). 54a. A B C D A  0 1 0 0  B  1 0 1 1   C 0 1 0 2   D  0 1 2 0   54b. No; since the matrix shows the number of nodes and the numbers of edges between each pair of nodes, only equivalent graphs will have the same matrix.  Chapter 2  2x  6y  8z  5 
2x  9y  12z  5 15y  20z  10 2(
2x  9y  12z)  2(5) → 
4x  18y  24z  10 4x  6y 
 4z  3 4x  6y 
 4z  3 24y  20z  13 15y  20z  10 → 15y  20z  10 
1(24y  20z)  
1(13) 
24y 
 20z  
13 
9y  
3 1 y  3 15y  20z  10 2x  6y  8z  5  x 
2 
1 0 1 2  f(x) 8 5 2 5 8  f (x)  f (x) 
 |3x |  2 x  O  59. Sample answer: using (60, 83) and (10, 65), 65 
 83   m 10 
 60 
18    
50 or 0.36  y 
 65  0.36(x 
 10) y  0.36x  61.4 7
4  3  y 
 7  4(x 
 5)   60. m   5
1 3  3  61. f(x)  5x 
 3 0  5x 
 3 3  5  x  62. [f g](x)  f(x) g(x)   5x(40x 
 10) 2   16x2 
 4x 63. f(x)  4  6x 
 x3 f(14)  4  6(14) 
 (14)3  
2656  40  1  y  4x  34   4  64.  2x 
 3  x  2d. Let A  a11 a12 , B  b11 b12 , a21 a22 b21 b22  3
x   2 2(2x 
 3)  x(3 
 x) 4x 
 6  3x 
 x2 x2  x 
 6  0 (x  3)(x 
 2)  0 x  3  0 or x 
 2  0 x  
3 x2 The correct choice is A.  Page 86  and C  c11 c12 . c21 c22 (AB)C  a11b11  a12b21 a11b12  a12b22 c11 c12 a21b11  a22b21 a21b12  a22b22 c21 c22  a11b11c11  a11b12c21  a12b21c11  a21b22c21 a21b11c11  a21b12c21  a22b21c11  a22b22c21 a11b11c12  a11b12c22  a12b21c12  a12b22c22 a21b11c12  a21b12c22  a22b21c12  a22b22c22  Graphing Calculator Exploration  A(BC)  a11 a12 b11c11  b12c21 b11c12  b12c22 a21 a22 b21c11  b22c21 b21c12  b22c22  1. All of the properties except for the Commutative Property of Multiplication hold true. When multiplying matrices, the order of the multiplication produces different results. However, in addition of matrices, order is not important. 2a. Let A  a11 a12 and B  b11 b12 . a21 a22 b21 b22 a a b b 11 12 11 12 AB  a21 a22 b21 b22  a11b11c11  a11b12c21  a12b21c11  a21b22c21  a b c a b c a b c a b c 21 11 11 21 12 21 22 21 11 22 22 21 a11b11c12  a11b12c22  a12b21c12  a12b22c22 a21b11c12  a21b12c22  a22b21c12  a22b22c22 Therefore (AB)C  A(BC). 3. All properties except the Commutative Property of Multiplication will hold for square matrices. A proof similar to the ones in Exercises 2a-2d can be used to verify this conjecture.   a11  b11 a12  b12 a21  b21 a22  b22   b11  a11 b21  a21  b12  a12 b22  a22    b11 b21  a11 a12 b21 b12 a21 a22  2-4A Transformation Matrices  2b. Let A  a11 a12 , B  b11 b12 , a21 a22 b21 b22 and C  c11 c21 a11  b11 (A  B)  C  a21  b21   Page 87 1. The new figure is a 90° counterclockwise rotation of LMN. 2. The new figure is an 180° counterclockwise rotation of LMN.  c12 c22 a12  b12  c11 c12 a22  b22 c21 c22  (a11  b11)  c11 (a12  b12)  c12 (a21  b21) c21 (a22  b22)  c22   (b11  c11) a12  (b12  c12)  a11 a21  (b21  c21) a22  (b22  c22)  a11 a12  b11  c11 b12  c12 b21  c21 b22  c22 a21 a22  A  (B  C) 2c. Let A  a11 a12 and B a21 a23 a b  a12b21 AB  11 11 a21b11  a22b21 BA  Thus, AB  3. The new figure is an 270° counterclockwise rotation of LMN.  b11 b12 . b21 b22 a11b12  a12b22 a21b12  a22b23  b11a11  b12a21 b11a12  b12a22 b21a11  b22a21 b21a12  b22a22  BA, since a12b21  b12a21.  41  Chapter 2  4. See students' work for graphs. Multiplying a vertex matrix by 0 
1 results in a vertex 1 0 matrix for a figure that is a 90° counterclockwise rotation of the original figure. Multiplying a 
1 0 vertex matrix by results in a vertex 0 
1 matrix for a figure that is a 180° counterclockwise rotation of the original figure. Multiplying a 0 1 vertex matrix by results in a vertex 
1 0 matrix for a figure that is a 270° counterclockwise rotation of the original figure.  2 
2  
2 2 1 1 
3 
3 A
(
2, 1), B
(2, 1), C
(2, 
3), D
(
2, 
3) y A  Pages 92–93  O  C x  D D  7.  Check for Understanding  1. Translation, reflection, rotation, dilation; translations do not affect the shape, size, or orientation of figures; reflections and rotations do not change the shape or size of figures; dilations do not change the shape, but do change the size of figures. 2. 90° counterclockwise  (360 
 90)° or 270° clockwise; 180° counterclockwise  (360 
 180)° or 180° clockwise; 270° counterclockwise  (360 
 270)° or 90° clockwise. 3. Sample answer: the first row of the reflection matrix affects the x-coordinates and the second row affects the y-coordinates. A reflection over the x-axis changes (x, y)  0 (x, 
y), so the first row needs to be [1 0] so the x is unchanged and the second row needs to be [0 
1] so the y-coordinates are the opposite. Similar reasoning can be used for a reflection over the y-axis, which changes (x, y) to (
x, y) and a reflection over the line y  x, which interchanges the values for x and y. 4a. 6 4b. 2 4c. 3 4d. 4  C  
1 0 
1 
4 
3 0 0 1 2 1 
2 
1 3 0  1 4 2 1 
2 
1 A
(1, 2), B
(4, 1), C
(3, 
2), D
(0, 
1)  y A  A  B  B  O C  C  3 
1 1  2 4 
2 2 4 
2 
3 1 
1  P
(2, 
3), Q
(4, 1), R
(
2, 
1) y Q P  O  Q  x  R R  P  0 
1 0 1 0 9. 
1  0 
1 0 1 0 
1 1 0 
6 
3 
1 
6 
3 
1  0 
1 4 2 
2 
4 
2 2 L
(
6, 
4), M
(
3, 
2), N
(
1, 2)  y  y  L  J  x  D D  0 1 8. 
1 0  0  
3 1.5 0 5. 1.5 
2 1 5 3 
2 7.5 4.5 
3 J
(
3, 7.5), K
(1.5, 4.5), L
(0, 
3) J  B B  A  Modeling Motion with Matrices  2-4  
1 3 3 
1 
1 
1 
1 
1  3 3 
1 
1 
2 
2 
2 
2  6.  K  M N  K N  x  O x  M  L L  L  10b. x  3 y4  10a. Landing 4N 3E Ball  Chapter 2  42  Pages 93–96  0 
3 0 3 0 14b. 3 
1 0 1  0 1 0 
1 0 3 0 
3  Exercises  1 1 5 3 3 15  1 4 1 3 12 3 A
(3, 3), B
(3, 12), C
(15, 3)  11. 3  0  
6 0 6 0 2 
3 0 3 0 3 0 
3 0 6 0 
6 A
(
3, 0), B
(0, 3), O
(3, 0), D
(0, 
3); A(
6, 0), B(0, 6), C
(6, 0), D(0, 
6)  y  B 12 10 8 6 B A 4 2 A C  y B B  C  2 4 6 8 10 12 14 x  3  12. 4  0 
5 
3  8 9 2  A  0  
15  4  6  27  4  3    14c. The final results are the same image. 3  3 3 3  1 4 15. 
2 1 0 5 
1 
2 
2 
2 
2 3 W
(1, 
2), X
(4, 3), Y
(6, 
3)  X  y  6 
3  X X  Z Z  x  O  W O  
3 
2 1 4 
6 
4 2 8 13. 2  0 2 3 2 0 4 6 4 P
(
6, 0), Q
(
4, 4), R
(2, 6), S
 (8,4)  y  W  16. 0 1 4 3  
2 
2 
2 
2 0 5 7 2 
1 
1 
1 
1 
2 
1 2 1  
1 4 6 1 Q
(
2, 
1), P
(
1, 4), Q
(2, 6), R
(1, 1)  S R Q  S  y x  O  P  x  Y  Y  R  Q  P  x  D  X  Y  C  D  3  2 1 1 
24, 12  y  Y  C  C  D  
9  4  X
(0, 6), Y

34, 64, Z
 3  B  A  A  Q  Q P  P  0 
2 0 2 0 14a. 2 
1 0 1  0 1 0 
1 0 2 0 
2 0  
6 0 6 0 3 
2 0 2 0 2 0 
2 0 6 0 
6 A
(
2, 0), B
(0, 2), C
(2, 0), D
(0, 
2); A(
6, 0), B(0, 6), C
(6, 0), D(0, 
6)  R R O  17.  y B  x  O  
3 1 5 1 3 3 3 3 0 4 8 4   1 5 1 
3 4 4 4 4 5 9 5 1 C
(0, 5), D
(4, 9), E
(8, 5), F
(4, 1)  A  A A D  B B C C D  y  D  C x  C  D  C  D  E  F  E  x  O F  43  Chapter 2  4 0 2 
6 
6 
6 
2 
6 
4   1 3 
1 
2 
2 
2 
1 1 
3  18a.  22.  F
(
2, 
1), G
(
6, 1), H
(
4, 
3) 18a-b. y  0 
1 1 2 3 1 2 1  1 0 
1 
2 
1 1 2 3 L
(1, 1), M
(2, 2), N
(1, 3)  y N  G  M  F L  G G  O L  F  H F O  x  N M  x H  23.  H  
2 
6 
4 1 1 1 
1 
5 
3   
1 1 
3 5 5 5 4 6 2 F (
1, 4), G(
5, 6), H(
3, 2) 18c. translation of 5 units left and 3 units up 1 0 
1 0 2 
1 0 2 19.  0 
1 
2 
4 
3 2 4 3 A
(
1, 2), B
(0, 4), C
(2, 3) 18b.  
1 0 0 4 4 0 0 
4 
4 0  0 
1 0 0 4 4 0 0 
4 
4 O
(0, 0), P
(
4, 0), Q
(
4, 
4), R
(0, 
4)  y R  Q  P  O O  P  x  y B  C  R  Q  A  24.  x  O A    
1 0 2 6 3 
1 
2 
6 
3 1  0 1 4 2 
4 
2 4 2 
4 
2 D
(
2, 4), E
(
6, 2), F
(
3, 
4), G
(1, 
2) D  y  V E  S T  x  U  S  Ox  G  G  V  F  F  b  Rx-axis d a b 1 
2 
1  1 
2 
1 c d 3 
1 
3 
3 1 3 a  3b 
2a 
 b 
a 
 3b  1 
2 
1 c  3d 
2c 
 d 
c 
 3d 
3 1 3  y  a  3b  1 
2a 
 b  
2 
a 
 3b  
1 c  3d  
3 
2c 
 d  1 
c 
 3d  3 Thus, a  1, b  0, c  0, and d  
1. By 0 . substitution, Rx-axis  1 0 
1  J K K I  x O  W  25a. Let a c  
1 
3 
1 2  
2 1 5 4 21. 0 1 1 0 
2 1 5 4 
1 
3 
1 2 H
(
2, 
1), I
(1, 
3), J
(5, 
1), K
(4, 2)  Chapter 2  T  W  O  H  y  U  D  E  H  
1 
3 
5 
4 
2 
2 
1 
2 
4 
4  
2 
1 
2 
4 
4 1 3 5 4 2 S
(
2, 1), T
(
1, 3), U
(
2, 5), V
(
4, 4), W
(
4, 2)  C B  20.  0 1 
1 0  J  I  44  0 
6 
3 
1 6 3 1 26. 
1  0 
1 4 2 
2 
4 
2 2 6 3 1  
2 
2 
2  4 1 
1 
4 
2 2 5 5 5 1 3 7 J(4, 1), K(1, 3), L(
1, 7)  25b. Let a c  b R y-axis. d a b 1 
2 
1 
1 2 1  c d 3 
1 
3 3 
1 
3 a  3b 
2a 
 b 
a 
 3b 
1 2 1  c  3d 
2c 
 d 
c 
 3d 3 
1 
3  a  3b  
1 
2a 
 b  2 
a 
 3b  1 c  3d  3 
2c 
 d  
1 
c 
 3d  
3 Thus, a  
1, b  0, c  0, and d  1. By 
1 0 . substitution, Ry-axis  0 1 25c. Let a b  Ry  x. c d a b 1 
2 
1  3 
1 
3 c d 3 
1 
3 1 
2 
1 a  3b 
2a 
 b 
a 
 3b  3 
1 
3 c  3d 
2c 
 d 
c 
 3d 1 
2 
1  L  y  J K L  K  J  x  O L  K J  a  3b  3 
2a 
 b  
1 
a 
 3b  
3 c  3d  1 
2c 
 d  
2 
c 
 3d  
1 Thus, a  0, b  1, c  1, and d  0. By 0 1 . substitution, Ry  x  1 0 a b 25d. Let  Rot90 c d a b 1 
2 
1 
3 1 3  c d 3 
1 
3 1 
2 
1 a  3b 
2a 
 b 
a 
 3b 
3 1 3  c  3d 
2c 
 d 
c 
 3d 1 
2 
1  0 
6 
3 
1  
6 
3 
1 27. 1 0 
1 4 2 
2 
4 
2 2 
1 0 
6 
3 
1  6 3 1 0 1 
4 
2 2 
4 
2 2 J(6, 
4), K(3, 
2), L(1, 2)  y J K  a  3b  
3 
2a 
 b  1 
a 
 3b  3 c  3d  1 
2c 
 d  
2 
c 
 3d  
1 Thus, a  0, b  
1, c  1, and d  0. By 0 
1 substitution, Rot90  1 0 25e. Let a b  Rot180. c d a b 1 
2 
1 
1 2 1  c d 3 
1 
3 
3 1 3 a  3b 
2a 
 b 
a 
 3b 
1 2 1  c  3d 
2c 
 d 
c 
 3d 
3 1 3  L L  x  O K  K  L  J  28.  J  
1 0 
6 
3 
1 6 3 1  0 1 4 2 
2 4 2 
2 0 
1 6 3 1 
4 
2 2  1 0 4 2 
2 6 3 1 J(
4, 6), K(
2, 3), L(2, 1)  y  a  3b  
1 
2a 
 b  2 
a 
 3b  1 c  3d  
3 
2c 
 d  1 
c 
 3d  3 Thus, a  
1, b  0, c  0, and d  
1. By 
1 0 . substitution, Rot180  0 
1 25f. Let a b  Rot270. c d  J J  J K  K  K L  x  O  a b 1 
2 
1  3 
1 
3 c d 3 
1 
3 
1 2 1 a  3b 
2a 
 b 
a 
 3b  3 
1 
3 c  3d 
2c 
 d 
c 
3d 
1 2 1  L  L  29a. The bishop moves along a diagonal until it encounters the edge of the board or another piece. The line along which it moves changes vertically and horizontally by 1 unit with each square moved, so the translation matrices are scalars. Sample matrices are 1 , c 
1 
1 , and c 1 1 ,c 1 1 1 
1 
1 1 1  a  3b  3 
2a 
 b  
1 
a 
 3b  
3 c  3d  
1 
2c 
 d  2 
c 
 3d  1 Thus, a  0, b  1, c  
1, and d  0. By 0 1 . substitution, Rot270  
1 0  45  Chapter 2  37. Let x represent hardback books and y represent paperback books. 4x  7y  5.75 3x  5y  4.25 3(4x  7y)  3(5.75) 12x  21y  17.25 → 
4(3x  5y)  
4(4.25) 
12x 
 20y  
17 y 0.25 4x  7y  5.75 4x  7(0.25)  5.75 x1 hardbacks $1, paperbacks $0.25 y 38. x  y  3 y  
x  3 (2, 4) (3, 2)  c 
1 
1 , where c is the number of squares 
1 
1 moved. 29b. The knight moves in combinations of 2 vertical-1 horizontal or 1 vertical-2 horizontal squares. These can be either up or down, left or right. Sample matrices are 1 1 , 2 2 1 1 
1 
1 
1 
1 2 2 , , , , 
2 
2 2 2 
2 
2 1 1 2 2 , 
2 
2 , and 
2 
2 . 
1 
1 1 1 
1 
1 29c. The king can move 1 unit in any direction. The matrices describing this are 1 1 , 0 0 
1 
1 , 0 0 , 0 0 , 1 1 , 1 1 , 0 0 1 1 
1 
1 1 1 
1 
1 
1 
1 
1 
1 , and . 1 1 
1 
1  (4, 2)  O (0, 0)  xy
3  a b . c d Dilation with scale factor 
1 a b 
a 
b 
1  c d 
c 
d Rotation of 180° 
1 0 a b 
a 
b  0 
1 c d 
c 
d The vertex matrices for the images of a dilation with scale factor 
1 and a rotation of 180° are the same, so the images are the same. 31. (0, 
125); (125, 0), (0, 125), (
125, 0) 32. Sample answer: There is no single matrix to achieve this. You could reflect over the x-axis and then translate 2(4) or 8 units upward. 33. See students' work; the repeated dilations animate the growth of something from small to large, similar to a lens zooming into the origin. 2 34. 
1 1 4 
1 2 0 
3 34a. Sample answer: the figure would be enlarged disproportionally. 3 y0 
1 1 4 2 
3 3 12 6 34b.  0 62 B
1 2 0 
3 
2 4 0 
6  30. Consider  4 B 2  A A 
4 
6  (3, 2) y 
 1  4(x 
 2)) y 
 1  4x  8 4x 
 y  9  0 40. y 
 6  2(x 
 1) y  2x  4 41. (f g)(x)  f(x) g(x)  x3(x2 
 3x  7)  x5 
 3x4  7x3 39.  g(x)  g(x) f  f(x)  x3    x2 
 3x  7  42. 2x  y  12 2x  y  12 → 
2(x  2y)  
2(
6) 
2x 
 4y  12 
3y  24 y  
8 2x  y  12 2x  (
8)  12 x  10 2x  2y  2(10)  2(
8) 4 The correct choice is B.  C  C  2 4 6 8 10 12 x D  Page 96 1.  D  1 x 2  Mid-Chapter Quiz 1   5y  17  3  y  
2x  9  y 3x  2y 
 18  34c. See students' work; the figure appears as if blown out of proportion. 3 8 1 5 31 85 35.   
2 4 
2 8 
2  (
2) 4  8  36. x  y  z  1.8 y 
 z  
5.6 x  2y  
3.8 x  2y  
3.8 Chapter 2  (4, 3)  4 13 
4 12 x 
 2y  4.6 x  2y  
3.8 2x  0.8 x  0.4 y 
 z  
5.6  1 2 x  5y 
 17  O  46  2   y  
1 0x  35  →  3x  2y  18    x  x  2. 4x  y  8 y  8 
 4x  6x 
 2y  
9 6x 
 2(8 
 4x)  
9 1 x  2  10. The result is the original figure. The original figure is represented by a b c . The d e f reflection over the x-axis is found by 1 0 a b c  a b c . The 0 
1 d e f 
d 
e 
f reflection of the image over the x-axis is found by 1 0 a b c  a b c . The matrix 0 
1 
d 
e 
f d e f for the final image is the same as that of the original figure.  4x  y  8  42  y  8 1  y6 2, 6 3. Let x represent trucks and y represent cars. x  4y 6x  5y  29,000 x  4y 6(4y)  5y  29,000 x  4(1000) y  1000  4000 4000 trucks, 1000 cars 4. 2x  y  4z  13 3x 
 y 
 2z  
1 5x  2z  12 1  2(3x 
 y 
 2z)  2(
1) 4x  2y  z  19  →  
2(5x  2z)  
2(12) 10x 
 3z  17  →  5x  2z  12 5x  2(1)  12 x2 (2, 5, 1) 5. x  y  1 2x 
 y  
2 3x  
1 1 x  
3 4  4x  y  z  8   113  z  8 z8  1 
3  6. y 
 3  x y  2x y  2x y  2(3) 6  2-5  6x 
 2y 
 4z  
2 4x  2y  z  19 10x 
 3z  17  Page 102  Determinants and Multiplicative Inverses of Matrices Check for Understanding  1. Sample answer: a matrix with a nonzero determinant 2 0 is not a square 2. Sample answer: 3 4 
3 5 2 
5 matrix. 1 
1 also has no determinant. 0 9 1 0 0 0 0 1 0 0 3. Sample answer: 0 0 1 0 0 0 0 1  
10x 
 4z  
24 10x 
 3z  17 
7z  
7 z1 2x  y  4z  13 2(2)  y  4(1)  13 y5  xy1 1 
3  y  1 1 y  13  4. Sample answer: The system has a solution if ad 
 bc 0, since you can use the inverse of the matrix a c to find the solution. b d 4 
1  4(3) 
 (
2)(
1) or 10 
2 3 12 
26 6.  12(32) 
 (
15)(
26) or 
6 
15 32 4 1 0 7. 5 
15 
1 
2 10 7 
15 
1 5 
1 5 
15 4 
1 0 10 7 
2 7 
2 10  4(
95) 
 1(33)  0(20)  
413 6 4 
1 3 3 0 3 0 3 
4  (
1) 8. 0 3 3 6 0 0 
9 0 
9 0 
9 0 0  6(0) 
 4(27)  (
1)(27)  
135 9. 
2 3  
2(7) 
 5(3) or 
29 5 7 1 7 
3 
2 9 
5 
2 10. 4 6  4(9) 
 6(6) or 0 6 9 does not exist 5.  
13, 113, 8 y
3x 2x 
 3  x x3  (3, 6) 5  8 
7  6 7. A  B  3  (
2) 
1  5 0  (
9) 4  10 1 13 
1  4 
9 14 8. impossible 9. B 
 3A  B  (
3A) 
3A  
3 3 5 
7 
1 0 4 
3(3) 
3(5) 
3(
7)  
3(
1) 
3(0) 
3(4) 
9 
15 21 or 3 0 
12 
2 8 6 
9 
15 21 B  (
3A)   5 
9 10 3 0 
12 6  21  
2  (
9) 8  (
15) 53 
9  0 10  (
12)  
11 
7 27 8 
9 
2  47  Chapter 2  4 
1 
2 2 1 0 1 0 2 
 (
1)  (
2) 20. 0 2 1 4 1 3 2 3 2 1 2 1 3  4(5)  1(
2) 
 2(
4)  26 2 
1 3 21. 3 0 
2 1 
3 0 0 
2 3 
2 3 0 
 (
1) 3 2 
3 0 1 0 1 
3  2(
6)  1(2)  3(
9)  
37 8 9 3 5 7 3 7 3 5 
9 3 22. 3 5 7 8 2 4 
1 4 
1 2 
1 2 4  8(6) 
 9(19)  3(11)  
90 4 6 7 
2 
4 3 
4 3 
2 23. 3 
2 
4  4 
6 7 1 1 1 1 1 1 1 1 1  4(2) 
 6(7)  7(5) 1 25 36 15 24. 31 
12 
2 17 15 9  5 4 x  
3 
3 
5 y 
24 1 
5 
4  
1 
5 
4 13 5 4 3 5 3 5  11.  
3 
5  
5 
4 3 5  1 
1 3  5 4 
3 
5  x y 1 
5 
4  
1 3 3 5  
3 
24  111 x  
13 y 129  13    111 129 
13, 1 3   6 
3 x 63  5 9 y 85 1 
9 3 1 
9 3  
3 9 
5 6 
3 
5 6  12.  6 5 
9  
9 3 
5 6  1  
3 9  6 
3 5 
9  1 
9 3 x  
3 9 
5 6 y x 8  y 
5  63 85  (8, 
5) 13. Let x represent the amount of metal with 55% aluminum content, and let y represent the amount of metal with 80% aluminum content. x  y  20 0.55x  0.8y  0.7(x  y) 0.55x  0.8y  0.7x  0.7y 0.15x 
 0.1y  0 15x 
 10y  0 x  y  20 15x 
 10y  0 1 1 15 1  
2 5  1 
10  →  1 1 15 
10   25 
12 
2 
 36 31 
2  15 31 
12 15 9 17 9 17 15  25(
78) 
 36(313)  15(669)  
3183 1.5 
3.6 2.3 25. 4.3 0.5 2.2 
1.6 8.2 6.6  1.5 0.5 2.2 
 (
36) 4.3 2.2  8.2 6.6 
1.6 6.6 4.3 0.5 23 
1.6 8.2  1.5(
14.74)  3.6(31.9)  2.3(36.06)  175.668 0 1 
4 2 3 3 3 3 2 
1  (
4) 26. 3 2 3 0 
3 4 8 4 8 
3 8 
3 4  x 20  y 0  1 
10 
1 
10 
1  
2 5 
15 
15 1 1  
10 
1 
15 1  1 1 15 
10  x y  1 
10 
1  
2 5 
15 1  20 0   0(17) 
 1(
12) 
 4(
25)  112  x 8  y 12 8 kg of the metal with 55% aluminum and 12 kg of the metal with 80% aluminum  Pages 102–105 14. 15. 16. 17. 18. 19.  2 
3  2(
2) 
 (
2)(
3) or 
10 
2 
2 1 
2 3 
1 0 2 2 2 0  2(0) 
 1(0) or 0 28. 1 0 does not exist 29. 4 2  4(2) 
 1(2) or 6 1 2 1 2 
2  6 
1 4 6 7 30.  6(7) 
 (
6)(7) or 84 
6 7  27.  Exercises  3 4  3(5) 
 2(4) or 7 2 5 
4 
1  
4(
1) 
 0(
1) or 4 0 
1 9 12  9(16) 
 12(12) or 0 12 16 
2 3  
2(1) 
 (
2)(3) or 4 
2 1 13 7  13(
8) 
 (
5)(7) or 
69 
5 
8 
6 5  
6(
8) 
 0(5) or 48 0 
8  Chapter 2  7 
7 6 6 
4 6  
4(
12) 
 8(6) or 0 31. 8 
12 does not exist 1  84  48  32.  9 13  9(36) 
 27(13) or 
27 27 36 1 36 
13 
2 7 
27 9 3  4  
8  
5  1  2  33.  1  2  1  1  34. 1 4 
1 1 2  4 
1 1 2  40.  1 x 2 1  9 y 
1 4 x 1  y 3  1 7  3 1 
2 
4 1  
3 9 5  
2 3 2 2 1 
1 1 
10 3 
3 1 8  (1, 3)  1 9 
6 4 6 1 6  78 
4  1 1 5 3 
2 1 
2 
1 7 
3  6 9 
6 
 9 4 6  1 x 6 6  7 8 
4 9 y x 0  y 
2  1 5 3 
2 
2 
5  
1 17 
3 1 
5 1  12 
12  x y  z  x 26  y 
41 
2 
5 
3 1    2 , 3  41.  1 
2 
5 x 26  
1 7 
3 y 1 
41 x 
9  y 7  1 5 3 
2  1 4 8 3 
3 1 
3 
3 6 
3  4 8 3 
3 
3 
8  
1 36 
3 4 
8 4  4 8 3 
3  1  1 
3 
8 x  
3 6 
3 y 4  x 1 y  
9 z x 1 y  
9 z  7 0  7  12 7  12  1 3 
5 5 4 1  37  4 5 
5 3  x  1 4 5 37 
5 3 y x  
3 y 3  2  3  1 4  
6 5 3 x 9 
2 
1 y 3 1 1 z 
1 
2 1 
9 
12 
15 21 5 15 21 
33 
1 
1(
9)  (
2)(5)  1(
1) 
12(
9)  (
15)(5)  21(
1) 15(
9)  21(5)  (
33)(
1) 
2 12 3  2 x  9 y  4  
 z 3  x  
24 y 
3 1 4 5 4 5  3 7 
5 3 
5 3  3 
5 5 4  1  3 2 
3  x 
9 y  5 z 
1   
9  3 
5 5 4  38.    
6 5 3 9 
2 
1 3 1 1  x 7  y 0 
3 
8 
3 4  x  y  172 , 172   1,  1 1  
3  4 
3  
1 
2 1 1 
9 
12 
15 21 15 21 
33  (
9, 7) 37.  1 x 1 
3  
6 y 
5 9  x 
4 y  0 z 1 3 
2 3 x 1 2 2 y 
2 1 
1 z 
4 1 
10 
4 1  9 
3 3 
3 0 5 1 8 1 
4(
4)  1(0)  (
10)(1) 1  9 
3(
4)  3(0)  (
3)(1) 5(
4)  1(0)  8(1) 6 1  9 9 
12  x y z x y z  9 
6 x 12  4 6 y 
12 1 6 6 6 6  7 8 
4 9 
4 9  (0, 
2) 36.  9 3 5 1  13, 
23  4 
1 x 1  1 2 y 7 2 1  1 2 1 9 
1 4 
1 4  2 1 
1 4  35.  1   
6  x 1  y 1 1 
3 
5 9  x  y  1  8 3  4  
5  1  9  1 1 
3 9 3 
5 9 5 1 1 1 
3 
6 
5 9   42 
 5
8 or 1 3 1  9 3 5 1  39.  1  
3  29, 
43, 
13 42. 
216 43. 30,143  
24 
3  (
3, 3)  49  Chapter 2  0.3 0.5 x 4.74  12 
6.5 y 
1.2 1 
6.5 
0.5  
1 
6.5 
0.5 7.95 
12 0.5 
12 0.3 0.3  44.  0.3 12 
6.5 1 
6.5  
 7.95 
12  
0.5 0.3  0.3 0.5 x 12 
65 y 1 
6.5 
0.5    
 7.95 
12 0.3 x 3.8  y 7.2  (3.8, 7.2) 45. 2(x 
 2y  z)  2(7) 6x  2y 
 2z  4  →  2(6x  2y 
 2z)  2(4) 4x  6y  4z  14  1 8 
2 16 10 1 10  112 
16  48. Let x represent the number of gallons of 10% alcohol solution, and let y represent the number of gallons of 25% alcohol solution. x  y  12 0.10x  0.25y  0.15(x  y) 0.10x  0.25y  0.15x  0.15y 
0.05x  0.10y  0 1 1 x  12 x  y  12 → 
0.05 0.10 y 0 
0.05x  0.10y  0 1 1 0.10 
1   0.10 
1 0.15 0.05 1 1 0.05 1 1  4.74 
1.2  
0.05  2x 
 4y  2z  14 6x  2y 
 2z  4 8x 
 2y  18  1  0.15  8 
2 16 10  → 12x  4y 
 4z  8 4x  6y  4z  14 16x  10y  22  1 x 10 2   112 
16 8 y x 2  y 
1  18 22  1   A
1   ad 
 bc  x 
 2y  z  7 2 
 2(
1)  z  7 z3 (2, 
1, 3) 46. Let x represent the number of cars produced in the first year, and let y represent the number of cars in the second year. x  y  390,000 → x  y  390,000 x  y  90,000 x 
 y  90,000 1 1 1 
1 1 1 1 1 
1 1  
2  
1 
1 
1 1  (A
1)2     Chapter 2  
b  a  d  ad 
 bc 
c  ad 
 bc  
b  ad 
 bc a  ad 
 bc 2 d 
ab  1  2  
 bd a2  bc  bc  
ac 
 cd  1  a2d2 
 2abcd  b2d2    (1(4)  3(
3)  1(
12))  1  x y   2 
17 or 8.5 1  82 or 8.5 square units 51. Let x represent the cost of complete computer systems, and let y represent the cost of printers. day 1: 38x  53y  49,109 day 2: 22x  44y  31,614 day 3: 21x  26y  26,353 using day 1 and day 2: 38 53 x 49,109  22 44 y 31,614 1 44 
53  1 44 
53 506 
22 38 53 
22 38 38  1 390,000  
2 
1 
1 
1 1 90,000 x 240,000  y 150,000 150,000 in the second year and 240,000 in the first year 47. Let A  a11 a12 and I  1 0 . 0 1 a21 a22 a22  d 
c    
1 
1  
1 
1 
1 2 
1 
1 1 1    a11a22 
 a21a12 A
1   
a21    a11a22 
 a21a12  a11a22 
 a21a12   a11a22 
 a21a12 AA
1   a21a22 
 a21a22   a11a22 
 a21a12 1 0 I  0 1 Thus, AA
1  I.  x y  Thus, (A2)
1  (A
1)2. a b 1 1 50. A  2 c d 1 e f 1 1 
3 1 1  2 0 4 1 3 0 1 1  2 1 4 1 
 (
3) 0 1  1 0 4 0 1 3 1 3 0  x 390,000  y 90,000  1 1 1 
1  1 1 
0.05 0.10  1 0.10 
1 12   0.15 0.05 1 0 x 8  y 4 8 gal of 10% and 4 gal of 25% 49. Yes A  a b . Does (A2)
1  (A
1)2 c d 2 2 A  a  bc ab  bd2 ac  cd bc  d 1 bc  d2 
ab 
 bd (A2)
1   a2d2 
 2abcd  b2d2 a2  bc 
ac 
 cd  8 
2 x  18 16 10 y 22 1 10 2 10 2    112 
16 8 
16 8 2 8  0.10  0.10 
1 0.05 1  
a12  a11a22 
 a21a12     a 11  a11a22 
 a21a12     22  44  44 
53 
22 38  
a11a12  a12a11   a11a22 
 a21a12   1  506  a11a22 
 a21a12   a11a22 
 a21a12  44 
53 
22 38 x  959 y 239 computer system: $959, printer: $239      38 53 22 44  x y 1    506  50  49,109 31,614  60. [f 
 g](x)  f(g(x))  f(x 
 1)  (x 
 1)2  3(x 
 1)  2  x2 
 2x  1  3x 
 3  2  x2  x [g 
 f ](x)  g(f(x))  g(x2  3x  2)  x2  3x  2 
 1  x2  3x  1 61. No, more than one element of the range is paired with the same element of the domain. 62. The radius of circle E is 3, so the circumference is 2(3) or about 18.85. The diagonal of the square 6 6
2 has length 6, so each side has length     2 
2 3
 2. The perimeter of the square is 4(3
 2) or 12
 2  16.92. The difference between the circumference of the circle and the perimeter of the square is approximately 18.85 
 16.92  1.93. The correct choice is B.  52. Let x represent Jessi's first test score, and let y represent Jessi's second test score. x  y  179 x  y  179 → yx7 x 
 y  
7 1 1 x  179 1 
1 y 
7 1 
1 
1  
1 
1 
1 2 
1 1 1 
1 1 1 1 
1  1 
1 
1 x 179  
2 y 
1 1 
7 x 86  y 93 first test: 86, second test: 93 53. 8 4 0 4  
3 
3 
3 
3 5 1 5 9 4 4 4 4 8  (
3) 4  (
3) 0  (
3) 4  (
3)  54 14 54 94 5 1 
3 1  9 5 9 13 H
(5, 9), I
(1, 5), J
(
3, 9), K
(1, 13) 1  
2  54.  3  4  
1 
1 
1 1  1 1 1 
1  8 
7  
4 0  21  6 
4 
3  55. x 
 3y  2z  6 2(4x  y 
 z)  2(8)  0 →  Graphing Calculator Exploration: 2-5B Augmented Matrices and Reduced Row-Echelon Form  x 
 3y  2z  6 8x  2y 
 2z  16 9x 
 y  22  4(4x  y 
 z)  4(8) → 16x  4y 
 4z  32 
7x 
 5y  4z  
10 
7x 
 5y  4z  
10 9x 
 y  22 
1(9x 
 y)  
1(22) → 
9x  y  
22 9x 
 y  22 9x 
 y  22 0 0 infinitely many solutions g (x) 56. x f(x) 
6  x  
5 2 O 
5  x  
4 0 
4  x  
3 
2 
3  x  
2 
4 g (x) 
 2x  5 
2  x  
1 
6  Page 106 2 1 
2 : 7 1 0 0 : 
1 1. 1 
2 
5 : 
1 , 0 1 0 : 5 , 4 1 1 : 
1 0 0 1 : 
2 (
1, 5, 
2) 1 1 1 : 6 1 0 0 : 7 2. 2 
3 4 : 3 , 0 1 0 : 1 , 4 
8 4 : 12 0 0 1 : 
2 (7, 1, 
2) 1 1 1 1 : 0 1 0 0 0 : 1 2 1 
1 
1 : 1 0 1 0 0 : 
1 3. , 
1 
1 1 1 : 0 0 0 1 0 : 2 0 2 1 0 : 0 0 0 0 1 : 
2 (1, 
1, 2, 
2) 4. Exercise 1: x  
1, y  5, z  
2; Exercise 2: x  7, y  1, z  
2; Exercise 3: w  1, x  
1, y  2, z  
2; They are the solutions for the system. 5. The calculator would show the first part of the number and follow it by ....  x  1  y 
 5  2(x 
 2)  57.  1  y 
 5  2x 
 1 2y 
 10  x 
 2 x 
 2y  8  0 3
5   m 2
1  58.  
2   1 or 
2 (y 
 5)  
2(x 
 1) or (y 
 3)  
2(x 
 2) y 
 5  
2(x 
 1) y 
 5  
2x  2 y  
2x  7 59a. 59b.  1  12 1  12  2-6  Solving Systems of Linear Inequalities  Page 109–110  Check for Understanding  1a. the sum of twice the width and twice the height 1b. Sample answer: skis, fishing rods 2. Tomas is correct. There are functions in which the coordinates of more than one vertex will yield the same value for the function.  or approximately 0.0833 x   18 18  12x 1.5  x; 1.5 ft  51  Chapter 2  8. Let x represent the number of greeting cards sold, and let y represent the income in dollars. x0 y0 y  0.45x  1500 y  1.70x  3. You might expect five vertices; however, if the equations were dependent or if they did not intersect to form the sides of a convex polygon, there would be fewer vertices. y 4. x  2y 
 4  O  y y 
 1.70x x  (3 13, 13)  6000 4000  xy
3  y 
 0.45x  1500  2000  y  5. (1, 3) (1, 0)  (1200, 2040) 2000 4000 6000 x  O (0, 4)  at least 1200 cards (7, 0.5)  O (7, 0)  Pages 110–111  x  9.  Exercises  y  (
1, 0), (
1, 3), (0, 4), (7, 0.5), (7, 0) 6. y  y  x 
 1 (1,0)  (0, 2)  (4, 3)  x  O O  (213, 13 )  x yx
1  f(x, y)  4x  3y f(0, 2)  4(0)  3(2) or 6 f(4, 3)  4(4)  3(3) or 25 1 1 1 1 1 f 23, 
3  423  3
3 or 83  y  10. 3  (0, 1)  25, 6 y 7.  ( 23, 1) y 
1  1 y 
3x1 1 O  (7, 7)  (3, 5)  y 
 3x  3  2  11.  (5, 3)  1  2  x  y 2x 5y 
 25  x O f(x, y)  3x 
 4y f(3, 5)  3(3) 
 4(5) or 
11 f(7, 7)  3(7) 
 4(7) or 
7 f(5, 3)  3(5) 
 4(3) or 3 3, 
11  O  5x  7y 
 14  x  y 
 3x  2  12. yes, it is true for both inequalities: 1  y  3x  5 ?  52  y  2x  1 ?  5  
2  2(3)  1  
2  4 true  
2  7 true  
2   Chapter 2  1 (3) 3  y  13.  y  18. y 
 2x  2  y 
 0.5x  1  (0, 4)  (0, 2)  x2
2  ( 25, 45 )  (25, 1 15 )  x  O  y 
 3x  2  f(x, y)  y 
 x f(0, 0)  0 
 0 or 0 f(0, 4)  4 
 0 or 4 f(2, 0)  0 
 2 or 
2 4, 
2 4x  5y 
 10 19. y y
6 (5, 6)  y y
0  (0, 0)  x  O (3, 3) x3
0  y
0 y 
 2x  4  35, 315, 
25, 115, 135, 15 14.  (2, 0) x  (0, 0) O  (0, 3)  (10, 6)  x
y  2x  5y 
 10  (0, 0), (0, 
3), (
3, 
3) y 15.  (0, 2)  (1, 5)  O  f(x, y)  x  y f(0, 2)  0  2 or 2 f(5, 6)  5  6 or 11 f(10, 6)  10  6 or 16 16, 2 y 20.  yx
7  5x  3y 
 20  x  (4, 0) (7, 0)  (2, 5), (7, 0), (4, 0), (1, 5) 16. f(x, y)  8x  y f(0, 0)  8(0)  0 or 0 f(4, 0)  8(4)  0 or 32 f(3, 5)  8(3)  5 or 29 f(0, 5)  8(0)  5 or 5 32, 0 y 17. x
5 2x  5y 
 10  x  O  (2, 5) y  5 
 0  (0, 4)  x
0 (0, 1)  xy
4 (3, 1) y
1  O  x  f(x, y)  4x  2y  7 f(0, 1)  4(0)  2(1)  7 or 9 f(0, 4)  4(0)  2(4)  7 or 15 f(3, 1)  4(3)  2(1)  7 or 21 21, 9 y 21.  (5, 4)  (0, 2) (5, 2) y 
 2 x  O  y 
 4x  6 x  4y 
 7  f(x, y)  3x  y f(0, 2)  3(0)  2 or 2 f(5, 4)  3(5)  4 or 19 f(5, 2)  3(5)  2 or 17 19, 2  (1, 2) (2, 2) O x  6y 
 10  (3, 1)  x  (4, 1) 2x  y 
 7  f(x, y)  2x 
 y f(
2, 
 2)  2(
2) 
 (
2) or 0 f(
1, 2)  2(
1) 
 2 or 
4 f(3, 1)  2(3) 
 1 or 5 f(4, 
1)  2(4) 
 (
1) or 9 9, 
4  53  Chapter 2  22. y  2x  y  2 
 16  (3, 8)  (1, 8)  25b. f(x, y)  5x  6y  f 52, 0  552  6(0) or 272 1  y
8  1  1  f 62, 0  562  6(0) or 322 1  1  f 3, 3  53  63 or 863 29 19  (1, 4)  y
5x  1  y
2  x  y  2x 
 13 0  2x 
 13 1 62  1  1  1  6  23  1  1  1  max at 72, 82  882; min at 22, 2  242 1  1  1  1  1  26. x  y  200 2x  y  300 x0 y0  y 300 (0, 200) 200 x
0 100 (0, 0)  512, 0  2x  y 
 300 (100, 100) x  y 
 200 (150, 0)    x  0  f(x, y)  $6.00x  $4.80y f(0, 0)  $6.00(0)  $4.80(0) or 0 f(0, 200)  $6.00(0)  $4.80(200) or $960 f(100, 100)  $6.00(100)  $4.80(100) or $1080 f(150, 0)  $6.00(150)  $4.80(0) or $900 $1080 27a. Let x represent the Main Street site, and let y represent the High Street site. x  20 y  20 10x  20y  1200 y x 
 20  y  16 
 x 2x 
 13  16 
 x 29 x  3 y  16 
 x  239 , 139   2y  17 2(16 
 x)  17 1  x  72 2y  17 1  y  82  712, 812  (20, 50 )  2y  17  10x  20y 
 1200  1  y  82  y 
 20  y  3x  1 1 82  3x  1  (20, 20 ) (80, 20 )  O  22  x 22, 82 y  7 
 2x 3x  1  7 
 2x 6 x  5 y  7 
 2x 6 y  7 
 25 23  5 65, 253  3y  
2x  11 3(7 
 2x)  
2x  11 1 22  x 1  y  7 
 2x 1 y  7 
 222 2 Chapter 2  1  1  1 62,  19  3y  
2x  11 y  7 
 2x  1  O y 
 0 200 300 x   3  y  7 
 2x y  3x  1  1  f 22, 2  522  6(2) or 242  29  y  3x  1  1  f 5, 5  55  65 or 335  y  16 
 3  2y  17  1  6 23  1  y  16 
 x 2y  17  1  1  52  x  y  16 
 x y  2x 
 13  19  f 22, 82  522  682 or 632  f(x, y)  
2x  y  5 f(1, 4)  
2(1)  4  5 or 7 f(1, 8)  
2(1)  8  5 or 11 f(3, 8)  
2(3)  8  5 or 7 f(6, 2)  
2(6)  2  5 or 
5 f(3, 2)  
2(3)  2  5 or 1 11, 
5 23. x  4, x  
4, y  4, y  
4 24. Sample answer: y  3, x  4, 4x  3y  12 25a. 3y  
2x  11 3y  
2x  11 y0 3(0)  
2x  11 y  2x 
 13 y0  29  f 72, 82  572  682 or 882  (6, 2) (3, 2)  O  1  1  1  x  27b. f(x, y)  30x  40y 27c. f(x, y)  30x  40y f(20, 20)  30(20)  40(20) or 1400 f(20, 50)  30(20)  40(50) or 2600 f(80, 20)  30(80)  40(20) or 3200 80 ft2 at the Main St. site and 20 ft2 at the High St. site 27d. Main Street: $1200 $10  120 ft2 120 30  3600 customers High Street: $1200 20  60 ft2 60 40  240 customers The maximum number of customers can be reached by renting 120 ft2 at Main St.  212, 2 54  28a. 3 is $3 profit on each batch of garlic dressing and 2 is $2 profit on each batch of raspberry dressing. 28b. 2x  3y  18 2x  y  10 x0 y0  2. Sample answer: In an infeasible problem, the region defined by the constraints contains no points. An unbounded region contains an infinite number of points. 3. Sample answer: First define variables. Then write the constraints as a system of inequalities. Graph the system and find the coordinates of the vertices of the polygon formed. Then write an expression to be maximized or minimized. Finally, substitute values from the coordinates of the vertices into the expression and select the greatest or least result.  y 2x  y 
 10 (0, 6)  4.  (3, 4) 2x  3y 
 18  x
0 (0, 0)  (5, 0)  O  x  y
0  f(x, y)  3x  2y f(0, 0)  3(0)  2(0) or 0 f(0, 6)  3(0)  2(6) or 12 f(3, 4)  3(3)  2(4) or 17 f(5, 0)  3(5)  2(0) or 15 3 batches garlic dressing, 4 batches raspberry dressing  5a. 25x  50y  4200 5b. 3x  5y  480 5c. y  160 140 120 (0, 84) 3x  5y 
 480 100 25x  50y 
 4200 80 60 (120, 24) 40 (160,0) 20 (0,0) x  2 1  2(2) 
 (
3)(1) or 7 
3 2 1  2 
1 7 3 2 y 30.  29.  O  y  2x  8  80 60 40 20  31.  d  O  
3
2
1
20 
40 
60 
80  1 2 3 p  32. {16}, {
4, 4}; no, two y-values for one x-value wxyz  4  33.  20  60  100 140  5d. P(x, y)  5x  8y 5e. P(x, y)  5x  8y P(0, 0)  5(0)  8(0) or 0 P(0, 84)  5(0)  8(84) or 672 P(120, 24)  5(120)  8(24) or 792 P(160, 0)  5(160)  8(0) or 800 160 small packages, 0 large packages 5f. $800 5g. No; if revenue is maximized, the company will not deliver any large packages, and customers with large packages to ship will probably choose another carrier for all of their business. 6. Let x  the number of brochures. Let y  the number of fliers. 3x  2y  600 y x  50 x 
 50 300 y  150  x  O  y  8 7 6 5 0.5x  1.5y 
 7 4 3 2 3x  9y 
 2 1 O x 
2 2 4 6 8 10 12 14   15  w  x  y  z  60  200 100  2-7  Linear Programming  Pages 115–116  O  (50, 225) (100, 150)  y 
 150 (50, 150) 3x  2y 
 600 100 200 300 x  C(x, y)  8x  4y C(50, 150)  8(50)  4(150) or 1000¢ C(50, 225)  8(50)  4(225) or 1300¢ C(100, 150)  8(100)  4(150) or 1400¢ 50 brochures, 150 fliers  Check for Understanding  1. Sample answer: These inequalities are usually included because in real life, you cannot make less than 0 of something.  55  Chapter 2  11. y  7. Let x  the number of Explorers. Let y  the number of Grande Expeditions. x  y  375 y 400 2x  3y  450 x0 300 y0 x  y 
 375 x
0 200 (0, 150) 100 (0, 0)  O  O  (3, 0)  (225, 0)  f 4, 3  3  3(3) or 12 f(4, 3)  3  3(3) or 12 f(4, 0)  3  3(0) or 3 f(3, 0)  3  3(0) or 3 alternate optimal solutions 12a. Let g  the number of cups of Good Start food and s  the number of cups of Sirius food. 0.84g  0.56s  1.54 12b. 0.21g  0.49s  0.56 12c. s 3  3 (0, 2.75)  2  y
0  5x  3y 
 15  x  56 48 40 32 24 16 8  O  0.21g  0.49 s 
 0.56 (1.5, 0.5) (2.67, 0) 1  2  g  12d. C(g, s)  36g  22s 12e. C(g, s)  36g  22s C(0, 2.75)  36(0)  22(2.75) or 60.5 C(1.5, 0.5)  36(1.5)  22(0.5) or 65 C(2.66, 0)  36(2.66)  22(0) or 95.76 0 cups of Good Start and 2.75 cups of Sirius 12f. 60.5¢ 13a. Let d  the number of day-shift workers and n  the number of night-shift workers. d5 n6 d  n  14 13b. d  y
6  10.  0.81y  0.56 s 
 1.54  O  infeasible  y  x  f(x, y)  3  3y  x  100 200 300 y
0  Exercises  O  (4, 0)  4x  3y 
 12  P(x, y)  1x  1.50y P(0, 0)  1(0)  1.50(0) or 0 P(0, 30)  1(0)  1.50(30) or 45 P(45, 0)  1(45)  1.50(0) or 45 alternate optimal solutions  9. y  y
3  2x  3y 
 450  1  Pages 116–118  3)  (4, 3)  R(x, y)  250x  350y R(0, 0)  250(0)  350(0) or 0 R(0, 150)  250(0)  350(150) or 52,500 R(225, 0)  250(225)  350(0) or 56,250 225 Explorers, 0 Grande Expeditions 8. Let x  the number of loaves of light whole wheat. Let y  the number of loaves of regular whole wheat. y 60 2x  3y  90 x  2y  80 40 x  2y 
 80 x0 (0, 30) 2x  3y 
 90 y0 20 x
0 (45, 0) (0, 0) x 20 40 60 80 O 
20  x
4  ( 34 ,  unbounded  (6, 8)  2x  y 
 48  d  n 
 14  d
5 x  2y 
 42  (9, 5)  x n
6  8 16 24 32 40 48 56  O  n  13c. $5.50 4  $7.50 4  $52 $7.50 8  $60 C(n, d)  52d  60n 13d. C(n, d)  52d  60n C(6, 8)  52(8)  60(6) or 776 C(9, 5)  52(5)  60(9) or 800 8 day-shift and 6 night-shift workers  Chapter 2  56  17. Let x  amount to deposit at First Bank. Let y  amount to deposit at City Bank. x  y  11,000 y x 
 7500 0  x  7500 12,000 1000  y  7000 x  y 
 11,000  13e. $776 14a. Let x  the number of acres of corn. Let y  the number of acres of soybeans. x  y  180 y x  40 200 x 
 40 y  20 x  2y x  y 
 180 x 
 2y  100 (40, 20)  O  y 
 20 100  (120, 60) (160, 20)  O  x 200  x 
 2y (33.3, 16.7) y 
 12 (24, 12) (38, 12) 20  O  40  60  x  C(x, y)  35,000x  18,000y C(24,12)  35,000(24)  18,000(12) or 1,056,000 C(33.3, 16.7)  35,000(33.3)  18,000(16.7) or 1,466,100 C(38, 12)  35,000(38)  18,000(12) or 1,546,000 24 nurses, 12 nurse's aides 19. Let x  units of snack-size bags. Let y  units of family-size bags. x  y  2400 y x  600 x 
 600 2400 y  900  S(x, y)  10x  15y S(0, 2)  10(0)  15(2) or 30 S(0, 6)  10(0)  15(6) or 90 S(10, 2)  10(10)  15(2) or 130 10 section I questions, 2 section II questions 16. Let x  the number of food containers. Let y  the number of drink containers. x  y  1200 y x  300 1200 y  450 (300, 900)  O  4000 8000 12,000x y 
 1000  40 x  y 
 20 20  8 (0, 6) 6x  15y 
 90 4 (10, 2) x (0, 2) y 
 2 O 4 8 12 16  x  y 
 1200 (750, 450) x 
 300  (7500, 3500) (7500, 1000)  I(x, y)  0.06x  0.065y I(0, 1,000)  0.06(0)  0.065(1,000) or 65 I(0, 7000)  0.06(0)  0.065(7,000) or 455 I(4000, 7000)  0.06(4000)  0.065(7000) or 695 I(7500, 3500)  0.06(7500)  0.065(3500) or 677.5 I(7500, 1000)  0.06(7500)  0.065(1000) or 515 $4000 in First Bank, $7000 in City Bank 18. Let x  the number of nurses. Let y  the number of nurse's aides. x  y  50 y x  y  20 60 y  12 x  y 
 50 x  2y  P(x, y)  150x  250y P(40, 20)  150(40)  250(20) or 11,000 P(120, 60)  150(120)  250(60) or 33,000 P(160, 20)  150(160)  250(20) or 29,000 120 acres of corn, 60 acres of soybeans 14b. $33,000 15. Let x  the questions from section I. Let y  the questions from section II. 6x  15y  90 y y2 12 x 
 0 x0  800 (300, 450) 400  (4000, 7000) y 
 7000  8000 (0, 7000) 4000 (0, 1000)  y 
 450  1600  400 800 1200 x  (600, 1800) x  y 
 2400 (1500, 900)  800  P(x, y)  17.50x  20y P(300, 450)  17.50(300)  20(450) or 14,250 P(300, 900)  17.50(300)  20(900) or 23,250 P(750, 450)  17.50(750)  20(450) or 22,125 $23,250  (600, 900)  O  y 
 900  800 1600 2400 x  P(x, y)  12x  18y P(600, 900)  12(600)  18(900) or 23,400 P(600, 1800)  12(600)  18(1800) or 39,600 P(1500, 900)  12(1500)  18(900) or 34,200 600 units of snack-size, 1800 units of family-size  57  Chapter 2  20. Let x  batches of soap. Let y  batches of shampoo. 12x  6y  48 y 20x  8y  76 (0, 8) x0 y0  23b. Sample answer: Spend more than 30 hours per week on these services. y 24. 20  20x  8y 
 76  y 
 x  5 10  12x  6y 
 48  x
0  O  
20 
10 O 10 20 x 
10 (15, 10) y  5 
 5 
20 (15, 10)  (3, 2) (3.8, 0)  (0, 0)  y
0  x  1  O  1  8x  3y 
 33 4 A (0, 1)  O  1  1  1  25. 4x  y  6 x  2y 
 12  (9, 0)  4x  y  6 4(2y 
 12)  y  6 8y 
 48  y  6 y6  x  4 8 12 (0, 0) y 
 0  x  12y 
 12 x  2(6) 
 12 or 0 26.  x  y  
1  9  0  6  1  3  2  0  3  3  (0, 6)  y y 
 3|x  2|  O  x  27. Sample answer: C  $13.65  $0.15(n 
 30); C  $13.65  $0.15(n 
 30) C  $13.65  $0.15(42 
 30)  $15.45  x 1  Area of trapezoid ACDE  2(12)(13  1)  84  28.  (40, 10) (60, 0) 20  40 60 (25, 0)  3
x   2  Chapter 2 Study Guide and Assessment  (25, 10) 20  2x 
 3  x  2(2x 
 3)  x(3 
 x) 4x 
 6  3x 
 x2 x2  x 
 6  0 (x  3)(x 
 2)  0 x  3  0 or x 
 2  0 x  
3 x2 The correct choice is A.  1  Area of ABF  2(10)(3)  15 Area of shaded origin  84 
 15  69 square units 23a. Let x  oil changes. Let y  tune-ups. y x 
 25 x  25 60 0  y  10 30x  60y  30(60) 30x  60y 
 1800 40  Page 119 Understanding and Using the Vocabulary  x  1. 3. 5. 7. 9.  P(x, y)  12x  20y P(25, 0)  12(25)  20(0) or 300 P(25, 10)  12(25)  20(10) or 500 P(40, 10)  12(40)  20(10) or 680 P(60, 0)  12(60)  20(0) or 720 $720 Chapter 2  1  minimum: 
22, maximum: 10  C (12, 9)  y 
 10 y 
 0O  1  f(15, 
10)  3(15) 
 2(
10) or 10  2x  3y 
 3 B (3, 3) x 
 12 8 12 y
0  1  f(0, 5)  3(0) 
 2(5) or 
22  2x  6y 
 84 D (12, 10)  4  1  f(
15, 
10)  3(
15) 
 2(
10) or 0  P(x, y)  40x  40y P(0, 0)  40(0)  40(0) or 0 P(0, 6)  40(0)  40(6) or 240 P(4, 5)  40(4)  40(5) or 360 P(9, 0)  40(9)  40(0) or 360 alternate optimal solutions y E (0, 14) 22. 12 F (0, 11) 8  1  f(x, y)  3x 
 2y  3 batches of soap and 2 batches of shampoo 21. Let x  the number of small monitors. Let y  the number of large monitors. x  2y  16 y xy9 xy
9 12 x  4y  24 x  2y 
 16 x0 x  4y 
 24 8 y0 (4, 5) (0, 6) 4 x
0  y
x5 (0, 5)  58  translation determinant scalar multiplication polygonal convex set element  2. 4. 6. 8. 10.  added inconsistent equal matrices reflections multiplied  Pages 120–122  18. 
x  2y 
 6z  4 x  y  2z  3 3y 
 4z  7  Skills and Concepts  2y  
4x 2(
x 
 2)  
4x 
2x 
 4  
4x x2 (2, 
4) 12. 6y 
 x  0 6(x 
 5) 
 x  0 6x 
 30 
 x  0 x6 (6, 1) 13. 3y  x  
1 x  
1 
 3y 11.  y  
x 
 2 y  
2 
 2 y  
4  yx
5 y6
5 y1  2x  5y 2(
1 
 3y)  5y 
2 
 6y  5y 19.  2  
11  y 2x  5y  2x  5
11 2  5  x  
11  
151, 
121  2y 
 15x  
4 y  6x  1 2(6x  1) 
 15x  
4 y  6(2)  1 12x  2 
 15x  
4 y  13 (2, 13) x2 15. 5(3x 
 2y)  5(
1) 15x 
 10y  
5 → 4x  10y  24 2(2x  5y)  2(12) 19x  19 x1 2x  5y  12 2(1)  5y  12 y2 (1, 2) 16. x  5y  20.5 x  5y  20.5 → 
x  3y  13.5 3y 
 x  13.5 8y  34 x  5y  20.5 y  4.25 x  5(4.25)  20.5 x  
0.75 (
0.75, 4.25) 17. 3(x 
 2y 
 3z)  3(2) 3(x 
 4y  3z)  3(14) 
3x  5y  4z  0 
3x  5y  4z  0 ↓ ↓ 3x 
 6y 
 9z  6 3x 
 12y  9z  42 
3x  5y  4z  0 
3x  5y  4z  0 
y 
 5z  6 
 7y  13z  42 
7(
y 
 5z)  
7(16) 7y  35z  
42 → 
7y  13z  42 
7y  13z  42 48z  0 z0 
y 
 5z  6 x 
 2y 
 3z  2 
y 
 5(0)  6 x 
 2(
6) 
 3(0)  2 y  
6 x  
10 (
10, 
6, 0) 14.  20.  21.  22.  23.  24.  25. 26.  59  2(
x  2y 
 6z)  2(4) 2x  3y 
 4z  5 ↓ 
2x  4y 
 12z  18 2x  3y 
 4z  15 7y 
 16z  13 
4(3y 
 4z)  
4(7) → 
12y  16z  
28 7y 
 16z  13 7y 
 16z  13 
5y  
15 y3 3y 
 4z  7 x  y  2z  3 3(3) 
 4z  7 x  3  2(0.5)  3 z  0.5 x  
1 (
1, 3, 0.5) x 
 2y  z  7 2(3x  y 
 z)  2(2) 3x  y 
 z  2 2x  3y  2z  7 4x 
 y 9 ↓ 6x  2y 
 2z  4 2x  3y  2z  7 8x  5y  11 5(4x 
 y)  5(9) → 20x 
 5y  45 8x  5y  11 8x  5y  11 28x  56 x2 4x 
 y  9 x 
 2y  z  7 4(2) 
 y  9 2 
 2(
1)  z  7 y  
1 z3 (2, 
1, 3) 8  (
5) A  B  7  (
3) 0  2 
4  (
2) 4 3  2 
6 
3 
 7 
5 
 8 B
A 2 
 0 
2 
 (
4)  
10 
13 2 2 3(
3) 3(
5) 3B  3(2) 3(
2) 
9 
15  6 
6 
4(2) 
4C  
4(
5)  
8 20 8 
3 
5 AB  7 0 
4 2 
2 7(
3)  8(2) 7(
5)  8(
2)  0(
3)  (
4)(2) 0(2)  (
4)(
2) or 
5 
51 
8 8 impossible 4A  4B  4A  (
4B) 4(7) 4(8) 
4(
3) 
4(
5) 4A  
4B  4(0) 4(
4) 
4(2) 
4(
2) 28 32 12 20   0 
16 
8 8 28  12 32  20 4A  (
4B)  0  (
8) 
16  8  40 52 
8 
8  Chapter 2  0 
4 2 5 
4 2 5 32. 1  0 
1 3 1 
3 
3 1 3 0 
1 
4 2 5  3 
1 
3 1 0 
3 1 3 
4 
2 5 A
(3, 
4), B
(1, 2), C
(
3, 5)  27. impossible 
4 2 5 
3 
3 
3 
7 
1 2 28.   3 1 
3 
4 
4 
4 
1 
3 
7 A
(
7, 
1), B
(
1, 
3), C
(2, 
7)  y  A  C  B  y  A A  B  B C  C A  1 0 
2 
1 0 1 
2 
1 0 1  0 
1 
3 2 4 
2 3 
2 
4 2 W
(
2, 3), X
(
1, 
2), Y
(0, 
4), Z
(1, 2)  33. 34.  y Y  W X  35.  Z  36.  x  O X  Z  W Y  30.  
1 0 2 2 
1 
1 
2 
2 1 1  0 
1 3 
5 
5 3 
3 5 5 
3 D
(
2, 
3), E
(
2, 5), F
(1, 5), G
(1, 
3)  y  37.  F  E  D  G  38. 39.  x  O D  G F  40.  E  3 1 
1 1.5 0.5 
0.5  
4 2 1 
2 1 0.5 P
(6, 
8), Q
(2, 4), R
(
2, 2)  41.  31. 0.5  4  R  2  Q  y  x  
2 O 
2 
4  2  4  6  43. 2 
5  2(1) 
 6(
5) or 32 6 1 1 1 5  32 
6 2 2 
4  2(2) 
 (
1)(
4) or 0 44. 
1 2 no inverse  P  
6 
8  Chapter 2  5 
4 
4 
4 5 5   
3 
3 
3 6 6 6 
3 5  
3(7) 
 (
4)(5) or 
1 
4 7 8 
4  8(3) 
 (
6)(
4) or 0 
6 3 3 
1 4 5 
2 6 7 3 
4 
2 6 5 6 5 3 
 (
1) 4 3 
4 7 
4 7  3(
10)  1(
62)  4(29)  24 5 0 
4 7 3 
1 2 
2 6 3 
1 5 
 0 7 
1  (
4) 7 
2 6 2 6 2  5(16) 
 0(44) 
 4(
20)  160 no, not a square matrix 3 8  3(5) 
 (
1)(8) or 23 
1 5 1  5 
8 23 1 3 5 2  5(4) 
 10(2) or 0 10 4 no inverse 
3 5  
3(
4) 
 1(5) or 7 1 
4 1  
4 
5 7 
1 
3  42. 3 2  3(7) 
 5(2) or 11 5 7 1 7 
2  11 
5 3  Q  R  x  O  C  29.  B  x  O  P  60  1 1 1 3 3 3  
2 3  3 
2  2 5 x 1  
1 
3 y 2 1 
3 
5  
1 
3 
5 2 5 1 2 1 2  45.  50.  
1 
1  
3 
5 
1 1 2  (13, 
5)  x 
3 
5  
1 y 1 2 x 13  y 
5  1 2  
6 4  4 
2 6 3  x  1 4 
2 24 6 y 3 x 
1  y 0  3 2 
6 4  
3 6  
3 5 x  1 
2 4 y 
2 1 4 
5  
1 4 
5 2 2 
3 
3 5 2 
3 1  4 
5 2 
3  
3 5 
2 4  1 4 
5 x  
2 y 2 
3 x  
7 y 
4  (
7, 
4) 48.  1 4.6 
2.7 2.9 8.8 1 8.8  48.31 
2.9  4.6 
2.7 2.9 8.8 1    48.31  x y 8.8 2.7 
2.9 4.6  1 
2  8.4 74.61  Page 123  x
1  (25, 3)  10 20 30 (0, 0) (25, 0)  x  Applications and Problem Solving  2 5 5 5 2(5)  5(3)  5(1) 30 52. 8 2 3 3  8(5)  2(3)  3(1)  49 6 4 1 1 6(5)  4(3)  1(1) 43 Broadman 30; Girard 49; Niles 43  y 
 10  2x (1, 5)  (22, 6)  m(x, y)  22x  42y m(0, 0)  22(0)  42(0) or 0 m(0, 6)  22(0)  42(6) or 252 m(22, 6)  22(22)  42(6) or 746 m(25, 3)  22(25)  42(3) or 676 m(25, 0)  22(25)  42(0) or 550 22 gallons in the truck and 6 gallons in the motorcycle  (5.7, 6.6)  y  x  y
0O  x 5.7  y 6.6 49.  12  10 (0, 6) y
6  1 8.8 2.7 8.8 2.7   48.31 
2.9 4.6 
2.9 4.6  2.7 4.6  8  20  
2 4  
2  4  f(x, y)  3x  2y  1 f(0, 4)  3(0)  2(4)  1 or 9 f(0, 9)  3(0)  2(9)  1 or 19 f(4, 7)  3(4)  2(7)  1 or 27 f(6, 5)  3(6)  2(5)  1 or 29 f(6, 4)  3(6)  2(4)  1 or 27 29, 9 51. Let x  gallons in the truck. Let y  gallons in the motorcycle. y x  y  28 0  x  25 30 x  y 
 28 x 
 25 0y6  (
1, 0) 47.  x  y 
 11 12 (0, 9) x 
 6 (4, 7) 2y  x 
 18 8 (6, 5) (0, 4) 4 (6, 4) y
4  O  3 2 x  
3 
6 4 y 6 1 4 
2  1 4 
2 24 6 3 2 6 3 3  46.  1  24  2 5 
1 
3  y  yx
6 (4, 2)  x O  (1, 2) y 
 2  (6, 2)  f(x, y)  2x  3y f(1, 
2)  2(1)  3(
2) or 
4 f(1, 5)  2(1)  3(5) or 17 f(4, 2)  2(4)  3(2) or 14 f(6, 
2)  2(6)  3(
2) or 6 17, 
4  61  Chapter 2  53. Let x  the shortest side. Let y  the middle-length side. Let z  the longest side. x  y  z  83 x  y  z  83 z  3x x  y  3x  83 1 z  2(x  y)  17 4x  y  83  Chapter 2 SAT & ACT Preparation Page 125  1  z  2(x  y)  17  1  (1  2)(2  3)(3  4)  2(20  x)  1  3x  2(x  y)  17 5x 
 y  34  1 4 5 
1  1 4 1 5 
1 1  
9  
1 
1 
5 4  1  (3)(5)(7)  2(20  x) x  83 y 34  1  105  2(20  x) 210  20  x x  190 The correct choice is D. 2. First convert the numbers to improper fractions.  
1 
1  
1 
1 
1 9 
5 4 
5 4  4 1 5 
1  1 
1 
1 x  
9 y 
5 4  83 34  1  z  3x z  3(13) z  39 13 in., 31 in., 39 in. 54a. Let x  number of Voyagers. Let y  number of Explorers. y 5x  6y  240 3x  2y  120 60 5x  18y  540 x0 40 y0 (0, 30) 20 y
0  O  16  25  Express both fractions with a common denominator. Then subtract. 1  1  16  25  64  75  53 
 64  3 
 4  12 
 12 11   
12 The correct choice is A. 3. You can solve this problem by writing algebraic expressions. Amount of root beer at start: x Amount poured into each glass: y Number of glasses: z Total amount poured out: yz Amount remaining: x 
 yz The correct choice is D. 4. 2x 
2 
 1  5 2x 
2  6 x 
2  3  3x  2y 
 120 5x  6y 
 240 (18, 25) (30, 15) 5x  18y 
 540 (40, 0)  20 (0, 0)  40  60  x  P(x, y)  2.40x  5.00y P(0, 0)  240(0)  5.00(0) or 0 P(0, 30)  2.40(0)  5.00(30) or 150 P(18, 25)  2.40(18)  5.00(25) or 168.20 P(30, 15)  2.40(30)  5.00(15) or 147 P(40, 0)  2.40(40)  5.00(0) or 96 18 Voyagers and 25 Explorers 54b. $168.20  x 
2  3 or x 
2  
3 x  5 or xor x  
1 The correct choice is D. 5. The total amount charged is $113. Of that, $75 is for the first 30 minutes. The rest (113 
 $75  $38) is the cost of the additional minutes. At $2 per minutes, $38 represents 19 minutes. (19 $2  $38). The plumber worked 30 minutes plus 19 minutes, for a total of 49 minutes. The correct choice is C.  Open-Ended Assessment  1a. A(
2, 
2), B(1, 
2), C(2, 1), and D(
3, 0) Sample answer: Two consecutive 90° rotations is the same as one 180° rotation. An additional 180° rotation will return the image to its original position. 1b. Two consecutive 90° rotations is the same as one 180° rotation. 2. No; such a coefficient matrix will not have an inverse. Consider the matrix equation 2 4 x  12 . The coefficient 2 4 has a 4 8 y 24 4 8 determinant of 0, so it has no inverse.  Chapter 2  1  53 
 64  3 
 4  x 13  y 31  Page 123  SAT and ACT Practice  1. Translate the information from words into an equation. Then solve the equation for x. Use the correct order of operations.  62  8. Start by representing the relationships that are given in the problem. Let P represent the number of pennies; N the number of nickels; D the number of dimes; and Q the number of quarters. He has twice as many pennies as nickels. P  2N Similarly, N  2D and D  2Q. You know he has at least one quarter. Since you need to find the least amount of money he could have, he must have exactly one quarter. Since he has 1 quarter, he must have 2 dimes, because D  2Q. Since he has 2 dimes, he must have 4 nickels. Since he has 4 nickels, he must have 8 pennies. Now calculate the total amount of money. 1 quarter  $0.25 2 dimes  $0.20 4 nickels  $0.20 8 pennies  $0.08 The total amount is $0.73. The correct choice is D.  6. Start by simplifying the fraction expression on the right side of the equation. 2x  5x 2x  5x  2  2   5  5 4   5  To finish solving the equation, treat it as a proportion and write the cross products. 2x  5x  4   5  5(2  x)  4(5  x) 10  5x  20  4x x  10 The correct choice is E. 7. Notice that the question asks what must be true. There are two ways to solve this problem. The first is by choosing specific integers that meet the criteria and finding their sums. I. 2  3  5, 3  4  7 Choose consecutive integers where the first one is even and where the first one is odd. In either case, the result is odd. So statement I is true. Eliminate answer choice B. II. 2  3  4  9, 3  4  5  12 One sum is odd, and the other is even. So statement II is not always true. Eliminate answer choices B, C, and E. III. 2  3  4  9, 3  4  5  12, 10  11  12  33 Statement III is true for these examples and seems to be true in general. Eliminate answer choice A. Another method is to use algebra. Represent consecutive integers by n and n  1. Represent even integers by 2k, and odd integers by 2k  1. I. n  (n  1)  2n  1 2n  1 is odd for any value of n. So statement I is always true. II. n  (n  1)  (n  2)  3n  3 If n is even, then 3n  3  3(2k)  3 is odd. If n is odd, then 3n  3  3(2k  1)  3  6k  3  3  6k  6 is even. So statement II is not always true. III. By the same reasoning as in II, the sum is a multiple of 3, so statement III is always true. n  (n  1)  (n  2)  3n  3  3(n  1) The correct choice is D.  3  2  9.  2     3  2  3 2  9 4  3   2  4  9  12  2    1 8  3  The correct choice is C. 10. There are two products, CDs and tapes. You need to find the number of tapes sold. You also have information about the total sales and CD sales. You might want to arrange the information in a table. Let t be the number of tapes sold. Price each  Number Sold  Total Sales  CDs  40  $480  Tapes  t  Total  $600  You can calculate the price of each CD. Since 40 CDs sold for $480, each CD must cost $12 ($480 40  $12). You know that the price of a CD is three times the price of a tape. So a CD 1 costs 3 of $12 or $4. You can calculate the total sales of CDs by subtracting $480 from $600 to get $120. Price each  Number Sold  Total Sales  CDs  $12  40  $480  Tapes  $4  t  $120  Total  $600  Now you can find t using an equation. 4t  120 t  30 Thirty cassette tapes were sold. The answer is 30.  63  Chapter 2  Chapter 3 The Nature of Graphs 3-1 Page 133  Algebraically: Substituting (
x, 
y) into the equation followed by substituting (
x, y) is the same as substituting (x, 
y). 6. f(x)  x6  9x f(
x)  (
x)6  9(
x) 
f(x)  
(x6  9x) f(
x)  x6 
 9x 
f(x)  
x6 
 9x no  Symmetry and Coordinate Graphs Graphing Calculator Exploration  1. f(
x)  f(x) 2. f(
x)  
f(x) 3. even; odd 4. f(x)  x8 
 3x4  2x2  2 f(
x)  (
x)8 
 3(
x)4  2(
x)2  2  x8 
 3x4  2x2  2  f(x) f(x)  x7  4x5 
 x3 f(
x)  (
x)7  4(
x)5 
 (
x)3  
x7 
 4x5  x3  
(x7  4x5 
 x3)  
f(x) 5. First find a few points of the graph in either the first or fourth quadrants. For an even function, a few other points of the graph are found by using the same y-values as those points, but with opposite x-coordinates. For an odd function, a few other points are found by using the opposite of both the x- and y-coordinates as those original points. 6. By setting the INDPNT menu option in TBLSET to ASK instead of AUTO, you can then go to TABLE and input x-values and determine their corresponding y-values on the graph. By inputting several sets of opposite pairs, you can observe whether f(
x)  f(x), f(
x)  
f(x), or neither of these relationships is apparent.  7.  1  f(x)  5x 
 x19 1  19  f(
x)   5(
x) 
 (
x) 1  
f(x)  
5x  x19 6a2  b 
 1 6a2  (
b) 
 1 6a2  
b 
 1 no 6(
a)2  b 
 1 6a2  b 
 1 yes 6(b)2  a 
 1 6b2  a 
 1 no 6(
b)2  (
a) 
 1 6b2  
a 
 1 no y-axis a3  b3  4 a3  (
b)3  4 a3 
 b3  4 no (
a)3  b3  4 
a3  b3  4 no (b)3  (a)3  4 a3  b3  4 yes 3 (
b)  (
a)3  4 
a3 
 b3  4 no yx  y-axis yx y  
x  →  y-axis yx y  
x  Pages 133–134  Check for Understanding  1. The graph of y  
x2  12 is an even function. The graph of xy  6 is an odd function. The graphs of x  y2 
 4 and 17x2  16xy  17y2  225 are neither. 2. The graph of an odd function is symmetric with respect to the origin. Therefore, rotating the graph 180° will have no effect on its appearance. See student's work for example. 3a. Sample answer: y  0, x  0, y  x, y  
x 3b. infinitely many 3c. point symmetry about the origin 4. Substitute (a, b) into the equation. Substitute (
b, 
a) into the equation. Check to see whether both substitutions result in equivalent equations. 5. Alicia Graphically: If a graph has origin symmetry, then any portion of the graph in Quadrant I has an image in Quadrant III. If the graph is then symmetric with respect to the y-axis, the portion in Quadrants I and II have reflections in Quadrants II and IV, respectively. Therefore, any piece in Quadrant I has a reflection in Quadrant IV and the same is true for Quadrants II and III. Therefore, the graph is symmetric with respect to the x-axis.  Chapter 3  10.  y  (2.5, 3) (4, 2)  (2.5, 3)  (1, 2)  (1, 2)  (4, 2)  O  x  2 
 x2 11. y  
 x-axis y-axis  b  
 2 
 a2 
b  
 2 
 a2 no b  
 2 
 (
 a)2 b  
 2 
 a2 yes  →  y 2 (1,1)  1  (0√2) (1,1)  (√2,0)  (√2,0)  
2 
1 O 
1 
2  64  1  1  f(
x)  
5x  x19 yes 8. 6x2  y 
 1 → x-axis  9. x3  y3  4 x-axis  
f(x)  
5x 
 x19  1  2 x  12. y  x3 x-axis  →  b  a3 
b  a3 b  a3 yes b  (
a)3 b  
a3 no  y-axis  18.  y  19.  f(x)  
7x5  8x f(
x)  
7(
x)5  8(
x) f(
x)  7x5 
 8x 
f(x)  
(
7x5  8x) 
f(x)  7x5 
 8x yes 1  f(x)  x 
 x100 1   
 (
x)100 f(
x)   (
x)  (1,1)  O  (1,1)  1  f(
x)  
x 
 x100  x  
f(x)  
x 
 x100 1  1  
f(x)  
x  x100  no 20. yes; x2 
 1  g(x)  x 13. x-intercept:  x2  25  y2  9  
  x2  25  
 9  1  1  g(
x)   02  
g(x)   x2   1 25 x2  25  36  25  36  25  
 9  1   y2  
 9  1 y2 
9  11 
25  y2   99  25      y-axis  yx  99   3
11  3
11   y   5  y  
x  y   5  3
11  3
11     6, 
  5 , 
6, 
 5 , 
6, 5  Pages 134–136    11 
25  y2  25   3
11  (
x)2  x2  when x  
6  y2  y2 
9  Replace x with 
x.  x2 
 1 
x x2 
 1 
 x  g(
x)  
g(x) 21. xy  5 → x-axis  x  5 (5, 0) other points: when x  6  (
x)2 
 1   g(
x)   (
x)  22. x  y2  1 x-axis  Exercises  →  y-axis  f(x)  3x f(
x)  3(
x) 
f(x)  
(3x) f(
x)  
3x 
f(x)  
3x yes 15. f(x)  x3 
 1 f(
x)  (
x)3 
 1 
f(x)  
(x3 
 1) 3 f(
x)  
x 
 1 
f(x)  
x3  1 no 16. f(x)  5x2  6x  9 f(
x)  5(
x)2  6(
x)  9 f(
x)  5x2 
 6x  9 
f(x)  
(5x2  6x  9) 
f(x)  
5x2 
 6x 
 9 no 14.  17.  yx y  
x  23. y  
8x x-axis y-axis yx y  
x  1   f(x)   4x7 1   f(
x)   4(
x)7 1   f(
x)  
 4x7  →  Determine the opposite of the function. ab  
5 a(
b)  
5 
ab  
5 ab  5 no (
a)b  
5 
ab  
5 ab  5 no (b)(a)  
5 ab  
5 yes (
b)(
a)  
5 ab  
5 yes y  x and y  
x a  b2  1 a  (
b)2  1 a  b2  1 yes (
a)  b2  1 
a  b2  1 no (b)  (a)2  1 b  a2  1 no (
b)  (
a)2  1 
b  a2  1 no x-axis b  
8a (
b)  
8a 
b  
8a no b  
8(
a) b  8a no (a)  
8(b) a  
8b no (
a)  
8(
b) 
a  8b no none of these   
f(x)  
 4x7  1  1   
f(x)  
 4x7  yes  65  Chapter 3  1  24. y  x2  →  28.  1  b  a2 1  
b  a2  (2, 1)  no  1  b  a2  (4, 4)   (a)   (b)2  (4, 4) y  1  a  b2  no   (
a)   (
b)2 1  
a  b2 →  y-axis yx y  
x 4x2  26. y2  9 
 4 x-axis  29.  1  y  
x  →  (2, 1)  a2  b2  4 a2  (
b)2  4 a2  b2  4 (
a)2  b2  4 a2  b2  4 (b)2  (a)2  4 a2  b2  4 (
b)2  (
a)2  4 a2  b2  4 all 4a2 b2  9 
 4 4a2 (
b)2  9 
 4  (2, 1)  O  x  yes 30. Sample answer:  yes  y yes  (4, 4) (1, 2)  yes  (2, 1)  O  x  (2, 3)  4a2  4(
a)2  →  31. y2  x2 x-axis  b2  9 
 4 4a2  b2  9 
 4 yes 4(b)2  b2  a2 (
b)2  a2 b2  a2 b2  (
a)2 b2  a2 yes; both  y-axis  (a)2  9 
 4  yx  (4, 4)  (1, 2) (1, 2)  no; y-axis  b2  9 
 4 yes y-axis  x  (1, 2)  yes  1  yx  O  (2, 1)  1   b (
a)2  y-axis  (4, 4) (1, 2)  1  (
b)  a2  x-axis  25. x2  y2  4 x-axis  y  4b2  a2  9 
 4 no  y  4(
b)2  (
a)2  9 
 4  y  
x  4b2  a2  9 
 4 no x-axis and y-axis 27.  x2    1 y2  x-axis  →  1  1  (
a)2  a2   yx  (b)2    b2  a2  y  
x  (
b)2  b2  a2   1 b2 1 b2 1 (a)2 1 a2 1 b2 1 (
a)2 1 a2 1 b2  yes  32. x  
3y x-axis  yes  →  a  
3b a  
3(
b) a  3b no (
a)  
3b a  
3b yes y-axis  y-axis  y  yes  O yes  x-axis, y-axis, y  x, y  
x  Chapter 3  x   a2   (
b)2  a2  b2 y-axis  O  1  a2  b2  66  x  33. y2  3x  0 x-axis  →  37. y  x3 
 x x-axis  b  a3 
 a (
b)  a3 
 a b  a3 
 a yes y-axis b  (
a)3 
 (
a) b  
a3  a no x-axis The equation y  x3 
 x is symmetric about the x-axis. y  b2  3a  0 (
b)2  3a  0 b2  3a  0 yes 2 b  3(
a)  0 b2 
 3a  0 no x-axis  y-axis  y  →  1  x  O  O  
1  34. y  2x2 x-axis  →  b  2a2 (
b)  2a2 b  2a2 yes b  2(
a)2 b  2a2 yes; both  y-axis  x  1  
1  38a.  x2  8  y2  a2  8  →   10  1  a2  8  x-axis  y  x  O  origin  (
b)2   10  1  a2  8 (
a)2  8 a2  8 (
a)2  8  y-axis  b2   1 0 1 b2   1 0  1 yes b2   1 0 1 b2   1 0  1 yes (
b)2   10  1 a2  8  b2   1 0  1 yes  x- and y-axis symmetry 
 12 
  8y2  35. x  x-axis  →  y-axis  
 12 
  8b2  a a  
 12 
 8(
b) 2 a  
 12 
  8b2 yes (
a)  
 12 
  8b2 a  
 12 
  8b2 yes; both  38b.  y  x  O  y  x  O  36. y  xy x-axis  ), (
2, 
5 ), (
2, 

5 ) 38c. (2, 

5 39. Sample answer: y  0 40. Sample answer: yx1 y  
x  1 y  
x 
 1 y  
2x  4 y  
2x 
 4 y  2x 
 4  →  y  b  ab (
b)  a(
b) b  
ab no b  (
a)b b  
ab no neither  y-axis  yx
1 y  2x  4  2 1 
2 
1 O 
1  y  1  2  x  
2 1 (1, 0)  O  (1, 0)  x  
1  67  Chapter 3  41.  y2  12 (6)2  12  x2  
 16  1 
 16  1 x2  y  2x 
 2 7  y  2x 
 2  consistent and dependent 47.  x2  
16  
2  O y
x2  16 
 2  14    
2 or 
7  y 
 2  
7(x 
 0) y  
7x  2 49. [f 
 g](x)  f(g(x))  f(x 
 6)  
2(x 
 6)  11  
2x  23 [g 
 f ](x)  g(f(x))  g(
2x  11)  (
2x  11) 
 6  
2x  5 50. 753 757  7537  7510 The correct choice is B.  1 y
x  x  43. Let x  number of bicycles. Let y  number of tricycles. 3x  4y  450 y 5x  2y  400 5x  2y 
 400 150 x0 (0, 112.5) y0  O  x   48. m   
2 
 0  y  100 x
0 50 (0, 0)  y y
x  x2  32 x  4
2  (4
2 , 6) or (
4
2 , 6) 42. No; if an odd function has a y-intercept, then it must be the origin. If it were not, say it were (0, 1), then the graph would have to contain (
1, 0). This would cause the relation to fail the vertical line test and would therefore not be a function. But, not all odd functions have a y-intercept. 1 Consider the graph of y  x.  (50, 75) 3x  4y 
 450  y
0 50 100 150 x (80, 0)  3-2  P(x, y)  6x  4y P(0, 0)  6(0)  4(0) or 0 P(0, 112.5)  6(0)  4(112.5) or 450 P(50, 75)  6(50)  4(75) or 600 P(80, 0)  6(80)  4(0) or 480 50 bicycles, 75 tricycles 8 5  4(8)  3(9) 4(5)  3(6) 44. 4 3 7 2 9 6 7(8)  2(9) 7(5)  2(6) 59 38  74 47 45. 3(2x  y  z)  3(0) 6x  3y  3z  0 → 3x 
 2y 
 3z  
21 3x 
 2y 
 3z  
21 9x  y  
21 3x 
 2y 
 3z  
21 4x  5y  3z  
2 7x  3y  
23 
3(9x  y)  
3(
21) 
27x 
 3y  63 → 7x  3y  
23 7x  3y  
23 
20x  40 x  
2 9x  y  
21 2x  y  z  0 9(
2)  y  
21 2(
2)  (
3)  z  0 y  
3 z7 (
2, 
3, 7)  Chapter 3  →  
12x  6y  
21  3 
 16  1  O  7  46. 4x 
 2y  7  x2  Page 142  Families of Graphs Check for Understanding  1. y  (x  4)3 
 7 2. The graph of y  (x  3)2 is a translation of y  x2 three units to the left. The graph of y  x2  3 is a translation of y  x2 three units up. 3. reflections and translations 4. When c 1, the graph of y  f(x) is compressed horizontally by a factor of c. When c  1, the graph of y  f(x) is unchanged. When 0  c  1, the graph is expanded 1 horizontally by a factor of c. 3 5a. g(x)  
x  1 3 5b. h(x)  

 x
1 3 5c. k(x)  
 x21 6. The graph of g(x) is the graph of f(x) translated left 4 units. 7. The graph of g(x) is the graph of f(x) compressed 1 horizontally by a factor of 3, and then reflected over the x-axis. 8a. expanded horizontally by a factor of 5 8b. translated right 5 units and down 2 units 8c. expanded vertically by a factor of 3, translated up 6 units  68  9a. translated up 3 units, portion of graph below x-axis reflected over the x-axis 9b. reflected over the x-axis, compressed 1 horizontally by a factor of 2 9c. translated left 1 unit, compressed vertically by a factor of 0.75 y y 10. 11.  O  x  20a. reflected over the x-axis, compressed horizontally by a factor of 0.6  20b. translated right 3 units, expanded vertically by a factor of 4 1 20c. compressed vertically by a factor of 2, translated down 5 units 21a. expanded horizontally by a factor of 5 21b. expanded vertically by a factor of 7, translated down 0.4 units 21c. reflected across the x-axis, translated left 1 unit, expanded vertically by a factor of 9 22a. translated left 2 units and down 5 units 22b. expanded horizontally by a factor of 1.25, reflected over the x-axis 3 22c. compressed horizontally by a factor of 5, translated up 2 units 23a. translated left 2 units, compressed vertically by 1 a factor of 3 23b. reflected over the y-axis, translated down 7 units 23c. expanded vertically by a factor of 2, translated right 3 units and up 4 units 24a. expanded horizontally by a factor of 2 1 24b. compressed horizontally by a factor of 6, translated 8 units up 24c. The portion of parent graph on the left of the yaxis is replaced by a reflection of the portion on the right of the y-axis. 2 25a. compressed horizontally by a factor of 5, translated down 3 units 25b. reflected over the y-axis, compressed vertically by a factor of 0.75 25c. The portion of the parent graph on the left of the y-axis is replaced by a reflection of the portion on the right of the y-axis. The new image is then translated 4 units right.  O x  12a.  x 0x1 1x2 2x3 3x4 4x5  f(x) 50 100 150 200 250  $250 $200 $150 $100 $50 0  0  1  2 3 4 Time (h)  5  12b. $250 $200 $150 $100 $50 0  26. y  x2 0  1  2  3 4 5 Time (h)  y  6  12c. $225  Pages 143–145  Exercises  13. The graph of g(x) is a translation of the graph of f(x) up 6 units. 14. The graph of g(x) is the graph of f(x) compressed  O  x  3  vertically by a factor of 4. 15. The graph of g(x) is the graph of f(x) compressed  0.25   27. y   x
4  3  1  16. 17. 18. 19.  horizontally by a factor of 5. The graph of g(x) is a translation of f(x) right 5 units. The graph of g(x) is the graph of f(x) expanded vertically by a factor of 3. The graph of g(x) is the graph of f(x) reflected over the x-axis. The graph of g(x) is the graph of f(x) reflected over the x-axis, expanded horizontally by a factor of 2.5, and translated up 3 units.  28.  29.  y  O  y  x O  69  x  Chapter 3  y  30.  31.  O  36a. 0  y  x O  y  32.  y  33.  O  x  x  4  
4 O 
4  4  [
7.6, 7.6] scl:1 by [
5, 5] scl:1 36b. 0.66   8x  
8 
12  34.  y  8 4  
8 
4 O 
4  y 
 f (|x |) 4  [
7.6, 7.6] scl:1 by [
5, 5] scl:1 36c. 
0.25  x  y 
 f (x ) 
8  35a. 0  [
7.6, 7.6] scl:1 by [
5, 5] scl:1 37a. 0  [
7.6, 7.6] scl:1 by [
5, 5] scl:1 35b. 0.5  [
7.6, 7.6] scl:1 by [
5, 5] scl:1 37b. 
2.5  [
7.6, 7.6] scl:1 by [
5, 5] scl:1 35c. 
1.5  [
7.6, 7.6] scl:1 by [
5.5] scl:1  [
7.6, 7.6] scl:1 by [
5, 5] scl:1  Chapter 3  70  42a. (1) y  x2 (3) y  
x2 42b. (1) y  37c. 0.6  (2) y  x3 (4) y  
x3 (2) y  4 
4 O 
4  [
7.6, 7.6] scl:1 by [
5, 5] scl:1 38a. The graph would continually move left 2 units and down 3 units. 38b. The graph would continually be reflected over the x-axis and moved right 1 unit.  (3)  5 4 Price 3 (dollars) 2 1 0  1  2  3  4  5  Fare Units  41a.  y  1  A  2bh 1  8   2(10)(5)  4   25 units2  O  4  8  12  x  8x  
8  O  y 8 4 8  12  46.  x 1  41c. The area of the triangle is A  2(10)(5)  25 units2. Its area is the same as that of the original triangle. The area of the triangle formed by y  f(x  c) would be 25 units2.  47. 48.  y 8 4  O  4  8  12  8x  
4  1  4  
4 O  (30, 20) (30, 0) x  y 
 50  20 (0, 0)  41b. The area of the triangle is A  2(10)(10) or 50 units2. Its area is twice as large as that of the original triangle. The area of the triangle formed by y  c f(x) would be 25c units2.  O  4  y  (4) 4  42c. (1) y  (x 
 3)2 
 5 (2) y  (x 
 3)3 
 5 (3) y  
(x 
 3)2 
 5 (4) y  
(x 
 3)3 
 5 43a. reflection over the x-axis, reflection over the y-axis, vertical translation, horizontal compression or expansion, and vertical expansion or compression 43b. horizontal translation 44. f(x)  x17 
 x15 f(
x)  (
x)17 
 (
x)15 
f(x)  
(x17 
 x15) f(
x)  
x17  x15 
f(x)  
x17  x15 yes; f(
x)  
f(x) 45. Let x  number of preschoolers. Let y  number of school-age children. x  y  50 y x  3(10) 60 (0, 50) y  5(10) y 
 50 x0 40 x 
 30 y0  40b.  0  8x  
8  
12  0.25[[x 
 1]]  1.50 if [[x]]  x  y 1.50 1.75 2.00 2.25 2.50  8x  
8  0.25[[x]]  1.50 if [[x]]  x  4  
4  
4  b  x 0x1 1x2 2x3 3x4 4x5  4  y 
4 O  39. The x-intercept will be 
a. 40a. y   
4 O  x  49.  40  60  x  I(x, y)  18x  6y I(0, 0)  18(0)  6(0) or 0 I(0, 50)  18(0)  6(50) or 300 I(30, 20)  18(30)  6(20) or 660 I(30, 0)  18(30)  6(0) or 540 30 preschoolers and 20 school-age 0 
1 5 1 
2  4 
3 1 1 0 
4 3 
1 5 1 
2 A
(4, 5), B
(
3, 1), C
(1, 
2) x2  25 9y 12  2z x  5 6z 
5(6x  5y)  
5(
14) 
30x 
 25y  70 → 6(5x  2y)  6(
3) 30x  12y  
18 
13y  52 y  
4 5x  2y  
3 5x  2(
4)  
3 x1 (1, 
4) The graph implies a negative linear relationship.  50. 3x 
 4y  0  →  perpendicular slope:  71  20 y
0  3  y  4x  4 
3  Chapter 3  y  7.  5  51. 5d 
 2p  500 → p  2d 
 250 
250 52. [f 
 g](x)  f(g(x))  f(x2 
 6x  9)  O  2   3(x2 
 6x  9) 
 2  8.  y  x  2   3x2 
 4x  4  O  x  [g 
 f](x)  g(f(x))   g3x 
 2  9.  2  y   3x 
 2 
 63x 
 2  9 2  2  2  4  8  4  20   9x2 
 3x  4 
 4x  12  9  O   9x2 
 3x  25  x  50  53. If m  1; d  1 
 1 or 
49. 50  If m  10; d  10 
 10 or 5.  10. Case 1 x  6 4 
(x  6) 4 
x 
 6 4 
x 10 x  
10 {xx  
10 or x 
2} 11. Case 1 3x 
 4  x 
(3x 
 4)  x 
3x  4  x 
4x  
4 x1 {x 1  x  2} 12a. x 
 12  0.005 12b. Case 1 x 
 12  0.005 
(x 
 12)  0.005 
x  12  0.005 
x  
11.995 x  11.995 12.005 cm, 11.995 cm  50  If m  50; d  50 
 50 or 49. 50   If m  100; d  100 
  100 or 99.5 50   If m  1000; d  1000 
  1000 or 999.95.  The correct choice is A.  3-3 Page 149  Graphs of Nonlinear Inequalities Check for Understanding  1. A knowledge of transformations can help determine the graph of the boundary of the shaded region, y  5  
x. 
2 2. When solving a one variable inequality algebraically, you must consider the case where the quantity inside the absolute value is nonnegative and the case where the quantity inside the absolute value is negative. 3. Sample answer: Pick a point not on the boundary of the inequality, and test to see whether it is a solution to the inequality. If that point is a solution, shade all points in that region. If it is not a solution to the inequality, test a point on the other side of the boundary and shade accordingly. 4. This inequality has no solution since the two graphs do not intersect  Pages 150–151  17.  x  x2 
 6  y  x 
9  y  
5x4  7x3  8 ? 
5(
1)4  7(
1)3  8 
3  
3  
4; yes 6. y  3x 
 4 
 1 ? 3(0) 
 4 
 1 3 3  3; no  5.  Chapter 3  4 4 
2  Case 2 3x 
 4  x 3x 
 4  x 2x  4 x2  Case 2 x 
 12  0.005 x 
 12  0.005 x  12.005  Exercises  13. y  x3 
 4x2  2 14. y  x 
 2  7 ? (1)3 
 4(1)2  2 ? 3 
 2  7 0 8 0  
1; no 8  8; no 15. y 
 
x  11  1  
1 ? 
 
(
2)   11  1 
1 
2; yes 16. y  
0.2x2  9x 
 7 ? 
0.2(10)2  9(10) 
 7 63  63  63; no  y  O  Case 2 x  6 x6 x  (
6)2 
 6 ?    
6  
9  
5; yes 18. y  2x3 
 7 ? 203 
 7 0  0  
7; yes  72  19. y  
x  2 y  
x  2 ? 
0 ? 
1 0 2 4 2 0  2; yes 4  3; no y  
x  2 y  
x  2 ? 
1 ? 

1 1 2 0 2 1  3; yes 0  i  2; no y  
x  2 ? 
1 
1  2 
1  3; yes (0, 0) (1, 1) and (1, 
1); if these points are in the shaded region and the other points are not, then the graph is correct.  y  20.  O  22.  x  y  O  32.  x  12 8 4  O 24.  4  8  12  O  x  y  y  25.  O O 26.  x  x y  O  27.  y  x  x  O 28.  x  y  29.  y  O  O  x  x  no solution 38. Case 1 2x  9 
 2x  0 
(2x  9) 
 2x  0 
2x 
 9 
 2x  0 
4x  9 9 x  
4 all real numbers  x  73  x  x  33. Case 1 x  4 5 
(x  4) 5 
x 
 4 5 
x 9 x  
9 {xx  
9 or x 1} 34. Case 1 3x  12  42 
(3x  12)  42 
3x 
 12  42 
3x  54 x  
18 {xx  
18 or x  10} 35. Case 1 7 
 2x 
 8  3 
(7 
 2x) 
 8  3 
7  2x 
 8  3 2x  18 x9 {x
2  x  9} 36. Case 1 5 
 x  x 
(5 
 x)  x 
5  x  x 
5  0; true {xx 2.5} 37. Case 1 5x 
 8  0 
(5x 
 8)  0 
5x  8  0 
5x  
8 8 x 5  y  23.  y  y  O  O  31.  O  y  21.  y  30.  Case 2 x  4 x4 x  5 5 1  Case 2 3x  12  42 3x  12  42 3x  30 x  10  Case 2 7 
 2x 
 8  3 7 
 2x 
 8  3 
2x  4 x 
2  Case 2 5 
 x  x 5
xx 
2x  
5 x  2.5 Case 2 5x 
 8  0 5x 
 8  0 5x  8 8 x 5  Case 2 2x  9 
 2x  0 2x  9 
 2x  0 9  0; true  Chapter 3  39. Case 1 2 
3x  5  
8 2  
3(
(x  5))  
8 2 
3(
x 
 5) 2 10 x   3 3 2 x 3  40.  41.  42.  43.  45b. The shaded region shows all points ( x, y) where x represents the number of cookies sold and y represents the possible profit made for a given week. 46. The graph of g(x) is the graph of f(x) reflected over the x-axis and expanded vertically by a factor of 2.  Case 2 2 
3x  5  
8   
8  
8  2  
3(x  5)  
8 2  10  
3x 
 3  
8 2  14  
3x  
3  1  34  1  →  47. y  
a4   
3 x7 x  
17 {x
17  x  7} x 
 37.5  1.2 Case 1 Case 2 x 
 37.5  1.2 x 
 37.5  1.2 
(x 
 37.5)  1.2 x 
 37.5  1.2 
x  37.5  1.2 x  38.7 
x  
36.3 x  36.3 36.3  x  38.7 Case 1 Case 2 3x 
 7  x 
 1 3x 
 7  x 
 1 3(
(x 
 7))  x 
 1 3(x 
 7)  x 
 1 3(
x  7)  x 
 1 3x 
 21  x 
 1 
3x  21  x 
 1 2x  20 
4x  
22 x  10 x 5.5 {x5.5  x  10} 30 units2 The triangular region has vertices A(0, 10), B(3, 4), and C(8, 14). The slope of side AB is 
2. The slope of side AC is 0.5, therefore AB is perpendicular to AC. The length of side AB is 3
5 . The length of side AC is 4
5  the area of the triangle is 0.5(3
5 )(4
5 ) or 30. 0.10(90)  0.15(75)  0.20(76)  0.40(80)  0.15(x)  80 0.15x  12.55 2 x  833  b  
a4 1  (
b)  
a4  x-axis  1  
b  
a4 b  y-axis  b yx  (a)  a  y  
x  (
a)  
a  y-axis  1 8 
3 4 
5  48.  49.  3  4  
5 
4  3  
1 
5 28 
4 8  
7  0  8 
4    3 (8) 4 3 (
4) 4  6 
3  
21  4  0  50. x 
2 
1 0 1 2  3 8  3 (
7) 4 3 (0) 4  f(x) 11 8 5 8 11  44.  f (x)  O 51. 50 45 40 States 35 with 30 Teen 25 Courts 20 15 10 5 0  [
1, 8] scl:1 by [
1, 8] scl:1 44a. b  0 44b. none 44c. b  0 or b 4 44d. b  4 44e. 0  b  4 45a. P (x )  300 200 100  Chapter 3  0  1980 1990 2000 2010 Year  52. [f 
 g](4)  f(g(4))  f(0.5(4) 
 1)  f(1)  5(1)  9  14 [g 
 f ](4)  g(f(4))  g(5(4)  9)  g(29)  0.5(29) 
 1  13.5  400  O  no  1  
 (
a)4 1 
a4 yes 1  
 (b)4 1  
 no (b)4 1   
 (
b)4 1  
 no (b)4  900 1000 1100 1200 1300 1400 1500 x  74  x  4. y  x  1 x-axis  53. Student A  15 1  Student B  3(15)  15 or 20 Let x  number of years past. 20 
 x  2(15 
 x) 20 
 x  30 
 2x x  10  →  y-axis yx y  
x  Page 151  y  x  1 → f(
x)  
 x  1 f(
x)  x  1 y-axis 5a. translated down 2 units 5b. reflected over the x-axis, translated right 3 units 1 5c. compressed vertically by a factor of 4, translated up 1 unit 6a. expanded vertically by a factor of 3 6b. expanded horizontally by a factor of 2 and translated down 1 unit 6c. translated left 1 unit and up 4 units  Mid-Chapter Quiz  1. x2  y2 
 9  0 x-axis  →  a2  b2 
 9  0 a2  (
b)2 
 9  0 a2  b2 
 9  0 yes y-axis (
a)2  b2 
 9  0 a2  b2 
 9  0 yes yx (b)2  (a)2 
 9  0 a2  b2 
 9  0 yes y  
x (
b)2  (
a)2 
 9  0 a2  b2 
 9  0 yes x2  y2 
 9  0 → f(x)  
 
x2  9 2 f(
x)  

(
x)   9 
f(x)  
 (
 
x2  9) 2 f(
x)  
 
x  9 
f(x)   
 
x2  9 yes x-axis, y-axis, y  x, y  
x, origin 2. 5x2  6x 
 9  y → 5a2  6a 
 9  b x-axis 5a2  6a 
 9  (
b) 5a2  6a 
 9  
b no y-axis 5(
a)2  6(
a) 
 9  b 5a2 
 6a 
 9  b no yx 5(b)2  6(b) 
 9  (a) 5b2  6b 
 9  a no y  
x 5(
b)2  6(
b) 
 9  (
a) 5b2 
 6b 
 9  
a no 5x2  6x 
 9  y → f(x)  5x2  6x 
 9 f(
x)  5(
x)2  6(
x) 
 9  5x2 
 6x 
 9 
f(x)  
(5x2  6x 
 9) 
f(x)  
5x2 
 6x  9 no none of these 7  7.  x  9. Case 1 2x 
 7  15 
(2x 
 7)  15 
2x  7  15 
2x  8 x 
4 
4  x  11 10. x 
 64  3 Case 1 x 
 64  3 
(x 
 64)  3 
x  64  3 
x  
61 x 61 61  x  67  7   a (
b) 7  a  
b no 7  (
a)  b  y-axis  y  O  a  b  x-axis  8.  y  O  7  →  3. x  y  b  a  1 (
b)  a  1 
b  a  1 no b  (
a)  1 b  a  1 yes (a)  (b)  1 a  b  1 no (
a)  (
b)  1 
a  b  1 no f(x)  x  1 
f(x)  
(x  1) 
f(x)  
x 
 1 no  x  Case 2 2x 
 7  15 2x 
 7  15 2x  22 x  11  Case 2 x 
 64  3 x 
 64  3 x  67  7  
a  b no 7  yx  (b)  (a)  3-4  7  a  b yes y  
x 7  →  x  y f(
x)  f(
x)   7  (
x) 7 
x  7  (
b)   (
a) 7 a  b yes 7 f(x)  x 7 
f(x)  
 x 7 
f(x)  
x yes  Inverse Functions and Relations  Pages 155–156  Check for Understanding  1. Sample answer: First, let y  f(x). Then interchange x and y. Finally, solve the resulting equation for y. 2. n is odd 3. Sample answer: f(x)  x2 4. Sample answer: If you draw a horizontal line through the graph of the function and it intersects the graph more than once, then the inverse is not a function.     y  x, y  
x, origin  75  Chapter 3  5. She is wrong. The inverse is f 
1(x)  (x 
 3)2 
 2, which is a function. 6. f
1(x) f(x)  x  1 x f
1(x) x f(x) 3 
2 
2 3 2 
1 
1 2 1 0 0 1 2 1 1 2 3 2 2 3  f (x )  9.  f(x)  
3x  2 y  
3x  2 x  
3y  2 x 
 2  
3y 1  2  1  2  y  
3x  3 f
1(x)  
3x  3; f
1(x) is a function. 10.  1  f(x)  x3 1  y  x3 1  x  y3  f (x )  1  y3  x y  f 1(x )  O  x  f
1(x)   3  1  x  1  3 
x  or  1  
1 3 
x ; f (x)  is a function.  f(x)  (x  2)2  6 y  (x  2)2  6 x  (y  2)2  6 x 
 6  (y  2)2 
x  
6y2 y  
2 
x 
6  
1 f (x)  
2  
x; 
 6 f
1(x) is not a function. 12. Reflect the graph of y  x2 over the line y  x. Then, translate the new graph 1 unit to the left and up 3 units. 11.  7.  f(x)  x3  1 x f(x) 
2 
7 
1 0 0 1 1 2 2 9  x 
7 0 1 2 9  f
1(x) f
1(x) 
2 
1 0 1 2  y f (x )  f (x ) f 1(x )  8.  x  O  x  O  13. f(x)  
(x 
 3)2  1 x f(x) 1 
3 2 0 3 1 4 0 5 
3  x 
3 0 1 0 
3  f
1(x) f
1(x) 1 2 3 4 5  1  f(x)  2x 
 5 1  y  2x 
 5 1  x  2y 
 5 1  x  5  2y y  2x  10 f
1(x)  2x  10 [f 
 f
1](x)  f(2x  10) 1  2(2x  10) 
 5 x  f (x )  [f
1 
 f ](x)  f
12x 
 5 1  f 1(x )   22x 
 5  10 x Since [f 
 f
1](x)  [f
1 
 f ](x)  x, f and f
1 are inverse functions. 14a. B(r)  1000(1  r)3 B  1000(1  r)3 1  O  x f (x )  B  1000 B 3  1000   (1  r)3 1r 3  r  
1   Chapter 3  76  
B   10  14b. r  
1   18.  3  
B   10 3   
1   
1100   10  Pages 156–158 15.  or 0.0323; 3.23%  Exercises  f(x)  x  2 x f(x) 
2 4 
1 3 0 2 1 3 2 4  f (x )  x 4 3 2 3 4  f (x ) 10 
10 O  O  19.  x  f(x)  2x x f(x) 
2 4 
1 2 0 0 1 2 2 4  x 4 2 0 2 4  f
1(x) f
1(x) 
2 
1 0 1 2  f 1(x )  O  f 1(x )  f (x ) f 1(x )  x  f
1(x) x f
1(x) 
10 
2 
3 
1 
2 0 
1 1 6 2  f(x)  x3 
 2 x f(x) 
2 
10 
1 
3 0 
2 1 
1 2 6  f (x )  20.  f (x ) f 1(x )  O  f(x) 
2 
1 0 1 2  f
1(x) f
1(x) 
2  x  
1 
1  x  0 0x1 1x2 2x3  O f (x ) 17.  x  10  f(x)  [x] x 
2  x  
1 
1  x  0 0x1 1x2 2x3  x 
2 
1 0 1 2  f (x ) f (x )  f (x )  
10  f 1(x )  f (x )  16.  f
1(x) f
1(x) 
2 
1 0 1 2  f
1(x) x f
1(x) 
42 
2 
11 
1 
10 0 
9 1 22 2  f(x)  x5 
 10 x f(x) 
2 
42 
1 
11 0 
10 1 
9 2 22  x  f(x)  3 x f(x) 
2 3 
1 3 0 3 1 3 2 3  f (x )  x 3 3 3 3 3  f
1(x) f
1(x) 
2 
1 0 1 2  f (x )  x O  x f 1(x )  77  Chapter 3  21.  f(x)  x2  2x  4 x f(x) 
3 7 
2 4 
1 3 0 4 1 7  x 7 4 3 4 7  24.  f
1(x) f
1(x) 
3 
2 
1 0 1  8  f (x )  
8  
4 O  f 1(x ) 8x  4  
4  f (x ) f (x )  
8  22.  4  f(x)  
(x  2)2 
 5 x f(x) 
4 
9 
3 
6 
2 
5 
1 
6 0 
9  x 
9 
6 
5 
6 
9  f
1(x) f
1(x) 
4 
3 
2 
1 0  
4 O  f 1(x )  O  25.  x  f(x)  2x  7 y  2x  7 x  2y  7 x 
 7  2y x
7  y  2  x  x
7  f
1(x)  2; f
1(x) is a function. 26.  f (x )  f(x)  (x  1)2 
 4 x f(x) 
4 5 
2 
3 
1 
4 0 
3 2 5  f (x )  4  f 1(x )  
4  f (x )  23.  f
1(x)  
x 
4 x f
1(x) 8 2 5 1 4 0  f(x)  x2  4 x f(x) 
2 8 
1 5 0 4 1 5 2 8  f (x ) 4  f(x)  x2  4 y  x2  4 x  y2  4 x 
 4  y2 y  
x ; 
 4 f
1(x)  
x  
4  x 5 
3 
4 
3 5  f
1(x) f
1(x) 
4 
2 
1 0 2  27.  f(x)  
x 
 2 y  
x 
 2 x  
y 
 2 y  
x 
 2 f
1(x)  
x 
 2; f
1(x) is a function. 1  f(x)  x 1  y  x 1  x  y 1  y  x 1  f
1(x)  x; f 
1(x) is a function. 28.  1  f(x)  
x2 1  f (x )  y  
x2  f 1(x )  1  x  
y2 1  y2  
x  O  1  y   
x 1  f
1(x)   
x; f 
1(x) is not a function. 29.  Chapter 3  78  f(x)  (x 
 3)2  7 y  (x 
 3)2  7 x  (y 
 3)2  7 x 
 7  (y 
 3)2 
x  
7y
3 y  3 
x 
7  f
1(x)  3 
x ; 
 7 f 
1(x) is not a function.  30.  31.  f(x)  x2 
 4x  3 y  x2 
 4x  3 x  y2 
 4y  3 x  1  y2 
 4y  4 x  1  (y 
 2)2 
x  1y
2 y  2  
x 1 f
1(x)  2  
x;  1 f 
1(x) is not a function.  35. Reflect the part of the graph of x2 that lies in the first quadrant about y  x. Then, translate 5 units to the left.  f (x )  x  O  1   f(x)   x2 1   y x2 1   x y2  36. Reflect the graph of x2 about y  x. Then, translate 2 units to the right and up 1 unit. f (x )  1  y  2  x 1  y  x 
 2 1  f
1(x)  x 
 2; f 
1(x) is not a function. 32.  1   f(x)   (x 
 1)2  y x ( y 
 1)2  y
1 y  x O  1  (x 
 1)2 1  ( y 
 1)2 1  x 1  x 1 1  
x  37. Reflect the graph of x3 about y  x to obtain the 3 3 graph of 
x. Reflect the graph of 
x about the x-axis. Then, translate 3 units to the left and down 2 units. f (x )  1  ; f 
1(x) is not a function. 
x 2  f(x)  
 (x 
 2)3 2  y  
 (x 
 2)3 2  x  
 ( y 
 2)3 2 ( y 
 2)3  
x 2 y 
 2  
 3 x 2 y  2 
 3 x 2 f
1(x)  2 
 3 x; f 
1(x) is not a function. 3  g(x)   x2  2x 3  y x2  2x 3  x y2  2y 3 y2  2y  x 3 y2  2y  1  x  1 3 (y  1)2  x  1 3 y  1   x  1 3 y  
1  x  1 3 g
1(x)  
1  x  1  f
1(x)  1   33.  34.  O  x  38. Reflect the graph of x5 about y  x. Then, translate 4 units to the right. Finally, stretch the translated graph vertically by a factor of 2. f (x )  O  79  x  Chapter 3  39.  2  1  2  1   42a. v  
2gh v2  2gh  2  1  h  f(x)  
3x  6 y  
3x  6 x  
3y  6 1  h  2  x 
 6  
3y 3  1  3  1  h  y  
2x  4 [f 
 f
1](x)  f 
2x  4 1   
3
2x  4  6 2  3  1  1  1  1   x 
 6  6 x  x 0x1 1x2 2x3 3x4 4x5  [f
1 
 f ](x)  f
1
3x  6 2  1   
2
3x  6  4 3  2  1  1  1  1   x 
 4  4 x Since [f 
 f
1](x)  [f
1 
 f ](x)  x, f and f
1 are inverse functions. 40. f(x)  (x 
 3)3  4 y  (x 
 3)3  4 x  (y 
 3)3  4 x 
 4  (y 
 3)3 3 
 x
4y
3 3 y  3  
 x
4 3 f
1(x)  3  
 x
4 3 [f 
 f
1](x)  f(3  
) x
4 3  [(3  
) x 
 4 
 3]3  4 x
44 x [f
1 
 f ](x)  f
1[(x 
 3)3  4] 3  3  
 [(x 
  3)3   4] 
 4 3x
3 x Since [f 
 f
1](x)  [f
1 
 f ](x)  x, f and f
1 are inverse functions. 41a. d(x)  x 
 4 d
1(x) x d(x) x d
1(x) 6 2 2 6 5 1 1 5 4 0 0 4 3 1 1 3 2 2 2 2  C (x ) $0.80 $0.70 $0.60 $0.50 $0.40 $0.30 $0.20 $0.10 1 2 3 4 5 6 7 8x  44b. positive real numbers; positive multiples of 10 44c. C
1(x) 0x1 1x2 2x3 3x4 4x5  x $0.10 $0.20 $0.30 $0.40 $0.50  8 7 6 5 4 3 2 1  C 1(x )  O  x 20¢ 40¢ 60¢ 80¢  44d. positive multiples of 10; positive real numbers 44e. C
1(x) gives the possible lengths of phone calls that cost x. 45. It must be translated up 6 units and 5 units to the 5 left; y  (x 
 6)2 
 5; y  6  
x. 46a.  1  KE  2mv2 2KE  mv2 2KE  m   v2  v  46b. v   v  2KE  m 2(15)  1  v  5.477; 5.5 m/sec 2KE  m  46c. There are always two velocities. 47a. Yes; if the encoded message is not unique, it may not be decoded properly. 47b. The inverse of the encoding function must be a function so that the encoded message may be decoded. 47c. C(x)  
2  
x 3 y  
2  
x 3 x  
2  
y 3 x  2  
y 3 (x  2)2  y  3 y  (x  2)2 
 3 
1 C (x)  (x  2)2 
 3  x  41b. No; the graph of d(x) fails the horizontal line test. 41c. d
1(x) gives the numbers that are 4 units from x on the number line. There are always two such numbers, so d
1 associates two values with each x-value. Hence, d
1(x) is not a function.  Chapter 3  C(x) $0.10 $0.20 $0.30 $0.40 $0.50  O  d 1(x)  O  (75)2  h  64 h  87.89 Yes. The pump can propel water to a height of about 88 ft.  v2  2g v2  2(32) v2  64  43a. Sample answer: y  
x 43b. The graph of the function must be symmetric about the line y  x. 43c. Yes, because the line y  x is the axis of symmetry and the reflection line. 44a.  f
1(x)  
2x  4 3  v2  42b. h  64  80  47d. C
1(x)  (x  2)2 
 3 C
1(1)  (1  2)2 
 3 or 6, F C
1(2.899)  (2.899  2)2 
 3 or 21, U C
1(2.123)  (2.123  2)2 
 3 or 14, N C
1(0.449)  (0.449  2)2 
 3 or 3, C C
1(2.796)  (2.796  2)2 
 3 or 20, T C
1(1.464)  (1.464  2)2 
 3 or 9, I C
1(2.243)  (2.243  2)2 
 3 or 15, O C
1(2.123)  (2.123  2)2 
 3 or 14, N C
1(2.690)  (2.690  2)2 
 3 or 19, S C
1(0)  (0  2)2 
 3 or 1, A C
1(2.583)  (2.583  2)2 
 3 or 18, R C
1(0.828)  (0.828  2)2 
 3 or 5, E C
1(1)  (1  2)2 
 3 or 6, F C
1(2.899)  (2.899  2)2 
 3 or 21, U C
1(2.123)  (2.123  2)2 
 3 or 14, N FUNCTIONS ARE FUN 48. Case 1 Case 2 2x  4  6 2x  4  6 
(2x  4)  6 2x  4  6 
2x 
 4  6 2x  2 
2x  10 x1 x  
5 {x
5  x  1} 49. both 50a. a  0, b  0, 4a  b  32, a  6b  54 50b.  53.  O 54.  Page 165  (8, 0)  1 
2 
1 4  2.  a  4x  2y  10 xy6  4 2 x  10 1 1 y 6 1 1 
2   1 
2 2 
1 
1 4 4 1 x 1 
2  2 y 
1 4 x 
1  y 7  4 2 1 1  52.  9 
3  
6 6  10 6    1 (
3) 2 1 (6) 2  1 (9) 2 1 (
6) 2 9  2  
3  y 
 7  
1(x 
 0)  Check for Understanding  an  n  x→  p(x) →  positive positive positive positive  even even odd odd   
  
     
  an  n  x→  p(x) →  negative negative negative negative  even even odd odd   
  
  
 
 
   3. Infinite discontinuity; f(x) →  as x → 
, f(x) → 
 as x → . 4. f(x)  x2 is decreasing for x  0 and increasing for x 0, g(x)  
x2 is increasing for x  0 and decreasing for x  0. Reflecting a graph switches the monotonicity. In other words, if f(x) is increasing, the reflection will be decreasing and vice versa. 5. No; y is undefined when x  
3. 6. No; f(x) approaches 6 as x approaches 
2 from the left but f(x) approaches 
6 as x approaches 
2 from the right.  (
1, 7) 1  2  y 
 y1  m(x 
 x1)  1. Sample answer: The function approaches 1 as x approaches 2 from the left, but the function approaches 
4 as x approaches 2 from the right. This means the function fails the second condition in the continuity test.  a
0  1  2  neither  Continuity and End Behavior  3-5  4a  b 
 32  1 4 2 1 1  x  1 4; 4 4 
1; 2
7   5
0 
5  5 or 
1  y  
x  7 56. b  c  180 If  PQ  is perpendicular to Q R , then m∠PQR  90. Since the angles of a triangle total 180, a  d  90  180. a  d  90 a  b  c  d  180  90 or 270 The correct choice is C.  (6, 8)  G(a, b)  a  b G(0, 0)  0  0 or 0 G(0, 9)  0  9 or 9 G(6, 8)  6  8 or 14 G(8, 0)  8  0 or 8 14 gallons 51. 4x  2y  10 → y6
x  1  4  55. m  b a  6b 
 54 (0, 9)  O (0, 0) b 
 0  y  3  
2 3  81  Chapter 3  7. an: positive, n: odd y →  as x → , y → 
 as x → 
. 8. an: negative, n: even y → 
 as x → , y → 
 as x → 
. 9.  24. x 
10,000 
1000 
100 
10 0 10 100 1000 10,000  f (x )  x O  decreasing for x  
3; increasing for x y 10.  1  y  x2  
3  y 1 10
8 1 10
6 1 10
4 0.01 undefined 0.01 1 10
4 1 10
6 1 10
8  y → 0 as x → , y → 0 as x → 
. 25.  1  f(x)  
x3  2  O  x  decreasing for x  
1 and x 
1  x  1 11a. t  4 11b. when t  4  Pages 166–168  x 
10,000 
1000 
100 
10 0 10 100 1000 10,000  1; increasing for 11c. 10 amps  Exercises  2 2.000000001 2.000001 2.001 undefined 1.999 1.999999 1.999999999 2  f(x) → 2 as x → , f(x) → 2 as x → 
.  12. Yes; the function is defined when x  1; the function approaches 
3 as x approaches 1 from both sides; and y  
3 when x  1. 13. No; the function is undefined when x  2. 14. Yes; the function is defined when x  
3; the function approaches 0 as x approaches 
3 from both sides; and f(
3)  0. 15. Yes; the function is defined when x  3; the function approaches 1 (in fact is equal to 1) as x approaches 3 from both sides; and y  1 when x  3. 16. No; f(x) approaches 
7 as x approaches 
4 from the left, but f(x) approaches 6 as x approaches 
4 from the right. 17. Yes; the function is defined when x  1; f(x) approaches 3 as x approaches 1 from both sides; and f(1)  3. 18. jump discontinuity 19. Sample answer: x  0; g(x) is undefined when x  0. 20. an: positive, n: odd y →  as x → , y → 
 as x → 
. 21. an: negative, n: even y → 
 as x → , y → 
 as x → 
. 22. an: positive, n: even y →  as x → , y →  as x → 
. 23. an: positive, n: even y →  as x → , y →  as x → 
.  Chapter 3  f(x)  26.  [
6, 6] scl:1 by [
30, 30] scl:5 increasing for x  
3 and x 1; decreasing for 
3  x  1 27.  [
7.6, 7.6] scl:1 by [
5, 5] scl:1 decreasing for all x  82  28.  33b. Since f is odd, its graph must be symmetric with respect to the origin. Therefore, f is increasing for 
2  x  0 and decreasing for x  
2. f must have a jump discontinuity when x  
3 and f(x) →  as x → 
.  f (x)  [
7.6, 7.6] scl:1 by [
8, 2] scl:1 decreasing for x  
1 and x 
1  O  29.  x  34a. polynomial 34b. [
25, 25] scl:5 by [
25, 25] scl:5 increasing for x  
1 and x 5; decreasing for 
1  x  2 and 2  x  5 30.  [
5, 80] scl:10 by [
500, 12000] scl:1000 0.5  t  39.5 34c. 0  t  0.5 and t 39.5 35a. 1954-1956, 1960-1961, 1962-1963, 1966-1968, 1973-1974, 1975-1976, 1977-1978, 1989-1991, 1995-1997 35b. 1956-1960, 1961-1962, 1963-1966, 1968-1973, 1974-1975, 1976-1977, 1978-1989, 1991-1995, 1997-2004 36a.  [
7.6, 7.6] scl:1 by [
1, 9] scl:1 decreasing for x  
2 and 0  x  2; increasing for 
2  x  0 and x 2 31.  [
7.6, 7.6] scl:1 by [
1, 9] scl:1 3 3 decreasing for x  
2 and 0  x  2; increasing 3 3 for 
2  x  0 and x 2 32. As the denominator, r, gets larger, the value of U(r) gets smaller. U(r) approaches 0. 33a. Since f is even, its graph must be symmetric with respect to the y-axis. Therefore, f is decreasing for 
2  x  0 and increasing for x  
2. f must have a jump discontinuity when x  
3 and f(x) → 
 as x → 
.  36b. 36c.  f (x)  O  36d. 36e. 37a.  x  83  [
1, 8] scl:1 by [
10, 1] scl:1 x4 Answers will vary. The slope is positive. In an interval where a function is increasing, for any two points on the graph, the x- and y-coordinates of one point will be greater than that of the other point, ensuring that the slope of the line through the two points will be positive. See graph in 36a. x 4 The slope is negative; see students' work. The function has to be monotonic. If the function were increasing on one interval and decreasing on another interval, the function could not pass the horizontal line test.  Chapter 3  37b. The inverse must be monotonic. If the inverse were increasing on one interval and decreasing on another interval, the inverse would fail the horizontal line test. That would mean the function fails the vertical line test, which is impossible.  3-5B Gap Discontinuities Page 170 1. {all real numbers xx  3}  38a. 40 35 Percent 30 with 25 Similar 20 Computer 15 Usage 10 5 0  0 2 4 6 8 10 12 14 16  [
10, 10] scl:1 by [
6, 50] scl:10 2. {all real numbers x
2  x  4}  38b. 0  x  1, 1  x  2, 2  x  4, 4  x  6, 6  x  8, x 8 39. For the function to be continuous at 2, bx  a and x2  a must approach the same value as x approaches 2 from the left and right, respectively. Plugging in x  2 to find that common value gives 2b  a  4  a. Solving for b gives b  2. For the function to be continuous at 
2, 

b 
  and x bx  a must approach the same value as x approaches 
2 from the left and right, respectively. Plugging in x  
2 gives 

b  2  
2b  a. We already know b  2, so the equation becomes 0  
4  a. Hence, a  4. 40. f(x)  (x  5)2 y  (x  5)2 x  (y  5)2 
x y5 y  
5  
x f 
1(x)  
5  
x 41. The graph of g(x) is the graph of f(x) translated left 2 units and down 4 units. 42. f (x, y)  x  2y f(0, 0)  0  2(0) or 0 f(4, 0)  4  2(0) or 4 f(3, 5)  3  2(5) or 13 f(0, 5)  0  2(5) or 10 13, 0 
4  5(2) 
 8(
4) or 42 43. 5 8 2 44a. c  47.5h  35  [
9.4, 9.4] scl:1 by [
6.2, 6.2] scl:1 3. {all real numbers xx  
3 or x  1}  [
18.8, 18.8] scl:1 by [
12.4, 12.4] scl:1 4. {all real numbers xx  
3 or x 
2}  44b. c  47.5h  35 1 c  47.5 24  35  [
4.7, 4.7] scl:1 by [
25, 25] scl:10 5. {all real numbers xx  
1 or x 1}     c  $141.875 f(x)  2x2 
 2x  8 f(
2)  2(
2)2 
 2(
2)  8  8  4  8 or 20 46. The volume of the cube is x3. The volume of the other box is x(x 
 1)(x  1)  x(x2 
 1) or x3 
 x. The difference between the volumes of the two boxes is x3 
 (x3 
 x) or x. The correct choice is A. 45.  Chapter 3  [
4.7, 4.7] scl:1 by [
3.1, 3.1] scl:1  84  6. {all real numbers xx  
6 or x  
2}  12. Yes; sample justification: if f(x) is a polynomial function, then the graph of y   f(x)  (x 
 [[x]]  0.25)  is like the graph of f(x), but with an infinite number of "interval bites" removed. 13. Yes; sample justification: the equation y  x2(x  
2)  (2x 
 4)(x  4)  ((x  
2) or (x  4))  is a possible equation for  the function described. 14a. [
9.4, 9.4] scl:1 by [
6.2, 6.2] scl:1 7. {all real numbers xx  3 or x  4}  [
15, 15] scl:2 by [
10, 20] scl:2 14b. [
9.4, 9.4] scl:1 by [
3, 9.4] scl:1 8. {all real numbers xx  
2 or 
1  x  1 or x  2}  [
9.1, 9.1] scl:1 by [
6, 6] scl:1  [
4.7, 4.7] scl:1 by [
2, 8] scl:1 9. {all real numbers xx  1 or 2  x  3 or x  4}  [
3, 6.4] scl:1 by [
2, 8] scl:1 10. {all real numbers xx  1 or x  3-6  Critical Points and Extrema  Page 176  Check for Understanding  1. Check values of the function at x-values very close to the critical point. Be sure to check values on both sides. If the function values change from increasing to decreasing, the critical point is a maximum. If the function values change from decreasing to increasing, the critical point is a minimum. If the function values continue to increase or to decrease, the critical point is a point of inflection. 2. rel. min.; f(0.99)  
3.9997 f(1)  
4 f(1.01)  
3.9997 By testing points on either side of the critical point, it is evident that the graph of the function is decreasing as x approaches 1 from the left and increasing as x moves away from 1 to the right. Therefore, on the interval 0.99  x  1.01, (1, 4) is a relative minimum.  2}  [
9.4, 9.4] scl:1 by [
6.2, 6.2] scl:1 11. Sample answer: y  x2  ((x  2) or ((x  5) and (x  7)) or x  8))  85  Chapter 3  12c. $58.80 per acre 12d. Rain or other bad weather could delay harvest and/or destroy part of the crop.  3. Sample answer:  y (4, 6)  Pages 177–179  (3, 1)  O  Exercises  13. abs. max.: (
4, 1) 14. abs. max.: (
1, 3); rel. min.: (0.5; 0.5); rel. max.: (1.5, 2) 15. rel. max.: (
2, 7): abs. min.: (3, 
3) 16. rel. max.: (
6, 4), rel. min.: (
2, 
3) 17. abs. min.: (3, 
8); rel. max.: (5, 
2); rel. min.: (8, 
5) 18. no extrema 19. abs. max.: (1.5, 
1.75)  x  (0, 4)  4. rel. min.: (
3, 
2); rel. max.: (1, 6) 5. rel. min.: (
1, 
3); rel. max.: (3, 3) 6. rel. max.: (0, 0); rel. min.: (2, 
16)  [
5, 5] scl:1 by [
8, 2] scl:1 20. rel. max.: (
1.53, 1.13); rel. min.: (1.53, 
13.13)  [
4, 6] scl:1 by [
20, 20] scl:5 7. rel. min.: (
2.25, 
10.54)  [
6, 4] scl:1 by [
14, 6] scl:2 8. f(
1.1)  0.907 f(
1)  1 f(
0.9)  0.913 max. 9. f(
2.6)  
12.24 f(
2.5)  
12.25 f(
2.4)  
12.24 min. 10. f(
0.1)  
0.00199 f(0)  0 f(0.1)  0.00199 pt. of inflection 11. f(
0.1)  
0.97 f(0)  
1 f(0.1)  
0.97 min. 12a. P(x)  (120  10x)(0.48 
 0.03x)  [
5, 5] scl:1 by [
16, 4] scl:2 21. rel. max.: (
0.59, 0.07), rel. min.: (0.47, 
3.51)  [
5, 5] scl:1 by [
5, 5] scl:1 22. abs. min.: (
1.41, 
6), (1.41, 
6); rel. max.: (0, 
2)  P (x ) 80 70 60 Profit 50 (dollars) 40 30 20 10 0  x 0 2 4 6 8 10 12 14 16 18 Time (weeks)  [
5, 5] scl:1 by [
8, 2] scl:1  12b. 2 weeks Chapter 3  86  23. rel. max.: (
1, 1); rel. min.: (0.25, 
3.25)  35b.  [
1, 12] scl:1 by [
200, 500] scl:100 2.37 cm by 2.37 cm 35c. See students' work. 36a. P  sd 
 25d  s(
200s  15,000) 
 25(
200s  15,000)  
200s2  20,000s 
 375,000  [
5, 5] scl:1 by [
5, 5] scl:1 24. no extrema  [
5, 5] scl:1 by [
5, 5] scl:1 25. abs. min.: (
3.18, 
15.47); rel. min. (0.34, 
0.80); rel. max.: (
0.91, 3.04)  [0, 100] scl:10 by [0, 130,000] scl:10,000 abs. max.: (50, 125,000) $50 36b. Sample answer: The company's competition might offer a similar product at a lower cost. 37a. AM 2  MB2  AB2 AM 2  x2  22 AM  
 x2  4 f(x)  5000(
 x2  4)  3500(10 
 x)  [
5, 5] scl:1 by [
16, 5] scl:2 26. f(
0.1)  
0.001 f(0)  0 f(0.1)  0.001 pt of inflection 27. f(3.9)  5.99 f(4)  6 f(4.1)  5.99 max. 28. f(
2.6)  
19.48 f(
2.5)  
19.5 f(
2.4)  
19.48 min. 29. f(
0.1)  6.98 f(0)  7 f(0.1)  6.98 max. 30. f(1.9)  
3.96 f(2)  
4.82 f(2.1)  
3.96 min. 31. f(2.9)  
0.001 f(3)  0 f(3.1)  0.001 pt. of inflection 32. f(
2.1)  4.32 f(
2)  4.53 f(
1.9)  4.32 max. 33. f(0.57)  2.86  37b.  [
10, 20] scl:2 by [0, 60,000] scl:10,000 abs. min.: (1.96, 42,141.4) 1.96 km from point B n 38. equations of the form y  xn or y  
x, where n is odd 39.  [
3, 7] scl:1 by [
50, 10] scl:10 The particle is at rest when t  0.14 and when t  3.52. Its positions at these times are s(0.14)  
8.79 and s(3.52)  
47.51.  f3  2.85 f(0.77)  2.86 min. 34. The point of inflection is now at x  
6 and there is now a minimum at x  
3. 35a. V(x)  2x(12.5 
 2x)(17 
 2x) 2  87  Chapter 3  47. Let x  number of 1-point free throws. Let y  number of 2-point field goals. Let z  number of 3-point field goals. 1x  2y  3z  32 x  y  z  17 y  0.50(18) 1x  2y  3z  32 → 1x  2y  3z  32 
1(x  y  z  17) 
x 
 y 
 z  
17 y  2z  15 y  0.50(18) y  2z  15 y9 9  2z  15 z3 x  y  z  17 x  9  3  17 x5 5 free throws, 9 2-point field goals, 3 3-point field goals  40. If a cubicle has one critical point, then it must be a point of inflection. If it were a relative maximum or minimum, then the end behavior for a cubic would not be satisfied. If a cubic has three critical points, then one must be a maximum, another a minimum, and the third a point of inflection. 41. No; the function is undefined when x  5. 42.  y  x  O  48. y  6  4 y  
2  43. Let x  units of notebook paper. Let y  units of newsprint. x  y  200 y x  10 200 (10, 190) y  80 x  y 
 200  y  O  x  (120, 80) y 
 80  100 (10, 80)  x 
 10  O  2  49. 2x  3y  15 → y  
3x  5  x 100  200  3  6x  4y  16 → y  2x 
 4  P(x, y)  400x  350y P(10, 80)  400(10)  350(80) or 32,000 P(10, 190)  400(10)  350(190) or 70,500 P(120, 80)  400(120)  350(80) or 76,000 120 units of notebook and 80 units of newsprint  2  
3  A (3, 4)  B (2, 4)  x  O D (3, 1)  C (2, 1)  T  
3  x  2, 
1  y  4 3  1(5) 
 2(3) or 
1; yes 45. 1 2 5 4 
2 5 7 3(4) 3(
2) or  3(5) 3(7) 
3 5 2B  2 
4 3 2(
3) 2(5) or  2(
4) 2(3) 12 
6  3A  2B  15 21 12  (
6)  15  (
8) 6 4  7 27  S  46. 3A  3  Chapter 3  12 15   
1; perpendicular  50. A relation relates members of a set called the domain to members of a set called the range. In a function, the relation must be such that each member of the domain is related to one and only one member of the range. You can use the vertical line test to determine whether a graph is the graph of a function. 51. The area of PTX is equal to the area of RTY. The area of STR is 25% of the area of rectangle PQRS. The correct choice is D. X Q P  y  44.  3  2  
6 21  
6 10 
8 6 
6 10 
8 6 
6  10 21  6  88  Y  R  3-7  x3  6. x  2, x  
1  Graphs of Rational Functions   y (x 
 2)(x  1) x3   y x2 
 x 
 2  Pages 185–186 1.  Check for Understanding  f (x ) O  x3  x3  y   x2 x 2  
 x3 
 x3 x3  y  1  1 2 1  
  
 x3 x2 x  x  no horizontal asymptotes 1   7. f(x)   x1 
 2  8. The parent graph is translated 4 units right. The vertical asymptote is now at x  4. The horizontal asymptote, y  0, is unchanged.  1a. x  2, y  
6 1  y   1b. y   x
2 
 6  8  2. Sample graphs: Vertical Asymptote  4  Horizontal Asymptote  y  y  
8  
4 O 
4  4  8x  
8  O x  O  x  9. The parent graph is translated 2 units left and down 1 unit. The vertical asymptote is now at x  
2 and the horizontal asymptote is now y  
1.  y  Slant Asymptote  y O  x  O x  10. x (x  1)   3. Sample answer: f(x)   x1  4. False; sample explanation: if that x-value also causes the numerator to be 0, there could be a hole instead of a vertical asymptote. 5. x  5  3x  5 20  x 
 3
 3x2 
 4x   5 → 3x  5   x
3 2 3x 
 9x 5x  5 5x 
 15 20 y  3x  5  y  11.  x   f(x)   x
5  12.  y  x   y x
5 y(x 
 5)  x xy 
 5y  x xy 
 x  5y x(y 
 1)  5y  O  x  O  x  5y   x y 
 1; y  1  13a. P  O  89  V Chapter 3  13b. P  0, V  0 13c. The pressure approaches 0.  Pages 186–188  (x  1)2  19. x  1   y x2 
 1 x2  2x  1   y x2 
 1 x2  x2  2x  14. x  
4   f(x)   x4  1  y(x  4)  2x xy  4y  2x xy 
 2x  
4y x(y 
 2)  
4y 
4y  x y 
 2; y  2  x3   y (x 
 2)4  y  x2   y x6 x2  x2  y  1  y  y  6 1    x2 x  x5  y y  0  1   22. f(x)   x
2 
 3 1  23. f(x)  
x  1 24. The parent graph is translated 3 units up. The vertical asymptote, x  0, is unchanged. The horizontal asymptote is now y  3.  1  x2  
 y   2x2 9x 5    x2 
 x2 
 x2 1  x  x3  x4  x4 8x3 24x2 32x 16     
 x 4  x4 
 x4  x4 x4 1  x  ; y  8 24 32 16 1 
    
 x3  x4 x x2 1  x
1  (2x  1)(x 
 5) x
1   2x2 
 9x 
 5 x  x2  x3  x4 
 8x3  24x2 
 32x  16   21. f(x)   x3  1  no horizontal asymptote 16. x   1  20. x  2  y  6 x2   x2 x2  1 
2,  2   x  x2  y ;y1 1 1 
  x2  2x   y x4  15. x  
6  1  2x   x2  x2 y   1 x2  
 x2 x2  Exercises  y  1  
 x2  y 5 ;y0 9 2 
  
  x2 x 17. x  
1, x  
3  O  x
2   y x2 
 4x  3 x  x2  x  2   x2  2 y x 4x 3   x2  x2 x2 1  x  25. The parent graph is translated 4 units right and expanded vertically by a factor of 2. The vertical asymptote is now x  4. The horizontal asymptote, y  0, is unchanged.  2   x2  y 4 3 ;y0 1     x x2 18. no vertical asymptote,  8  x2   y x2  1  4  x2  O   x2  
8  y  x2 1   x2 x2  
4  4 
4 
8  1  y  1 ;y1 1   x2  Chapter 3  y  90  8x  x
1 1  x  4
 x2  3 x
3 → x
1 x4 2 x  4x 
x 
 3 
x 
 4 1 yx
1 31. x3 4 x
 x2  3 x
4 → x  3 
 x x2 3x 
 4 3x 
4 yx3 32. x
2 2  x2  1
 x3 
 2 x2  x 
4 → x
2
 x2  1 3 x x 
 2x2 
4 
 2x2 
2 
2 yx
2 30.  26. The parent graph is translated 3 units left. The translated graph is then expanded vertically by a factor of 2 and translated 1 unit down. The vertical asymptote is now x  
3 and the horizontal asymptote is now y  
1.  y 4  O 
4  x  4 
4  27. The parent graph is expanded vertically by a factor of 3, reflected about the x-axis, and translated 2 units up. The vertical asymptote, x  0, is unchanged. The horizontal asymptote is now y  2.  y  1 x 2  33. 2x 
  8  3
 x2 
 4 x1 x2  4 
8  
4 O 
4  8x  4  
8  →  3 
 2x 5 
2x  1 5 15 
2x  4 11 
4  1 x 2  
  5  4  1  
  11  4  2x 
 3  5  y  2x 
 4  34. No; the degree of the numerator is 2 more than that of the denominator. y y 35. 36.  28. The parent graph is translated 3 units right. The translated graph is expanded vertically by a factor of 10 and then translated 3 units up. The vertical asymptote is x  3 and the horizontal asymptote is y  3. 12  5  
 4  (2, 4)  x  O  y  O  x  8 4  y  37. 4 O 4  4  8  12  38.  y  x  8  (2, 0)  O  (2,  12  29. The parent graph is translated 5 units left. The translated graph is expanded vertically by a factor of 22 and then translated 4 units down. The vertical asymptote is x  
5 and the horizontal asymptote is y  
4.  14  )  x  39.  x  O  40.  y  y  y 8  O  x  4 
16 
12 
8  x  O  
4  O  4  (3, 0) x  
4 
8 
12 
16  91  Chapter 3  480  3t  48. abs. max.: (2, 1)   41a. C(t)   40  t  480  3t   C(t)  40  t  41b.  480  3t   10   40  t  400  10t  480  3t 7t  80 t  11.43 L 42. Sample answer: The circuit melts or one of the components burns up. 43. To get the proper x-intercepts, x 
 2 and x  3 should be factors of the numerator. The vertical asymptote indicates that x 
 4 should be a factor of the denominator. To get point discontinuity at (
5, 0), make x  5 a factor of both the numerator and denominator with a bigger exponent in the numerator. Thus, a sample answer is f(x)  44a.  (x 
 2)(x  3)(x  5)2  . (x 
 4)(x  5)  V  x2 h  A(x)  4x h  2x2  2 A(x)  4xx 2   2x 120  120  x2 h 120  x2  [
1, 6] scl:1 by [
5, 2] scl:1 49. x2 
 9  y y2 
 9  x y2  x  9 y  
x 9 50. f(x, y)  y 
 x f(0, 0)  0 
 0 or 0 f(4, 0)  0 
 4 or 
4 f(3, 5)  5 
 3 or 2 f(0, 5)  5 
 0 or 5 5; 
4 5  
4(
6) 
4(5) 51. 
4 
6 8 
4 
4(8) 
4(
4) 24 
20  
32 16 52. Let x  price of film and y  price of sunscreen. 8x  2y  35.10 3x  y  14.30 8x  2y  35.10 y  14.30 
 3x 8x  2(14.30 
 3x)  35.10 2x  28.60  35.10 x  3.25 y  14.30 
 3x y  14.30 
 3(3.25) y  4.55 $3.25; $4.55 53. x  y  3 xy3 ? ? 003 323 0  3 no 5  3 yes xy3 xy3 ? ? 
4  2  3 
2  4  3 
2  3 no 2  3 no (3, 2)  480  h  A(x)  x  2x2  44b. A(x ) 320 300 280 260 240 220 200 180 160 140  O  2 4 6 8 10 12 14 1618 x  44c. The surface area approaches infinity. 45. If the degree of the denominator is larger than that of the numerator, then y  0 will be a horizontal asymptote. To make the graph intersect the x-axis, the simplest numerator to use is x. x  Thus, a sample answer is f(x)   x2  1 . 46a. A vertical asymptote at r  0 and a horizontal asymptote at F  0. 46b. The force of repulsion increases without bound as the charges are moved closer and closer together. The force of repulsion approaches 0 as the charges are moved farther and farther apart. 47a. 47b.  1  55. [f 
 g](x)  f(g(x))  f(2 
 x2)  8(2 
 x2)  16 
 8x2 [g 
 f ](x)  g(f(x))  g(8x)  2 
 (8x)2  2 
 64x2  a2 
 9  a
3  x m  2.9 5.9  2.99 5.99  3 —  3.01 6.01  3.1 6.1  The slope approaches 6.  Chapter 3  1   54. 15y 
 x  1 → y  1 5 x  15  92  56. Let x  the width of each card and y  the height of each card. The rectangle has a base of 4x or 5y. The rectangle has a height of x  y. A  bh 180  4x(x  y) 4x  5y 180  4xx  5  y  5  180  5  y  5  4x  V  khg2 288  k(40)(1.5)2 3.2  k V  3.2hg2 12b. V  3.2hg2 V  3.2(75)(2)2 V  960 50 960  48,000 m3 12a.  4x  36x2  4(5)  25  x2 y4 5x Perimeter  2(4x)  2(x  y) P  2(4 5)  2(5  4) P  58 in. The correct choice is B.  Pages 194–196 13.  14.  Direct, Inverse, and Joint Variation  3-8  Pages 193–194  15.  Check for Understanding  16.  1a. inverse 1b. neither 1c. direct 2. Sample answer: Suppose y varies directly as xn. Then y1  kx1n and y2  kx2n y1  kx1n y1  y2 y1  y2  kx1    kx 2    x1  x2  17.  y  
54  k(9)2 2 
3  7.  8.  y  k  19.  y  w 2  1  12  20.  k(3)(10)  y  4 2  21.  22.  24.  93  y  
48  1  6  k(20)  y  5 2 y  0.168  kxz  y  w  k(2)(
3)   y 
4 y  14  kz2  y  x 2  
6  3 3  k(9)2  y  6 3  
2  k  y  
2 7  y  w  y  x 3  k(6)2  10. A varies jointly as 
 and w; 1 11. y varies inversely as x; 
3.  1  3 2 y  1 2 (
4) (
3)  y  z 2  kb2  y2  1  3 2 y  1 2x z  kx  23. a  c  1 x4; 7.  1  
x 1  
0.16   y  z2  
3  4 2k  0.4(4)(20)  0.4  k  k   6 2 0.3  k  0.4xz  3  2 2  y  kx3z2 
9  k(
3)3(2)2  4 y  5 2  
3x2 2  
3(6)2  kxz  9. y varies directly as  y  0.484 y  1.21  y  0.5xz3 y  (0.5)(
8)(
3)3 y  108  y  w 2  r1  
xy  k 
1.21 (0.44)  k 0.484  k 
xy  0.484 or y  0.484  y  
24  y  kxz3 16  k(4)(2)3 0.5  k  1 2  16  k 18.  Simplify.  y  kx2  r  164  1 2  3. The line does not go through the origin, therefore its equation is not of the form y  kxn. 4a. Sample answer: The amount of money earned varies directly with the number of hours worked. 4b. Sample answer: The distance traveled by a car varies inversely as the amount of gas in the car. 4c. Sample answer: The volume of a cylinder varies jointly as its height and the radius of its base. 5. xy  k xy  12 4(3)  k 15y  12 6.  y  0.2x y  0.2(6) y  1.2 xy  
50 x(
40)  
50 x  1.25 y  15xz y  15(0.4)(3) y  18 x2y  36 32y  36 y4 r  16t2  4  k2  Division property of equality.  12  k  Exercises  y  kx 0.3  k(1.5) 0.2  k xy  k 25(
2)  k 
50  k y  kxz 36  k(1.2)(2) 15  k x2y  k (2)2(9)  k 36  k r  kt2  0.3x  0.3(14)  2xz  2(4)(
7)  
2z2  
2(
4)2 4  15b2  a  c 15b2  45  12  96  1 0  15  k x2y  k 2 (4) (2)  k 32  k  8  b yx2  32 8x2  32 x  2  Chapter 3  25. C varies directly as d; . 26. y varies directly as x;  4  10
11  G; 6.67  6.67  4 . 3   (6.67  30. y varies inversely as the square root of x; 2. 31. A varies jointly as h and the quantity b1  b2; 0.5. 32. y varies directly as x and inversely as the square 1 of z; 3. 33. y varies directly as x2 and inversely as the cube of z; 7. 34. y varies jointly as the product of the cube of x and z and inversely as the square of w. 35a. Joint variation; to reduce torque one must either reduce the distance or reduce the mass on the end of the fulcrum. Thus, torque varies directly as the mass and the distance from the fulcrum. Since there is more than one quantity in direct variation with the torque on the seesaw, the variation is joint. 35b. T1  km1d1 and T2  km2d2 T1  T 2 km1d1  km2d2 Substitution property m1d1  m2d2 of equality 35c. m1d1  m2d2 75(3.3)  (125)d2 1.98  d2; 1.98 meters 36a. tr  k 36b. tr  k tr  36,000 45(800)  k t(1000)  36,000 36,000  k t  36 minutes 37. If y varies directly as x then there is a nonzero constant k such that y  kx. Solving for x, we find 1 1 x  ky k is a nonzero constant, so x varies directly as y.  10
11)  kL  1.68  1.07 1.68  k 2   10
2   (0.001)2  10
8  k  16  62 576  k 39. a is doubled  10
8(3)   R (0.003)2  R  1.78  10
3   k  y  42.  8 6 4 2  O  
8
6
4
2 2 4 6 8x 
2 
4 
6 
8  f(x)  (x 
 3)3  6 y  (x 
 3)3  6 x  (y 
 3)3  6 x 
 6  (y 
 3)3 3 
 x
6y
3 3 y  
 x
63 3 
1 f (x)  
 x 
 6  3; f
1(x) is a function. 
1 0 1 3 
1 
3 
1 
3 1 3 44.  0 1 2 
2 
4 0 2 
2 
4 0 A
(
1, 2), B
(
 
2), C
(1, 
4), D
(3, 0) 43.  45. 4x 
 2y  7 
12x  6y  
21  a2  b2  c2 (6)2  (25)2  c2 6.5  c  N m2   10
11  kg2  (5.98 1024)(1.99 1030)  (1.50 1011)2   R r2  41.  k  k  1022)   3.53 1022 N 40d. 3.53 1022  (1.99 1020)x 178  x; about 178 times greater  38a. I  d2 I  d2  1024)(7.36  m1 m2  40c. F  G d 2  5  29. y varies inversely as the square of x; 4.  38b.  (5.98  1020  G  (3.84 108)2  1.99  27. y varies jointly as x and the square of z; 3. 28. V varies directly as the cube of r;  m1 m2  F  G d 2  40b.  1 . 4  7  →  y  2x 
 2  →  y  2x 
 2  7  consistent and dependent y  576  I  d 2  46.  576   I (6.5)2 I  13.6 lux  x  kb2  a  c 3  O     1 k 2b  2  a   3 12c  a  1 kb2 4  1 c3 8  18.6 
 23.2   47. m   2000 
 1995  48.  kb2  a  2c 3 m1 m2  40a. F  Gd 2  Chapter 3  94  
4.6  5 144  42  or 
0.92  y 
 18.6  
0.92(x 
 2000) y  
0.92x  1858.60  9 or 122 1 12 is divisible by 3, 4, 6, and 12.  a2 
 b  4 no x-axis a  
2b a  
2(
b) a  2b no (
a)  
2b a  2b no (b)  
2(a) b  
2a no (
b)  
2(
a) b  
2a no; none  The correct choice is D. →  17. x  
2y x-axis y-axis yx y  
x 1  Chapter 3 Study Guide and Assessment Page 197 1. 4. 7. 10.  2. continuous 5. maximum 8. monotonic  a2  b 1   a2   (
b)  x-axis  Understanding and Using the Vocabulary  even decreasing inverse Joint  1  →  18. x2  y  1  a2  
b no  3. point 6. rational 9. slant  1  (
a)2  b  y-axis  1  a2  b yes 1  yx  (b)2  (a) 1  b2  a  Pages 198–200  11. f(
x)  
2(
x) f(
x)  2x 12. f(
x)  (
x)2  2 f(
x)  x2  2 13. f(
x)  (
x)2 
 (
x)  3 f(
x)  x2  x  3 
f(x)  
(x2 
 x  3) 
f(x)  
x2  x 
 3 no 14. f(
x)  (
x)3 
 6(
x)  1 f(
x)  
x3  6x  1 
f(x)  
(x3 
 6x  1) 
f(x)  
x3  6x 
 1 no 15. xy  4 → x-axis y-axis yx y  
x  16. x  y2  4 x-axis y-axis yx y  
x  y  
x  Skills and Concepts  →  
f(x)  
(
2x) 
f(x)  2x yes 
f(x)  
(x2  2) 
f(x)  
x2 
 2 no  19. 20. 21. 22.  (
b)2    no  1  (
a) 1 
ay  b2  no; y-axis y The graph of g(x) is a translation of the graph of f(x) up 5 units. The graph of g(x) is a translation of the graph of f(x) left 2 units. x graph of f(x) expanded The graph of O g(x) is the vertically by a factor of 6. O x The graph of g(x) is the graph of f(x) expanded 4  horizontallyy by a factor of 3 and translated down y 4 units. 23.  ab  4 a(
b)  4 
ab  4 no (
a)b  4 
ab  4 no (b)(a)  4 ab  4 yes (
b)(
a)  4 ab  4 yes y  x and y  
x a  b2  4 a  (
b)2  4 a  b2  4 yes (
a)  b2  4 
a  b2  4 no (b)  (a)2  4 a2  b  4 no (
b)  (
a)2  4  24.  x  O  x  O  25.  27. Case 1 4x  5 
(4x  5) 
4x 
 5  95  26.  7 7 7  Case 2 4x  5 4x  5 4x  7 7 2  Chapter 3  f(x)
4x  3x 
121 x  
3 x f(x) {xx  
3 or x 0.5} 
2 
7 28. Case 1 
1 
4 x 
 3  2  11 0 
1 
(x 
 3)  2  11 1
x  5 2 11  f (x )
x  6 x f (
6 x) 1 {x
6  x  12} f (x )  32.  x 0.5 f
1(x) x f
1(x) 
7 
2 Case 2 
4 
1 x 
 3  2  11 
1 0 x 
 3  2  11 2 x1  12  f(x)  (x  1)2 
 4 x f(x) 
3 0 
2 
3 
1 
4 0 
3 1 0 f 1(x )  f (x )  29.  f (x )  O  x  1  x 
2 
1 0 1 2  x 5.5 5.25 5 4.75 4.5  f(x) 5.5 5.25 5 4.75 4.5  f
1(x) f
1(x) 
2 
1 0 1 2  f(x)  (x 
 2)3 
 8 y  (x 
 2)3 
 8 x  (y 
 2)3 
 8 x  8  (y 
 2)3 3 
 x8y
2 3 y  
 x82 3 
1 f (x)  
 x  8  2; yes 34. f(x)  3(x  7)4 y  3(x  7)4 x  3(y  7)4  33.  f (x ) 6 4 2  30.  f (x )  
6
4 O 2 4 6 
4 1 
6 f (x )  x  3 x  4 3  x   (y  7)4 y7  y  
7  f
1(x)  
7   2  x 
3 
2 
1  f(x) 2.3 2 1  f
1(x) x f
1(x) 2.3 
3 2 
2 1 
1  1  
1  
1  0  —  7  1  2  1  2  7  1 2 3  5 4 3.7  5 4 3.7  1 2 3  f(x)  x  3  
2  x  3 4 x ; 3  4  no  35. Yes; the function is defined when x  2; the function approaches 6 as x approaches 2 from both sides; and y  6 when x  2. 36. No; the function is undefined when x  
1. 37. Yes; the function is defined when x  1; the function approaches 2 as x approaches 1 from both sides; and y  2 when x  1. 38. an: negative, n: odd y → 
 as x → , y →  as x → 
. 39. an: positive, n: odd y →  as x → , y → 
 as x → 
.  1  
2  40.  1  y  x2  1 x 
1000 
100 
10 1 10 100 1000  y f (x )  O f 1(x )  x  O  f(x)  
4x  5  31.  x 0 
3 
4 
3 0  f
1(x) f
1(x) 
3 
2 
1 0 1  x  y 1.000001 1.0001 1.01 2 1.01 1.0001 1.000001  y → 1 as x → , y → 1 as x → 
. 41. an: positive, n: odd y →  as x → , y → 
 as x → 
.  Chapter 3  96  x  52. x  1  42.   y x
1 x  x  y  x 1  
  x x 1  y  1 ; 1 
 x  x2  1  53. x  
2  [
5, 5] scl:1 by [
20, 10] scl:5 decreasing for x  
2 and x 1; increasing for 
2  x  1  y1   y x2 x2 1   x2  x2  y  x 2  43.    x2  x2 1 1  x 2  y  1 2    x x2  no horizontal asymptotes (x 
 3)2  54. x  
3,  44. 45. 46.  47.   y x2 
 9 x2 
 6x  9   y x2 
 9  [
6, 6] scl:1 by [
5, 20] scl:5 decreasing for x  
3 and 0  x  3; increasing for 
3  x  0 and x 3 abs. max.: (
2, 1) rel. max.: (0, 4), rel. min.: (2, 0) f(2.9)  0.029 f(3)  0 f(3.1)  0.031 min. f(
0.1)  6.996 f(0)  7 f(0.1)  7.004 pt of inflection  9 x2 6x    x2 
 x2  x2  y   x2 9   x2 
 x2  9 6 1 
 x  x 2  y   ;y1 9 1 
 x 2  x2 x2  2 x1 → x
 x2 2x 2x 1 56. y  kxz 5  k(
4)(
2) 0.625  k 55.  1  48. f(x)  x  1 2  49. f(x)  
x 50. The parent graph is translated 2 units left and expanded vertically by a factor of 3. The vertical asymptote is now x  
3. The horizontal asymptote, f(x)  0, is unchanged.  57.  f (x )  y 20   k  
x  k  
49   140  k  x  O  58.  51. The parent graph is translated 3 units right and then translated 2 units up. The vertical asymptote is now x  3 and the horizontal asymptote is f(x)  2.  O  97  y  140  
x  10   140  
x  kx2  y  z  320x2  7.2  4  k(0.3)2  y  40  320  k  y8  320(1)2  Applications and Problem Solving  59. x 
 6.5  0.2; Case 1 x 
 6.5  0.2 
(x 
 6.5)  0.2 
x  6.5  0.2 
x  
6.3 x  6.3  x  yes; y  x  2 y  0.625xz y  0.625(
6)(
3) y  11.25  
x  14 x  196  y  z  Page 201  f (x )  1  x  2  x  Case 2 x 
 6.5  0.2 x 
 6.5  0.2 x  6.7 6.3  x  6.7  Chapter 3  60a.  x 0x1 1x2 2x3 3x4 4x5 5x6 3.60 3.20 2.80 2.40 Cost 2.00 (dollars) 1.60 1.20 0.80 0.40 0  Page 201  C(x) 0.40 0.80 1.20 1.60 2.00 2.40  y  1b. Sample answer: y  x2  y  x  0123456789 Time (min)  O  1c. Sample answer: 
xy  1  y  C
1(x) 0x1 1x2 2x3 3x4 4x5 5x6  x 0.40 0.80 1.20 1.60 2.00 2.40  x O  1d. Sample answer: xy  1  9 8 7 6 Time 5 (min) 4 3 2 1 0 0 0.40 1.20 2.00 2.80 3.60 Cost (dollars)  y  x  O  1e. Sample answer: y  x3  60d. positive multiples of $0.40; positive real numbers 60e. C
1(x) gives the possible number of minutes spent using the scanner that cost x dollars. 61a.  x  O  60b. positive real numbers; positive multiples of $0.40 60c.  Open-Ended Assessment  1a. Sample answer: x  y2  y  h(t)  O  x  1 0.5  O  0.5  1  2. Sample answer: 
2(x 
 4)2  1  t  61b. 1.08 m  Chapter 3  98  3. Notice that 450 miles is the distance to Grandmother's house, not the round trip. This is a multiple-step problem. First calculate the number of gallons of gasoline used in each direction of the trip.  3a. Sample answer:  y  miles  miles per gallon 450  25  (0, 0)  x  O  (2, 1)  3b. abs. min.: (2, 
3); rel. max.: (0, 0); rel. min.: (
2, 
1)  Chapter 3 SAT & ACT Preparation  4.  y  SAT & ACT Practice  1. Always factor or simplify algebraic expressions when possible. Notice that the numerator in the problem is the difference of two squares, a2 
 b2. Factor it. y2 
 9  3y 
 9   18 gallons  On the trip to Grandmother's, the cost of gasoline is 18 gallons $1.25 per gallon or $22.50. On the trip back, the gasoline cost is 18 gallons $1.50 per gallon or $27.00. The difference between the costs is $4.50. A faster way to find the cost difference is to reason that each gallon cost $0.25 more on the trip back. So the total amount more that was paid was 18 gallons $0.25 or $4.50. The correct choice is B.  (2, 3)  Page 203   gallons  O  x  ( y  3)(y 
 3)    3y 
 9  Factor the denominator. Both the numerator and denominator contain the factor (y 
 3). Simplify the fraction. (y  3)(y 
 3)  3y 
 9  ( y  3)(y 
 3)  The portion of the graph of f(x) which is shown crosses the x-axis 3 times. The correct choice is D. 5. Notice that the denominators are all powers of ten. Carefully convert each fraction to a decimal. Then add the three decimals.  y3      3( y 
 3) 3  The correct answer is E. 2. You need to find the statement that is not true. Compare the given information with each answer choice. Choice A looks like x  y  z, except for the numbers. Multiply both sides of the equation x  y  z by 2. 2(x  y)  2z or 2x  2y  2z So choice A is true. For choice B, start with x  y and subtract y from each side. x
yy
y0 So choice B is true. For choice C, start with x  y and subtract z from each side. x
zy
z So choice C is true. For choice D, substitute y for x and x  y for z. z x  2 xy  900  10  90  9     100  1000  90  0.9  0.009  90.909  The correct choice is C. You could also use your calculator on this problem. 6. Combine like terms. (10x4 
 x2  2x 
 8) 
 (3x4  3x3  2x  9)  (10x4 
 3x4)  (
3x3)  (
x2)  (2x 
 2x)  (
8 
 9)  7x4 
 3x3 
 x2  0 
 17 The correct choice is A. 7. One method of solving this problem is to "plug in" a number in place of n. Choose a number that when divided by 8, has a remainder of 5. For example, choose 21. 21  2(8)  5 Then use this value for n in the answer choices. Find the expression that has a remainder of 7.  2y  y  2  2y So choice D is also true. For choice E, write each side of the equation in terms of y. z 
 y  (x  y) 
 x  y 2x  2y y 2y So choice E is not true. The correct choice is E.  n1  21  1  22  Choice A: 8  8  8  2R6 The remainder is 6. n2  21  2  23  Choice B: 8  8  8  2R7 You could also reason that since n divided by 8 has a remainder of 5, then (n  2) divided by 8 will have a remainder of (5  2) or 7. The correct choice is B.  99  Chapter 3  8.Simplify the expression inside the square root symbol. Factor 100 from each term. Then factor the trinomial. 
 100x2  60 0x  9 00  x3    
 100(x2  6x  9)  x3    10
 x2  6 x9  x3    (x  3 )(x   3) 10
  x3    10
 (x  3 )2  x3    10(x  3)  x3  10. There are two equations and two variables, so this is a system of equations. First simplify the equations. Start with the first equation. Divide both sides by 2. 4x  2y  24 2x  y  12 Now simplify the second equation. Multiply both sides by 2x. 7y  2x  7y  7(2x) 7y  14x Divide both sides by 7. y  2x You need to find the value of x. Substitute 2x for y in the first equation. 2x  y  12 2x  (2x)  12 4x  12 x3 The answer is 3.   10 The correct choice is B. 9. Since a  b  c, substitute a  b for c in a 
 c  5. So, a 
 (a  b)  5. Then 
b  5 or b  
5. Substitute 
5 for b in b 
 c  3. So, 
5 
 c  3. Then 
c  8 or c  
8. The correct choice is B.  Chapter 3  7  100  Chapter 4 Polynomial and Rational Functions 4-1  11. 2; x2 
 14x  49  0 (x 
 7)(x 
 7)  0 x
70 x7  Polynomial Functions  Pages 209–210  Check for Understanding  x
70 x7  f (x)  1. A zero is the value of the variable for which a polynomial function in one variable equals zero. A root is a solution of a polynomial equation in one variable. When a polynomial function is the related function to the polynomial equation, the zeros of the function are the same as the roots of the equation. 2. The ordered pair (x, 0) represents the points on the x-axis. Therefore, the x-intercept of a graph of a function represents the point where f(x)  0. 3. A complex number is any number in the form a  bi, where a and b are real numbers and i is the imaginary unit. In a pure imaginary number, a  0 and b 0. Examples: 
2i, 3i; Nonexamples: 5, 1  i 4. y  60 40  (0, 49)  f (x ) 
 x 2  14x  49  20 (7, 0)  O  4  8  x  12  12. 3; a3  2a2 
 8a  0 a(a2  2a 
 8)  0 a(a  4)(a 
 2)  0 a0 a40 a  
4 30 20  a
20 a2  f (a) f (a) 
 a 2  2a 2  8a  10 (4, 0) (0, 0) (2, 0)  O  x  
4  
2 O 
10  2  4a  t4 
 1  0 (t2 
 1)(t2  1)  0 (t 
 1)(t  1)(t2  1)  0 t
10 t10 t1 t  
1  13. 4; 5. 3; 1 6. 5; 8 7. no; f(x)  x3 
 5x2 
 3x 
 18 f(5)  (5)3 
 5(5)2 
 3(5) 
 18 f(5)  125 
 125 
 15 
 18 f(5)  
33 8. yes; f(x)  x3 
 5x2 
 3x 
 18 f(6)  (6)3 
 5(6)2 
 3(6) 
 18 f(6)  216 
 180 
 18 
 18 f(6)  0 9. (x 
 (
5))(x 
 7)  0 (x  5)(x 
 7)  0 x2 
 2x 
 35  0; even; 2 10. (x 
 6)(x 
 2i)(x 
 (
2i))(x 
 i) (x 
 (
i))  0 (x 
 6)(x 
 2i)(x  2i)(x 
 i)(x  i)  0 (x 
 6)(x2 
 4i2)(x2 
 i2 )  0 (x 
 6)(x2  4)(x2  1)  0 (x3 
 6x2  4x 
 24)(x2  1)  0 x5 
 6x4  5x3 
 30x2  4x 
 24  0; odd; 1  t2  1  0 t2  
1 t  i  f (t )  (1, 0)  (1, 0)  O  t  f (t ) 
 t 4 1  14a. x2  r2  62 r2  36 
 x2 14b. V(x)  (36 
 x2)(2x) V(x)  (36 
 x2)(2x) V(x)  72x 
 2x3 14c. V(x)  72x 
 2x3 V(4)  72(4) 
 2(4)3 V(4)  502.65 units3  Pages 210–212  V(x)  Bh V(x)  (36 
 x2)(2x)  Exercises  15. 4; 5 16. 7; 3 17. 3; 5 18. 5; 
25 19. 6; 
1 20. 2; 1 21. Yes; the coefficients are complex numbers and the exponents of the variable are nonnegative integers.  101  Chapter 4  37. (x 
 (
1))(x 
 1)(x 
 4)(x 
 (
4)(x 
 5)  0 (x  1)(x 
 1)(x 
 4)(x  4)(x 
 5)  0 (x2 
 1)(x2 
 16)(x 
 5)  0 (x4 
 17x2  16)(x 
 5)  0 x5 
 5x4 
 17x3  85x2  16x 
 80  0; odd; 5 38. (x 
 (
1))(x 
 1)(x 
 3)(x 
 (
3))  0 (x  1)(x 
 1)(x 
 3)(x  3)  0 (x2 
 1)(x2 
 9)  0 x4 
 10x2  9  0 39. 1; x  8  0 f (x) x  
8 (0, 8)  1 a  22. No;   a1, which is a negative exponent. 
  12a 23. yes; f(a)  f(0)  (0)4 
 13(0)2  12(0) f(0)  0 24. no; f(a)  a4 
 13a2  12a f(
1)  (
1)4 
 13(
1)2  12(
1) f(
1)  1 
 13 
 12 f(
1)  
24 25. yes; f(a)  a4 
 13a2  12a f(1)  (1)4 
 13(1)2  12(1) f(1)  1 
 13  12 f(1)  0 26. yes; f(a)  a4 
 13a2  12a f(
4)  (
4)4 
 13(
4)2  12(
4) f(
4)  256 
 208 
 48 f(
4)  0 27. no; f(a)  a4 
 13a2  12a f(
3)  (
3)4 
 13(
3)2  12(
3) f(
3)  81 
 117 
 36 f(
3)  
72 28. yes; f(a)  a4 
 13a2  12a f(3)  (3)4 
 13(3)2  12(3) f(3)  81 
 117  36 f(3)  0 29. f(b)  b4 
 3b2 
 2b  4 f(
2)  (
2)4 
 3(
2)2 
 2(
2)  4 f(
2)  16 
 12  4  4 f(
2)  12; no 30. f(x)  x4 
 4x3 
 x2  4x f(
1)  (
1)4 
 4(
1)3 
 (
1)2  4(
1) f(
1)  1  4 
 1 
 4 f(
1)  0; yes 31a. 3; 1 31b. 2; 2 31c. 4; 2 32. (x 
 (
 2))(x 
 3)  0 (x  2)(x 
 3)  0 x2 
 x 
 6  0; even; 2 33. (x 
 (
1))(x 
 1)(x 
 5)  0 (x  1)(x 
 1)(x 
 5)  0 (x2 
 1)(x 
 5)  0 x3 
 5x2 
 x  5  0; odd; 3 34. (x 
 (
2))(x 
 (
0.5))(x 
 4)  0 (x  2)(x  0.5)(x 
 4)  0 (x2  2.5x  1)(x 
 4)  0 3 2 x 
 4x  2.5x2 
 10x  x 
 4  0 x3 
 1.5x2 
 9x 
 4  0 2x3 
 3x2 
 18x 
 8  0; odd; 3 35. (x 
 (
3))(x 
 (
2i))(x 
 2i)  0 (x  3)(x  2i)(x 
 2i)  0 (x  3)(x2 
 4i2)  0 (x  3)(x2  4)  0 3 x  3x2  4x  12  0; odd; 1 36. (x 
 (
5i))(x 
 (
i))(x 
 i)(x 
 5i)  0 (x  5i)(x  i)(x 
 i)(x 
 5i)  0 (x  5i)(x 
 5i)(x 
 i)(x  i)  0 (x2  25)(x2  1)  0 x4  26x2  25  0; even; 0 a4  Chapter 4  13a2  f (x) 
 x  8  O x  (8, 0)  40. 2;  a2 
 81  0 (a 
 9)(a  9)  0 a
90 a9  (9, 0) f (a) 
20 
10 O 
20  a90 a  
9  (9, 0) 10  20a  2 
40 f (a) 
 a  81  
60 
80  (0, 81)  41. 2; b2  36  0 b2  
36 b  6i  80 60 f (b) 
 b 2  36 40  f (b)  (0, 36)  20 
4 
2 O  42. 3;  t3  2t2 
 4t 
 8  0 t2(t  2) 
 4(t  2)  0 (t  2)(t2 
 4)  0 (t  2)(t  2)(t 
 2)  0 t20 t20 t  
2 t  
2 4 (2, 0)  
4  
2 O 
4 (0, 8) 
12  102  f (t) (2, 0) 2  4 t  f (t) 
 t 3  2t 2  4t  8  2  4b  t
20 t2  n3 
 9n  0 n(n2 
 9)  0 n(n 
 3)(n  3)  0 n0 n
30 n3  43. 3;  20  47. 4;  n30 n  
3  4m4  17m2  4  0 (4m2  1)(m2  4)  0 4m  1  0 m2  4  0 
1 m  

4  m   4 m  0.5i  f (n)  m  2i  f (m)  10 (3, 0) 
4  (0, 0)  
2 O 
10 
20  44. 3;  (3, 0)  (0, 4)  4n  2  f (m) 
 4m 4  17m 2  4  f (n ) 
 n 3 9n  m  O  6c3 
 3c2 
 45c  0 c(6c2 
 3c 
 45)  0 c(c 
 3)(6c  15)  0 c0 c
30 c3  48. 6c  15  0 15  c  
6  (u  1)(u2 
 1)  0 (u  1)(u  1)(u 
 1)  0 u10 u10 u  
1 u  
1  u
10 u1  f (u)  c  
2.5 100  f (c)  (1, 0)  O  (2.5, 0) 50 (0, 0) 
4  
2 O 
50 
100  45. 4;  (0, 1)  (3, 0)  2  u  f (u) 
 (u  1)(u 2  1)  4c  f (c) 
 6c 3  3c 2  45c  a 4  a2 
 2  0 (a2  2)(a2 
 1)  0 (a2  2)(a 
 1)(a  1)  0 a2  2  0 a
10 a2  
2 a1 a  
2 i  49a.  a10 a  
1  y  49b.  x  O  f (a)  (1, 0)  (1, 0)  49c.  y  49d.  y  x  O y  (1, 0)  f (a) 
 O a 4  a 2  2 (0, 2)  a  x4 
 10x2  9  0 (x2 
 9)(x2 
 1)  0 (x 
 3)(x  3)(x 
 1)(x  1)  0 x
30 x30 x
10 x3 x  
3 x1  46. 4;  20  x  O  49e.  y  O  x  49f. not possible  x10 x  
1  f (x) O  x  10 (0, 9) (3, 0)(1, 0) (1, 0) (3, 0) 
4 
2 O 
10 
20  2  4x  f (x) 
 x 4  10x 2  9  103  Chapter 4  50.  56.  (x 
 B)(x 
 C)  0 x2 
 Cx 
 Bx  BC  0 x2 
 (C  B)x  BC  0 
C 
 B  B from x2  Bx  C  0 BC  C B1 
C 
 1  1 
C  2 C  
2 Sample answer: 1; 
2    [
5, 5] sc11 by [
2, 8] sc11 50a. 4 50b. 2; 
1, 1 50c. There are 4 real roots. However, there is a double root at 
1 and a double root at 1. 51a. V(x)  99,000x3  55,000x2  65,000x 51b. r  0.15 x1r x  1  0.15 x  1.15 V(x)  99,000x3  55,000x2  65,000x V(1.15)  99,000(1.15)3  55,000(1.15)2  65,000(1.15) V(1.15)  150,566.625  72,737.5  74,750 V(1.15)  298,054.125; about $298,054.13 52. 1 and 3 are two of its zeros. 53a.  1  d(t)  2at2 1  x
2   57. y   x(x  2)(x 
 2) 1  58a. Let x  the width. The length  2(52 
 2x) or 26 
 x. A(x)  x(26 
 x) 58b. A(x)  x(26 
 x) A(x)  26x 
 x2  1  d(t)  2at2 1  d(30)  2(16.4)(30)2  d(60)  2(16.4)(60)2  d(30)  7380 ft  d(60)  29,520 ft  [
5, 30] sc15 by [
2, 200] sc120 x  13 26 
 x  26 
 13  13 13 yd by 13 yd 59. The graph of y  2x3  1 is the graph of y  2x3 shifted 1 unit up. 60. (
6, 9)  1  d(t)  2at2 1  d(120)  2(16.4)(120)2 d(120)  118,080 ft 53b. It quadruples; (2t)2  4t2. 54. Let x  the width of the sidewalk. The length of the pool would be 70 
 2x feet. The width of the pool would be 50 
 2x feet. A  
w 2400  (70 
 2x)(50 
 2x) 2400  3500 
 240x  4x2 0  4x2 
 240x  1100 0  x2 
 60x  275 0  (x 
 55)(x 
 5) x 
 55  0 x
50 x  55 x5 Use x  5 since 55 is an unreasonable solution. 5 ft 55. Let x  the number of pizzas. (160  16x)(16 
 0.40x)  4000 
6.4x2  192x  2560  4000 
6.4x2  192x 
 1440  0 x2 
 30x 
 225  0 (x 
 15)(x 
 15)  0 x 
 15  0 x 
 15  0 x  15 x  15 16 
 0.40x  16 
 0.40(15)  $10  Chapter 4  61. 
15 5  
15(3) 
 (
9)(5) 
9 3  
45  45 or 0; no 2 
1 3 
9 2 62. AB  3 4 5 7 
6 2(3)  (
1)(5) 2(
9)  (
1)(7)  3(3)  4(5) 3(
9)  4(7) 2(2)  (
1)(
6) 3(2)  4(
6) 1 
25 10 29 1 
18 63. x  4y  9 4y  
x  9   1  y  9  y  
4x  4  x  O x  4y  9  64. Parallel; the lines have the same slope.  104  65. [f 
 g](x)  f(g(x))         Completing the Square x2  4x 
 5  0 x2  4x  5 2 x  4x  4  5  4 (x  2)2  9 x  2  3 x23 x1 Quadratic Formula     1 f 2x  6 1 x  6 2 
 4 2 1 x2  6x  36 4 1 x2  6x  32 4  
4  [g 
 f ](x)  g( f(x))  g(x2 
 4) 1  x   2(x2 
 4)  6 1  
4  
 42 
 4 (1)(
5)  2(1) 
4  
36    2x2 
 2  6  x  2  1   2x2  4  
4  6  x  2  66. The pictograph shows two more small car symbols in the row for 1999 than it does for 2000. These two small cars represent the 270 additional cars that were sold in 1999 compared to 2000. Since the two small cars represent 270 real cars, each  x  
2  3 x  
2  3 x1 See students' work. 5. x2  8x 
 20  0 x2  8x  20 x2  8x  16  20  16 (x  4)2  36 x  4  6 x46 x2 6. 2a2  11a 
 21  0  270  small car symbol must represent  or 135 real 2 cars. The correct choice is A.  4-2  Quadratic Equations  11  Pages 218–219  Check for Understanding  2 
 4(5)(7)   13  
(
13)  2(5)  3a. equals 0 3c. positive number 4. Graphing  or  21  11  21  121  21  11  121    a2  2a  1 6  2  16  a  1412  21869 11  17  a  4  4 11  17  11  a  4  4 3  7.  b2  17  a  4  
4  a  2  a  
7  
 4ac  
 4(1)(36)  0; 1 real 122  m  
12    
0  2(1)  12  m  
2 or 
6  13  
29   10  8. b2 
 4ac  (
6)2 
 4(1)(13)  
16; 2 imaginary  3b. negative number  t  
(
6)  

16   2(1) 6  4i   2  3  2i  x 2  4x  54 
6  x  4  
6 x  
10  a2  2a  2  f (x ) 
 f (x ) (5, 0)  x  
2 
 3 x  
5  a2  2a 
 2  0  1. Add 4 to each side of the equation to get t2 
 6t  4. Determine the value needed to make t2 
 6t a perfect square trinomial. Add this value (9) to each side. Take the square root of each side of the equation and solve the two resulting equations. t  3  
13  2. Quadratic Formula; Since the leading coefficient does not equal 1 and the discriminant equals 185 which is not a perfect square, the Quadratic Formula would be the best way to get an exact answer. Completing the square can also be used, but errors in arithmetic are more likely. A graph will give only approximate solutions. p  x  2  
3 x  
5  p2 
 6p  5  0 (p 
 5)(p 
 1)  0 p
50 p5 10. r2 
 4r  10  0 9.  (1, 0)  
4 
2 O 
4  2x  
8  r Factoring x2  4x 
 5  0 (x  5)(x 
 1)  0 x50 x  
5  r  p
10 p1  
(
4)  
 (
4)2  
 4(1 )(10)  2(1) 4  

24   2 4  2i
6   2  r r  2  i
6   x
10 x1  105  Chapter 4  11.  P  12I 
 0.02I 2 1600  12I 
 0.02I 2 0.02I 2 
 12I  1600  0 I 2 
 600I  80,000  0 (I 
 200)(I 
 400)  0 I 
 200  0 I 
 400  0 I  200 amps I  400 amps  17. t2 
 3t 
 7  0 t2 
 3t  7 9  t  3 2 
 2 3 t 
 2    9    p
    z 
 1  
5 z  
4  m m  21. b2 
 4ac  (
5)2 
 4(1)(9) or 
11; 2 imaginary s  361  4 3  s  19  p 
 2  
2  d d x x x 24.  1  d2 
 4d  
8 1  9  9    d2 
 4d  6 4  
 8  64  d 
 382  614 3  p  1  3  1  d  1  2  3  16.  3g2  p  1  d 
 8  
8 d  1  4  25.  
 12g  
4  g2  4  
 4g  
3 4  k  8  (g 
 2)2  3 g
2 g  
5  
97   2(2)  
5  
97  4   26. 
7  i
5  2
6   3 2
6  2  3  28. s  s s s  Chapter 4  
4  
112   2(3) 
4  4
7   6  
2  2
7  3 2k2  5k   9 2k2  5k 
 9  0 b2 
 4ac  52 
 4(2)(
9) or 97; 2 real k  g2 
 4g  4  
3  4  2  

140   2(4) 2  2i
35   8 1  i
35   4 3p2  4p   8 3p2  4p 
 8  0 b2 
 4ac  42 
 4(3)(
8) or 112; 2 real p  d 
 8  8 d 
 8  8  84  
0   2(36) 84 7  or  72 6  23. b2 
 4ac  (
2)2 
 4(4)(9) or 
140; 2 imaginary  1  3  5  

11   2(1) 5  i
11   2  22. b2 
 4ac  (
84)2 
 4(36)(49) or 0; 1 real  15. d2 
 4d  8  0 3   
7  
121  2(6) 
7  11  12 3 1 
2, 3  m  p  11 p  
8 14. x2 
 10x  21  0 x2 
 10x  
21 x2 
 10x  25  
21  25 (x 
 5)2  4 x 
 5  2 x
52 x 
 5  
2 x7 x3 3  1  19. b2 
 4ac  (6)2 
 4(4)(25)  
364 2 imaginary; the discriminant is negative. 20. b2 
 4ac  72 
 4(6)(
3) or 121; 2 real  9     
37  2  18. 
2  4  p2 
 3p  4  88  4 3 2 p 
 2 3 19    2 2  
37     2 3  Exercises  12. z2 
 2z 
 24  0 z2 
 2z  24 z2 
 2z  1  24  1 (z 
 1)2  25 z 
 1  5 z
15 z6 13. p2 
 3p 
 88  0 p2 
 3p  88  37   4  t  2  1 2  Pages 219–221  9  t2 
 3t  4  7  4  106  
b  
 b2 
 4ac   2a 5  
 (
5)2  
 4(3)(9)   2(3) 5  

83   6 5  i
83   6  27. 5  2i  29. x2 
 3x 
 28  0 (x 
 7)(x  4)  0 x
70 x7 30. 4w2  19w 
 5  0 (4w 
 1)(w 
 5)  0 4w 
 1  0 4w  1 1 w  4 31.  4r2 
 r  5 4r2 
 r 
 5  0 (4r 
 5)(r  1)  0 4r 
 5  0 4r  5 5 r  4  32. p  p p  b2 
 4ac  0 82 
 4(1)(c)  0 64 
 4c  0 
4c  
64 c 16 36a. A  bh A  12(16) A  192 1 (12 
 2x)(16 
 2x)  2(192) 35.  x40 x  
4  w50 w  
5  (12 
 2x)(16 
 2x)  96 36b. (12 
 2x)(16 
 2x)  96 192 
 56x  4x2  96 4x2 
 56x  96  0 x2 
 14x  24  0  r10 r  
1  f (x)  
b  
 b2 
 4ac   2a 
2  
 22 
 4 (1)(8)  2(1) 
2  

28   2 
2  2i
7   2  20 10  O  p p  
1  i
7   x  2 
 4( 2
6   
 (2
6  ) 1)(
2)  2(1) 2
6   
32   2  x  2
6   4
2   2  33. x   
20  A A A  
b  
 b2 
 4 ac  2a 
0.05  
 0.052  
 4(0.01)(
 18)   2(0.01) 
0.05  
0.722 5   0.02 5  
0.05  
0.722  or 0.02  A  40 40 years old 34c.  x  A  f (x ) 
 x 2  14x  24  1  37a. d(t)  v0t 
 2gt2  d (t)  1  d(t)  5t 
 2(32)t2  t  O  d(t)  5t 
 16t2 d (t) 
 5t  16t 2  37b. 0 and about 0.3 37c. The x-intercepts indicate when the woman is at the same height as the beginning of the jump. 37d. d(t)  5t 
 16t2 
50  5t 
 16t2 37e. 
50  5t 
 16t2 16t2 
 5t 
 50  0  
0.05 
 
0.722 5   0.02  A  
45  P  150 125 100 P 
 2 75 0.01A  0.05A  107 50 25  O  20  36c. roots: 2, 12 12 
 2x  12 
 2(2) or 8 16 
 2x  16 
 2(2) or 12 8 ft by 12 ft 12 
 2x  12 
 2(12) or 
12 16 
 2x  16 
 2(12) or 
8 ∅    2
2  x  
6 34a. P  0.01A2  0.05A  107 P  0.01(25)2  0.05(25)  107 P  6.25  1.25  107 P  114.5 mm Hg 34b. P  0.01A2  0.05A  107 125  0.01A2  0.05A  107 0  0.01A2  0.05A 
 18 A  10  
10  t t t t  25 50 75 100 A  
b 
 b2 
 4 ac  2a 2 5  
 (
5)  
 4(1 6)(
5 0)  2(16) 5  
3225   32 5  
3225   32  t  1.93 s about 1.93 s  As a woman gets older, the normal systolic pressure increases.  107  t  5 
 
3225   32  t  
1.62  Chapter 4  ax2  bx  c  0  38.  b  c  x2  ax  a  0 b  2  2  x2  ax  2a  
a  2a b  c  b  2  2  2  2  2  b  2a  
 b2 
 4ac      2a b  x  
2a  x  
 b2 
 4 ac  2a  b2 
 4 ac 
b  
  2a  18a2  3a 
 1  0 (3a  1)(6a 
 1)  0 3a  1  0 3a  
1  39. 2;  6a 
 1  0 6a  1  1  1  a  
3  a  6  f (a) f (a) 
 18a 2  3a  1  a  O 40.  x 
2 
1 0 1 2  y  y 0 
1 
2 
1 0  x  O y  |x |  2  f(x)  (x 
 9)2 y  (x 
 9)2 x  (y 
 9)2 
x y
9 y  
x  9 f1(x)  
x  9 42. 3x  4y  375 
2(5x  2y)  
2(345) 41.  3x  4y  375 3(45)  4y  375 y  60 43. m   →  3x  4y  375 
10x 
 4y  
690 
7x  
315 x  45  (45, 60)  619 – 595  2.8 – 2.4 24   m 0.4  m  60  60   619 – x  2.8 – 3.2  619 
 x  
24 x  $643  44. 3y  8x  12 3y  
8x  12 8 y  3x  4; 
3 45. x2  x 
 20  (x  5)(x 
 4) The correct choice is A. Chapter 4  Page 226  Check for Understanding  1. The Remainder Theorem states that if a polynomial P(x) is divided by x 
 r, the remainder is P(r). If a division problem has a remainder of 0, then the divisor is a factor of the dividend. This leads to the Factor Theorem which states that the binomial x 
 r is a factor if and only if P(r)  0. 2. (x3 
 4x2 
 7x  8) (x 
 5); x2  x 
 2; 
2 3. The degree of a polynomial is one more than the degree of its depressed polynomial. 4. Isabel; if f(
3)  0, then (x 
 (
3)) or (x  3) is a factor. 5. 2 1 
1 4 6. 
5 1 1 
17 15 2 2 
5 20 
15 1 1 6 1 
4 3 0 x  1, R6 x2 
 4x  3 7. f(x)  x2  2x 
 15 f(3)  (3)2  2(3) 
 15  9  6 
 15 or 0; yes 8. f(x)  x4  x2  2 f(3)  (3)4  (3)2  2  81  9  2 or 92; no 9. f(x)  x3 
 5x2 
 x  5 1 1 
5 
1 5 f(1)  (1)3 
 5(1)2 
 1  5 1 
4 
5  1 
 5 
 1  5 or 0 1 
4 
5 0 x 
 1 is a factor x2 
 4x 
 5  (x 
 5)(x  1) (x 
 5), (x  1), (x 
 1) 10. f(x)  x3 
 6x2  11x 
 6 f(1)  (1)3 
 6(1)2  11(1) 
 6  1 
 6  11 
 6 or 0 1 1 
6 11 
6 1 
5 6 1 
5 6 0 x 
 1 is a factor x2 
 5x  6  (x 
 2)(x 
 3) (x 
 1), (x 
 2), (x 
 3) 11. 
1 1 6k2 0 
7 k k  
4 
1 1 6 1 
1 
6 6k 12a. 12 12b. 12 12c. 11 12d. f(x)  x7  x9  x12 
 2x2  x12  x9  x7 
 2x2  x(x11  x8  x6 
 2x)  x2(x10  x7  x5 
 2) x, x2, x11  x8  x6 
 2x, or x10  x7  x5 
 2 13. h  r  4 V  r2h V  r2(r  4) 5  r2(r  4) 5  r3  4r2 0  r3  4r2 
 5 0  (r3  4r2 
 5) 1 1 4 0 
5 1 5 5 1 5 5 0 r
10 hr4 r  1 in. h  1  4 or 5 in.  x  2ba  
ac  4ba 2 
4ac  b  x  2ba   4a x  The Remainder and Factor Theorems  c  x2  ax  
a b  4-3  108  Pages 226–228 14. 
7 1  20 
7 13  91 
91 0  
9 27 3 
18 1 
6 9 x2 
 6x  9, R 
1  15. 3 1  
5 2 8 4 
4 
8 1 
1 
2 0 x2 
 x 
 2  (x 
 2)(x  1) (x 
 4), (x 
 2), (x  1) 2 1 
2 
4 8 2 0 
8 1 0 
4 0 x2 
 4  (x 
 2)(x  2) (x 
 2), (x 
 2), (x  2) 1 1 4 
1 
4 1 5 4 1 5 4 0 x2  5x  4  (x  1)(x  4) (x 
 1), (x  1), (x  4) 
1 1 3 3 1 
1 
2 
1 1 2 1 0 x2  2x  1  (x  1)(x  1) (x  1), (x  1), (x  1) 2 1 0 
9 0 24 0 
16 2 4 
10 
20 8 16 1 2 
5 
10 4 8 0 x5  2x4 
 5x3 
 10x2  4x  8 2 1 2 
5 
10 4 8 2 8 6 
8 
8 1 4 3 
4 
4 0 x4  4x3  3x2 
 4x 
 4 2 1 4 3 
4 
4 2 12 30 
52 1 6 15 
26 
56 2 times 
1 1 2 
1 
2 
1 
1 2 1 1 
2 0 x2  x 
 2  (x  2)(x 
 1) 1 time; 
2, 1 f(x)  2x3 
 x2  x  k f(1)  2(1)3 
 (1)2  1  k 02
11k 
2  k f(x)  x3 
 kx2  2x 
 4 f(2)  (2)3 
 k(2)2  2(2) 
 4 0  8 
 4k  4 
 4 0  
4k  8 2k f(x)  x3  18x2  kx  4 f(
2)  (
2)3  18(
2)2  k(
2)  4 0  
8  72 
 2k  4 0  
2k  68 34  k f(x)  x3  4x2 
 kx  1 f(
1)  (
1)3  4(
1)2 
 k(
1)  1 0  
1  4  k  1 0k4 
4  k  31. 4 1  Exercises 
28 27 
1  1 x  13 0 
1 16. 2 1 1 0 2 6 12 24 1 3 6 12 23 x3  3x2  6x  12, R23 0 
8 0 16 17. 
2 1 
2 4 8 
16 1 
2 
4 8 0 x3 
 2x2 
 4x  8 5 
4 
2 18. 
1 3 
2 
3 5 
10 14 3 
5 10 
14 12 3x3 
 5x2  10x 
 14, R12 19. 1 2 0 
2 
3 2 2 0 2 2 0 
3 2x2  2x, R 
3 20. f(x)  x2 
 2 21. f(x)  x5  32 f(1)  (1)2 
 2 f(
2)  (
2)5  32  1 
 2 or 
1; no  
32  32 or 0; yes 22. f(x)  x4 
 6x2  8 f(
2 )  (
2 )4 
 6(
2  )2  8  4 
 12  8 or 0; yes 23. f(x)  x3 
 x  6 f(2)  (2)3 
 2  6  8 
 2  6 or 12; no 24. f(x)  4x3  4x2  2x  3 f(1)  4(1)3  4(1)2  2(1)  3  4  4  2  3 or 13; no 25. f(x)  2x3 
 3x2  x f(1)  2(1)3 
 3(1)2  1  2 
 3  1 or 0; yes 26a-d. r 1 3 
2 
8 1 1 4 2 
6 
1 1 2 
4 
4 2 1 5 8 8 
2 1 1 
4 0 d  32.  33.  34.  35.  36.  37.  38.  27. (
6 )4 
 36  36 
 36 or 0 28. 
1 1 7 
1 
7 
1 
6 7 1 6 
7 0 x2  6x 
 7  (x 
 1)(x  7) (x 
 1)(x  1)(x  7) 29. 2 1 1 
4 
4 2 6 4 1 3 2 0 x2  3x  2  (x  1)(x  2) (x 
 2), (x  1), (x  2) 30. 1 1 
1 
49 49 1 0 
49 1 0 
49 0 x2 
 49  (x 
 7)(x  7) (x 
 1), (x 
 7), (x  7)  39.  40.  109  Chapter 4  1  44b.  41. d(t)  v0t  2at2 1 25  4t  2(0.4)t2 0  0.2t2  4t 
 25 5 0.2 4 
25 1 25 0.2 5  0 t
50 t5s 42. 1 1 1 
7 a b 1 2 
5 
5  a 1 2 
5 
5  a 
5  a  b  V (x )  20O 15O 10O  V (x ) 
 2x 3  38x 2  180x  50 4 8 12 x O 3 2 44c. V(x)  2x 
 38x  180x 224  2x3 
 38x2  180x 44d. 224  2x3 
 38x2  180x 0  2x3 
 38x2  180x  224 0  x3 
 19x2  90x  112 2 1 
19 90 112 2 
34 
112 1 
17 56 0 2 in. 45. P(3  4i)  0 and P(3 
 4i)  0 implies that these are both roots of ax2  bx  c. Since this polynomial is of degree 2 it has only these two roots. x  3  4i x 
 3  4i (x 
 3)2  
16 2 x 
 6x  9 16 x2 
 6x  25  0 a  1, b  
6, c  25 46. r2  5r 
 8  0 r2  5r  8 25 25 2 r  5r  4  8  4  
2 1  2 
5 
5  a 
5  a  b 
2 0 10 
10 
 2a 1 0 
5 5  a 
15 
 a  b 
5  a  b  0 
5  a  b  0 
15 
 a  b  0 
5  a  10  0 
20  2b  0 a50 2b  20 a  
5 b  10 43a. V(x)  (3 
 x)(4 
 x)(5 
 x) V(x)  (12 
 7x  x2)(5 
 x) V(x)  
x3  12x2 
 47x  60 43b. V (x ) V (x ) 
 x 3  12x 2  47x  60  O  25O  x  2  r  52  43c. V  
 w h V  3 4 5 or 60 3 3 V  (60) 5 5  5  57   4  
57   r  2    2 5  
57   r  
 2  2   36 V(x)  
x3  12x2 
 47x  60 36  
x3  12x2 
 47x  60 43d. 36  
x3  12x2 
 47x  60 0  
x3  12x2 
 47x  24  47a. f(x)  x4 
 4x3 
 x2  4x f(2)  (2)4 
 4(2)3 
 (2)2  4(2) f(2)  16 
 32 
 4  8 or 
12; no 47b. f(0)  (0)4 
 4(0)3 
 (0)2  4(0) f(0)  0 
 0 
 0  0 or 0; yes 47c. f(
2)  (
2)4 
 4(
2)3 
 (
2)2  4(
2) f(
2)  16  32 
 4 
 8 or 36; no 47d. f(4)  (4)4 
 4(4)3 
 (4)2  4(4) f(4)  256 
 256 
 16  16 or 0; yes 48. f(x)  x5 
 32  [
3, 10] sc11 by [
200, 100] sc125 about 0.60 ft 1  44a. 
  2(20 
 2x) or 10 
 x w  18 
 2x h x V(x)  (10 
 x)(18 
 2x)(x) V(x)  (180 
 38x  2x2)(x) V(x)  2x3 
 38x2  180x  Chapter 4  (0, 
32); point of inflection [
4, 4] sc11 by [
50, 10] sc110 49. wider than parent graph and moved 1 unit left  110  50. Let x  number of 100 foot units of Pipe A and y  number of 100 foot units of Pipe B. 4x  6y  48 2x  2y 
 18 y 2x  2y  18 2x  y 
 16 (0, 8) 2x  y  16 (3, 6) 4x  6y 
 48 x0 y0  4-4 Page 232  O (0, 0) y 
 0 (8, 0) x P(x, y)  34x  40y P(0, 0)  34(0)  40(0) or 0 P(0, 8)  34(0)  40(8) or 320 P(3, 6)  34(3)  40(6) or 342 P(7, 2)  34(7)  40(2) or 318 P(8, 0)  34(8)  40(0) or 272 3 
 100 foot units of A, or 300 ft of A 6 
 100 foot units of B, or 600 ft of B 51. 4x  2y  3z  6 4x  2y  3z  6 2x  7y  3z → 2x  7y 
 3z  0 
3x 
 9y  13  
2z 
3x 
 9y  2z  
13 4x  2y  3z  6 2x  7y 
 3z  0 6x  9y 6 2(2x  7y 
 3z)  2(0) 3(
3x 
 9y  2z)  3(
13) ↓ 4x  14y 
 6z  0 
9x 
 27y  6z  
39 
5x 
 13y  
39 5(6x  9y)  5(6) 6(
5x 
 13y)  6(
39) ↓ 30x  45y  30 
30x 
 78y  
234 
33y  
204 68 y  11 6x  9y  6  6x  9  68  11  Page 233  6 546  6x  
1 1 91  x  
11  Check for Understanding  1. possible values of p: 1, 2, 3, 6 possible values of q: 1 possible rational roots: 1, 2, 3, 6 2. If the leading coefficient is 1, then q must equal 1. p p Therefore, q becomes 1 or p, and p is defined as a factor of an. 3. Sample answer: f(x)  x3 
 x2  x 
 3; f(
x)  (
x)3 
 (
x)2  (
x) 
 3 f(
x)  
x3 
 x2 
 x 
 3; 0 3 or 1 possible positive zeros and no possible negative zeros 4. Sample answer: You can factor the polynomial, graph the function, complete the square, or use the Quadratic Formula if it is a second-degree function, or use the Factor Theorem and the Rational Root Theorem. I would factor the polynomial if it can be factored easily. If not and it is a second-degree function, I would use the Quadratic Formula. Otherwise, I would graph the function on a graphing utility and use the Rational Root Theorem to find the exact zeros.  4x  2y  3z  6  91 4(
11)  Graphing Calculator Exploration  1. 3; 1, 
1, 
2 2. 2; 
1, 2 3. (1) 1 positive; f(
x)  (
x)4  4(
x)3  3(
x)2 
 4(
x) 
 4 f(
x)  x4 
 4x3  3x2  4x 
 4; 3 or 1 (2) 1 positive; f(
x)  (
x)3 
 3(
x) 
 2 f(
x)  
x3  3x 
 2; 2 or 0 4. In the first function, there are 2 negative zeros, but according to Descartes' Rule of Signs, there should be 3 or 1 negative zeros. This is because the 
2 is a double zero. In the second function, there is one negative zero, but according to Descartes' Rule of Signs, there should be 2 or 0 zeros. This is because 
1 is a double root. 5. One number represents two zeros of the function.  (7, 2)  x
0  The Rational Root Theorem   211  3z  6 68  294  3z  11 98  z  11  
9111 , 6181, 9118   5.  
7 
 2 2  6  52. M2, 2 or (
4.5, 4) 
2 
 2 6 
 3  N2, 2 or (
2, 1.5) 4 
 1.5   slope of  MN  
4.5 
 (
2) or 
1 2 
 (
3)   slope of  RI   
7 
 (
2) or 
1  p 1, q  2  r 1 
1 2 
2  1 1 1 1 1  
4 
3 
5 
2 
6  1 
2 6 
3 13  2 0 
4 
4 
24  rational root1  RI. Since the slopes are the same, M N L 53. a b a b a b ac  bc ac bc a
c b
c I. true II. true III. false The correct choice is D.  111  Chapter 4  6. p: 1, 3 q: 1, 2 P : q  Pages 234–235 1  10.  3  1, 3, 2, 2 r 1 
1  2 2 2  3 5 1  
8 
3 
9  3 0 12  1  2  2  4  
6  0  
2  2  2  
6  
1.5  3 
3  2 2  9 
3  19 1  60 0  1  p  q:  Exercises  1, 2, 3, 6 r 1 2  1 1 1  11.  p : q  1, 2, 3, 6, 9, 18  1  r 
1 
2  7. 2 or 0; f(
x)  8(
x)3 
 6(
x)2 
 23(
x)  6 f(
x)  
8x3 
 6x2  23x  6; 1 1  3  1  3  1  3  1, 2, 3, 6, 2, 2, 4, 4, 8, 8 r 1 2  
6 2 10  8 8 8  
23 
21 
3  8x2  10x 
 3  0 (4x 
 1)(2x  3)  0 4x 
 1  0 4x  1 1 x  4 1 1 
12, 4,  6 
15 0  12.  2x  3  0 2x  
3 3 1 x  
2 or 
12  7 8 6 4 2  7 15 1 
5 
3  18 14 0  
5 
4  9 5  
7 
2  5 1  
2 0  1, 2  r 2  2  1 1 1 1 1  1 4 9  1 1  2 0  x3 
 4x2  5x 
 2 
4 
2  1 1  x2 
 2x  1  0 (x 
 1)(x 
 1)  0 x  1, x  1 rational roots: 1, 2  
15 0 
16 0 0  13.  p : q  1, 2, 4, 5, 10, 20 r 1 
1 2  
5, 
3, 1 9. r2  152 
 x2  
5 
4 
6 
3  1 1 1 1  
4 
8 2 
10  20 12 18 0  x2 
 3x 
 10  0 (x 
 5)(x  2)  0 x  5, x  
2 rational roots: 
2, 2, 5 14. p: 1, 3 q: 1, 2  1  V  3r2h 1  1152  3(152 
 x2)(15  x) 3456  (152 
 x2)(15  x) 3456  3375  225x 
 15x2 
 x3 3 2 x  15x 
 225x  81  0 Possible rational roots: 1, 9, 81 f(x)  x3  15x2 
 225x  81  0 f(1)  
128 f(
1)  320 f(9)  0 f(
9)  2592 f(81)  611,712 f(
81)  
414,720 x represents 9 cm.  Chapter 4  p : q  r 1  1, 3, 5, 15 r 1 
1 
3 
5  
2 
3 
4  1 1 1  x2 
 4x  9  0 does not factor rational root: 
2  8. 1; f(
x)  (
x)3  7(
x)2  7(
x) 
 15 f(
x)  
x3  7x2 
 7x 
 15; 2 or 0 p : q  
6 
8 0  x2  4x  3  0 (x  3)(x  1)  0 x  
3, x  
1 rational roots: 
3, 
1, 2  rational roots: 
3, 2, 1  p : q  
5 
2 3  2 3 4  p : q  1  r  2  
1  0  
6  3  1  2  2  0  0  
6  0  2x3 
 6  0 x3  3 3 x  
3  1 rational root: 2  112  3  1, 3, 2, 2  15. p: 1 q: 1, 2, 3, 6 p : q  1  1  20. 2 or 0 positive f(
x)  
10x3 
 17x2  7x  2 1 negative  1  1, 2, 3, 6 r  6  35  
1  
7  
1  1  2  6  38  18  2  0  
17  
7  2  10  
22  4  0  6  38  18  2  6  36  6  0  rational zeros: 
2, 5, 2  1 1  1  
3  21. 2 or 0 positive f(
x)  x4 
 2x3 
 9x2  2x  8 2 or 0 negative  6x2  36x  6  0 x2  6x  1  0 does not factor 1 1 rational roots: 
3, 2  r 1  16. 4; 3 or 1; f(
x)  (
x)4 
 2(
x)3  7(
x)  4(
x) 
 15 f(
x)  x4  2x3 
 7x 
 4x 
 15; 1 negative 1 positive 17. f(
x)  
x3  7x 
 6 0 or 2 negative r 1 
1  1 1 1  0 1 
1  
7 
6 
6  1 1  3 6  
10 8  1 1  
9 
6  2 3  
2 
8  8 0  x3  3x2 
 6x 
 8 r 
1  1 1  3 2  
6 
8  
8 0  x2  2x 
 8  0 (x  4)(x 
 2)  0 x  
4, x  2 rational zeros: 
4, 
1, 1, 2 22. 2 or 0 positive f(
x)  x4 
 5x2  4 2 or 0 negative  
6 
12 0  x2 
 x 
 6  0 (x 
 3)(x  2)  0 x  3, x  
2 rational zeros: 
2, 
1, 3 18. 1 positive f(
x)  
x3 
 2x2  8 1 negative f(x)  x3 
 2x2 
 8x 0  x(x2 
 2x 
 8)  x(x 
 4)(x  2) x  0, x  4, x  
2 rational zeros: 
2, 0, 4 19. 1 positive f(
x)  
x3  3x2  10x 
 24 2 or 0 negative r 3  10  1 
2  10x2 
 22x  4  0 5x2 
 11x  2  0 (5x 
 1)(x 
 2)  0 1 x  5, x  2  6x3  38x2  18x  2 r  r  r 1  1 1  
5 
4  0 1  0 
4  4 0  x3  x2 
 4x 
 4 r 
1  1 1  1 0  
4 
4  x2 
 4  0 (x 
 2)(x  2)  0 x  2, x  
2 rational zeros: 
2, 
1, 1, 2 23a. f(x)  (x 
 2)(x  2)(x  1)2 0  (x 
 2)(x  2)(x  1)2 x
20 x20 x2 x  
2 
24 14  
4 0  (x  1)2  0 x10 x  
1  23b. f(x)  (x 
 2)(x  2)(x  1)2 f(x)  (x2 
 4)(x2  2x  1) f(x)  x4  2x3 
 3x2 
 8x 
 4 23c. 1 positive f(
x)  x4 
 2x3 
 3x2  8x 
 4 3 or 1 negative 23d. There are 2 negative zeros, but according to Descartes' Rule of Signs, there should be 3 or 1. This is because 
1 is actually a zero twice. 24a. Let 
  the length. w

4 h  2
 
 1 V(
)  
 w h V(
)  
(
 
 4)(2
 
 1) V(
)  (
2 
 4
)(2
 
 1) V(
)  2
3 
 9
2  4
  x2  6x  8  0 (x  4)(x  2)  0 x  
4, x  
2 rational zeros: 
4, 
2, 3  113  Chapter 4  24b. V(
)  2
3 
 9
2  4
 2208  2
3 
 9
2  4
 24c. 2208  2
3 
 9
2  4
 0  2
3 
 9
2  4
 
 2208 r 12  2 2  
9 15  
2208 0  4 184  w

4 h  2
 
 1 w  12 
 4 or 8 h  2(12) 
 1 or 23 12 in. 8 in. 23 in. Sample answer: x4  x3  x2  x  3  0 Sample answer: x3 
 x2 
 2  0 Sample answer: x3 
 x  0 Let 
  the length. h

9 1 V(
)  3Bh  Page 235  1 1  V(
)  3
3 
 3
2 1  26b. V(
)  3
3 
 3
2  25  2  x  52  1  26c. 6300  3
3 
 3
2  5  0 630  
5  25  
5  25  
18,900 0  0 
10,000  b  39  
441   2(6) 39  21  b  1 2 39  21  b  1 2  39 
 21  b  1 2  or  3  b5  b  2  5. 
2 1  3 
2 
8 
2 
2 8 1 1 
4 0 x2  x 
 4 6. 4 1 
4 2 
6 4 0 8 1 0 2 2 2; no 6 7. 1 1 
2 
5 1 
1 
6 1 
1 
6 0 x2 
 x 
 6  (x 
 3)(x  2) (x 
 3)(x 
 1)(x  2)  1,000,000 0  x  100 ft 28. The graphs are reflections of each other over the x-axis. The zeros are the same. 29. 
7 1 
1 
56 
7 56 1 
8 0 x
8 30. b2 
 4ac  62 
 4(4)(25)  
364; 2 imaginary 31. (x 
 1)(x 
 (
1))(x 
 2)(x 
 (
2))  0 (x 
 1)(x  1)(x 
 2)(x  2)  0 (x2 
 1)(x2 
 4)  0 x4 
 5x2  4  0 32. y  4.3x 
 8424.3 y  4.3(2008) 
 8424.3 y  $210.10  Chapter 4  x  2  x  10 x  
15 2 2 4. b 
 4ac  (
39) 
 4(6)(45)  441; 2 real roots  h

9 h  30 
 9 or 21 base: 30 in. by 30 in., height: 21 in. 27. d  0.0000008x2(200 
 x) 0.8  0.0000008x2(200 
 x) 0  0.00016x2 
 0.0000008x3 
 0.8 0  8x3 
 1600x2 
 8,000,000 0  x3 
 200x2 
 1,000,000 
200 
100  
5 
 25  x  2  
  30  1 1  25  x  2  0  
3 
 9
2 
 18,900  r 100  625   4  x  2   2  1  0  3
3 
 3
2 
 6300 
9 21  25  x2  5x  4  150  4  1  6300  3
3 
 3
2  1 1  Mid-Chapter Quiz  1. (x 
 1)(x 
 (
1))(x 
 2i)(x 
 (
2i))  0 (x 
 1)(x  1)(x 
 2i)(x  2i)  0 (x2 
 1)(x2  4)  0 x4  3x2 
 4  0 3 2 2. 3; x 
 11x  30x  0 x(x2 
 11x  30)  0 x(x 
 6)(x 
 5)  0 x0 x
60 x
50 x6 x5 3. x2  5x  150  V(
)  3(
2)(
 
 9)  r 30  3
x   2  2(2x 
 3)  x(3 
 x) 4x 
 6  3x 
 x2 x2  x 
 6  0 (x  3)(x 
 2)  0 x30 x
20 x  
3 x2 The correct choice is A.  
  12  25a. 25b. 25c. 26a.  2x 
 3  x  33.  8.  p : q  1, 3 r 
3  1 1  6 3  x2  3x  1  0 does not factor rational root: 
3  114  10 1  3 0  4. Nikki; the sign changes between 
2 and 
1. 5. r 1 
4 
2 
2 1 
6 10 
1 1 
5 3 0 1 
4 
2 1 1 
3 
5 2 1 
2 
6 3 1 
1 
5 4 1 0 
2 5 1 1 3 4 and 5, 
1 and 0 6. r 1 
3 
2 4 
2 1 
5 8 
12 
1 1 
4 2 2 0 1 
3 
2 4 1 1 
2 
4 0← 2 1 
1 
4 
4 3 1 0 
2 
2 4 1 1 2 12 
2 and 
1, at 1, 3 and 4 2 
4 0 
3 7. r 0 2 
4 0 
3 1 2 
2 
2 
5 2 2 0 0 
3 3 2 2 6 15 approximate zero: 2.3 1 3 2 8. r 
2 1 1 0 
1 1 2 0 zeros: 
2, 
1 9. Sample answer: r 1 0 0 
8 2 1 1 1 1 
7 
5 2 1 2 4 0 2 upper bound: 2 f(
x)  x4  8x  2 r 1 0 0 8 2 0 1 0 0 8 2 lower bound: 0 10. Sample answer: r 1 0 1 0 
3 1 1 1 2 2 
1 2 1 2 5 10 17 upper bound: 2 f(
x)  x4  x2 
 3 r 1 0 1 0 
3 1 1 1 2 2 
1 2 1 2 5 10 17 lower bound: 
2 11a. Let x  amount of increase. V(x)  (25  x)(30  x)(5  x) V(x)  (750  55x  x2)(5  x) V(x)  x3  60x2  1025x  3750  9. 1 positive F(
x)  x4 
 4x3  3x2  4x 
 4 3 or 1 negative r 1  1 1  4 5  
4 4  3 8  
4 0  x3  5x2  8x  4  0 r 
1  1 1  5 4  8 4  4 0  x2  4x  4  0 (x  2)(x  2)  0 x  
2, x  
2 rational zeros: 
2, 
1, 1 10. Let r  radius. hr6 1 V  3r2h 1  27  3r2(r  6) 1  0  3r3  2r2 
 27 0  r3  6r2 
 81 r 3  1 1  6 9  
81 0  0 27  r3  hr6 h  3  6 or 9  r  3 cm, h  9 cm  Locating Zeros of a Polynomial Function  4-5  Pages 239–240  Check for Understanding  1. If the function is negative for one value and positive for another value, the function must cross the x-axis in at least one point between the two values. f (x ) (b, f (b ))  f (b )  O  a  b  x  f (a ) (a, f (a))  2. Use synthetic division to find the values of the polynomial function for consecutive integers. When the values of the function change from positive to negative or from negative to positive, there is a zero between the integers. 3. Use synthetic division to find the values of the polynomial function for consecutive integers. An integer that produces no sign change in the quotient and the remainder is an upper bound. To find a lower bound of a function, find an upper bound for the function of 
x. The lower bound is the negative of the upper bound for the function of 
x.  115  Chapter 4  11b. V  
 w h 1.5V  1.5(3750) V  25(30)(5) 1.5V  5625 V  3750 V(x)  x3  60x2  1025x  3750 5625  x3  60x2  1025x  3750 11c. 5625  x3  60x2  1025x  3750 0  x3  60x2  1025x 
 1875 r 1 60 1025 
1875 1 1 61 1086 
789 2 1 62 1149 423 x  1.7 25  x  25  1.7 30  x  30  1.7  26.7  31.7 5  x  5  1.7  6.7 about 26.7 cm by 31.7 cm by 6.7 cm  Pages 240–242  17. r 2 0 1 
3 3 
3 2 
6 19 
60 183 
2 2 
4 9 
21 45 
1 2 
2 3 
6 9 0 2 0 1 
3 3 1 2 2 3 0 3 2 2 4 9 16 35 no real zeros 18. r 6 24 
54 
3 
6 6 
12 18 
111 
5 6 
6 
24 117 yes; f(
6)  
111, f(
5)  117 19–25. Use the TABLE feature of a graphing calculator. 19. 
0.7, 0.7 20. 
2.6, 
0.4 21. 
2.5 22. 
0.4, 3.4 23. 
1, 1 24. 
1.3, 0.9, 7.4 25. 
1.24 26. Sample answers: r 3 
2 5 
1 1 3 1 6 5 upper bound: 1 f(
x)  
3x3 
 2x2 
 5x 
 1 r 
3 
2 
5 
1 0 
3 
2 
5 
1 lower bound: 0 27. Sample answers: r 1 
1 
1 1 1 0 
1 2 1 1 1 upper bound: 2 f(
x)  x2  x 
 1 r 1 1 
1 1 1 2 1 lower bound: 
1 28. Sample answers: r 1 
6 2 6 
13 1 1 
5 
3 3 
10 2 1 
4 
6 
6 
25 3 1 
3 
7 
15 
58 4 1 
2 
6 
18 
85 5 1 
1 
3 
9 
58 6 1 0 2 18 95 upper bound: 6 f(
x)  x4  6x3  2x2 
 6x 
 13 r 1 6 2 
6 
13 1 1 7 9 3 
10 2 1 8 18 30 47 lower bound: 
2  Exercises  12. r 1 0 0 
2 
1 1 
1 1 
3 0 1 0 0 
2 1 1 1 1 
1 2 1 2 4 6 3 1 3 9 25 1 and 2 13. r 2 
5 1 
1 2 
7 8 0 2 
5 1 1 2 
3 
2 2 2 
1 
1 3 2 1 4 0 and 1, 2 and 3 1 
2 0 1 
2 14. r 
2 1 
4 8 
15 28 
1 1 
3 3 
2 0 0 1 
2 0 1 
2 1 1 
1 
1 0 
1 2 1 0 0 1 0 at 
1, at 2 15. r 1 0 
8 0 10 
3 1 
3 1 
3 19 
2 1 
2 
4 8 
6 
1 1 
1 
7 7 3 0 1 0 
8 0 10 1 1 1 
7 
7 3 2 1 2 
4 
8 
6 3 1 3 1 3 19 
3 and 
2, 
2 and 
1, 1 and 2, 2 and 3 1 0 
3 1 16. r 
2 1 
2 1 
1 
1 1 
1 
2 3 0 1 0 
3 1 1 1 1 
2 
1 2 1 2 1 3 
2 and 
1, 0 and 1, 1 and 2  Chapter 4  116  32f. 
1.4, 3.4 (Use TABLE feature of a graphing calculator.) 33a. 1890: P(0)  
0.78(0)4  133(0)3 
 7500(0)2  147,500(0)  1,440,000  1,440,000 1910: P(20)  
0.78(20)4  133(20)3 
 7500(20)2  147,500(20)  1,440,000  2,329,200 1930: P(40)  
0.78(40)4  133(40)3 
 7500(40)2  147,500(40)  1,440,000  1,855,200 1950: P(60)  
0.78(60)4  133(60)3 
 7500(60)2  147,500(60)  1,440,000  1,909,200 1970: P(80)  
0.78(80)4  133(80)3 
 7500(80)2  147,500(80)  1,440,000  1,387,200 The model is fairly close, although it is less accurate at for 1950 and 1970. 33b. 1980 
 1890  90 P(90)  
0.78(90)4  133(90)3 
 7500(90)2  147,500(90)  1,440,000 P(90)  
253,800 33c. The population becomes 0. 33d. No; there are still many people living in Manhattan. 34. Sample answer: f(x)  (x 
 
2 )(x  
2 )(x  1) f(x)  (x2 
 2)(x  1) f(x)  x3  x2 
 2x 
 2; 
2 , 
1  29. Sample answers: r 1 5 
3 
20 1 1 6 3 
17 2 1 7 11 2 upper bound: 2 f(
x)  
x3  5x2  3x 
 20 r 
1 5 3 
20 1 
1 4 7 
13 2 
1 3 9 
2 3 
1 2 9 7 4 
1 1 7 8 5 
1 0 3 
5 6 
1 
1 
3 
38 lower bound: 
6 30. Sample answers: r 1 
3 
2 3 
5 1 1 
2 
4 
1 
6 2 1 
1 
4 
5 
15 3 1 0 
2 
3 
14 4 1 1 2 11 39 upper bound: 4 f(
x)  x4  3x3 
 2x2 
 3x 
 5 r 1 3 
2 
3 
5 1 1 4 2 
1 
6 2 1 5 8 13 21 lower bound: 
2 31. Sample answers: r 1 5 
3 20 0 
15 1 1 6 3 23 23 8 upper bound: 1 f(
x)  
x5  5x4  3x3  20x2 
 15 r 
1 5 3 20 0 
15 1 
1 4 7 27 27 8 2 
1 3 9 38 76 137 3 
1 2 9 47 141 408 4 
1 1 7 48 192 753 5 
1 0 3 35 175 860 6 
1 
1 
3 2 12 57 7 
1 
2 
11 
57 
399 
2808 lower bound: 
7 32a. 4 32b. 1, 5 32c. 3 or 1; f(
x)  x4  3x3 
 2x2 
 3x 
 5 1 negative real zero r 1 
3 
2 3 
5 32d. 5 1 2 8 43 210 4 1 1 2 11 39 3 1 0 
2 
3 
14 2 1 
1 
4 
5 
15 1 1 
2 
4 
1 
6 0 1 
3 
2 3 
5 
1 1 
4 2 1 
6 
2 1 
5 8 
13 21 
3 1 
6 16 
45 130 
2 and 
1, 3 and 4 32e. Sample answers: upper bound: 4 (See table in 32d.) f(
x)  x4  3x3 
 2x2 
 3x 
 5 r 1 3 
2 
3 
5 1 1 4 2 
1 
6 2 1 5 8 13 21 lower bound: 
2  f (x ) f (x ) 
 x 3  x 2  2x  2  x  O  35a. 37.44  60x3  60x2  60x 35b. f(x)  60x3  60x2  60x 
 37.44 35c.  f (x )  f (x ) 
 4O 60x 3  60x 2  60x  37.44 2O 
2 
1 O 
20  1  2x  
40 1  about 2 35d. 0.4 (Use TABLE feature of a graphing calculator.) 36. Sample answer: f(x)  x2 
 1  117  Chapter 4  37a.  f (x ) 100,000  f (x ) 
 0.125x 5  3.125x 4  4000  42.  7 9  7(6) 
 3(9) or 15 3 6 
3  8 
2  4  43. (x, y)  2, 2  80,000   (2.5, 1) 44. x 
 2y 
 4  0  60,000  1  40,000 2O,000  O  45. 4  8  x  Page 247  4.  
b  
 b2 
 4 ac  2a 4  
 (
4)2  
 4(4.9)(
1 750)   2(4.9) 4  
34,31 6   9.8 4  
34,31 6  4 
 
34,31 6   t   9.8 9.8  41.  2  x(x 
 2)  2  2(x 
 2)  2 x2 
 2x  2  2x 
 4  2 x2 
 4x  4  0 (x 
 2)(x 
 2)  0 x
20 x
20 x2 x2 If you solve the equation, you will get x  2. However, if x  2, the denominators will equal 0.  4.9t2 
 4t 
 1750  0  t  19.3 about 19.3 s  2    x x
2  2  x
2  2 2   x   x 
 2 (x 
 2)  2  x 
 2 (x 
 2)  1  t  Check for Understanding  1. Multiply by the LCD, 6(b 
 2). Then, solve the resulting equation. 2. If a possible solution causes a denominator to equal 0, it is not a solution of the equation. 3. Decomposing a fraction means to find two fractions whose sum or difference equals the original fraction.  
1750  4t 
 2(9.8)t2  t  Rational Equations and Partial Fractions  4-6  1  t  y  C D B y  x cannot be true. The correct choice is B.  d(t)  v0t 
 2gt2  40.  A  12 16 20 24 x  37b. f(0)  
0.125(0)5  3.125(0)4  4000 f(0)  4000 deer 37c. 1920 
 1905  15 f(15)  
0.125(15)5  3.125(15)4  4000 f(15)  67,281.25 about 67,281 deer 37d. in 1930 38a. 81.58  6x4  18x3  24x2  18x 38b. 81.58  6x4  18x3  24x2  18x 0  6x4  18x3  24x2  18x 
 81.58 about 1.1 (Use TABLE feature of a graphing calculator.) 38c. x  rate  1 1.1  rate  1 0.1  rate about 10% 39. 2 or 0; f(
x)  
2x3 
 5x2  28x  15 1 negative zero r 2 
5 
28 15 
3 2 
11 5 0 2x2 
 11x  5  0 (2x  1)(x 
 5)  0 x  0.5 x5 rational zeros: 
3, 0.5, 5  t  1  y  2x 
 2; 2; 
2  5.  5  b 
 b  4  b 
 5bb  (4)b b2 
 5  4b b2 
 4b 
 5  0 (b 
 5)(b  1)  0 b
50 b5  t  
18.5  y 6.  9  b5  b10 b  
1 3    b
3  9 3    b  5 (b  5)(b 
 3)   b 
 3 (b  5)(b 
 3)  O  Chapter 4  y 
 x 4x 1  9(b 
 3)  3(b  5) 9b 
 27  3b  15 6b 
 42  0 b7  x  118  7. t4   t   3 t4    t
4 t 3  (t)(t 
 4) t
4    
16  Pages 247–250    t2 
 4t  
16     t2 
 4t (t)(t 
 4)  12.    (t  4)(t 
 4)  3(t)  
16 t2 
 16  3t  
16 t2  3t  0 t(t  3)  0 t0 t30 t  
3 But t 0, so t  
3. 3p 
 1   p2 
 1 3p 
 1   p2 
 1  8.  1  1  16 ; x  exclude: 0  1  But m  1  16  1  6  16 false  1  16  4  1  5  2y  3(y  2)  
y2 5y  6  
y2 y2  5y  6  0 ( y  3)( y  2)  0 y30 y20 y  
3 y  
2 But y 
2, so y  
3.  10  (2n 
 5)(n  1)  (2n  5)(n 
 1) 2n2 
 3n  5  2n2  3n 
 5 
6n  
10 5  n  3  3  5  7  6 7  6  false  11b. 3  3a    2(a  1)  4(a 
 1)7a 
 3(a  1)5  6(a 
 1)3a 28a2 
 28a 
 15a 
 15  18a2 
 18a 10a2 
 25a 
 15  0 2a2 
 5a 
 3  0 (2a  1)(a 
 3)  0 2a  1  0 a
30  true   57.14  3 60  20  3x  3a    2a  2  7a 5    3(a  1) 
 4(a 
 1) (12)(a 
 1)(a  1)  3a   2(a  1) (12)(a 
 1)(a  1)  Solution: a  1, a  31 3 60  20  3x  7a 5  
  3a  3 4a 
 4 7a 5  
  3(a  1) 4(a 
 1)  17.  7  7 5    36 
 1  6 1 7 1  7  6 48 49    42 42  3  b  1  b  1  3(b  2) 2b  2  3b  6 
4  b  7  3  1     b2  b1  1 1 3     b  2  b  2 (b  2)(b  1)   b  1 (b  2)(b  1)  
2  6 true  6  1  b2  16.  7  5  2
1  2n  5  2    Test a  
1: 1   
1 
 1  6  11a.  2n 
 5     n
1  n1  10 2n 
 5 2n  5     n 
 1  n 
 1 (n 
 1)(n  1)   n  1 (n 
 1)(n  1)  6(a 
 1)  30  7(a 
 1) 6a 
 6  30  7a 
 7 31  a  Test a  36: 1  10   n2 
 1  15.    1 a 
 1  6 ; exclude: 1  Test a  2: 1   
y  4 true  Solution: x  0, x 10.  3    y   y2  2 
y 3     y  2  y (y)(y  2)   y  2 (y)(y  2)  
16 true  4  m  34  0 m  
34  0, so m  
34. 2  y2  14.  16  (
1)   Test x  
1: 5   (
1)  54  m 
 34    2m2  m2 
 34m  2m2 0  m2  34m 0  m(m  34) m0  5x  1  16 5x  15 x3  Test x  4: 5  4  t
20 t2  2  2  Test x  1: 5  1   (0)t  m 
 34 2  m12m2   2m 2m  B     p1  p
1     p1  p
1  9. 5  x  1  m  13.  3p 
 1  3p 
 1  A(p 
 1)  B(p  1) Let p  1. 3(1) 
 1  A(1 
 1)  B(1  1) 2  2B 1B Let p  
1. 3(
1) 
 1  A(
1 
 1)  B(
1  1) 
4  
2A 2A 3p 
 1   p2 
 1    0  0 t2 
 8t  12  0 (t 
 6)(t 
 2)  0 t
60 t6    (p  1)(p 
 1) A  Exercises  12   t 
 8 t 12   t 
 8 t t 12  t2 
 8t  1  a  
2   57.14  a3  60  20  57.14(3  x) 200  171.42  57.14x 0.50  x; 0.50 h  119  Chapter 4  1  a    1 1
a  a
1  18.  1(1 
 a)(a 
 1)    1  1
a    a  a
1  23.  (1 
 a)(a 
 1)  a 
 1 
 a2  a  a 
 1  a(1 
 a) 
a2  2a 
 1  
a2  2a 
 1 00 all reals except 1 19.    2q(2q 
 3) 
 2q(2q  3)  (2q  3)(2q 
 3) 4q2 
 6q 
 4q2 
 6q  4q2 
 9 0  4q2  12q 
 9  q     6m 
 9 1    3m 3m 6m 
 9  (12m) 3m  31m     24.  3m 
 3   4m  3m 
 3 (12m)   4m  
4 7 3      x
1 2
x x1 
4  (x 
 1)(2 
 x)(x x
1      25. 7 3     1)   2
x x1  x
2  5m 
 4   (m  2)(m 
 2)   A  B m2  m2  
4y   3y2 
 4y  1 
4y   3y2 
 4y  1     m
2  
4y  (3y 
 1)(y 
 1) A B    3y 
 1 y
1  2A 
2   2    
4y   3y2 
 4y  1  26.  9 
 9x   x2 
 9 9 
 9x   x2 
 9     3y 
 1  y
1  9 
 9x  (x  3)(x 
 3) A B    (x  3) (x 
 3)  Let x  3. 9 
 9(3)  A(3 
 3)  B(3  3) 
18  6B 
3  B Let x  
3. 9 
 9(
3)  A(
3 
 3)  B(
3  3) 36  
6A 
6  A 
6 
3 9 
 9x       2  1  
1 (2)(
1 
 48)   2 2 1  
145   4  x3 x 
9 
6 
3 ,  x3 x
3  27a. a(a 
 6)  Chapter 4  m
2  
43  
23A  22b. 
1, 2  (n  1)(n 
 2)  (n 
 2)(n  6)  4(n  1) n2 
 n 
 2  n2  4n 
 12  4n  4 2n2 
 n 
 18  0   x  
413  A13 
 1  B313 
 1  n6 4 (n  1)(n 
 2)  (n  1)(n 
 2) 1   n1 n
2  n  x
2  
4y  A(y 
 1)  B(3y 
 1) Let y  1. 
4(1)  A(1 
 1)  B(3(1) 
 1) 
4  2B 
2  B Let y  13.  n6 4    1 n1 n
2  22c.  5m 
 4   m2 
 4 5m 
 4   m2 
 4  m 
4  (x 
 1)(2 
 x)(x  1) 
4(2 
 x)(x  1)  7(x 
 1)(x  1)  3(x 
 1)(2 
 x) 
4(
x2  x  2)  7(x2 
 1)  3(
x2  3x 
 2) 4x2 
 4x 
 8  4x2  9x 
 13 5  13x 5   x 13  22a. (n  1)(n 
 2)  x  5m 
 4  A(m 
 2)  B(m  2) Let m  2. 5(2) 
 4  A(2 
 2)  B(2  2) 6  4B 1.5  B Let m  
2. 5(
2) 
 4  A(
2 
 2)  B(
2  2) 
14  
4A 3.5  A 3.5 1.5 5m 
 4       2  4  4(6m 
 9)  3(3m 
 3) 4  24m 
 36  9m 
 9 15m  23 m  2135 21.  x(x 
 2)   A  B  x 
 2x  
12  
144  
9) 4(4)(
  2 4 
12  
288   8 
12  12
2   8  
3  3
2  2  20.  x
6    x 
 6  A(x 
 2)  B(x) Let x  2. 2 
 6  A(2 
 2)  B(2) 
4  2B 
2  B Let x  0. 0 
 6  A(0 
 2)  B(0) 
6  
2A 3A x
6 
2     3   2  2q 2q  
   1 2q  3 2q 
 3 2q 2q  
  (2q  3)(2q 
 3)  1(2q  3)(2q 
 3) 2q  3 2q 
 3    x
6   x2 
 2x x
6   x2 
 2x  120  x
3  a
2  a  27b.    a
2  a  a
4   a
6  (a)(a 
 6)    a
4  a
6  30.  (a)(a 
 6)  (a 
 2)(a 
 6)  (a 
 4)(a) a2 
 8a  12  a2 
 4a 12  4a 3a 
1 
 2  
1  1
2  Test a  1:  1  
1 4
2  Test a  4:  4  Test a  7:  1  2 7
2  7 5  7  
1 
 4   
1 
 6 5 3  7 false 1
4   1
6 3  5 true 4
4   4
6  Test x  Test x  Test x    0 false 7
4   7
6  Test x    3 true  Solution: 0  a  3, 6  a 28.  2  w  3  29 ; w  Test w  1:  2  1  3  31. 29  
1  9   0 true 40 (
2)2 
 16 
2:  0 (
2 
 5)(
2  1) 
12   0 false 7 0 
 16   0 0:  02 
 4(0) 
 5 
16   0 true 
5 4.52 
 16 4.5:  0 (4.5 
 5)(4.5  1) 4.25   0 false 
2.75 2 6 
 16 0 6:  62 
 24 
 5 20   0 true 7  1  4a 1  4a     5  8a 5  8a    1 ; 2 1  2  7  4  29  1  Solution: w  0, w 29.  (x 
 3)(x 
 4)  (x 
 5)(x 
 6)2    Test a  
1:  4(
1)  8(
1)  1  2  7  1  2  1  29  10  Test x   Test x   Test x   5  
8 Test a  1:  true  9   0; exclude 5, 6  Test a  2:  (x 
 3)(x 
 4)  0 x
30 x
40 x3 x4 (0 
 3)(0 
 4) Test x  0: 2  0 Test x   exclude: 0  a  5 29 false 2 29 3  Test w  10:  (10) 10 32  10  5  2  5  4a  
29 true  1   0; exclude 5, 
1  Solution: x  
4, 
1  x  4, x  exclude: 0  2  3w  29 w9 2 3 Test w  
1:  
1  0  0 x2  16 x  4 (
5)2 
 16 Test x  
5:  0 (
5 
 5)(
5  1)  27c. 0, 6 27d. Test a  
1:  x2 
 16   x2 
 4x 
 5 x2 
 16  (x 
 5)(x  1) x2 
 16  Solution: 0  5  8(1)  1  2  7  8 1 5    4(2) 8(2) 7  16 7  a  4  1  2 1  2 1  2  1  4(1)    false  true  false  (0 
 5)(0 
 6) 12   0 true 
180 (3.5 
 3)(3.5 
 4) 3.5: 2  0 (3.5 
 5)(3.5 
 6) 
0.25   0 false 
9.375 (4.5 
 3)(4.5 
 4) 4.5: 2  0 (4.5 
 5)(4.5 
 6) 0.75   0 true 
1.125 (5.5 
 3)(5.5 
 4) 5.5: 2  0 (5.5 
 5)(5.5 
 6) 3.75   0 false 0.125 6.5 
 3)(6.5 
 4) 6.5:( 2  0 (6.5 
 5)(6.5 
 6) 8.75   0 false 0.375  Solution: x  3, 4  x  5  121  Chapter 4  1  2b  1 1  2b  1  32.  1    b1    1  b1    8 ; 15 8  15  1  exclude: 
2, 
1  15(b  1)  15(2b  1)  8(2b  1)(b  1) 45b  30  16b2  24b  8 0  16b2 
 21b 
 22 0  (16b  11)(b 
 2) 16b  11  0 b
20 11 b2 b  
16 1  1  8  15 8  15    Test b  
2:  2(
2)  1  
2  1 4  
3 1  1    Test b  
0.8:  2(
0.8)  1  (
0.8)  1 10  3 1  2(
0.6)  1  1   Test b  
0.6: (
0.6)  1 5 
2 1 1 8    Test b  0:  2(0)  1  0  1 15 8 2 1 5 true 1 1 8       Test b  3: 2(3)  1  3  1 15 11 8   false 28 15 1 11 Solution: 
1  b  
16, 
2  b  2  33.  7  y1  
6  2  14  0.30 false  62  0.30  8 0.30 true Solution: x  
5 or x 5  true  36a.  false  1  8  1  1   d  3 2 i  1  8  36b.  1  1   d  3 2 i  18(32di)  d1  312 (32di) i  4di  32  di 3di  32 2 di  103 cm x  1    37. Sample answer:  x
3  x2  5(x 
 3)  2x 5x 
 15  2x 3x  15 x  5 tons  7  39a.  7 true  1  1  1   2r  r  2 0 1  10  1  1  1   2r  r  2 0  110 (20r)  21r  1r  210 (20r)  7 false  2r  10  20  r r  30 2r  2(30) or 60; 60 ohms, 30 ohms 40. Let x  the number of quiz questions to be answered.  4x  x  105 2  20  5x2  52x 5x2 
 52x  20  0 (5x 
 2)(x 
 10)  0 5x 
 2  0 2 x  5  1  10  39b.  7  Solution: 
1  y  0 34. Let x  the number. 1  
0.4  Test x  6:  6
5  7 false  7  Test y  1:  0.30  38. Let x  capacity of larger truck. 5 x    2 x
3   Test y  
0.5:  
0.5  1 7  11 7  2  0.30 true  02  0
5  Test x  0: false 8  15 8  15 8  15 8  15  0.30  0.36  7  
7   0.30; exclude 5   Test x  
6:  
6 
 5  7  7(y  1) 1y1 0y Test y  
2:  0.30  x  2  0.30(x 
 5) x  2  0.30x 
 1.5 0.7x  
3.5 x  
5  7; exclude 
1  7  
2  1  x2  x
5 x2  x
5  35.  11  x  20  x   0.70  11  x  0.70(20  x) 11  x  14  0.70x 0.3x  3 x  10 questions 41. Let x  the speed of the wind.  x 
 10  0 x  10  1062  200  x  738    200 
 x  1062(200 
 x)  738(200  x) 212,400 
 1062x  147,600  738x 64,800  1800x 36  x; 36 mph  Chapter 4  122  42.  48. 
5  1 1 1      b c a 1 1    (a)(b)(c) b a       c(a)(b)(c) 1  43a.  1  x 1  x  a   2y  z 1 1  1  1  1  1  x 1  x  43b.    23 0  45  1  1  1  1  1    6 0  90   (360x)   1  x  1  60    (360x)  
3x  12 x  
4  ? 2(6)  3  3  6 5  3  2 false no 52. y2  121x2 → b2  121a2 52a. (
b)2  121a2 52b. b2  121(
a)2 b2  121a2 yes b2  121a2 yes 2 2 52c. (a)  121(b) 52d. (
a)2  121(
b)2 a2  121b2 no a2  121b2 no 53a. Let x  short answer questions and y  essay questions. y 20 x  y  20 2x  12y  60 16 x0 x  y 
 20 y0 12  15,000 m  20x  15,000 x  750 gallons 750 $1.20  $900 x $1.20  $900 
 $200 1 x  5833 gallons Let y  number of miles per gallon. 15,000 m ym    1 g 5833 g  8 2x  12y 
 60 (0, 5) 4 x
0  1  5833y  15,000 y  25.7; about 25.7 mpg d  T  s 2  2x  3  0 3 x  
2  2x  3   x  g  45.  0 25 25  51. y  x  1  90  360  6x  4x 360  10x 36  x 44. Let x  number of gallons of gasoline. 20 m  g  
30 25 
5  50. 3x 
 12  0 3x  12 3x  12 x4    23 0  45  1  0 
5 
5  1 no 49. 2; 12x2  8x 
 15  0 (6x 
 5)(2x  3)  0 6x 
 5  0 5 x  6  bc  ac  ab bc  ab 
 ac bc  a(b 
 c) bc  b
c  1  O  26  26    103   s5  s
5  32(s  5)(s 
 5)  26(3)(s 
 5)  26(3)(s  5) 32s2 
 800  78s 
 390  78s  390 2 32s 
 156s 
 800  0 8s2 
 39s 
 200  0 (8s  25)(s 
 8)  0 8s  25  0 s
80 25  12  x 16 20 (20, 0)  S(x, y)  5x  15y S(0, 0)  5(0)  15(0) or 0 S(0, 5)  5(0)  15(5) or 75 S(18, 2)  5(18)  15(2) or 120 S(20, 0)  5(20)  15(0) or 100 18 short answer and 2 essay for a score of 120 points  26 26   1032(3)(s  5)(s 
 5)   s  5  s 
 5 (3)(s  5)(s 
 5)  s  
8  4 8 (0, 0) y 
 0  (18, 2)  s8  8 mph 46.  3x 
 5y  5y  3x  5y   5y 
 5y  11 
 1  10  47.  r 
3 
2 
1 0 1 2  1 1 1 1 1 1 1  2 
1 0 1 2 3 4  
3 0 
3 
4 
3 0 5  
5 
5 1 
1 
5 
5 5  
3 and 
2, 
2 and 
1, 1 and 2  123  Chapter 4  53b. Let x  short answer questions and y  essay questions. y 20 x  y  20 x  y 
 20 2x  12y  120 16 x0 2x  12y 
 120 y0 12  4-7  Pages 254–255  (12, 8)  x O  4 8 (0, 0) y 
 0  12  16 20 (20, 0)  S(x, y)  5x  15y S(0, 0)  5(0)  15(0) or 0 S(0, 10)  5(0)  15(10) or 150 S(12, 8)  5(12)  15(8) or 180 S(20, 0)  5(20)  15(0) or 100 12 short answer and 8 essay for a score of 180 points 1 1 3 5 x 54. 1 1 
3 
5 1(3)  1(
3) 1(5)  1(
5)  x 1(3)  1(
3) 1(5)  1(
5) 0 0 x 0 0 55. y 
 y1  m(x 
 x1) y 
 1  2(x 
 (
3)) y 
 1  2x  6 2x 
 y  7  0 3000 
 5000   56a. m   20 
 60  56b. $2000; $50  m  50 y 
 3000  50(x 
 20) y  50x  2000 C(x)  50x  2000 56c.  C (x ) $4000 $3000  C (x ) 
 50x  Cost   2000  62 
 4  22  10 7  $2000  0  2 4 6 8 Televisions Produced  x  1  57. A of JKL  2(9)(7) or 31.5 1  A of small triangle  2(5)(3) or 7.5 A of shaded region  31.5 
 7.5 or 24 The answer is 24.  Chapter 4  7  
21 4 
   
7 0  1 
17   
17  8. 
a  4  
a   
37 
a 
 4  7 
 
a   
3 a 
 4  49 
 14
a  
3a
3 
42  
14
a 
3  1764  196(a 
 3) 9a
3 12  a Check: 
a  4  
a   
37 
12    
12 4 3 
7 
16   
9 7 437  $1000 0  Check for Understanding  1. To solve the equation, you need to get rid of the radical by squaring both sides of the equation. If the radical is not isolated first, a radical will remain in the equation. 2. The process of raising to a power sometimes creates a new equation with more solutions than the original equation. These extra or extraneous solutions do not solve the original equation. 3. When solving an equation with one radical, you isolate the radical on one side and then square each side. When there is more than one radical expression in an equation, you isolate one of the radicals and then square each side. Then you isolate the other radical and square each side. In both cases, once you have eliminated all radical signs, you solve for the variable. 4. 
1 
 4  2 t Check: 
1 
 4  2 t 3 1 
 4t  4 1 
 4
4  2 
4t  3 3  
1 32 t  
4 22 3 3 5. 
 x  4  12  3 Check: 
 x  4  12  3 3 3 
 x  4  
9 

733   4  12  3 3 x  4  
729 

729   12  3 x  
733 
9  12  3 33 6. 5  
x 
42 Check: 5  
x 
42 
x 
 4  
3  5  
13 4 
2 x
49 5  
9 2 x  13 53 2 no real solution 7. 
6x 
 4  
2x  0  1 6x 
 4  2x  10 4x  14 x  3.5 Check: 
6x 
 4  
2x  0  1  (0, 10)  8 x
0 4  Radical Equations and Inequalities  124  16. 4
 3m2 
 15  4 
 3m2 
 15  1 3m2 
 15  1 3m2  16  9. 
5x  48 5x  4  0 5x  4  64 5x  
4 5x  60 x  
0.8 x  12 Test x  
1: 
5(
1)  48 

1   8 meaningless Test x  0: 
5(0)   48 
4   8 true Test x  13: 
5(13)   48 
69   8 false Solution: 
0.8  x  12 10. 3  
4a 
   10 5 4a 
 5  0 
4a 
7 5 4a  5 4a 
 5  49 a  1.25 4a  54 a  13.5 Test a  0: 3  
4(0) 
  5  10 3  

5   10 meaningless Test a  2: 3  
4(2) 
  5  10 4  
3   10 true Test a  14: 3  
4(14) 
  5  10 3  
51   10 false Solution: 1.25  a  13.5 11a. v  
 v02   64h  16  m2  3 4  m  3
3  Check:  4  Check:  14. 
8n 

12 5  Check: 
x 85 
17 8 5 
25 5 55 3 Check: 
 y
74 3 
71 
74 3 
64 4 44 Check: 
8n 5 

12  
8n 
3 5  84 
 5 
 1  2  3  13. 
 y
74 y 
 7  64 y  71  8n 
 5  9 8n  14 7 n  4  4 4
1 44 3m2 
 15  4 4
 2   
 15  4 4 3
3
3 4  17.  18.  Exercises  12. 
x 85 
x  8  25 x  17  2   
 15  4 4 33
3  90  
 102  64h 90  
100   64h 11b. 90  
100   64h Check: 90  
100   64h 8100  100  64h 90  
100  5) 64(12 8000  64h 90  
8100  1125  h; 125 ft 90  90   Pages 255–257  4
 3m2 
 15  4  19.  7  5 
14 

12 3
12 22  20.   15. 
x  16  
x  4 x  16  x  8
x  16 0  8
x 0  
x 0x Check: 
x  16  
x  4  
0  1  
0 6 4 
16 4 44  125  4 4
1 44 
9u 
   
7u 4 20 
 9u 
 4  7u 
 20 2u  
16 u  
8 Check: 
9u 
   
7u 4 20 
 
9(
8)  
 4  
7(
8)  
 20 

76   

76  no real solution 3 
5 6u 
   2  
3 3 
5 6u 
   
5 6u 
 5  
125 6u  
120 u  
20 3 Check: 
5 6u 
   2  
3 3 
6(
20 ) 
 5  2  
3 3 

125   2  
3 
5  2  
3 
3  
3  
 4m2 
 3m  2 
 2m 
 5  0 
 4m2 
 3m  2  2m  5 4m2 
 3m  2  4m2  20m  25 
23  23m 
1  m Check: 
 4m2 
 3m  2 
 2m 
 5  0 2 
 3( 
 4(
1) 
1)  2 
 2(
1) 
 5  0 
9 2
50 3
30 00 
k  9 
 
k    
3  
k   9  
3   
k  k  9  3  2
3k k 6  2
3k  36  4(3k) 36  12k 3k Check: 
k  9 
 
k    
3  
3   9 
 
3   
3  
12  
 
3   
3  2
3  
 
3   
3  
3   
3   Chapter 4  26. 
2x  1  
2x   65 
2x  1  5 
 
2x   6 2x  1  25 
 10
2x   6  2x  6 
30  
10
2x 6  3  
2x  6 9  2x  6 3  2x  
a 1  2 
 1  
a 2  1 a  21 
 2
a 1  2  1  a  12 
2
a  2  
10 1 4(a  21)  100 a  21  25 a4 Check: 
a  2 
 1  
a 1 2  1 
4 1  2 
 1  
4 2  1 
25  
 1  
16  5
14 44 22. 
3x  4 
 
2x   
73 
3x  4  3  
2x   
7 3x  4  9  6
2x  
 7  2x 
 7 x  2  6
2x 
7  x2  4x  4  36(2x 
 7) x2 
 68x  256  0 (x 
 4)(x 
 64)  0 x
40 x 
 64  0 x4 x  64 Check: 
3x  4 
 
2x   
73 
3(4)   4 
 
2(4)  
 73 
16  
 
1 3 4
13 33 Check: 
3x  4 
 
2x   
73 
3(64)   4 
 
2(64) 
  73 
196  
 
121 3 14 
 11  3 33 3 23. 2
1 7b 
  
 4  0 3 2
1 7b 
   4 8(7b 
 1)  64 7b 
 1  8 7b  9 21.  3  2  Check:  3  3  2
1 7b 
  
 4  0 2 77 
 1 
 4  0 9  3  
40 2
8 4
40 00 4 24. 
3t 
20 Check: 4  
3t 2  4  
3t 
20 4  3t  16 t 25.  16  3  33 
 2  0 16  4  
20 
16 2
20 00  
x  2 
 7  
x 9 x  2 
 14
x  2  49  x  9 
14
x  2  
42 196(x  2)  1764 x29 x7 Check: 
x  2 
 7  
x 9 
7   2 
 7  
7  9 3
74 
4 4 no real solution  Chapter 4  22  6  5 3    
9 5 
4 235 55 27. 
3x  1  
x 0  11 
 1 3x  10  x  11 
 2
x  11  1 2x 
 2  
2
x  11 
x  1  
x  11 x2 
 2x  1  x  11 x2 
 3x 
 10  0 (x 
 5)(x  2)  0 x
50 x20 x5 x  
2 Check: 
3x  1  
x 0  11 
 1 
3(5)   10  
5 1  1 
 1 
25   
16 
1 54
1 5 3 Check: 
3x  1  
x 0  11 
 1 
3(
2)   10  

2 11 
1 
4   
9 
1 23
1 22 Solution: x  
2 28a. 
3t 
 1  t  6 4 
3t 
 1  6 
 t 4 3t 
 14  36 
 12t  t2 0  t2 
 15t  50 0  (t 
 5)(t 
 10) t
50 t 
 10  0 t5 t  10 Check: 
3t 
 1  t  6 4 
3(5) 
  14  5  6 
1 56 156 66 Check: 
3t 
 1  t  6 4 
3(10) 
  14  10  6 
16   10  6 4  10  6 14 6 10 28b. 5  9  3    65 
2x  1  
2x 22  1   b  7 Check:  x  126  29. 
2x  
75 2x 
 7  25 2x  32 x  16 2x 
 7  0 2x  7 7  x  2  Test x  0: 
2(0)  
 75 

7 5 meaningless Test x  4: 
2(4) 
  75 
8 
75  15 false Test x  17: 
2(17) 
  75 
27 5 true  34. 
m   2  
3m   4 m  2  3m  4 
2m  2 m  
1 m20 m  
2 3m  4  0 3m  
4 4  m  
3 Test m  
3: 

3 2    
3(
3)  4 

1   

5  meaningless Test m  
1.6: 

1.6   2  
3(
1. 6)  4 
0.4   

0.8  meaningless Test m  
1.2: 

1.2   2  
3(
1. 2)  4 
0.8   
0.4  false Test m  0: 
0  2  
3(0)    4 
2   
4  true Solution: m  
1 35. 
2c 
5 7 Test c  0: 
2(0) 
   57 2c 
 5 49 

5 7 2c 54 meaningless c 27 Test c  5: 
2(5) 
  57 
5 7 2c 
 5  0 false 2c  5 Test c  28: 
2(28) 
  5  7 c  2.5 
51 7 true Solution: c 27  Solution: x  16 30. 
b 46  b  4  36 b  32 b40 b  
4  Test b  
5: 

5 6 4 

1 6 meaningless Test b  0: 
0 46  
4 6 26 true Test b  33: 
33 6 4 
37 6 false Solution: 
4  b  32 Test a  0: 
0 
54  31. 
a 
54  

5 4 a 
 5  16 meaningless a  21 Test a  6: 
6 
54  a
50 
1 4 a5 14 true Test a  22: 
22 
4 5 
17 4 false Solution: 5  a  21 Test x  0: 
2(0) 
  56 32. 
2x 
56  

5 6 2x 
 5  36 meaningless 2x  41 Test x  5: 
2(5) 
  56 x  20.5 
5 6 2x 
 5  0 true 2x  5 Test x  22: 
2(22) 
  56 x  2.5 
39 6 false Solution: 2.5  x  20.5 4 4 Test y  0: 
 5(0) 
 92 33. 
 5y 
 9  2 4 

9 2 5y 
 9  16 meaningless 5y  25 4 Test y  2: 
 5(0) 
 92 y5 4 
1 2 5y 
 9  0 true 5y  9 4 Test y  6: 
 y  1.8 5(6) 
 92 4 
21 2 false Solution: 1.8  y  5  36a. t   37.  2s  g  3  2(7.2)  g  3  14.4  g  3  36b.  9  14.4  g 14.4  g  9g  14.4 g  1.6 m/s2 3  
x 
 5  
 x
3 3 x 
 5  
 (x 
 3 )2 3 2 x 
 5  
 x 
 6 x9 (x 
 5)3  x2 
 6x  9 x3 
 15x2  75x 
 125  x2 
 6x  9 x3 
 16x2  81x 
 134  0 Use a graphing calculator to find the zero.  [
2, 10] sc11 by [
10, 10] sc11 about 7.88 38a. s  
30fd  s  
30(0.6 )(25)  s  
450  s  21.2 mph 38b. s  
30fd  35  
30(0.6 )d  1225  18d 68.06  d; about 68 ft  127  Chapter 4  38c. No; it is not a linear function. 
  g  39a. T  2  
  g  39b. t  2  1  9.8  T  2  44.  T2  22 4 2    2  
  g   2  
  g 
 4g  r 1  r 
1  x  g x  g  Ta  Tb  x   g    ra 3  r b  p  r  141,433,433.8; about 141,433,434 mi 41. 
2x 
9
ab  
2x 
9ab  2x  9  0, so a  b  0 no real solution when a  b  0 tc  T  2  t  (
200)  2  t
c   2   t 
 200  
t  416  2 2 t 
 832t  173,056  4  t2      p2  2  t2 
 400t  40,000   4 t2  400t  50,000  4 t2  400t  50,000  2500  
 832t  173,056  123,056  1232t 99.88  t; about 99.88 psi a2  2a  1  43.  a  3  1     3   4a  2 ; exclude: 
 2  a2 3 a     2a  1 (6)(2a  1)   3  2(2a  1) (6)(2a  1)  6(a  2)  a(2)(2a  1)  3(3) 6a  12  4a2  2a  9 0  4a2 
 4a 
 3 0  (2a 
 3)(2a  1) 2a 
 3  0 2a  1  0 3  a  2  6 5  11 6  6 0  1056  w  p 
 1056 w  O  10Ow  
10O  4  3  y  
7x  7  1  a  
2  7  perpendicular slope: 4  3  2  7  y 
 5  4(x 
 2) 7  3  y  4x  2 51. A  r2  A  r2  1 2  2      1  4     (1)2  1  4   1    1 5  4  The correct choice is C. Chapter 4  p 10O  46b. x- and y-axes 46c. It increases. 46d. It is halved. 0 3 47. 4 
1 6 2 
2  4(0)  (
1)(2)  6(5) 4 0 2 4(0)  0(2)  2(5) 5 1 4(3)(
1)(
2)  6(1) 4(3)  0(
2)  2(1) 28 20  10 14 48. 
4(a  b  c)  
4(6) → 
4a 
 4b 
 4c  
24 2a 
 3b  4c  3 2a 
 3b  4c  3 
2a 
 7b  
21 
4(a  b  c)  
4(6) → 
4a 
 4b 
 4c  
24 4a 
 8b  4c  12 4a 
 8b  4c  12 
12b  
12 b1 
2a 
 7b  
21 abc6 
2a 
 7(1)  
21 71c6 a7 c  
2 (7, 1, 
2) 49. y  
3.54x  7125.4 y  
3.54(2010)  7125.4 y  10 students 50. 7y  4x 
 3  0  t 
 (
200)   502  2 t  200 2   2500  2    
6 0  
10O  108  2  108 
 2   1 1  v    42.  
5 6  5 11  46a. p  w  225 67,200,000 3    687 rb 50,625 3.03 1023    471,969 rb 3 50,625rb3  1.43 1029 rb3  2.83 1024    5 6  x2  5x  6  0 (x  3)(x  2)  0 x30 x20 x  
3 x  
2 
3, 
2, 
1, 1 45a. point discontinuity 45b. jump discontinuity 45c. infinite discontinuity  4
  x It must be multiplied by 4. 40.  1 1  x3  6x2  11x  6  0  x  g    6, 3, 2, 1  1  8.9  T  2.01 s T  2.11 s 39c. Let x  the new length of the pendulum. 
  g  p : q  128  4-8  12. f(x)  
1.25x  5 13. f(x)  8x2 
 3x 
 9 14. Sample answer: f(x)  1.03x4 
 5.16x3  6.08x2  0.23x  0.94 15. Sample answer: f(x)  0.09x3 
 2.70x2  24.63x 
 65.21 16. Sample answer: f(x)  4.05x4 
 0.09x3  6.69x2 
 222.03x  2697.74 17. Sample answer: f(x)  
0.02x3  8.79x2  3.35x  27.43 18a. Sample answer: f(x)  1.99x2 
 1.74x  2.76 18b. Sample answer: f(x)  
0.96x3  0.56x2  0.36x  4.05 18c. Sample answer: Cubic; the value of r2 for the cubic function is closer to 1. 19a. Sample answer: f(x)  0.126x  22.732 19b. Sample answer: 2010 
 1900  110 f(x)  0.126x  22.732 f(110)  0.126(110)  22.732 f(110)  36.592 37 19c. Sample answer: 2025 
 1900  125 f(x)  0.126x  22.732 f(125)  0.126(125)  22.732 f(125)  38.482 38 20. Sample answer:  Modeling Real-World Data with Polynomial Functions  Pages 261-262  Check for Understanding y  1a. Sample answer:  O  x  y  1b. Sample answer:  O  x  y  1c. Sample answer:  O  x  2. You need to recognize the general shape so that you can tell the graphing calculator which type of polynomial function to use as a model. 3. Sample answer: If companies use less packaging materials, consumers keep items longer, and old buildings are restored instead of demolished, the amount of waste will decrease more rapidly. If consumers buy more products, companies package items in larger containers, and many old buildings are destroyed, the amount of waste will increase instead of decrease. 4. quartic 5. Sample answer: f(x)  1.98x4  2.95x3 
 5.91x2  0.22x  4.89 6. Sample answer: f(x)  3.007x2  0.001x 
 7.896 7a. Sample answer: f(x)  0.48x  58.0 7b. Sample answer: 2010 
 1950  60 f(x)  0.48x  58.0  0.48(60)  58.0  86.8% 7c. Sample answer: f(x)  0.48x  58.0 89  0.48x  58.0 64  x 1950  64  2014  Pages 262-264 8. cubic 10. linear  x f(x)  1 1  2 3  3 6  4 3  5 6 7 8 9 –13 –49 –112 –209 –347  21a. Sample answer: f(x)  0.008x4 
 0.138x3  0.621x2  0.097x  18.961 21b. Sample answer: 1994 
 1992  2 f(x)  0.008x4 
 0.138x3  0.621x2  0.097x  18.961 f(2)  0.008(2)4 
 0.138(2)3  0.621(2)2  0.097(2)  18.961 f(2)  20.663 about 21% 22. A sixth-degree polynomial; there are 5 changes in direction. 23a. Sample answer: f(x)  0.109x2 
 0.001x  48.696  Exercises 9. quadratic 11. quadratic  129  Chapter 4  27.  23b. Sample answer: f(x)  0.109x2 
 0.001x  48.696 100  0.109x2 
 0.001x  48.696 0  0.109x2 
 0.001x 
 51.304  r 
1 0 1  2 2 2 2  
1 
3 
1 1  0 3 0 1  1 
2 1 2  
2 0 
2 0  
1, 1 28a. Let x  number of weeks. P  (120  10x)(0.48 
 0.03x) P  57.6  1.2x 
 0.3x2  23c.  24a. 24b.  24c.  [
5, 25] sc11 by [
50, 10] sc15 root: (21.7, 0) 1985  22  2007 Sample answer: 1998 
 1985  13 f(x)  0.109x2 
 0.001x  48.696 f(13)  0.109(13)2 
 0.001(13)  48.696 f(13)  67.104 No; according to the model, there should have been an attendance of only about 67 million. Since the actual attendance was much higher than the projected number, it is likely that the race to break the homerun record increased the attendance. Sample answer: f(x)  
0.033x3  1.471x2 
 1.368x  5.563 Sample answer: 1996 
 1990  6 f(x)  
0.033x3  1.471x2 
 1.368x  5.563 f(6)  
0.033(6)3  1.471(6)2 
 1.368(6)  5.563 f(6)  43.183 about 43.18 million Sample answer: f(x)  
0.033x3  1.471x2 
 1.368x  5.563 200  
0.033x3  1.471x2 
 1.368x  5.563 0  
0.033x3  1.471x2 
 1.368x 
 194.437  [
20, 20] sc12 by [
40, 60] sc15 maximum: (2, 58.8) 2 weeks 28b. $58.80 per tree 29. x 
 0.10x  0.90x 0.90x  0.10(0.90x)  0.90x  0.09x  0.99x The correct choice is B.  4-8B Fitting a Polynomial Function to a  Set of Points  Page 266 1. y  7x3 
 4x2 
 17x  15 2. y  7x3 
 4x2 
 17x  15; yes 3. Sample answer: y  
5x6 
 2x5  40x4 
 2x3  x2  8x 
 4 4. Infinitely many; suppose that you are given a set of n points in a coordinate plane, no two of which are on the same vertical line. You can pick an infinite number of other points with different xcoordinates. You could find polynomial functions that went through the original n points and any number of the other points. 5. There is no problem with using L10 with list L1 for the example. However, if you are using a different list which happens to have 0 as one of its elements, using L10 will result in an error message, since 00 is undefined.  [
5, 25] sc15 by [
300, 50] sc150 root: (14.8, 0) 14  1990  2004 about 2004 25. 5 
 
b 20  Check: 5 
 
b  20 5  
b 2  5 
 
23 2 0 25  b  2 5 
 
25 0 23  b 5
50 00 26.  6  p3  Chapter 4 Study Guide and Assessment  p    p
3  1  6 p    p  3  p 
 3 (p  3)(p 
 3)  1(p  3)(p 
 3)  Page 267  6(p 
 3)  p(p  3)  (p  3)(p 
 3) 6p 
 18  p2  3p  p2 
 9 9p  9 p1  Chapter 4  Understanding and Using the Vocabulary  1. Quadratic Formula 3. zero 5. polynomial function  130  2. Integral Root Theorem 4. Factor Theorem 6. lower bound  7. Extraneous 9. complex numbers  20. b2 
 4ac  42 
 4(1)(4)  0; 1 real  8. complex roots 10. quadratic equation  a  Pages 268-270  a  Skills and Concepts  a  
2 21. b2 
 4ac  (
1)2 
 4(5)(10)  
199; 2 imaginary  11. no; f(a)  a3 
 3a2 
 3a 
 4 f(0)  (0)3 
 3(0)2 
 3(0) 
 4 f(0)  
4 12. yes; f(a)  a3 
 3a2 
 3a 
 4 f(4)  (4)3 
 3(4)2 
 3(4) 
 4 f(4)  0 13. no; f(a)  a3 
 3a2 
 3a 
 4 f(
2)  (
2)3 
 3(
2)2 
 3(
2) 
 4 f(
2)  
18 14. f(t)  t4 
 2t2 
 3t  1 f(
3)  (
3)4 
 2(
3)2 
 3(
3)  1 f(
3)  73 no 15. 3; x3  2x2 
 3x  0 x(x2  2x 
 3)  0 x(x  3)(x 
 1)  0 x0 x30 x
10 x  
3 x1  4  1  7  25. f(x)  x4 
 10x2  9 f(3)  (3)4 
 10(3)2  9  81 
 90  9 or 0; yes p : q  1, 2  r 1  x2 
 x 
 2  0 (x 
 2)(x  1)  0 x
20 x2 rational roots: 
1, 1, 2  x  27.  79  x  4 7
9  x  4  x  4  x4  x  
2  1  p : q  1  1 1  
2 
1  
1 
2  2 0  x10 x  
1  r  1  0  
1  1  
1  
1 1 rational root: 1 28. p: 1, 2, 4 q: 1, 2  1 1  
1 1  0 0  1 1  
2 0  2 2 2  
2 0 2  
2 
2 2  
4 
6 0  p : q  17. b2 
 4ac  (
10)2 
 4(3)(5)  40; 2 real  m  1 3   
8  2  1 or 4; no  7  
81   2(2)  m  1  i
199   10  1  16. b2 
 4ac  (
7)2 
 4(2)(
4)  81; 2 real  m  r  f 
2  4
2 
 7
2  1  f (x) 
 x 3  2x 2  3x  79  1  

199   2(5)  f(x)  x3 
 x2 
 10x 
 8 f(
2)  (
2)3 
 (
2)2 
 10(
2) 
 8  
8 
 4  20 
 8 or 0; yes 23. f(x)  2x3 
 5x2  7x  1 f(5)  2(5)3 
 5(5)2  7(5)  1  250 
 125  35  1 or 161; no 24. f(x)  4x3 
 7x  1  26.  x  r  22.  f (x)  O  
4  
0   2(1) 4 
2  1  1, 2, 4; 2  r 1 2  2x2  2x  2  0 x2  x  1  0 does not factor  10  
40   2(3) 10  2 
10   6 5  
10   3  rational root: 2  18. b2 
 4ac  (
1)2 
 4(1)(6)  
23; 2 imaginary 1  

23   2(1)  1  i
23 x   2 b2 
 4ac  32  x  19.  
 4(
2)(8)  73; 2 real  y  
3  
73   2(2)  y  3  
73   4  131  Chapter 4  29. p: 1, 3 q: 1, 2 p : q  33. 1  1  2 1 
2 3  2 3  
 2  3  2 2 2 2 2  3 5 1 9 
3  
6 
1 
7 21 3  
11 
12 
4 52 
20  
3 
15 1 153 57  2  4  
4  
13  
2  2  2  
7  
2  15  2  6  
3  
2  3  4 105   
 4  0  
6  
2  0  2  rational root: p : q  r 1  r 
1  19  31  3 
2  1  0  
7  1  12  
4  1 2  1 1  1 2  
6 
3  
5 
5  7 2  3 0  1 1  
3 
3  2 0  
5 1  1 1 1 1 1  
3 1 
2 13 13  0 
2 
1 4 
4  2 0  1 
1 3 53 
51  2  4  8  1, 2, 4, 8, 3, 3, 3,  3 r 1  3 3  
2 8  7 10  x20 x  
2  1  
4  1  8  2  4  0  8  0  4x2  8  0 x2  2  0 does not factor 1 rational root: 
4 Chapter 4  
34 8  
56 0  2  
11  12  9  2  
12  18  0  1 1  0 2  
13 
9  1 1  2 5  
9 6  x2  5x  6  0 (x  3)(x  2)  0 x30 x20 x  
3 x  
2 rational zeros: 
3, 
2, 2, 3  1  1, 2, 4, 2 4  
1 6  1 1  1 
2  r 3  32. p: 1, 2 q: 1, 2, 4  r  5 0  0 
18  x3  2x2 
 9x 
 18  0  4  1  1 0  36. 2 or 0 positive f(
x)  x4 
 13x2  36 2 or 0 negative  
8 0  rational roots: 
2, 
3, 1  p : q  5 5  
5 0  r  r 2  3x2  10x  8  0 (3x  4)(x  2)  0 3x  4  0 4 x  
3  0 5  2x2 
 12x  18  0 x2 
 6x  9  0 (x 
 3)(x 
 3)  0 x
30 x
30 x3 x3 1 rational zeros: 
2, 3  rational roots: 
2, 2 31. p: 1, 2, 4, 8 q: 1, 3 1  4 5  x2  6x  8  0 (x  4)(x  2)  0 x40 x20 x  
4 x  
2 rational zeros: 
4, 
2, 7 35. 2 or 0 positive f(
x)  
2x3 
 11x2 
 12x  9 1 negative  x3 
 3x  1  0 r 
2 
1 4 
4  1 1  r 7  r  0 1  x2  5  0 does not factor rational roots: 
1, 1 34. 1 positive f(
x)  
x3 
 x2  34x 
 56 2 or 0 negative  1, 2, 4  r 
2  1 1  x3  x2  5x  5  0  x4  2x3 
 3x2 
 5x  2  0  p : q  1, 5  1, 3, 2, 2 r 1 
1 3 
3  30.  p : q  132  
18 0  36 0  37.  r 
2 
1 0  3 3 3 3  0 
6 
3 0  0 12 3 0  r 0 1 2 3 4  2  2x  x  3 x3  
4 
4 
3 
2 
1 0  1 1 1 1 1 1  r 
1 0 1 2 3 4  
3 
4 
3 
2 
1 0 1  r 
2 
1 0 1  1 1 1 1 1  m  r 
2 
1 0 1 2  47.  3 
 2y  5 y  
1 3  5    Test y  
2:  
2 
 2  
2 7  5  
2  
2 true 3  5    Test y  
0.5:  
0.5 
 2  
0.5  
1 
3 
2 
1 0  0 6 2 0 0  1 
11 
1 1 1  
8  
10 false 3 5 Test y  1: 1 
 2  1 1  5 true Solution: y  
1, y 0 2  x1  48.  
11 3 
8 
11 
6 7  2 
7 
3 1 5 9  3 
3 11 3 
3 17  x0  1   1
 x 
 1 ; exclude 
1, 1  
9 
9 
9 
9 
9  25 34 25 16 7  2  1  4  
2  3 
24 
58 
24 
8 
10  Test x  
0.5:  6 64 6 
2 
14  2  
0.5  1  true 1   1
 
0.5  1  4  
1 false 2 1   1 
  Test x  0.5:  0.5  1 0.5 
 1 4  3  Test x  2: Test x  4:  6  n 
 n  5  0  n 
 n6  5(n)  0(n)  5n 
 6  0 (n  6)(n 
 1)  0 n60 n  
6  2(x 
 1)  (x  1)(x 
 1) 
 (x  1) 2x 
 2  x2 
 x 
 2 0  x2 
 3x 0  x(x 
 3) x
30 x3    Test x  
2:  
2  1  1 
 
2 
 1  0 and 1 43. Use the TABLE feature of a graphing calculator. 
4.9, 
1.8, 2.2 44.  5  
 2  y; exclude: 0  2 1    x  1 (x  1)(x 
 1)  1 
 x 
 1 (x  1)(x 
 1)  4 4 4 4 4 4  r 
1 0 1 2  3  y  3y 
 2y  5yy  
2 and 
1, 0 and 1, 1 and 2 42.  8  
 (
8)2  
 4(1)(3)  2(1) 8  
 52  2   m  4  
13  
3 1 
3 
5 
5 
3 1  
1 and 0 41.  1  6(m  1)(m 
 1) 5(m  1)(m 
 1)  (2m)(3)(m 
 1) 
 2(m  1) 5m2 
 5  6m2 
 6m 
 2m 
 2 0  m2 
 8m  3  m  1 1 1 1 1 1 1  2m     2m  2 
 3m 
 3  2m 1   566(m  1)(m 
 1)   2(m  1) 
 3(m 
 1)   2 2 
1 
2 
1 2  
1 and 0, 3 and 4 40.  5  6  46.  0 and 1, 3 and 4 39.  x3    2x2  x3  (2x2) 1x(2x2)   2x   
1 and 0 38.  1  x  45.  1 
23 
2 1  2  21 2  3 2  41 2  5   3 true 1   1
 2
1   0 false 1   1
 4
1 2   3 true  Solution: x  
1, 0  x  1, x 3 49. 5 
 
x 20 Check: 5 
 
x 20 5  
x 2 5 
 
23 2 0 25  x  2 5 
 
25 0 23  x 5
50 00  n2  n
10 n1  133  Chapter 4  3  3  50. 
1 4a 
   8  5 Check: 
1 4a 
   8  5 3 3 
1 4a 
   
3 
4(
6. 5)  
185 3 4a 
 1  
27 

27 85 4a  
26 
3  8  5 a  
6.5 55 51. 3  
x  8  
x  35 9  6
x  8  x  8  x  35 6
x  8  18 
x 83 x89 x1 Check: 3  
x  8  
x  35 3  
1   8  
1 5  3 3  
9   
36  336 66 52. 
x 
57 x
5 0 x 
 5  49 x 5 x  54 Test x  0: 
0 
57  

5   7 meaningless Test x  10: 
10 
7 5 
5   7 true Test x  60: 
60 
7 5 
55   7 false Solution: 5  x  54 53. 4  
2a 6 7 2a  7 0 
2a 2 7 2a 
7 2a  7  4 a 
3.5 2a  
3 a  
1.5 Test a  
5: 4  
2(
5)  76 4  

3   6 meaningless Test a  
2: 4  
2(
2)  76 4  
3   6 false Test a  0: 4  
2(0)   76 4  
7   6 true Solution: a  
1.5 54. cubic 55. f(x)  2x2 
 x  3  Page 271  58a.  g (x ) 
 0.006x 4  0.140x 3  0.053x 2 100  1.79x  O 10  x  20  58b. g(x)  
0.006x4  0.140x3 
 0.053x2  1.79x  x(
0.006x3  0.140x2 
 0.053x  1.79)  x(x3 
 23.3 x2  8.83 x 
 298.3 ) r 1 
23.333 8.833 
298.333 1 1 
22.333 
13.503 
311.836 5 1 
18.333 
82.835 
712.508 23.5 1 0.167 12.758 0 rational zeros: 0, about 23.5 
  g  59. T  2 1.6  2 0.25  0.06   
  9.8  
  9.8 
  9.8  0.64  
; about 0.64 m  Page 271  Open-Ended Assessment x  2    1. Sample answer:  x  3  2x  1  x 2    x  3 (x  3)(2x  1)   2x  1 (x  3)(2x  1)  x(2x  1)  2(x  3) 2x2  x  2x  6 2 2x 
 x 
 6  0 (2x  3)(x 
 2)  0 2x  3  0 x
20 3 x  
2 x2 2a. Sample answer: x 
 4  
x 
2 2b. Sample answer: x 
 4  
x 
2 (x 
 4)2  x 
 2 x2 
 8x  16  x 
 2 x2 
 9x 
 18  0 (x 
 6)(x 
 3)  0 x
60 x
30 x6 x3 Check: x 
 4  
x 
2 x 
 4  
x 
2 6 
 4  
6  
2 3 
 4  
3  
2 2  
4  
1  
1  22 
1 1 The solution is 6. Since 
1 1, 3 is an extraneous root. 3a. Sample answer:  Applications and Problem Solving  56. Let x  width of window. Let x  6  height of window. A  
w 315  x(x  6) 315  x2  6x 0  x2  6x 
 315 0  (x  21)(x 
 15) x  21  0 x 
 15  0 x  
21 x  15 Since distance cannot be negative, x  15 and x  6  21. the window should be 15 in. by 21 in. 57. Let x  width. Let x  6  length. (x  12)(x  6) 
 (x  6)(x)  288 x2  18x  72 
 x2 
 6x  288 12x  72  288 x  18 x  6  24 18 ft by 24 ft Chapter 4  g (x )  200  x f(x)  
3 
12  
2 0  
1 2  
0.5 1.125  x f(x)  0 0  0.5 
0.625  1 0  2 8  3b. Sample answer: f(x)  x3  x2 
 2x 3c. Sample answer: 
2, 0, 1  134  Chapter 4 SAT & ACT Preparation Page 273  4.You may want to draw a diagram.  SAT and ACT Practice  s  1. There are two ways to solve this problem. You can use the distance formula or you can sketch a graph. d  
 (x2 
  x1)2  (y2 
 y1)2 2  
 (1 
 ( 
2)  (3 
  (
1))2  
 32  42  
9 6  1  
25  or 5 y  s6 Use the formula for the perimeter of a rectangle, where 
 represents the length and w represents the width. 2
  2w  P Replace w with s. Replace 
 with s  6. 2(s  6)  2s  60 The correct choice is E. 5. First find the slope of the given line. Write the equation in the form y  mx  b. 3x 
 6y  12 6y  3x 
 12 1 1 y  2x 
 2 The slope is 2.  (1, 3) 4  O  x  (2, 1) 3  So the slope of the line perpendicular to this line is the negative reciprocal of this slope. The slope of the perpendicular line is 
2. The correct choice is A. 6. Be sure to notice the small piece of given information: x is an integer. You need to find the number, written in scientific notation, that could be x3. This means that the cube root of the number is an integer. Take the cube root of each of the answer choices and see which one is an integer. You can use your calculator or do the calculations by hand. Notice that 2.7 is one-tenth of 27, which is 33. 2.7 1013  27 1012 3 
 27  1012  3 104 or 30,000. 30,000 is an integer. When you try the same calculation with each of the other answer choices, the resulting power of 10 has a fractional exponent. So the number cannot be an integer. The correct choice is C. 7. This is a system of equations, but you do not need to solve for x or y. You need to find the value of 6x  6y. Notice that the first equation contains 5y and the second contains 
1y. If you subtract the second from the first, you have 6y. Similarly, subtraction of the x values gives a result of 6x. Use the same strategy that you would for solving a system. Subtract the second equation from the first. 10x  5y  14 
4x  5y  
2 6x  6y  12 The correct choice is C.  When you sketch the points and draw a right triangle as shown above, you can see that this is a 3-4-5 right triangle. Using the Pythagorean Theorem, you can calculate that the length of the hypotenuse is 5. 52  42  32 The correct choice is C. 2. Points on the graph of f(x) are of the form (x, f(x)). To move the entire graph of f(x) up 2 units, 2 must be added to each of the second coordinates. Points on the translated graph are of the form (x, f(x)  2). The function which represents the translation of the graph up 2 units is f(x)  2. The correct choice is E. 3. You need to find both the x- and y-coordinates of point C. Use the properties of a parallelogram. First find the y-coordinate. Since opposite sides of a parallelogram are parallel and side AD is on the x-axis, point C must have the same y-coordinate as point B. So the y-coordinate is b. This means you can eliminate answer choices A and B Now find the x-coordinate. Since opposite sides of a parallelogram have equal length and side AD has length d, side BC must also have length d. Point B is a units from the y-axis, so point C must be a  d units from the y-axis. The x-coordinate of point C is a  d. So point C has coordinates (a  d, b). The correct choice is E.  135  Chapter 4  (x 
 a2  x2 
 2ax  a2  (x2  a2) 
 2ax So the quantity in Column A equals the quantity in Column B plus the sum of the squares of x and a. Since neither x nor a equal 0, their squares must be greater than 0. So the quantity in Column A is always greater than the quantity in Column B. The correct choice is A. 10. Since the problem does not include a figure, draw one. Label the four points.  8. You can solve this problem using the midpoint formula or by sketching a graph. The midpoint formula: x1  x2 y1  y2  3  (
4) 5  3  
1 8  2, 2  2, 2  2, 2  
2, 4 1  y  (   12,  4)  (3, 5)  (4, 3)  O  x  3  2  E  F  8  G  One method of solving this problem is to "plug-in" numbers for the segment lengths. Since 5 EG  3 EF, let EF  3. Then EG  5. This means that FG must equal 2, since EF  FG  EG.  The correct choice is B. 9. The expression x2 
 2ax  a2 is a perfect square trinomial and can be factored as (x 
 a)2. The square of a real quantity is never negative. The correct choice is A.  HF  5FG  5(2)  10 HG  HF 
 FG  10 
 2  8 EF  HG  3   8  The answer is .375 or 3/8.  Chapter 4  H  136  Chapter 5 The Trigonometric Functions Pages 281–283  Angles and Degree Measure  5-1  Pages 280–281  Check for Understanding  1. If an angle has a positive measure, the rotation is in a counterclockwise direction. If an angle has a negative measure, the rotation is in a clockwise direction. 45  26   2. Add 29, 60, and  3600 .  3. 270°  360k° where k is an integer 4. y  x  O  1260° 5. 34.95°  34°  (0.95 60)
  34°  57
 34° 57
 6. 
72.775°  
(72°  (0.775 60)
)  
(72°  46.5
)  
(72°  46
  (0.5 60))  
(72°  46
  30) 
72° 46
 30    24. 23° 14
 30  23°  14
 60
   30 3600   23.242° 1°  1°  1°  13.  453  360   1.26  1°  3600    26. 233° 25
 15  233°  25
 60
   15 3600   233.421° 1°  1°  1 (360°)  15° 24 1 1   360°  60 24 1 1 1    360° 60 60 24  1°    28. 
405° 16
 18  
405°  16
 60
   18 3600  1°  1°   
405.272°    29. 1002° 30
 30  1002°  30
 60
   30 3600   1002.508° 1°  1°  3 
360°  
1080° 31. 2 360°  720° 1.5 360°  540° 33. 7.5 (
360°)  
2700° 2.25 360°  810° 35. 5.75 (
360°)  
2070° 4 360°  1440° 30°  360k°; Sample answers: 30°  360k°  30°  360(1)° or 390° 30°  360k°  30°  360(
1)° or 
330° 38. 
45°  360k°; Sample answers: 
45°  360k°  
45°  360(1)° or 315° 
45°  360k°  
45°  360(
1)° or 
405° 39. 113°  360k°; Sample answers: 113°  360k°  113°  360(1)° or 473° 113°  360k°  113°  360(
1)° or 
247° 30. 32. 34. 36. 37.   14. 
 360  
2.22  
2.22  2  
0.22 
0.22 360°  
78° 360° 
 78°  282°; IV  15.  
 180°  227° 
 180°  47° 16. 360° 
 210°  150° 180° 
   180° 
 150°  30° 17.  1°    27. 173° 24
 35  173°  24
 60
   35 3600   173.410°  798    360(1)°  453°   360  453°   93°; II  1°   
14.089°  8. 29° 6
 6  29°  6
   6   29.102° 9. 2 (
360°)  
720° 10. 4.5 360°  1620° 11. 22°  360k°; Sample answers: 22°  360k°  22°  360(1)° or 382° 22°  360k°  22°  360(
1)° or 
338° 12. 
170°  360k°; Sample answers: 
170°  360k°  
170°  360(1)° or 190° 
170°  360k°  
170°  360(
1)° or 
530° 1°  60
  1°    25. 
14° 5
 20  
14°  5
 60
   20 3600     7. 
128° 30
 45
  
128°  30
 60
   45 3600   
128.513° 1°  Exercises  18. 
16.75°  
(16°  (0.75 60)
)  
(16°  45
) 
16° 45
 19. 168.35°  168°  (0.35 60)
  168°  21
 168° 21
 20. 
183.47°  
(183°  (0.47 60)
)  
(183°  28.2
)  
(183°  28
  (0.2 60)  
(183°  28
  12) 
183° 28
 12 21. 286.88°  286°  (0.88 60)
  286°  52.8
  286°  52
  (0.8 60)  286°  52
  48 286° 52
 48 22. 27.465°  27°  (0.465 60)
  27°  27.9
  27°  27
  (0.9 60)  27°  27
  54 27° 27
 54 23. 246.876°  246°  (0.876 60)
  246°  52.56
  246°  52
  (0.56 60)  246°  52
 
 33.6 246° 52
 33.6     0.25°, or 0.25(60
)  15
      0.0042°,  or 0.0042(60)(60)  15  137  Chapter 5  40. 217°  360k; Sample answers: 217°  360k°  217°  360(1)° or 577° 217°  360k°  217°  360(
1)° or 
143° 41. 
199°  360k°; Sample answers: 
199°  360k°  
199°  360(1)° or 161° 
199°  360k°  
199°  360(
1)° or 
559° 42. 
305°  360k°; Sample answers: 
305°  360k°  
305°  360(1)° or 55° 
305°  360k°  
305°  360(
1)° or 
665° 43. 310°  360k°  310°  360(0)° or 310° 44. 60°  360k°  60°  360(2)° or 780° 60°  360k°  60°  360(
3) or 
1020° 45.  400°  360°   1.11  46.    360(1)°  400°   360°  400°   40°; I 47.  940°  360°   2.61  48.  
624°  360°  1059°  360°  36,000,000 or 1.08 63.  52. 53. 54. 55.  57.  1275°  360°    360(2)°  1059°   720°  1059°   339°; IV  Chapter 5  1  2    360°  rotation  60 seconds  minute  60 minutes  hour    1 rotation  70 minutes  360°   revolution 1  2 minute  1°  1°  60 minutes  hour  24 hours  day    day  1 revolution   San Antonio:  1 hour  20.6 rotations  24 hours  day  24 revolutions    day  24 
 20.6  3.4 revolutions about 3.4 revolutions 66b. Sydney: 20.6 rotations  day  7 days  week  360°  rotation  24 revolutions   San Antonio:  day   51,840° 7 days  week  360°  revolution    60,480° 60,480° 
 51,840°  8640° 67a. Use graphing calculator to find cubic regression. Sample answer: f(x)  
0.00055x3  0.0797x2 
 3.7242x  76.2147 67b. 2010 
 1950  60 f(x)  
0.00055x3 
 0.0797x2 
 3.7242x  76.2147 f(60)  
0.00055(60)3  0.0797(60)2 
 3.7242(60)  76.2147  20.8827 Sample answer: about 20.9%   32,400°/second 360°  revolution  1°  66a. Sydney:    1,944,000°/minute  34,200°  60 seconds  minute    81° 45
 34.4  81°  45
 60
   34.4 3600   81.760°   2.90  30 seconds   22,320° second  1°   3.54  95 revolutions  minute  360°  rotation    65b. 24° 33
 32  24°  33
 60
   32 3600   24.559°  60. 90k, where k is an integer 61.  62 rotations  second  107 degrees  64. 25°  120k°, where k is an integer 65a. 44.4499°  44°  (0.4499 60)
  44°  26.994
  44°  26
  (0.994 60)  44°  26
  59.64 44° 26
 59.64 68.2616°  68°  (0.2616 60)
  68°  15.696
  68°  15
  (0.696 60)  68°  15
  41.76 68° 15
 41.76   2.94   
2.75  90 revolutions 360°    second revolution 90 revolutions 60 seconds    second 1 minute  360°  rotation  62 rotations 60 seconds 60 minutes 360°     second minute hour rotation hours   24 day  1,928,448,000°/day    360(2)°  1045°   720°  1045°   325° 360° 
   360° 
 325° or 35° 58. 20° 180° 
   180° 
 20° or 160° 180°    180°  20° or 200° 360° 
   360° 
 20° or 340° 59.  107 to 3.6    80,352,000°/hour   
0.78    360(3)°  1275°   1080°  1275°   195°; III 360° 
   360° 
 327° or 33° 180° 
   180° 
 148° or 32° 563° 
 360°  203°  
 180°  203° 
 180° or 23° 
420°  360°  
60° 56. 360° 
 197°  163° 360° 
 60°  300° 180° 
 163°  17° 360° 
 300°  60° 1045°  360°  360°  revolution 3.6 107  62 rotations  second  62 rotations  second  
2.75  2  
0.75 
0.75 360°  
269° 360° 
 269°  91°; II 51.    107  1,339,200°/minute   
1.73  
989°  360°  360°  revolution  100,000 revolutions  minute  
1.73  1  
0.73 
0.73 360°  
264° 360° 
 264°  96°; II 50.  30,000 revolutions  minute  10,800,000 or 1.08    360(
1)°  
280°  
 360°  
280°   80°; I    360(2)°  940°   720°  940°   220°; III 49.  
280°  360°  62.  34,200°/minute   17,100°  138  3  68. 
5 6n   
 15  
10 3 
5 6n    5 6n  5  125 6n  120 n  20 3 Check: 
5 6n   
 15  
10 3 
6(20)   5 
 15  
10 3 
125  
 15  
10 5 
 15  
10 
10  
10  x3  x2 x3  x2 x3  x2  69.  (x  2)(x  3)  (3, 5) (1, 5) (0, 3)  77. [f g](x)  f(g(x))  f(x 
 0.3x)  (x 
 0.3x) 
 0.2(x 
 0.3x)  x 
 0.3x 
 0.2x  0.06x  0.56x 78. m∠EOD  180° 
 m∠EOA 
 m∠BOD  180° 
 85° 
 15°  80° m∠OED  m∠EDO 1 m∠OED  2(180° 
 m∠EOD)  3   2
 x2  5x  6 3   2
 (x  2)(x  3)    (x  2)(x  3)(2) 
 3  (x  2)(x  3) (x  2)(x  3)   1   2(180° 
 80°)  50° m∠ECA  180° 
 m∠EOC 
 m∠OED  180° 
 (80°  15°) 
 50°  35° The correct choice is D.  5-2  52  y  20  y 
 xx 11  26 10  2  48  2. R1   x  R1  R2  R2   f (x)  R3  R3  f (x) 
|(x  1)2  2|  O  Graphing Calculator Exploration  1. Sample answers:  O  74.  Trigonometric Ratios in Right Triangles  Page 284   r2  4.91  r2 about 4.91 73. x  1  0 x 
1 Point discontinuity  x  O  (x  3)(x  3)  (x  2)(x  3)(2) 
 3 x2  6x  9  2x2  10x  12 
 3 0  x2  4x 0  x(x  4) x  0 or x  4  0 x  
4 70. 2 1 1 0 8 1 2 4 24 1 2 12 25 25 71. (x 
 (
5))(x 
 (
6))(x 
 10)  0 (x  5)(x  6)(x 
 10)  0 (x2  11x  30)(x 
 10)  0 x3  x2 
 80x 
 300  0 72. r1t1  r2t2 18(
3)  r2(
11) 18(
3)  
11  y  76.  3. R1  R1   x  R2  R2   decreasing for x  
1, increasing for x 
1 75. expanded vertically by a factor of 3, translated down 2 units  R3  R3   5  13 15  39 12  13 36  39 5  12 15  36 12  13 36  39 5  13 15  39 12  5 36  15  24  or about 0.3846 or about 0.3846 or about 0.9231 or about 0.9231 or about 0.4167 or about 0.4167 or about 0.9231 or about 0.9231 or about 0.3846 or about 0.3846 or 2.4 or 2.4  4. Each ratio has the same value for all 22.6° angles. 5. yes 6. Yes; the triangles are similar.  139  Chapter 5  Pages 287–288  11. (AC)2  (CB)2  (AB)2 82  52  (AB)2 89  (AB)2 
89   AB side opposite  sin A   hypotenuse  Check for Understanding  1. The side opposite the acute angle of a right triangle is the side that is not part of either side of the angle. The side adjacent to the acute angle is the side of the triangle that is part of the side of the angle, but is not the hypotenuse. 2. cosecant; secant; cotangent a  b  c , a  c , b  a  tan A   b  a  tan A   3. sin A  c, cos A  c, tan A  b, csc A   sec A   cot A   side opposite  side adjacent  6.  15
514  15 17  or  cos T   
514  514 
514  side opposite  tan T   side adjacent 15  tan T  1 7 1 1    csc v  sin v 7. tan v   cot v  1 5  csc v   2 or 2  12   cos T   hypotenuse  sin T   or  17
514   514  or  1  cot v   cot P   2
91   6  1   csc v   or 3  1  3  1  cos v   1  5  9  9  1  1  1   18. tan v   cot v   19. sec v   cos v  1   tan v   0.75 or about 1.3333  side opposite  side adjacent   sin R   hypotenuse  
91   3   cos R   hypotenuse  
I t 
Io  sec R   I  sec R   or   sin R    It o  0.5I0  It  Exercises  10. (AC)2  (CB)2  (AB)2 802  602  (AB)2 10,000  (AB)2 100  AB side opposite  sin A   hypotenuse 60  3    sin A   100 or 5  side adjacent   cos A   hypotenuse 80  4    cos A   100 or 5  side opposite   tan A   side adjacent 60  3   tan A  8 0 or 4  Chapter 5  1   sec v   0.125 or 8  20. (RT )2  (TS)2  (RS)2 142  (TS)2  482 (TS)2  2108 TS  
2108  or 2
527   
2   2  Pages 288–290  7  or 3   sin v   2.5 or 0.4  or 5  tan R   2  4  3  7   17. sin v   csc v  cos 45°   or  1  1   16. cos v   sec v  2
527   14    
91   10  1  tan R   or  or   15. csc v   sin v   14. cot v   tan v  
I t Io  cos v  20  2
91  I  It o   4
91  40  3
91   91   
527  2
527  or  48 24 side opposite  side adjacent  sec P  9.  10
91   91  cos A   13. tangent  cot P   csc P   12  4
91   tan A   2
91  
91   or  20 10 hypotenuse  side opposite 20 10  or  6 3 side adjacent  side opposite  csc P   8
89   89  or  side opposite  8. (PS)2  (QS)2  (QP)2 (PS)2  62  202 (PS)2  364 PS  
364  or 2
91  side opposite side adjacent   sin P   cos P hypotenuse hypotenuse cos P   8  
89    tan A   side adjacent  1  6 3  sin P  2 0 or 10 side opposite  tan P   side adjacent 3
91  6  tan P   2
91  or 91 hypotenuse  sec P   side adjacent  3  sin A  40 or 1 0   tan v   1.5 or about 0.6667   5  cos A   12. (AC)2  (AC)2  122  402 (AC)2  1456 AC  
1456  or 4
91  side opposite side adjacent   sin A   cos A  hypotenuse hypotenuse  4. sin A  cos B, csc A  sec B, tan A  cot B 5. (TV)2  (VU)2  (TU )2 172  152  (TU )2 514  (TU )2 
514   TU  sin T   hypotenuse  5 5
89   or  
89  89 side opposite  side adjacent 5  8 (BC)2  (AB)2  sin A   side adjacent   cos A   hypotenuse  140  14  7  cos R   or  csc R   48 24 hypotenuse  side opposite  csc R   48  2
527   hypotenuse  side adjacent  cot R   48  24  14  7  cot R   side adjacent  side opposite 14 7
527   or  527 2
527   or  
527   7  or  24
527   527  21. (ST)2  (TR)2  (SR)2 382  (TR)2  402 (TR)2  156 TR  
156  or 2
39  side opposite side adjacent   sin R   cos R hypotenuse hypotenuse 38  19  sin R  40 or 20 side opposite   tan R   side adjacent  tan R  sec R  sec R   38 19
39   or  39 2
39  hypotenuse  side adjacent  40 20
39  or  2
39  39  cos R  csc R  csc R  cot R  cot R   side opposite   28. sin R   hypotenuse 3  sin R  7 a2  b2  c2 32  b2  72 b2  40 b  
40  or 2
10  side adjacent  cos R   hypotenuse  2
39  
39   or  40 20 hypotenuse  side opposite 40 20  or  38 19 side adjacent  side opposite  
39  2
39  or  38 19  cos R  csc R   side opposite   cos R   hypotenuse  
154  
7   or  2
22  44 side opposite  side adjacent  9
22  9 cos R   or  2
22  44 hypotenuse  csc R   side opposite   sin R   hypotenuse  sin R  tan R  tan R  sec R  sec R   
7   9 hypotenuse  side adjacent 2
22   9  23. cot (90° 
 v)  tan v cot (90° 
 v)  1.3 24b. 0.186524036 24d. 1.37638192 25. v sin cos v sin cos  72° 0.951 0.309 82° 0.990 0.139  74° 0.961 0.276 84° 0.995 0.105  25a. 1 26.  cot R  cot R   14° 0.242 0.970 0.249  12° 0.208 0.978 0.213  v sin cos tan  8° 0.139 0.990 0.141  6° 0.105 0.995 0.105  4° 0.070 0.998 0.070  2° 0.035 0.999 0.035  v2   tan 13°   9.8(15.5)  35.07  v2 5.9  v about 5.9 m/s v2  29d. increase  v2  9.8(15.5) v2  40.70  6.4  v about 6.4 m/s  side opposite  side adjacent   30. sin v   hypotenuse  80° 0.985 0.174   cos v   hypotenuse  side opposite  hypotenuse —— side adjacent  hypotenuse  sin v  cos v    sin v  cos v    hypotenuse  sin v  cos v    hypotenuse  sin v  cos v sin v  cos v    side adjacent  side opposite side opposite  side adjacent  hypotenuse hypotenuse  side adjacent  side opposite   tan v  31a. ∠  90° 
 L 
 23.5°  (N  10)360  365 (172  10)360   cos 365  cos  ∠  90° 
 26° 
 23.5° (
0.99997) ∠  87.5° (N  10)360  ∠  90° 
 L 
 23.5° cos  365  10° 0.174 0.985 0.176  ∠  90° 
 26° 
 23.5°  cos  ∠  90° 
 26° 
 23.5° ∠  40.5°  1  31b. ∠  90° 
 L 
 23.5° ∠  90° 
 64° 
 23.5°  (355  10)360  365  (N  10)360  365 (172  10)360   cos 365  cos  n  ∠  90° 
 64° 
 23.5° 
0.99997 ∠  49.5° (N  10)360  ∠  90° 
 L 
 23.5° cos  365  n  ∠  90° 
 64° 
 23.5°  ] cos[ 365  1.5103  n  ∠  90° 
 64° 
 23.5° ∠  2.5°  1  26a. 0 26c. 0 sin v i sin vr sin 45°  sin 27° 55
   7
10  20  v2  ∠  90° 
 26° 
 23.5° 16° 0.276 0.961 0.287  or  29b. tan v  gr  tan v  gr tan 15°   25b. 0 18° 0.309 0.951 0.325  27.  v2  tan v  gr  29c.  88° 0.999 0.035  v sin cos tan  7  2
10   29.53  v2 5.4  v about 5.4 m/s  24c. 35.34015106  86° 0.998 0.070  sec R   v2  2
154  2
22   or  
7  7 side adjacent  side opposite 9
7  9  or  
7  7  78° 0.978 0.208  2
10   3  sec R   3
10  3  or  2
10  20 hypotenuse  side adjacent  tan R    tan 11°   9.8(15.5)  24a. 0.7963540136  76° 0.970 0.242  cot R  29a.  side adjacent  csc R   cot R   7  3 side adjacent  side opposite  csc R   22. (ST )2  (TR)2  (SR)2 (
7 )2 
 92  (SR)2 88  (SR)2 
88   SR; 
88  or 2
22   2
10   7 hypotenuse  side opposite  side opposite   tan R   side adjacent  26b. 1  141  (355  10)360  Chapter 5  31c. 87.5° 
 40.5°  47° 49.5° 
 25°  47° neither  2.  y  sin(B 
 A)   32. x  t c os A  sin(60° 
 41°)  cos 41°  x  10  x  O    x  10(0.4314) x  4.31; about 4.31 cm 33. 88.37° 88°  (0.37 60)
  88°  22.2
  88°  22
  (0.2 60)  88°  22
  12 88° 22
 12 34. positive: 1 f(
x)  x4 
 2x3  6x 
 1 negative: 3 or 1 35. 35a. 23 employees 35b. $1076  As v goes from 0° to 90°, the y-coordinate increases. As v goes from 90° to 180°, the y-coordinate decreases. x  cos v   3. cot v  y   sin v  4.  1  1  y  1x  O 1  Function sin  or cos  cos  or sec  tan  or cot   [
10, 50] scl:10 by [
10, 1200] scl:100 7 
3 5 0 
1 4 
1 4 0 36. 4 
 (
3) 5 0 
1  7 2 0 8 0 8 2 8 2 0  7(2) 
 (
3)(8)  5(8)  78  y  1  or  7.   1 
3   2 , 2  sin 30°  y sin 30°   1 
2  y  tan 30°   1  2  
3   2  tan 30°   1  
3   tan 30°   
3   3  1  y 
 3  
2(x 
 6) y  6  1  1  12  2(2x)(3x)  3x  3(2) or 6  12  3x2 4  x2 2x  a2  b2  c2 42  62  c2 52  c2 
52   c; 
52  or 2
13   1  sec 30°  x  The correct choice is C.  5-3 Page 296  Trigonometric Functions on the Unit Circle Check for Understanding  1. Terminal side of a 180° angle in standard position is the negative x-axis which intersects the unit 1 1 circle at (
1, 0). Since csc v  y, csc 180°  0 which is undefined.  Chapter 5  cos 30°  x cos 30°   
3   2  1  csc 30°  y csc 30°   1  1  2  csc 30°  2  2x  2(2) or 4  38. A  2bh  1  2  tan 30°  x  y 
 y1  m(x 
 x1) 1 
2x  1  6. (0, 
1); sec(
90°)  x or 0; undefined  3
5  m  0   5. (
1, 0); tan 180°  x or  
1 ; 0   37. m   6
2 
2  4  Quadrant II III  
 
 
 
   I     142  x  cot 30°  y 
 3  2  1  2  sec 30°   1   cot 30°   sec 30°   2  
3   cot 30°  
3   sec 30°   2
3   3  
 3  2  IV 
  
  x  8. terminal side — Quadrant III reference angle: 225° 
 180°  45°  12. cos v  r 1  cos v  
2  
2   
2  
  2 ,
 2   sin 225°  y  cos 225°  x   
2  sin 225°  
  2 y  
 2 
2  
 2 
 2  csc 225°  csc 225°   1  y  1  sin v  r  1  
2  
 2 2 
 
2   sin v  sec v  sec v   x  cot 225°   cos v  r  x  tan v  x  sin v  5  4  cos v  5  3  tan v  3  csc v  y  r  sec v  x  r  cot v  y  5  sec v  3  5  cot v  4  or 

3   csc v   cot v  or 
2  cot v   2  
3  2
3   3  x  y 
1  
3  

3   3  Exercises  4  15. (1, 0); tan 360°  x or 1; 0  y  x  0  
1  x  16. (
1, 0); cot(
180°)  y or 0; undefined  3  1  1   17. (0, 
1); csc 270°  y or  
1 ; 
1  18. (0, 1); cos(
270°)  x or 0 1  1   19. (
1, 0); sec 180°  x or  
1 ; 
1 y  x  cos v  r v v  y  11. tan v  x tan v  
1 x  1, y  
1 y  sin v  r 
1  
2  
2  sin v  
  2 r csc v  y  
2 csc v   or 

2  
1 x cot v  y 1  cot v   
1 or 
1  sin v   r  x 2  
1  14. (0, 1); sin 90°  y or 1  10. r  r  
 (
6)2   62 r  
72  or 6
2  6  cos 6
2   
2 sin v   cos 2 r csc v  y  6
2 csc v    6 or 
2 x cot v  y 
6 cot v  6 or 
1  
3   
1  y  
 x2  y2  sin v   tan v   Pages 296–298  y  y   
3  2  13. C  2r cos L C  2r cos L C  2(3960) cos 0° C  2(3960) cos 90° C  24,881.41 C0 The circumference goes from about 24,881 miles to 0 miles.  cot 225°  1  sin v  r  r  csc v  y  cot v   
2  
 2  
2  
 2  sec 225°  

2  9. r  
 x2  y2 r  
 32  42 t  
25  or 5  sin v  r  y  tan v  x  csc v   cot 225°  y  1  
2  
2 2 
 
2   csc v  4  3  y2 
3 y Quadrant II, so y  
3   
2   csc 225°  

2   sec 225°  x  sec 225°   x  
1, r  2  csc 225°  y  tan 225°  1  sec 225°   22  (
1)2  y2  cos 225°  
  2  tan 225°  x tan 225°   r 2  x2  y2  20. Sample answers: 0°, 180°  tan v  x  
6  6
2  
2  
 2  sec v  sec v   6  22.   tan v   
6 or 
1  r  x 6
2   
6     
2  
2 ,  2 2    sin 45°  y  cos 45°  x  sin 45°   cos 45°   tan 45°   or 

2   tan 45°   r2  12  (
1)2 r2  2 r  
2  x cos v  r cos v  sec v  sec v   1  
2  
2   2 r  x 
2   1   
2  2 y  x 
2   2  
2   2  csc 45°  or 1  csc 45°  csc 45°   
2   2 1  y 1  
2   2 2  
2    csc 45°  
2  r 2  x2  y2  cos v   21. undefined  sec 45°   1  x  sec 45°   1  
2   2 2  
2   sec 45°   x  cot 45°  y cot 45°   
2   2  
2   2  or 1  sec 45°  
2   or 
2   143  Chapter 5  23. terminal side — Quadrant II reference angle: 180° 
 150°  30°  25. terminal side — Quadrant III reference angle: 210° 
 180°  30° 1 
3   
  2 , 
2  
3  1  
  2 , 2  sin 150°  y  cos 150°  x  1 sin 150°  2 y tan 150°  x  cos 150°   1  2  tan 150°  
3 
 2 1  tan 150°  
 
3 
3  tan 150°  
  3 1 sec 150°  x 1 sec 150°   
3  
 2  csc150°   
3  
 2 1  csc 150°  y 1  1  2  x  cot 150°  y  2  
3  
2  1  2  1   
2  tan 315°   y  x  tan 315°   
 2 
 2  
2   2  or 
1  cos 315°   
2   2  csc 315°   1  y  csc 315°   1  
2  
 2 2 
 
2   csc 315°   cot 210°   
3  
2  
 1 
 2  cot 210°  
3   1  cot 315°  y  1  
2   2 2  
2   cot 315°   y  x  
2   2  
2  
 2  cos 330°  x 1  sin 330°  
2  csc 315°  

2  sec 315°  x  x  cot 210°  y   2
3 
 3  sin 330°  y cos 315°  x  sin 315°  
  2  csc 210°  
2  1  
3   2 , 
2  
2   
2   2 ,
 2   sin 315°  y  1  1 
2  26. terminal side — Quadrant IV reference angle: 360° 
 330°  30°  24. terminal side — Quadrant IV reference angle: 360° 
 315°  45°  or 
1  sec 315°  
2   Chapter 5  csc 210°   sec 210°   2
3   sec 315°   1 
2  
3  
 2 1 tan 210°   
3  
3   tan 210°  3 1 sec 210°  x 1 sec 210°   
3  
2  
3   csc 210°  y  sec 210°  
  
3   sec 150°  
  3  sec 315°   cos 210°  
  2  2  cot 150°  

3   sec 150°  
  
3   cos 210°  x  1 sin 210°  
2 y tan 210°  x  tan 210°   csc 150°  2  cot 150°   sin 210°  y  144  cos 330°   tan 330°  x  csc 330°   1 
 2  tan 330°  
3  2 1 tan 330°  
  
3  
3  tan 330°  
  3 1 sec 330°  x 1 sec 330°   
3   2 2 sec 330°   
3  2
3   sec 330°  3  csc 330°   
3   2 1  y 1  1 
 2  csc 330°  
2 x  cot 330°  y cot 330°   
3   2  1 
 2  cot 330°  

3   31. r  
 x2  y2 r  
 (
6)2   62 r  
72  or 6
2   27. terminal side — Quadrant I reference angle: 420° 
 360°  60° 
3  12,  2   sin 420°  y  cos 420°  x  sin 420°   cos 420°  2  tan 420°  tan 420°   
3   2 y  x  1  1  csc 420°  csc 420°   tan 420°  
3  sec 420°  sec 420°   sin v   csc 420°  y  
3   2  1  2  csc 420°   1  x 1  1  2  cot 420°  cot 420°   sec 420°  2  cot 420°  cot 420°   sin v   1  
 3  2 2  
3   csc v  csc v   
6  6
2   6   tan v   
6 or 
1  
2   cos v  
  2 x  r  sec v  x sec v   cot v  y 
6  6
2   
6  cot v  6  or 

2   or 
1  cos v  r  y  0  cos v  2 or 1  r  2  sec v  2 or 1  x  tan v  x  2  tan v  2 or 0  sec v  x  r  cot v  y  2  cot v  0  0  x  2  undefined 33. r  
 x2  y2 r  
 12  ( 
8)2 r  
65   undefined  y  sin v  r or 
1  or 2 34.  y  x  tan v  x  1  
65  
65   65  tan v  1 or 
8  cos v  r  
8  cos 
65  8
65  sin v  
  cos 65 r   csc v  y  
65 
65  csc v   or 
  
8 8 x cot v  y 1 1   cot v   
8 or 
 8 r  
 x2 
 y2  sin v   
3  1   2 , 2  v v  
8  r  sec v  x sec v   
65   1  or 
65   r  
 52  ( 
3)2 r  
34   r  
 (
4)2   (
3 )2 r  
25  or 5  y  sin v   y  r  cos v   sin v   3 
5 r  y 5 
3  cos v  
5  csc v   tan v  x  y  csc v  0  29. terminal side — Quadrant 1 reference angle: 390° 
 360°  30°  csc v   
2   2 r  y  6
2  6  csc v  y  x  30.  cos v   sin v  2 or 0  cot (
45°)  y  1 csc 390°  y 1 csc 390°   1  2 r  
 x2  y2  6  6
2   y  x  sin v  r  
2  
2    2 ,
 2   cot (
45°)   cos v  r  or 
2  32. r  
 x2  y2 r  
 22  02 r  
4  or 2  2
3   3 x  y 1  2  
3   2 1  
3   
3  3  28. terminal side — Quadrant IV reference angle: 45°  
2  
 2 
 
2   2  y  sin v  r  x  r 4  r  sec v  x 5  sec v  
4  sin v  r  y tan v  x 
3 3   tan v   
4 or 4 x cot v  y 
4 4   cot v   
3 or 3  
3  cos v 
34  3
34  sin v  
  cos v 34 r csc v  y 
34 
34    csc v   
3 or 
 3 x cot v  y 5 5   cot v   
3 or 
 3 r  
 x2  y2  sin v   35.  y  x  tan v  x  5  
34  5
34   34  tan v  5 or 
5  cos v  r    r  sec v   y  cos v  r  15  cos v  17 or 
17  r  sec v  x  17    sec v   
8 or 
 8  sin v  1 7 csc v  y csc v  1 5  145  tan v  x 8  r  17  
34   5  y  x  
8  3  sec v  x  r  
 (
8)2   152 r  
289  or 17 sin v  r  
3  15  15    tan v   
8 or 
 8 x  cot v  y 17  
8  8  cot v  15 or 
1 5  Chapter 5  36. r  
 x2  y2 r  
 52  ( 
6)2 r  
61   r  
 x2  y2 r  
 (
5)2   62 r  
61   y  sin v  r  
6  
61  6
61   sin v  
  61  6  
61   sin v  r  sin v    6
61  61  sin v  sin v   The sine of one angle is the negative of the sine of the other angle. 37. If sin v  0, y must be negative, so the terminal side is located in Quadrant III or IV 38. cos v  cos v   r2  12 
1 3    x2    sec v  sec v   y2  y  
5  
5  5  y  sin v   5    tan v   
12 or 12  r  sin v   r  csc v  y  sec v  x 13  13  csc v   13  13    csc v   
5 or 
 5    sec v   
12 or 
 12  csc v   r  cot v  y 
12  csc v   12    cot v   
5 or 5 r  39. csc v  y  sin v  sin v   y  r 1  2  x  cos v  r cos v    

3  2  or  
3  
 2  tan v   y  x  tan v   1  

3  
3   
 3  tan v  sec v   r  x  cot v   sec v   2  

3  2
3   
 3 y  r 1 
5  cot v   sec v  40. sin v  sin v   y  
1, r  5  x  cos v  r cos v   2
6   5  sec v   r  x  sec v   5  2
6  5
6   12  Chapter 5  52  tan v        (
1)2  sin v  sin v  csc v   tan v   y2  csc v   or 

2   cot v  cot v   1  

2  
2   
 2  r2  (
1)2  (
1)2 r2  2 r  
2  y  x  cos v  r  
1  
2   
2  
 2 r  y 
2   or 
1  cos v  cos v   tan v  x  
1  
2  
2   
 2  
1   tan v   
1 or 1  r  sec v  x 

2   sec v   
2   
1  or 

2   r 2  x2  y2  y  22  x2  (
1)2 3  x2 
3 x Quadrant III, so x  

3   tan v  x tan v   r  5  
1  

2   1  x  r  tan v   
1  

3  
3   3  45. g sin v cos v  0 sin v  0 or cos v  0 v  0° v  90° 46a. k is an even integer.  or 
5  x  cot v  y cot v   1  
3  
3   3  cot v  y  44. csc v  y  csc v  y  2
6   
1  cos v   csc v  
2 r  2, y  
1  x2  
1  2
6  
6  
 12  cos v   r 2  x2  y2  y  or 

3     y  

2   
3   
6 
 3 r  y 
3   

2   
6  
 2  sin v  r  csc v   x2  tan v  x  tan v  x  cot v  1 (Quadrant III) x  
1, y  
1  24  x2 2
6 x Quadrant IV, so x  2
6   tan v   sec v   r2  x  y 

3   1  y  x  cos v  r  43. cot v  y  22  x2  12 3  x2 
3 x Quadrant II, so x  

3   
5   2  (
3 )2  12  y2 2  y2 
2 y Quadrant IV, so y  

2   x  r 2  x2  y2  csc v  2 r  2, y  1  
5   csc v   r 2  x2  y2  sin v  r  y  tan v  x  sin v  13 or 
1 3  1  
5  
5  cos v   5 x cot v  y 1 cot v  2  sec v  
3  r  
3 , x  1  25  5  y Quadrant III, so y  
5  r  csc v  y  cos v   r  y2  sin v  r  2  
5  2
5   5 r  x 
5   or 1  42. sec v  x  132  (
12)2  y2  x  
12, r  13  r2  12  22 r2  5 r  
5  x cos v  r  y  sin v   x  r  r 2  x2  y2  tan v  2 y  2, x  1  y  sin v  r sin v   y  41. tan v  x  47. cos v   or 
2
6   cos v  
1  cos v  1 v  0°  146  46b. k is an odd integer.  I t Io  It  Io  48. Let x  
1. y  
3(
1) y3 r2  x 2  y 2 r2  (
1)2  (3)2 r2  10 r  
10  sin v   y  r  3 sin v   
10  3
10  sin v   10 r csc v  y  
10 csc v   3 x cot v  y 
1 cot v  3 or  x cos v  r 
1 cos v   
10  
10  cos v  
  10 r sec v  x 
10  sec v   
1 or  tan v  tan v   y  x 3  
1  56. 

10   C  sin v  sin v  51.  
840  360    9n 
 212  
4 2  n  
3  34, 
23, 12 2x 
 4y  7 ? 2(9) 
 4(3)  7 6  7; yes 2x 
 4y  7 ? 2(2) 
 4(
2)  7 12  7; no  2x 
 4y  7 ? 2(
1) 
 4(2)  7 
10  7; yes  1  59. absolute value; f(x)  22 
 x    360(
2)°  
840°  60. A of square 
 A of circle  A s2 
 r2  A 2 2 
 (1)2  A 0.86  A The correct choice is C.   
 720°  
840°   
120°  5-4  
b  
 b2 
 4 ac  2a  9  
 (
9)2  
 4(4 )(5)  2(4) 9  
1  x  8 91 9
1 x  8 or x  8 10 8 x  8 or 1.25 x  8 or 1 y y  kx k  x 9  y  (
0.6)(21) k 
15  k  
0.6  4  1  x  54.   
 11p  23  4m  9n 
 2p  
4  3  4  p  2  58.  360° 
 120°  240°; III 52. 5 
 
b 20  5  
b 2  25  b  2 23  b 53. x   38  3  4  1  csc v 1  7  5 5  7   
2.33  2 3  1  2(2) 
 (
3)1 2 7 
1 2  38m 
 11p  23  4  50. sin v   5x  
15 
10 
5 O 
2  3  Since AC  36, AD  18. 18  4  22 ft 49c. Refer to 49b for diagram and reasoning. Since AC  30, AD  15. 15 4  19 ft 1 r 2  2  m  4  4  49d.  4  57. 3(8m 
 3n 
 4p)  3(6) 24m 
 9n 
 12p  18 → 4m  9n 
 2p  
4 4m  9n 
 2p  
4 28m 
 14p  14 4(8m 
 3n 
 4p)  4(6) 32m 
 12n 
 16p  24 6m  12n  5p  
1 → 6m  12n  5p  
1 38m 
 11p  23 11(28m 
 14p)  11(14) 
14(38m 
 11p)  
14(23) ↓ 308m 
 154p  154 
532m  154p  
322 
224m  
168  36  D  2 
3 1  7  60˚  A  f (x) 
 x 2  16  f (x)  
6  ABC is equilateral. m∠BCA  60° m∠ACD  m∠BCA  90° m∠ACD  60°  90° m∠ACD  30°  300˚  6  
4  1  B  f(x)  x2 
 16 y  x2 
 16 x  y2 
 16 x  16  y2 
x   16  y  or 
3  
3  49a. 4  2(36)  76 ft 49b.  36  55.  Applying Trigonometric Functions  Pages 301–302  Check for Understanding  1a. cos or sec 1b. tan or cot 1c. sin or csc A 2. Sample answer: Find a. 10  C  y  
12.6  147  a  38˚  B  Chapter 5  3. ∠DCB; ∠ABC; the measures are equal; if parallel lines are cut by a transversal, the alternate interior angles are congruent. 4. Sample answer: If you know the angle of elevation of the sun at noon on a particular day, you can measure the length of the shadow of the building at noon on that day. The height of the building equals the length of the shadow times the tangent of the angle of elevation of the sun. 5.  a  tan A  b  10  tan 49° 13
  —a— a tan 49° 13
  10 10 — a— tan 49° 13
 a  8.6 a  — sin 16° 55
  — 13.7  b  a  13.7 sin 16° 55
  a 4.0  a  18   tan 76°   113  sin 26°  c c sin 26°  18  52.1  a  c  cos B   7.  cos 16° 45
   22.3  cos 47° 18
  —c—  8b. Let x   of the base.  cos 55° 30
   x  sin 55° 30
  10  c  32.9  10 sin 55° 30
  x 8.2  x about 8.2 cm 8c.  x  10  10 cos 55° 30
  x 5.66  x base  2x base  2(5.66) base  11.3 cm 9.  c cos 47° 18
  22.3 22.3 — c— cos 47° 18
  8a. Let x  altitude.  13 cos 16° 45
  a 12.4  a 1  2  a  cos B  —c—  18.  18  sin 26°  c  41.1 a  c a —— 13  a  sin A  —c—  17.  sin B  c  6.  13 tan 76°  a  b  tan B  —a—  16.  h  1 A  —2—bh 1 A  —2—(11.3)(8.2) A  46.7 cm2  n  cos 30°  —12—  sin 30°  —12—  19.  12 sin 30°  h 6h  12 cos 30°  n 10.4  n  6  sin 45°  —p—  m tan 45°  6  p sin 45°  6  6  tan 45°  m m  6  tan 45°  6   p sin 45°  m6 20a.  p  8.5  10.8  cos 36°  —x— x cos 36°  10.8  175  tan 13° 15
  —x—  10.8  — x— cos 36°  x tan 13° 15
  175 175  x tan 13° 15
  x  13.3 cm  x  743.2 ft  tan 36°   20b.  1  s 2  10.8 1  Pages 302–304 10.  Exercises a  tan A  —b—  11.  a  cos 67°  —16—  6 tan 37°  a 4.5  a b  sin B  —c— sin 62°   a  cos B  —c— a  tan 37°  6  12.  10.8 tan 36°  2 s 2 10.8 tan 36°  s 15.7  s about 15.7 cm 20c. P  5s P  5(15.7) P  78.5 cm  16 cos 67°  a 6.3  a 13.  b —— 24  a  sin A  —c— sin 29°   24 sin 62°  b 21.2  b  4.6 —— c  21a.  cos 42° 30
   1  c sin 29°  4.6 4.6 — c— sin 29°  x cos 42° 30
  —2—(14.6)  c  9.5 14.  a  cos B  —c— 17.3  15.  c cos 77°  17.3  a tan 61°  33.2 — a— tan 61°  c  76.9  a  18.4  tan 42° 30
   21b.  33.2  — c— cos 77°  1 ——(14.6) 2  x  1 (14.6) 2  tan 42° 30
  x 6.7  x  about 6.7 m  Chapter 5  1 (14.6) 2  cos 42° 30
  x  9.9 m  33.2  tan 61°  —a—  17.3  x  b  tan B  —a—  cos 77°  —c—  1 (14.6) 2  x  148  1  1  21c. A  —2—bh  28. Let M represent the point of intersection of the altitude and E F . Since GEF is isosceles, the altitude bisects  EF . EMG is a right triangle. a a — Therefore, sin v  —s— or s sin v  a and tan v  — 0.5b or 0.5b tan v  a. 29. Latasha: Markisha:  22a. r  —2—(6.4) or 3.2  1  a  A  —2—(14.6)(6.7)  — cos 30°  — 3.2  3.2 cos 30°  a 2.771281292  a about 2.8 cm 22b. Let x  side of hexagon. 22c. P  6s 1 P  6(3.2) x 2 P  19.2 cm  sin 30°  3.2 A  48.8 m2  x  1  2 3.2 sin 30°  x 3.2  x; 32 cm 1  22d. A  —2—pa 1  y tan 47° 30
  x  A  —2—(19.2)(2.771281292)  y  A  26.6 cm2 195.8 sin 10° 21
 36  —x— x sin 10° 21
 36  195.8 195.8 — x— sin 10° 21
 36 1  V  —3— area of base height  tan   1 ——s 2  x  1 s 2  1  1  V  —6— s3 tan   tan   x   
3 
—12—,  2   84 ft 8 ft 25b. 84 
 8  76 76  x  87.8 ft  sec 120°   csc 120°   csc 120°  cot 120°   1  1 
 2  cot 120°   sec 120°  
2  x tan 6°  3900 3900 —— tan 6°  32.  x  37,106.0 ft 3900  sin 6°  —x— x sin 6°  3900 3900  — x— sin 6°  sin P   Barge:  tan 20°  —x—  tan 12° 30
  —x—  tan P   x tan 20°  208  x tan 12° 30
  208  tan P   208  208 —— tan 20°  x  571.5 938.2 
 571.5  366.8 ft; no  x  cot 120°     72  22  (PQ)2 53  (PQ)2 
53   PQ P — — sin P  r (PR)2  sin P   x  37,310.4 ft  x  1  csc 120°  y  1  3900  208  1  sec 120°  —x—  tan 6°  —x—  27. Yacht:  
3   2 y  x 
3   2  1 
 2  tan 120°  

3   x  43.9 ft  26b.  cos 120°  
—2—  tan 120°   76  x sin 60°  76 76 — x— sin 60°  x tan 60°  76 76 — x— tan 60°  cos 120°  x  sin 120°   sin 60°  —x—  25c.  tan 60°  —x—  sin 120°  y tan 120°   60˚  x  40  x  — — tan 54° 54
  40 tan 47° 30
  25a.  26a.  40  x  — y— tan 54° 54
  x  ——— tan 54° 54
 
 tan 47° 30
 x  131.7 ft 31. terminal side — Quadrant II reference angle: 180° 
 120°  60°  V  —3— (s2)—2— s tan  1  y tan 54° 54
  40  x  x —— tan 47° 30
 x  tan 47° 30
  tan 54° 54
(x)  tan 47° 30
(40  x) x tan 54° 54
  40 tan 47° 30
  x tan 47° 30
 x(tan 54° 54
 
 tan 47° 30
)  40 tan 47° 30
  x  1088.8 ft 24. height:  — sin 42°  — 225  250 sin 35°  x 225 sin 42°  x 143.4  x 150.6  x 1506 
 143.4  7.2 Markisha's; about 7.2 ft 30. Let x  the height of the building. Let y  the distance between the buildings. x 40  x tan 47° 30
  —y— tan 54° 54
  y  32 sin 30°  —2—x  23.  x  — sin 35°  — 250  (RQ)2  2  
53  2
53   53 P —— q 2 —— 7  1  
3   2 2
3   3 x —— y 1 
 2  
 3  2 
3  
 3  (PQ)2  q  cos P  —r— cos P  cos P   7  
53  7
53   53    33. 43° 15
 35  43°  15° 60
   35 3600   43.260° 1°  208 —— tan 12° 30
  x  938.2  149  1°  Chapter 5  4. r  
 x2  y2 r  
 22  ( 
5)2 r  
29   y  34.  y  sin v  —r—  y  |x  2|  sin v   x  O  36.  m miles —— h hours  
5  
29  5
29   sin v  
  29  tan v  x  cos v   2  
29   tan v  —2— or 
2  cos v   2
29   29  r  35. Let x  the cost of notebooks and y  the cost of pencils. 3x  2y  5.89 4x  y  6.20 3x  2y  5.80 → 3x  2y  5.80 
2(4x  y)  
2(6.20) 
8x 
 2y  
12.40 
5x  
6.60 4x  y  6.20 x  $1.32 4(1.32)  y  6.20 y  $0.92  
5  5  r  csc v  —y—  sec v  x  
29 
29    or 
  5 
5 x cot v  —y— 2 2 —  cot v  — 
5 or 
 5 550 tan 27.8°  —x—  csc v   5.  y  x  cos v  r  sec v   
29   2  x tan 27.8°  550 550   x tan 27.8°  mx  x  1043.2 ft  x hours  —h— miles  The correct choice is E.  Page 304  1. 34.605 °  34°  (0.605 60)
  34°  36.3
  34°  36
  (0.3 60)  34°  36
  18 34° 36
 18 2.  
400° —— 360°  Pages 308–309  
1.11  1  
0.11 
0.11 360°  
40° 360° 
 40°  320°; IV 3. (GI )2  (IH )2  (GH )2 (GI )2  102  122 (GI )2  44 GI  
44  or 2
11   5. 60°, 300°  cos G  csc G   10  2
11  5
11   tan G  11 hypotenuse — sec G  — side adjacent 12 sec G   2
11   csc G   tan G   Chapter 5  1  7.  side adjacent  side opposite   
11  2
11  or  12 6 hypotenuse —— side opposite 12 6  or —— 10 5  8.  side adjacent 2
11   10  or  6. 150°, 330°  
3  sin sin
1  2  
3  
3  Let A  sin
1  2 . Then sin A  2 .  
3  
3 sin sin
1    2 2 3 3 Let A  cos
1 —5—. Then cos A  —5— r 2  x2  y2          52  32  y2 16  y2 4y 4 3 4 tan A  —3—; tan cos
1 —5—  —3—  — cot G  — side opposite  cot G   30˚  — not — cos .  10  5  1b. angle  4. Marta; they need to find the inverse of the cosine,  — cos G  — hypotenuse  — tan G  — side adjacent  sec G   60˚  side opposite   sin G  —1— 2 or 6  Check for Understanding  1a. linear 2. They are complementary. 3. Sample answer:   
1.11  — sin G  — hypotenuse  Solving Right Triangles  5-5  Mid-Chapter Quiz  
11   5  6
11   11  150  r  r  9. tan R  —s—  4  10. cos S  —t—  7  cos arccos —5—  —5—  12  tan R  —1— 0  4  cos S  —20— 7  12  R  tan
1 —1— 0 R  35.0° 11. A  78°  90° A  12° Find b.  S  cos
1 —20—  2  tan tan
1 —3—  —3— 2  Find c.  — sec A  — cosA  41  sec A   cos 78°  —c—  41 tan 78°  b  c cos 78°  41  sec A   41  192.9  b  — c— cos 78°  2  A  12°, b  192.9, c  197.2 a2  b2  c2 Find B.  tan B   
562 c  tan B   1  — csc A  — sin A  b —— a 21 —— 11  1  csc A  —1— or 1 csc (arcsin 1)  1 21  5  r 2  x2  y2 132  52  y2 144  y2 12  y  Find A. A  62.35402464  90 A  27.64597536 c  23.7, A  27.6°, B  62.4° 13. 3.2°  B  90° B  58° Find a. Find b.  —— tan A  —5—; tancos
1 —1— 3  5 12  2  cos A  —c—  a  cos 32°  —1— 3  b   
21 ; 5 n  —m— 15  —9—  cos A  28. tan N  1280  — 14a. tan x  — 2100  tan N 1280  — x  tan
1 — 2100  cos sin
1 —5—  2  1280  tan 38°  —x—  m 8  sin M  —1— 4 M  sin
1 —14—  N  59.0°  M  34.8°  cos M   1280 —— tan 38°  n —— p 22 —— 30  n  31. tan N  —m— 18.8  — tan N  — 14.3 22  M  cos
1 —30—  x  1638.3 ft  M  42.8° m 32. cos N  —p—  1280  tan 65°  —x— x tan 65°  1280  32.5  7.2  — sin M  — 54.7 7.2  x  596.9 ft  Exercises  — M  sin
1 — 54.7  N  65.1°  M  36.5°  18 —— 24  24  tan B  1 8 18  151  32.5  — N  cos
1 — 17.1  34. tan A  90° 16. 120°, 300° 30°, 330° 18. 90°, 270° 225°, 315° 20. 135°, 315° Sample answers: 30°, 150°, 390°, 510°  18.8  — N  tan
1 — 14.3  N  52.7° m 33. sin M  —p—  — cos N  — 17.1  1280  — x— tan 65°  8  N  tan
1 —9— 30. cos M   x tan 38°  1280  
21   5  29. sin M  —p— 15  x  31.4°  15. 17. 19. 21.  2  r 2  x2  y2 52  x2  22 21  x2 
21 x  b  13 sin 32°  a 13 cos 32°  b 6.9  a 11.0  b B  58°, a  6.9, b  11.0  Pages 309–312  12  5  27. Let A  sin
1 —5—. Then sin A  —5—  a  sin 32°  —13—  5  —— 26. Let A  cos
1 —1— 3 . Then cos A  13  B  62.4°  sin A  —c—  5  25. Let A  arcsin 1. Then sin A  1.  B  tan
1 —11—  23.7  c  1  2  5 5 —— 2  sec cos
1 —5—  —2—  c  197.2  112  212  c2  2  1  a  b  tan 78°  —4— 1  14c.  2  2  cos B  —c—  x  2  24. Let A cos
1 —5—. Then cos A  —5—.  b  14b.  4  23. Let A  tan
1 —3—. Then tan A  —3—.  S  53.1°  tan B  —a—  12.  4  22. Let A  arccos —5—. Then cos A  —5—.  24  A  tan
1 —24—  B  tan
1 —18—  A  36.9°  B  53.1°  Chapter 5  42. A  33°  90° A  57°  1  35. —2—(14)  7 vertex angle  2m∠B  base angles: tan A   8 —— 7  b  8  b  B  tan
1 —8—  A  48.8°  15.2 sin 33°  b 8.3  b 43. 14°  B  90° B  76°  B  41.2° 2m∠B  82.4°  about 48.8°, 48.8°, and 82.4° 36. a2  b2  c2 212  b2  302 b  
459  b  21.4 44.427004  B  90 B  45.6° 37. 35°  B  90° B  55°  9.8 sin 14°  a 2.4  a 647  — v  sin
1 — 1020  v  39.4° 44b.  c cos 35°  8 8 — c— cos 35°  647  — x— tan 39.4°  x  788.5 ft 45a. Since the sine function is the side opposite divided by the hypotenuse, the sine cannot be greater than 1. 45b. Since the secant function is the hypotenuse divided by the side opposite, the secant cannot be between 1 and 
1. 45c. Since cosine function is the side adjacent divided by the hypotenuse, the cosine cannot be less than 
1. 46. 10 
 6  4  b  sin B  —c— 12.5  tan 47°  —a—  sin 47°  —c—  a tan 47°  12.5  c sin 47°  12.5  a  11.7 a2  b2  c2  12.5  — c— sim 47°  c  17.1 a  tan A  —b— 3.8  3.82  4.22  c2  — tan A  — 4.2  4  tan v  —1— 5  3.8  — A  tan
1 — 4.2  
32.08 c 5.7  c 42.13759477  B  90 B  47.9°  4  v  tan
1 —1— 5  A  42.1°  v  14.9° 8  a2  b2  c2 sin B  a2  3.72  9.52 sin B  a  
76.56  3.7 — a  8.7 B  sin
1 — 9.5 A  22.92175446  90 B  22.9° A  67.1° 41. 51.5°  B  90° B  38.5° tan A  tan 51.5°   a —— b 13.3 —— b  b tan 51.5°  13.3 13.3  Chapter 5  sin A  sin 51.5°   8  c  17.0  v  4.6°  v  2.9°  45 —— 2200 45  v  1.2° 49.  65 miles —— hour  tan v  tan v  v  13.3  b  10.6  — v  tan
1 — 100  — v  tan
1 — 2200  c sin 51.5°  13.3 — c— sin 51.5°  5  — v  tan
1 — 100  48. tan v   a —— c 13.3 —— c  — b— tan 51.5°  5  — 47b. tan v  — 100  — 47a. tan v  — 100  b —— c 3.7 —— 9.5  40.  647  tan 39.4°  —x— x tan 39.4°  647  c  9.8  12.5 —— tan 47°  9.8 cos 14°  b 9.5  b  647  8  12.5  b  — cos 14°  — 9.8  — 44a. sin v  — 1020  cos 35°  —c—  a  a  A  44.4°  a  b  b  cos A  —c—  — sin 14°  — 9.8  21  A  sin
1 —30—  b  tan B  —a—  15.2 cos 33°  a 12.7  a  a  cos A  —c—  38. A  47°  90° A  43°  a  — cos 33°  — 15.2  sin A  —c—  21  sin A  —30—  a  8 tan 35°  a 5.6  a  39.  a  sin A  c  tan A  —b— tan 35°  —8—  cos B  —c—  — sin 33°  — 15.2  7  A  tan
1 —7—  a  sin B  —c—  7  tan B  —8—  5280 feet 1 hour —— —— mile 3600 seconds v2 —— gr 95.32 —— 32 (1200) 95.32 — tan
1 — 32(1200)  v  13.3°  152  feet  —  95.3 — second  50.  sin v ——i sin vr sin 60° —— sin vr sin 60° —— 2.42   2.42 y-axis   sin vr  0.3579  sin vr sin
1 0.3579  vr 21.0°  vr 51. Draw the altitude from Y to XZ. Call the point of intersection W. m∠X  m∠XYW  90° 30°  m∠XYW  90° m∠XYW  60° In XYW: XW WY cos 30°  —1— sin 30°  —16— 6 16 cos 30°  XW 16 sin 30°  WY 13.9  XW 8  WY In ZYW: 8  yx y  
x  y-axis 56. 
1 0 0 1  — tan 19.5°  — WZ 8  Z  sin
1 —2— 4  8  — WZ  — tan 19.5°  WZ  22.6 cos m∠WYZ   8  m∠WYZ  cos
1 —2— 4 m∠WYZ  70.5° Y  m∠XYW  m∠WYZ y  XW  WZ Y  60°  70.5° y  13.9  22.6 Y  130.5° y  36.5 52. baseball stadium: football stadium: 1000  x tan 63°  1000 1000 — x— tan 63°  y tan 18°  1000 1000 — y— tan 18°  x  509.5 distance  x  y distance  509.5  3077.7 distance  3587.2 ft 53. (FD)2  (DE)2  (FE)2 72  (DE)2  152 (DE)2  176 DE  
176  or 4
11  side opposite  — sin F  — hypotenuse  tan F  tan F  sec F  sec F   4
11   15 side opposite —— side adjacent 4
11   7 hypotenuse —— side adjacent 15  7  22.2 
 42.5  
20.3  m  —7— 0 or 
0.29  y  3077.7  y 
 22.2  
0.29(x 
 1950) y  
0.29x  587.7 59. 2x  5y 
 10  0 5y  
2x  10 2  y  
—5—x  2 2  
—5—; 2 side adjacent  csc F  csc F  cot F  cot F   
5 
5 
3 
1 
2 
3 4 6 3 
2   58. m   1950 
 1880  — cos F  — hypotenuse  cos F   no  4  (
2) 
3  2 2  (
2)  8  (
5) 
2  1 01 9  (
7) 6  2 
3  (
2) 2 
1 0  3 
1 1 2 8 
5  1000  tan 18°  —y—  no  (5, 
3), (5, 4), (3, 6), (1, 3), (2, 
2) 4 
3 2 
2 2 
2 57. 8 
2 0  
5 1 1 9 6 
3 
7 2 
2  8 —— 24  tan 63°  —x—  yes  
1(
2)  0(
2) 0(
2)  1(
2) 5 5 3 1 2  
3 4 6 3 
2  WZ tan 19.5°  8  Z  19.5°  no   
1(
5)  0(
3) 
1(
5)  0(4) 0(
5)  1(
3) 0(
5)  1(4) 
1(
3)  0(6) 
1(
1)  0(3) 0(
3)  1(6) 0(
1)  1(3)  8  sin Z  —2— 4  sin F   y3 
 x2  2 (
y)3 
 x2  2 
y3 
 x2  2; y 3 
 x2  2 3 y 
 (
x)2  2 y3 
 x2  2; y 3 
 x2  2 3 (x) 
 (y)2  2 x3 
 y2  2; y3 
 x2  2 (
x)3 
 (
y)2  2 
x3 
 y2  2;  55. x-axis  n  60.  7  15 hypotenuse —— side opposite 15 15
11   or  4
11  44 side adjacent —— side opposite 7 7
11   or  4
11  44  a —— ac  ab  cd  c  — —— b— a  c d  10  a  c  10  The correct choice is A.  The Law of Sines  5-6 Page 316  54. Use TABLE feature of a graphing calculator. 
0.3, 1.4, 4.3  1.  x —— sin 30° x  1  2  Check for Understanding    2x   153  x
3   sin 60°    2x  sin 90°   x
3  
 3  2    2x  1  2x    2x  60˚  x  2x  x 
3  30˚  Chapter 5  Pages 316–318  2. Sample answer:  A b  C  b —— sin B b —— sin 70°  c  35˚  40˚ 10  B  1  a —— sin A a —— sin 30°  1  — K— 2ab sin X  b —— sin B b —— sin 59°  14  — — sin 81° 14 sin 40°  a —— sin A a —— sin 25°  8.6  8.6 sin 55.9°  — b— sin 27.3°  a  — — sin A  8.6 sin 96.8°  — c— sin 27.3°  8.2  — — sin 93.9°  a —— sin 39
 51
  19.3  — — sin 64° 45
 19.3 sin 39° 15
  — a— sin 64° 45
  78°  sin A sin C  sin 22° sin 53°  — K  —2—(14)2 — sin 105°  b —— sin 51° 30
  K  30.4 units2 10. Let d  the distance from the fan to the pitcher's mound. v  180° 
 (24° 12
  5° 42
) or 150° 6
  c  18.11843058  b —— sin B b —— sin 50°  c  — — sin C 12  — — sin 65° 12 sin 50°  — b— sin 65°  c —— sin 61.3°  8.2  — — sin 93.9° 8.2 sin 61.3°  — c— sin 93.9°  b —— sin 76°  19.3  — — sin 64° 45
 19.3 sin 76°  — b— sin 64° 45
  125  — — sin 91° 10
 125 sin 51° 31
  — b— sin 91° 10
  b  97.8 18. A  180° 
 (29° 34
  23° 48
) or 126° 38
 a b ——  —— sin A sin B  60.5  — — sin 5° 42
  a —— sin 126° 38
  60.5 sin 150° 6
  — d— sin 5° 42
  11  — — sin 29 ° 34
 11 sin 126° 38
  — a— sin 29° 34
  d  303.7 ft  a  17.9  Chapter 5  12 sin 120°  a  13.50118124 b  20.7049599 B  76°, a  13.5, b  20.7 17. C  180° 
 (37° 20
  51° 30
) or 91° 10
 a c ——  —— sin B sin C   K  —2— b2  sin B  d —— sin 150° 6
  12  — — sin 35°  b  3.447490503 c  7.209293255 A  93.9°, b  3.4, c  7.2 16. B  180° 
 (39° 15
  64° 45
) or 76° a c b c ——  —— ——  —— sin A sin C sin B sin C  17  — — sin 63° 50
  K 9. C  180° 
 (22°  105°) or 53° 1  b  — — sin B  — c— sin 35°  8.2 sin 24.8°  c  18.7  1  12  — b— sin 93.9°  a  17 sin 98° 15
  K  30 sin 100°  c —— sin C c —— sin 120°  — — sin 35°  b —— sin 24.8°  — c— sin 63° 50
  8. K   30  — — sin 50
  a  12 b  10.14283828 C  65°, a  12, b  10.1 15. A  180° 
 (24.8°  61.3°) or 93.9° b a c a ——  —— ——  —— sin B sin A sin C sin A  8.6  — — sin 27.3°  — — sin A  1 —— bc sin A 2 1 ——(14)(12) sin 2 82.2 units2  b  — — sin B  a c ——  —— sin A sin C a 12 ——  —— sin 65° sin 65° 12 sin 65° — a— sin 65°  b  15.52671055 c  18.61879792 C  96.8°, b  15.5, c  18.6 7. A  180° 
 (17° 55
  98° 15
) or 63° 50
 c —— sin C c —— sin 98° 15
  c  — — sin C  — b— sin 50°  a  8.84174945 C  120°, a  8.8, c  18.1 14. C  180° 
 (65°  50°) or 65°  14  — — sin 81°  c —— sin C c —— sin 96.8°  — — sin 27.3°  30  — — sin 50°  12 sin 25°  a  9.111200533 b  12.14992798 C  81°, a  9.1, b  12.1 6. C  180° 
 (27.3°  55.9°) or 96.8° a  b —— sin B b —— sin 100°  — a— sin 35°  c  — — sin C  — b— sin 81°  — — sin A  c  — — sin C  a  19.58110934 b  38.56725658 A  30°, a  19.6, b  38.6 13. C  180° 
 (25°  35°) or 120°  14 sin 59°  — a— sin 81°  b —— sin B b —— sin 55.9°  20 sin 70°  30 sin 30°  K  ab sin X 4. Both; if the measures of two angles and a nonincluded side are known or if the measures of two angles and the included side are known, the triangle is unqiue. 5. C  180° 
 (40°  59°) or 81° c  20  — — sin 40°  — c— sin 40°  — a— sin 50°  1  K  —2—ab sin X  2ab sin X  — — sin C  20  — — sin 40°  a  — — sin A  b  29.238044 c  29.238044 B  70°, b  29.2, c  29.2 12. A  180° 
 (100°  50°) or 30°  k  —2—ab sin X  1  c —— sin C c —— sin 70°  20 sin 70°  1  — K— 2ab sin Z  a  — — sin A  — b— sin 40°  3. Area of WXYZ  Area of triangle ZWY  Area of triangle XYW. m∠X  m∠Z triangle ZWY: triangle XYW:  a —— sin A a —— sin 40°  Exercises  11. B  180° 
 (40°  70°) or 70°  154  28c. P  112.7  72.7  80 P  265.4 ft  1  19. K  —2— bc sin A K  1 ——(14)(9) 2  sin 28°  m  1  r s n sin N ——  ——. Thus sin M  —— and sin R sin R sin S n r sin S ——. Since ∠M  ∠R, sin M  sin R and s m sin N r sin S ——  ——. However, ∠N  ∠S and n s m r m n sin N  sin S, so —n—  —s— and —r—  —s—. Similar  sin B sin C  — K  —2—a2 — sin A 1  sin 37° sin 84°  — K  —2—(5)2 — sin 59°  K  8.7 units2 21. C  180° 
 (15°  113°) or 52° 1  MNP  RST. 30. 360° 5  72°  sin 15° s in 52°  — K  —2—(7)2 — sin 113°  K  5.4 units2 22. K  K K  triangle:  1 ——bc sin A 2 1 ——(146.2)(209.3) sin 2 13,533.9 units2  62.2°  1 1  K  —2—(12.7)(5.8) sin 42.8° K  25.0 units2 24. B  180° 
 (53.8°  65.4°) or 60.8°  4 sin 62° 30
  — x— sin 97° 15
  x  3.6 mi 31b. v  180° 
 (20° 15
  62° 30
) or 97° 15
 Let y  the distance from the balloon to the football field. 4 4 ——  —— sin 20° 15
 sin 97° 15
  sin B sin C  — K  —2—a2 — sin A  K  1 sin 60.8° sin 65.4° — (19.2)2 ——— 2 sin 53.8° 181.3 units2  K 25. K  ab sin X (formula from Exercise 3) K  (14)(20) sin 57° K  234.8 cm2 26. Area of pentagon  5 Area of triangle 360° 5  72° 9  K  72˚  1 ——(9)(9) 2  K  38.51778891 27.  5  4 sin 20° 15
  — y— sin 97° 15
  y  1.4 mi 32. 180° 
 30°  150° v  180° 
 (26.8°  150°) or 3.2° Let x  the length of the track. x —— sin 26.8°  9  sin 72°  45˚  1  K  —2—(300)(300)sin 72°  K  42,797.54323 pentagon: 5K  5(42,797.54323) 5K  213,987.7 ft 31a. v  180° 
 (20° 15
  62° 30
) or 97° 15
 Let x  the distance from the balloon to the soccer fields. x 4 ——  —— sin 62° 30
 sin 97° 15
  23. K  —2—ac sin B  1    proportions can be derived for p and t. Therefore,  sin A sin C  — K  —2— b2 — sin B 1  n  — —— 29. Applying the Law of Sines, — sin M  sin N and  K  29.6 units2 20. A  180° 
 (37°  84°) or 59°  100  — — sin 3.2° 100 sin 26.8°  — x— sin 3.2°  x  807.7 ft 33a. Let x  the distance of the second part of the flight. v  180° 
 (13°  160°) or 7°  5K  5(38.51778891) 5K  192.6 in2 Area of octagon  8 Area of triangle 360° 8  45°  x —— sin 13°  80  — — sin 7° 80 sin 13°  — x— sin 7°  x  147.6670329 distance of flight  80  147.7 or about 227.7 mi 33b. Let y  the distance of a direct flight.  5  y —— sin 160°  80  — — sin 7° 80 sin 160°  — y— sin 7° 1  K  —2—(5)(5) sin 45°  K  8.838834765 8K  70.7 28a. 180° 
 (95°  40°)  45° 28b.  x —— sin 95°  y  224.5 mi  8K  8(8.838834765)  80  — — sin 45° 80 sin 95°  — x— sin 45°  x  112.7065642 about 112.7 ft and 72.7 ft  y —— sin 40°  ft2 80  — — sin 45° 80 sin 40°  — y— sin 45°  y  72.72311643  155  Chapter 5  x  34.  55˚  y  90° 
 63°  27° 180° 
 (55°  63°)  62° Let x  the vertical distance. Let y  the length of the overhang.  63˚ 62˚  13 ft  sin v   cos v  13  y —— sin 63°  — — sin 63° 13 sin 27° —— sin 63°  35b.  a —— sin A a —— b a —— sin A a —— c  a —— c a —— c  
35   6  
1  
35  
35   tan v  
 35 x cot v  —y— 
35  cot v   
1  r  sec v   y  6.684288563  sec v   6  
35  6
35   35  y  sin A  — — sin B c  — — sin C sin A  — — sin C  8  sin A sin A  sin C  — —— 
 —c—  — sin C 
 sin C    sin A 
 sin C —— sin C a
c  O  sin A 
 sin C  — 35c. From Exercise 34b, —c—  — sin C sin A 
 sin C sin C —  ——. or — a
c c a c ——  —— sin A sin C a sin A ——  —— c sin C a sin A ——  1  ——  1 c sin C sin A a c sin C ——  ——  ——  —— sin C c c sin C ac sin A  sin C ——  —— c sin C sin C sin A  sin C ——  —— c ac sin A 
 sin C — Therefore, — a
c ac sin A  sin C — —— or — a 
 c  sin A 
 sin C . a —— sin A a —— b a ——  1 b a b ——  —— b b ab —— b b —— ab  or 

35   sin A  sin C  — — ac  b  — — sin B sin A  — — sin B sin A  — — sin B  1 sin B  sin A  (2, 11)  (8, 7) y
4  (2, 4)  (8, 4)  4  12  x 16  M(x, y)  100x  250y M(2, 4)  100(2)  250(4) or 1200 M(2, 11)  100(2)  250(11) or 2950 M(4, 11)  100(4)  250(11) or 3150 M(8, 7)  100(8)  250(7) or 2550 M(8, 4)  100(8)  250(4) or 1800 4 standard carts, 11 deluxe carts 40. 4x  y  2z  0 3x  4y 
 2z  20 7x  5y 
 2z  20 3(3x  4y 
 2z)  3(20) 2(
2x  5y  3z)  2(14) ↓ 9x  12y 
 6z  60 
4x  10y  6z  28 5x  22y  88 
5(7x  5y)  
5(20) → 
35x 
 25y  
100 7(5x  22y)  7(88) 35x  154y  616 129y  516 y4 7x  5y  20 4x  y  2z  0 7x  5(4)  20 4(0)  4  2z  0 x0 z  
2 (0, 4, 
2) 41. 
6  3x 
 y 3x 
 y  12 y  3x  6 y  3x 
 12  — —— — sin B  sin B sin A  sin B  — — sin B sin B  — — sin A  sin B 45  36. tan v  —20— 45  v  tan
1 —20—  y  v  66.0°  6  3x  y  12  O  Chapter 5  6  — csc v  — 
1 or 
6  38. 83°  360k° 39. Let x  standard carts and let y  deluxe carts. 2x8 y x
2 x
8 16 4  y  11 x  y 
 15 x  y  15 (4, 11) y 
 11 12  b  — — sin B  c  r  csc v  —y—  tan v   — 
1— sin C 
 1  a
c —— c  35d.  tan v  —x—  sec v  —x—  6.6 sin 63° —— sin 62°  35  x2 
35 x  Quadrant IV, so x  
35  x  66  — — sin 62°  y  x  6.623830843 about 6.7 ft 35a.  62  x2  (
1)2  cos v  —r—  63˚  x  r 2  x2  y2  1 
—6—  y  
1, r  6  27˚  x —— sin 27°  y  37. sin v  —r—  156  x  5. Since 44°  90°, consider Case I. a sin B  23 sin 44° a sin B  23 (0.6947) a sin B  15.97714252 12  16.0; 0 solutions 6. Since 17°  90°, consider Case I. a sin C  10 sin 17°  2.923717047 2.9  10  11; 1 solution  42. Area of one face of the small cube  12 or 1 in2. Surface area of the small cube  6 1 or 6 in2. Area of one face of large cube  22 or 4 in2. Surface area of large cube  6 4 or 24 in2. Surface area of all small cubes  8 6 or 48 in2. The difference in surface areas is 48 in2 
 24 in2 or 24 in2. The correct choice is A.  Page 319  c  sin C 11  sin 17°  History of Mathematics  1. See students' work; the sum is greater than 180°. In spherical geometry, the sum of the angles of a triangle can exceed 180°. 2. See students' work. Sample answer: Postulate 4 states that all right angles are equal to one another. 3. See students' work.  Page 323  10 sin 17°  A  sin
1 11 10 sin 17°  A  15.41404614 B  180° 
 (15.4°  17°) or about 147.6° c  sin C 11  sin 17°  b  20.16738057 A  15.4°, B  147.6°, b  20.2 7. Since 140°  90°, consider Case II. 3  10; no solutions 8. Since 38°  90°, consider Case I. b sin A  10 sin 38° b sin A  6.156614753 6.2  8  10; 2 solutions a  sin A 8  sin 38°  Check for Understanding  30˚  A 6  sin 30°  10    sin B  B  93.6˚  C  A  10  sin B 10 sin 30°  6 10 sin 30° sin
1 6      10    sin B 10 sin 38°  B  sin
1 8 10 sin 38°  B  50.31590502 180° 
   180° 
 50.3° or 129.7° Solution 1 C  180° 
 (50.3°  38°) or 91.7°  B 6  b    sin B  sin B  8  1. A triangle cannot exist if m∠A  90° and a  b sin A or if m∠A  90° and a  b. 2. B 123.6˚ 6 30˚ 26.4˚ 10  b    sin 147.6° 11 sin 147.6°  Graphing Calculator Exploration  56.4˚  b    sin B   b sin 17°  1. B  44.1°, C  23.9°, c  1.8 2. B  52.7°, C  76.3°, b  41.0; B  25.3°, C  103.7°, b  22.0 3. The answers are slightly different. 4. Answers will vary if rounded numbers are used to find some values.  Page 324  10    sin A  sin A  11  The Ambiguous Case for the Law of Sines  5-7  a    sin A  C  a  sin A 8  sin 38°  c    sin C c    sin 91.7° 8 sin 91.7°  180° 
   180° 
 56.4°  123.6° C  180° 
 (30°  123.6°)  26.4°   c sin 38°  c  12.98843472 Solution 2 C  180° 
 (129.7°  38°) or 12.3°  B  56.44269024 C  180° 
 (30°  56.4°)  93.6° 3. Step 1: Determine that there is one solution for the triangle. Step 2: Use the Law of Sines to solve for B. Step 3: Subtract the sum of 120 and B from 180 to find C. Step 4: Use the Law of Sines to solve for c. 4. Since 113°  90°, consider Case II. 15  8; 1 solution  a  sin A 8  sin 38°  c    sin C c    sin 12.3° 8 sin 12.3°   c sin 38°  c  2.768149638 B  50.3°, C  91.7°, c  13.0; B  129.7°, C  12.3°, c  2.8  157  Chapter 5  13. Since 61°  90°, consider Case I. a sin B  12 sin 61° a sin B  10.49543649 8  10.5; 0 solutions 14. two angles are given; 1 solution 15. Since 100°  90°, consider Case II. 15  18; 0 solutions 16. Since 37°  90°, consider Case I. a sin B  32 sin 37° a sin B  19.25808074 27 19.3; 2 solutions 17. Since 65°  90°, consider Case I. b sin A  57 sin 65° b sin A  51.65954386 55 51.7; 2 solutions 18. Since 150°  90°, consider Case II. 6  8; no solution 19. Since 58°  90°, consider Case I. b sin A  29 sin 58° b sin A  24.59339479 26 24.6; 2 solutions  9. Since 130°  90°, consider Case II. 17 5; 1 solution c  sin C 17  sin 130°  b    sin B 5    sin B 5 sin 130°  sin B  17 B  sin
117 5 sin 130°  B  13.02094264 A  180° 
 (13.0  130°) or 37.0° c  sin C 17  sin 130°  a    sin A a    sin 37.0° 17 sin 37.0°   a sin 130°  a  13.35543321 A  37.0°, B  13.0°, a  13.4 10a.  45 ft  70 ft  a  sin A 26  sin 58°  10˚  90° 
 10°  80° 180° 
 80°  100°  10b.  b    sin B 29    sin B 29 sin 58°  sin B  2 6 B  sin
12  6 29 sin 58°  45  70 100˚ 80˚  70  sin 100°  B  71.06720496 180° 
   180° 
 71.1° or 108.9° Solution 1 C  180° 
 (58°  71.1°) or 50.9°  x˚ 10˚    sin x  x  a  sin A 26  sin 58°  45  sin x 45 sin 100°  70 45 sin 100° sin
1 7 0    26 sin 50.9°  c  23.80359004 Solution 2 C  180° 
 (58°  108.9)° or 13.1°  x  39.3° v  180° 
 (100° 
 39.3°) or about 40.7° 45  c    sin 50.9°   c sin 58°    10c.  c    sin C  a  sin A 26  sin 58°  70  c    sin C c    sin 13.1° 26 sin 13.1°  100˚ 80˚ y  sin 40.7°  39.3˚ y   c sin 58°  c  6.931727606 B  71.1°, C  50.9°, c  23.8; B  108.9°, C  13.1°, c  6.9  10˚  70    sin 100° 70 sin 40.7°   y sin 100°  y  46.4 ft  Pages 324–326  Exercises  11. Since 57°  90°, consider Case I. b sin A  19 sin 57° b sin A  15.93474079 11  15.9; 0 solutions 12. Since 30°  90°, consider Case I. c sin A  26 sin 30° c sin A  13 13  13; 1 solution. Chapter 5  158  24. Since 36°  90°, consider Case I. c sin B  30 sin 36° c sin B  17.63355757 19 17.6; 2 solutions  20. Since 30°  90°, consider Case I. b sin A  8 sin 30° b sin A  4 4  4; 1 solution a  sin A 4  sin 30°  b  b  sin B 19  sin 36°    sin B 8    sin B 8 sin 30°  sin C  1 9  B  sin
14  C  sin
11  9  8 sin 30°  30 sin 36°  B  90 C  180° 
 (30° 
 90°) or 60°  C  68.1377773 180° 
   180° 
 68.1° or 111.9° Solution 1 A  180° 
 (36°  68.1°) or 75.9°  c    sin C c    sin 60°  C  b  sin B 19  sin 36°  4 sin 60°  sin 30°  C  6.92820323 B  90°; C  60°, c  6.9 21. Since 70°  90°, consider Case I. a sin C  25 sin 70° a sin C  23.49231552 24 23.5; 2 solutions c  sin C 24  sin 70°     sin A  A  a  sin A 25  sin A 25 sin 70°  24 25 sin 70° sin
1 2 4    a  31.34565276 Solution 2 A  180° 
 (36°  111.9°) or 32.1° b  sin B 19  sin 36°  a  17.1953669 A  75.9°, C  68.1°, a  31.3; A  32.1°, C  111.9°, a  17.2 25. Since 107.2°  90°, consider Case II. 17.2 12.2; 1 solution    a  sin A 17.2  sin 107.2°   C  sin
1 17.2 12.2 sin 107.2°  b  13.46081025 Solution 2 B  180° 
 (70°  101.8°) or 8.2°  C  42.65491459 B  180° 
 (107.2°  42.7°) or 30.1° a  sin A 17.2  sin 107.2°  b  sin B b  sin 8.2° 24 sin 8.2°  sin 70°  a  a    sin 40° 20 sin 40°   a sin 80°  b    sin B b    sin 30.1° 17.2 sin 30.1°   b sin 107.2°  b  9.042067456 B  30.1°, C  42.7°, b  9.0 26. Since 76°  90°, consider Case I. b sin A  20 sin 76° b sin A  19.40591453 5  19.4; no solution  b  3.640196918 A  78.2°, B  31.8°, b  13.5; A  101.8°, B  8.2°, b  3.6 22. C  180° 
 (40°  60°) or 80°   sin A  12.2    sin C   sin C   17.2  24 sin 31.8°  c  sin C 20  sin 80°  c    sin C  12.2 sin 107.2°  b    sin 31.8°  b  a    sin 32.1° 19 sin 32.1°  b    a    sin A   a sin 36    sin B    a    sin 75.9° 19 sin 75.9°   b sin 70°  c  sin C 24  sin 70°  a    sin A   a sin 36°  A  78.1941432 180° 
   180° 
 79.2° or 101.8° Solution 1 B  180° 
 (70°  79.2°) or 31.8° c  sin C 24  sin 70°  30    sin C 30 sin 36°  sin B  4  a  sin A 4  sin 30°  c    sin C  c  sin C 20°  sin 80°  b    sin B b    sin 60° 20 sin 60°   b sin 80°  a  13.05407289 b  17.58770483 C  80°, a  13.1, b  17.6 23. Since 90°  90°; consider Case II. 12  14; no solution  159  Chapter 5  27. Since 47°  90°, consider Case I. 16  10; 1 solution c  sin C 16  sin 47°  a  a  sin A 15  sin 29°  10  sin B  1 5  31.    sin A  A  B  sin
11  5 20 sin 29°  10 sin 47°  16 10 sin 47° 
1  sin 16    B  40.27168721 180° 
   180° 
 40.3° or 139.7° Solution 1 C  180° 
 (29° 
 40.3°) or 110.7°    A  27.19987995 B  180° 
 (47° 
 27.2) or 105.8° c  sin C 16  sin 47°  a  sin A 15  sin 29°  b    sin B b    sin 105.8°  c  28.93721187 Perimeter  a  b  c  15  20  28.9 or about 63.9 units Solution 2 C  180° 
 (29° 
 139.7°) or 11.3° a  sin A 15  sin 29°  c    sin C    sin C  C       a  c  6.047576406 Perimeter  a  b  c  15  20  6.0 or about 41.0 units   32.  15 sin 55°  A  70.93970395 180° 
   180° 
 70.9° or 109.1° Solution 1 C  180° 
 (70.9°  55°) or 54.1° b c    sin B sin C 13  sin 55°  a  c    sin 54.1° 13 sin 54.1°    sin 26.7°   c sin 55°  42 sin 26.7°  c  12.8489656 Perimeter  a  b  c  15  13  12.8 or about 40.8 Solution 2 c  180° 
 (109.1°  55°) or 15.9° b c    sin B sin C  a  29.33237132 A  73.3°, C  66.7°, a  62.6; A  26.7°, C  113.3°, a  29.3 29. Since 125.3°  90°, consider Case II. 32  40; no solution  13  sin 55°  19.3    sin x  13 sin 15.9°  c  4.35832749 Perimeter  a  b  c  15  13  4.4 or about 32.4 A  70.9°, B  55°, C  54.1°   x  sin
1 21.7 19.3 sin 57.4°  x  48.52786934 v  180 
 (57.4°  48.5°) or 74.1° y    sin 74.1° 21.7 sin 74.1°   y  sin 57.4°  c    sin 15.9°   c sin 55°  19.3 sin 57.4°   sin x   21.7  19.3 cm  74.1˚  21.7 cm  y  24.76922417 48.5˚ 57.4˚ 24.8 cm  Chapter 5  15    sin A  A  sin
113  a  21.7  sin 57.4°  a    sin A  15 sin 55°    sin A  21.7  sin 57.4°  b  sin B 13  sin 55°  sin A  1 3  a  sin A a  sin 73.3° 42 sin 73.3°  sin 40°   a sin 40°  30.  c    sin 11.3° 15 sin 11.3°  a  62.58450564 Solution 2 A  180° 
 (40°  113.3°) or 26.7° b  sin B 42  sin 40°  c    sin C   c sin 29 °  C  66.67417652 180° 
   180° 
 66.7° or 113.3° Solution 1 A  180° 
 (40°  66.7°) or 73.3° b  sin B 42  sin 40°  c    sin 110.7°   c sin 29 °  b  21,0506609 A  27.2°, B  105.8°, b  21.1 28. Since 40°  90°, consider Case I. c sin B  60 sin 40° c sin B  38.56725658 42 38.6; 2 solutions 60  sin C 60 sin 40°  42 60 sin 40° sin
1 4 2  c    sin C  15 sin 110.7°  16 sin 105.8°   b sin 47°  b  sin B 42  sin 40°  20    sin B 20 sin 29°    sin A  sin A   b    sin B  160  37. Distance from satellite to center of earth is 3960  1240 or 5200 miles. angle across from 5200 mi side  45°  90° or 135° 5200 3960    sin 135° sin x  33. side opposite 37°  15  18 or 33 side between v and 37°  15  22 or 37 Let x  the measure of the third angle. 33  sin 37°  37    sin x  3960 sin 135°  37 sin 37°   sin x   5200  sin x  3 3  x  sin
133 x  42.43569405 v  180 
 (37°  42.4°) or about 100.6° 34a. a  b sin A 34b. a  b sin A a  14 sin 30° a  14 sin 30° a7 a  7 or a  14 34c. a b sin A a 14 sin 30° a 7 and a  14 7  a  14   x  sin
1 5200  37 sin 37°  35.  184.5  sin 59°  3960 sin 135°  x  32.58083835 v  180° 
 (135°  32.6°) or about 21.4° 21.4°  (2 hours)  0.0689953425 hours or about 4.1 360° minutes 38. P turns 20(360°) or 7200° every second which equals 72° every 0.01 second. PQ  sin O 15  sin 72°  140  sin Q  15 Q  sin
11  5  140 sin 59°  5 sin 72°   sin x   184.5  x  sin
1 184.5  140 sin 59°  Q  18.48273235 m∠P  180° 
 (72°  18.5°) or about 89.5°  x  40.57365664 v  180° 
 (59°  40.6°) or about 80.4° 90° 
 80.4°  9.6° 36a. 12°  90° and 316 450 sin 12°; 2 solutions  QO  sin P QO  sin 89.5°  QO  15.77133282 QO 
 5  15.8 
 5 or about 10.8 cm 39a. b  c sin B 12  17 sin B   v  sin
1 316 450 sin 12°  v  17.22211674 180° 
   180° 
 17.2° or 162.8° turn angle  180° 
 162.8° or 17.2° about 17.2° east of north 36b. v  180° 
 (162.8°  12°) or about 5.2° x  12  17 12 sin
1 1 7  12  17 12 sin
1 1 7  x  138.3094714 d  rt 138.3  23t 6.013455278  t; about 6 hr 36c. 180° 
 20°  160° 180° 
 (160°  12°)  8° 200 c    sin 8° sin 160°  Port  12  17 12 sin
1 1 7  20˚ 160˚   sin B B  sin B B  44.90087216 B B  44.9° 40. Area of rhombus  2(Area of triangle) triangle:  c 200 mi.  B  44.90087216  B B  44.9° 39c. b c sin B 12 17 sin B  200 sin 160°  c  491.5032301 Since 491.5 450, the ship will not reach port.   sin B  44.90087216  B B 44.9° 39b. b  c sin B 12  17 sin B  316  sin 12° 316 sin 5.2°  sin 12°   c sin 8°  15    sin 72° 15 sin 89.5°  450 sin 12°    PQ    sin O   QO   sin 72°  450    sin v   sin v   316  x  sin 5.2°  5    sin Q 5 sin 72°    sin x  316  sin 12°  OP    sin Q  12˚  1  K  2bc sin A 1  K  2(24)(24) sin 32° K  152.6167481 rhombus: A  2(152.6) or about 305.2 in2  161  Chapter 5  4. Sample answers:  75  41. tan 22°  x  1  42. 3; 2  x x x x  c  C  53˚ a  10  sin 53°  c 10   c sin 53°  B  c  12.52135658 b  tan B  a 10  tan 53°  a  2  

188   8 2  2i
47   8 1  i
47   4  3x  x
1  10   a tan 53°  x
1  3x     1 x
1 x
1     9x 3x     x
1   3  x
1  5x 
 2y  9 5x 
 2(3x 
 1)  9 5x 
 6x  2  9 x  
7 (
7, 
22) 45. 
2x  5y  7 2  4x 
 1  x
1  9x  x
1  a  7.535540501 A  37°, a  7.5, c  12.5 C  180° 
 (5.5°  45°) or 80° A b a    sin B sin A 4x 
 1   9 x  x  b  55˚  b  sin 45°  c 45˚  C  10   b sin 55°  B  b  8.6321799 c a    sin C sin A c 10    sin 80° sin 55° 10 sin 80°  c sin 55°  7  c  12.0222828 C  80°, b  8.6, c  12.0 c2  a2  b2 
 2ab cos C c2  102  82 
 2(10)(8) cos 50° c2  61.15398245 c  7.820101179  y  5x  5 5  perpendicular slope: 
2 5  y 
 4  
2(x 
 (
6)) 5  y 
 4  
2x 
 15 2y 
 8  
5x 
 30 5x  2y  
22 46. Perimeter of XYZ  4  8  9 or 21  A 8  1  length of A B   3 of perimeter 1   3(21) or 7  C  c  a c    sin A sin C 10 7.8    sin A sin 50° 10 sin 50°  sin A   7.8 10 sin 50°  B A  sin
1  7.8    50˚ 10  The Law of Cosines  322 
 382 
 462  
2(38)(46)  322 
 382 
 462 cos
1  
2(38)(46)    Check for Understanding  1. The Law of Cosines is needed to solve a triangle if the measures of all three sides or the measures of two sides and the included angle are given. 2. Sample answer: 1 in., 2 in., 4 in. 3. If the included angle measures 90°, the equation becomes c2  a2  b2 
 2ab cos C. Since cos 90°  0, c2  a2  b2 
 2ab(0) or c2  a2  b2.   cos A  A  43.49782861  A a  sin A 32  sin 43.5°  b    sin B 38    sin B 38 sin 43.5°  sin B  3 2 B  sin
132 38 sin 43.5°  B  54.8 C  180° 
 (43.5°  54.8°) or 81.7° A  43.5°, B  54.8°, C  81.7°  Chapter 5    A  78.4024367 B  180° 
 (78.4°  50°) or 51.6° A  78.4°, B  51.6°, c  7.8 5. a2  b2  c2 
 2bc cos A 322  382  462 
 2(38)(46) cos A  The answer is 7.  Pages 330–331  10    sin 55° 10 sin 45°  y  3x 
 1 y  3(
7) 
 1 y  
22  44.  5-8  10  2  
 (
2)2  
 4(4)(12)   2(4)  43. no    A  180° 
 (90°  53°) or 37° b sin B  c  A  75   x tan 22° x  185.6 m 4 
4 13 
6 2 
1 6 4 
2 12  0 4x2 
 2x  12  0  162  6. c2  a2  b2 
 2ab cos C c2  252  302 
 2(25)(30) cos 160° c2  2934.538931 c  54.1713848  Pages 331–332  c a    sin C sin A 54.2 25    sin 160° sin A 25 sin 160°  sin A   54.2 25 sin 160°  A  sin
1  54.2    7.8  sin 51°    7 sin 51°     B  sin
1 7.8  7 sin 51°  B  44.22186872 C  180° 
 (51°  44.2°) or 84.8° B  44.2°, C  84.8°, a  7.8 12. c2  a2  b2 
 2ab cos C 72  52  62 
 2(5)(6) cos C 72 
 52 
 62  
2(5)(6) 72 
 52 
 62  cos
1  
2(5)(6)   cos v    v   cos C   C  78.46304097  C  81.0  v about 81.0° 1 8. s  2(2  7  8)  8.5 K  
8.5(8. 5 )(8.5 
 2 
 7)(8.5 
 8)  6.4 units2 1 9. s  2(25  13  17)  27.5 K  
27.5(2 7.5  
7.5 25)(2 
7.5 13)(2 
 17)  102.3 units2 10. a2  b2  c2 652  652  c2 65 ft 8450  c2 91.92388155  c c 65 ft  7    sin B   sin B   7.8  A  9.1 B  180° 
 (9.1°  160°) or 10.9° A  9.1°, B  10.9°, c  54.2 7. The angle with greatest measure is across from the longest side. 212  182  142 
 2(18)(14) cos v 212 
 182 
 142  
2(18)(14) 212 
 182 
 142  cos
1  
2(18)(14)  Exercises  11. a2  b2  c2 
 2bc cos A a2  72  102 
 2(7)(10) cos 51° a2  60.89514525 a  7.803534151 a b    sin A sin B  a  sin A 5  sin A  c    sin C 7    sin 78.5° 5 sin 78.5°  sin A  7 A  sin
17 5 sin 78.5°  A  44.42268919 B  180° 
 (44.4°  78.5°) or 57.1° A  44.4°, B  57.1°, C  78.5° 13. c2  a2  b2 
 2ab cos C 72  42  52 
 2(4)(5) cos C 72 
 42 
 52  
2(4)(5) 
  
   cos C   cos
1 
2(4)(5)   C 72    42  52  101.536959  C a c    sin A sin C 4 7    sin A sin 101.5° 4 sin 101.5° sin A  7 4 sin 101.5° A  sin
1 7    
 2bc cos v 652  652  91.92 
 2(65)(91.9) cos v a2  652 
 652 
 91.92  
2(65)(91.9)  b2  c2   cos v  652 
 652 
 91.92  cos
1  v 
2(65)(91.9)    45.01488334  v    A  34.05282227 B  180° 
 (34.1°  101.5°) or 44.4° A  34.1°, B  44.4°, C  101.5° 14. b2  a2  c2 
 2ac cos B b2  162  122 
 2(16)(12) cos 63° b2  225.6676481 b  15.02223845  c 50 ft 65 ft  a  sin A 16  sin A  45.0˚  c2  a2  b2 
 2ab cos C c2  652  502 
 2(65)(50) cos 45.0° c2  2128.805922 c  46.1 ft  b    sin B 15.0    sin 63° 16 sin 63°   sin A   15.0  A  sin
1 15.0  16 sin 63°  A  71.62084388 C  180° 
 (71.6°  63°) or 45.4° A  71.6°, C  45.4°, b  15.0  163  Chapter 5  b2  a2  c2 
 2ac cos B 13.72  11.42  12.22 
 2(11.4)(12.2) cos B  15.  13.72 
 11.42 
 12.22  
2(11.4)(12.2) 13.72 
 11.42 
 12.22 cos
1  
2(11.4)(12.2)    a  sin A 11.4  sin A  b  1  23. s  2(174  138  188)  250 76 12 12 6 K  
250  11,486.3 units2   cos B  1  24. s  2(11.5  13.7  12.2)  18.7  B  8.7  
(18.7 11.5)
 7)(18. 13.7 2.2) 
 1 K  
187(1  66.1 units2 25a. d2  302  482 
 2(30)(48) cos 120° d2  4644 d  68.1 in. 25b. Area of parallelogram  2(Area of triangle)  70.8801474  B    sin B 13.7    sin 70.9° 11.4 sin 70.9°   sin A   13.7  1  K  2(30)(48) sin 120°   A  sin
1 13.7 11.4 sin 70.9°  K  623.5382907 2K  2(623.5382907) or about 1247.1 in2  A  51.84180107 C  180° 
 (51.8°  70.9°) or 57.3° A  51.8°, B  70.9°, C  57.3° 16. c2  a2  b2 
 2ab cos C c2  21.52  132 
 2(21.5)(13) cos 79.3° c2  527.462362 c  22.96654876 a c    sin A sin C 21.5 23.0    sin A sin 79.3° 21.5 sin 79.3°  sin A   23.0 21.5 sin 79.3°  A  sin
1  23.0    1  26a. s  2(15  15  24.6)  27.3 7.3  
7.3 15)(2 
7.3 15)(2 
 24.6) K  
27.3(2  105.6 Area of rhombus  2(105.6)  211.2 cm2 26b. 24.62  152  152 
 2(15)(15) cos v 24.62 
 152 
 152  
2(15)(15)  14.92 
 23.82 
 36.92  
2(23.8)(36.9)     cos v  
  152  74 ft  v  x  38 ft    13.75878964  v about 13.8° 18. d12  402  602 
 2(40)(60) cos 132° d12  8411.826911 d1  91.71601229 180° 
 132°  48° d22  402  602 
 2(40)(60) cos 48° d22  1988.173089 d2  44.58893461 about 91.7 cm and 44.6 cm  88 ft  382  742  882 
 2(74)(88) cos v  382 
 742 
 882  
2(74)(88) 
  
   cos v   cos
1 
2(74)(88)   v 382  742  882  25.28734695  v side opposite   sin v   hypotenuse x  sin 25.3°  74  1  19. s  2(4  6  8)  9  31.60970664  x about 31.6 ft 29a. x2  1002  2202 
 2(100)(220) cos 10° x2  15,068.45887 x  122.7536511 about 122.8 mi 29b. (100  122.7536511) 
 220  2.7536511 about 2.8 mi   4)(9 
 6)(9 
 8) K  
9(9 
  11.6 units2 1  20. s  2(17  13  19)  24.5 4.5  
4.5 17)(2 
4.5 13)(2 
 19) K  
24.5(2  107.8 units2 1  21. s  2(20  30  40)  45 )(45 
 20 
5 30)(40) 
 4 K  
45(45  290.5 units2 1  22. s  2(33  51  42)  63 )(63 
 33 
3 51)(62) 
 4 K  
63(63  690.1 units2  Chapter 5  152  110.1695875  v 180° 
 110.2°  69.8° about 110.2°, 69.8°, 110.2°, 69.8° 27. The angle opposite the missing side  45°. x2  4002  902 
 2(400)(90) cos 45° x2  117,188.3118 x  342.3 ft 28.    A  66.90667662 B  180° 
 (66.9° 
 79.3°) or 33.8° A  66.9°, B  33.8°, c  23.0 17. 14.92  23.82  36.92 
 2(23.8)(36.9) cos v  14.92 
 23.82 
 36.92 cos
1  
2(23.8)(36.9)  
   cos v  cos
1  v 
2(15)(15) 24.62  164  30.  side opposite  202 ft 82.5˚ I 201.5 ft  x  y  II  125 ft   34. tan v   side adjacent  180.25 ft  570   tan v   700 570   v  tan
1  700  III 75˚  v  39.2° 35. 
775°  2(360°)  
55° reference angle  
55° or 55° 36. 3 1 
7 
k 6 3 
12 
36 
 3k 
30 
 3k 1 
4 
12 
 k 
30 
 3k  0 k  
10  158 ft  1  I: K  2(201.5)(202) sin 82.5° K  20,177.3901 II: x2  201.52  2022 
 2(201.5)(202) cos 82.5° x2  70,780.6348 x  266.046302 y2  1582  180.252 
 2(158)(180.25) cos 75° y2  42.711.98851 y  206.6687894 206.72  266.02  1252 
 2(266.0)(125) cos v 206.72 
 266.02 
 1252  
2(266.0)(125)  5t 
 t   37. m   5t 
 2t 4t  4  3  23x6  m  3t or 3  2yx 2  38.   cos v   y 3 8x6   y 3 The correct choice is A.  206.72 
 266.02 
 1252  cos
1  v 
2(266.0)(125) 48.93361962  v K  1 (266.0)(125) 2  5-8B Graphing Calculator Exploration:  sin 48.9°  Solving Triangles  K  12,536.58384 III: K   1 (180.25)(158) 2  sin 75°  Page 334  K  13,754.54228 Area of pentagon  I  II  III  20,177.4  12,536.6  13,754.5  46,468.5 ft 31. I: 242  352  402 
 2(35)(40) cos v 242 
 352 
 402  
2(35)(40) 242 
 352 
 402  
2(35)(40)  cos
1   1. AB  12.1, B  25.5°, C  119.5° 2.  C   cos v  v  242 
 302 
 202  
2(30)(20) 242 
 302 
 202  cos
1  
2(30)(20)    40 51˚ O A (0, 0)   cos v  v  x 50  B (50, 0)  51˚ O A (0, 0)  x 50  B (50, 0)  Find y using A C . y  (tan 51°)x Find y using  BC . 40 50    sin 51° sin C  52.89099505  v the player 30 ft and 20 ft from the posts 20,000  32a. sin 6°  x  50 sin 51°  sin C  4 0  20,000   x sin 6°  C  sin
140 50 sin 51°  x  191,335.4 ft 15,000  C  76.27180414 B  80 
 51 
 76.27180414  52.72819586 y  tan(180 
 52.72819586)   x 
 50  32b. sin 3°  y y  C  40  36.56185036  v 242  302  202 
 2(30)(20) cos v  II:  y  y  15,000  sin 3°  y  286,609.8 ft 32c. 6° 
 3°  3° d2  191,335.42  286,609.82 
 2(191,335.4)(286,609.8) cos 3° d2  9,227,519,077 d  96,060.0 ft 33. Since 63.2°  90°, consider Case I. b sin A  18 sin 63.2° b sin A  16.06654473 17 16.1; 2 solutions  (x 
 50) tan(127.2718041)  y Set the two values of y equal to each other. (tan 51°)x  (x 
 50), tan 127.2718041° (tan 51°)x  x(tan 127.2718041° 
 50(tan 127.2718041°) x  
50(tan 127.2718041°)  tan 51° 
 tan 127.271804°  x  25.77612538 y  (tan 51°) (25.77612538)  31.83086394  165  Chapter 5  C could also equal 180 
 76.27180414 or 103.728195° B  180 
 51 
 103.7281959°  25.2718041 y  tan (180 
 25.2718041)   x 
 50  15.  16.  17.  
50(tan 154.7281959°)  tan 51° 
 tan 154.7281959°  18.  Pages 336–338  2. 4. 6. 8. 10.  19.  860°  360°  20.  false; arcosine false; adjacent to false; coterminal false; Law of Cosines true    360(2)°  860°   720°  860°   140°; II  Chapter 5  14.  1146°  3 60°  1072°  360°   2.98  654°  360°   1.82  
832°  360°   
2.31    360(
2)°  
832°  
 720°  
832°   
112° 360° 
 112°  248°; III 21. 
284° has terminal side in first quadrant. 360° 
 284°  76° 22.  592°  360°   1.64    360(1)°  592°   360°  592°   232° terminal side in third quadrant 232° 
 180°  52° 23. (BC)2  (AC)2  (AB)2 152  92  (AB)2 306  (AB)2 
306   AB 3
34   AB  Skills and Concepts   2.39   
0.83    360(1)°  654°   360°  654°   294°; IV  11. 57.15°  57°  (0.15 60)
  57°  9
 57° 9
 12. 
17.125°  
(17°  (0.125 60)
)  
(17°  7.5
)  
(17°  7
  (0.5 60))  
(17°  7
  30) 
17° 7
 30 13.  
300°  360°    360(2)°  1072°   720°  1072°   352°; IV  Understanding and Using the Vocabulary  false; depression true true true false; terminal side   2.77    360(
1)°  
300°  
 360°  
300°   60°; I  Chapter 5 Study Guide and Assessment  1. 3. 5. 7. 9.  998°  360°    360(2)°  998°   720°  998°   278°; IV  x  13.82829048 y  (tan 51°) (13.82828048)  17.07651659 B  52.7°, C  76.3°, b  40.9; B  25.3°, C  103.7°, b  220 3. Law of Cosines 4. Sample answer: put vertex A at the origin and vertex C at (3, 0).  Page 335   
0.43    360(
1)°  
156°  
 360°  
156°   204°; III  (x 
 50) tan 154.7281959°  y Set the two values of y equal to each other. (tan 51°) x  (x 
 50)tan 154.7281959° (tan 51°) x  x(tan 154.7281959°) 
 50(tan 154.7281959°) x  
156°  360°  opposite side   3.18   cos A   hypotenuse  sin A   cos A   tan A     360(3)°  1146°   1080°  1146°   66°; I  tan A   166  adjacent side   sin A   hypotenuse  5
34  15  or  34 3
34  side opposite  side adjacent 15 5  or  9 3  9  3
34   or  3
34   34  29. r  
 x2  y2 r  
 82  ( 
2)2 r  
68  or 2
17   24. (PM)2  (PN)2  (MN)2 82  122  (MN)2 208  (MN)2 
208   MN 4
13   MN  y  sin v  x  opposite side  sin M  tan M   3
13  12  or  13 4
13  side opposite  side adjacent 12  cos M  csc M   3  tan M  8 or 2  csc M   hypotenuse  cot M    sec M   side adjacent  4
13  
13  or  2 8 (PN)2  (MN)2  sec M   
2  cos v  2
17  
17  sin v  
  cos v  17 r csc v  y 2
17  csc v    
2 or 

17 x cot v  y 8  cot v   
2 or 
4 2 x  y2 30. r  
  sin v   adjacent side   cos M   hypotenuse   sin M   hypotenuse   2
13 8  or  13 4
13  hypotenuse  side opposite   4
13 
13  or  12 3 side adjacent  side opposite 2  8   cot M  1 2 or 3  25. (MP)2  (MP)2  102  122 (MP)2  44  or 2
11  MP  
44 opposite side  26.  10 5  sin M  1 2 or 6 side opposite  tan M   side adjacent 10 5
11  tan M   or  11 2
11  hypotenuse  sec M   side adjacent 12 6
11  sec M   or  11 2
11  1  sec v   cos v 1  cos v   secv   
11  2
11  cos M   12 or 6 hypotenuse  csc M   side opposite 12 6  csc M  1 0 or 5 side adjacent  cot M   side opposite  cot M   28.  
2  sin v   2 r csc v  y 3
2  csc v   3 or x cot v  y 3 cot v  3 or 1 x2  y2 r  
  cos v  sec v   2
11   10  or  
11   5  
2   tan v  tan v   y  x 3  3  y  sin v  csc v  csc v   12  13 r  y 13  12  sec v  sec v   x  or  or  13 
5  tan v  cot v  cot v   12  
5 x  y 
5  12  cot v  0 undefined  x  tan v  x  4  
41   4
41  41 r  x 
41   4  tan v  4  cos v  r  5  
41   5
41  sin v  41 r csc v  y  
41 csc v   5 r  
 x2  y2  cos v  cos v  sec v  sec v   y  5  x  cot v  y 4  cot v  5  
9  
106   
106   or 
9 x cot v  y 
5 5   cot v   
9 or 9 r  
 x2  y2  y  tan v  x  
5  
106   cos v   
9  9    tan v   
5 or 5   5
106  cos v  
  106 r  sec v  x 
106   
 9  sec v   
106   
5  
106   or 
  5  r  
 (
4)2   42 r  
32  or 4
2   tan v  x 5 
1 3  
2  2  csc v  y  y  x  0   tan v   
2 or 0  cot v  y  r  cos v  r cos v    sec v   
2 or 
1   9
106  33.  
5  13 r  x 13  
5  2  csc v   
17   4  y  r  sin v  
  106 or 
2   or  tan v  x  sin v  r  sin v   r  
 (
5)2   122 r  
169  or 13 sin v  r  2
17   8  r  
 (
5)2   (
9 )2 r  
106  y x sin v  r cos v  r  
2   2 r  x  sec v   sec v   
2  y  or 1  1  r  undefined 31. r  
 x2  y2 r  
 42  52 r  
41   32.  3
2   3  
2  sec v  x  sec v  x  csc v  0   5  cos v   tan v  8 or 
4  r  csc v  y  1 5  cos v   7 or 7  sin v   8  2
17  4
17   17  cos v  2 or 
1  sin v   3  3
2   y  0  sin v  2 or 0  adjacent side   cos M   hypotenuse  3  3
2   tan v  x  r  
 (
2)2   02 r  
4  or 2 y x sin v  r cos v  r   sin M   hypotenuse  27. r  
 x2  y2 r  
 32  32 r  
18  or 3
2  y x sin v  r cos v  r  x  cos v  y  or  y  cos v  r  4  4
2   cos v   sin v  r  12 
5  sin v  sin v   5  or 
1 2  csc v  csc v   
2   2 r  y  4
2  4   csc v  
2  167  x  y  tan v  x  
4  4
2   4   tan v   
4 or 
1   
2  cos v  
  2 r  sec v  x sec v   4
2   
4  x  cot v  y 
4  cot v  4 or 
1  sec v  

2  Chapter 5  34. r  
 x2  y2 r  
 52  02 r  
25  or 5 y sin v  r 0  csc v   tan v  x  y  5  tan v  5 or 0  cos v  r  sin v  5 or 0 csc v   x  0  cos v  5 or 1  r  y 5  0  sec v  sec v   r  x 5  5  cot v   or 1  cos v   undefined  x  r  r2    x2  3 
8  82    (
3)2  x  
3, r  8 y  sin v  r 
55   8 r csc v  y 8  csc v   
55  8
55  csc v  55 y tan v  x  sin v   36.  sin v  sin v  sec v  sec v  37.  sin 42°     tan v  sec v  sec v   
55  
55   or 
  
3 3 r  x 8 8  or 
 3 
3  cot v  cot v  cot v   cos v  cos v  cos v   x —— csc r 
1  csc 
 10 
 10 
1 0  x2  v v  b —— sin B b —— sin 70°  x  y 
3  
 55 3
55   
 55  15 sin 42°  b 10.0  b  cot v  38.  b  tan B  —a— tan 67°   r —— x 
 10  
3  a —— sin A a —— sin 74°  or  or  sin A  sin 38°   a  — — sin A 84  — — sin 52° 84 sin 58°  — c— sin 52°  c  — — sin C 8  — — sin 49°  b —— sin B b —— sin 57°  sin B sin C  1  sin 96° sin 64°  — K  —2—(19)2 — sin 20°  a —— c 24 —— c  K  471.7 units2 48. C  180° 
 (56°  78°) or 46° 1  c sin 38°  24 24 — c— sin 38°  sin A sin C  — K  —2—b2 — sin B 1  sin 56° sin 46°  — K  —2—(24)2 — sin 78°  K  175.6 units2 1  49. K  —2— bc sin A 1  K  —2—(65.5)(89.4) sin 58.2° K  2488.4 units2 1  50. K  —2— ac sin B 1  K  —2—(18.4)(6.7) sin 22.6°  41. 180° 42. A  49°  90° A  41°  K  23.7 units2 a  cos B  —c— 16  cos 49°  —c— c cos 49°  16 16 — c— cos 49° c  24.4  A  41°, b  18.4, c  24.4  Chapter 5  84  — — sin 52°  c —— sin C c —— sin 58°  — K  —2—a2 — sin A  1  3  a  10.2  b  a  — — sin A  c  — — sin C 8  — — sin 49° 8 sin 57°  — b— sin 49°  a  10.1891739 b  8.889995197 A  74°, a  10.2, b  8.9 47. B  180° 
 (20°  64°) or 96° 1  
1  
3  b  8 sin 74°  a tan 67°  24 24  a tan 67°  tan 49°  —1— 6 16 tan 49°  b 18.4  b  cos 64°  —2— 8  — a— sin 49°  
10  
3  40. 30°, 210°  b  a  b  100.1689124 c  90.39983243 A  52°, b  100.2, c  90.4 46. A  180° 
 (57°  49°) or 74°  y2  24 —— a  tan B  —a—  b  84 sin 70°  c  39.0 39.  cos A  —c—  — b— sin 52°  x  

10   a  28 sin 64°  a 28 cos 64°  b 25.2  a 12.3  b B  26°, a  25.2, b  12.3 45. A  180° 
 (70°  58°) or 52°  cot v  —y—  b —— c b —— 15  15  A  cos
1 —20—  sin 64°  —28—    r2  (
1)2  (
3)2 r2  10 r  
10  r2  15  cos A  —20—  sin A  —c—  y2  55  
55 y Quadrant 11, so y  
55  y tan v  x  y —— r 
3  
 10 3
10  
10 r —— x 
 10  or 
1  sin B   y2  y2  tan v  3; Quadrant III y  
3, x  
1 sin v     b  cos A  —c—  A  41.4° 41.40962211°  B  90° B  48.6° a  13.2, A  41.4°, B  48.6° 44. 64°  B  90° B  26°  x  y 5  0  cot v   undefined 35. cos v   43. a2  b2  c2 a2  152  202 a  
175  a  13.2  168  B  113.7°, C  37.3°, b  22.7; B  8.3°, C  142.7°, b  3.6 54. Since 45°  90°, consider Case I. 83 79; 1 solution  51. Since 38.7°  90°, consider Case I. c sin A  203 sin 38.7° c sin A  126.9242592 172 126.9; 2 solutions a —— sin A 172  sin 38.7°  c  a —— sin A 83 —— sin 45°  — — sin C 203  — — sin C  203 sin 38.7° — sin C  — 172 203 sin 38.7° — C  sin
1 — 172    79 sin 45°    B  sin
1 —8—  3 79 sin 45°  B  42.30130394 C  180° 
 (45°  42.3°) or 92.7° a c ——  —— sin A sin C 83 —— sin 45°  a  — — sin A  83 sin 92.7°  c  117.2495453 B  42.3°, C  92.7°, c  117.2 55. a2  b2  c2 
 2bc cos A a2  402  452 
 2(40)(45) cos 51° a2  1359.446592 a  36.87067388  172 sin 93.7°  b  274.5059341 Solution 2 B  (180°
 (38.7°  132.4°) or 8.9° b a    sin B sin A b 172 ——  —— sin 8.9° sin 38.7° 172 sin 8.9° — b— sin 38.7°  a —— sin A 36.9 —— sin 51°  sin  — B  sin
1 — 36.9  40 sin 51°  B  57.39811237 C  180° 
 (51°  57.4) or 71.6° a  36.9, B  57.4°, C  71.6° 56. b2  a2  c2 
 2ac cos B b2  512  612 
 2(51)(61) cos 19° b2  438.9834226 b  20.95193124 b a ——  —— sin B sin A 21.0 51 ——  —— sin 19° sin A 51 sin 19° — sin A  — 21.0 51 sin 19° — A  sin
1 — 21.0  c        202 
 112 
 132 —— 
2(11)(13) 202 
 112 
 132 — cos
1 — 
2(11)(13)  b  — — sin B    b    b sin 113.7° 12 sin 113.7° —— sin 29 °   cos C  C  112.6198649  C a c ——  —— sin A sin C 11 20 ——  —— sin A sin 112.6° 11 sin 112.6° sin A  —20— 11 sin 112.6° A  sin
1 —20—  b  22.6647614 Solution 2 B  180° 
 (29°  142.7°) or 8.3° a —— sin A 12 —— sin 29°    A  52.4178316 C  180° 
 (52.4°  19°) or 108.6° b  21.0, A  52.4°, C  108.6° 57. c2  a2  b2 
 2ab cos C 202  112  132 
 2(11)(13) cos C  C  37.30170167 180 
   180° 
 37.3° or 142.7° Solution 1 B  180° 
 (29°  37.3°) or 113.7° a —— sin A 12 sin 29°  40  — — sin B 40 sin 51°  — — sin C    b  — — sin B  — sin B  — 36.9  b  42.34881128 B  93.7°, C  47.6°, b  274.5; B  8.9°, C  132.4°, b  42.3 52. Since 57°  90°, consider Case I. b sin A  19 sin 57° b sin A  15.93474074 12  15.9; no solution 53. Since 29°  90°, consider Case I. c sin A  15 sin 29° c sin A  7.272144304 12 7.3; 2 solutions 15 sin C 15 sin 29° C  —1— 2 15 sin 29° 
1 —— C  sin 12  c  — — sin 92.7°  — c— sin 45°  172  — — sin 38.7°  — b— sin 38.7°  a —— sin A 12 sin 29°  79  — — sin B  sin B  —8— 3  C  47.55552829 180° 
   180° 
 47.6° or 132.4° Solution 1 B  180° 
 (38.7°  47.6°) or 93.7° b —— sin B b —— sin 937°  b  — — sin B  b  — — sin B    b  — — sin 8.3°    A  30.51023741 B  180° 
 (30.5°  112.6°) or 36.9° A  30.5, B  36.9°, C  112.6°  12 sin 8.3°  — b— sin 29°  b  3.573829815  169  Chapter 5  y2  1 y  1 Since y is a length, use only the positive root. Another method is to use the Triangle Inequality Theorem. The hypotenuse must be shorter than the sum of the lengths of the other two sides. 5y  3  4 5y  7 Which of the answer choices make this inequality true? 5(1)  5  7 5(2)  10 7 The correct choice is A. 2. If you recall the general form of the equation of a circle, you can immediately see that this equation represents a circle with its center at the origin. (x 
 h)2  (y 
 k)2  r2 If you don't recall the equation, you can try to eliminate some of the answer choices. Since the equation contains squared variables, it cannot represent a straight line. Eliminate choice D. Similarly, eliminate choice E. Since both the x and y variables are squared, it cannot represent a parabola. Eliminate choice C. The choices remaining are circle and ellipse. This is a good time to make an educated guess, since you have a 50% chance of guessing correctly. It represents a circle. The correct choice is A. 3. Use factoring and the associative property. 999 111  3 3 n2 (9 111) 111  3 3 n2, 3 3 (111)2  3 3 n2. So n must equal 111. The correct choice is C. 4. A  58. b2  a2  c2 
 2ac cos B b2  422  6.52 
 2(42)(6.5) cos 24° b2  1307.45418 b  36.15873588 b c ——  —— sin B sin C 36.2 6.5 ——  —— sin 24° sin C 6.5 sin 24° — sin C  — 36.2 6.5 sin 24° — C  sin
1 — 36.2      C  4.192989407 A  180° 
 (24°  4.2°) or 151.8° b  36.2, A  151.8°, C  4.2°  Page 339  Applications and Problem Solving 8  59a. sin v  —1— 2 8  v  sin
1 —1— 2 v  41.8° x  cos v  —12—  59b.  x  cos 41.8°  —12— 12 cos 41.8°  x 8.94427191  x about 8.9 ft 60a. x2  4.52  8.22 
 2(4.5)(8.2) cos 32° x2  24.9040505 x  5.0 mi 60b.  8.2 —— sin v  5.0  — — sin 32° 8.2 sin 32°  — sin v  — 5.0 — v  sin
1 — 5.0 8.2 sin 32°  v  60.54476292 180 
 v  180 
 60.5 or about 119.5°  45˚  D  Page 339 1.  435.86  ab Sample answer: about 40 cm and 10.9 cm 2a. Sample answer: a  10, b  24, A  30°; 10  24, 10  24 sin 30° 2b. Sample answer: b  18; 10  18, 10 18 sin 30°  Chapter 5 SAT & ACT Preparation SAT and ACT Practice  1. There are several ways to solve this problem. Use the Pythagorean Theorem on the large triangle. (2y  3y)2  42  32 (5y)2  16  9 25y2  25 25y2 —— 25  Chapter 5  7  7  B 7 C Since ABC is an equilateral triangle and one side is 7 units long, each side is 7 units long. so AC  7. A D  is the hypotenuse of right triangle ACD. One leg is 7 units long. One angle is 45°, so the other angle must also be 45°. A 45°
45°
90° triangle is a special right triangle. Its hypotenuse is 
2  times the length of a leg. (The SAT includes this triangle in the Reference Information at the beginning of the mathematics sections.) The hypotenuse is 7
2 . The correct choice is B.  Open-Ended Assessment  1 K  —2— ab sin C 1 125  —2—ab sin 35°  Page 341  45˚  25   —25—  170  8. Factor the polynomial in the numerator of the fraction. Simplify the fraction. Solve for x.  5. You need to find the fraction's range of values, from the minimum to the maximum. The minimum value of the fraction occurs when a is as small as possible and b is as large as possible. Since the smallest value of a must be slightly greater than 4, and the largest value of b must be slightly less than 9, this minimum value of the 4 fraction must be larger than —9—. The maximum value of the fraction occurs when a is as large as possible and b is as small as possible. This 7 maximum must be smaller than —7— or 1. The correct choice is A. 6. Start by making a sketch of the situation.  x2  7x  12 ——  5 x4 (x  3)(x  4) ——  5 x4  x35 x2 The correct choice is B. 9.  O  3  C 30 T  9000  A  T  is a radius of the circle. T is on the circle, so O The length of O T  is 3. Since 
 TA is tangent to the circle, OTA is a right angle, and OTA is a right triangle. In particular, OTA is a 30°-60°-90° right triangle. In a 30°-60°-90° right triangle, the length of the hypotenuse is 2 times the length of the shorter leg.  1:00 P.M. 9000 36,000 9000 10:00 A.M. 9000  By 1:00 P.M. the pool is three-fourths full. Three fourths of 36,000 gallons is 27,000 gallons. The pool contained 9,000 gallons at the start. So 27,000 
 9,000 or 18,000 gallons were added in 3 hours. The constant rate of flow is 18,000 gallons 3 hours or 6,000 gallons per hour. To fill the remaining 9,000 gallons at this same rate will take 9,000 gallons 6,000 gallons per hour or 1.5 hours. One and a half hours from 1:00 P.M., is 2:30 P.M. The correct choice is C. 7. There are two right triangles in the figure. You need to find the length of one leg of the larger triangle, but you don't know the length of the other leg. Use the Pythagorean Theorem twice— once for each triangle. Let y represent the length of side AC. In the smaller right triangle, y2  42  62 y2  16  36 x2  52 You do not need to solve for y. In the larger triangle, 102  x2  y2 100  x2  52 x2  48 x  
48  x  4
3  The correct choice is B.  OA  2(OT ) OC  CA  2(OT ) 3  CA  2(3) or 6 CA  3 The correct choice is B. 10. Draw a diagram from the information given in the problem. Drawing a valid diagram is the most difficult part of solving this problem. Your diagram could be different from the one below and still be valid.  C 5  A  4 4  X  B  5  D Since two segments bisect each other, you know the length of each half of the segment. Notice that B D  is a side of a right triangle. It is a 3-4-5 right triangle. So BD  3. The answer is 3.  171  Chapter 5  Chapter 6 Graphs of Trigonometric Functions Pages 348–351  Angles and Radian Measure  6-1  16. 135°  135°  Exercises   180°  3  Pages 347–348 1.    180°  5  5   
2   180°  20. 
75°  
75° 5  22.  7  12  7   12  125   18 11  3  3. Divide 10 by 8. 4. Let R  2r. For the circle with radius R, s
  Rv or 2rv which is 2(rv). Thus, s
  2s. For the circle 1 1 with radius R, A
  2R2v or 2(2r)2v which is  23.  180°    25. 
3.5 
3.5  1 (4r2)v 2  or 42r2v. Thus, A
  4A.   180°  4  3  2  9. reference angle: sin  3  4  
2   tan  
 3  11  6  s  15  3  4  
  or   ; 4  Quadrant 2  1  11  6  
  or  5 ; 6  7  32. Quadrant 3  2  1  A  2.1 units2    180°  1  14  3  2  is coterminal with 3 2    14   tan 3  

3 19  3  A  5  33. 
6 is coterminal with 6 5    reference angle:  
 6 or 6; Quadrant 2 
3   19  cos 
6  
2 34. s  rv  2  3  s  14    180°  36. 150°  150°  1 r2v 2 1 3 (62)  2 10  5   6     s  rv  A  17.0 units2  5  s  146 s  36.7 cm     6  38. s  rv  s  rv  s  1.46 s  0.7 m  35. s  rv    s  29.3 cm   —10— A  15. 30°  30°    reference angle:  
 3 or 3; Quadrant 2    180°  14. 54°  54°  A  2(1.42)3  7  sin 6  
2  s  rv 77  s  15 180  s  20.2 in.  13. A  2r2v    31. reference angle: 6 
  or 6; Quadrant 3  77    5  
 2  5    180  s  39.3 in.    cos 4  
2  12. 77°  77° 5  6  7  30. reference angle: 4 
  or 4; Quadrant 3  180°     
3  11. s  rv   1002.7°  
3   7   2  10. reference angle:    tan 6  3   270°  100.3°  5  180°    29. reference angle: 6 
  or 6; Quadrant 3  180°    8. 
1.75  
1.75  27. 17.5 17.5  sin 3  
2  19  3   
200.5° 180°    
3   5    180°   6   2    180°    28. reference angle: 2 
 3 or 3; Quadrant 4  6. 570°  570°   3    
 6.2   
29.0°  1  5. 240°  240°  7.  26.  180°     3  660°   974.0°    
 6.2  11  180°     105° 24. 17  17    180°  21. 1250°  1250°   
12  x  1  5  s  1412    180°  s  18.3 cm 37. 282°  282° 47  s  rv  47  s  143 0   s  68.9 cm 39. 320°  320°  3  s  1411 s  12.0 cm  172    180°   30  16   9 s  rv  16  s  149 s  78.2 cm  Chapter 6    180°  19. 
450°  
450°   3  3
 4  O   6  18. 300°  300°    2. 90°; —4—  y  7   4  Check for Understanding    180°  17. 210°  210°    180°  1  78°  78°  1   30  40. r  2d    180°  53b.  13  r  2(22) r  11  s  rv  s  rv  5  4  70.7  r 18.00360716  r   180°  42. 60°  60°  11  s  rv    s  2  d  36.0 m  s  11.5 in.  A  A  65.4  44. 90°  90°       units2  A A    180°  45. A  A  46. A      A  A  9.6 units2   —— 180°  47. 225°  225°  A A  5 —— 4  A       s  rv  s  rv  s  3960—6— 0 s  207.3 mi 57. 84.5°  84.5°      52a.  1  58a. r  2d 3  r  52c. s  rv s  3(3.4) s  10.2 m  53a. 225°  225°  s  rv  5  4  1.5 rotations  1.5 2 radians  3 radians     1  2  1 22  r  1.25 s  rv s  1.25(3) s  11.8 ft  mm2  58b.  s  rv  3.6  3.6  1  42  1.25v  180°     206.3°  3.6  v   3  59a. v  2 
 2 or 2  180°    1  A  2r2v   194.8°  3.4  v; 3.4 radians  s  rv 4 s  0.679  s  1.03 mi s  0.94 mi 1.03  1.46  0.94  1.8  5.23 mi  1  52b. 3.4  3.4  4   9  s  rv 169  s  0.70 360   48.38  r; about 48.4 mm 1 51a. A  2r2v 51b. s  rv 1 2 s  12.2(0.2)   15  2 r (0.2) s  2.4 in. 150  r2 12.247  r about 12.2 in. 1 A  2r2v 1 15.3  2(32)v    180°  80°  80°  169  50b. A  2r2v A  2757.8    180°    360  A  15 ft2 1  114  r    180°   —— 180°  A  2(48.42)4 3  4  s  967.6 mi    1  3 —— 4  s  rv 7 s  396090   60  A  2(52)(1.2)   180°  7    180°   —9— 0  3°  3°  49b. A  2r2v  50a. 135°  135°  14°  14°  s  760.3 mi 56c. 34° 
 31°  3°  1  49a. s  rv 6  r(1.2) 5  r; 5 ft  56b. 45° 
 31°  14°    180°       A  70.7 units2   26.3°  s  rv 11  s  3960 180   1 r2v 2 4 1 (12.52) —— 2 7 140.2 units2  180°    0.5  0.5  11  41  —90— 1 A  —2—r2v 1 41 A  2(7.32) 90 A  38.1 units2  1 ——r2v 2 1 5 (62)  2 4  1.44  v; about 1.4 radians    180     48. 82°  82°  11.5  8v  11°  11°    2 1 r2v 2 1  (222)  2 2  A  380.1 units2 1 r2v 2  1 (72) —— 2 8    10.5  22.9v 0.46  v; about 0.5 56a. 45° 
 34°  11°    13.56  r; about 13.6 cm 43. A   s  rv  s  rv  55.  14.2  r3  1 r2v 2 5 1 (102) —— 2 12  11  6  d  2(18.0) s  rv   3    180°   6  s  15.0 in. d  2r     143.2°  54. 330°  330°  13  180°    2.5  2.5  5  2v 2.5  v  s  113 0   41.  s  rv  3  A  2(152)—2— 1  A  530.1 ft2    180°  59b.  1  A  2r2v 3  750  2r22 1  318.3098862  r2 17.84124116  r; about 17.8 ft  5  s  24 s  7.9 ft  173  Chapter 6  60. 3.5 km  350,000 cm s  rv 350,000  32v 10,937.5  v; 10,937.5 radians 61. Area of segment  Area of sector 
 Area of triangle  yx  y  
x  1  1  A  2r2a 
 —2—r r sin a  all 70. 4x 
 2y  3z  
6 5x 
 4y 
 3z  
75 9x 
 6y  
81 2(4x 
 2y  3z)  2(
6) → 3(3x  3y 
 2z)  3(2)  1  A  2r2(a 
 sin a) 1  62. s  2(6  8  12)  13 K  
s(s 
 a)(s 
 b)(s 
 c) K  
13(13 13 
 6)(8)(13 
 
  12) K  
455  K  21.3 in2 63. Since 152°  90°, consider Case II. 10.2  12, so there is no solution. 64. C  180° 
 38° 
 27°  115° 560  sin 115°  5(9x 
 6y)  5(
81) 6(17x  5y)  6(
6)  x   sin 38°   280.52  x  172.7 yd 65. I, III 66a. Find a quadratic regression line using a graphing calculator. Sample answer: y  102x2 
 505x  18,430 66b. 2020 
 1970  50 y  102x2 
 505x  18,430 y  102(50)2 
 505(50)  18,430 y  248,180 Sample answer: about 248,180 67. r 1 
3 
2 6 10 1 1 
2 
4 2 12 2 1 
1 
4 
2 6 3 1 0 
2 0 10 4 1 1 2 14 66  3 4 5  
2 2 8  
6 
4 10  Linear and Angular Velocity  Page 355  Check for Understanding  2.  5 rev  1 min  2 radians  1 rev  1 min  60 s  3. Linear velocity is the movement along the arc with respect to time while angular velocity is the change in the angle with respect to time. 4. Both individuals would have the same change in angle during the same amount of time. However, an individual on the outside of the carousel would travel farther than an individual on the inside during the same amount of time. v 5. Since angular velocity is —t—, the radius has no effect on the angular velocity. Let R  2r. For v v a circle with radius R, v
  Rt or (2r)t which is  10 6 30  Sample answers: 4; 
2 68. 
2 1 6 12 12 
2 
8 
8 1 4 4 4 No; there is a remainder of 4. 69. x2  y2  16 → a2  b2  16 x-axis a2  b2  16 a2  (
b)2  16 a2  b2  16; yes y-axis a2  b2  16 (
a)2  b2  16 a2  b2  16; yes  Chapter 6  6-2  1.  f(
x)  (
x)4 
 3(
x3 
 2(
x)2  6(
x)  10 f(
x)  x4  3x3 
 2x2 
 6x  10 1 1 1  8x 
 4y  6z  
12 9x  9y 
 6z  6 17x  5y  
6 45x 
 30y  
405 → 102x  30y  
36 147x  
441 x  
3 4x 
 2y  3z  
6 4(
3) 
 2(9)  3z  
6 z8  9x 
 6y  
81 9(
3) 
 6y  
81 y9 (
3, 9, 8) 71. b 72. Since q  0, 
q 0. Given that p 0, p 
 q  p q and p q 0. So the expression p 
 q is nonnegative. The correct choice is B.  a    sin 27°  a  280.52  r 1 2  a2  b2  16 (b)2  (a)2  16 a2  b2  16; yes a2  b2  16 (
b)2  (
a)2  16 a2  b2  16; yes  2r t. Thus v
  2v. v  2  11.6 or about 36.4 radians 2  1420  or about 4461.1 radians 2  6.4 9. 700 2  1400  6. 5.8 7. 710 8. 3.2 v  q  t  q  7  6.4  q  15  q  2.9 radians/s  q  293.2 radians/min  q  t  174  v  1400  10. v  rq 11. v  rq v  12(36) v  7(5) v  432 in./s v  110.0 m/min 12a. r  3960  22,300 or 26,260 mi s  rv s  26,260(2) s  164,996.4 mi v 12b. v  r t 2 v  26,26024 v  6874.9 mph  1  35a. In 1 second, the second hand moves 6 0 (360°) or 6°.   180°  6°  6°  v  rv  v  3030 v  3.1 mm/s 35b. In 1 second, the minute hand moves     1 1   60 60  (360°) or 0.1°.  0.1°  0.1°  Pages 355–358 13. 14. 15. 16. 17. 18. 19.  3 2  6 or about 18.8 radians 2.7 2  5.4 or about 17.0 radians 13.2 2  26.4 or about 82.9 radians 15.4 2  30.8 or about 96.8 radians 60.7 2  121.4 or about 381.4 radians 3900 2  7800 or about 24,504.4 radians 1.8 2  3.6 20. 3.5 2  7 v  q  t  3.6  q  3  v  1  v  404 5  v  2  v  5.6 ft/s  27.  85 radians  1 second  60 seconds  1 minute  28. v  rq v  8(16.6) v  132.8 cm/s 30. v  rq v  1.8(6.1) v  34.5 m/min 32. v  rq v  39(805.6) v  31,418.4 in./min 60 seconds  1 minute   120°  120°  180° 2  3 v q  t 120°  1 second  2  3  1 2  3  36b. v   2  t  31 s 37a. 3 2  6 radians v  1 revolution  2 radians  1 6  v  7.1 ft/s v  37c. 7.1 
 3.1  4 ft/s  37b. v  r t   52.4 radians/s  6  60  3.1  r    9.87  r; about 9.9 ft   811.7 rpm  38a. 35°  35°  29. v  rq v  4(27.4) v  109.6 ft/s 31. v  rq v  17(75.3) v  4021.6 in./s 33. v  rq v  88.9(64.5) v  18,014.0 mm/min 1 revolution  360°  1 minute  60 seconds  v r t  v  22—2—60  0.1 radian/s 2 radians —— 1 revolution  v r t  8  40t  q  28.5 radians/min  26.  q  v  r t  r  2(80) or 40  v q  t 245.2 q  27  1 revolution 2 radians    50 seconds 1 revolution 500 revolutions 1 minute —— —— 1 minute 60 seconds  q  1  36a. r  2d  q  9.4 radians/s 24. 122.6 2  245.2  q  39.3 radians/min  34b.  v  0.003 mm/s  56.8  v  t 200  16  0.008   or  180  v  rv 0.008  v  18 180   q  1 9  q  9.0 radians/s 23. 100 2  200  34a.    180°  v  34.4  25.  0.008°  0.008°  q  t  q  1 2  q   610  (360°) or about 0.008°.  1 1   12 60  q  7.3 radians/min 22. 28.4 2  56.8  q  t  q  35c. In 1 second, the hour hand moves  7  q  1.3 radians/s 21. 17.2 2  34.4  0.1   or  180  v  0.05 mm/s  v  q  9    180°  v  rv 0.1  v  27 180   Exercises  q  t    or 30    180°  7   36 v  lighter child: q  t q  7  36  1  2  q  1.2 radians/s v  heavier child: q  t   20 rpm  q  7  36  1  2  q  1.2 radians/s  v  rq 2 v  53 v  10.5 in./s  175  Chapter 6  38b. lighter child: v  rq v  9(1.2) v  11.0 ft/s heavier child: v  rq v  6(1.2) v  7.3 ft/s 39a. 3 miles  190,080 inches  42c. 3960  500  4460; C  2(4460) or 28023.00647 t  28,023.00647 17,000 or 1.648412145 v  q  t 2   q 1.65  1  r  2d r  q  3.8 Its angular velocity is between 3.8 radians/h and 4.1 radians/h. 43a. B clockwise; C counterclockwise  s  rv  1 (30) 2  190,080  15v  r  15  43b. vA  rAtA v  12,672  v  vA  3.01 120  1 revolution   2017 revolutions 2 2 radians 2.75 revolutions 60 seconds 60 minutes     1 revolution second 1 minute 1 hour  12,672 39b.  vA  360 The linear velocity of each of the three rollers is the same.   19,800 radians/hour v  rq v  15(19,800) v  933,053.0181 933,053.0181 inches   vB  rBtB  1 mile   40a. Mercury:   14.7 mph  v  v  10.9 km/h Earth:  v  6.5 km/h Mars:  v  v r t  v 2  23.935  v  6356   12 1  A  2r2v  v  3375  1  A  47.5 cm2    1  r  2d  45.  v  1668.5 km/h v  861.2 km/h 40b. The linear velocity of Earth is about twice that of Mars. 41a. v  vm cos qt  1  r  2(7.3) r  3.65    x    v  360°  41b. v  4 cost 0  4 cos t 0  cos t  t  2 1  t  2 or 0.5 s  sin v  or  A A    46. 35°20
55  35°  20
 60
   55 3600  1°  2   35.349° 47. 10  
k 
58  
k 
 5  
2  
k 
54  k9  q  4.1 radians/h  t  tC   2   —— (17,000) 2   (17,000) 2  2   2r  speed 17,000   2r r  176  1°  Check: 10  
k  
58 10  
9  
58 10  
4 8 10  2  8 12 8  no real solution 48. (x 
 (
4))(x 
 3i)(x 
 (
3i))  0 (x  4)(x 
 3i)(x  3i)  0 (x  4)(x2  9)  0 3 x  4x2  9x  36  0  4250  r 4250 
 3960  290; about 290 mi Chapter 6  y  2.952912029  1 bh 2 1 (2.15)(2.95) 2  A  3.16761261 Area of pentagon  10(3.17) or about 31.68 cm2  17,000   q 1.54  2  t  y   cos 36°   3.65  x  2.145416171  3  t  2 or 1.5 s  v  v  y  cos v  r   sin 36°   3.65  q  t  42b. q  t  10 or 36° x  r x  3  t  2  42a. 3960  200  4160 miles C  2r t  C speed C  2(4160) t  26,138.05088 C  26138.05088 t  1.537532405   —— 2  
 y  r    v  4 cos t  4  7  A  2(7.22)12  v r t 2  24.623      180°  7  2   v  6052 5832.5   v C  1  75  vC 75 rpm  44. 105°  105°  v  2  360  4.8  180  vB 180 rpm  v  r t   v  2440 1407.6   v  v B  1  360  2.0  Venus:  v  r t  vC  rCtC  v  y  49. x3  y  maximum value of the cosine function occurs when x  n, where n is an even integer, and its minimum value occurs when x  n, where n is an odd integer. 5. yes; 4 6. 0 7. 1  1  x  O  8.  3  2  y  9. 0
5  1   50. m   
6 
 8  m  
5  
14  or  5  14  O  y 
 y1  m(x 
 x1) y
0 y  5 (x 
 (
6)) 14 5 15 x   14 7  y 
 cos x  5  y  10.    2b  3 b 4  P P 2P  7  3 b 2 7 b 2  y 
 sin x  1   2b 
4  b  
3  O  
2  x  The correct choice is D. 
1  Graphing Sine and Cosine Functions  6-3  7 x  6  
1  51. P  2a  2b P  2   2n, where n is an integer  Page 363  11. Neither; the period is not 2. 12. April (month 4):  y  49  28 sin 6(t 
 4)  Check for Understanding  y  49  28 sin  1. Sample answer:  
 4)  y  49 October (month 10):  y  49  28 sin 6(t 
 4)  y O   (4 6  x  y  49  28 sin   (10 6  
 4)  y  49 The average temperatures are the same.  period: 6 3  5  2. Sample answers: 
2, 2, 2  Pages 363–366  3. cos x  cos(x  2) 4. y  yes; 6 14. no 15. yes; 20 no 18. no 19. 1 0 22. 1 23. 
1 sin   cos   0  (
1)  
1 26. sin 2 
 cos 2  0 
 1  
1 27.   2n, where n is an integer  y 
 sin x  1  O    
1  2  Exercises  13. 17. 21. 25. 3  4  5 x  y 
 cos x  28.  Both functions are periodic functions with the period of 2. The domain of both functions is the set of real numbers, and the range of both functions is the set of real numbers between 
1 and 1, inclusive. The x-intercepts of the sine function are located at n, but the x-intercepts of  the cosine function are located at 2  n, where n is an integer. The y-intercept of the sine function is 0, but the y-intercept of the cosine function is 1. The maximum value of the sine function occurs  when x  2  2n and its minimum value occurs  29.    2   2  16. no 20. 0 24. 
1   2n, where n is an integer  n, where n is an integer  30. v  2n, where n is an integer  y  31. y 
 sin x  
5  
4  1  
3  O  x  
1  3  when x  2  2n, where n is an integer. The  177  Chapter 6  32.  y 1   43b. csc v   sin v  1  1   
1   sin v  8  1  — 1— sin v  y 
 cos x  O  10 x  9  sin v  1  sin v  
1   2  3   2   2n, where n  33. y 
 cos x  
5  
4  
3  y   44a. sec v   cos v  1   1 cos v  O  1   44b. sec v   cos v  1  1   
1   cos v  1  cos v  1 cos v  
1 2n, where n   2n, where n is an integer is an integer 44c. sec v is undefined when cos v  0.   n, where n is an integer 2  x  
1  45.  y 1  y 
 sin x  O  5  x  6  
1    [0, 2] sc1—2— by [
2, 2] sc11 3 7  x  4, 4  y  35. y 
 cos x  
3  46.  1  O  
2  x  
1  36.   2n, where n  is an integer is an integer 43c. csc v is undefined when sin v  0. n, where n is an integer  
1  34.  1   43a. csc v   sin v  y    [0, 2] sc1—2— by [
2, 2] sc11  y 
 sin x  1   5  0  x  4, 4  x  2 47.  O  4  5  x  
1  37. y  cos x; the maximum value of 1 occurs when x  4, the minimum value of 
1 occurs when 7 9 11 x  5, and the x-intercepts are 2, 2, and 2. 38. Neither; the graph does not cross the x-axis.    [0, 2] sc12 by [
2, 2] sc11 none  39. y  sin x; the maximum value of 1 occurs when 11 x  
2, the minimum value of 
1 occurs 13 when x  
2, and the x-intercepts are 
7, 
6, and 
5.  48.    40. Sample answer: a shift of 2 to the left   41. x  2  n, where n is an integer 42. x  n, where n is an integer    [0, 2] sc12 by [
2, 2] sc11   3  x  0, 2  x  , 2  x  2  Chapter 6  178  54a. v  3.5 cos t  49.  k —— m  v  3.5 cos 0.9     19.6 —— 1.99  v  
3.3 cm v  3.5 cos t  k  m  v  3.5 cos 1.7    0  3.5 cos t 0  cos t    1.570796327   v  3.5 cos t  54c.   5  x  4, 4 (t 
 4) 6 (7 
 4) 6  1  cos t cos1 1  t  y  74 January (month 1):  2  t  (t 
 4) 6 (1 
 4) 6  55a.   4  n   2, where n is an integer 55c. 
1  55b. 1 55e.  55d.   y 1  
2  O  
  y 
 cos 2x    2 x  
1  y 
 2 sin x  1  O 
1    19.6 —— 1.99 19.6 —— 1.99  y 2  
  19.6 —— 1.99    2.00206591  t; about 2.0 s  y  12 74 
 12  62; it is twice the coefficient. 51b. Using answers from 51a., 74  12  86; it is twice the constant term. 52a. n, where n is an integer 52b. 2 52c. 
2 52d. 2  
2    k —— m 19.6 —— 1.99  3.5  3.5 cos t  51a. July (month 7):  52e.    19.6 —— 1.99 19.6 — t — 1.99 19.6 — t — 1.99  0.5005164776  t; about 0.5 s    [0, 2] sc12 by [
2, 2] sc11  y  43  31 sin    k —— m 19.6 —— 1.99  cos1 0   y  43  31 sin    v  3.5 cos t  54b.  50.  y  43  31 sin  19.6 —— 1.99  v  2.0 cm  [0, 2] sc12 by [
2, 2] sc11  x  0, 2, 2  y  43  31 sin      56a. P  500  200 sin [0.4(t 
 2)] P  500  200 sin [0.4(0 
 2)] or about 357 pumas D  1500  400 sin (0.4t) D  1500  400 sin (0.4(0)) or 1500 deer 56b. P  500  200 sin [0.4(t 
 2)] P  500  200 sin [0.4(10 
 2)] or about 488 pumas D  1500  400 sin (0.4t) D  1500  400 sin (0.4(10)) or about 1197 deer 56c. P  500  200 sin [0.4(t 
 2)] P  500  200 sin [0.4(25 
 2)] or about 545 pumas D  1500  400 sin (0.4t) D  1500  400 sin (0.4(25)) or about 1282 deer  2 x  
2  52f. It expands the graph vertically. 53a. P  100  20 sin 2t P  100  20 sin 2(0) or 100 P  100  20 sin 2(0.25) or 120 P  100  20 sin 2(0.5) or 100 P  100  20 sin 2(0.75) or 80 P  100  20 sin 2(1) or 100 53b. 0.25 s 53c. 0.75 s  57.  500 revo lutions  1 minute  1 min ute   60 seconds  58. 
1.5  
1.5  2 rad ians   1 revolution   52.4 radians per second  18 0°     
85.9° 59. 45°, 135°  179  Chapter 6  2 x x2  4       x2 2
x x2 
 4 2   
1(x  2)(x 
 x2  60.  
1(x  2)(x 
 2)    68. Perimeter of square RSVW  RS  SV  VW  WR  5  5  5  5 or 20 Perimeter of rectangle RTUW  RT  TU  UW  WR  (5  2)  5  (5  2)  5  24 24 
 20  4 The correct choice is B.   2) 2 
 x x    (
1)(x  2)(x 
 2) x2 
 4  x2  4  
1(x 
 2)(2)  (x  2)(x)  (
1)(x2  4) 
2x  4  x2  2x 
 x2 
 4 x2 But, x 2, so there is no solution. 61. 1 positive real zero f(
x)  
2x3  3x2  11x 
 6 2 or 0 negative real zeros 2 2 3 
11 
6 4 14 6 2 7 3  0 2x2  7x  3  0 (2x  1)(x  3)  0 2x  1  0 or x30 1  x  
2 
3,  1 
2,  62. 1  1  1 12; no  n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30  2 
9 18 3 
6 
6  12  2 1 3  g (x ) 
  O  x2 x2  x  x  vertical: x2  x  0 x(x  1)  0 x  0 or x  1  0 x  
1 horizontal: y1  64. reflected over the x-axis, expanded vertically by a factor of 3 
2 4 
1 65. 1 
1 0 
3 4 5  
2 
1 0 
 4 1 0  (
1) 1 
1 4 5 
3 5 
3 4  
2(
5) 
 4(5) 
 1(1)  
11 3 2 1 66. 
1 0 0 1 2 
4 6 
1(3)  0(2) 
1(2)  0(
4) 
1(1)  0(6) 0(3)  1(2) 0(2)  1(
4) 0(1)  1(6) 
3 
2 
1  2 
4 6 A
(
3, 2), B
(
2, 
4), C
(
1, 6)   3  67. x  2y  y  2 x 3  ART  y  O  Chapter 6  History of Mathematics  1.  x  
3  g (x )  63.  Page 367  n 2  n3 2 12 36 80 150 252 392 576 810 1100 1452 1872 2366 2940 3600 4352 5202 6156 7220 8400 9702 11,132 12,696 14,400 16,250 18,252 20,412 22,736 25,230 27,900  [0, 30] sc15 by [0, 30,000] sc15000  The graph is not a straight line. It curves upward, increasing more rapidly as the value of n increases. 2. See students' work.  x 
 23 y  x  180  Amplitude and Period of Sine and Cosine Functions  6-4 Page 369  7.  2  4     2  y  y 
 sin 4   1  Graphing Calculator Exploration  1.  O     2  3 2    
1 2  8. 10  10; 2    y  2. The graph is shrunk horizontally. 3. The graph of f(x)  sin kx for k  0 is the graph of f(x)  sin kx reflected over the y-axis.  y 
 10 sin 2  8 4  O  Pages 372–373  1. Sample answer: y  5 sin 2v 2. The graphs are a reflection of each other over the v-axis.  y  2  y 
 3 cos 2  3 2 1  or 4  D: period  2 C has the greatest period. 4. Period and frequency are reciprocals of each other.  O    
1 
2 
3  y y 
 3 cos   O    
2  10. 0.5  0.5;  y 
 cos 3  2  2  3  2  2  1  6  y      3   12  y 
 0.5 sin  1   6  y 
 cos     O All three graphs are periodic and curve above and below the x-axis. The amplitude of y  3 cos v is 3, while the amplitude of y  cos v and y  cos 3v is 2 1. The period of y  cos 3v is 3, while the period of y  cos v and y  3 cos v is 2.  2  4  6  8  10  
1 1  1  11. 
5  5;  6. 
2.5  2.5  2  1  4   8  y  y  y 
  15 cos 4  0.2  y 
 2.5 cos  2  O    2  B: period  5  5.  3  9. 3  3; 2    2  2  1  2  2  
8  3. A: period  2 or   C: period     
4  Check for Understanding     2  3  4  O    
2  2  4  6  8  
0.2  12. A  0.8 A  0.8  2  k   2  k   or 2  y  0.8 sin 2v  181  Chapter 6  2  k  13. A  7 A  7     3  k  y  7 sin 6v  2  k  14. A  1.5  2    3  20.   5  O  2  2  y  1.5 cos 5v 2  k  3  y  3 4    21.   v 3 2  k  16. A  0.25 A  0.25 y  0.25 sin (588  2  1  4  y  1    294  y 
 cos 4  1  k  588 t)    Exercises  2  4  6  8  
1  17. 2  2 y 2    3   8  O  Pages 373–377  2  
1  k  6 or 3  cos    6 2  3  A  4  y 
 cos 2  1  k  5 or 5  15. A  4    y  or 6  2  A  1.5  2  2  y 
 2 sin   22.  2  6     3  y O    2  3  y 
 sin 6  1    
2  O  3     2    
1  3  18. 
4  4  y  2  y 
  34 cos   23. 5  5; 1  2  0.5  y  O  4  y 
 5 cos    2  3    
0.5  O    y  
1    2  y   2  3    2  y 
 2 cos 0.5     
2  O 
2  Chapter 6  4   24. 
2  2;  0.5  4  y 
 1.5 sin   1  O  3  
4  19. 1.5  1.5 2  2  182    2  3  4  2 2  2  2   30. 3  3;  0.5  4  25. 
5  5; 9  y  y y
  0.4  2 5   4  O  y 
 3 cos 0.5  sin 9  2     2  O    2  3  4  5   4  5   4  5  
2  
0.4  1 2  1  2  31. 
3  3; 3   26. 8  8;  0.5  4  y  y y 
 8 sin 0.5  8  1 1  y 
  3 cos 3  4  O    2  3  4  5  O      2  3  
4 
1  
8  27. 
3  3;  2    2  4  32.  1  3  1   3;  2  1  3  y  y 4  
 sin 2   y 
 3  1   2  O 
2    3 2  2  5  2  O  2  3  2   3;    2   3  
1  
4  28.    1  y 
 3 sin 3  2  
2   6  2  3  7  14  33. 
4  4;   3  y  y
  0.8  2 3  2  1  2   4  y  3
  cos 7     y 
 4 sin 2  4 2  0.4    O 
0.4   2    3 2  2  5 2  O  3  
2    2  4  5   
4  
0.8 2  34. 
2.5  2.5;  29. 3  3; 2    x  y 
 3 sin 2  2  1  5   10  y   2  y 
 2.5 cos 5  2  O  3    2  3  4  5   O  
2    2  3  4  5   
2  2  1    35. 0.5  0.5;  698  349  183  Chapter 6  2  k  36. A  0.4 A  0.4  k 2  k  37. A  35.7 A  35.7 y  35.7 sin 8v A y  1  4 1 4  1 4    1  5    1  v  y  
1.5 sin 2   0.75  A  3.8 y  3.8 sin(240  k  3 2  k  A  15 y  15 cos (72  5   4 8  A  4.5  k  5  1    120  k  240 t) 2  k  54. A  15  8  y  0.34 sin 3v 40. A  4.5  2  k  53. A  3.8  8  A  0.34   4  k  2  sin 6v  39. A  0.34  2  k  52. sine curve A  
1.5    3   2  k1  y  
3 cos v  k6 2  k  2  k  A  
3  51. cosine curve    4    k2  y  0.5 sin 2v  k8 2  k  2  k  50. sine curve A  0.5  v  5  y  0.4 sin  38. A    10  1   36  k  72 t)  55.  8  y  4.5 sin 5v 2  k  41. A  16   30   A  16  k  15   y  16 sin 15v 2  k  42. A  5 A  5 y  5 cos v  2  k  5  5  A   8 y  2  k  y  7.5 cos  A   y  0.5 cos    2  k 2  3   5  57c.  10 v 3  47. A  17.9  37d.   2  y y 
 2  sin    16  k  8  2   v 8  48. A  1.5  2  k     2  A  1.5 k4 y  1.5 sin 4v, y  1.5 cos 4v 49. cosine curve A  2  2  k  O    2  3  A  0.2 y  0.2 sin (524  1  v  y  2 cos 2  184  4 2  k  58a. A  0.2   4  k  2  Chapter 6  2  1    A  17.9 y  17.9 cos  y  
1.5 sin 4(12)  y  
1.1 ft y  0 ft 57a. Maximum value of sin v  1. Maximum value of 2  sin v  2  1 or 3 57b. Minimum value of sin v  
1 Minimum value of 2  sin v  2  (
1) or 1  20  3  k  3 2  k    y  
1.5 sin 4(3)  10  A  5    56c. y  
1.5 sin 4t  56b. y  
1.5 sin 4t   0.3  k  2  cos    y  
1.5 sin 4t  1  3  20 v 3  46. A  5  3  2    2  k  A  0.5  8  A  1.5; down first, so A  
1.5  v  3  45. A  0.5  2  k     6  k  A 
 (
A)  2  k  4  cos 14v  A  7.5  y  56a. A      7  k  14  44. A  7.5  2 5  All the graphs have the same shape, but have been translated vertically.  k1  43. A  8 5 8   2  5  1    262  k  524 t)  2  k  1  58b. A  2(0.2) A  0.1 y  0.1 sin (1048  59a. y  A cos t    2  k  65.  k  262  9.8  6  0.6525352667     9.8  6  0.458022743     y y 
 cos(  
)    15.1   A  tan1  19.5  2  
  g 
  9.8  4  3  
  9.8 
  9.8  4.17  
; about 4.17 m 67. b2 
 4ac  52 
 4(3)(10)  
95 2 imaginary roots 68a. Let x  the number of Model 28 cards and let y  the number of Model 74 cards. y 30x  15y  240 40 20x  30y  360 12x  10y  480 32 12x  10y 
 480 x0 x
0 24 y0   2  O  15.1   tan A   19.5  4.1  2   n, where n is an integer  1  a  tan A  b  T  2  66.  60b. 
1 60d.    c2  A  37.75273111 B  180° 
 (90°  37.8°) or 52.2° c  24.7, A  37.8°, B  52.2°  9.8  6  y  
1.2 about 1.2 m to the left  2  1    b2  24.66292764  c  y  1.5 cos 7.9  60c.  a2  15.12  19.52  c2  t)   9.8 y  1.5 cos t 6 9.8 y  1.5 cos 4 6    2  73   s  9 180   s  11.5 in.  y  0.6 about 0.6 m to the right  60a.  s  rv  73  180  1    131    59c. y  1.5 cos t    180°  t)  g  
  y  1.5 cos t 59b.  64. 73°  73°  k  1048  58c. A  2(0.2) A  0.4 y  0.4 sin (262  1    524  16 30x  15y 
 240 (3, 10) (0, 12) 20x  30y 
 360 8 (8, 0) y 
 0 (0, 0)  5   
1  O 61a. y  1.5 cos t y  1.5 cos t  k  m  2  k    18.5  0.4    61b. y  1.5 cos t y  1.5 cos t  k  m  18.5  0.6  2  k    y  1.5 cos 5.6t 61c. y  1.5 cos t y  1.5 cos t  k  m  k  0.9 s/cycle 1     5.6  k  1.1 s/cycle frequency: 2  k    18.5  0.8  y  1.5 cos 4.8t    1  1.1  0 69. 1 0 
1   0.9 hertz  q  24  32  x 40  
2 1 3 
3 
1 
1 
4 
2  1(1)  0(
1)  1(
2)  0(
1) 0(
2)  (
1)(
1) 0(1)  (
1)(
1) 1(3)  0(
4) 1(
3)  0(
2) 0(3)  (
1)(
4) 0(
3)  (
1)(
2)  
2 1 3 
3 1 1 4 2 (
2, 1), (1, 1), (3, 4), (
3, 2)   4.8  k  1.3 s/cycle 1   frequency:  1.3  0.8 hertz  61d. It increases. 61e. It decreases. 62. 0 63. 84 2  168 radians q  16  P(x, y)  100x  60y P(0, 0)  100(0)  60(0) or 0 P(0, 12)  100(0)  60(12) or 720 P(3, 10)  100(3)  60(10) or 900 P(0, 8)  100(0)  60(8) or 480 3 of Model 28, 10 of Model 74 68b. $900   6.8   frequency:  0.9  1.1 hertz  y  1.5 cos 6.8t  8  70.  g (x )  v  t 168  6  q  88.0 radians/s  O  185  x  Chapter 6  71. y  14.7x  140.1 y  14.7(20)  140.1 y  $434.10 72.  x2 (
4)2 (
3)2 (
2)2  x 
4 
3 
2  Translations of Sine and Cosine Functions  6-5 Page 378  y 16 9 4  Graphing Calculator Exploration  1.  {(
4, 16), (
3, 9), (
2, 4)}; yes 73.  1  radius  2(10) or 5  A  s2  A  r2 A  (5)2 or 25 4(25)  100  100  s2 10  s  2. The graph shifts farther to the left. 3. The graph shifts farther to the right.  The correct choice is C.  Page 377 1.  5  6  Mid-Chapter Quiz  5  Page 383  180°     6  150° 1  s  rv  2. r  2d r  1 (0.5) 2  5  s  0.253  or 0.25  s  1.3 m 1  3. A  2r2v 2  A  2(82)5 1  A  40.2 ft2 4. 7.8 2  15.6 or about 49.0 radians 5. 8.6 2  17.2 6. v  rq v v  3(8) q  t v  75.4 meters/s 17.2 q  7  5. Jamal; 6.  q  7.7 radians/s 7. 1 8.  Check for Understanding  1. Both graphs are the sine curve. The graph of y  sin x  1 has a vertical shift of 1 unit upward, while the graph of y  sin (x  1) has a horizontal shift of 1 unit to the left. 2. sine function 3a. increase A 3b. decrease h 3c. increase k 3d. increase c 4. Graph y  sin x and y  cos x, and find the sum of their ordinates. c 
k    2     
2 
—   6  or 3  y  
  y 
 3 cos (  2 )  2  y y 
 cos x  1  O    2  3  4  5   
2  O  9 x  8  7  7. 3; y  3 A  1;  
1  9. 
7  7;  2  1  3  2  2  2  y    y 
 7 cos 3  O  4  O  2  4  6  8  10 12   
8  10. A  5 A  5 y  5 sin 6v Chapter 6  y 
 sin 2  3  4   6  8  
4    y  2  k     3  k6  186    2  3  4  5   2    13c.  8. 2  2; 2  ; 
2; 
5  y O    
2  80  5   4  3  2  P 
 30 sin 2
t  100  P 120  y 
 2 sin (2  
)  5  P 
 30 sin 2
t  100  40  
4  O  1  
6    1  1  9. 
2  2;  2  1  2  Pages 383–386  
4   4; 
   2; 3 1  14.   2  c 
k    1  5  6  t  Exercises  2  or 2; A  1; 1  2  y  y  y 
 sin(  2
)  1  
  y 
 3  2 cos ( 2  4 )  4  
  
2  1  4  3  2  O  2    2  3  4  5   4  5   
1  O    5   4  3  2    c  2  k  10. A  20  1  A  20 k  2 y  20 sin 2v  100  2  k  A  0.6  11.   
2.13 c 
   
2.13   12.4  O  h7  16.  2.13      y  0.6 cos  6.2 v  6.2  7  12.  sin x 
 cos x sin 0 
 cos 0    2  sin 2 
 cos 2  1    sin  
 cos   1  3  2  3  2  sin  2  sin 2 
 cos 2  y  c 
k  3      2 
— 1  4  2 or 
2; A  2;  1  8  4  2  y 
1    
 cos  2  y  x 0      
1  2.13   c 6.2  y 
 sin(2  
)  1   k 6.2  c 
k   6.2  y  h  100    A  0.6  2  15. 
k  
 2; A  1; 2    c0  3  2  1   O  2  3  4 5  6 7   
1  
1  y 
 2 cos ( 4  
2 )  
2  
1  1  y 
 sin x  cos x  1  17. 2; y  2 2 A  1;  1  4  1   2  O    y  x  2  2  
1    1  y 
 sin 2  2 1  13a.  130  70  2  13b. A    100; P  100  130 
 70  2  A  30 A  30  2  k  O 1    2  3  4  5   
1  k  2 P  30 sin 2t  100  187  Chapter 6  2  18. 
4; y  
4 A  5;  2  1   2  y  y  30  O    
2    24. 5  5; 3; 
3; 20  2  4  3  5   y 
 20  5 cos (3  
)  20  
4  10  
6  O  
8  y 
 5 cos   4  A  1;   2   2    y  y  O  y 
 7  cos 2  8    
4  O 20.     
4 
2  21. 3  3;  2  1  y  2  4  3  5   2 
4   16; 
5 26. 10  10;  1  8; 
 1  4  or 2; 
3  
 2  y     2;  1  2; 0    2  4  3  
15  5   2 
 (
6)    3  A  4 A  4; 4    y 
 6 sin (  
 3  28. A  7  29. A  50   2  4  3  k  c 
 2   h  
7   3  2  3  2  c  
3  2  2  k  A  50  5   3   4  k  c   
 8  2  3  8  3  h  
25 4  c  
3  4  y  50 sin 3v 
 3 
 25 8   3  3  A  4  1  3  y  y   2  3  2  k  3  30. A  4    
12 2  23. 1  1;  1  6; 
   4; 
2  4  3 4  5   y 
 2  sin (  c  
1 0   k  10  c  
10  2  k    c   2  2  k  32. A  5 4  A  5  
4  y  188  4 5     6  c    c  
4 7  5  h7  c  
  
12  3  k  12  cos (12v 
 4)     
4  4  A  3.5 k4 y  3.5 cos (4v 
 )  7  )  1  h  4  1  4   
 3  12     5  sin (10v 
 10)  4  31. A  3.5  
2  Chapter 6   3  2  
4  O  2  k  y  7 sin 3v 
 3 
 7  2    4; 2 to the left or 
2; down 2, or 
2  A  7  ) 2  6 4  O    27. A  2  2  
2  5   4  3  2  
10   22. 6  6; 1  2; 
  1  
3; 2  y    
5  
2  8    O  
  y 
 3 cos (  2 )   4  y 
 10 sin ( 4  4
)  5  5  2  O    1  y 
 4 cos 4  3  4 2  5   4  3  2  
2  6  c 
k  5   4  3  2  0 1 1 2  25. 4  4;  1  4; 
 1  0; 
3  19. 7; y  7  2  2    7  h  5  2  k  33. A  100  c  45 
  2  0 2  A  100  k  45  38.  h  
110   45  2  y  100 cos 45v 
 110 34. A   1 
 (
3)  2  2  k  A  2   4  k  h  
1  1  2  y  
2 cos 2 
 1 v  3.5 
 (2.5)  2  k  A  0.5 A  0.5; 0.5 y  0.5 sin 2v  3 36.  x 0  1    0  3  2  
1  2  0  y 8  c    
2  0  k2  c0  sin 2x 0  sin x  sin 2x 0  0.71    2  1  1.71  1    0  1    0  2  0  0  3  2  
1  3  0  
1  h3  2  0  4  0  0  y y 
 sin x  sin 2x  2 1  sin x 0    2  2x 0    4   2  A  2; 
2  35. A  2  sin x 0  x 0  c0  sin x  x 0  
1   1  2.57  
2    2  O    3.14 3  2  39.  
 1  3.71    y 4  2  6.28  O 
4  40.   x 0  cos x 1  3  2  sin x 0  x  3  y  y 
 2 sin x  y 
 3 cos x  y 
 cos 2x  cos 3x  2  x  cos x 
 sin x 1    2  0  1  
1    
1  0  
1  3  2  0  
1  1  2  1  0  1  O 
2   2    3 2  y 
 cos 3x  2  y 
 cos 2x  41a. 2000  1000  3000 2000 
 1000  1000 41b. 10,000  5000  15,000 10,000 
 5000  5000  y  O  2  4  O  
1    
2  2  1  y 
 2 sin x  3 cos x  x y 
 x  sin x  4  2  x  2  6  37.  2  41c. y 
 cos x  sin x   2  x  
2  [0, 24] sc11 by [0, 16,000] sc11000 41d. months number 3 and 15 41e. months number 0, 12, 24 41f. When the sheep population is at a maximum, the wolf population is on the increase because of the maximum availability of food. The upswing in wolf population leads to a maximum later.  189  Chapter 6  y  42.  3   f(x)   x
1  49.  y
x  3  3   y x
1  2  3   x y
1    x(y 
 1)  3  x O    
  2  3  3  y 
 1  x  4  3  
2  y  x  1  y 
 cos x  y 
 x cos x  1 1 3 5 X 1 1 
3 
5 1(3)  1(
3) 1(5)  1(
5)  X 1(3)  1(
3) 1(5)  1(
5) 0 0 X 0 0 51. 7(3x  5y)  7(4) 21x  35y  28 → 14x 
 35y  21 14x 
 35y  21 35x  49 x  1.4 3x  5y  4 3(1.4)  5y  4 y  
0.04 (1.4, 
0.04) 50.  
3  43a. 46 
 42  4 ft 1  t  21  4  1 (42) 2  t  25  43b. r  2d r  r  21 43c.  3 revolutions  60 seconds  1 revolution    x seconds  x  20 s 2  k  43d. A  21 A  21; 21 h  25  21 sin 43e.   20  k  t  10  h  25  52.    10  x  4 
6  4 
4  4 
2  4 0  4  x 
6 
4 
2 0  t  10 t  10  h  25  21 sin    46  25  21 sin   t 1  sin 10 t sin
1  sin 10 5  t; 5 s  y  53. 3x 
 y  7  0 y  3x  7 slope: 3  t  43f. h  25  21 sin 1 0   10  h  25  21 sin 1 0  0   
2  
 2  There is a 4  or  75  cos x  45c. y  cos x2  45d. y  sin 
x     6-6  1  294  k  588 47. v  rq v  7(19.2) v  134.4 cm/s 48. asymptote: x  2 x
3   y x
2  y(x 
 2)  x 
 3 yx 
 2y  x 
 3 
2y 
 3  x 
 yx 
2y 
 3  x(1 
 y) 
2y 
 3  1
y  42  1  3   1050 cubic feet  Check for Understanding  1. any function that can be written as a sine function or a cosine function 2. Both data that can be modeled with a polynomial function and data that can be modeled with a sinusoidal function have fluctuations. However, data that can be modeled with a sinusoidal function repeat themselves periodically, and data that can be modeled with a polynomial function do not. 3. Sample answers: the amount of daylight, the average monthly temperatures, the height of a seat on a Ferris wheel  y x3  y
 x2  O  Modeling Real-World Data with Sinusoidal Functions  Pages 390–391  y  0.25 sin 588t  x  x  asymptote: y  1  Chapter 6  y 
 y1  m(x 
 x1) y 
 (
2)  3(x 
 3) y  2  3x 
 9 3x 
 y 
 11  0  phase difference. 45b. y  x  46.  x  1050 7.48  7854 gal The correct answer is 7854.    4  x 45a. y  
sin 2  k  O  1  44. 
k  
2 or 0 c 
k  y  |x  4|  54. 4 inches  3 foot  h  25 ft c  y 2 0 2 4  190    4a. y  
5 cos6t  5  8d. h  
3 cos 3t  3.5  4b. 5 units above equilibrium    y  
5 cos 6 0  5  h  
3 cos 3(25)  3.5  y  
5 5 units below equilibrium  t 6   6  4c. y  
5 cos  y  
5 cos     h  2 units    R  1200  300 sin 2 0  7  R  1200  y  
4.33 about 4.33 units above equilibrium 140 
 80    1  6a. A   P  30 sin 2t  110  66° 
 41°  2  6b. h   A  12.5° 6c. 12 months    H  232 9c. R: 1200  300  1500 H: 250  25  275 no  h  110  k  2    H  250  25 sin 2 0 
 4  140  80  A  30    9b. H  250  25 sin 2t 
 4  h  2  5. A  2 2  k    9a. R  1200  300 sin 2t    R  1200  300 sin 2t  9d.  66°  41°  2    1500  1200  300 sin 2t  h  53.5°    300  300 sin 2t   2  k  6d. A  12.5   12  h  53.5    sin1 1  2t   k  6  cos 6t  c  cos 6 1   cos 6  c   41  
12.5  
12.5  
12.5   1  cos 6  c y  
12.5  1  sin 2t sin11  t 2    53.5 c  53.5   1t January 1, 1971 9e. 250 
 25  225          
25  25 sin 2t 
 4    cos1 1 
 6  c      
1  sin 2t 
 4  
0.5  c    sin1 
 1  4  t    2      
0.5  t 2 2 July 1, 1969; k     6e. y  
12.5 cos 6t 
 0.5  53.5  y  
12.5 cos 6(2) 
 0.5  53.5  y  42.82517529 Sample answer: About 42.8°; it is somewhat close to the actual average.  t 
 0.5  6  (10) 
 0.5 6      sin1 
 1  2t 
 4    Sample answer: y  
12.5 cos 6t 
 0.5  53.5  y  
12.5 cos     225  250  25 sin 2t 
 4    cos1 1  6  c  6f. y  
12.5 cos     H  250  25 sin 2t 
 4   2  k4 next minimum: July 1, 1973 9f. See students' work.  53.5    53.5  2  k  4  10. A  2  y  53.20504268 Sample answer: About 53.2°; it is close to the actual average.   10   A2  k  5   y  2 cos 5t  c 2 11. h  4.25; A  3.55; k  12.40; 
    
4.68  Pages 391–394  Exercises  k  7a. 0.5 7b.  2  k     6.2    6.2  2.34   c 3.1  2.34    y  3.55 sin  6.2 t  3.1   4.24   660  c   2  1   k 330 1 7c.  1  330 hertz  12a. h  47.5; A  23.5; k  12; 
   4 6     330    k  6  2  c  
3  2  y  23.5 sin 6t 
 3  47.5  8a. 3.5  
3  6.5 units 8b. 3.5 
 3  0.5 units 2 2 6  8c. k   5 or 5  3  191  Chapter 6    2  12b. y  23.5 sin 6t 
 3  47.5   
  c  14. 
k  
 1 or   2      3  y  23.5 sin 6 3 
 3  47.5  increase shift by 2;   2  2  y  35.75 about 35.8°  
k  2  t 6   6  12c. y  23.5 sin  y  23.5 sin   y  67.9°  81° 
 73°  3  Sample answer: y  3 cos x 
 2  5 13.25 
 1.88  2  k   12  13.25  1.88  15b. h  2  15a. A  2  A  5.685 ft h  7.565 ft 15c. 4:53 P.M. 
 4:30 A.M.  12:23 or about 12.4 h  h  77°  13d. A  4  3  3  81°  73°  A  4° 13c. 12 months  c  c  
2  13b. h  2  13a. A  2  3  
1  2  2  3    47.5 2 8 
 3  47.5  
  c  2  k  15d. A  5.685  h  77  k    k  6 73  
4 cos   1  c  77     13.25  5.685 sin  6.2 4.5  c 4.5   5.685  5.685 sin  6.2  c    6  1  cos   c  cos1 1  cos1 1 
    6     h  5.685 sin  6.2 t  c  7.565  
4  
4 cos   c   6  h  7.565    6.2  4:30 A.M.  4.5 hrs    y  
4 cos 6t  c  77   6   6   12.4  4.5   1  sin  6.2  c  c  4.5   sin1 1   6.2  c  c  4.5   sin1 1 
  6.2  c  
0.5235987756  c  Sample answer: y  
4 cos 6t 
 0.5  77  
0.7093918895  c   Sample answer: h  5.685 sin  6.2 t 
 0.71  7.565    13e. y  
4 cos 6t 
 0.5  77   15e. 7:30 P.M.  19.5 hrs  y  
4 cos 6 8 
 0.5  77     h  5.685 sin  6.2 t 
 0.71  7.565  y  80.41594391 Sample answer: About 80.4°; it is very close to the actual average.     h  5.685 sin  6.2 19.5 
 0.71  7.565  h  8.993306129 Sample answer: about 8.99 ft 16a. Table at bottom of page.    13f. y  
4 cos 6t 
 0.5  77   y  
4 cos 6 5 
 0.5  77  y  79.08118409 Sample answer: About 79.1°; it is close to the actual average.  Month January February March April May June July August September October November December Chapter 6  Sunrise A.M. 7:19 6:56 6:16 5:25 4:44 4:24 4:33 5:01 5:31 6:01 6:36 7:08  A.M. Time in Decimals 7.317 6.933 6.267 5.416 4.733 4.4 4.55 5.017 5.517 6.017 6.6 7.133  Sunset P.M. 4:47 5:24 5:57 6:29 7:01 7:26 7:28 7:01 6:14 5:24 4:43 4:28  192  P.M. Time in Decimals 16.783 17.4 17.95 18.483 19.017 19.433 19.467 19.017 18.233 17.4 16.717 16.467  Daylight Hours (P.M.-A.M.) 9.47 h 10.47 h 11.68 h 13.07 h 14.28 h 15.03 h 14.92 h 14 h 12.72 h 11.38 h 10.12 h 9.33 h  15.03 
 9.33  15.03  9.33  16b. A  2  24. 402  322  202 
 2(32)(20) cos v  16c. h  2  A  2.85 h 16d. 12 months  402 
 322 
 202   cos v   
2(32)(20)  h  12.18 h  402 
 322 
 202  2  k  16e. A  2.85   12  k   v  cos1 
2(32)(20)   h  12.18  v  97.9° 180° 
 97.9°  82.1° about 97.9°, 82.1°, 97.9°, 82.1°    6   y  
2.85 cos 6t  c  12.18   6   6  9.47  
2.85 cos   25.  1  c  12.18  
2.71  
2.85 cos   c   cos1 0.950877193  6  c   cos1 0.950877193 
 6  c 
0.2088597251  c  Sample answer: y  
2.85 cos 6t 
 0.21  12.18   y  70.5  19.5 sin 6t  c  A  m
4    51  70.5  19.5 sin 6 1  c  26. 
2    6  
19.5  19.5 sin   c   sin1 
 1  6  c   sin1 
 1 
 6  c  18a.  B  B  
1  3      m4  m
4  m4  2  
1 
6 
2k  8 4k 
 14 
2k  7  4k 
 20  k 
4 k
4  f (x ) 
 2 x  1  5  
2.094395102  c Sample answer: about 
2.09 1 minute  60 seconds 7 t 15 7 t 15 7  4 15  A     m
4  m4  2 4k 
 20  0 k5 f (x ) 27.    
1  sin 6  c  14 revolutions  1 minute  2m  16    (m 
 4)(m  4)  2m  16  A(m  4)  B(m 
 4) Let m  
4. 2(
4)  16  A(
4  4)  B(
4 
 4) 8  
8B 
1  B Let m  4. 2(4)  16  A(4  4)  B(4 
 4) 24  8A 3A    0.950877193  cos 6  c  17. 70.5 
 19.5  51  2m  16   m2 
 16 2m  16  (m 
 4)(m  4)  x  O  2 radians  1 revolution    7  15  rad/s  y  
3.5 cos  18b.   y  
3.5 cos   y  
3.5 cos    increasing: x 
1; decreasing: x  
1 28. The correct choice is E.  y  
3.197409102 about (4, 
3.20) 120 
 (
120)  2  k  19. A  2 A  120   60  k  t 30  VR  120 sin   6-7    30    Page 400  20. See students' work. 2    y 
 3 cos(2  
)  5  8  Check for Understanding  1. Sample answers: 
, , 2 2. The asymptotes of y  tan v and y  sec v are the same. The period of y  tan v is  and the period of y  sec v is 2.  21. 
3  3; 2  ; 
2; 5  y  Graphing Other Trigonometric Functions    3  3. Sample answers: 2, 
2  6 4  4. 0 5. 1 6. n, where n is an odd integer  2  7.  O    2  3  4    4   n, where n is an integer  5   22. 2n where n is an integer 23. 800°    180°  40   9  193  Chapter 6  8.    1   
4  1   ;    24.   
4  25.  y  26.  
  y 
 tan (  4 )  4  27.  2 
2  
  2  2   n, where n is an integer   
4  n, where n is an integer 3   n, where n is an integer 4  n, where n is an odd integer 2  28. n, where n is an integer  O    
2  2   29.  
4  9.    4    1   ;   
 2 
 1  y 
  y 
 cot(  2 )  8     ; 
2; h  
1  4  y 
 sec(2  
)  1  y  
2  4  
  
  O    
4    2  
8  2 
2     2    O  
2  2   30.  
4  2  1  3   6; no phase shift; no vertical shift  y 10. k:  2  k   3  k  c:  2  3  c 
 2  3      3  4  h: h  
4 2  c  
9  
2  2  y  csc 3v 
 9 
 4 2    11. k: k  2  c: 
  1  k  2  c  1  2     
4  
  31.  2  1  12c.  y    8  12b. F  f sec v F  715.4 sec v  6 4   2; no phase shift, no vertical shift 800  
2  
   2;   
 4 
 1  2  F 
 715.4 sec  400    O  32.  2  y 
 csc   5  2  y  
2    1  2  O     y    12d. 715.4 N 12e. The tension becomes greater.    23.   2n, where n is an integer  Chapter 6  
  
2 
4  Exercises  0 14. 0 undefined 16. 
1 
1 18. undefined undefined 20. 0 n, where n is an integer n, where n is an even integer 3  2  
  y 
 tan ( 2  4 )  1  2 
2  13. 15. 17. 19. 21. 22.    2   2; h  1  4  Pages 400–403  2   2; no phase shift; h  5  y  cot 2v  8  2  1    
2 
4    12a. f  ma f  73(9.8) f  715.4 N    O  h: h  0  c  8 1    y 
 sec 3  2  194  O    2    33.  2  2    2   ; 
2; h  
3  O  
    
2  2  k  
6 
8  34.   6;    6 
 1  3       
2;  2    O  
    
2  2      40    2    d 40  
6  
40    d  10 tan 40(3)  y 
 sec   d  2.4 ft from the center   y 
 cos   O  44c. d  10 tan 40t   2     
2  d  10 tan 40(15)  
4  d  24.1 ft from the center y  45.  4  
2, 
, 0, , 2 c    c: 
  1 0  36. k: k  2   2  1  k  —2—  2  h: h  
6 
2  c0  
    37. k: k  2  c    c   
—4—  c: 
2  8  k2  h: h  7  2    c  38. k: k    c: 
2  
4  k2  c  h: h  
10    2  2    2  c  c: 
  2   39. k: k  3   3  2  k  3  c  c:  1  v  F  2(68.6) sec 2 v  F  34.3 sec 2 46c.  2 
3  2  1  5  v  46b. F  2f sec 2  h: h  
1  y  csc 3v 
 3 
 1  5  1  46a. f  m 9.8 f  7 9.8 f  68.6 N    y  sec 2v  2 
 10  2  1  2   4  F  c 
 1  5   
  40  h: h  12    F 
 34.3 sec 2    20  c  5   y  cot 5  5  12 v    The graph of y  csc v has no range values between 
1 and 1, while the graphs of y  3 csc v and y  
3 csc v have no range values between 
3 and 3. The graphs of y  3 csc v and y  
3 csc v are reflections of each other.    y  cot 2v 
 4  7  k  O  
2 
4  v  y  tan2 
 6  40. k:  t  20  44b. d  10 tan 40 t  2    k  
  d 
 10 tan 40 t  20  O 10 
20 
10 
20  4  2  h: h  7  7  no phase shift no vertical shift  y    c   
2  
4  35.  
     40  
8  
2  c  c: 
2  4  y  tan 2v 
  y 
 sec ( 3  6 )  2  2    k2 44a.  
  2  c  3  43. k: k  2  h  
2  y 
2  2  3 2  y 
 csc(2  
)  3  h: h  
8   3  y  sec 3v  3 
 8  
12  2  1  3  c  c: 
  2  
  42. k: k  3  
4  h: h  
5  c  3  y  csc (6v  3) 
 5    2    c  c: 
6  
2  k6  y 
2    41. k: k  3  O  195     Chapter 6  52. C  180° 
 (62°31
  75°18
) or 42°11
  46d. 34.3 N 46e. The tension becomes greater. 47a. 220 A 47b. 47c.  2  60  a  sin A 57.3  sin 62°31
  1   3 0 s   
 6 
 60  57.3 sin 75°18
  b  62.47505783  1    360    6    a  sin A 57.3  sin 62°31
    I  220 sin 60 60 
 6 I  
110 A  49b. h   c  43.37198044 C  42°11
, b  62.5, c  43.4  3.99  0.55  2  53a.  A  1.72 ft h  2.27 ft 49c. 12:19 P.M. 
 12:03 A.M.  12:16 or about 12.3 hr 2  k  49d. A  1.72   12.3  k  h  2.27  2  12.3  12:03  0.05 hr since midnight  73˚  2  t  c 12.3 2  0.05 12.3 0.1   c 12.3  h  1.72 sin   3.99  1.72 sin  1.72  1.72 sin  0.1   1  sin  12.3  c      2.27  c  2.27  4m x  54.  0.1   sin1 1 
  12.3  c  72    b2    42    sin A     c2  7 
 65 7
65   sin A  65 a tan A  b 7 tan A  4  sin A    b  4 
 65 4
65   65  cos A     1.55  2.27  cos A   12  1.55  2.27  55.  h  3.964014939 Sample answer: 3.96 ft  4  x2 
 4 ——  0 x2 
 3x 
 10 (x 
 2)(x  2) ——  0 (x 
 5)(x  2)  Test 
3:  zeros: 2, 
2  (
3)2 
 4  (
3)2 
 3(
3) 
 10  y 2  y 
 2 cos  
2  
  O 
1  5  5   2  1  Test 0:     2  Test 3:  
2  51. s  rv  Test 6:   s  183  
4   02 
 3(0) 
 10 
4  
10 32 
 4   32 
 3(3) 
 10 5  
10 2 6 
4   62 
 3(6) 
 10 32  8 02  excluded values: 5, 
2 0  0 false  0  0 false 0  0 true 0  0 false  
2  x  5  s  6 cm  6   56. k   0.5  k  12  Chapter 6  y  13.7 m a  c  c2  cos A  c  49e. noon  12 hrs since midnight  2  1  2  a2   y cos 73°  
65 c  1.545254923  c 2  Sample answer: h  1.72 sin  12.3 t  1.55  2.27  h  1.72 sin   4  x  4 tan 73° x  13.1 m  0.1  h  1.72 sin   4  53c. cos 73°  y  53b. tan 73°  4   sin1 1   12.3  c  2 t 12.3 2  12.3  c    sin 42°11
 57.3 sin 42°11
    3.99 
 0.55  2  c    sin C   c sin 62°31
  48. y  
1 tan v  2  50. 2  2;  b    sin 75°18
   b sin 62°31
  47d. I  220 sin 60t 
  49a. A   b    sin B  196  t  kr 10  12r r  0.83  57. 3x  2y  8 y  3 
2x  y  y  2x  1 
2y  
x  4 y  1 x 2  
2  O  Trigonometric Inverses and Their Graphs  6-8  4  Page 410  x  Check for Understanding  1. y  sin1 x is the inverse relation of y  sin x, y  1 1  (sin x)1 is the function y   sin x , and y  sin(x ) 1  is the function y  sin x. 58. y  17.98x  35.47; 0.88 59. A: impossible to tell B: 6(150)  10(90) 900  900; true C: impossible to tell D: 150  30  2(90) 150  210; false E: 3(90)  30  2(150) 270  330; false The correct choice is B.  2. For every y value there are more than one x value. The graph of y  cos1 x fails the vertical line test. 3. The domain of y  Sin x is the set of real numbers   between 
2 and 2, inclusive, while the domain of y  sin x is the set of all real numbers. The range of both functions is the set of all real numbers between 
1 and 1, inclusive. 4. Restricted domains are denoted with a capital letter. 5. Akikta; there are 2 range values for each domain value between 0 and 2. The principal values are between 0 and , inclusive. 6. y  Arcsin x x  Arcsin y Sin x  y or y  Sin x  6-7B Sound Beats Page 404 1.   2  y  y 1  y 
 Sin x  y 
 Arcsin x  O  
1    1 x    
2  the third graph  x  
1   y  Cos x  2  7.  2.   2  O  
2    x  Cos y  2   Cos1 x  y  2   y  Cos
1 x 
 2  y   2  1 
 y 
 Cos (x  2 )  3. 4. 5.  6.  7. 8.  Sample answer: The graph seems to stay above the x-axis for an interval of x values, and then stay below the x-axis for another interval of x values. 0.38623583 no 
1.78043; yes; the value for f(x) is negative and corresponds to a point not graphed by the calculator. Sample answer: As you move 1 pixel to the left or right of any pixel on the screen, the x-value for the adjacent pixel decreases or increases by almost 7. Thus, the "find" behavior of the function cannot be observed from the graph unless you change the interval of numbers for the x-axis. See students' work. Yes; no, they only provide plausible visual evidence.    
2  O 
1  8. Let v  Arctan 1. Tan v  1  v  4   2  y 
  y 
 Cos1x  2  x  
1  O  1 x    
2  9. Let v  Tan1 1. Tan v  1  v  4 cos (Tan1 1)  cos v   cos 4 
2    2  197  Chapter 6  16.  10. Let v  Cos1 2. 
2   
2   Cos v  2  y  arctan x x  arctan y tan x  y or y  tan x    y  v  —4— cos Cos1   
 2  2  
    2   v 
 2    
  4 2  
 —4—   cos    cos   cos      y 
 arctan x  O  
2  11. true 12. false; sample answer: x  1; when x  1, Cos1(
1)  , 
Cos1(1)  0 13a. C  2r 13b. C  40,212 cos v C  2(6400) C  40,212 km 13c. C  40,212 cos v 3593  40,212 cos v cos1   2 
  x  2  
  
2    2  3593  40,212 3593  40,212  y y 
 tan x  O  1  2  
2  1  Cos x  y or y  2 Cos x  y  y    1  y 
 12 Cos x   cos v  O  
1   x  O  y 
 Arccos 2x  1.48  v; about 1.48 radians 13d. C  40,212 cos v C  40,212 cos 0 C  40,212 km  x  y  Arccos 2x x  Arccos 2y Cos x  2y  17.  v    1 x  
1    y  2  Arcsin x  18.    x  2  Arcsin y  Pages 410–412 14.    x 
 —2—  Arcsin y  Exercises  y  arccos x x  arccos y cos x  y or y  cos x    y 2  y  y  1  
1  y 
 cos x    y 
 arccos x    Sin x 
 2  y or y  Sin x 
 2    O  2  x  y  y    1   2 
 2  1 x   x   2  O   Arcsin x  O  y 
 Sin (x  
2 )  
1  x  y  tan 2  19.  y  
1  15.  O  1 x  x  tan 2  
1  y  tan1 x  2  y  Sin x x  Sin y Sin1 x  y or y  Sin1 x  2 tan 1 x  y; y  2 tan1 x  y  y   2  1  y 
 Sin x   
1  Chapter 6   2  x  
  O  
1  x  y 
 tan 2  2  y 
 Sin1 x  O  
2  y  y  O 
2  1 x    
2  198  y 
 2 tan1 x     x  O  
2 
  2  x    y  Tan x 
 2  20.  1  30. Let a  Sin1 1 and   Cos1 2.    x  Tan y 
 2  1  Sin a  1  Cos   2      Tan1 x  y 
 2    a  2    3  sin Sin1 1 
 Cos1 2  sin (a 
 )    1  Tan1 x  2  y    21. y     sin 6  y y 
 Arccot x  4    y 
 Cot x  2  1   2   2  2  O 
2  31. No; there is no angle with the sine of 2. 32. false; sample answer: x  2; when x  2, Cos1(cos 2)  Cos1 1, or 0, not 2. 33. true 34. false; sample answer: x  
1; when x  
1, Arccos (
1)   and Arccos (
 (
1))  0. 35. true 36. true   x 
4 
2 O  4 x  2  
4  22. Let v  Sin1 0. Sin v  0 v0  23. Let v  Arccos 0. Cos v  0  v  —2—  
3   25. If y  tan 4, then  
3   Tan v  3 v  y  1.    6  tan   4  Sin1   26. If y  sin   then y    Sin1 1  O   ——. 4    2  2    2  0  sin 6t 
 3   2  sin1 0  6t 
 3   2  3  1  2  Tan a  1  Sin   1    4    2    cos (Tan1 1 
 Sin1 1)  cos (a 
 )      cos 4 
 2  cos 
4    41.  
2   2 1  Sin1 2.  42.    5  3   6t    4  2     6t   n, where n is an integer I  I0 cos2 v 1  8 cos2 v  1  Sin   2     6  cos1  cos Cos1 0  Sin12  cos (a  ) 1      or   6t 
 3  4t 10  t April and October 40. P  VI Cos v 7.3  122(0.62) Cos v 0.0965097832  Cos v Cos1 0.0965097832  v 1.47  v; about 1.47 radians  28. Let a  Tan1 1 and   Sin1 1.      0  23.5 sin 6t 
 3  0  6t 
 3  a  2  2  54.5  54.5  23.5 sin 6t 
 3    2   2  Cos a  0    y  54.5  23.5 sin 6t 
 3  39.  )  cos y cos (Tan1 
3   cos 3  29. Let a  Cos1 0 and   x   —— 2  1  27. If y  Tan1 
3 , then y  3.  a      Sin1 y    sin (2y)   sin 2 4  sin    y 
 tan(Tan1x )   
 2  Cos1 2, 
2  2 Cos1 2    37. false; sample answer: x  2; when x  2, cos1 2 is undefined. y 38.    24. Let v  Tan1 3.     sin 2 
 3    No; the inverse is y  Tan1 x  2.  1  8 1  8 1  8   cos2 v  cos v v  1.21  v; about 1.21 radians 43a. 6:18  12:24  18:42 or 6:42 P.M. 43b. 12.4 h     cos 2  6 2   cos 3  7.05 
 (
0.30)  43c. A  2  1   
2  A  3.675 ft  199  Chapter 6  2  k  43d. A  3.675  7.05  (
0.30)   12.4    k 6.2  y  48.  h  2  1  y 
 cos x  h  3.375  6:18  6.3 h   y  3.675 sin  6.2 t  c  3.375   6.2 6.3  6.2  7.05  3.675 sin   3.675  3.675 sin  6.3  6.2  1  sin   sin1 sin1  1  6.3  6.2  1
  6.3  6.2   c  
11  6.3  c  3.375   c  49.  30  c  x  30  sin 25°  
     sin  6.2 t 
 1.62   t 6.2  2.625  3.675 sin      sin1       
 1.62  30  sin 65°  
 1.62  1  a  2(180° 
 50°) or 65° 30  y    sin 50° 30 sin 50°   y sin 65°  
 1.62  y  25.4 units 50. 210° 
 180°  30° 51. p: 1, 2, 3, 6 q: 1, 2  4.767243867  t 0.767243867 60  46.03463204; Sample answer: about 4:46 A.M.  p ——: q  52.  y  1  3  1, 2, 3, 6, 2, 2  g (x ) 1  y 
 sin (Tan1 x ) 1  g (x ) 
 x  2  3  O  2  v  2(25°) or 50°  30  y 30    1.62   1.62  t  44.    30  6  3.375  3.675  2.625  sin 1  3.675 6.2 2 . 6 2 5  sin1   3.675  x    sin 130°  x  54.4 units     — — 6.2 t    6.2 t  
 30  30 sin 130°   sin  6.2 t 
 1.62   sin      x sin 25°  y  3.375  3.675    t 6.2  v  25° a  180° 
 (25°  25°) or 130°  30  25˚ 30  
1.621467176  c Sample answer:   y  3.375  3.675 sin  6.2 t 
 1.62  2.625  3.675 2.625  3.675  x  
1  c  43e.  
9 O  
10  x  2  O  x  
1  45a. v  v  decreasing for x  2 and x 2 53. [f 
 g](x)  f(g(x))  f(3x)  (3x)3 
 1  27x3 
 1 [g 
 f ](x)  g(f(x))  g(x3 
 1)  3(x3 
 1)  3x3 
 3 54. D  4, F  6, G  7, H  8 value: (4  6  7  8)4  (25)4 or 100 The correct choice is D.  D
d cos1 2 c 6
4 1   cos 2(10)  v  1.47 radians 45b. L  D  (d 
 D)v  2C sin v L  (6)  (4 
 6)1.47  2(10) sin 1.47 L  35.81 in. 46. n, where n is an integer 47. A  5  2  k  A  5   3  k  2 —— 3  c  
 2  
  3  h  
8  2  c  3  2  y  5 sin 3v  3 
 8 2  Chapter 6  200  2  Chapter 6 Study Guide and Assessment Page 413 1. 3. 5. 7. 9.  y  Understanding and Using the Vocabulary 2. 4. 6. 8. 10.  radian the same angle radian sunusoidal  1  y 
 0.5 sin 4  angular amplitude phase frequency domain  O    11. 60°  60°     180°  12. 
75°  
75°    180°  13. 240°  240°  14.  5  6  4   3 15.  7 
4        180°  180°     
315° 17. s  rv 3 s  154 s  35.3 cm  16. 2.4  2.4  O  q  q  2.3 radians/s 27. 15.4 2  30.8 q q  q q  q  6.5 radians/s 29. 
1 30. 0      c   2  
4  
2  A  4 k4 y  4 sin (4v  8) 
 1 2  k  37. A  0.5  2  k  3  38. A  —4— 3  A  4  h  
1  c  8    c    c  2 
3  
2  3  A  0.5 k2 2 y  0.5 sin 2v 
 3  3   h3  c   4  
8  0  k8  c0  h5  3  y  4 cos 8v  5 120 
 80  2  k  39. A  2 A  20 y  20 sin 2t  100 130 
 100  40. A  2  v  t 7.2  2  120  80  1  h  2  k  2 2 —— k  h  100 130  100  1  h  2  A  15 k  2 y  15 sin 2t  115  h  115  2  41. period: 1 or 2, no phase shift, no vertical shift  q  11.3 radians/min 28. 50 2  100  v  t 30.8  15  2  k  36. A  4  s  19.6 cm 20. s  rv  s  155 s  9.4 cm  q  6  
1  5  q  2  180°    19. 150°  150° 5  6 s  rv 5 s  156 s  39.3 cm 21. 5 2  10 or about 31.4 radians 22. 3.8 2  7.6 or about 23.9 radians 23. 50.4 2  100.8 or about 316.7 radians 24. 350 2  700 or about 2199.1 radians 25. 1.8 2  3.6 26. 3.6 2  7.2 q  4   4  y 
  13 cos 2  s  1512  v  t 3.6  5    1   137.5°   18. 75°  75°  180° 5    12 s  rv    180°  3  y  5  
12 5 180°   6    150°  7 
4  2  1  2  1  35. 
3  3;  Skills and Concepts    3  2  
1 1  Pages 414–416    34. 0.5  0.5; 4  2  y 1  v  t 100  12  y 
 13 csc   q  26.2 radians/min 31. 1 32. 0  O    2  3  4   2  33. 4  4; 2    y  
1  y 
 4 cos 2  4 2  O    2  3   
2 
4  201  Chapter 6  42.   ; 3    2 
 3  Page 417     
6; no vertical shift  Applications and Problem Solving   50a. A  11.5  2k  12  y  k  8  y 
 2 tan ( 3  
2 )  4  O    h  64   6    6    c  
2    y  11.5 sin 6t 
 2  64 50b. April: month 4  2     c  
  3      y  11.5 sin 6t 
 2  64  
4      y  11.5 sin 6 4 
 2  64  
8  y  69.75; about 69.8° 50c. July: month 7  43. vertical shift: 4      y  11.5 sin 6t 
 2  64  y      y  11.5 sin 6 7 
 2  64  6 4  y  74.0°  y 
 sec   4  F   B IL sin v  51.  2  0.2  O    2   0.04   5.0(1) sin v    3  
2  0.04(5.0(1) sin v)  0.2 0.2   sin v   0.04(5.0)(1)  44. vertical shift: 
2  sin v  1  v  2  y 2  O    2    3  Page 417  
2  1.  y 
 tan   2  A 26.2   
6  Open-Ended Assessment 1 r2v 2 1 r2v 2 2  Sample answer: r  5 in., v  3 46. Let v  Sin1 1. Sin v  1  v  2  45. Let v  Arctan 
1. Tan v  
1  v  
4  2a. Sample answer: If the graph does not cross the y-axis at 1, the graph has been translated. The first graph has not been translated and the second graph has been translated.    47. If y  tan 4, then y  1.  y    Cos1tan 4  Cos1y   Cos1 1 Let v  Cos1 1. Cos v  1 v0 
3   1  O    2  x    2  x    48. If y  Sin1 2, then y  —3—.  
1  sinSin1 2  sin y 
3     y   sin —3—  1  
3    2 1  49. Let a  Arctan 
3  and   Arcsin 2. Tan a  
3  a  1    3  cos (Arctan 
3   Arcsin  O  Sin   2  1  2    6  
1   cos (a  )      cos 3  6  cos  2b. Sample answer: If the equation does not have the form y  A cos kv, the graph has been translated. The graph of y  2 cos 2v has not been translated. The graph of y  2 cos (2v  ) 
 3 has been translated vertically and horizontally.    2  0  Chapter 6  202  Chapter 6 SAT & ACT Preparation Page 419  If either of the two factors equals 0, then the statement is true. Set each factor equal to 0 and solve for x. x
40 or x20 x4 x  
2 The solutions of the equation are 4 and 
2. To find the sum of the solutions, add 4  
2  2. The correct choice is D. 6. You may want to label the triangle with opposite, adjacent, and hypotenuse.  SAT and ACT Practice  1. Since there is no diagram, draw one. Sketch a right triangle and mark the information given.  4 
  A  Notice that this is one of the "special" right triangles. Its sides are 3-4-5. So the hypotenuse is 5. The sine is opposite over hypotenuse (SOH).  
  C  5 hypotenuse  4  sin v  5 The correct choice is B. 2. Let x be the smaller integer. The numbers are two consecutive odd integers. So, the larger integer is 2 more than the first integer. Represent the larger integer by x  2. Write an equation that says that the sum of these two integers is 56. Then solve for x. x  (x  2)  56 2x  2  56 2x  54 x  27 Be sure to read the question carefully. It asks for the value of the larger integer. The smaller integer is 27 and the larger integer is 29. The correct choice is C. 3. Factor the numerator. a2 
 b2  (a  b)(a 
 b) (sin v  cos v)(sin v 
 cos v)  sin v 
 cos v  opposite 3  adjacent  3  B  To find cos v, you need to know the length of the adjacent side. Notice that the hypotenuse is 5 and one side is 3, so this is a 3-4-5 right triangle. The adjacent side is 4 units. Use the ratio for cos v. adjacent  4    cos v   hypotenuse  5  The correct choice is C. 7. Look at the powers of the variables in the equation. There is an x2 term, an x term, and a y term, but no y2 term. It cannot represent a line, because of the x2 term. It cannot represent a circle or an ellipse or a hyperbola because there is no y2 term. So, it must represent a parabola. The general form of the equation of a parabola is y  a(x 
 h)2  k. The correct choice is A. 8. Factor each of the numerators and determine if the resulting expression could be an integer, that is, the numerator is a multiple of the denominator.   sin v  cos v  The correct choice is B. 4. First find the coordinates of point B. Notice that there are two right triangles. One has a hypotenuse of length 15 and a side of length 12. This is a 3-4-5 right triangle. The coordinates of point B are (9, 12). Since point A has coordinates (0, 0), each point on side AB must have coordinates in the ratio of 9 to 12 or 3 to 4. The only point among the answer choices that has this ratio of coordinates is (6, 8). A slightly different way of solving this problem is to write the equation of the line containing points A and B.  I II III  16n  16  n1 16n  16  16n 16n2  n  16n  16(n  1)    n  1  16; an integer 16(n  1)  n1    ; not an integer  16n n n(16n  1)  16n  1    ; not an integer  16n 16  Only expression I is an integer. The correct choice is A. 1  9. Since x 1, 1 
 x  0. So x1 
 x   x
 1. x 1  Since x 1, x x 
 1 1. So   1. x 
 1 x The correct choice is D.  12  y  9 x Then test each point to see whether it makes the equation a true statement. You could also plot each point on the figure and see which point seems to lie on the line segment. The correct choice is E. 5. Factor the polynomial on the left side of the equation. x2 
 2x 
 8  0 (x 
 4)(x  2)  0  203  Chapter 6  Add the two equations. m∠CAD  m∠BAD  m∠ACD  m∠BCD  160, so two of the angles in ABC have the combined measure of 160°. Therefore, the third angle in this triangle, ∠B, must measure 20°. The correct answer is 20.  10. Notice that the triangles are not necessarily isosceles. In ADC, the sum of the angles is 180°, so m∠CAD  m∠ACD  80. Since segment AD bisects ∠BAC, m∠BAD  m∠CAD. Similarly, m∠BAC  m∠ACD. So, m∠BAD  m∠BCD  80.  Chapter 6  204  Chapter 7 Trigonometric Identities and Equations 12.  Basic Trigonometric Identities  7-1 Page 427  1. Sample answer: x 
 45° 2. Pythagorean identities are derived by applying the Pythagorean Theorem to a right triangle. The opposite angle identities are so named because A is the opposite of A. 3.  1  
 
  
 cot v, 14.  sin A  cos A sin A   cos A  1  3  2  tan v 
 tan v 
  10.  v  cos2    15. cos x csc x tan x 
 cos x sin x  cos x  1   16. cos x cot x  sin x 
 cos x  sin x   sin x cos x  cos2  cos2 x  sin2 x   
 sin x 1   
 sin x  
 csc x 17.  B
  F csc v  I
  BI
 
 F csc v BI
   F
 csc v  F 
 BI
 csc v  1  1  
 5 2 2  
5  2
5 5  F 
 BI
sin v  Pages 427–430  v
1   cos2 v 
 1  
2  
2    2  2  
 1  24  1  2  2
6  cos v 
 5  1  19. Sample answer: 45°  2
6  Quadrant III, so 5  sec v  tan v sec 45°  tan 45° 
 2  1  11. tan2 v  1 
 sec2 v  472  1 
 sec2 v  1 
 sec2 v  65  49 
65   7  Exercises  18. Sample answer: 45° sin v cos v 
 cot v sin 45° cos 45° 
 cot 45°  cos2 v 
 25  16  49  x   
 sin x  sin x  152  cos2 v 
 1 1  25  sin x  
1  1  tan v 
  1  sin v — cos v  sin v 1 sin v    sin v cos v 1  cos v  
 sec v   9. tan v 
  cot v  3  sin2  
  
  
 tan A 5. Rosalinda is correct; there may be other values for which the equation is not true. 6. Sample answer: v 
 0° sin v  cos v 
 tan v sin 0°  cos 0° 
 tan 0° 01
0 10 7. Sample answer: x 
 45° sec2 x  csc2 x 
 1 sec2 45°  csc2 45° 
 1 (
2 )2  (
2 )2 
 1 22
1 41  sec v 
  csc v  cot v  
   8. sec v 
  cos v 1 2 sec v 
    1  sin (360°  30°) 1  sin 30°  
 csc 30°  sin (A)  
  
   csc (330°) 
  sin (330°)   4. tan(A) 
  cos (A)  
  2
 
3  
 cos 3 13. 330° 
 360°  30°  Check for Understanding  1 1 cos v    tan v 
  cot v , cot v 
 tan v , sin v 1  cot2 v 
 csc2 v  7
 
  
 2
   3 3 7
   cos 3 
 cos  
 sin v 
 sin 45° 
 2  
 2 
 2  
2   2  
 sec2 v 
 sec v 
65   Quadrant IV, so  7  205  Chapter 7  29. 1  cot2 v 
 csc2 v  20. Sample answer: 30° sec2  x1
  sec2 30°  1 
    2
 3 2  3  2  cos x  csc x cos 30°  csc 30°  
11   1  cot2 v 
  3   1  cot2 v 
 191  
3   2    1
  2  12  9  1
  
 3  4  1  3  
 3  4    cot2 v 
 29 
2   cot v 
   3 
2  Quadrant II, so   3  30. tan2 v  1 
 sec2 v  2  tan2 v  1 
 54  21. Sample answer: 30° sin x  cos x 
 1 sin 30°  cos 30° 
 1 1  2  tan2 v  1 
 2156 tan2 v 
 19 6 tan v 
 34  
3    2 
 1  1  
3    2  1  Quadrant II, so 34  22. Sample answer: 0° sin y tan y 
 cos y sin 0° tan 0° 
 cos 0° 00 
1 0 1 23. Sample answer: 45° tan2 A  cot2 A 
 1 tan2 45°  cot2 45° 
 1 11
1 21 24. Sample answer: 0  31. sin2 v  cos2 v 
 1 2  13  1  9  cos2 v 
 89 2
2  cos v 
   3 2
2  Quadrant III, so cos v 
  3 sin v  tan v 
  cos v  
  
  
  
  1    cos v  2  cos v  cos 2  3 tan v 
 — 2
2 3  cos 0  2  cos 0  cos 2 
    1 2
2   tan v 
   
  cos 2  cos 0  cos 2  25. csc v 
  2  26. cot v 
  1 csc v 
  2  23  1  tan v  4  9  1 cot v 
  
3    5  4  5  cot v 
  
3 4
3  cot v 
  3     cos v 
 cos v 
   cos2 v 
 1 15  
15    cos v 
  
 15 Quadrant I, so   sec2 v sec v  1  cos v 
  sec v   cos2 v 
 1  cos2 v 
 1 6   1 
 sec2 v  
13   Quadrant III, so sec v 
  3  27. sin2 v  cos2 v 
 1 1 2  4 1  16  
2   4   1 
 sec2 v  13  
 9 
 13 3 
   4  csc v 
 2  or  32. tan2 v  1 
 sec2 v  010 01 1  sin v   cos2 v 
 1  cos2 v 
 1  4  4  3
13  13  or    1  33. cos v 
  sec v  sin2 v  cos2 v 
 1  1 cos v 
  7 5  sin2 v  57 
 1  cos v 
  28. sin2 v  cos2 v 
 1  1  
 13  3 3  
13   57  2 2  sin2 v  3 
 1 2
6  Quadrant III, so   7  sin2 v  49 
 1 sin2 v 
 59 
5 sin v 
   3 Quadrant II, so  Chapter 7  
5  3  206  2  sin2 v  2459 
 1 sin2 v 
 2449 2
6  sin v 
   7  1  34. sec v 
  cos v  tan2 v  1 
 sec2 v tan2 v  1 
 82 tan2 v  1 
 64 tan2 v 
 63 tan v 
  3
7   1 sec v 
  1  8  sec v 
 8 Quadrant IV, so 3
7   40.  19
  5  
  
 2(2
)  5 19
  sin 5 19
 
 — tan  19
 5 cos 5  1  35. 1  cot2 v 
 csc2 v   sin v 
  csc v  2 43  1 sin v 
  53  sin 5 
 — 
 cos 5  sin v 
 35  
 tan 5  1    
 csc2 v  6 1  19 
 csc2 v 25  9  53  
  
  
 csc2 v  41.  
 csc v  Quadrant IV, so 53 36. 1  cot2 v 
 csc2 v 1  (8)2 
 csc2 v 1  64 
 csc2 v 65 
 csc2 v 
65  
 csc v Quadrant IV, so 
65   10
 —— 3  
  
 3
  3 1 10
 csc —3— 
  10
 sin 3 1  
 sin 3
  
    3  1 
  
 sin 3    
  
 csc 3  1  37. sec v 
  cos v  42. 1290° 
 7(180°)  30°  sin2 v  cos2 v 
 1 
3 2  sin2 v   4 sin2 v  13 6 sin2 v    1 sec v 
  
3    4  4
3  or   sec v 
  3 
 3 4    1   sec (1290°) 
  cos (1290°)  
1  
  
1  
  13  
 1 6  1  cos (7(180°)  30°) 1  cos 30°  
 sec 30° 43. 660° 
 2(360°)  60°  
13   sin v 
  4 
13   cos (660°)  Quadrant II, so 4   cot (660°) 
  sin (660°)  sin v  tan v 
  cos v  
  
 13   
  4 tan v 
 — 
3    4  cos (2(360°)  60°)  sin (2(360°)  60°) cos 60°  sin 60°  
 cot 60°  
 13 
3   
39  3  tan v 
  or  sec2 A  tan2 A  2sin2 A  2cos2 A  
  sin 2(2
)  5 —— 
 
 cos 2(2
)  5  
  44.  
39  2 4
3 2    3 3  
 
3 2 13 2 2   2  4 4                  
 
  
 csc x  48  9  9  39 
  21136  213 6  
  sec x —— tan x  1  cos x — sin x  cos x 1  sin x  45.  cot v —— cos v  cos v  sin v  
  cos v 1   
 sin v  9  9  32  16  
 csc v  46.  
 12  sin (v  
) —— cos (v  
)  sin v  — 
— cos v  
 tan v 47. (sin x  cos x)2  (sin x  cos x)2 
 sin2 x  2sin x cos x  cos2 x  sin2 x  2sin x cos x  cos2 x 
 2sin2 x  2 cos2 x 
 2(sin2 x  cos2 x) 
2  38. 390° 
 360°  30° sin 390° 
 sin (360°  30°) 
 sin 30° 27
 3
  
 3
   39.  8 8  27
 3
  
 cos 3
   cos  8 8  
 cos 38
  207  Chapter 7  — —— 48. sin x cos x sec x cot x 
 sin x cos x— cos x  sin x  cos x  1  
 cos x  eAs 
 W sec v  — —— 49. cos x tan x  sin x cot x 
 cos x— cos x   sin x sin x  cos x  sin x  eAs  
 sec v  
 sin x  cos x  W  W 
 eAs cos v 56b. W 
 eAs cos v W 
 0.80(0.75)(1000) cos 40° W  459.6266659 459.63 W 57. FN  mg cos v 
 0 FN 
 mg cos v mg sin v  mkFN 
 0 mg sin v  mk(mg cos v) 
 0 mk(mg cos v) 
 mg sin v  — —— 50. (1  cos v)(csc v  cot v) 
 (1  cos v)— sin v  sin v  1  W sec v  e 
 A s  56a.  cos v  1  cos v  — 
 (1  cos v)— sin v  1  cos2 v  — 
— sin v sin2 v   
 sin v  
 sin v 51. 1  cot2 v  cos2 v  cos2 v cot2 v 
 1  cot2 v  cos2 v(1  cot2 v) 
 csc2 v  cos2 v(csc2 v) 
 csc2 v(1  cos2 v) 
 csc2 v(sin2 v)  mg sin v   mk 
  mg cos v sin v   mk 
  cos v  1  mk 
 tan v  —(sin2 v) 
— sin2 v  
1 52.  58.  sin x sin x    1  cos x 1  cos x sin x  sin x cos x 
  1  cos2 x    sin x  sin x cos x  1  cos2 x  
  2 sin x   
 1  cos2 x  h  2 sin x  a   
 sin2 x 2   
 sin x  
 2csc x 53. cos4 a  2cos2 a sin2 a  sin4 a 
 (cos2 a  sin2 a)2 
 12 or 1 2 54. I 
 I0 cos v 0 
 I0 cos2 v 0 
 cos2 v 0 
 cos v cos1 0 
 v 90° 
 v 55. Let (x, y) be the point where the terminal side of A intersects the unit circle when A is in standard position. When A is reflected about the x-axis to obtain A, the y-coordinate is multiplied by 1, but the x-coordinate is unchanged. So, sin (A) 
 y 
  sin A and cos (A) 
 x 
 cos A.  v
  360°  2n  
  180° , n  a  2  a a   tan v 
  , so h 
  2 tan v 
 2 cot v. h  The area of the isosceles triangle is 2(a)2 cot n 1  a  180°  
 4 cot n. There are n such triangles, so a2  180°  A 
 4na2 cot n. 1  180°  y  59.  A  B  C  E  x  
  F D  O  y sin v 
 EF and cos v 
 OF since the circle is a unit  (x, y)  CD  CD    circle. tan v 
  OD 
 1 
 CD.  A  O  CO  CO  
A    sec v 
  OD 
 1 
 CO. EOF  OBA, so  x  OF  EF  BA  BA  cos v  EO  (x, 
y)  OB  1  OB      Also by similar triangles,  EF 
 OA , or EF 
 1 . 1  1  OB     Then csc v 
  sin v 
 EF 
 1 
 OB. 
2   60. Cos1   2  
 135°  Chapter 7  OF      
 OA 
 1 
 BA. Then cot v 
 sin v 
 EF 
 BA.  208  61.  42  y 1  y  cos (x   
6   68. m 
  4  5  )  
  2  9  or  y  y1 
 m(x  x1) 2 9  2  y  4 
 9(x  (4)) 2  28  y 
 9x  9  x O  2 3  6  7 6  5 3  69. m∠BCD 
 40° 1 40 
 m (BC)  13 6  2  80 
 m (BC) 1 m∠BAC 
 2mBC  1  1  
  62. 2(3° 30
) 
 7°  m∠BAC 
 40° The correct choice is C.   7° 
 7°   180°  
 s 
 rv  m∠BAC 
 2(80)  7
  180  7
   s 
 20 180   s  2.44 cm 63. B 
 180°  (90°  20°) or 70° a  b  sin A 
 c  cos A 
 c  a  cos 20° 
 3 5  35 sin 20° 
 a 35 cos 20° 
 b 11.97070502  a 32.88924173  b a 
 12.0, B 
 70°, b 
 32.9 64. 2 2 1 8 4  4 1 0 4  2 5 2  0 2x2  5x  2 
 0 (2x  1)(x  2) 
 0 2x  1 
 0 or x  2 
 0 1  x 
 2 65.  2,  1 2,  2x2   7x  4 
 0  Verifying Trigonometric Identities  Page 433  Graphing Calculator Exploration  1. yes 2. no 3. no 4. No; it is impossible to look at every window since there are an infinite number. The only way an identity can be proven is by showing algebraically that the general case is true.  b  sin 20° 
 35  7-2  5.  x 
 2  2  
  [2
, 2
] sc12 by [2, 2] sc11 sin x  7  x2  2x  2 
 0 7  x2  2x 
 2 7  49  Pages 433–434  49  x2  2x  16 
 2  16  x  742 
 8116 7  9  x  4 
 4 7  Check for Understanding  1. Answers will vary. 2. Sample answer: Squaring each side can turn two unequal quantities into equal quantities. For example, 1  1, but (1)2 
 12. 3. Sample answer: They are the trigonometric functions with which most people are most familiar. 4. Answers will vary.  9  x 
 4  4 x 
 0.5 or 4 66. continuous 67. 4(x  y  2z) 
 4(3) 4x  4y  8z 
 12 4x  y  z 
 0 → 4x  y  z 
 0 3y  9z 
 12 x  y  2z 
 3 x  5y  4z 
 11 4y  2z 
 14 4(3y  9z) 
 4(12) → 12y  36z 
 48 3(4y  2z) 
 3(14) 12y  6z 
 42 30z 
 90 z 
 3 3y  9z 
 12 x  y  2z 
 3 3y  9(3) 
 12 x  (5)  2(3) 
 3 y 
 5 x 
2 (2, 5, 3)  cot x   5. cos x 
  csc x  cos x 
 cos x 
  cos x  sin x — 1  sin x cos x  1  cos x 
 cos x  209  Chapter 7  6.  1  tan x  sec x  Pages 434–436  cos x  
 sin x  1  13. tan A 
  1 cos x   sin x 1 
 sin x  1    cos x cos x 1 cos x   sin x  1 
  sin x  1  cos x cos x  sin x  1  tan A 
 tan A 
  cos x   
 sin x  1  tan A 
 tan A 14. cos v 
 sin v cot v  1   7. csc v  cot v 
  csc v  cot v 1  csc v  cot v  cos v   cos v 
 sin v  sin v    csc v  cot v 
  csc v  cot v  csc v  cot v csc v  cot v  cos v 
 cos v   csc v  cot v 
  csc2 v  cot2 v  csc v  cot v 
 csc v  cot v 
  1  sin x  csc v  cot v  (1  cot2 v)  cot2 v csc v  cot v  1   15. sec x  tan x 
  cos x 1  sin v tan v 
 sin v tan v 
 sin v tan v 
 sin v tan v 
  sec x  tan x 
 sec x  tan x  sin x  cos x  cos x  sin A  
 sec x  1  cos x  
 sec x  sin x  cos A  cos x  sin x  sin x  cos x  cos x      sec x csc x 
  cos x  sin x  sin x  cos x  1  2 sin2 A cot A 
 1  2 sin2 A cot A  sin2 x  cos2 x    sec x csc x 
  cos x sin x  sin x cos x  1  4  sin2 x  cos2 x   sec x csc x 
  cos x sin x  sec x  1   sec x csc x 
  cos x sin x 1  1    sec x csc x 
  cos x  sin x  sec x csc x 
 sec x csc x  1  
 4  2 sin2 v  1   18. sin v  cos v 
  sin v  cos v  1  sin x 
 4  sin v  cos v 
  11. Sample answer: cos x 
 1 cot x  sin x 
 cos x cot x cos x  sin x  sin2 v  cos2 v  sin v  cos v 
  cos x    sin x 
 cos x  sin x  I cos v  R2   
 R2 csc v  I cos v  R2   I sin v 
 — 1  R2  sin v  I cos v  R2   I sin v sin v  
 —  sin v 1 2   R sin v  I cos v  R2  
 R 2  2 sin2 v  (sin2 v  cos2 v)  sin v  cos v   sin v  cos v 
  sin v  cos v  (sin v  cos v)(sin v  cos v)  sin v  cos v  sin v  cos v 
 sin v  cos v  cos x  sin2 x 
 cos2 x cos2 x  sin2 x 
 cos x 1 
 cos x cos x 
 1  2  sec A csc A   19. (sin A  cos A)2 
  sec A csc A sec A csc A  2    (sin A  cos A)2 
  sec A csc A  sec A csc A 1  1    (sin A  cos A)2 
 2 sec A  csc A  1  I cot v  (sin A  cos A)2 
 2 cos A sin A  1 (sin A  cos A)2 
 2 cos A sin A  sin2 A  cos2 A (sin A  cos A)2 
 (sin A  cos A)2  cos v  cos v  Chapter 7  cos x  sin x  cos x(sin x  cos x)    sec x csc x 
  cos x  sin x  2  1  2 sin2 A  sin A 
 1  2 sin A cot A  12.   1 cos x 
 sec x   sin x  cos x  sec x 
 sec x 17. sec x csc x 
 tan x  cot x  2  1  2 sin A cos A  sin A 
 1  2 sin A cot A  sin x  cos x — 1  cos x  
 sec x   
 sec x sin x  cos x  sin v tan v 
 sin v tan v 9. (sin A  cos A)2 
 1  2 sin2 A cot A sin2 A  2 sin A cos A  cos2 A 
 1  2 sin2 A cot A 1  2 sin A cos A 
 1  2 sin2 A cot A  1 tan x 
 4 tan x 1  
  sec x 4  1  tan x  sin x  cos x sin x  1 cos x  16.  1   cos v cos v 1 cos2 v    cos v cos v 1  cos2 v  cos v sin2 v  cos v sin v  sin v  cos v  10. Sample answer: sin x 
  sin x    sec x  tan x 
  cos x  cos x  csc v  cot v 
 csc v  cot v 8. sin v tan v 
 sec v  cos v sin v tan v 
  Exercises  sec A  csc A 1  cos A — 1  sin A sin A  cos A  I cos v  210  20.  (sin v  1)(tan v  sec v) 
 cos v sin v tan v  tan v  sin v sec v  sec v 
 cos v sin v  sin v  1  cos x  cos x sin x  sin x  cos x 
  cos x sin x    1 1 sin x cos x  1      sin v  cos v  cos v  sin v cos v  cos v 
  cos v sin2 v  sin v  sin v  1  cos v sin2 v  1  cos v cos2 v  cos v  
  1  sin y  cos y  
  1  sin y  cos y  cos y(1  sin y)  1  sin2 y  
  1  sin y  cos y  cos y(1  sin y)  cos2 y   
 cos y  1  sin y  cos y  1  sin y  cos y  cos y  1  sin y  21. cos y  1  sin y    1  sin y  1  sin y  
 cos v  cos x sin x cos x sin x    sin x  cos x 
  cos x  sin x sin x  cos x      1  cos x 1  sin x  
 cos v  cos2 x  
 cos v  cos v 
 cos v  cos2 x  sin2 x  cos2 x   sin x  cos x 
  sin x  cos x (sin x  cos x)(sin x  cos x)  sin x  cos x 
  sin x  cos x sin x  cos x 
 sin x  cos x 28. sin v  cos v  tan v sin v 
 sec v  cos v tan v sin v   sin v  cos v   cos v sin v 
 sec v  cos v tan v sin2 v   sin v  cos v   cos v 
 sec v  cos v tan v cos2 v  sin2 v    sin v   cos v  cos v 
 sec v  cos v tan v cos2 v  sin2 v   
 sec v  cos v tan v sin v   cos v 1  cot2 x   sin v   cos v 
 sec v  cos v tan v  csc2 x  1  sin v  sec v 
 sec v  cos v tan v   23. csc x  1 
  csc x  1   csc x  1 
  csc x  1  cos v   sin v  cos v  sec v 
 sec v  cos v tan v  (csc x  1)(csc x  1)  csc x  1  sin v   cos v  cos v  sec v 
 sec v  cos v tan v  csc x  1 
 csc x  1 24. cos B cot B 
 csc B  sin B cos B cot B 
 cos B cot B 
 cos B cot B 
 cos B cot B 
 cos B cot B 
  cos v tan v  sec v 
 sec v  cos v tan v sec v  cos v tan v 
 sec v  cos v tan v 29. Sample answer: sec x 
 
2   1   sin B sin B sin2 B 1    sin B sin B 1  sin2 B  sin B cos2 B  sin B cos B  cos B  sin B  csc x  cot x 1  sin v — cos v  sin v 1  cos x  cos B cot B 
 cos B cot B 25. sin v cos v tan v  cos2 v 
 1 sin v  sin2  v  cos2  v 
1 1 
1  1  cos x  1  cos x  x  cos x  sin x  cos x —— sin x  cos x  sin x sin x  cos x  1  cos x       2 sin x  sin x  sin2 x 
 1  cos x 1  2 cos x  cos2 x  sin2 x (1  cos x)2  1  cos2 x (1  cos x)2  (1  cos x)(1  cos x) 1  cos x  1  cos x  
 
2   
2  —— 
 2 cos x  1 sin x  1  cos x   csc2 x  2 csc x cot x  cot2 x 
  1  cos x cos2  
 
2   1  tan x  1  cot x sin x  1 cos x   (csc x  cot x)2 
  1  cos x  26.  
 
2   sec x 
 
2  30. Sample answer: tan x 
 2  2  sin v cos v  cos v  cos v 
 1  1  sin2 x  sin2 x    sin x  cos x 
  sin x  cos x  sin x  cos x  22. cos v cos (v)  sin v sin (v) 
 1 cos v cos v  sin v(sin v) 
 1 cos2 v  sin2 v 
 1 1
1  csc x  1 
  sin2 x    sin x  cos x 
  cos x  sin x  sin x  cos x  1  sin y  
  sin x    27. sin x  cos x 
  1  tan x  1  cot x  1  cos x   
 1  cos x 1  cos x   
 1  cos x  
2 
2  tan x 
 2 31. Sample answer: cos x 
 0  1  cos x   
 1  cos x  1  cot x  1  cos x   
 1  cos x  sec x    csc x 
 cos x 1  cos x — 1  sin x sin x   cos x  tan x   
 cos x  tan x  
 cos x  tan x  tan x 
 cos x 0 
 cos x  211  Chapter 7  38. yes  1  32. Sample answer: sin x 
 2 1  cos x sin x    sin x 1  cos x 1  2 cos x  cos2 x sin2 x    sin x(1  cos x) sin x(1  cos x) 1  2 cos x  cos2 x  sin2 x  sin x(1  cos x) 2  2 cos x  sin x(1  cos x) 2(1  cos x)  sin x(1  cos x) 2  sin x  
4 
4 
4 
4 
  
4  [2
, 2
] sc12 by [4, 4] sc11  
4  39. no  2 
 4 sin x 1  2  
 sin x  33. Sample answer: sin x 
 1 cos2 x  2 sin x  2 
 0 1  sin2 x  2 sin x  2 
 0 0 
 sin2 x  2 sin x  1 0 
 (sin x  1)2 0 
 sin x  1 sin x 
 1 34. Sample answer: cot x 
 1 csc x 
 sin x tan x  cos x  
  [2
, 2
] sc12 by [4, 4] sc11 40a. P 
 I 02 R sin2 2
 ft P 
 I 02 R(1  cos2 2pft) 40b. P 
 I 02 R sin2 2
ft  sin x  I 02 R   csc x 
 sin x  cos x  cos x sin2  x  cos2   P
 csc2 2
ft  x    csc x 
  cos x  cos x  x  41. f(x) 
  
 1  4 x2  1   csc x 
  cos x 1  sin x cos x  sin x  1   tan v 2 f(x) 
   1   
 cos x  tan v 1 4   cot x 
 1 35.  2  1  2  
1  tan3 v  1   sec2 v tan v  1 (tan v  1)(tan2 v  tan v  1)   (tan2 v  1) tan v  1 tan2 v  tan v  1  tan2 v  1  1  2  tan v  f(x) 
 
 1  ta n2 v  1
0  1
0 1
0 tan v  1 
 0 tan v 
 1  1   tan v 2  f(x) 
 
 sec2 v 1  2  tan v f(x) 
  sec v  1   cot v 
  tan v 1  cot v 
 1  sin v  1  2    cos v f(x) 
  1  cot v 
 1 36. no   cos v  1  f(x) 
 2 sin v sin a   42. sin a 
 sin a sin c ⇒ sin a 
  sin c cos b   cos b 
  sin a ⇒ cos b 
 sin a cos b cos c   cos c 
 cos a cos b ⇒ cos b 
  cos a  Then cos b 
 sin a cos b  
  [2 
, 2 
] sc12 by [2, 8] sc11  sin a  cos c  sin a  cos c    
 sin c  cos a  37. yes    
 cos a  sin c  
 tan a cot c 43. y 
  gv2   2v02 cos2 v  x sin v    cos v  gv2   sec2 v  x tan v y
 2v 2 0  [2 
, 2 
]  Chapter 7  
 sc12  g x2   (1  tan2 v)  x tan v y 
  2v 2  by [4, 4] sc11  0  212  51. Let x 
 the number of shirts and y 
 the number of pants. y 100 x  1.5y 100 2.5x  2y  180 2.5x  2y 180 (0, 65) 80 1.5x  3y 195 x  1.5y  100 x 0 60 (40, 40) y 0 1.5x  3y  195  44. We find the area of ABTP by subtracting the area of OAP from the area of OBT. 1 OB 2  1  1  1   BT  2OA  AP 
 2  1  tan v  2 cos v sin v  
 2 cos v  cos v sin v 1 sin v   
 2 sin v cos v  cos v 1  1    
 2 sin v cos v  cos v  1  
 
 
 
  1  2 1  2 1  2 1  2  cos2 v  1  sin v  40 x0 20 (0, 0)    1 cos2 v  cos v sin2 v  cos v  sin v    sin v  a sin b     45. By the Law of Sines,  sin b 
 sin a , so b 
 sin a . Then 1  A 
 2ab sin   A 
 2a sin a  sin  1  a sin b  a2 sin  sin    A
 2 sin a  A
 A
 46.  53.  sin b sin   2 sin (180°  (b  )) a2 sin b sin   2 sin (b  ) a2  tan x  cos x  sin x tan x  sec x  tan x  sin x  cos x  sin x  cos2 x  sin2 x  cos x 1  sin x  cos x sin x  1 cos x    cos x 1  sin x  15
  ab  ba    
 ab  ba  ab  ab  Sum and Difference Identities Check for Understanding  1. Find a counterexample, such as x 
 30° and y 
 60°. 2. Find the cosine, sine, or tangent, respectively, of the sum or difference, then take the reciprocal. 3. The opposite side for 90°  A is the adjacent side for A, so the right-triangle ratio for sin (90°  A) is the same as that for cos A.  2 
 45° c 
 90°  180°   
 1 6  
  
 168.75°  ba  ba  Pages 441–442  c  
 180°  A 
 2 k
2 y 
 2sin (2x  90°) 15
  16  ab  ab  7-3  
1  48.  80 100  
 1 The correct choice is D.  sin x    cos x  sin x  cos x 
  1 sin x    cos x cos x  
 360°  k  40 60 y0    
 a  b  1(a  b)  
  47. A 
 2  20  P(x, y) 
 5x  4.5y P(0, 0) 
 5(0)  4.5(0) or 0 P(0, 65) 
 5(0)  4.5(65) or 292.50 P(40, 40) 
 5(40)  4.5(40) or 380 P(72, 0) 
 5(72)  4.5(0) or 360 40 shirts, 40 pants 52. {16}, {4, 4}; no, 16 is paired with two elements of the range  tan v sin2 v a  x  O  2   cos v sin v  b  (72, 0)  60
  168.75° 
 168°  0.75°  1 ° 
 168°  45
 168° 45
 3 49. 
 3y  1  2 
 0 3 
 3y  1 
 2 3y  1 
 8 y
3 50. x  1 
 0 x 
 1 f(x) 
  90˚ 
 A  3  Check: 
 3y  1  2 
 0 3 
 3(3)  12 
0 3 
8 2 
0 22 
0  A 1   4. cot (a  b) 
  tan (a  b)  1 
  tan a  tan b  1  tan a tan b  3x  x1  1  tan a tan b  3x   y 
 x1  y(x  1) yx  y y y y  3y   
 tan a  tan b  
 3x 
 3x 
 3x  yx 
 x(3  y)  1  1    1 cot a  cot b cot a cot b  
   1 1 cot a cot b    cot a cot b cot a cot b  1   
 cot a  cot b  
x  3y
0 y
3  213  Chapter 7  10.  5. cos 165° 
 cos (45°  120°) 
 cos 45° cos 120°  sin 45° sin 120° 
   
2  2   2  1   
2  2    
3   2  
2   
6   4 
 
 tan 3  4 
 
 tan 3  tan 4  
 
    6. tan 12 
  
  sin v  2 —— 
 cot v 
 cos v  2    
  (sin v)  0  (cos v)  1  (cos v)  0  (sin v)  1 cos v   sin v   
 2  
3 7. 795° 
 2(360°)  75° sec 795° 
 sec 75° cos 75° 
 cos (30°  45°) 
 cos 30° cos 45°  sin 30° sin 45° 
 
 sec 795° 
    
3 
2    2 2 
6   
2   4 4  
6   
2   1   2   
  sin v cos 2  cos v sin 2 
 cot v ——— 
 
 cos v cos 2  sin v sin 2  
3 1  1  
3 1 4  2
3   2  
  
  tan v  2 
 cot v  11.  
 —— 
 
 1  tan 3 tan 4 
  sin (90°  A) 
 cos A sin 90° cos A  cos 90° sin A 
 cos A 1  cos A  0  sin A 
 cos A cos A 
 cos A  cot v 
 cot v 12. sin (x  y) 
  
2   2  sin (x  y) 
 —— 1 1    sin x cos y sin y    1 sin x  cos y sin x cos y  sin (x  y) 
 ——  sin x cos y 1 1    sin x cos y  
 
6   
2  
   2 1  
   2 1  
  
65  65    81 or 9  
  
15  15    16 or 4  4  9  sin (x  y) 
 sin (x  y) 13. sin (nq0t  90°) 
 sin nq0t cos 90°  cos nq0t sin 90° 
 sin nq0t  0  cos nq0t  1 
 cos nq0t  1  4  sin (x  y) 
 sin x cos y  cos x sin y 
65   sin x cos y  cos x sin y  1  sin (x  y) 
  cos y 
 
 1  si n2 y  
15   
 cot v  1  cot x tan y  csc x sec y cos x sin y   1 sin x  cos y  cos x  8. cos x 
 
 1  si n2 x  
 cot v    
 9  4    9  4  4  1  Pages 442–445  4
15   
65    
 36  cos x 
 
 1  si n2 x  1   9. csc x 
  sin x 1  5  3   
 sin x  
  3  
  sin x 
 5    1  
  3  2 5  
  1265 or 45  
  or  
  sin y  16. cos   tan y 
  cos y    1  5 2  13  
  144 12  or   169 13  12  13 — 5  13  12  or 5  tan x  tan y   tan (x  y) 
  1  tan x tan y 3  4  17. sin  12  5   
 —— 3 12 1  45 
 
  Chapter 7    
3   2  
3 
2    1 
2       2 2 2 2   
2  
6 
  4 7
 
 
  
 cos    12 4 3 
 
 
 
 
 cos 4 cos 3  sin 4 sin 3 
2 
2   
3  1    
  2  2  2  2 
2   
6  
  4 
 
 
  
 sin    12 3 4 
 
 
 
   
 sin 3 cos 4  cos 3 sin 4   
2  
3 
2 1    
  2  2  2  2 
6   
2  
  4     
  3  4  sin y 
 
 1  co s2 y 
  
2 
2   1      2 2 2 
2   
6   4  15. sin 165° 
 sin (120°  45°) 
 sin 120° cos 45°  cos 120° sin 45°  sin x   tan x 
  cos x 3  5  4  5  Exercises  14. cos 105° 
 cos (45°  60°) 
 cos 45° cos 60°  sin 45° sin 60°  63  20 — 4 5 63 16  214            18. tan 195° 
 tan (45°  150°) 
  25.  tan 45°  tan 150°  1  tan 45° tan 150°   3  1  
3 
   3 1  1
3 
  113 17  
 4(2)   12 12 113 17 cot 12 
 cot 1 2 17 7     tan 12 
 tan 6  4   tan 6  tan 4  
  
 —— p  1  tan 6 tan 4  3   3  3 — 3   3  3   3  12  63   or 2  3  6    12 
 cos 4  3    
 cos 4 cos 3  sin 4   3  2 2 1    
  2  2  2  2    1 3 —— 
  3 1  3  1  
  19. cos 
    
  
    
  p  sin 3 cot  2   6   4  
 
  
 
       2  
  225 15  or   289 17  
  
 
  
  1  3  
   3  9 3    25 or 5  3  8  17  
 
  8  17 — 15  17 8  15  tan (x  y) 
 
    2  
  
 2  2  2  2   6  2   
 4  
  4  csc 2 
   
  225 15  or   289 17 sin x   2  sin y 
  1  co s2 y  2 
  1  
 sin 6 cos 4  cos 6 sin 4  5  4   tan x 
  cos x     2  4  28. cos x 
  1  si n2 x  
 2   6   1  4  5  24  
    
  2 
  1  
 25  4    
  16 4    25 or 5  6   2     sin y 
  1  co s2 y  3 2  5  
 
5
5  
5
5  sec 1275° 
    1225 35     1369 or 37  12  8  cos (x  y) 
 cos x cos y  sin x sin y    2  6  
 sin 
6  4  35  27. sin x 
  1  co s2 x   
  4  24. sin  12  37  621  
 2  2  2  2  5  12  2 1
      
 629  1  2  1  1
    15  23. 1275° 
 3(360°)  195° sec 1275° 
 sec 195° cos 195° 
 cos (150°  45°) 
 cos 150° cos 45°  sin 150° sin 45°  3  cos y 
  1  si n2 y  8 2  17     
 
1 7 
 37   
 17 
 37   22. 735° 
 2(360°)  15° sin 735° 
 sin 15° sin 15° 
 sin (45°  30°) 
 sin 45° cos 30°  cos 45° sin 30°  2  2  1   
 3 2  sin (x  y) 
 sin x cos y  cos x sin y     3  2    2 2   2  6   4  
 3 2  26. sin x 
  1  co s2 x  tan 45°  tan 120°  1  tan 45° tan 120° 1  (3 ) 
  1  1 (3 ) 1  3   
 1  3 4  2 3  or 2  3 
  2 23  5  
 tan    12 4 3  5 tan 4  tan 3 
 ——  5 1  tan 4 tan 3 1  (3 ) 
  1  1 (3 ) 4  2 3   
 or 2  3  2  
  
  113  12  33   3 — 3   3  3  
 2  3   20. tan 165° 
 tan (45°  120°)  21. tan    
 sin y  2 
  1 3  5  16 4    25 or 5   tan y 
  cos y  
  tan x  tan y  1  tan x tan y 4 8    3 15  
  4  5 — 3  5 4  3   8 4 1  15  3  12 1 5 — 77  45 36 77    6  2  
 6   2   215  Chapter 7  29. sec x 
  tan2 x 1 
 
 cos x 
 tan x 
 5  3  cos y 
  1  si n2 y  2 
  1  5  3  
   34 34  or   3 9  
  34 3 3   or  34 34  sin x  cos x  1  2 
  1 1  3  2 2  89 or  3  5  3 3  5  
  3 34   34 5 34   34  

  
  
  1  4  5  
  2 1
    
   or 16  25  
  
   
   5 5  13  5  13  144 12  or   169 13    
 
5
1 3   
 5 
 13  5  4  
 sin y 
 
 
  1   sec (x  y) 
  cos (x  y)  3  2 2  3  1 
 — 56  55   1  co s2 y  65  
 56  
  1  32. cos a 
  1  si n2a    
  5 5   or  9 3 sin y  cos y   5  3 — 2  3  12  
 65  2 2  3  tan y 
  3  56  1  
 
  sin b 
  1  co s2b  2 
  1 1  5  
  24 2 6    25 or 5  
  2 
  1 2  7  45 3 5    49 or 7  sin (a  b) 
 sin a cos b  cos a sin b  5   or 2  2 6  3 5  
 
5
7  
5
7 1  tan x  tan y  2   2  630   tan (x  y) 
  1  tan x tan y 5  2 1
    cos(x  y) 
 cos x cos y  sin x sin y  cos y 
  sec y 1 
 —  6  5 5  6  
  35   5  33. sin x 
  1  co s2 x     2 6 
 ——  5 5 1  
6
2  
  169 13  or   25 5  3  5  
  5 34 1    3 34  8 4 66 53     102  102 1217   534   102   30. tan x 
  cot x 1 
 —  
  
  12  5  sin y 
  1  co s2 y 3 34 2 2    34 3  
  2 
  1    cos y 
  sec y 1 — 
 13  cos (x  y) 
 cos x cos y  sin x sin y 
  
  1  cos x 
  1  si n2 x  sin x 
 —  sin x 
  sec y 
  tan2 y 1   31. sin x 
  csc x 1 
 —  
 
   10  65  12 —— 12  5 5  12  sin y 
  1  co s2 y  2 
  1 1  3  
  2 2  89 or  3  
  2 
  1 3  4   7 7    16 or 4  cos (x  y) 
 cos x cos y  sin x sin y 2 2   7  
 
3
4  
3
4 1   10  6 5   
 12  5 5   3  3  2  14   
 12   270  1225  
 1 9    cos 
2  x 
 sin x  34.     cos 2 cos x  sin 2 sin x 
 sin x 0  cos x  1  sin x 
 sin x sin x 
 sin x 35. cos (60°  A) 
 sin (30°  A) cos 60° cos A  sin 60° sin A 
 sin 30° cos A  cos 30° sin A 1  2  36.  Chapter 7  216   3  1   3  cos A  2 sin A 
 2 cos A  2 sin A  sin (A  ) 
 sin A sin A cos   cos A sin  
 sin A (sin A)(1)  (cos A)(0) 
 sin A sin A 
 sin A  37.    43. VL 
 I0qL cos 
qt  2  cos (180°  x) 
 cos x cos 180° cos x  sin 180° sin x 
 cos x 1  cos x  0  sin x 
  cos x cos x 
 cos x    VL 
 I0qL(cos qt  0  sin qt  1) VL 
 I0qL(sin qt) VL 
 I0qL sin qt  1  tan x   38. tan (x  45°) 
  1  tan x tan x  tan 45°  1  tan x tan 45° tan x  1  1  (tan x)(1) 1  tan x  1  tan x  1  tan x   
 1  tan x  sin 2(a  b) 44. n 
  b sin 2 1  1  tan x   
 1  tan x 1  tan x   
 1  tan x  sin 2(a  60°) n 
  60° sin 2 1  tan A  tan B   39. sin (A  B) 
  sec A sec B sin A  sin B   +  cos A cos B sin (A  B) 
 —— 1 1    cos A cos B sin A  cos A  sin 
2  30° n 
  sin 30° a  sin B   + cos B cos A cos B ——  sin (A  B) 
 1  1 cos A cos B    cos A cos B  sin (A  B) 
  a   2  sin A cos B  cos A sin B  1  n 
 2
   3  2  a   
cos 2  2 a  1  a  45. The given expression is the expanded form of the     sine of the difference of 3  A and 3  A. We have  cos (A  B) 
 —— 1 1    cos A cos B      sin 
3  A  
3  A 
 sin (2A) 
 sin 2A  sin B    1 cos A  cos B cos A cos B  cos (A  B) 
 ——  1 1 cos A cos B    cos A cos B  cos (A  B) 
    a sin2  n 
 3  sin 2  cos2  1  tan A tan B  sec A sec B sin A sin B   1 cos A  cos B  sinA  a  sin 2 cos 30°  cos 2 sin 30° n 
  1  sin (A  B) 
 sin (A  B) 40. cos (A  B) 
    VL 
 I0qL
cos qt cos 2  sin qt sin 2  46a.  f(x  h)  f(x)  h  
 
  cos A cos B  sin A sin B  1  46b.  cos (A  B) 
 cos (A  B)  sin (x  h)  sin x  h sin x cos h  cos x sin h  sin x  h  y y 
 sin x cos 0.1  cos x sin 0.1  sin x 0.1  1  sec A sec B   41. sec (A  B) 
  1  tan A tan B 1  cos A  1  1  cos A  1  cos B  0.5    cos B sec (A  B) 
 —— sin A sin B   1 cos A  cos B  O   cos A cos B  sec (A  B) 
 —— sin A sin B  cos A cos B     1  cos A  cos B sec (A  B) 
 sec (A  B) 
  1 2 3 4 5 6 7 8 x  0.5 1  1  cos A cos B  sin A sin B 1  cos (A  B)  46c. cos x sin (a  b)   47. tan (a  b) 
  cos (a  b)  sec (A  B) 
 sec (A  B) 42. sin (x  y) sin (x  y) 
 sin2 x  sin2 y (sin x cos y  cos x sin y)(sin x cos y  cos x sin y) 
 sin2 x  sin2 y 2 2 (sin x cos y)  (cos x sin y) 
 sin2 x  sin2 y sin2 x cos2 y  cos2 x sin2 y 
 sin2 x  sin2 y sin2 x cos2 y  sin2 x sin2 y  sin2 x sin2 y  cos2 x sin2 y 
 sin2 x  sin2 y 2 2 2 sin x(cos y  sin y)  sin2 y(sin2 x  cos2 x) 
 sin2 x  sin2 y (sin2 x)(1)  (sin2 y)(1) 
 sin2 x  sin2 y sin2 x  sin2 y 
 sin2 x  sin2 y  tan (a  b) 
  sin a cos b  cos a sin b  cos a cos b  sin a sin b sin a cos b cos a sin b    cos a cos b cos a cos b  tan (a  b) 
 ——— cos a cos b sin a sin b    cos a cos b cos a cos b tan a  tan b   tan (a  b) 
  1  tan a tan b  Replace b with b to find tan(a  b). tan (a  (b)) 
 tan (a  b) 
  tan a  tan (b)  1  tan a tan (b) tan a  tan b  1  tan a tan b  48a. Answers will vary.  217  Chapter 7  48b. tan A  tan B  tan C 
 tan A tan B tan C tan A  tan B  tan (180°  (A  B)) 
 tan A tan B tan(180°  (A  B)) tan A  tan B  
 tan A  tan B  
  56.  1  s 
 rv  A 
 2r2 v  18 
 r(2.9)  tan 180°  tan (A  B)  1  tan 180° tan (A  B) tan 180°  tan (A  B) tan A tan B 1  tan 180° tan (A  B) 0  tan (A  B)  1  0  tan (A  B) 0  tan (A  B) tan A tan B 1  0  tan (A  B)  A  6.2 r; 6.2 ft A 57. c2 
 702  1302  2(70)(130) cos 130° c2 33498.7345 c 183 miles 58. 120°  90°, consider Case 2. 4  12, 0 solutions  tan A  tan B  tan (A  B) 
 tan A tan B tan (A  B)  59.   1  tan A tan B   (tan A  tan B)   1  tan A tan B  tan (A  B) 
 tan A tan B (A  B) tan (A  B)(1  tan A tan B)  tan (A  B) 
 tan A tan B (A  B) (1  tan A tan B  1) tan (A  B) 
 tan A tan B (A  B) tan A tan B tan (A  B) 
 tan A tan B (A  B)  35 ft  1  cos2 x  37˚12′  2 2  49. sec2 x 
  1  sin2 x  csc x  cot x  
    x  6˚40′  1  cos2 x  2 2  sec2 x 
  cos2 x  1  cot x  cot x  v 
 37° 12
  6° 40
 or 30° 32
 a 
 90°  6° 40
 or 96° 40
 b 
 180°  (30° 32
  96° 40
) or 52° 48
  cos2 x  1    sec2 x 
  cos2 x  cos2 x  1  sec2 x 
 sec2 x  1  1 sec2 x 
 sec 2 x  35  sin 30° 32
 sin v  50. sin2 v  cos2 v 
 1  51. Arctan 3 
   x
 sin 30° 32
  1 8 — 
 3 7 8 1  
 37  7   
 21  63 cos2 v 
 64 3 7 cos v 
 8 3 7 Quadrant III, so  8  60.   3  1  61. Case 1 x  1  4 (x  1)  4 x  1  4 x  5 x 5    3  
 2 52. k, where k is an integer 86  50  2  4  53. A 
 2 A 
 18 
 68    h 
 2   t  c 2    1  2    c 2  y 
 18 sin 
  50 
 18 sin 
  18 
 18 sin 
   2  1 
 sin 
  c  sin1  86  50  
 2      68 c  68  
 68    c  
c   y 
 18sin 
2t    68 360  Case 2 x  1  4 x14 x3 {xx  8  30°  54. 8 
 8; 1 
 360; 1 
 30°  The correct choice is A.  55. sin (540°) 
 sin (360°  180°) 
0  Chapter 7  x 
 1  5 or x  3}  1 2 62. 
 1(6)  3(2) 3 6 
 6  6 or 12 63. fg(4) 
 f(g(4)) 
 f(5(4)  1) 
 f(21) 
 3(21)2  4 
 1319 gf(4) 
 g(f(4)) 
 g(3(4)2  4) 
 g(44) 
 5(44)  1 
 221 (8)62 8 62 64. (8)62 862 
 6 
 
 
 (1)62 
 1 2  ( 1) 
 2  c 3   
   2 2  x 54.87 ft 4x3  3x2  x 
 0 x(4x2  3x  1) 
 0 x(4x  1)(x  1) 
 0 x 
 0 or 4x  1 
 0 or x  1 
 0 x 
 4    3  sin (Arctan 3 ) 
 sin  x   
 sin 52° 48
 35 sin 52° 48
   tan v 
  cos v  
182  cos2 v 
 1  1 (6.2)2(2.9) 2 55.7 ft2  218  8  Page 445 1. csc v 
  1  
  2  7 7  2  1  v1
  v
  cot2  v
  cot v 
  sec2  cos v 
  v  4 2  1
 25  9 5  3  csc2  
  sec2  v  sec2  v  v  
  7 2  2 49  4 45  4 5 3  2    
  2 
  1 2  3   5  tan y 
  sec2 y 1     3 4 
  5 1  
4
 3 5  
   
    5.  6.  v v  sec2 v 2 2 sec v csc v   sec2 v  sec2 v 1  sin2 v — 1 1  cos2 v cos2 v  sin2 v  1 cot2 v  1   80  413  7-3B Reduction Identities  
 csc2 v  Page 447  
 csc2 v  1. sin, cos, sin 2. cot, tan, cot 3. tan, cot, tan 4. csc, sec, csc 5. sec, csc, sec 6a. (1) cos, sin, cos (2) sin, cos, sin (3) cot, tan, cot (4) tan, cot, tan (5) csc, sec, csc (6) sec, csc, sec 6b. Sample answer: If a row for sin a were placed above Exercises 1-5, the entries for Exercise 6a could be obtained by interchanging the first and third columns and leaving the middle column alone. 7a. (1) cos, sin, cos (2) sin, cos, sin (3) cot, tan, cot (4) tan, cot, tan (5) csc, sec, csc (6) sec, csc, sec 7b. Sample answer: The entries in the rows for cos a and sec a are unchanged. All other entries are multiplied by 1. 8a. Sample answer: They can be used to reduce trigonometric functions of large positive or negative angles to those of angles in the first quadrant. 8b. Sample answer: sum or difference identities  
 csc2 v 
 csc2 v  1  sin x  1      cos x  sin x 
 2  cos x  cos x  sin x  1
21 1
1  1  cot a tan b   7. tan (a  b) 
  cot a  tan b 1   1 tan a  tan b —— tan (a  b) 
 1   tan b tan a 1   1 tan a  tan b tan a  tan (a  b) 
 ——  1 tan a   tan b tan a tan a  tan b   tan (a  b) 
  1  tan a tan b  tan (a  b) 
 tan(a  b) 8. cos 75° 
 cos(30°  45°) 
 cos 30° cos 45°  sin 30° sin 45°  3   2   80  413  
1  
1 1
1  
 csc2 v csc2 v 
 csc2 v cot x sec x sin x 
 2  tan x cos x csc x  cos x  sin x   5  43  4  4  5 3  4 5  43   4  53    or  
 59 59  1    1  cot2 x 
 1  sec2  
  22 1 
 3   tan x  tan y  
 cos 4  csc2  3   tan (x  y) 
  1  tan x tan y  3 5    1 1   sec2 x  csc2 x cos2 x  sin2 x  2  10. tan x 
 4  1  sec v  
 5  4  1  1  tan2 x  7 7     16 or 4  35  6   1 
  5 3 
  3  4   
 12  cos 4 
 cos 
5  4 4.   7  2 
  1  
 
3
4  
3
4  
 sec v  19  
  cos (x  y) 
 cos x cos y  sin x sin y  5  19  4  
  59 or 35  Quadrant II, so 3 3.  cos y 
  1  si n2 y  5  
3  1 
 sec2 v 16  9  v
  cot2  5 3  2  Quadrant 1, so 2.  cot2  1  cot2 v 
 
  1 
 —  tan2  9. cos x 
  1  si n2 x  Mid-Chapter Quiz 1  sin v   2  
 
2
2  
2
2 1    2  6   
 4  219  Chapter 7  Double-Angle and Half-Angle Identities  7-4 Page 453  2 2  5  
   Check for Understanding  cos2 v 
 25 21   cos v 
  5  4 21   
 25  cos 2v 
 cos2 v  sin2 v  2 21  1  cos   tan 2v 
  1  tan2 v  21 2  
  ,   2
 21  
 —— 21 2 2  1  
 21   1  cos 2 2  2  3c. I, II, III or IV 
    4 2  3  
   0 
 2(1) 0 2  Sample answer: v 
 2 5. Both answers are correct. She obtained two different representations of the same number. One way to verify this is to evaluate each expression with a calculator. To verify it algebraically, square each answer and then simplify. The same result is obtained in each case. Since each of the original answers is positive, and they have the same square, the original answers are the same number.  cos v  21 4   or  17  sin2 v  cos2 v 
 1 3 2  sin2 v  
5 
 1   1 
 sec2 v  25  
 9 5 3 
 1  
 sec v  16  sin2 v 
 2 5  sec2 v sec v (Quadrant III)   3  sin 2v 
 2 sin v cos v 
 2
5
5 4  3  24  
 25 cos 2v 
 cos2 v  sin2 v 3 2  4 2  
 
5  
5  
 sin  2  7  
 2 5     4  1  cos  2  2 tan v   tan 2v 
  1  tan2 v  (Quadrant I)  2
3 
 — 4 2 1  
3 4    2 1  2  2    2  2   
 2  
  330°  7. tan 165° 
 tan 2 1  cos 330°    1  co s 330°  (Quadrant II)  8  3 — 7 9  24  or 7 2   10. tan 2v 
  cot v  tan v   3   1  3 —  3 1  3  2  tan v    tan 2v 
  cot v  tan v  tan v  tan 2v 
  
 (2  3 ) 
 3 2  tan 2v 
  2 tan v  cot v tan v  tan2 v 2 tan v  1  tan2 v  tan 2v 
 tan 2v  220  4  sin v 
 5  (Quadrant III)  1 3 
  5 or 5    4  
  4 21   21 — 17  21  9. tan2 v  1 
 sec2 v    Chapter 7  2 2  2 tan v   2 .  
   21  2  sin  
 2 sin 2  
  (Quadrant I)  
 2
5
 5   sin 2
2 
 2 sin 2  
  2 21   or  21  17  a    2  5 — 21    5 2  21   sin 2v 
 2 sin v cos v  
 25  3a. III or IV 3b. I or II 4. sin 2v 
 2 sin v  6. sin  
  
 sin2 v  a    8  
   
 
 5   
5  Letting v 
 2 yields sin 2 
 a   cos2 v 
 1  
 sin2 v   
 sin v or sin 2 
   tan v 
  cos v  21  1. If you are only given the value of cos v, then cos 2v 
 2 cos2 v  1 is the best identity to use. If you are only given the value of sin v, then cos 2v 
 1  2 sin2 v is the best identity to use. If you are given the values of both cos v and sin v, then cos 2v 
 cos2 v  sin2 v is just as good as the other two. 2. cos 2 
 1  2 sin2 v cos 2v  1 
 2 sin2 v cos 2v  1  2 1  cos 2v  2 1 cos 2v  2  sin v  8. sin2 v  cos2 v 
 1  sec A  sin A  1   11. 1  2 sin 2A 
  sec A  1  1  2   sin A sin 2A 
 ——  1  1  12.  1  cos A   sin A —— 1  cos A  cos A    cos A  
   2 sin A cos A sin 2A  
    2  3  2 —  2  3  2  (2  3 )(2  3 )  (2  3 )(2  3 )  (2  3  )2  43  
 2  3  3  17. sin 8 
 sin  cos 2v v1 cos 2v  1 
 2 cos2 v  3  4  2   
  —  3  
  1  cos 2v  2 
 cos2 v  1  cos 4 —— 2  (Quadrant I)   2  P 
 I02 R sin2 qt P 
 I02 R (1  cos2 qt)  
  P 
 I02 R
1 
2 cos 2qt  2 1  1  
  P 
 I02 R
2  2 cos 2qt 1     
   sin x 
 2 
 2 cos2  1  (Quadrant I)  1  
2   3 1  
2  
  sin x  2  1  2   1  cos 6 —— 5 1  cos 6  3  
   13.  
 tan  2 
  sin 2A 
 1  sin A cos A  1 1  sin 2A 
 1  2 1 1  sin 2A 
 1  2 x x sin x sin 2 cos 2 
 2 x x 2 sin 2 cos 2 sin x —— 
   2 2 x sin 2 2 sin x — 
  2 2  5  6  5  1  cos A  1  2 sin 2A 
 1  2 1  2 1  2  16. tan  1  cos A  5  12  1  1  P 
 2 I02 R  2 I02 R cos 2qt  7  18. cos 12 
  1  2 2     2  2  2 7  6  cos 2  7π  Pages 454–455    
  Exercises 30°  14. cos 15° 
 cos 2 
  1  cos 30°    2  
 15. sin 75° 
 
    45°  19. tan 22.5° 
 tan 2  
  
  1  cos 45°   1  cos 45°  
    (Quadrant I)   2  (Quadrant I)    3  
  1  2  
  2      3  2    2  3   1  2 — 2  2  3   2 150° sin 2 1  cos 150°  2  (Quadrant II)  
    
  (Quadrant I)   3  
  1  cos 6 —— 2  1  
2  2   2   3   2  
  1  2 —  2 1  2    2  2  2 — 2   2  2   
  
  
  (2  2 )(2  2 )  (2  2 )(2  2 )  )2 (2  2  42  
   2  2  2  2 22  2  
 2 1  221  Chapter 7  v  20. tan 2 
 
  
 
  23. tan2 v  1 
 sec2 v (2)2  1 
 sec2 v 5 
 sec2 v  
 sec v 5 cos v 
 sec v  1  cos v    1  cos v   1 1  4  1 1   4 3  4 — 5  4  
    v
     2 5  20  sin2 v 
 25  sin v 
  2 5  sin v   tan v 
  cos v  
1  v
  sin2  9  25 3  5  (Quadrant I)  
  3  5 — 4  5  
  3  4  sin v 
 5 2 5  4  
 5 cos 2v 
 cos2 v  sin2 v  2 5  4  3  
 5  24  2 tan v   tan 2v 
  1  tan2 v  cos 2v 
 cos2 v  sin 2 v  
   22.  sin2  2(2)   
 1  (2)2  
   4 2 3 2 
 5  5 7 
 2 5 2 tan v  2v 
  1  tan2 v  
 v  
   1  2 3    4  1   24. cos v 
  sec v  
  
  
   cos2  v
1  cos2  v
1  
 sin2  7  tan v 
 
  2 2  cos v 
 3  
  sin v  cos v 1  3 — 2 2  3 1  22   sin2 v 
 1 6  7  sin v 
 4 tan v 
 
   2  or 4  (Quadrant I)  
  sin 2v 
 2 sin v cos v  
 cos 2v 
  
  tan 2v 
     7  
 2
4
4 7 3  cos 2v 
 cos2 v  sin2 v  1 2  3 2  1  2   
 1 6 or 8  2 tan v  1  tan2 v  2 tan v   tan 2v 
  1  tan2 v  
   
   2  2    
37   
   7  2 1  
 3   
    2 1  4  Chapter 7   2 7  
 
4  
4  7  9   2  2  4  
  3  
 8  v  sin2 v  2 2 2   2  2 — 14  16  (Quadrant II)  sin v  cos v  7  4 — 3 4  7 3  sin 2v 
 2 sin v cos v  
 
3  
3 
  v  cos2 v 
 1 3 2  8  4 2  9 cos2  1  4 3 3 4  sin2 v  
4 
 1  cos2 v 
 9  1 2 2   3 3  4    
 3 or 3  3 2 4  3 2 1  4 3  2 24 — or  7 7  16  
 2
  2 5 2  
 
5  
5  
 25  
   5  
 2
5
5  
 2
5
5  tan  (Quadrant II)  sin 2v 
 2 sin v cos v  sin 2v 
 2 sin v cos v 3  or 5  sin2 v  
5 
 1  21. sin2 v  cos2 v 
 1 sin2   5  1  5   sin2 v  cos2 v 
 1  15   35 or  5 4  2 5  (Quadrant II)  4 2  or 7  
  222  2 7  3 — 2  9  or 37   1  25. 1  cot2 v 
 csc2 v 1  
2 
 csc2 v 3 2  sin  13  
 4 13   2 
 1  v
 csc v  
 
  2 tan v   tan v 
  cot v  
 
  csc2 v csc v   tan 2v 
  1  tan2 v  1  3  2 2  3  2  1  13     2 2  13   
   2 21 2  1  
 21   
  4 21   21 —— 17  21  (Quadrant III) sin2 v  cos2 v 
 1   213 2 2  
 13   cos v 
 1  cos2  2 13   or  13  v
  cos v 
  sin2   2 2 3  a
  3 13   7  sin a 
 3  12  tan 2a 
  cos 2v 
 cos2 v  sin2 v  tan    
  
    
  
   12  5  
252  cos2 v 
 1  2  sin v  
 
  30.  2 21   or  21  1  sec v csc v 
 2 sec v csc v  (cos A  sin A)(cos A  sin A)  cos A  sin A  cos A  sin A 
 cos A  sin A (sin v  cos v)2  1 
 sin 2v 2 sin v  2 sin v cos v  cos2 v  1 
 sin 2v 2 sin v cos v  1  1 
 sin 2v 2 sin v cos v 
 sin 2v sin 2v 
 sin 2v   31. cos x  1 
  2(cos x  1)   21  
 2
5
 5   2 cos2 x  1  1   cos x  1 
  2(cos x  1)  4 21  2 cos2 x  2   
  25   cos x  1 
  2(cos x  1)  2(cos2 x  1)  cos 2v 
 cos2 v  sin2 v 
  1  csc v sec v 
 2 sec v csc v  cos 2x  1  sin 2v 
 2 sin v cos v    1  
 2 sec v csc v  cos A  sin A 
  tan v 
  cos v  

  1  
 2 sec v csc v   cos A  sin A 
  cos A  sin A  (Quadrant IV)   21   2 5 17  25  1  
 2 sec v csc v  cos2 A  sin2 A  21   2  1  sin 2v 1  2 sin v cos v 1 1    sin v cos v  cos 2A  cos v 
  5  2 5 — 21    5 2  21   2 14  or  5   29. cos A  sin A 
  cos A  sin A  21  cos2 v 
 25  
 5    1  2 1  2  sin2 v  cos2 v 
 1   14  5    2  1  1  2  1    csc 2v 
 2 sec v csc v  28.   26. sin v 
  csc v 1 
 — 5 2  (Quadrant II)  
  14  2 1  
 2  2 2 1  3  or   14  7     
  2  or  2  14  
 2   4  3  5  9  
  7   3   2  3  2 tan a  1  tan2 a 2  213  2 13  
   
  
1 7  
 1 3  
    sin a   tan a 
  cos a  sin2 a 
 9     
 2
 13 
 13   313  2 
 13 5 
 1 3 2 tan v  2v 
  1  tan2 v 2 2 3  4 21   or  17  27. sin2 a  cos2 a 
 1  117  16 9 3 13   13  (Quadrant III) sin 2v 
 2 sin v cos v 2 13    221  
 21   
   cos x  1 
  2(cos x  1)    2 2 5  cos x  1 
  2(cos x  1)(cos x  1)  2(cos x  1)  cos x  1 
 cos x  1 cos2 v  sin2 v   32. sec 2v 
  cos2 v  sin2 v 1   sec 2v 
  cos 2v  sec 2v 
 sec 2v A  sin A   33. tan 2 
  1  cos A  223  Chapter 7  1  
  A 1  cos 2
2 A sin 2 2  A  L tan 45°  tan 2  39a. tan
45°  2 
  L 1  tan 45° tan 2 L 1  tan 2  tan 2 
   
  L 1  1  tan 2  A A 2 sin  cos   A  2 2 tan 2 
  A A  tan 2 
  tan  A  2  
  A  tan 2 
  1  2 cos2 2  1 A A 2 sin 2 cos 2  A 2 cos2 2 A sin 2  A cos 2 A tan 2  
  39b.  1  cos L  1 1  cos L  1  cos L  1  1  cos L     
  
  sin 3x 
 3 sin x  4 sin3 x sin(2x  x) 
 3 sin x  4 sin3 x sin 2x cos x  cos 2x sin x 
 3 sin x  4 sin3 x 2 sin x cos2 x  (1  2 sin2 x) sin x 
 3 sin x  4 sin3 x  34.  1  cos L  1 1  cos L  1  cos L  1  1  cos L     1  cos 60˚ 1   1  cos 60˚  1  cos 60˚  1  1  cos 60˚ 1 1  2  1 1 1   2     1  2 sin x(1  sin2 x)  (1  2 sin2 x) sin x 
 3 sin x  4 sin3 x  2 sin x  2 sin3 x  sin x  2 sin3 x 
 3 sin x  4 sin3 x 3 sin x  4 sin3 x 
 3 sin x  4 sin3 x 35. cos 3x 
 4 cos3 x  3 cos x cos(2x  x) 
 4 cos3 x  3 cos x 2 (2 cos x  1)cos x  2 sin2 x cos x 
 4 cos3 x  3 cos x  
  2 cos3 x  cos x  2 cos x  2 cos3 x 
 4 cos3 x  3 cos x 4 cos3 x  3 cos x 
 4 cos3 x  3 cos x v2  sin2 2v 2g  
 40.  sin2 2v   36.  
 sin2 v v2  R
 R
  21  4 sin2 v cos2 v   3  3  tan a   trigonometry,  1 tan2v  
   3   3  (1  3 ) tan a 
 3  3  PA  BA  cos v sin(v  a)  g cos2 a 2v2 cos v sin(v  45°)  g cos2 45° 2v2 cos v(sin v cos 45°  cos v sin 45°)  g cos2 45°   2 2   2v2 cos v (sin v)  2  (cos v) 2  tan a 
  3  3   3  1  3   tan a 
  9  3   3 33   
    6  53  tan a 
  3   
        2   3      1   2  
 2  2  2  2   6  2   
 4  
     sec 12 
   2  2 2 2 v cos v(sin v  cos v)  R 
  1 g  2  
  2 v2 (2 cos v sin v  2 cos2 v)  g  1    cos  12 1  2   6    4 4 2  4 6   or 6 
   2  4  v2  2  R 
 g(2 cos v sin v  (2 cos2 v  1) 1)  Chapter 7    
 cos 3 cos 4  sin 3 sin 4   2  2 g  2  v2  2 (sin g    41. cos 12 
 cos 
3  4  R 
   R
  
 3  3  tan a  2v2  
  R
  
3   tan a 
 3  3 tan a  3  
 4 cos2 v 37. ∠PBD is an inscribed angle that subtends the same arc as the central angle ∠POD, so m∠PBD  38. R 
  or 2  3   tan (a  30°) 
 7  (2 sin v cos v)2   
 sin2 v  
    3  3   3  3  3   3  12  6 3  6  tan a  tan 30°  1  tan a tan 30°   
 sin2 v  
    tan (a  30°) 
 3   sin2 v 2g  1 v. By right triangle 2 PA sin v   
 . 1  OA 1  cos v  1 1   2  1 1  2  1 1   3  1 1  3  
  (2 cos2 x  1)cos x  2(1  cos2 x)cos x 
 4 cos3 x  3 cos x     42. Sample answer: sin( )2  cos( )2 
 sin   cos  
 0  (1) 
 1 1  2v  cos 2v  1)  224  17  10  43. s 
 rv 17 
 10  v 17  10  17  180°   
 1 0    7-5  Solving Trigonometric Equations  Page 458  Graphing Calculator Exploration  97.4°  
v  44. Let x 
 the distance from A to the point beneath the mountain peak. h  tan 21°10
 
  570  x h 
 (570  x) tan 21°10
 h tan 36°40
 
 x h 
 x tan 36°40
 (570  x) tan 21°10
 
 x tan 36°40
 570 tan 21°10
 
 x tan 36°40
  x tan 21°10
 570 tan 21°10
 
 x(tan 36°40
  tan 21°10
) 570 tan 21°10
  tan 36°40
  tan 21°10
  
x  617.7646751  x  1.  2.  h  tan 36°40
 
 x tan 36°40
  h  617.8  h 460 ft 45. (x  (3))(x  0.5)(x  6)(x  2) 
 0 (x  3)(x  0.5)(x  6)(x  2) 
 0 (x2  2.5x  1.5)(x2  8x  12) 
 0 x4  5.5x3  9.5x2  42x  18 
 0 2x4  11x3  19x2  84x  36 
 0 46.  y 
 2x  5 x 
 2y  5 x  5 
 2y x5  2  3. Exercise 1: (1.1071, 0.8944), (4.2487, 0.8944) Exercise 2: (5.2872, 0.5437), (0.9960, 0.5437) 4. The x-coordinates are the solutions of the equations. Substitute the x-coordinates and see that the two sides of the equation are equal.  y  5.  y 
 2x  5  
y  O  x y
  x5 2    47. x  2y 
 11 x 
 11  2y  x  2y 
 11 x  2(2) 
 11 x 
7 48. ab 
 3 3 b 
 a a2  [0, 2] sc14 by [3, 3] sc11 5a. The x-intercepts of the graph are the solutions of the equation sin x 
 2 cos x. They are the same. 5b. y 
 tan 0.5x  cos x or y 
 cos x  tan 0.5x  3x  5y 
 11 3(11  2y)  5y 
 11 33  6y  5y 
 11 11y 
 22 y
2  Page 459  (7, 2)  (a  b)2 
 64  2ab  b2 
 64  a2  2a
a  
a 
 64 3  Check for Understanding  1. A trigonometric identity is an equation that is true for all values of the variable for which each side of the equation is defined. A trigonometric equation that is not an identity is only true for certain values of the variable. 2. All trigonometric functions are periodic. Adding the least common multiple of the periods of the functions that appear to any solution to the equation will always produce another solution. 3. 45°  360x° and 135°  360x°, where x is any integer  3 2  a2  6  
a 
 64 3 2  a2  
a 
 70 3 2  a2  b2 
 70 The correct answer is 70.  225  Chapter 7  12. tan2 x  2 tan x  1 
 0 (tan x  1)(tan x  1) 
 0 tan x  1 
 0 tan x 
 1  4. Each type of equation may require adding, subtracting, multiplying, or dividing each side by the same number. Quadratic and trigonometric equations can often be solved by factoring. Linear and quadratic equations do not require identities. All linear and quadratic equations can be solved algebraically, whereas some trigonometric equations require a graphing calculator. A linear equation has at most one solution. A quadratic equation has at most two solutions. A trigonometric equation usually has infinitely many solutions unless the values of the variable are restricted. 5. 2 sin x  1 
 0 6. 2 cos x  3 
0 2 sin x 
 1   2 cos x 
 3  12  cos x 
 2  sin x 
  sin x 
  
  cos x  sin x  cos x 
  cos2 x  3 cos x 
 2 cos2 x  3 cos x  2 
 0 (cos x  1)(cos x  2) 
 0 cos x  1 
 0 or cos x  2 
 0 cos x 
 1 cos x 
 2 x 
 (2k  1) no solutions 14. sin 2x  cos x 
 0 2 sin x cos x  cos x 
 0 cos x (2 sin x  1) 
 0 cos x 
 0 or 2 sin x  1 
 0 13.   3  x 
 30° 7. sin x cot x 
  3  x 
 4  k    x 
 30°   3  2  3  2  3  2    5  or x 
 6  2k  11  x 
 6 or x 
 6 11.  x
   k  0 1  cos v  2  1  2  1  2  3  4  4  v 3 16. W 
 Fd cos v 1500 
 100  20 cos v 0.75 
 cos v v 41.41°  Pages 459–461  Exercises  17. 2  sin x  1 
 0 2  sin x 
 1  18. 2 cos x  1 
 0 2 cos x 
 1  1   2  2  1  cos x 
 2  sin x 
  2 sin x 
 19.  no solutions  cos x 
  15. 2 cos v  1 2 cos v  cos v 
 2 at 3 and 3  sin2 2x  cos2 x 
 0 1  cos2 2x  cos2 x 
 0 1  (2 cos2 x  1)2  cos2 x 
 0 1  (4 cos4 x  4 cos2 x  1)  cos2 x 
 0 4 cos4 x  5 cos2 x 
 0 cos2 x(4 cos2 x  5) 
 0 cos2 x 
 0 or 4 cos2 x  5 
 0 5 cos x 
 0 cos2 x 
 4   2  sin x 
 2 x 
 6  2k  x 
 30° or x 
 330° 8. cos 2x 
 sin2 x  2 2 cos2 x  1 
 (1  cos2 x)  2 2 cos2 x  1 
 cos2 x  1 3 cos2 x 
 0 cos2 x 
 0 cos x 
 0 x 
 90° or x 
 270° 9. 3 tan2 x  1 
 0 3 tan2 x 
 1 1 tan2 x 
 3  3 tan x 
 3  5 7 11 x 
 6 or x 
 6 or x 
 6 or x 
 6 10. 2 sin2 x 
 5 sin x  3 2 2 sin x  5 sin x  3 
 0 (2 sin x  1)(sin x  3) 
 0 2 sin x  1 
 0 or sin x  3 
 0 sin x 
 12 sin x 
 3 7  1  x 
 2  k  x 
 120°  x 
 45° sin 2x  1 
 0 2 sin x cos x  1 
 0 sin2 x cos2 x 
 14 1  sin2 x (1  sin2 x) 
 4 1  sin2 x  sin4 x  4 
 0 1  sin4 x  sin2 x  4 
 0  
sin2 x  12
sin2 x  12 
 0 1  sin2 x  2 
 0   5  2  1  sin2 x 
 2  no solutions  1    2  sin x 
  or 2 2 x 
 45°  Chapter 7  226  tan 2x  3 
0 tan 2x 
 3   20.  2 tan x  1  tan2 x  28.  
 3    2  1 3  3   3  tan x 
 21.   2   2   cos x  2  sin x  2 
 2  2  cos x 
 2  cos x 
 1 x 
 0° 29. 2 sin v cos v  3  sin v 
 0 sin v (2 cos v  3 ) 
 0 sin v 
 0 or 2 cos v + 3 
0 v 
 0° or v 
 180° 2 cos v 
 3   tan x 
 3  x 
 60°  x 
 30° cos2 x 
 cos x cos2 x  cos x 
 0 cos x(cos x  1) 
 0 cos x 
 0 or x 
 90°   3  cos v 
 2 v 
 150° or v 
 210°  cos x  1 
 0 cos x 
 1 x 
 0° 2 22. sin x 
 1  cos x sin x 
 1  1  sin2 x sin2 x  sin x  2 
 0 (sin x  1)(sin x  2) 
 0 sin x  1 
 0 or sin x  2 
 0 sin x 
 1 sin x 
 2 x 
 90° no solution 23. 2  cos x  1 
 0 2  cos x 
 1  30. (2 sin x  1)(2 cos2 x  1) 
 0 2 sin x  1 
 0 or 2 cos2 x  1 
 0 2 sin x 
 1 2 cos2 x 
 1 1  1  cos2 x 
 2  sin x 
 2   x 
 6  cos x 
  5   2  2    or x 
 6  3  x 
 4 or x 
 4 5  7  or x 
 4 or x 
 4 31.  2   cos x 
 2 x 
 135° or x 
 225° 1 24. cos x tan x 
 2 sin x   2  cos x  2  sin x  2  2 tan x 
 3  (1  tan2 x)   3  tan2 x 2 tan x 
 3 2 3  tan x  2 tan x  3 
0 (3  tan x  1)(tan x  3 ) 
 0 3  tan x  1 
 0 tan x  3 
0 tan x 
   cos (x  45°)  cos (x  45°) 
 2  cos x cos 45°  sin x sin 45°  cos x cos 45°  sin x sin 45° 
 2   4 sin2 x  1 
 4 sin x 4 sin2 x  4 sin x  1 
 0 (2 sin x  1)(2 sin x  1) 
 0 2 sin x  1 
 0 2 sin x 
 1 1  sin x 
 2 7  11  x 
 6 or x 
 6  1    cos x  cos x 
 2  32. 2  tan x 
 2 sin x  1  sin x 
 2  sin x    2 cos x 
 2 sin x  x 
 30° or x 
 150° 25. sin x tan x  sin x 
 0 sin x (tan x  1) 
 0 sin x 
 0 or tan x  1 
 0 x 
 0° or x 
 180° tan x 
 1 x 
 45° or x 
 225° 2 26. 2 cos x  3 cos x  2 
 0 (2 cos x  1)(cos x  2) 
 0 2 cos x  1 
 0 or cos x  2 
 0 2 cos x 
 1 cos x 
 2 1 cos x 
 2 no solution   
 2 cos x 2  2  2  
 cos x   7  x 
 4 or x 
 4 2  tan x 
 2 sin x would also be true if both tan x sin x  and sin x equal 0. Since tan x 
  cos x , tan x equals 0 when sin x 
 0. Therefore x can also equal 0 and .   7  0, 4, , 4 33.  x 
 60° or x 
 300° 27. sin 2x 
 sin x 2 sin x cos x 
 –sin x 2 sin x cos x  sin x 
 0 sin x (2 cos x  1) 
 0 sin x 
 0 or 2 cos x  1 
 0 x 
 0° or x 
 180° 2 cos x 
 1  sin x 
 cos 2x  1 sin x 
 1  2 sin2 x  1 2 2 sin x  sin x 
 0 sin x(2 sin x  1) 
 0 sin x 
 0 or 2 sin x  1 
 0 x 
 0 or x 
  2 sin x 
 1 1  sin x 
 2 7  x 
 6 or  1  cos x 
 2  11  x 
 6  x 
 120° or x 
 240°  227  Chapter 7  34.  cot2 x  csc x 
 1 csc2 x  1  csc x 
 1 csc2 x  csc x  2 
 0 (csc x  2)(csc x  1) 
 0 csc x  2 
 0 or csc x  1 
 0 csc x 
 2 or csc x 
 1 1  sin x 
 2 x
 35.  or x 
  5  6  3  x 
 2  sin x  cos x 
 0 sin x 
 cos x sin2 x 
 cos2 x sin2 x  cos2 x 
 0 sin2 x  1  sin2 x 
 0 2 sin2 x  1 
 0 1  sin2 x 
 2 sin x 
  1  1   2  or  36.  v
 37. sin x 
 x
 38.  or v 
  1  sin x cos x 
 2 1  sin2 x cos2 x 
 4 1  sin2 x(1  sin2 x) 
 4 1  sin2 x  sin4 x) 
 4 1  sin4 x  sin2 x  4 
 0  
sin2 x  12
sin2 x  12 
 0  11  6  1  sin2 x  2 
 0 1  2 k  or  sin2 x 
 2  11  x 
 6  2k  sin x 
   sin x  2(1  sin2 x) 
 1 2 sin2 x  sin x  1 
 0 (2 sin x  1)(sin x  1) 
 0 2 sin x  1 
 0 or sin x  1 
 0 2 sin x 
 1 sin x 
 1 x
  1  2   6  x
  2k or x 
  5  6  3  2  44.   3   3  2 cos2 x  1 
 2 2  3   2 cos2 x 
 2   2k  2  3   cos2 x 
 4   2k   2   3  2  cos x 
   tan x 3 tan2 x 
 3 2 3 tan x  3  tan x 
 0 tan x(3 tan x  3 ) 
 0 tan x 
 0 or 3 tan x  3 
0 x 
 k 3 tan x 
 3     11  x 
 12  k or x 
 12  k sin4 x  1 
 0 (sin2 x  1)(sin2 x  1) 
 0 sin2 x  1 
 0 or sin2 x  1 
 0 2 sin x 
 1 sin2 x 
 1 sin x 
 1 no solutions  x 
 2  k 46. sec2 x  2 sec x 
 0 sec x(sec x  2) 
 0 sec x 
 0 or sec x  2 
 0 45.   3   x 
 6  k 2(1  sin2 x) 
 3 sin x 2  2 sin2 x 
 3 sin x 2 2 sin x  3 sin x  2 
 0 (2 sin x  1)(sin x  2) 
 0 2 sin x  1 
 0 or sin x  2 
 0 2 sin x 
 1 sin x 
 2 1 sin x 
 2 no solution  1  cos x 
 0  sec x 
 2  no solution  cos x 
 2  1  2  x 
 3  2k or    4  x 
 6  2k or  x 
 3  2k  5  x 
 6  2k Chapter 7   3  cos2 x  sin2 x 
 2 cos2 x  (1  cos2 x) 
 2  tan x 
 3 40.   2  2  x 
 4  k  sin x  2  cos x  cos x  2 cos x 
 1  39.  5  x 
 3  2k  43.  cos x tan x  2 cos2 x 
 1  sin x 
    x 
 3  2k or  no solution  1  3 sin v 
 cos 2v 1  3 sin v 
 1  2 sin2 v 2 2 sin v  3 sin v  2 
 0 (2 sin v  1)(sin v  2) 
 0 2 sin v  1 
 0 or sin v  2 
 0 2 sin v 
 1 sin v 
 2 1 sin v 
 2 no solution 1 2 7   6  cos x 
 2  cos x 
 2  3  4  7 . 4  7  6  1  sec x 
 2   2  2  sin x and cos x must be opposites, so x 
 or x 
  
 cos x  sin x  (cos x  sin x)(cos x  sin x) 
 1 cos2 x  sin2 x 
 1 cos2 x  (1  cos2 x) 
 1 2 cos2 x  1 
 1 2 cos2 x 
 2 cos2 x 
 1 cos x 
 1 x 
 k 42. 2 tan2 x  3 sec x 
 0 2(sec2 x  1)  3 sec x 
 0 (2 sec x  1)(sec x  2) 
 0 2 sec2 x  3 sec x  2 
 0 2 sec x  1 
 0 or sec x  2 
 0 2 sec x 
 1 sec x 
 2  sin x 
 1    6  1  cos x  sin x  41.  228  47.  48.  sin x  cos x 
 1 sin2 x  2 sin x cos x  cos2 x 
 1 sin2 x  2 sin x cos x  1  sin2 x 
 1 2 sin x cos x 
 0 sin x cos x 
 0 sin2 x cos2 x 
 0 sin2 x (1  sin2 x) 
 0 2 sin x 
 0 or 1  sin2 x 
 0 sin x 
 0 sin2 x 
 1 x 
 2k sin x 
 1  x 
 2  2k  58a.  1.00 sin 35°   sin r 
  2.42  sin r 0.2370150563 r 13.71° 58b. Measure the angles of incidence and refraction to determine the index of refraction. If the index is 2.42, the diamond is genuine. 59. D 
 0.5 sin (6.5 x) sin (2500t) 0.01 
 0.5 sin (6.5(0.5)) sin (2500t) 0.02 
 sin 3.25 sin 2500t 0.1848511958 sin 2500t 0.1859549654 2500t The first positive angle with sine equivalent to sin (0.1859549654) is   0.1859549654 or 3.326477773.  2 sin x  csc x 
 3 2 sin2 x  1 
 3 sin x 2 2 sin x  3 sin x  1 
 0 (2 sin x  1)(sin x  1) 
 0 2 sin x  1 
 0 or sin x  1 
 0 2 sin x 
 1 sin x 
 1    1  x 
 2  2k  sin x 
 2   x 
 6  2k or x
 49.  50.  5  6   2k    4  3  4  0v  or  1  sin(bx  c) 
 2 360°  The period of the function sin(bx  c) is b, so the given interval consists of 1  61.  
 b periods.   xy 
 34 3 cos v sin v 17   
  sin v cos v 4 22  cos v  4 sin v 17  
  33 sin v  4 cos v 22  3 cos v  4 sin v 
 17  3 sin v  4 cos v 
 22  ↓ 9 cos v  12 sin v 
 317  16 cos v  12 sin v 
 82  25 cos v 
 82   317   2 53. 0, 1.8955  5.5  107   sin v 
  0.003    317  82   cos v 
  25  sin v v  0.0001833333333 0.01° 56. sin 2x sin x 2 sin x cos x sin x 2 sin x cos x – sin x 0 sin x(2 cos x  1) 0 The product on the left side of the inequality is  5 equal to 0 when x is 0, 3, , or 3. For the product to be negative, one factor must be positive and the  5 other negative. This occurs if 3 x  or 3 x 2. 57.  360°  360°  b  The equation sin (bx  c) 
 2 has two solutions per period, so the total number of solutions is 2b.  3  v  0.0013 s a  sin v 
 2 at 4 and 4 52. 0.4636, 3.6052 54. 0.3218, 3.4633  55. sin v 
 D  t  a sin(bx  c) 
 2   2  2    3.326477773  2500 a  0 1   2  t  60. a sin(bx  c)  d 
 d  2   3 cos v  2  5 7 3 cos v 
 2 at 6 and 6 5 7   v   6 6 1   cos v  2  0 1 cos v  2 1  5 cos v 
 2 at 3 and 3  5 0  v 3 or 3 v 2   sin v  1 51. 2 2  sin v sin v  n1 sin i 
 n2 sin r 1.00 sin 35° 
 2.42 sin r  360  v  v 341.32°  18.68020037  v2  R 
 g sin 2v 152   20 
  9.8 sin 2v  0.8711111111 sin 2v 2v 60.5880156 or v 30.29°  2v v  119.4119844 59.71°  229  Chapter 7  135°  
  
  0 3 2 2 4 2 1 2 1  0 x2  2x  1 
 0 (x  1)(x  1) 
 0 x1
0 x 
 1 (x  2)(x  1(x  1)   cot v 
  tan v  1 – co s 135°    1  co s 135°  135°  tan 2 
  67. 2   1  62. cot 67.5° 
 cot 2  (Quadrant 1)   
    
     2 1   2  2  1   2   2  2   2  2 – 2   2    
  (2   2)    4  2  (2  2 )(2  2 )  (2  2 )(2  2 ) 2  cot 67.5 
  22   2  1  2  2    2  
  2   2  2   
  2  (2  2 )  (2  2 )(2  2 )  
  22 2  42  [5, 5] sc11 by [2, 8] sc11 max: (1, 7), min: (1, 3) 69. 3x  4 
 16 6 
 2y x
4 y
3 (4, 3) 70. x  y  z 
 1 xyz
1 2x  y  3z 
 5 x  y  z 
 11 3x  4z 
 6 2x 
 12 x
6 3x  4z 
 6 x  y  z 
 11 3(6)  4z 
 6 6  y  (3) 
 11 4z 
 12 y
2 z 
 3 (6, 2, 3)  
 2 1 63.   2  
 5  2  
 5  71.   2  sin x 
 5   2  Sample answer: sin x 
 5 64. A 
  2 , 3  2  y  x 7 5 3 1 1  g(x) 4 2 0 2 4  2 3  g (x)  g (x ) 
 |x  3|  O 2 3  y 
 cos 1  72. A 
 2bh  O  90˚  1  A 
 2(6)(1)  180˚ 270˚ 360˚  A
3 The correct choice is C.   23  65.  x1
0 x 
 1  68.  
  
  tan x  sec x sin x  cos x — 1  cos x  1  45 miles  hour  5280 ft  12 inches  1 hour      
 792 in/sec  mile  ft 3600 sec v  v 
 r t v  792 
 7 1 792  
  7 792  radians 7  2  18 rps  66. undefined  Chapter 7  230  x  Page 462  History of Mathematics  4.  1. x2 
 52  52  2(5)(5) cos 10° x 0.87 x2 
 52  52  2(5)(5) cos 20° x 1.74 x2 
 52  52  2(5)(5) cos 30° x 2.59 x2 
 52  52  2(5)(5) cos 40° x 3.42 x2 
 52  52  2(5)(5) cos 50° x 4.23 x2 
 52  52  2(5)(5) cos 60° x
5 x2 
 52  52  2(5)(5) cos 70° x 5.74 x2 
 52  52 2(5)(5) cos 80° x 6.43 x2 
 52  52  2(5)(5) cos 90° x 7.07 Angle Measure 10° 20° 30° 40° 50° 60° 70° 80° 90°  5.  Slope-Intercept Form: y 
 mx  b, displays slope and y-intercept Point-Slope Form: y  y1 
 m(x  x1), displays slope and a point on the line Standard Form: Ax  by  C 
 0, displays no information Normal Form: x cos f  y sin f  p 
 0, displays length of the normal and the angle the normal makes with the x-axis See students' work for sample problems. x cos f  y sin f  p 
 0 x cos 30°  y sin 30°  10 
 0  3 x 2  6.  1   2y  10 
 0  x  y  20 
 0 3 x cos f  y sin f  p 
 0 x cos 150°  y sin 150°  3 
0  3  1  
0 2x  2y  3  Length of Chord (cm) 0.87 1.74 2.59 3.42 4.23 5.00 5.74 6.43 7.07  x  y  23 
0 3 x cos f  y sin f  p 
 0  7.  7  7  x cos 4  y sin 4  52 
0  2 x 2  8.  
   2     2 y  52 
0  x  2 y  102 
0 2 x  y  10 
 0 4x  3y 
 10  A2   B2 
  42  32 or 5 4 x 5  4x  3y  10 
 0  3  10     5 y  5 
 0 4  3  5x  5y  2 
 0 3  4  sin f 
 5, cos f 
 5, p 
 2; Quadrant III  7-6 Page 467  tan f 
  Normal Form of a Linear Equation 9.  Check for Understanding  1. Normal means perpendicular 2. Compute cos 30° and sin 30°. Use these as the coefficients of x and y, respectively, in the normal  3  1  sin f  1 2   10 10       10 y  5 
 0 10 10 3     
 10 , cos f 
 10 , y   
0  10 10  tan f 
 f  231  3  or 4  f y 
 3x  2 3x  y  2 
 0  A2   B2 
  32  12 or 10  3  x 10  3 10  x 10  form. The normal form is 2x  2y  10 
 0. 3. The statement is true. The given line is tangent to the circle centered at the origin with radius p.  3 5  4 5  10    10 — 3 10   10  10   p 
  5 ; Quadrant I  1  or 3  18°  Chapter 7  10.  2 x  2 y 
 6 2 x  2 y  6 
 0  A2   B2 
  2 2  (2  )2 or 2   6 2  2y  2 
 0  2  2y  3 
 0   2 2 sin f 
 2, cos f 
 2, 2  2 tan f 
 — or 1  2  2  17.   3  1  2x  2y  5 
 0   2 x 2  2 x 2  18.  x  y  10 
 0 3 x cos f  y sin f  p 
 0 4  4  x cos 3  y sin 3  5 
 0  p 
 3; Quadrant IV   3  1  2x  2y  5 
 0 y  10 
 0 x  3 x cos f  y sin f  p 
 0  19.  3  x cos 300°  y sin 300°  2 
 0  f 315° 11a. 3x  4y 
 8 y
  x cos f  y sin f  p 
 0 x cos 210°  y sin 210°  5 
 0  1 x 2  y  3 x 4  2   3  y  3 
 0 x  3 x cos f  y sin f  p 
 0  20.  11  O  x   3 x 2  4  p
  5  12  12  Exercises 22.   3   2y  15 
 0      sin   12 
 0  x  2 y  24 
 0 2 14. x cos f  y sin f  p 
 0 x cos 135°  y sin 135°  32 
0  2 2x     2 y 2  x cos  5  6   y sin   3 2x    5  6 1 y 2   32 
0  x cos   y sin    2  or 1  3x  4y 
 15  23.  3x  4y  15 
 0  A2   B2 
  32  ( 4)2 or 5   23 
0  23 
0  3 4 15 x  y   
 5 5 5 3 4 x  y  3 
 5 5 4 sin f 
 5, cos  2
0  0x  1y  2 
 0 y2
0  tan f 
  f Chapter 7   2  2 —  2  2   2  p 
 2; Quadrant I  f 
 45°  x  y  43 
0 3 16. x cos f  y sin f  p 
 0   2  1 1     2 2  2y  2 
 0   2 2 f 
 2, cos f 
 2,    y   
0 2 2  tan f 
  x  2 y  62 
0 2 xy6
0 x cos f  y sin f  p 
 0  15.  f 247° xy
1 xy1
0  A2   B2 
  12  12 or 2  1  x  2  2 x 2  x cos 4  y sin 4  12 
 0  2 y 2  5  –1 12 3 tan f 
 — or 5 5 –1 3  y  30 
 0 x  3 x cos f  y sin v  p 
 0  2 x 2  12   sinf 
 1 3 , cos f 
  13 , p 
 5; Quadrant III  x cos f  y sin f  p 
 0 x cos 60°  y sin 60°  15 
 0    65   1 3 x  13 y  5 
 0  or 1.6 miles  1 x 2  13.  12     13 y  13 
 0  8  Pages 467–469 12.  5 x 13   5y  5 
 0 8  5  1   2y  43 
0  x  y  83 
0 3 21.  A2   B2 
  52  1 22 or 13  3x  4y 
 8 3x  4y  8 
 0  A2   B2 
  32  ( 4)2 or 5 3 x 5  11  x cos 6  y sin 6  43 
0  3x  4y 
 8  11b.  3   2y  2 
 0  232  4 5 — 3  5  307°  0 0 3  f 
 5, p 
 3; Quadrant IV 4  or 3  24.  y 
 2x – 4 2x  y  4 
 0  A2   B2 
  (2)2   12 or 5  2  4    5 4 5     5 y  5 
 0  5 2 5 5, cos f 
 5, p  5 –5 1 — or 2 2 5  5  x  3  sin f 
 tan f 
    3  1  5 4  
 5; Quadrant IV  1  f   3  1  2  f   3  30.  3  4  8  x cos f  y sin f  p 
 0 3 x 5  4   5y  10 
 0  3x  4y  50 
 0  1  y  2 
 4(x  20)  31.  1   A2   B2 
  (4)2   42 or 42 ; p 
 42   2  4  y  2 
 4x  5  4   2    or , sin f 
  or  cos f 
  2 2 4 2 4 2    x  4y  28 
 0  x cos f  y sin f  p 
 0   A2   B2 
  (1)2   42 or 17    2   2  
0 2x  2y  42 xy8
0 32. 22 x 
 y  18 22 x  y  18 
 0  A2   B2 
  2(2)2  (1)2 
 9 
3   2817   p
 17 ; Quadrant II  2 1 18 2 x  y   
 3 3 3 18 p 
 3 
 6 units  tan f 
 — or 4 17   – 17 f  40°     cos f 
 1 0 or 5 , sin f 
 10 or 5  210°  1 4 28 x   y   
0  17  7 17  1  4 2817 17 17         17 x  17 y  17 
 0 4 17 17     sin f 
  17 , cos f 
  17 , 4 17   17   12061  
  61 ; Quadrant I   A2   B2 
  62  82 or 10; p 
 10 6  1  –2  3 tan f 
 — or 3  3 –2 27.  
0  6 5 124  x   y   
0 61 61 61     6 5 12061 61 61      
0 61 61 x  61 y  61 61 5 6 , cos f 
   sin f 
  61 61 , p 5 61   61 5 — tan f 
 6 61 or 6   61  sin f 
 2, cos f 
 2, p 
 1; Quadrant III  f  
1  6x  5y  124 
 0  A2   B2 
  62  52 or 61   2x  2y  2 
 0 1  108°  x y    20 24 x y     1 20 24  29.  
0  2x  2y  1 
 0   610  p 
  5 ; Quadrant II  tan f 
 — or 3 10   – 10  x3
0 sin f 
 0, sin f 
 1, p 
 3 0 tan f 
 1 or 0 f 
 0° 26. 3 x  y 
 2 x  y  2 
 0 3  A2   B2 
  (3  )2  ( 1)2 or 2  3  y4
0  1 3 12  x   y   
0 10 10  10    10 3 10 610         10 x  10 y  5 
 0 3 10 10     sin f 
  10 , cos f 
  10 , 3 10   10  f 333° 25. x
3 x3
0  A2   B2 
  12  02 or 1 1 x 1  
y4  x  3y  12 
 0  A2   B2 
  12  ( 3)2 or 10   1  x  y   
 0   5  5  5   5 2 x 5  x  3  28.  104°  33a.  0  y  45˚  O  x  1.25 ft  233  Chapter 7  33b. p 
 1.25, f 
 45° x cos (45°)  y sin (45°)  1.25 
 0  2 x 2  y  36a.   2   2y  1.25 
 0  x  2 y  2.5 
 0 2  2  y  34a.  1  x  O The angles of the quadrilateral are 180°  a, 90°, f2  f1, and 90°. Then 180°  a  90°  f2  f1  90° 
 360°, which simplifies to f2 
 f1  a. If the lines intersect so that a is an interior angle of the quadrilateral, the equation works out to be f2 
 180°  f1  a. 36b. tan f2 
 tan(f1  a)      x  O   tan f  tan a  f and the supplement of v are complementary angles of a right triangle, so f  180°  v 
 90°. Simplifying this equation gives v 
 f  90°. 34b. tan v. The slope of a line is the tangent of the angle the line makes with the positive x-axis 34c. Since the normal line is perpendicular to 
, the slope of the normal line is the negative 1  reciprocal of the slope of 
. That is,  tan v 
 cot v. 34d. The slope of 
 is the negative reciprocal of the 1  slope of the normal, or  tan f 
 cot f.  1  
 1  tan f tan a 1  If the lines intersect so that a is an interior angle of the quadrilateral, the equation works tan f1  tan a  out to be tan f2 
  1  tan f1 tan a . 37. 5x  y 
 15 5x  y  15 
 0  A2   B2 
  52  ( 1)2 or 26  5 1 15  x   y    26  6  26 2 6  26 5 1526 2  x   y    26 26 26  35b.  5x  2y 
 20 5x  2y  20 
 0  52  ( 2)2 
 25 4   or 29   Quadrant I  5 2 20  x   y      9 29 29 2  29 29 5 2 2029  x   y    29 29 29  f 67° f  90° 
 67°  90° or 157° x cos 157°  y sin 157°  3 
 0 12 1 3x    5 y 13   1526   26  3
0  x  5 y  13   2029  36    5   29  
0  2029   
 0, p 
  29  13.85564879  13.85564879  500 6927.824395; $6927.82 38. 2 cos2 x  7 cos x  4 
 0 (2 cos x  1)(cos x  4) 
 0 2 cos x  1 
 0 or cos x  4 
 0 2 cos x 
 1 cos x 
 4  y 12 13   1526   
 0, p 
  26  3x  4y 
 36 3x  4y – 36 
 0  A2  B2 
  32  42 or 5 3 4 36 36 x  y   
 0, p 
  5 5 5 5  A2   B2 
  52  1 22 or 13 35a.  5 12 39 x  y   
 0 13 13 13 5 12 x  y  3 
 0 13 13 12 5  sin f 
 1 3 , cos f 
 13 ; 12  13 12 tan f 
 — 5 or 5  13  
0  3
0  1  cos x 
 2 x
    3  no solution or x 
  5  3  39. sin x 
  1  co s2 x  O  x  35c. See students' work. 35d. The line with normal form x cos f  y sin f  p 
 0 makes an angle of f with the positive x-axis and has a normal of length p. The graph of Armando's equation is a line whose normal makes an angle of f  d with the x-axis and also has length p. Therefore, the graph of Armando's equation is the graph of the original line rotated d° counterclockwise about the origin. Armando is correct. See students' graphs. Chapter 7  
  1 1  
6    
  36 35      36 or 6  sin y 
  1  co s2 y 2  
3 1   5 5   
  9 or 3  2  
  sin(x  y) 
 sin x cos y  cos x sin y  2 35   5     
 
 6 
 3   
 6 
 3     5  235   
 18  234  1  2  2    40. A 
 1, 4 
 2 or 90°  y 1  1 1 2 4 3  34  y 
 sin 4        O  45˚        1 3 2 2 
 5 4 1 1 0 1 2 x  
 15 4 3 y 0 1 3 2 1 3 2 1 2 x   5 4 1  4 3  y 
 5 4 1  15 1 30 x 
 5 y 15 x 6 
 y 3 (6, 3) 46. The value of 2a  b cannot be determined from the given information. The correct choice is E.  45.  90˚  1 d  41. r 
 2 13.4                   r 
 2 or 6.7 x2 
 6.72  6.72  2(6.7)(6.7) cos 26°20
 x2 9.316604344 x 3.05 cm x  x5 x  x5  42.  17  1  17  1     25  x2 
 x  5  Page 474     x2  25 
 x  5  x    (x  5)(x  5)
 (x  5)(x  5)  
 (x  5)(x  5)
 x  5  1  x(x  5)  17 
 x  5 x2  5x  17 
 x  5 x2  4x  12 
 0 (x  6)(x  2) 
 0 x  6 
 0 or x  2 
 0 x 
 6 x
2 43. original box: V 
 
wh 
462 
 48 new box: V 
 
wh 1.5(48) 
 (4  x)(6  x)(2  x) 72 
 x3  12x2  44x  48 0 
 x3  12x2  44x  24 x 0.4 0.5  P  Q 4. The formula is valid in either case. Examples will vary. For a vertical line, x 
 a, the formula subtracts a from the x-coordinate of the point. For a horizontal line, y 
 b, the formula subtracts b from the y-coordinate of the point. 5. 2x  3y 
 2 → 2x  3y  2 
 0  V(x) 4.416 1.125  V(0.5) is closer to zero, so x 
 0.5. 4  x 
 4  0.5 or 4.5 6  x 
 6  0.5 or 6.5 2  x 
 2  0.5 or 2.5 4.5 in. by 6.5 in. by 2.5 in.  y  44.  x
2  d
  Ax1  By1  C   A2   B2  d
  2(1)  (3)(2)  2   22  ( 3)2 2  13 2   or   d
 13  13  (2, 6)    6. 6x  y 
 3 → 6x  y  3 
 0  xy
8 (2, 3)  Check for Understanding  1. The distance from a point to a line is the distance from that point to the closest point on the line. 2. The sign should be chosen opposite the sign of C where Ax  By  C 
 0 is the standard form of the equation of the line. 3. In the figure, P and Q are any points on the lines. The right triangles are congruent by AAS. The corresponding congruent sides of the triangles show that the same distance is always obtained between the two lines.   (x  5)(x  5)
 x  5  17  Distance From a Point to a Line  7-7  (5, 3) y
3  d
  x  d
  O  Ax1  By1  C   A2   B2 6(2)  (1)(3)  3   62  ( 1)2 12   1237    or  d
 37  37  f(x, y) 
 3x  y  4 f(2, 3) 
 3(2)  3  4 or 7 f(2, 6) 
 3(2)  6  4 or 4 f(5, 3) 
 3(5)  3  4 or 16 16, 4  235  Chapter 7  7. 3x  5y 
 1 When x 
 2, y 
 1. Use (2, 1). 3x  5y 
 3 → 3x  5y  3 
 0 Ax1  By1  C   A2   B2 3(2)  (5)(1)  3 d 
   32  ( 5)2 2 34 4  d 
  or  17  34 2 34   17 1 y 
 3x  3 Use (0, 3). 1 y 
 3x  7 → x  3y  21 Ax1  By1  C d 
   A2   B2 1(0)  3(3)  21 d 
   12  32 30 d 
  or 310   10  2  14. y 
 4  3x → 2x  3y  12 
 0 d
  d
  8.  310   6x1  8y1  5   62  82  9. d1 
  d
  25  13  2513   13  15. y 
 2x  5 → 2x  y  5 
 0  16.  2x1  3y1  4   22  ( 3)2 6x1  8y1  5 2x1 3y1  4  
  10 13   
  17.  2x1 3y1 
 4   13   613 x  813 y  513  
 20x  30y  40 (20  613 )x  (813   30)y  40  513 
0 10. (2000, 0)  d
 d
  11. d 
 d
 d
  10    10  d
 d
  3  5(2000)  (3)(0)  0   52  ( 3)2 10,000  or about 1715  34  d
 ft  d
 d
  Exercises  d
 d
  d
    Ax1  By1  C   A2   B2 4(3)  (5)(0)  (6)   42  ( 5)2 6 41   d 
  or  41 41  20. y 
 2x  1 Use (0, 1). 2x  y 
 2 → 2x  y  2 
 0 d
  534   or 17  d
 d
  Ax1  By1  C   A2   B2 2(0)  (1)(0)  3  (2)2   (1 )2  3 5 3  or  5 5   Chapter 7  
  6   Ax1  By1  C   A2   B2 5(3)  (3)(5)  10   52  ( 3)2  13. 2x  y 
 3 → 2x  y  3 
 0 d
  Ax1  By1  C   A2   B2 3 6(0)  (8) 8  1   62  ( 8)2 4 8  or  5 10  19. 4x  5y 
 12 When x 
 3, y 
 0. Use (3, 0). 4x  5y 
 6 → 4x  5y  6 
 0  534   17  d
  3  4  5  Ax1  By1  C   A2   B2 3(2)  (4)(0)  15   32  ( 4)2 21  5  10  34   10   1  10  18. 6x  8y 
 3 When x 
 0, y 
 8. Use 
0, 8. 6x  8y 
 5 → 6x  8y  5 
 0  21  5  12. d 
   18 
 0   d 
  or  10  Ax1  By1  C   A2   B2  Pages 475–476  Ax1  By1  C   A2   B2 2(3)  3(1)(1)  (5)  d
  22  ( 1)2 0   d
 or 0  5 4 y 
 3x  6 → 4x  3y Ax1  By1  C d 
   A2   B2 4(1)  3(2)  (18) d 
   42  32 16 16   d 
 5 or 5 16  5 Ax1  By1  C d 
   A2   B2 3(0)  (1)(0)  1  d
  32  ( 1)2  d
  
0  613 x  813 y  513  
 20x  30y  40 (20  613 )x  (30  813 )y  40  513  
 0;  d
  2513   d 
  or 13  d2 
  6x1  8y1  5   10  Ax1  By1  C   A2   B2 2(2)  3(3)  (12)  22  32   3 5  5  236  Ax1  By1  C   A2   B2 2(0)  (1)(1)  (2)   22  ( 1)2 3  5   35   or   5  2  21. y 
 3x  6 Use (0, 6). 3x  y 
 4 → 3x  y  4 
 0 Ax1  By1  C   A2   B2 3(0)  1(6)(1)  (4) d 
   32  12 2 10    d 
 10 or  5  8 y 
 5x 1 Use  27. y 
 3x  1 → 2x  3y  3 
 0  y 
 3x  2 → 3x  y  2 
 0 3x1  y1  2 2x1  3y1  3 d1 
  d2 
   22  ( 3)2  32  12  d
  22.  2x1  3y1  3 
   13  (210   313 )x  (13   310 )y  310   213 
0 2x1  3y1  3 3x1  y1  2 
    13 10   Ax1  By1  C   A2   B2 8(0)  (5)(1)  15   82  ( 5)2  d
  210 x  310 y 310  
 313 x  13 y  213  (210   313 )x  (13   310 )y  310   213 
0  28a. Linda: (19, 112)   2089  20   or   d
 89  89   2089   89  d
    d
  3  23. y 
 2x  Use (0, 0).  d
  3  y 
 2x  4 → 3x  2y  8 
 0  d
  8 d 
   13 8 13   13  or  d
 d
 d
  d
 d
  813  13  24. y 
 x  6 xy1
0  d
  Linda 4x  3y  228 
 0 4x  3(140)  228 
 0 4x 
 192 x 
 48 29. Let x 
 1. y  tan v 
 x y  tan 40° 
 1  3x1  4y1  10 5x1  12y1  26  d2 
   32  42  52  ( 12)2 3x1  4y1  10 5x1  12y1  26  
  5 13  y  0.839  0  m  5x  12y 26  d
  1 1  
  13  39x  52y  130 
 25x  60y  130 64x  8y  260 
 0 16x  2y  65 
 0 26.  0.8390996312   m
 10  39x  52y  130 
 25x  60y  130 14x  112y 
 0 x  8y 
 0 3x1  4y1  10   5  Ax1  By1  C   A2   B2 4(45)  (3)(120)  228   42  ( 3)2 48  or 9.6 5  28b.  Use (0, 6).  Ax1  By1  C   A2   B2 1(0)  1(6)  (1)   12  12 5 5 2  or  2  2  25. d1 
  Ax1  By1  C   A2   B2 4(19)  (3)(112)  228   42  ( 3)2 32  or 6.4 5  Father: (45, 120)  Ax1  By1  C   A2   B2 3(0)  2(0)  8   32  22  d
  10   210 x  310 y  310  
 313 x  13 y  213   (0, 1).  8x  15 
 5y → 8x  5y  15 
 0 d
  3x  y  2  1 1    d  0.839  Ax1  By1  C   A2   B2 0.839(16)  1(12)  0  2  12 0.839   y  y1 
 m(x  x1) y  0.839 y 0.839x  y  0.839(x  1) 0.839x 0  d  1.092068438 1.09 m 30. The radius of the circle is  [(5)  (2)]2  (6  2)2 or 5. Now find the distance from the center of the circle to the line.  4x1  y1  6 15x1  8y1  68 d1 
  d2 
  2 2 2  82  4  1 (15)  4x1  y1  6 15x1  8y1  68  
   17 17   68x  17y  102 
 1512 x  817 y  6817   d
  (68  1517 )x  (17  817 )y  102  6817 
0 4x1  y1  6 15x1  8y1  68  
   17 17   d
  68x  17y  102 
 1512 x  817 y  6817   d
  (68  1517 )x  (17  817 )y  102  6817 
0  Ax1  By1  C   A2   B2 5(5)  (12)(6)  32   52  ( 12)2 65  13  d
5 Since the distance from the center of the circle to the line is the same as the radius of the circle, the line can only intersect the circle in one point. That is, the line is tangent to the circle.  237  Chapter 7  47  3  33.    31. m1 
  3  1 or 4  y7
  3 (x 4   1)  3x  4y  25 
 0  Ax1  By1  C a1 
   A2   B2 3(1)  (4)(3)  25 a1 
   32  ( 4)2 34 a1 
 5 3  4 7  or  m2 
  1  (3) 2 7 y  4 
 2(x   2x  7y 
 5 2x  7y  5 
 0  A2   B2 
  22  ( 7)2 or 53  2  x 53   34.  5  53 53 2 5     53 x  53 cos 2A 
 1  2 sin2 A  2 3 
 1  2 6 5 
 6 2 60°  
 2,  
 60° 1 1  a2 
  35.  (3))  y  Ax1  By1  C   A2   B2 7(1)  2(7)  13   72  22  34  or a2 
   53 7  (3)  m3 
  1  (1)  
0  
   7x  2y  13 
 0 a2 
  7  53 7   53 y  y   
0   53  53  1   3453   53  1  y 
 csc (  60˚) ˚ 300˚ 480˚  O 120  or 5  y  7 
 5(x  1) 5x  y  2 
 0 a3 
 a3 
  Ax1  By1  C   A2   B2 5(3)  (1)(4)  2   52  ( 1)2  36. 110  3 
 330 180°  (60°  40°) 
 80° x2 
 3302  3302  2(330)(330) cos 80° x2 179979.4269 x 424.24 miles   1726   1726  34 3453 ,     5 53 , 26 17  
  37. T 
 2 g   or   a3 
  26  26  32.   T 
 2   9.8 2  T 2.8 s 38.  2 1 8 k 2 20 1 10 20  k 20  k 
 0 k 
 20 39. 2x  y  z 
 9 2x  y  z 
 9 → 2(x  3y  2z) 
 2(10) 2x  6y  4z 
 20 7y  5z 
 11 x  2y  z 
 7 x  3y  2z 
 10 y  2z 
 3 5(y  z) 
 5(3) 5y  5z 
 15 → 7y  5z 
 11 7y  5z 
 11 2y 
 4 y 
 2 yz
3 x  2y  z 
 7 2  z 
 3 x  2(2)  (5) 
 7 5 
 z x 
 6 (6, 2, 5)  y 10 8 6 4 2  O 2 4 6 8  x  The standard form of the equation of the line through (0, 0) and (4, 12) is 3x  y 
 0. The standard form of the equation of the line through (4, 12) and (10, 0) is 2x  y  20 
 0. The standard form for the x-axis is y 
 0. To find the 3x  y bisector of the angle at the origin, set  
 y  10 3 x. To find the and solve to obtain y 
  1  10   bisector of the angle of the triangle at (10, 0), set 2x  y  20  5   
 y and solve to obtain 2x  (1  5 )y   20 
 0. The intersection of these two bisectors is the center of the inscribed circle. To solve the  40. square: A 
 s2  3  x into system of equations, substitute y 
  1  10    16 
 s2  the equation of the other bisector and solve for x to  4
s AE 
 s  h AE 
 4  3 or 7 EF 
 AE EF 
 7 The correct choice is C.  20(1  10 ) 20(1  10 )  . Then y 
   5  3 5   210  5  3 5   210  3 60   
  . This y-coordinate is the 1   10 5  35   210   get x 
  inradius of the triangle. The approximate value is 3.33.  Chapter 7  238  1  triangle: A 
 2bh 1  6 
 2(4)h 3
h  Chapter 7 Study Guide and Assessment Page 477  19.  Understanding and Using the Vocabulary  1. b 5. i 9. e  2. g 6. j 10. c  Pages 478–480  3. d 7. h  4. a 8. f  sin4 x  cos4 x  sin2 x (sin2 x  cos2 x)(sin2 x  cos2 x)  sin2 x sin2 x  cos2 x  sin2 x cos2 x  1 sin2 x 1  cot2 x   3   2  17    2     2    2  6   
 4  5 2  11  2  9  
  tan2 v 
 1 6  
  1  
  1  1     sin x 
 sin x sin x   4  23  24. cos x 
  1  si n2 x  16. cos2 x  tan2 x cos2 x 
 1  18.  x
1  
  sin2 x 
 1 1
1   
 1  cos v  2  tan   tan  3 4  2  1  tan  tan  3 4  31  1  (3 )(1) 1  3   1 3    or 2  3 
  2  
 sin x  1  cos v  1  cos v sec v  1  tan v sec v  1  tan v sec v  1  tan v sec v  1  tan v      23. tan 1 2 
 tan 
 3  4   1   (1  sin2 x)
 15. csc x  cos2 x csc x 
  sin x sin x  
  1    2  6  3  
   2   
 
 4  tan v 
 4  
   3  4  tan2 v  1 
 
4  
 (csc v  cot v)2  7  
 

2
2  
2
2  tan2 v  1 
 sec2 v  1  cos v  1  cos v 1  cos v  1  cos v 1  cos v  1  cos v 1  cos v  1  cos v 1  cos v  1  cos v    16  cos v 
 5  cos2  7  
 
sin 4 cos 6  cos 4 sin 6   cos2 v 
 1 cos2 v 
 2 5  1  4  5 5  4  7  
 sin 
4  6  sin2 v  cos2 v 
 1  
35  sin y 
  1  co s2 x  
  1 7  2 25  
   or  
  625 25 576  2 2  1  
3    5   or  
  9 3  24  5  cos (x  y) 
 cos x cos y  sin x sin y  5  
 
25
3  
25
3  1 cos v 2    sin v sin v (1  cos v)2  sin2 v (1  cos v)2  1  cos2 v (1  cos v)2  (1  cos v)(1  cos v)  

  1  17   22. sin 
1 2  
 sin 12  1  17.   2    2  6  1  5  3 3  5  sin2 x  x cos2 x cos2 x    3   
 4   14. sec v 
  cos v  cos2   2  
 2  2  2  2  v1
 v 42  1 
 sec2 v 17 
 sec2 v 17  
 sec v 1  13. sin v 
  csc v  
  1  21. cos 15° 
 cos (45°  30°) 
 cos 45° cos 30°  sin 45° sin 30° sec2  
   2    2  6  
2  
  
 1  cot2 x   
 4  1  1  2  tan2  
  
 1  cot2 x  
 
2
2  2  2  Skills and Concepts  1  12.  
 1  cot2 x  
 1  cot2 x 20. cos 195° 
 cos (150°  45°) 
 cos 150° cos 45°  sin 150° sin 45°   11. csc v 
  sin v  
  
 1  cot2 x  24    2  7   48  75  
 7 5  1  cos v  tan v  
 sec v  1  tan v(sec v  1)   
 sec2 v  1 tan v(sec v  1)   
 tan2 v sec v  1  
 tan v  239  Chapter 7  1  sin y 
  1  co s2 y   25. cos y 
  sec y  
   2 2  
  2   or  
  9 3  
 tan (x  y) 
  5   3  2  3  
 
   
 
 
 26. cos 75° 
 cos  5  2 5  4  8  5 5  8 10  45   8  5 5  2  3    2    3  
    
    30.  27. sin  7  8  v  cos2 v 
 1  16  or  4  sin v 
 5 sin 2v 
 2 sin v cos v 
 2
5
5 4  (Quadrant I)  3  24  
 25     2  31. cos 2v 
 2 cos2 v  1 3 2  
 2
5  1    2   3  2  7  
 2 5  7  4  2  2 tan v  sin v   32. tan v 
  cos v  7 1  cos 4  
    
     2  
  (Quadrant II)   2  1 2  2  
   
  4 2 1  3 24 7  
   
 2
25
2 5 24  45°  28. sin 22.5° 
 sin 2 1  cos 45°  or  4 2 3  4  3  33. sin 4v 
 sin 2(2v) 
 2 sin 2v cos 2v  
 2   
  2  4  5  3  5   tan 2v 
  1  tan2 v  
    2  2   7  336   
  625  (Quadrant I)  tan x  1 
 sec x (tan x  1)2 
 sec2 x tan2 x  2 tan x  1 
 tan2 x  1 2 tan x 
 0 tan x 
 0 x 
 0° 35. sin2 x  cos 2x  cos x 
 0 1  cos2 x  2 cos2 x  1  cos x 
 0 cos2 x  cos x 
 0 cos x (cos x  1) 
 0 cos x 
 0 or cos x  1 
 0 x 
 90° or x 
 270° cos x 
 1 x 
 0°  34.    2  1 2  2  2    2 
 2  Chapter 7  (2  3  )2  43  sin2 v 
 25    
  sin2  180  255  61   3  1   2  
 sin  (2  3 )(2  3 )  (2  3 )(2  3 )  3 2  
   
   3  1  2  sin2 v  
5 
 1  1  cos 150°  2  
   3  1  2    
 2  3   180  825   61 150°  2  
  (Quadrant I)  
  
   5 5  1  4  2       5  or 2  tan x  tan y  1  tan x tan y  5 5    4 2   1  cos  6   1  cos 6  
  sin y   tan y 
  cos y    6  2  
 tan 
   5  5  
 3  29. tan  1  
3  1  3  2    12  240  36.  cos 2x  sin x 
 1 1  2 sin2 x  sin x 
 1 2 sin2 x  sin x 
 0 sin x (2 sin x  1) 
 0 sin x 
 0 or x 
 0° or x 
 180°  45.  6x 
 4y  5 6x  4y  5 
 0  A2   B2 
  62  ( 4)2 or 213  6 4 5  x   y   
0 2 13 2 13 2 13 13 13 13 3 2 5       13 x  13 y  26 
 0 13 13 2 3    sin f 
  13 , cos f 
  13 , p  2 sin x  1 
 0 1 sin x 
 2 x 
 30° or x 
 150°   2  37. sin x tan x  2 tan x 
 0  tan f 
  tan x 
sin x  2 
 0 2   tan x 
 0 x 
 k  2   sin x 
 2 
 0  or  3  x 
 4  2k or 4  2k 38.  sin 2x  sin x 
 0 2 sin x cos x  sin x 
 0 sin x (2 cosx  1) 
 0 sin x 
 0 or x 
 k  9 5 3 x  y   
 0 1  06 1  06 1  06 91  06 51  06 3 106 x  y   
 0 106 106 106 51  06 91  06 31  06   , p 
 ; Quadrant I sin f 
 106 , cos f 
  106 106  2 cos x  1 
 0 1  cos x 
 2 2  tan f 
  x 
 3  2k 4  or x 
 3  2k cos2 x 
 2  cos x x  cos x  2 
 0 (cos x  1)(cos x  2) 
 0 cos x  1 
 0 or cos x  2 
 0 cos x 
 1 cos x 
 2 x 
 2k no solution 40. x cos f  y sin f  p 
 0 39.  47.  cos2       3  tan f 
  y  43 
0 x  3 41. x cos f  y sin f  p 
 0 x cos 90°  y sin 90°  5 
 0 0x  1y  5 
 0 y5
0 42. x cos f  y sin f  p 
 0 x cos  43.  2  y sin 3  1 3 2x  2y  48. d 
 d
  3
0  f  6 13  49. 2y 
 3x  6 → 3x  2y  6 
 0 Ax  By  C  1 1  d
 2 2  B A   3(3)  2(4)  (6)   32  22 23 2313  d 
  or 13 13   2313   13  d
  xy8
0  tan f 
  2(5)  (3)(6)  2   22  ( 3)2     A2   B2 
  72  32 or 58   358   58  758   58  Ax1  By1  C  2 A B2   6   2  7 3 8  x   y   
0 58 58  58   758  358  458  x  y   
 0 58 58 29 358  758    sin f 
 58 , cosf 
 58,  or 7   or   d
 13  13  3
0  
0 2 x  
2y  42  44.  72   10 — 2   10  2   
 2; Quadrant II  98°  f  y  6 
 0 x  3 x cos f  y sin f  p 
 0 x cos 225°  y sin 225°  42 
0  2  or 9  f 29° x  7y 
 5 x  7y  5 
 0  A2   B2 
  12  ( 7)2 or 52    2y  23 
0  2  3  5  106  91  06  106  1 7 5 x  y   
 0 52  52  52  2  72  2  10x  10y  2 
 0 72  2  sin f 
 10, cos f 
 10, p  x cos 3  y sin 3  23 
0 1 x 2  2  or 3  f 146° 46. 9x 
 5y  3 9x  5y  3 
 0  A2   B2 
  92  52 or 106   2   sin x 
 2   2 13   13 —— 13 3   13  13 5   
 26 ; Quadrant II  50. 4y 
 3x  1 → 3x  4y  1 
 0 Ax1  By1  C  4 58  p 
 29;  d 
 2  2 B A    Quadrant I  d
  3(2)  (4)(4)  (1)   32  ( 4)2 23  3  23  d 
 5 or 5  or 7  23  5  23°  241  Chapter 7  57. x  3y  2 
 0  1  51. y 
 3x  6 → x  3y  18 
 0 d
 d
  3  Ax1  By1  C  2 A B2  1(21)  (3)(20)  18   12   (3)2  y 
 5x  3 → 3x  5y  15 
 0 d1 
   2110 21  10 x y 
 3  6 Use (0, 6). x y 
 3  2 → x  3y  6 
 Ax1  By1  C d 
 2  A  B2 1(0)  (3)(6)  6 d 
   12  ( 3)2  d 
  or  10  52.  12 10 24  10 12 10 d 
  5 3 y 
 4x  3 Use (0, 3). 3 1 y 
 4x  2 → 3x  4y  2 Ax1  By1  C d 
 2  A  B2  3(0)  (4)(3)  (2) d 
   32  ( 4)2 14 14 d 
 5 or 5 14 d 
 5  34 x  334 y  234  
 310 x  510 y  1510  (34   310 )x  (334   510 )y  234   1510  
0  0  x1  3y1  2  10   
0  Page 481  Ax  By  C  A   B 1(0)  1(1)  (5)   12  12 4  or 22  2   59. d 
 d
  d 
 22  2  55. y 
 3x  2 d
 d
  d
  Use (0, 2). 60.  sin2 v v0 2    cos2 v cos2 v  
   1 cos2 v 2g   cos2 v v02 sin2 v 
 2 g Ax1  By1  C  2 A  B2 4(1600)  (2)(0)  0   42  ( 2)2 6400   20  x   sin 30° 
  100  13 9  50  50  3   y
 cos 15°  y 
 2  2 → x  2y  3 
 0 x1  2y1  3 3x1  y1  2  d2 
   12  22  32  12 3x1 y1 2 x1 2y1  3   
  10 5    y  51.76 yd  Page 481  35 x  5 y  25  
 10 x  210 y  310  (35   10 )x  (5   210 )y  25   310 
0 3x1 y1 2   10   Open-Ended Assessment 30°  1. Sample answer: 15°; 15° 
 2   sin 2 
  2 1  cos 30°  30°  x 2y  3  1 1  
  5    35 x  5 y  25  
 10 x  210 y  310  (35   10 )x  (5   210 )y  25   310 
0  
    3  1 2  2   2    3  
 2 30°  cos 2 
 
  1  cos 30°    2    3  1 2  2    2  3   
 2 Chapter 7  v 
 15°  cos 15° 
 y  56. y 
 3x  2 → 3x  y  2 
 0 d1 
  30°  45°  v 
 90°  100 sin 30° 
 x 50 
 x x cos v 
 y   or  d 
  13 13 x  sin2 v    cos2 v 
  1 2g   cos2 v v0 2  d 
 1431 ft  Ax1  By1  C   2 B2 A  2(0)  (3)(2)  3   22  ( 3)2 9   Applications and Problem Solving  v02 tan2 v   2g sec2 v  1 1  d
 2 2  d
  3x1  5y1  15   34  58. The formulas are equivalent.  54. x  y 
 1 Use (0, 1). xy
5→xy5
0  d
  
  310 x  510 y  1510  
 34 x  334 y  234  (34   310 )x  (334  510 )y 234   1510 
0  d 
  or  5  53.  x1  3y1  2 3x1  5y1  15  d2 
  2  (   (1)2   32 3 5)2 3x1  5y1  15 x1  3y1  2  
   34  10   242  tan 2 
  1  cos 30°   1  cos 30°  
    
    30°  
 
  3. One way to solve this problem is to label the three interior angles of the triangle, a, b, and c. Then write equations using these angles and the exterior angles. a  b  c 
 180 x  a 
 180 y  b 
 180 z  c 
 180 Add the last three equations. x  a  y  b  z  c 
 180  180  180 x  y  z  a  b  c 
 180  180  180 Replace a  b  c with 180. x  y  z  180 
 180  180  180 x  y  z 
 180  180 or 360 The correct choice is D. 4. Since x  y 
 90°, x 
 90°  y. Then sin x 
 sin (90°  y). sin (90°  y) 
 cos y   3  1  2    3  1  2  2  3   2  2  3   2   (2  3 )(2  3 )  (2  3 )(2  3 )    4 3 (2  3)2  
 2  3   1  cos2 x   2. Sample answer: sin x tan x 
  cos x 1  cos2 x   sin x tan x 
  cos x sin x  sin2 x  sin2 x  cos x   
 cos x  sin x  cos y    sin x  cos x 
 cos x sin2 x  cos y  The correct choice is D. Another solution is to draw a diagram and notice b b that sin x 
 c and cos y 
 c.  SAT & ACT Preparation Page 483  sin(90°  y)   
  
 1 
 cos y cos y  sin x  cos y  SAT and ACT Practice  1. The problem states that the measure of ∠A is 80°. Since the measure of ∠B is half the measure of ∠A, the measure of ∠B must be 40°. Because ∠A, ∠B, and ∠C are interior angles of a triangle, the sum of their measures must equal 180°. m∠A  m∠B  m∠C 
 180 80  40  m∠C 
 180 120  m∠C 
 180 m∠C 
 60 The correct choice is B. 2. To find the point of intersection, you need to solve a system of two linear equations. Substitution or elimination by addition or subtraction can be used to solve a system of equations. To solve this system of equations, use substitution. Substitute 2x  2 for y in the second equation. 7x  3y 
 11 7x  3(2x  2) 
 11 7x  6x  6 
 11 x
5 Then use this value for x to calculate the value for y. y 
 2x  2 y 
 2(5)  2 or 8 The point of intersection is (5, 8). The correct choice is A.  
  b  c — b  c  
1  x˚  c  a b  y˚  5. In order to represent the slopes, you need the coordinates of point A. Since A lies on the y-axis, let its coordinates be (0, y). Then calculate the two y0 y   slopes. The slope of  AB  is  0  (3) 
 3 . The slope y0 y   of A D is  0  3 
  3 . The sum of the slopes is y y    
 0. 3 3 The correct choice is B. 6. Since PQRS is a rectangle, its angles measure 90°. The triangles that include the marked angles are right triangles. Write an equation for the measure of ∠PSR, using expressions for the unmarked angles on either side of the angle of x°. 90 
 (90  a)  x  (90  b) 0 
 90  a  b  x a  b 
 90  x The correct choice is A.  243  Chapter 7  7. Simplify the fraction. One method is to multiply both numerator and denominator by 1 y  y  y2   
 y2 1 2 1  y   y2  
  
  1 y  y    1 2 1  y   y2 y3  y  y2  2y  1  9. Since the volume V varies directly with the temperature T, the volume and temperature satisfy the equation V 
 kT, where k is a constant. 1 When V 
 12, T 
 60. So 12 
 60k, or k 
 5. 1 The relationship is V 
 5T. To find the volume when the temperature is 70°, 1 substitute 70 for T in the equation V 
 5T. 1 V 
 5(70) or 14. The volume of the balloon is 14 in3. The correct choice is C. 10. Two sides have the same length. The lengths of all sides are integers. The third side is 13. From Triangle Inequality, the sum of the lengths of any two sides must be greater than the length of the third side. Let s be the length of the other two sides. Write and solve an inequality. 2s  13 s  6.5 The length of the sides must be greater than 6.5. But the length of the sides must be an integer. The smallest integer greater than 6.5 is 7. The answer is 7. If you answered 6.5, you did not find an integer. If you answered 6, you found a number that is less than 6.5.  y2 . y2    y(y2  1)   
 (y  1)(y  1) y(y  1)(y  1)   
 (y  1)(y  1) y2  y   
 y1  Another method is to write both the numerator and denominator as fractions, and then simplify. 1 y  y   
 1 2 1  y   y2  y2  1         y —— y2  2y  1  y2 y2  1  y2   
 y
 y2  2y  1  y(y  1)(y  1)   
 (y  1)(y  1) y2  y   
 y1  The correct choice is A. 8. Since the triangles are similar, use a proportion with corresponding sides of the two triangles. BC BD  
  AC AE 2 4  
  23 AE  2AE 
 4(2  3) AE 
 10 The correct choice is E.  Chapter 7  244  Chapter 8 Vectors and Parametric Equations 11.  Geometric Vectors  8-1  
13 x 
  2z 
 2.9 cm  Page 490  Check for Understanding  12˚  2.9 cm, 12° 12. h  2.9 cos 55° h 
 1.66 cm  1. Sample answer:  
b  
a  2z 
 
 13
x  13a.  
a 
 b
  v  2.9 sin 55° v 
 2.38 cm 100  5  13b. Use the Pythagorean Theorem. c2  a2  b2 c2  (100)2  (5)2 c2  10,025 c  10,02 5 or about 100.12 m/s  Draw u a . Then drawu b so that its initial point (tip) is on the terminal point (tail) of u a . Draw a dashed line from the initial point ofu a to the terminal point ofu b . The dashed line is the resultant. 2. Sample answer: A vector has magnitude and direction. A line segment has only length. A vector can be represented by a directed line segment. 3. Sample answer: the velocities of an airplane and a wind current 4. No, they are opposites. 5-11. Answers may vary slightly. 5. 1.2 cm, 120° 6. 2.9 cm, 55° 7. 1.4 cm, 20°  Pages 491–492  Exercises  14. 2.6 cm, 128° 15. 1.4 cm, 45° 16. 2.1 cm, 14° 17. 3.0 cm, 340° 18-30. Answers may vary slightly.  
r  
s  18.  
x 
 y
  8.  3 cm 3.5 cm  
r  
s  
y  101˚  3 cm, 101°  
x  70˚  19.  3.5 cm, 70°  
x  9. 2.6 cm  
s  210˚  
s  
t  3.4 cm  
t 25˚  3.4 cm, 25° x 
 
 
y  
y 
  20.  
s  2.6 cm, 210°  
z  10.  359˚  4y 
  
z  u 
 3.8 cm  
s  u
  3.8 cm, 359°  4y 
 12.9 cm  51˚  12.9 cm, 51°  245  Chapter 8  21. 324˚  
t  26.  
u  u 
 
r
  
r  5.5 cm  3.5 cm  
r 
 
t 
 u
  22˚  3.5 cm, 22° 27. 
r  
s 
 
 u  
u 
 
r 5.5 cm, 324° 22.  
u 
 5.4 cm s 
  
r  
r 
 
t 3.9 cm  155˚  
r 133˚  
t 3.9 cm, 155°  5.4 cm, 133°  23.  2s 
  28. 2r
  u 
  5.2 cm  
 12 
r 358˚  5.5 cm 2s 
  u
 
 12 
r  5.5 cm, 358°  
r  128˚  29. Draw to scale: 
2t 
  24.  
r  
s 
  5.2 cm, 128°  301˚  3s 
  
s  3.4 cm  3u 
  4.2 cm  
r 
 2t
 
 
s  3u
 3.4 cm, 301°  45˚  30. Draw to scale: 4.2 cm, 45°  3t
  25.  
 2u 11.7 cm  322˚  3u 
  357˚  11.7 cm, 357° 31. h  2.6 cos 128° h  1.60 cm 32. h  1.4 cos 45° h  0.99 cm 33. h  2.1 cos 14° h  2.04 cm  8.2 cm 
2s 
 3u 
 
 2s
  8.2 cm, 322°  Chapter 8  246  3t 
  2u
  v  2.6 sin 128° v  2.05 cm v  1.4 sin 45° v  0.99 cm v  2.1 sin 14° v  0.51 cm  34. h  3.0 cos 340° v  3.0 sin 340° h  2.82 cm v  1.03 cm 35. c2  a2  b2 c2  (29.2)2  (35.2)2 c2  2091.68 c  2091. 68  or about 45.73 m 36. The difference of the vectors; sample answer: The other diagonal would be the sum of one of the vectors and the opposite of the other vector, so it would be the difference. 37. Yes; sample answer:  41. h  47 cos 40° v  47 sin 40° h 
 36 mph v 
 30 mph 42. It is true when k  1 or whenu a is the zero vector. 2 2 2 43. c  a  b c2  (50)2  (50)2 c  5000  or about 71 lb 44.  
a  60˚  2.4 cm  u u b  24 a equilateral triangleu a u b  24 lb 45. The origin is not in the interior of the acute angle. d1  
d2  
r 
 s
 
r  
s 
 
r  60˚  60˚ 60˚  
s 
r  b 
  60˚ 60˚  
s  x
y2  x
y2   
1  (
1)  
2    or  d1   2 2  38.  y
5  d2  2 2 or y 
 5  35 N  0  1   x
y2  
 2   
( y 
 5)  (y 
 5) x 
 y  2  2 x 
 y  2  2 y 
 52  x 
 y  2 
 2 y  52 0 x 
 (1  2 )y  2  52 0  40 N  1  sin v    46. csc v cos v tan v   sin v  cos v  cos v  60 N  sin v  cos v     sin v  cos v  47.  6.1  
  4  1  
n where n is an integer  48. 23˚  61 N, 23° north of east 39. Sometimes;  

2, 52
 scl

2 by [
3, 3] scl
1  a 
 
 
b  5  b 
  3
  a 
  2.3 
 2.7 
 
 b  x  , 2 for 0  x  2
  5  5   49. tan 18°29
   0.5b  
a  5   b 0.5 tan 18°29
  b  29.9 cm 5  sin 18°29
  h  
a 
 b
  
a  5  a 
  1.5 
 
 b   h sin 18°29
  a 
  2.3 
 2.7 
 
 b  h  15.8 cm  5 b 
  40a. v  1.5 sin 52° v 
 1.18 N 40b. h  1.5 cos 78° h 
 0.31 N  h  1.5 cos 52° h 
 0.92 N v  1.5 sin 78° v 
 1.47 N  247  Chapter 8  u 2. Use XY    (x2 
  x1)2  (y2 
 y1)2 and replace the values for x and y. x(
5, 
6), y(3, 
4) u XY    [3 
 ( 
5)]2   [
4 
 (
 6)]2 2 2   (8)  (
2)  64 4   or 68  3. Jacqui is correct. The representation is incorrect. 2, 0  0, 
5 is not equal to 51, 0  (
2)0, 1. u 
 5j u. The correct expression is 2i u 4. MP  
3 
2, 4 
 (
1) or 
5, 5 u (
5)2   (5)2 MP     50  or 52  units u 5. MP  0 
 5, 5 
 6 or 
5, 
1 u MP    (
5)2   (
1 )2  26  units u 6. MP  4 
 (
19), 0 
 4 or 23, 
4 u MP    (23)2  (
4)2  545  units 7. u t u u u v  
1, 4  3, 
2  
1  3, 4  (
2) or 2, 2 u u 
u v 8. t  1u  50. vo  volume of original box vn  volume of new box v o  
o  w o  h o (w  1)  w  2w (w  1)2w2  2w3  2w2 vn  
n  w n  h n  (w  2)  (w  1)  (2w  2)  (w2  3w  2)(2w  2)  2w3  8w2  10w  4 2w3  8w2  10w  4  160 w 
1 1 2 3  2 2 2 2 2  8 6 10 12 14  10 4 20 34 52  
156 
160 
136 
88 0  wo  3 
o  2w  2  3 or 6 ho  w  1  3  1 or 4 So, the dimensions of the original box are 3 ft  4 ft  6 ft x2   51. g(x)   (x 
 1)(x  3)    vertical: As x approaches 1 and 
3, the expression approaches  or 
. So, x  1 and x  
3 are vertical asymptotes.     y  2 x    x2 x2  x2 2x 3    
  x2 x2 x2 1 2    x x2  2 3 1   
  x x2   
8
1, 4  
8(
1), 
8(4) or 8, 
32 11. 8, 
6   82  ( 
6)2  100  or 10 u 
 6j u 8i 12. 
7, 
5   (
7)2   (
5 )2  74  u 
 5j u 
7i u 13. Let T represent the force Terrell exerts. u Let W represent the force Mr. Walker exerts. u u Tx  400 cos 65° Wx  600 cos 110° 
 169.05 
 
205.21 u u Ty  400 sin 65° Wy  600 sin 110° 
 362.52 
 563.82 u u T  169.05, 362.52, W  
205.21, 563.82 u u T  W  
36.16, 926.34 u u T  W   (
36. 16)2  (926.3 4)2 
 927 N  As x increases positively or negatively, the expression approaches 0. So, y  0 is a horizontal asymptote. 52. Let x, x  2, and x  4 be 3 consecutive odd intergers. 3x  2(x  4)  3 x  4  15 3x  2x  8  3 3x 
 2x  11 x  11 The correct answer 15.  8-2  Algebraic Vectors  Pages 496–497  or 
312, 4  u  6u 9. u t  4u v  4 
1, 4  63, 
2  
4, 16  18, 
12  
4  18, 16  (
12) or 14, 4 u u 10. t  
8u  x2   horizontal: y   x2  2x 
 3  y  2 1 
1, 4 
 3, 
2 2 
12, 2 
 3, 
2 
12 
 3, 2 
 (
2)  Check for Understanding  1. Sample answer:u a  8, 6,u b  6, 8; equal vectors have the same magnitude and direction.  Pages 497–499  Exercises  u 14. YZ  2 
 4, 8 
 2 or 
2, 6 u YZ   (
2)2   62  40  or 210   Chapter 8  248  u 15. YZ  
1 
 (
5), 2 
 7 or 4, 
5 u YZ   42  ( 
5)2  41  u 16. YZ  1 
 (
2), 3 
 5 or 3, 
2 u YZ   32  ( 
2)2  13  u 17. YZ  0 
 5, 
3 
 4 or 
5, 
7 u YZ   (
5)2   (
7 )2  74  u 18. YZ  0 
 3, 4 
 1 or 
3, 3 u (
3)2   32 YZ    18  or 32  u 19. YZ  1 
 (
4), 19 
 12 or 5, 7 u YZ   52  72  74  u 20. YZ  7 
 5, 6 
 0 or 2, 6 u YZ   22 62  40  or 210  u 21. YZ  23 
 14, 
14 
 (
23) or 9, 9 u YZ   92  92  162  or 92  u 22. AB  36 
 31, 
45 
 (
33) or 5, 
12 u AB   52  ( 
12)2  169  or 13 23. u a u b u c  6, 3  
4, 8  6  (
4), 3  8 or 2, 11 u u 24. u a  2b c  26, 3  
4, 8  12, 6  
4, 8  12  (
4), 6  8 or 8, 14 u u c 25. a  b  2u  6, 3  2
4, 8  6, 3  
8, 16  6  (
8), 3  16 or 
2, 19 u  3u 26. u a  2b c  26, 3  3
4, 8  12, 6  
12, 24  12  (
12), 6  24 or 0, 30 u  4c u u 27. a  
b  
6, 3  4
4, 8  
6, 
3  
16, 32  
6  (
16), 
3  32 or 
22, 29 28. u a u b 
 2u c  6, 3 
 2
4, 8  6, 3 
 
8, 16  6 
 (
8), 3 
 (16) or 14, 
13 u u 29. a  3b  36, 3  3  6, 3  3 or 18, 9 u c 30. a  
1u   2 
12
4, 8 
12  (
4),    u 32. u a  0.4b 
 1.2u c  0.46, 3 
 1.2
4, 8  2.4, 1.2 
 
4.8, 9.6  2.4 
 (
4.8), 1.2 
 9.6 or 7.2, 
8.4 1 u 33. u a  (2b 
 5u c)     3 1 (26, 3 
 5
4, 8) 3 1 (12, 6 
 
20, 40) 3 1 (12 
 (
20), 6 
 40) 3 1 32 34 32, 
34 or , 
 3 3 3     u u u u 34. a  (3b  c )  5b  36, 3  
4, 8  56, 3  18, 9  (
4, 8  30, 15  18  (
4)  30, 9  8  15 or 44, 32 u 
 2.5n u  3
5, 
6 
 2.56 
 9 35. 3m  [
15, 
18 
 15, 
22.5  [
15 
 (15), 
18 
 (
22.5  
30, 4.5 36. 3, 4   32  42  25  or 5 u  4j u 3i 37. 2, 
3   22  ( 
3)2  13  u 
 3j u 2i 38. 
6, 
11   (
6)2   (
1 1)2  157  u 
 11j u 
6i 39. 3.5, 12   (3.5)2   122  156.2 5  or 12.5 u  12j u 3.5i 40. 
4, 1   (
4)2   12  17  u u 
4i j 41. 
16, 
34   (
16)2  (
 34)2  1412  or 2353  u 
 34j u 
16i u 42. ST  
4 
 (
9), 
3 
 2 or 5, 
5 u 
 5j u 5i 43. Student needs to show that (u v 1 u v 2) u v 3 u v 1  (u v 2 u v 3) u u u ( v 1  v 2)  v 3  [a, b  c, d]  e, f  a  c, b  d  e, f  a  c  e, b  d  f  a  c  e, b  d  f  a, b  c  e, d  f  a, b  [c, d  e, f] u v 1  (u v 2 u v 3) 44a. 100 N  Fy  20˚  Fx  
12   8 or 2, 
4  u F y   44b. sin 20°   100 u Fy  100 sin 20°  u u 31. u a  6b  4c  66, 3  4
4, 8  36, 18  
16, 32  36  (
16), 18  32 or 20, 50  
 34 N  249  Chapter 8  45a.  53. Let a  400, b  600, C  46.3 ° c2  4002  6002 
 2(400)(600) cos 46.3 ° c2 
 18.8578.39 c 
 434 Pabc 
 400  600  434 
 1434 ft 1 s  2(a  b  c)  Surfer  Vk  Vs  15  30˚  Vx  Shore  45b. sin 30°  u Vk   15  u Vk 15  sin 30°  1  s 
 2(1434) or 717 k 
 s(s 
 a)(s 
 b)(s 
 c) k 
 717(7 17 400)(7 
 17 600)(7 
 17 434) 
 k 
 7,525 ,766,0 79  k 
 86,751 sq ft  
 30 mph u u u u 46a. Since QR  ST  0, QR  
ST . So, they are opposites. u u 46b. QR and ST have the same magnitude, but opposite direction. So, they are parallel. Quadrilateral QRST is a parallelogram.  54. Sample answer: f(x)  3x2 
 2x  1 r 3 
2 1 3 1 2 3 4  d  47a. t  r   150 m  5 m/s  or 30 s  47b. d  rt  (1.0 m/s)(30 s) or 30 m u u 47c. V B  V C  0.5  1.0   12  52  26  or about 5.1 m/s (x2 
 x1) v cos v 48. cos v   → (x 
 x ) u sin v  49.  u v (y2 
 y1)  u v  2  An upper bound is 2. 1 2 9  f(
x)  3x2 
 2x  1 r 3 2 1 1 3 5 6 2 3 8 17  1  A lower bound is 
1.  55.  → (y2 
 y1) u v sin v  u PQ  
2 
 8, 5 
 (
7)  
10, 12 u PQ    (
10)2  122  244  u RS  7 
 8, 0 
 (
7)  
1, 7 u RS    (
1)2   72  50  none  50. d  d  [
4, 4] scl
1 by [
4, 4] scl
1 max: (0, 3), min: (0.67, 2.85) 56.  Ax1  By1  C   A2   B2 3(
1) 
 7(4) 
 1   32  ( 
7)2 
3 322  d   or about 4.2  58  51. sin 255°  sin (225°  30°)  sin 225° cos 30°  cos 225° sin 30° 2    3  2  1  2   
2  2  
2 6  
 2    
  4  y →  as x → , y →  as x → 
 57. 7x  1 7x 
 1 1 
1 This statement is true regardless of the value of x, so it is true for all real values of x. The correct choice is A.  52. y  A sin (kx  c) A: A  17 A  17 or 
17 2
 
 k:    k 4 k8 c c: 
k  
60° c 
  
60° 8 c  480° y  17 sin (8x  480°) Chapter 8  f(x)  x2  3x  1 x f(x) 
10,000 99,970,001 
1000 997,001 
100 9701 
10 71 0 1 10 131 100 10,301 1000 1,003,001 10,000 100,030,001  250  8-3  11. 132, 3454, 0   1322  34542  02  11,947 ,540  
 3457 N  Vectors in Three-Dimensional Space  Pages 502–503  Check for Understanding  1. Sample answer: sketch a coordinate system with the xy-axes on the horizontal, and the z-axis u is two units along the pointing up. Then, vector 2i u x-axis, vector 3j is three units along the y-axis, u is four units along the z-axis. Draw and vector 4k broken lines to represent three planes.  Pages 503–504  Exercises  12.  z y  z x B(4,1,
3) u 42  12  (
 3)2 OB     26   y x 13. 2. Sample answer: To find the components of the vector, you will need the direction (angle) with the horizontal axis. Using trigonometry, you can obtain the components of the vector. 3. Sample answer: Neither is correct. The sign for theu j -term must be the same (
), and the coefficient for theu k -term is 0, so the correct way to express the vector as a sum of unit vectors is u u. i 
 4j  z  B(7, 2, 4) y x u OB    72  22  42  69  14. z  4. G(4,
1, 7)  B(10,
3,15)  16 14 12 10 8 6 4 2 
4  
12
10
8
6
4
22 2 4 y 
2 4 
4 6 8  u OG    42  ( 
1)2  72  66  u 5. RS  3 
 (
2), 9 
 5, 
3 
 8 or 5, 4, 
11 u RS    52  42  (
 11)2  162  or 92  u 6. RS  10 
 3, 
4 
 7, 0 
 (
1) or 7, 
11, 1 u RS    72  ( 
11)2  12  171  or 319  u u 7. u a  3f g  31, 
3, 
8  3, 9, 
1  3, 
9, 
24  3, 9, 
1  3  3, 
9  9, 
24  (
1) or 6, 0, 
25 u 8. u a  2u g 
 5f  23, 9, 
1 
 51, 
3, 
8  6, 18, 
2 
 5, 
15, 
40  6 
 5, 18 
 (
15), 
2 
 (
40) or 1, 33, 38 u 9. EF  6 
 (
5), 
6 
 (
2), 6 
 4  11, 
4, 2 u 
 4j u  2k u 11i u 10. EF  
12 
 (
12), 17 
 15, 
22 
 (
9)  0, 2, 
13 u 
 13k u 2j  x u OB    102  (
3)2  152  334  u 15. TM  3 
 2, 1 
 5, 
4 
 4 or 1, 
4, 
8 u TM    12  ( 
4)2  (
8)2  81  or 9 u 16. TM  
3 
 (
2), 5 
 4, 2 
 7 or 
1, 1, 
5 u TM    (
1)2   12  (
5)2  27  or 33  u 17. TM  3 
 2, 1 
 5, 0 
 4 or 1, 
4, 
4 u TM    12  ( 
4)2  (
4)2  33  u 18. TM  
1 
 3, 1 
 (
5), 2 
 6 or 
4, 6, 
4 u TM    (
4)2   62  (
4)2  68  or 217  u 19. TM  
2 
 (
5), 
1 
 8, 
6 
 3 or 3, 
9, 
 9 u TM    32  ( 
9)2  (
9)2  171  or 319   251  Chapter 8  u 20. TM  1 
 0, 4 
 6, 
3 
 3 or 1, 
2, 
6 u TM    12  ( 
2)2  (
6)2  41  u 21. CJ   3 
 (
1), 
5 
 3, 
4 
 10 or 4, 
8, 
14) u 42  ( 
8)2  (
14 )2 CJ     276  or 269  u  2u 22. u u  6w z  62, 6, 
1  23, 0, 4  12, 36, 
6  6, 0, 8  18, 36, 2 1 23. u u  u v 
u w  2u z  u 35. G1G2   (x2 
  x1)2  (y2 
 y1)2  (z2 
 z1)2 u   (x1 
  x2)2  (y1 
 y2)2  (z1 
 z2)2  G1G2 because (x 
 y)2  (y 
 x)2 for all real numbers x and y. 36. Ifu m  m1, m2, m3, then u u m    (m )2   (m  )2  (m )2. If 
m 1  Since m12  (
m1)2, m22  (
m2)2, and m32  u  m u. (
m3)2, 
m u u 37. 3, 
2, 4  6, 2, 5  F  O u u 9, 0, 9  F  O u F  
9, 0, 9 or 
9, 0, 
9 1 38. m  2(x1  x2, y1  y2, z1  z2)  2   2, 
2, 2 
 2, 6, 
1  6, 0, 8 3 5   6, 
 72, 112 1  1  24. u u  4 u v 
u w 3   2(2  4, 3  5, 6  2)  3   2(6, 8, 8)  1   44, 
3, 5 
 2, 6, 
1   (3, 4, 4) u 39a. OK  1 
 0, 4 
 0, 0 
 0 or 1, 4, 0 u u i  4j u 39b. TK  1 
 2, 4 
 4, 0 
 0 or 
1, 0, 0 u 
i u 40. u c  b 
u a u c  3, 1, 5 
 1, 3, 1 u c  2, 
2, 4 z 41a.   3, 
4, 4 
 2, 6, 
1 9 15   1, 
84, 44 1  3  2 25. u u  3u v 
 3 u w  2u z 2   34, 
3, 5 
 32, 6, 
1  23, 0, 4  12, 
9, 15 
 3, 4, 
3  6, 0, 8 4  2   163, 
13, 233 26. u u  0.75u v  0.25u w 2  2   0.754, 
3, 5  0.252, 6, 
1  3, 
2.25, 3.75  0.5, 1.5, 
0.25  3.5, 
0.75, 3.5 u u 27. u u  
4w z  
42, 6, 
1  3, 0, 4  
8, 
24, 4  (3, 0, 4  
5, 
24, 8 2 2 28. 3u f  3u g 
 5u h   3   (
m1)2  (
 m2 )2  (
m3  )2.  1 u u  24, 
3, 5 
 2, 6, 
1  23, 0, 4  1  2  u  
m1, 
m2, 
m3, then 
m  2 
3, 3  4.5, 
1  3
2, 1, 6  6  O 6  x  y  (15 
 0)2  (15 
 0)2  (15 
 0)2 d    675  or about 26 feet    15  41c. sin v  26    u 29. LB  5 
 2, 
6 
 2, 2 
 7 or 3, 
8, 
5 u 
 8j u 
 5k u 3i u 30. LB  
4 
 (
6), 5 
 1, 
1 
 0 or 2, 4, 
1 u  4j u 
u 2i k u 31. LB  7 
 9, 3 
 7, 
2 
 (
11) or 
2, 
4, 9 u 
 4j u  9k u 
2i u 32. LB  
8 
 12, 7 
 2, 
5 
 6 or 
20, 5, 
11 u  5j u 
 11k u 
20i u 33. LB  
8 
 (
1), 5 
 2, 
10 
 (
4) or 
7, 3, 
6 u  3j u 
 6u 
7i k u 34. LB  6 
 (
9), 5 
 12, 
5 
 (
5) or 15, 
 7, 0 u 
 7j u 15i 52 36 278 
5, 5, 1 5  Chapter 8  12  12  6  
  41b. Find distance between (0, 0, 0) and (15, 15, 15).  2 
 56, 
3, 
3 12 6 6 , 
, 
 5 5 5   
2, 3, 
3  
6, 3, 18 
  2  12  v  sin1  15   675  v  35.25° u (1 
 2 )2  (  
 0 3 )2  (0 
 0)2 42. AB     4  or 2 u BC    (1 
 1)2   1  3  
 3   2    22   3  
0  2   36 
  63    3 or 
 1.69 u AC    (1 
 2)2   1  3  
0  2    22   3  
0  2   2  or 
 1.41 No, the distances between the points are not equal. A and B are 2 units apart, B and C are 1.69 units apart, and A and C are 1.41 units apart. 43. 3, 5  
1, 2  3  (
1), 5  2  2, 7  252  44.  45.  u AB  
3 
 5, 3 
 2 or 
8, 1 u CD  d1 
 0, d2 
 0 or d1, d2 u u AB  CD 
8, 1  d1, d2 D  (
8, 1) sin 2X  1 
 cos 2X 2 sin X cos X  1 
 cos2 X  sin2 X 2 sin X cos X  2 sin2 X cos X  sin X  u i 2. u a u a  ax ax  ay ay  u u j k ay az ay az ay u az u i 
 ax az u j  ax k az ax az ax ay u 
 (a a 
 a a )j u  (ayaz 
 ayaz)i x z x z  (a a 
 a a )u k   cot X  x y   cot X  cot X  cot X  3.  cot X  cot X 46. cos v   4.  2  3  sin2 v  1 
 cos2 v sin2 v  1 
  47.  4  9  5.  5 sin2 v  9 5  sin v  3 v y  6 sin 2  6.  amplitude  6 or 6 2
  period  k  7.  2
  or 4
 1  2 2
 rad 1 min     1 rev 60 sec   rev   48. 16  min   8
   15 radians per second  13, 1, 
5  1, 
3, 2 13(1)  1(
3)  (
5)(2) 13 
 3 
 10  0 13, 1, 
5  
2, 1, 
5 13(
2)  1(1)  (
5)(
5) 
26  1  25  0 u i u j u k 8. 6, 2, 10  4, 1, 9  6 2 10 4 1 9 2 10 u 6 10 u 6 2u i 
 j  k  1 9 4 9 4 1 u 
 14j u 
 2k u or 8, 
14, 
2, yes  8i  49. Yes, because substituting 7 for x and 
2 for y results in the inequality 
2 180 which is true. y 4x2 
 3x  5 
2 4(7)2 
 3(7)  5 
2 180 3 31  4  3  2    50. 2
 21  3  4  3  So, A, C, and D are not correct. 2 21 
 3 31  3  2  3   4  3  4  So, B is not correct. The correct choice is E.  8-4  8, 
14, 
2  6, 2, 10 8(6)  (
14)(2)  (
2)(10) 48 
 28 
 20  0 8, 
14, 
2  4, 1, 9 8(4)  (
14)(1)  (
2)(9) 32 
 14 
 18  0 9. Sample answer: Let T(0, 1, 2), U(
2, 2, 4), and V(
1, 
1, 
1) u TU  
2, 1, 2 u UV  1, 
3, 
5 u u TU  UV  u u i j u k  
2 1 2 1 
3 
5 2u 2u 1u i 
 
2 j  
2 k  1 
3 
5 1 
5 1 
3 u  5k u or 1, 
8, 5 u i 
 8j  Perpendicular Vectors  Pages 508–509  x y  u 
 0j u  0k u  0i  0, 0, 0 u 0 Sample answer: No, because a vector cannot be perpendicular to itself. 5, 2  
3, 7  5(
3)  2(7)  
15  14  
1, no 
8, 2  4.5, 18  
8(4.5)  2(18)  
36  36  0, yes 
4, 9, 8  3, 2, 
2  
4(3)  9(2)  8(
2)  
12  18 
 16  
10, no u u u i j k 1, 
3, 2  
2, 1, 
5  1 
3 2 
2 1 
5 
3 2u 1 2u 1 
3 u  i 
 j  k 1 
5 
2 
5 
2 1 u 
u u or 13, 1, 
5, yes  13i j 
 5k  Check for Understanding  1. Sample answer: Vector  vw  is the negative of vector w  v .  u u u i j k u v u w  
1 0 3 1 2 4 0 3u  i 
 
1 3 u j  
1 0 u k 2 4 1 4 0 2 u  7j u 
 3k u  
6i u u u i j k u v u w  1 2 4 
1 0 3 2 4u  i 
 1 4u j  1 2u k 0 3 
1 3 
1 0 u u u  6i 
 7j  3k  253  Chapter 8  u 10. AB  (0.65, 0, 0.3) 
 (0, 0, 0)  0.65, 0, 0.3 u F  0, 0, 
32 u u u T  AB  F  u u u i j k  0.65 0 0.3 0 0 
32 0 0.3 u 0.65 0.3 u 0.65 0.3 u i 
 j  k  0 
32 0 
32 0 0 u 
 20.8j u  0k u  0i  u i u j u k 21. 0, 1, 2  1, 1, 4  0 1 2 1 1 4 1 2u 0 2u 0 1u  i 
 j  k 1 4 1 4 1 1 u  2j u 
u  2i k or 2, 2, 
1, yes 2, 2, 
1  0, 1, 2 2(0)  2(1)  (
1)(2) 22
20 2, 2, 
1  1, 1, 4 2(1)  2(1)  (
1)(4) 22
40  u T    02  (
20.8  )2  02  20.8 foot-pounds  Pages 509–511  u i u j u k 22. 5, 2, 3  
2, 5, 0  5 2 3 
2 5 0 2 3u 5 3u 5 2u  i 
 j  k 5 0 
2 0 
2 5 u 
 6j u  29k u or 
15, 
6, 29, yes  15i  Exercises  11. 4.8  6, 
3  4(6)  8(
3)  24 
 24  0, yes 12. 3, 5  4, 
2  3(4)  5(
2)  12 
 10  2, no 13. 5, 
1  
3, 6  5(
3)  (
1)(6)  
15 
 6  
21, no 14. 7, 2  0, 
2  7(0)  2(
2) 0
4  
4, no 15. 8, 4  (2, 4  8(2)  4(4)  16  16  32, no 16. 4, 9, 
3  
6, 7, 5  4(
6)  9(7)  (
3)(5)  
24  63 
 15  24, no 17. 3, 1, 4  2, 8, 
2  3(2)  1(8)  4(
2) 68
8  6, no 18. 
2, 4, 8  16, 4, 2  
2(16)  4(4)  8(2) 
32  16  16  0, yes 19. 7, 
2, 4  3, 8, 1  7(3)  (
2)(8)  4(1)  21 
 16  4  9, no 20. u a u b  3, 12  8, 
2  24 
 24  0, yes u b u c  8, 
2  3, 
2  24  4  28, no u a u c  3, 12  3, 
2  9 
 24  
15, no  
15, 
6, 29  5, 2, 3 (
15)(5)  (
6)(2)  29(3) 
75 
 12  87  0 
15, 
6, 29  
2, 5, 0 (
15)(
2)  (
6)(5)  29(0) 30 
 30  0  0 u i u j u k 23. 3, 2, 0  1, 4, 0  3 2 0 1 4 0 2 0u 3 0u 3 2u  i 
 j  k 4 0 1 0 1 4 u 
 0j u  10k u or 0, 0, 10, yes  0i 0, 0, 10  3, 2, 0 0(3)  0(2)  10(0) 0000 0, 0, 10  1, 4, 0 0(1)  0(4)  10(0) 0000  u u u i j k 24. 1, 
3, 2  5, 1, 
2  1 
3 2 5 1 
2 
3 2u 1 2u 1 
3 u  i 
 j  k 1 
2 5 
2 5 1 u  12j u  16k u or 4, 12, 16, yes  4i 4, 12, 16  1, 
3, 2 4(1)  12(
3)  16(2) 4 
 36  32  0 4, 12, 16  5, 1, 
2 4(5)  12(1)  16(
2) 20  12 
 32  0  u u i j u k 25. 
3, 
1, 2  4, 
4, 0  
3 
1 2 4 
4 0 
1 2 u 
3 2 u 
3 
1 u  i 
 j  k 
4 0 4 0 4 
4 u  8j u  16k u or 8, 8, 16, yes  8i 8, 8, 16  
3, 
1, 2 8(
3)  8(
1)  16(2) 
24 
 8  32  0 8, 8, 16  4, 
4, 0 8(4)  8(
4)  16(0) 32 
 32  0  0  Chapter 8  254  u u i u j k 26. 4, 0, 
2  
7, 1, 0  4 0 
2 
7 1 0 0 
2 u 4 
2 u i 
 j  4 0u k  1 0 
7 0 
7 1 u  14j u  4k u or 2, 14, 4, yes  2i  30. Sample answer: Let T(
2, 1, 0), U(
3, 0, 0), and V(5, 2, 0). u TU  
1, 
1, 0 u UV  8, 2, 0 u u j u k i u u TU  UV  
1 
1 0 8 2 0 
1 0 u i 
 
1 0 u j  
1 
1 u k  2 0 8 0 8 2 u  0j u  6k u or 0, 0, 6  0i  2, 14, 4  4, 0, 
2 2(4)  14(0)  4(
2) 80
80 2, 14, 4  
7, 1, 0 2(
7)  14(1)  4(0) 
14  14  0  0 27. Sample answer: u  
v , 
v , 
v  Letu v  v1, v2, v3 and 
v 1 2 3 u u u j k i u u)  v  (
v v v v 1  2  31. Sample answer: Let T(0, 0, 1), U(1, 0, 1), and V(
1, 
1, 
1). u TU  1, 0, 0 u UV  
2, 
1, 
2 u u u i j k u u TU  UV  1 0 0 
2 
1 
2 0u 0u 0u i 
 1 j  1 k  0 
1 
2 
2 
2 
2 
1 u  2j u 
u  0i k or 0, 2, 
1  3  
v1 
v2 
v3 v v3 u v v2 u 3 u 2  i 
 v1 j  v1 k 
v2 
v3 
v1 
v3 
v1 
v2 u 
 0j u  0k u0  0i u u u i j k u u 28. a  (b  c )  a1 a2 a3 (b1  c1) (b2  c2) (b3  c3) u u a3 a a3 a1 2 i 
 j  (b2  c2) (b3  c3) (b1  c1) (b3  c3) u
 2 u 2 (b3 a1 c3) 
 a3 a(b  [a ck )]i (b1  c1) (b2  2c2) 2 u [a1 (b3  c3) 
 a3 (b1  c1)]j  u [a1 (b2  c2) 
 a2 (b1  c1)]k u
  [(a2b3  a2c3) 
 (a3b2  a3c2)]i u [(a1b3  a1c3) 
 (a3b1  a3c1)]j u [(a1b2  a1c2) 
 (a2b1  a2c1)]k u  [(a2b3 
 a3b2)  (a2c3 
 a3c2)]i 
 u [(a1b3 
 a3b1)  (a1c3 
 a3c1)]j u [(a1b2 
 a2b1)  (a1c2 
 a2c1)]k u u  (a2b3 
 a3b2)i  (a2c3 
 a3c2)i 
 u 
 (a c 
 a c )u (a1b3 
 a3b1)j 1 3 3 1 j  u  (a c 
 a c )k u (a1b2 
 a2b1)k 1 2 2 1 u 
 (a b 
 a b )j u  [(a2b3 
 a3b2)i 1 3 3 1  u u
 (a1b2 
 a2b1)k ] [(a2c3 
 a3c2)i u u (a c 
 a c )j  (a c 
 a c )k ] 1 3         ab  1 2  33a.  elbow  forearm  2 1    a3 u i 
 a1 a3 u j  a1 a2 u k  b3 b1 b3 b1 b2  a2 a3 u i 
 a1 a3 u j  a1 c2 c3 c1 c3 c1 u (u a  b ) (u a u c)  a2 u k c2  0.04 m  u u u 33b. T  AB  F u AB  0.04 cos (
30°), 0, 0.04 sin (
30°)  0.02(3 ), 0, 
0.02 u F  0, 0, 600 u u u i j k u u AB  F  0.023  0 
0.02 0 0 600 u 
 123 u  0i u j  0k u u u T  AB  F   123  or about 21 N-m u i u j u k 34. u x u y  2 3 0 
1 1 4  3 0u i 
 2 0u j  2 3u k 1 4 
1 4 
1 1 u 
 8j u  5k u  12i    2  600 N  30˚  a2 a3 u i 
 a1 a3 u j  a1 a2 u k  b2 b3 b1 b3 b1 b2 2    3 1  32. The expression is false.u m u n andu n u m have the same magnitude but are opposite in direction.  1 u A  2x u y    1  122  (
8)2  (5)2  2 1   2233  u 35a. o  120, 310, 60 u c  29, 18, 21 35b. u o u c  120(29)  310(18)  60(21)  29. Sample answer: Let T(0, 
2, 2), U(1, 2, 
3), and V(4, 0, 
1) u TU  1, 4, 
5 u UV  3, 
2, 2 u u u i j k u u TU  UV  1 4 
5 3 
2 2 4 
5 u 1 
5 u 1 4u i 
 j  k  
2 2 3 2 3 
2 u 
 17j u 
 14k u or 
2, 
17, 
14  
2i   $10,320  255  Chapter 8  u u2 
 2a u cos v u2  b u b 40. BA 2 a  36a.   (a1 
  b1)2  (a2 
 b2)2 2 2 2 2  a  a2   b12  b22 2 1  2 2 
 2 a  a22 b  b22 cos v 1  1  (a1 
 b1)2  (a2 
 b2)2  F   a12  a22  b12  b22 
 2  a12  a22  45˚   b12  b22 cos v a12 
 2a1b1  b12  u u 36b. W  F  d  cos v W  120  4  cos 45° W 
 339 ft-lb u 37a. X  2 
 1, 5 
 0, 0 
 3 or 1, 5, 
3) u Y  3 
 2, 1 
 5, 4 
 0 or 1, 
4, 4 u u u i j k u u X Y  1 5 
3 1 
4 4 5 
3 u 1 
3 u 1 5u  i 
 j  k 
4 4 1 4 1 
4 u 
 7j u 
 9k u or 8, 
7, 
9  8i   a22 
 2a2b2  b22  a12  a22  b12  b22 
 2  a12  a22  b12  b22 cos v 
 2a1b1 
 2a2b2  
2  a12  a22  b12  b22 cos v a1b1  a2b2   a12  a22  b12  b22 cos v a1b1  a2b2 u cos v u u b  a a u b u cos v u b  a  u 41. AB  5 
 3, 3 
 3, 2 
 (
1) or 2, 0, 3 42. D(8, 3) E(0, 
2) u DE  0 
 8, 
2 
 3) or 
8, 
5 u (
8)2   (
5 )2 DE     89  43. 4x  y 
 6  0  A2   B2   42  12 or 17   37b. The cross product of two vectors is always a vector perpendicular to the two vectors and the plane in which they lie. u u 38a. v u p  (q r) u u u i j k u r  2 1 
4 q u 
3 1 
5 1 
4 u 2 
4 u 2 1u  i 
 j  k 1 
5 
3 
5 
3 1 u 
 22j u  5k u or 
1, 
22, 5  
i  417   17 617   x  y 
   17 17 17 617  p  17 
 1.46 units  17 sin f  1 7 1   tan f  4  u u u p  (q r )  0, 0, 
1  
1, 
22, 5  0(
1)  0(
22)  (
1)(5)  
5 or 5 units3 0 0 
1 2 1 
4 38b. 
3 1 
5 1 
4 2 
4 2 1  0
 0 (
1) 1 
5 
3 
5 
3 1  4 17  cos f  17  f  14° 44. A  36°, b  13, and c  6 a2  b2  c2 
 2 bc cos A a2  132  62 
 2(13)(6) cos 36° a 
 8.9 sin 36° sin B  
  8.9 13   
5 or 5 units3 They are the same. u u 39. Need (kv w ) u u  0. [k1, 2  
1, 2]  5, 12  0 [k, 2k  
1, 2]  5, 12  0 k 
 1, 2k  2  5, 12  0 (k 
 1)5  (2k  2)12  0 5k 
 5  24k  24  0 29k  19  0 19 k  
29  Chapter 8  0  B 
 sin1  13 sin 36°  8.9  B 
 59.41° or 59°25
 C 
 180° 
 36° 
 59°25
 C 
 84.59° or 84°35
 45.  h  tan 73°  4 4 tan 73°  h  13.1  h; 13.1 m 46. 3  3x 
 4 10  3x 
4 7  3x 
 4 49 x 17.67 4 47. 81  3 64  26  (22)3 or (23)2 4  22 2  21 9  32 So 64  43  82 The correct choice is B.  256  4  cos 73°  
  4   
 cos 73°  
  13.7 m  Page 511  Mid-Chapter Quiz  8-4B Graphing Calculator Exploration:  Finding Cross Products  1. 2.3 cm 46˚  Page 512  Fx 
 2.3 cos 46° 
 1.6 cm  Fy 
 2.3 sin 46° 
 1.7 cm  1. 49, 32 55 2. 168, 96, 76 3. 0, 0, 0 4. 11, 15, 3 5. 0, 0, 7 6. 0, 40, 0 u u 7. u  x 
 6, 6, 12 u u u v  
  62  62  ( 12)2 
 216  8. u u u v 
 1, 13, 20 u u u v  
  12  ( 13)2  ( 20)2 
 570  9. Sample answer: Insert the following lines after the last line of the given program. :Disp "LENGTH IS" :Disp ((BZ  CY)2  (CX  AZ)2  (AY  BX)2)  2. 115˚ 2.7 cm  Fy 
 27 sin 245° Fx 
 27 cos 245° 
 11.4 mm 
 24.5 mm u 3. CD 
 4  (9), 3  2 or 5, 5 u 52  ( 5)2 CD  
  
 50 or 52  u 4. CD 
 5  3, 7  7, 2  (1) or 2, 0, 3 u 2  32 CD  
  22  0 
 13  u u 5. u r 
 t  2s 
 6, 2  2 4, 3 
 6, 2  8, 6 
 6  8, 2  6 or 14, 8 u u 6. u r 
 3u v 
 3 1, 3, 8  3, 9, 1 
 3, 9, 24  3, 9, 1 
 3  3, 9  9, 24  (1) or 6, 0, 25 7. 3, 6  4, 2 
 3(4)  6(2) 
 12  12 
 0; yes 8. 3, 2, 4  1, 4, 0 
 3(1)  (2)(4)  4(0) 
38 
 11; no u 9. 1, 3, 2  2, 1, 1 
 u u u j k 1 3 2 2 1 1 3 2u 1 2u 1 3u 
 i  j  k 1 1 2 1 2 1 u  5j u  7k u or 1, 5, 7 , yes 
 i  8-5  Applications with Vectors  Pages 516–517  Check for Understanding  1. Sample answer: Pushing an object up the slope requires less force because the component of the weight of the object in the direction of motion is mg sin v. This is less than the weight mg of the object, which is the force that must be exerted to lift the object straight up. 2. The tension increases. 3. Sample answer: Forces are in equilibrium if the u resultant force is O . Current  4. 23 knots  17˚  u u 5. F1 
 300i u u F2 
 (170 cos 55°)u i  (170 sin 55°)j u u F1  F2 
 (300   170 c os 55° )2  (1 70 sin 55°)2 421.19 N  1, 5, 7  1, 3, 2 (1)(1)  5(3)  (7)(2) 1  15  14 
 0 1, 5, 7  2, 1, 1 (1)(2)  5(1)  (7)(1) 2  5  7 
 0 10. Let X(2, 0, 4) and Y(7, 4, 6). XY 
  (7  2 )2  (4  0)2  (6   4)2 
 45  or about 6.7 m  tan v 
  170 sin 55°  300  170 cos 55°  v 
 tan1 
  300  170 cos 55°  170 sin 55°  u u 6. F1 
 50i u u F2 
 100j u u F1  F2 
  502  1002 111.8 N 100 tan v 
 5 or 2 0 v 
 tan1 2 63.43° 7. horizontal 
 18 cos 40° 13.79 N vertical 
 18 sin 40° 11.57 N  257  Chapter 8  u u  (33 sin 90°)j u or 33j u 8. F1 
 (33 cos 90°)i u u  (44 sin 60°)j u F2 
 (44 cos 60°)i u u or 22i  22 3 j u u 2 F1  F2 
  22  (33  223  )2 74 N 33  223   u u v 1 
 (115 cos 60°)i  (115 sin 60°)j 15. u u u or 57.5i  57.53 j u  (115 sin 120°)j u u v2 
 (115 cos 120°)i u u or 57.5i  57.53 j 2  (1153 u u v v  
 0  )2 1 2 
 1153  199.19 km/h Since tan v is undefined and the vertical component is positive, v 
 90°. 16. The force must be at least as great as the component of the weight of the object in the direction of the ramp. This is 100 sin 10°, or about 17.36 lb. u u 17. F 
 105i  3  23   tan v 
 22 or 2 v 
 tan1 
2 3  23   73° A force with magnitude 74 N and direction 73°  180° or 253° will produce equilibrium. 4 mph  9a.  1  u u  (110 sin 50°)j u F2 
 (110 cos 50°)i u u F1  F2
  (105  110 cos  50° )2  (1 10 sin 50°)2 194.87 N  12 mph  110 sin 50°  105  110 cos 50° 110 sin 50° tan1  105  110 cos 50°  tan v 
 9b. If v is the angle between the resultant path of the ferry and the line between the landings, 1 4 1 1, or about  then sin v 
 1 3 2 or 3 . So v 
 sin 19.5°.  Pages 517–519  sin1 
7 5 
v 52.1  44° v u u  (250 sin 25°)j u 19. F1 
 (250 cos 25°)i u u u F2 
 (45 cos 250°)i  (45 sin 250°)j u u F1  F2
  Wind  11.  27˚  42 N  256 mph 53˚  (250 cos  25°  45  cos 25 0°)2  (250 s in 25°  45 sin  25 0°)2  342 lb  12.    25.62° F 
 w sin v 52.1 
 75 sin v 52.1  
 sin v 75  18.  Exercises  10.  
  v
  220.5 lb tan v 
  94˚  250 sin 25°  45 sin 250°  250 cos 25°  45 cos 250° 250 sin 25°  45 sin 250°  v 
 tan1 
  250 cos 25°  45 cos 250°  16.7° u u  (70 sin 330°)j u or 353 u 20. F1 
 (70 cos 330°)i u i  35j  454 lb  u u  (40 sin 45°)j u or 202 F2
 (40 cos 45°)i u i  202 u j u u u F 
 (60 cos 135°)i  (60 sin 135°)j or 30 2   302 u j 3  u u 13. F1 
 425i u u F 
 390j  tan v 
 v
  2  u u F1  F2 
  4252  3902 576.82 N 390    37.5°  78  58.6 lb u u u 21. F1 
 (23 cos 60°)i  (23 sin 60°)j u u or 11.5i  11.53j  u u  (23 sin 120°)j u F2 
 (23 cos 120°)i u u or 11.5i  11.53j  u u F1  F2 
  02  ( 233  )2 
 233  39.8 N Since tan v is undefined and the vertical component is positive , v 
 90°. A force with magnitude 39.8 N and direction 90°  180° or 270° will produce equilibrium.  78  v 
 tan1 85 42.5° u u 14. v 1 
 65i  u  (50 sin 300°)j u or 25i u 253 u v 2 
 (50 cos 300°)i u j  u u v v2 
  902  (253  )2 1 99.87 mph 53    tan v 
 9 0 or  18  v 
 tan1 
1 8  5 3  A positive value for v is about 334.3°.  Chapter 8  
  u u u F1  F2  F3 
  (353   10 )2   2 (35  502  )2    tan v 
  425 or 85  25 3  35  502   353   102  35  502  1  tan 353   102   258  22. a 
 g sin 40° 
 32 sin 40° 20.6 ft/s2 u u  (36 sin 20°)j u 23. F1 
 (36 cos 20°)i u u F 
 (48 cos 222°u i  (48 sin 222°)j  F1 
  760 lb cos 174.5°   F2 
  cos 6.2° F1  761 lb 30. Sample answer: Method b is better. Let F be the force exerted by the tractor, T be the tension in the two halves of the rope, and v be the angle between the original line of the rope and half of the rope after it is pulled. At equilibrium, F  2T sin v  F 
 0, or T 
  v 30°, 2 sin v . So, if 0° the force applied to the stump using method b is greater than the force exerted by the tractor. 31. Let T be the tension in each towline and suppose the axis of the ship is the vertical direction. 2T sin 70°  6000 
 0 6000  T
 2 sin 70°  2  u u F1  F2
   cos 20°  48 co s 222° )2  (3 6 sin 2 0°  4 8 sin 2 22°)2 (36 19.9 N  tan v 
 v
  36 sin 20°  48 sin 222°  36 cos 20°  48 cos 222° 36 sin 20°  48 sin 222° tan1  36 cos 20°  48 cos 222°  
    264.7° or 5.3° west of south 24a. 135 lb 165˚ 70 lb  75˚  120˚  3192.5 tons 32. Let T be the tension in each wire. The halves of the wire make angles of 30° and 150° with the horizontal. T sin 30°  T sin 150°  25 
 0  115 lb  u u 24b. F1 
 70i u u  (135 sin 165°)j u F2 
 (135 cos 165°)i u u u F1  F2  F3 
  1 T 2  134.5 lb 135 sin 165°  115 sin 240°  70  135 cos 165°  115 cos 240° 135 sin 165°  115 sin 240°  v 
 tan1 
  70  135 cos 165°  115 cos 240°   2v02  2  1002  
 3 sin 65° cos 65° 2 239.4 ft 36. Sample answer: A plot of the data suggests a quadratic function. Performing a quadratic regression and rounding the coefficients gives y 
 1.4x2  2x  3.9. 37. b  0.3 b (0.33, 0.33) p  0.2 0.5 (0.2, 0.46) b  p  0.66 bp 0.4 0.3  1   27a. tan v 
 1 8 or 6  0.2  1  v 
 tan1 6 9.5° south of east 27b. s 
  (0.3, 0.3)  p O 0.1 0.2 0.3 0.4 The vertices are at (0.2, 0.3), (0.3, 0.3), (0.33, 0.33) and (0.2, 0.46). cost function C(p, b) 
 90p  140b  32(1  p  b) 
 32  58p  108b C(0.2, 0.3) 
 32  58(0.2)  108(0.3) or 76 C(0.3, 0.3) 
 32  58(0.3)  108(0.3) or 81.8 C(0.33, 0.33) 
 32  58(0.33)  108(0.33) or 86.78 C(0.2, 0.46) 
 32  58(0.2)  108(0.46) or 93.28 The minimum cost is $76, using 30% beef and 20% pork.  18.2 mph 28. F cos v 
 100 cos 25° 90.63 N 29. F1 cos 174.5°  F2 cos 6.2° 
 0 F1 sin 174.5°  F2 sin 6.2°  155 
 0 cos 174.5°   The first equation gives F2 
  cos 6.2° F1. Substitute into the second equation. cos 174.5° sin 6.2°  cos 6.2°  (0.2, 0.3)  0.1   182  32  F1 sin 174.5°   0.  35. d 
 g sin v cos v  208.7° or 28.7° south of west u u u Since F1  F2  F3 0, the vectors are not in equilibrium. u 25. W 
 F u d u  (1600 sin 50°)j u]  1500i u 
 [(1600 cos 50°)i 
 (1600 cos 50°)(1500)  (1600 sin 50°)(0) 1,542,690 N-m 26a. Sample answer: The horizontal forward force is u F cos v. You can increase the horizontal forward force by decreasing the angle v between the handle and the lawn. 26b. Sample answer: Pushing the lawnmower at a lower angle may cause back pain. 3  1   2T  25 
 0 T 
 25 lb 33. u u  u v 
 9(3)  5(2)  3(5) 
 2 The vectors are not perpendicular sinceu u u v u 34. AB 
 0  12, 11  (5), 21  18 
 12, 6, 3  (70  135 c os 165 °  11 5 cos 2 40°)2  (135 s in 165 °  11 5 sin 2 40°)2   tan v 
  155  sin 174.5°  cos 174.5° tan 6.2°  F1  155 
 0  F1 (sin 174.5°  cos 174.5° tan 6.2°) 
 155  259  Chapter 8  11b. 50  10t 
 0 50 
 10t 5
t When t 
 5, the coordinates of the defensive player are (10  0.9(5), 54  10.72(5)) or (5.5, 0.4), so the defensive player has not yet caught the receiver.  38. *4  *(3) 
 (43  4)  [(3)3  (3)] 
 60  (24) 
 84 The correct choice is A.  Vectors and Parametric Equations  8-6  Pages 523–524  524–525  Check for Understanding  1. When t 
 0, x 
 3 and y 
 1. When t 
 1, x 
 7 and y 
 1. The graph is a line through (3, 1) and (7, 1). 2. Sample answer: For every single unit increment of t, x increases 1 unit and y increases 2 units. Then, the parametric equations of the line are x 
 3  t, y 
 6  2t. 3. When t 
 0, x 
 1 and y 
 0, so the line passes through (1, 0). When t 
 1, x 
 0 and y 
 1, so the line passes through (0, 1), its y-intercept. The 10   slope of the line is  0  1 or 1.  4. x  (4), y  11 
 t 3, 8 x  4, y  11 
 t 3, 8 x  4 
 3t y  11 
 8t x 
 3t  4 y 
 8t  11 5. x  1, y  5 
 t 7, 2 x  1 
 7t y  5 
 2t x 
 1  7t y 
 5  2t 6. 3x  2y 
 5 7. 4x  6y 
 12 2y 
 3x  5 6y 
 4x  12 3 5 2 y 
 2x  2 y 
 3x  2 x
t 3 5 y 
 2t  2  x
t 2 y 
 3t  2  x 
 4t  3 x  3 
 4t  8. 1  9. x 
 9t x  
 t 9  3  4x  4 
 t y 
 5t  3  y 
 4t  2 x y 
 4
9  2  y 
 5
4x  4  3 1  y
 10.  5 4x  t 1 0 1 2    3  4  y 
 9x  2  3  4  x 2 2 6 10  y 2 1 0 1  y 11a. receiver: x  5 
 0t y  50 
 10t x
5 y 
 50  10t O defensive player: x  10 
 0.9t y  54 
 10.72t x 
 10  0.9t y 
 54  10.72t  Chapter 8  Exercises  12. x  5, y  7 
 t 2, 0 x  5 
 2t y  7 
 0t x 
 5  2t y
7 13. x  (1), y  4 
 t 6, 10 x  1, y  4 
 t 6, 10 x  1 
 6t y  4 
 10t x 
 1  6t y 
 4  10t 14. x  (6), y  10 
 t 3, 2 x  6, y  10 
 t 3, 2 x  6 
 3t y  10 
 2t x 
 6  3t y 
 10  2t 15. x  1, y  5 
 t 7, 2 x  1 
 7t y  5 
 2t x 
 1  7t y 
 5  2t 16. x  1, y  0 
 t 2, 4 x  1, y 
 t 2, 4 x  1 
 2t y 
 4t x 
 1  2t 17. x  3, y  (5) 
 t 2, 5 x  3, y  5 
 t 2, 5 x  3 
 2t y  5 
 5t x 
 3  2t y 
 5  5t 18. x 
 t y 
 4t  5 19. 3x  4y 
 7 20. 2x  y 
 3 4y 
 3x  7 y 
 2x  3 3 7 y 
 2x  3 y 
 4x  4 x
t x
t y 
 2t  3 3 7 y 
 4t  4 21. 9x  y 
 1 y 
 9x  1 x
t y 
 9t  1  22. 2x  3y 
 11 3y 
 2x  11 2 11 y 
 3x  3 x
t 2 11 y 
 3t  3  23. 4x  y 
 2 y 
 4x  2 x
t y 
 4t  2  24. 3x  6y 
 8 6y 
 3x  8 1 4 y 
 2x  3 1  The slope is 2. 1  y  5 
 2(x  2) 1  x  y 
 2x  6 x
t 1  y 
 2t  6  260  25. x 
 2t x  
 t 2  1  x  7 
 2t 2x  14 
 t y 
 3t y 
 3(2x  14) y 
 6x  42  y
1t x y 
 1  2 1  y 
 2x  1 27.  x 
 4t  11 x  11 
 4t 1 11 x    
 t 4 4 y
t3 1 11 y 
 4x  4  3 1  23  y 
 4x  4 29.  33.  1  x 
 7  2t  26.  28.  x 
 4t  8 x  8 
 4t 1 x  2 
 t 4 y
3t 1 y 
 3  4x  2 1  y 
 4x  5  x 
 3  2t x  3 
 2t 1 3 x   
 t 2 2  [10, 10] Tstep1 [20, 20] Xscl2 [20, 20] Yscl2  y 
 1  5t 1 3 y 
 1  5 
2x  2 5  34.  17  y 
 2x  2 30. Regardless of the value of t, x is always 8, so the parametric equations represent the vertical line with equation x 
 8. 31a. x  11, y  (4) 
 t 3, 7 x  11, y  4 
 t 3, 7 31b. x  11 
 3t y  4 
 7t x 
 3t  11 y 
 7t  4 31c. x 
 3t  11 x  11 
 3t 1 x 3  11   3 
 t  y 
 7t  4 1 11 y 
 7
3x  3  4 7  89  y 
 3x  3  [10, 10] Tstep1 [10, 10] Xscl1 [10, 10] Yscl1 35a. x 
 2  3t and y 
 4  7t If t  0, then x  2 and y  4, so the part of the line to the right of point (2, 4) is obtained. 35b. x 0 2  3t 0 3t 2 2 t 3  32.  36. x  y 
 cos2 t  sin2 t 
1 0  cos2 t  1 and 0  sin2 t  1, so the graph is the segment of the line with equation x  y 
 1 from (1, 0) to (0, 1).  y [5, 5] Tstep1 [10, 10] Xscl1 [10, 10] Yscl1  1  ( 12 , 12 )  x  1  261  Chapter 8  37a. target drone: x 
 3  (1)t x
3t missile: x
2t 37b. 3  t 
 2  t 1 
 2t 1  
 t 2  45. The slope is 1. y  1 
 1[x  (3)] y1
x3 xy4
0 46. The linear velocity of the belt around the larger  y 
 4  0t y
4 y 
 2  2t  pulley is (120 rpm)
2  2 in./rev 
 1080 9  in./min. The linear velocity around the smaller pulley must be the same, so its angular velocity is  1  When t 
 2, the missile has a y-coordinate of 3, not 4, so it does not intercept the drone. 38a. Ceres: x 
 1  t, y 
 4  t, z 
 1  2t Pallas: x 
 7  2t, y 
 6  2t, z 
 1  t 38b. Adding the equations for x and y for Ceres gives x  y 
 3. Subtracting the equations for x and y for Pallas results in x  y 
 1. The solution of this system is x 
 1 and y 
 2. Eliminating t from the equations for y and z results in the system 2y  z 
 7, y  2z 
 4 which has solution y 
 2 and z 
 3. Hence, the paths cross at (1, 2, 3). 38c. 1  t 
 1 v t 
 2 7  2t 
 1 v t 
 4 Ceres is at (1, 2, 3) when t 
 2 but Pallas is at (1, 2, 3) when t 
 4. The asteroids will not collide.   (1080 in./min)
 2  3 in.  
 180 rpm. The correct 1 rev  choice is D.  8-6B Graphing Calculator Exploration:  Modeling with Parametric Equations  Page 526 1. 408.7t 
 418.3(t  0.0083) 408.7t 
 418.3t  3.47189 9.6t 
 3.47189 3.47189   t
 9.6  0.362 hr or 21.7 min 2. d 
 rt 3.47189  
 408.7
 9.6  147.8 mi  39. The line is parallel to the vector 0  
3, 5  1, 1  1 , 3  8  1 or  line is x  
  4, 9 . The vector equation of the  , y 
 1, z  1  1 3  1  x  3, y  1, z  1 
 t 1  1  x
  1 3  x  3 
 3t  1 , 3  
t  1 , 3  500   3. The time for plane 1 to fly 500 miles is  408.7 . The  4, 9 or  500   time for plane 2 is  418.3  0.0083. Suppose the speed of plane 1 is increased by a mph.  4, 9 .  y  1 
 4t   1 t 3  500  408.7  a  y 
 1  4t  500   
 418.3  0.0083  1 
  500   0.0083 418.3 500 a 
   408.7 500   0.0083 418.3  408.7  a  500  z  1 
 9t z 
 1  9t u u u 40. v 1 
 (150 cos 330°)i  (150 sin 330°)j u  (50 sin 245°)j u u v 
 (50 cos 245°)i 2  6.7 mph  u u v v2 
 1  (150 c os 330 °  50 cos 2 45°)2   (15 0 sin  330°  50 sin 245°)2   162.2 km/h tan v 
  150 sin 330°  50 sin 245°  150 cos 330°  50 cos 245°  8-7  150 sin 330°  50 sin 245°  v 
 tan1 
  150 cos 330°  50 cos 245°   Page 531  47°53
4 or 47°53
4 south of east 41. 1, 3  3, 2 
 1(3)  3(2) 
 3 Since the inner product is not 0, the vectors are not perpendicular. 42. Since A 90°, a b, and a b sin A, no solution exists. 43. A graphing calculator indicates that there is one real zero and that it is close to 1. f(1) 
 0, so the zero is exactly 1. 3 3  x  2 
 2y 2 x 3  4   3 
 y 2  4  y 
 3x  3 Chapter 8  Check for Understanding  1. Sample answer: a rocket launched at 90° to the horizontal; tip-off in basketball 2. Equal magnitude with opposite direction. 3. The greater the angle of the head of the golf club, the greater the angle of initial velocity of the ball. u  
 v u sin v u  
 v u cos v 4. v 5. v y x 
 50 sin 40° 
 20 cos 50° 32.14 ft/s 12.86 m/s u  
 v u cos v u  
 v u sin v 6. v v x y 
 45 cos 32° 
 45 sin 32° 38.16 ft/s 23.85 ft/s u  
 v u cos v u  
 v u sin v 7. v v x y 
 7.5 cos 20° 
 7.5 sin 20° 7.05 m/s 2.57 m/s  x 
 2y  2  44.  Modeling Motion Using Parametric Equations  262  16. To find the time the projectile stays in the air, set y 
 0 and solve for t. u sin v  1gt2 
 0 tv 2    8a. 300 mph 
 mile 
 3600 s  
 440 ft/s u cos v x 
 tv 5280 ft  h  x 
 t(440) cos 0° x 
 440t u sin v  1gt2  h y 
 tv  u sin v  gt) 
 0 t(v 2 1  u sin v  1gt 
 0 v 2 u sin v 
 1gt v  2  1  y 
 t(440) sin 0°  2(32)t2  3500  u sin v 2v  g   3500 y
 8b. Sample graph: y 16t2  4000 3000 Height (feet) 2000 1000  x  g  
  3500   t2 
  16  Pages 531–533     16 3500  14.8 s  1084 u 
   v 7 sin 78° u 158.32 ft/s v 1 u  cos v  50 yd 17b. x 
 3tv  Exercises  u  
 v u cos v 9. v x 
 65 cos 60° 
 32.5 ft/s u  
 v u cos v 10. v x 
 47 cos 10.7° 46.18 m/s u  
 v u cos v 11. v x 
 1200 cos 42° 891.77 ft/s u  
 v u cos v 12. v x 
 17 cos 28° 15.01 ft/s u u cos v 13. vx 
 v 
 69 cos 37° 55.11 yd/s u  
 v u cos v 14. v x 
 46 cos 19° 43.49 km/h u 15a. x 
 tv  cos v  
 3(7)(158.32) cos 78°  50 127 yd u cos v 18. x 
 tv x  
 t u v  cos v  u sin v  gt2 y 
 tv 2 2 x u sin v  1g 
x y 
  v 1  u cos v v  y 
 x tan v   2 g  u2 cos2 v 2v of the x2-term  u cos v v  The presence (due to the force of gravity) means that y is a quadratic function of x. Therefore, the path of a projectile is a parabolic arc. 19. To find the time the projectile stays in the air if the initial velocity isu v, set y 
 0 and solve for t. u sin v  1gt2 
 0 tv 2 u sin v  1gt 
 0 t 
v 2 u sin v  1gt 
 0 v 2  2  y 
 175t sin 35°  y
0 175t sin 35°  16t2 
 0 t(175 sin 35°  16t) 
 0 175 sin 35°  16t 
 0 175 sin 35° 
 16t 175 sin 35°  16  1  u  
 v u sin v v y 
 65 sin 60° 
 56.29 ft/s u  
 v u sin v v y 
 47 sin 10.7° 8.73 m/s u  
 v u sin v v y 
 1200 sin 42° 802.96 ft/s u  
 v u sin v v y 
 17 sin 28° 7.98 ft/s u u sin v vy 
 v 
 69 sin 37° 41.53 yd/s u  
 v u sin v v y 
 46 sin 19° 14.98 km/h 1 u y 
 tv  sin v  gt2  x 
 175t cos 35°  15b.  u2g sin 2v v  g  As the angle increases from 0° to 45°, the horizontal distance increases. As the angle increases from 45° to 90°, the horizontal distance decreases. 17a. y 
 300 when t 
 7 u sin 78°  1(32)72 
 300 7v 2 u sin 78°  784 
 300 7v u sin 78° 
 1084 7v  8c. 16t2  3500 
 0 16t2 
 3500  t 8d. x 
 440t 
 440(14.8) 
 6512 ft  
t  The greater the angle, the greater the time the projectile stays in the air. To find the horizontal distance covered, substitute the expression for t in the equation for x. u cos v x 
 tv u sin v 2v u cos v 
  v  2000 4000 6000 Horizontal Distance (feet)  t
  2  16t2  u sin v 
 gt v 2 1  u sin v 2v  g
t  
t  To find the range, substitute this expression for t in the equation for x. u cos v x 
 tv  
t  u  2v  sin v u  cos v 
 g v  x 
 175t cos 35°  u2 sin 2v v  
 175 
1  cos 35° 6 175 sin 35°  
 g  899.32 ft or 299.77 yd  263  Chapter 8  22b.  If the magnitude of the initial velocity is doubled u)2 sin 2v (2v u, the range becomes   or to 2v g  u2 sin 2v v  4 g. The projectile will travel four times as far.   20a. 800 km/h 
 km 
 3600 s  u x 
 tv  cos v h  1000 m  222.2 m/s  x 
 222.2 t cos 45°  23a. y 
 300 when t 
 4.8 u sin 82°  1(32)(4.8)2 
 300 4.8v 2 u sin 82°  368.64 
 668.64 4.8v  u sin v  gt2 y 
 tv 2 1  1  y 
 222.2t sin 45°  2(9.8)t2  668.64 u 
   v 4.8 sin 82° u 140.7 ft/s v  y 
 222.2t sin 45°  4.9t2 The negative coefficient in the t-term in the equation for y indicates that the aircraft is descending. The negative coefficient in the equation for x is arbitrary. 20b. y 
 222.2t sin 45°  4.9t2 
 222.2(2.5) sin 45°  4.9 (2.5)2 423.4 The aircraft has descended about 423.4 m. 20c.  423.4 m  2.5 s  u cos v  100 23b. x 
 3tv 1  131.3 yd u cos v 24a. x 
 tv  x 
 155t cos 22° y 
 155t sin 22°  16t2  3 24b. x 
 420 155t cos 22° 
 420 420  t
 155 cos 22°  169 m/s  or km 3600 s   169 m/s 
 1000 m 
 h  
 608.4 km/h 21a. 70 mph 
  5280 ft  mi  
  
  h  3600 s  y 
 155t sin 22°  16t2  3 420   2  t
 t  t 3.84 s u cos v x 
 tv 323.2 ft 21b. y
8 308 t
3 sin 35°  16t2  10 
 8 308  n 35°   4(16)2   
3038 si  2  t 
  t 3.71 s u cos v x 
 tv 312.4 ft 21c. From the calculations in part b, the time is about 3.71s. y 22a.  17.4   27. cos A 
  21.9 17.4   A 
 cos1  21.9  A  t  O  Chapter 8  155 sin 22°   (155 s in 22° )2  4 (16) 3   3.68 s u cos v x 
 tv 528.86 ft 25. x 
 11  t x  11 
 t x  11 
 t y 
 8  6t y 
 8  6(x  11) y 
 6x  58 26a. mg sin v 
 300(9.8) sin 22° 1101.3 N 26b. mg cos v 
 300(9.8) cos 22° 2725.9 N  16t2  t
3 sin 35°  2 
 0 308 3 sin 35°   420  36.04 ft Since 36.04  15, the ball will clear the fence. 24c. y
0 155t sin 22°  16t2  3 
 0  sin 35°  
3 sin 35°  4(16)10 t 
  308  2    
 155 
 155 cos 22°  sin 22°  16 
 155 cos 22°   3  308  3  ft/s y
0 308 t
3 sin 35°  16t2  10 
 0 308  3  u sin v  1gt2  h y 
 tv 2  x  264  37°  u 5a. BE 
 0  5, 2  5, 4  0 or 5, 3, 4 A(5, 5  (3), 0) 
 A(5, 2, 0) C(5  (5), 5, 0) 
 C(0, 5, 0) D(5  (5), 5  (3), 0) 
 D(0, 2, 0) F(5, 5  (3), 0  4) 
 F(5, 2, 4) G(5, 5, 0  4) 
 G(5, 5, 4) H(5  (5), 5, 0  4) 
 H(0, 5, 4) 5 5 0 0 0 5 5 0 The matrix is 2 5 5 2 2 2 5 5 . 0 0 0 0 4 4 4 4  28. 2(2x  y  z) 
 2(2) x  3y  2z 
 3.25 5x  y 
 0.75 1(2x  y  z) 
 1(2) 4x  5y  z 
 2.5 6x  4y 
 0.5 4(5x  y) 
 4(0.75) 6x  4y 
 0.5 14x 
 3.5 3.5 x 
 14    x 
 0.25 5x  y 
 0.75 2x  y  z 
 2 5(0.25)  y 
 0.75 2(0.25)  0.5  z 
 2 y 
 0.5 z
1 2 2 29.   5    3 
 25  9 
 16 The correct choice is B.  Page 534 1.  7°12
  360°  5b.  History of Mathematics  5c.  7.2°   
 360° 1  5000 stadia  
 x  x 
 50(5000) x 
 250,000 stadia 250,000(500) 
 125,000,000 ft 125,000,000 5280 23,674 mi The actual circumference of Earth is about 24,901.55 miles. 2. See students' work. No solution exists. 3. See students' work.  8-8      1 0 0 0 1 0 0 0 1 5 5 
 2 5 0 0    5 5 0 0 0 2 5 5 2 2 0 0 0 0 4 0 0 0 5 2 2 0 0 4  E  F C  B      G  x  5 2 4 5 2 4    5 0 5 5 4 4 0 5 5 5 4 4    z  O D  y  A  The image is the reflection over the xz-plane. 5d. The dimensions of the resulting figure are half the original. 6a. The scale factor of the dilation is 4. The translation increases x-coordinates by 2. The matrices are 4 0 0 D 
 0 4 0 and 0 0 4 2 2 2 2 2 2 2 2 T
 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0  Check for Understanding     1. Matrix T multiplies x-coordinates by 2 and y- and z-coordinates by 2, so it produces a reflection over the yz-plane and increases the dimensions two-fold. u 2. CC 
 
 8  6, 8  7, 2  3 or 2, 1, 1 2 2 2 2 2 2 The matrix is 1 1 1 1 1 1 . 1 1 1 1 1 1      H  Transformation Matrices in Three-Dimensional Space  Pages 539–540  54 54 04 04 2  (1) 5  (1) 5  (1) 2  (1) 02 02 02 02 04 54 54 04 2  (1) 2  (1) 5  (1) 5  (1) 42 42 42 42 9 9 4 4 4 9 9 4 
 1 4 4 1 1 1 4 4 2 2 2 2 6 6 6 6    
 5 0 1  50          1 0 0 0 1 0 
 T, so the transformations 0 0 1 are the same.  3. VU 
      4a-c. Transformation Reflection Translation Dilation  Orientation yes no no  Site no no yes  Shape no no no  265  Chapter 8  z  6b. Sample answer: If the original prism has vertices A(3, 3, 0) B(3, 3, 3), C(3, 3, 3), D(3, 3, 0), E(5, 3, 0), F(5, 3, 3), G(5, 3, 3), and H(5, 3, 0), then the image has vertices A
(10, 12, 0), B
(10, 12, 12), C
(10, 12, 12), D
(10, 12, 0), E
(22, 12, 0), F
(22, 12, 12), G
(22, 12, 12), and H
(22, 12, 0).  F  D  B D
 C G D A F
 F E H  G
  H
  H A  B  G  E  z C
  C  O y  x  B
  The result is a translation of 2 units along the yaxis and 4 units along the z-axis.  A
  01 01 01 01 11. 0  (2) 3  (2) 3  (2) 0  (2) 1  (2) 2  (2) 5  (2) 4  (2)    21 0  (2) 1  (2) 1 1 1 1 
 2 1 1 2 1 0 3 2 z  E
 x    Pages 540–542  Exercises  u 7. FB 
 3  3, 1  7, 4  4 or 0, 6, 0 A(2, 3,  (6), 2) 
 A(2, 3, 2 C(4, 7  (6), 1) 
 C(4, 1, 1) 2 3 4 4 2 3 The matrix is 3 1 1 7 3 7 . 2 4 1 1 2 4 u 8. AH 
 4  (3), 1  (2), 2  2 or 7, 3, 4 B(3, 2  3, 2) 
 B(3, 1, 2) C(3, 2  3, 2  (4)) 
 C(3, 1, 2) D(3, 2, 2  (4)) 
 D(3, 2, 2) E(3  7, 2, 2  (4)) 
 E(4, 2, 2) F(3  7, 2, 2) 
 F(4, 2, 2) G(3  7, 2  3, 2) 
 G(4, 1, 2) 3 3 3 3 4 4 4 4 The matrix is 2 1 1 2 2 2 1 1 2 2 2 2 2 2 2 2 u 9. CF 
 6  4, 0  (1), 0  2 or 2, 1, 2 D(2  2, 2  1, 3  (2)) 
 D(4, 1, 1) E(1  2, 0  1, 4  (2)) 
 E(3, 1, 2) 2 1 4 4 3 6 The matrix is 2 0 1 1 1 0 . 3 4 2 1 2 0    F
  A
      E
  Chapter 8    C
 G
 H
  B
  y        D
  00 00 00 00 10. 0  (2) 3  (2) 3  (2) 0  (2) 14 24 54 44 20 20 20 20 0  (2) 0  (2) 3  (2) 3  (2) 14 44 54 24 0 0 0 0 2 2 2 2 
 2 1 1 2 2 2 1 1 5 6 9 8 5 8 9 6        The result is a translation of 1 unit along the xaxis, 2 units along the y-axis, and 2 units along the z-axis. 01 01 01 01 12. 05 35 35 05 1  (3) 2  (3) 5  (3) 4  (3) 21 21 21 21 05 05 35 35 1  (3) 4  (3) 5  (3) 2  (3) 1 1 1 1 3 3 3 3 
 5 8 8 5 5 5 8 8 2 1 2 1 2 1 2 1 z      21 21 3  (2) 3  (2) 5  (2) 2  (2) 3 3 1 1 3 0  x      D
  21 0  (2) 4  (2) 3 3 2 2 1 2  F
 x  A
 E
    C
 G
 H
  y  B
  The results is a translation of 1 unit along the xaxis, 5 units along the y-axis, and 3 units along the z-axis.  266        1 0 0 0 0 0 13. 0 1 0 0 3 3 0 0 1 1 2 5 0 0 0 0 2 
 0 3 3 0 0 1 2 5 4 1    z F
  0 0 4 2 0 4  2 0 1 2 3 5    0.75 0 0 1 0 0 0 0.75 0 
 0 1 0 0 0 0.75 0 0 1 0.75 0 0 0 0.75 0 , so the figure is three-fourths 0 0 0.75 the original site and reflected over all three coordinate planes. 2x 2 0 0 x 19a. 2y 
 0 2 0 y , so the transformation can 5z 0 0 5 z 2 0 0 be represented by the matrix 0 2 0 . 0 0 5 19b. The transformation will magnify the x- and y-dimensions two-fold, and the z-dimension 5-fold. 23.6 23.6 23.6 23.6 23.6 20a. 72 72 72 72 72 0 0 0 0 0 20  23.6 136  23.6 247  23.6 20b. 58  72 71  72 74  72 27  0 53  0 59  0 302  23.6 351  23.6 83  72 62  72 37  0 52  0 43.6 159.6 270.6 325.6 374.6 
 14 1 2 11 10 27 53 59 37 52 20c. The result is a translation 23.6 units along the x-axis and 72 units along the y-axis. 1 0 0 21. The matrix 0 1 0 would reflect the prism 0 0 1 0.5 0 0 over the yz-plane. The matrix 0 0.5 0 0 0 0.5 would reduce its dimensions by half. 0.5 0 0 1 0 0 0.5 0 0 0 0.5 0 0 1 0 
 0 0.5 0 0 0 0.5 0 0 1 0 0 0.5 22a. Placing a non zero element in the first row and third column will skew the cube so that the top is no longer directly above the bottom. 1 0 1 Sample answer: 0 1 0 0 0 1 22b. Sample graphs:  2 2 2 0 3 3 4 5 2 2 3 2    18.  C
  D
  G
  A
  H
  E
  y  The transformation does not change the figure. 1 0 0 0 0 0 0 2 2 2 2 14. 0 1 0 0 3 3 0 0 0 3 3 0 0 1 1 2 5 4 1 4 5 2 0 0 0 0 2 2 2 2 
 0 3 3 0 0 0 3 3 1 2 5 4 1 4 5 2 z6 3 3 6 9 96 3O y  A 9 3 12 6 B  x E H D      F      The transformation results in reflections over the xy  and xz-planes. 1 0 0 0 0 0 0 2 2 2 2 15. 0 1 0 0 3 3 0 0 0 3 3 0 0 1 1 2 5 4 1 4 5 2 0 0 0 0 2 2 2 2 
 0 3 3 0 0 0 3 3 1 2 5 4 1 4 5 2 z    x  H
     G
 C
  F
                      D
  The transformation results in reflections over all three coordinate planes. 16. The matrix results in a dilation of scale factor 2, so the figure is twice the original size. 3 0 0 1 0 0 3 0 0 17. 0 3 0 
 0 1 0 0 3 0 , so the 0 0 3 0 0 1 0 0 3 figure is three times the original size and reflected over the xy-plane.          y  A
        E
  B
       C         G            B
  x           z F  G    C H  x  D  z  B E  y  F
  G
  A C
  E
  B
  y  H
  x  267  D
  A
  Chapter 8  80x3  80x2  80x 
 24.2 80x3  80x2  80x  24.2 
 0 A graphing calculator indicates that there is a solution between 0 and 1. By Descartes' Rule of Signs, it is the only solution. When x 
 0.2, 80x3  80x2  80x  24.2 
 4.36 and when x 
 0.3, 80x3  80x2  80x  24.2 
 9.16. So the solution to the nearest tenth is 0.2. 30. Divide each side of the equations by 2, 3, 4, and 6, respectively, so that the left side is x  2y. I. x  2y 
 4 II. y 
 4 8 III. x  2y 
 2 IV. x  2y 
 3  23. The first transformation reflects the figure over all three coordinate planes. The second transformation stretches the dimensions along the y- and z-axes and skews it along the xy-plane. (The first row of T changes the x-coordinate of (x, y, z) to x  2z.) 24. To multiply the x-coordinate by 3, the first row of the matrix must be 3 0 0. Since the y-coordinate is multiplied by 2, the second row is 0 2 0. To convert a z-coordinate to x  4z, use a third row of 1 0 4. 3 0 0 The matrix is 0 2 0 . 1 0 4 25a. The x-coordinates are unchanged, the y-coordinates increase, and the z-coordinates decrease, so the movement is dip-slip.    25b.    123.9 88.0 205.3  41.3 145.8 246.6  29.    201.7 28.3 261.5  73.8 82.6 212.0  129.4 36.4 97.1 123.9 166.4 85.3    123.9 41.3 201.7 73.8 129.4 36.4  86.4 144.2 29.9 84.2 95.5 125.5 206.5 247.8 262.7 213.2 165.2 84.1 0 0 0 0 0 0 
 1.6 1.6 1.6 1.6 1.6 1.6 1.2 1.2 1.2 1.2 1.2 1.2     Only I and II are equivalent, so the correct choice is A.  Chapter 8 Study Guide and Assessment    Page 543    26a. La Shawna x
0 y 
 16t2  150 16t2  150 
 0 150 
 16t2 150  16  1. 3. 5. 7. 9.  Jaimie x 
 35t y 
 16t2  150  3.06  t  x 
 35t  
p   
 35 16 150  y
  4.1 cm  4.1 cm, 23°  
q 2p 
  5.3 cm 25˚  
t  5.3 cm, 25°  x1   2  5 2 48 5x  5  
   2  15.  10  3p 
  q
  28. sec 
cos1 5 
  1  2 cos cos1  5 1 
 2  5  
    q 
  3p 
  2.5 cm 98˚  5  
 2  Chapter 8  p 
  
q  
q  14.  y 
 2t  10 y
  12. 2.9 cm, 10°  23˚  107 ft 26b. Since the stones have the same parametric equations for y, they land at the same time. In part a, it was calculated that the elapsed time is about 3.06 seconds. 27. x 
 5t  1 x  1 
 5t x1  5  unit cross vector standard components  Skills and Concepts  11. 1.3 cm, 50° 13.  150    16 
 t  2. 4. 6. 8. 10.  resultant magnitude inner parallel direction  544–546  
 t2  Understanding and Using the Vocabulary  2.5 cm, 98°  268  2p 
  q
  u  5v u 31. u u 
 2w u u 
 2 4, 1, 5  5 1, 7, 4 u u 
 8, 2, 10  (5, 35, 20 u u 
 8  (5), 2  35, 10  (20) u u 
 13, 37, 30 u u 32. u 
 0.25u v  0.4w u u 
 0.25 1, 7, 4  0.4 4, 1, 5 u u 
 0.25, 1.75, 1  1.6,  0.4, 2 u u 
 0.25  1.6, 1.75  (0.4), 1  2 u u 
 1.35, 1.35, 1 33. 5, 1  2, 6 
 5(2)  (1)6 
10  6 
16; no 34. 2, 6  3, 4 
 2(3)  6(4) 
6  24 
 18; no 35. 4, 1, 2  3, 4, 4 
 4(3)  1(4)  (2)4 
 12  4  8 
 0; yes 36. 2, 1, 4  6, 2, 1 
 2(6)  (1)(2)  4(1) 
12  2  4 
18; no 37. 5, 2, 10  2, 4, 4 
 5(2)  2(4)  (10)(4) 
 10  8  40 
 42; no u u i j u k 38. 5, 2, 5  1, 0, 3 
 5 2 5 1 0 3 2 5 u 5 5u 5 2 u u  j  k 
 0 3 1 3 1 0 u  10j u  2k u or 6, 10, 2 
 6i  q 
  16. 4p 
  q
  3.5 cm  4p 
  82˚  3.5 cm, 82° 17. h 
 1.3 cos 50° v 
 1.3 sin 50° h 
 0.8 cm v 
 1 cm 18. h 
 2.9 cos 10° v 
 2.9 sin 10° h 
 2.9 cm v 
 0.5 cm u 19. CD 
 7  2, 15  3 or 5, 12 u CD  
  52  1 22 
 169  or 13 u 20. CD 
 4  (2), 12  8 or 6, 4 u CD  
  62  42 
 52  or 213  u 21. CD 
 0  2, 9  (3) or 2, 12 u CD  
  (2)2   122 
148  or 237  u 22. CD 
 5  (6), 4  4 or 1, 8 u CD  
  12  ( 8)2 
 65  23. u u 
u v u w u u 
 2, 5  3, 1 u u 
 2  3, 5  (1) or 5, 6 24. u u 
u v u w u u 
 2, 5  3, 1 u u 
 2  3, 5  (1) or 1, 4 u u  2w u 25. u 
 3v u u 
 3 2, 5  2 3, 1 u u 
 6, 15  6, 2 u u 
 6  6, 15  (2) or 12, 17 u  2w u 26. u u 
 3v u u 
 3 2, 5  2 3, 1 u u 
 6, 15  6, 2 u u 
 6  6, 15  (2) or 0, 13 u 27. EF 
 6  2, 2  (1), 1  4 or 4, 1, 3 u EF  
  42  ( 1)2  (3)2 
 26  u 28. EF 
 1  9, 5  8, 11  5 or 10, 3, 6 u EF  
  (10)2  ( 3)2   62 
 145  u 29. EF 
 2  (4), 1  (3), 7  0) or 6, 2, 7 u 2  72 EF  
  62  2 
 89  u 30. EF 
 4  3, 0  7, 5  (8) or 7, 7, 13 u EF  
  (7)2   (7 )2  1 32 
 267   6, 10, 2  5, 2, 5 6(5)  10(2)  (2)(5) 30  20  10 
 0 6, 10, 2  1, 0, 3 6(1)  10(0)  (2)(3) 6  0  6 
 0 39. 2, 3, 1  2, 3, 4 
  u i 2 2  u u j k 3 1 3 4 3 1u 2 1u 2 3 u 
 i  j  k 3 4 2 4 2 3 u  6j u  0k u or 9,  6, 0 
 9i 9, 6, 0  2, 3, 1 9(2)  (6)(3)  0(1) 18  18  0 
 0 9, 6, 0  2, 3, 4 9(2)  (6)(3)  0(4) 18  18  0 
 0  269  Chapter 8  u u i u j k 1 0 4 5 2 1 0 4 u 1 4 u 1 0 u 
 i  j  k 2 1 5 1 5 2 u u u 
 8i  19j  2k or 8, 19, 2 8, 19, 2  1, 0, 4 (8)(1)  19(0)(2)(4) 808
0 8, 19, 2  5, 2, 1 (8)(5)  19(2)  (2)(1) 40  38  2 
 0 41. 7, 2, 1  2, 5, 3 
 u i u j u k 7 2 1 2 5 3 2 1u 7 1u 7 2u 
 i  j  k 5 3 2 3 2 5  47. x  4, y  0 
 t 3, 6 x  4, y 
 t 3, 6 x  4 
 3t y 
 6t x 
 4  3t 48. x 
 t 49. x 
 t 1 5 y 
 8t  7 y 
 2t  2 
 7  8t u  
 v u cos v u  
 v u sin v 50. v v x y 
 15 cos 55° 
 15 sin 55° 8.60 ft/s 12.29 ft/s u  
 v u cos v u  
 v u sin v 51. v v x y 
 13.2 cos 66° 
 13.2 sin 66° 5.37 ft/s 12.06 ft/s u  
 v u cos v u  
 v u sin v v 52. v x y 
 18 cos 28° 
 18 sin 28° 15.89 m/s 8.45 m/s u 53. CH 
 4  3, 2  4, 2  (1) or 7, 6, 3 A(3, 4  (6), 1  3) 
 A(3, 2, 2) B(3, 4  (6), 1) 
 B(3, 2, 1) D(3, 4, 1  3) 
 D(3, 4, 2) E(3  (7), 4, 1  3) 
 E(4, 4, 2) F(3  (7), 4, 1) 
 F(4, 4, 1) G(3  (7), 4  (6), 1) 
 G(4, 2, 1) The matrix for the figure is 3 3 3 3 4 4 4 4 2 2 4 4 4 4 2 2 . 2 1 1 2 2 1 1 2 The matrix for the translated figure is 5 5 5 5 2 2 2 2 2 2 4 4 4 4 2 2 . 5 2 2 5 5 2 2 5  40. 1, 0, 4  5, 2, 1 
  u  31k u or 1, 19, 31 
u i  19j 1, 19, 31  7, 2, 1 1(7)  (19)2  31(1) 7  (38)  31 
 0 1, 19, 31  2, 5, 3 1(2)  (19)5  31(3) 2  (95)  93 
 0 42. Sample answer: Let x(1, 2, 3), y(4, 2, 1) and z(5, 3, 0) u xy 
 4  1, 2  2, 1  3 or 5, 0, 4 u yz 
 5 (4), 3  2, 0  (1) or 9, 5, 1 u u 5, 0, 4  9, 5, 1 
 u i j k 5 0 4 9 5 1 0 4 u 5 4 u 5 0u 
 i  j  k 5 1 9 1 9 5 u  31j u  25k u or 20, 31, 25 
 20i u u 43. F1 
 320i u u F 
 260j          z E
  H
 G
  A
  D
  F
  C
  y  2  u u F1  F2 
  3202  2602 412.31 N 260  B
  x  13    tan v 
  320 or 16  The figure moves 2 units along the x-axis and 3 units along the z-axis.  13  v 
 tan1 1 6 39.09° u 44. u v1 
 12j u  (30 sin 116°)j u u v2 
 (30 cos 116°)i u u v v  
  (30 co s 116° )2  (1 2  3 0 sin  116°)2 1  2  41 m/s tan v 
  12  30 sin 116°  30 cos 116° 12  30 sin 116°   v 
 tan1 
 30 cos 116°   108.65° 45. x  3, y  (5) x  3, y  5 x  3 
 4t x 
 3  4t 46. x  (1), y  9 x  1, y  9 x  1 
 7t x 
 1  7t Chapter 8  
 t 4, 2 
 t 4, 2  y  5 
 2t y 
 5  2t 
 t 7, 5 
 t 7, 5 y  9 
 5t y 
 9  5t  270       1 0 0 54. 0 1 0 0 0 1 3 3 3 2 2 4 2 1 1 3 3 3 
 2 2 4 2 1 1 z    3 4 4 4 4 4 4 4 2 2 2 2 1 1 2 3 4 4 4 4 4 4 4 2 2 2 2 1 1 2      tan v 
  Page 547  7  18  73   7  Open-Ended Assessment  1a. Sample answer: X(4, 1), Y(1, 1) u XY 
 1  4, 1  (1) or 3, 2 u 1b. XY  
  (3)2   22 or 13  u The magnitude of XY only depends on the differences of the coordinates of X and Y, not the actual coordinates. 2a. Sample answer: P(1, 1), Q(3, 3), R(3, 1), S(5, 3) u PQ 
 3  1, 3  1 or 2, 2 u RS 
 5  3, 3  1 or 2, 2 u u PQ and RS are parallel because they have the same direction. In fact, they are the same vector. 2b. Sample answer:u a 
 8, 4 ,u b 
 3, 6 u u a  b 
 8(3)  (4)6 or 0 u a andu b are perpendicular because their inner product is 0.  G y  x B  The figure is reflected over the xz-plane.  Page 547  or  v 
 tan1 
  18  73   H  F A O  C  35  90  353   13.1°  E D  u 58. F1 
 90i u u u F2 
 (70 cos 30°) i  (70 sin 30°)j or u u 353  i  35j u u F1  F2 
  (90  353  )2  3 52 154.6 N  Applications and Problem Solving  1 u 3  55. AB 
 1 cos 120°, 0, 1 sin 120° or 2, 0, 2 u F 
 0, 0, 50 u u u T 
 AB  F 1 ft u u u i j k 50 lb 1 3  60˚ 
 2 0 2  0 0 50 3   3   1  1   u    u 2 j  2 0u 2 i  2 
 0 k 0 50 0 0 0 50 u  25j u  0k u or 0, 25, 0 
 0i u 02  ( 25)2  02 T  
  
 25 lb-ft u sin v  1gt2  h 56. y 
 tv 2  Chapter 8 SAT & ACT Preparation Page 549  1  
 0.5(38) sin 40°  2(32)(0.5)2  2 10.2 ft 57a.  16 km/h  3 km/h 35˚  250 m  u 
 (16 cos 55°)c u u  (16 sin 55°)j b u u c 
 3j u u 2  (1 b c  
  (16 co s 55°) 6 sin  55°  3)2 13.7 km/h 57b.  u  250  16 sin 55°  3   
 16 cos 55°  1  A 
 2bh  16 sin 55°  3  16 cos 55°  u 
 250 
 u  SAT and ACT Practice  1. Recall that the formula for the area of a parallelogram is base times height. You know the base is 5, but you don't know the height. Don't be fooled by the segment BD; it is not the height of the parallelogram. Try another method to find the area. The parallelogram is made up of two triangles. Find the area of each triangle. Since ABCD is a parallelogram, AB 
 DC and AD 
 BC. The two triangles are both right triangles, and they share a common side, BD. By SAS, the two triangles are congruent. So you can find the area of one triangle and multiply by 2. The hypotenuse of the triangle is 5 and one side is 3. Use the Pythagorean Theorem to find the other side. 52 
 32  b2 25 
 9  b2 16 
 b2 4
b The height is 4. Use the formula for the area of a triangle.    1  A 
 2(4)(3) or 6  275.3 m  Since the parallelogram consists of two triangles, the area of the parallelogram is 2  6 or 12. The correct choice is A.  271  Chapter 8  6. This figure looks more complex than it is. A semicircle is just one half of a circle. Notice that the answer choices include , so don't convert to decimals. Find the radius of each semi-circle. Calculate the area of each semi-circle. The area of the shaded region is the area of the large semi-circle minus the area of the medium semi-circle plus the area of the small semi-circle.  2. In order to write the equation of a circle, you need to know the coordinates of the center and the length of the radius. The general equation for a circle is (x  h)2  (y  k)2 
 r2, where the center is (h, k) and the radius is r. From the coordinates of points A and B, you know the length of the side is 4. So the center Q, has coordinates (0, 4). To calculate the length of the radius, draw the radius OB. This creates a 45°-45°-90° right triangle. The two legs each have length 2. The hypotenuse has length 22 . )2 (x  4)2  (y  0)2 
 (22 (x  4)2  y2 
 4(2) (x  4)2  y2 
 8 The correct choice is B. 3. Write the equation for the perimeter of a rectangle. then replace x with its value in terms of y. Solve the equation for y. p 
 2x  2y  9  1  Large semi-circle area 
 2 32 
 2 4  1  Medium semi-circle area 
 2 22 
 2 1  1  Small semi-circle area 
 2 12 
 2 9  4  1  6  Shaded area 
 2  2  2 
 2 
 3 The correct choice is A. 7. The only values for which a rational function is undefined are values which make the denominator 0. Since f(x) 
  x2 –3x  2  , x–1  the denominator is only 0 when x – 1 
 0 or x 
 1.  p 
 2
3 y  2y 2  The correct choice is D. 8. Start by sketching a diagram of the counter  4  p 
 3y  2y 10  p 
 3y 3p  10  
y  28  38  30  The correct choice is B. 4. Recall the triangle Inequality Theorem: the sum of the lengths of any two sides of a triangle is greater than the length of the third side. Let x represent the length of the third side. 40  80  x 120  x 40  x  80 x  40 Since x must be greater than 40, x cannot be equal to 40. The correct choice is A. To check your answer, notice that the other answer choices are greater than 40 and less than 120, so they are all possible values for x. 5. Since the answer choices have fractional exponents of x, start by rewriting the expression with fractional exponents. Simplify the fractions and use the rules for exponents to combine terms. 2 3 3 9    x2   x3 
 x 3  x 9 2 1   
 x3  x3 2 1   
x( 3  3 ) 
x1 or x The correct choice is E.  40  Use your calculator to find the area of the whole counter and then subtract the area of the white tiles in the center. The white tiles cover an area of (30  2)(40  2) or (28)(38). (30)(40) 
 1200 (28)(38) 
 1064 Red tiles 
 1200  1064 
 136 The correct choice is B. 9. First, find the slope of the line containing the points (–2, 6) and (4, –3). m
  –3 – 6  4 – –2 3  –9  m 
 6 or – 2 The point-slope form of the line is 3  y – 6 
 – 2(x – –2). 3  y – 6 
 – 2x – 3 3  y 
 – 2x  3 So the y-intercept of the line is 3. The correct choice is B.  Chapter 8  272  10. Write an expression for the sum of the areas of the two triangles. Recall the area of a triangle is one half the base times the height. 1 (AC)(AB) 2  1 (AC)(AB) 2  1  1  1   2(CE)(ED) 
 2(AC)2  2(CE)2 1  
 2[(AC)2  (CE)2]  1  Using the Pythagorean Theorem for ACE, you know that (AC)2  (CE)2 
 (AE)2 or 1.   2(CE)(ED)  From the figure, you know that ABC and CDE are both isosceles, because of the angles marked x° and because B C D  is a line segment. These two triangles have equal corresponding angles. Since they are isosceles triangles, AC 
 AB and CE 
 ED. Use these equivalent lengths in the expressions for the area sum.  1  1  So the sum of the two areas is 2(1) 
 2. You can grid the answer either as .5 or as 12.  273  Chapter 8  Chapter 9 Polar Coordinates and Complex Numbers 11.  Polar Coordinates  9-1  90˚  120˚  Check for Understanding  180˚  1. There are infinitely many ways to represent the angle v. Also, r can be positive or negative. 2. Draw the angle v in standard position. Extend the terminal side of the angle in the opposite direction. Locate the point that is r units from the pole along this extension. 3. Sample answer: 
60° and 300° Plot (4, 120) such that v is in standard position and r is 4 units from the pole. Extend the terminal side of the angle in the opposite direction. Locate the point that is 4 units from the pole along this extension. r  
4 v  120 
 180 or v  120  180  
60  300 4. The points 3 units from the origin in the opposite direction are on the circle where r  3. 5. All ordered pairs of the form (r, v) where r  0. 6.  90˚  120˚ 150˚  60˚  330˚  210˚ 240˚  8.  120˚  270˚ 90˚  C  180˚  1 2 3 4  240˚  270˚  19  6    13  7  , 
2,    13    25  → 2,    6  Chapter 9    5.25  15 cos 
12  5  90˚  120˚   (1) → 2,  (3) → 2,  19  6  60˚ 30˚  5 10 15 20  0˚  330˚ 300˚  270˚  15b. 210 
 (
30)  240 N  2  A 360 (r ) 240  2   360 (20 )   838 ft2  11 6 5 3  Pages 558–560  7  6  , 2, ,  16.  120˚  Exercises 90˚  0˚  330˚ 240˚  274  30˚  1 2 3 4  210˚     17.  60˚  E  180˚  (
r, v  (2k  1)) 7  6    5  6.25  9  15 cos 
12  180˚  →
2, 6  2(2)) → 
2, 6 → 2,    2.52 (
3)2 
 2(2.5)(
3) cos 
4 
 6  150˚  →
2, 6  2(1) → 
2, 6    6  300˚  270˚  150˚  (r, v  2k)  0˚    15a.  0  25  6  5 3  3 2   4.37   3  3 2  11 6  7 6  330˚  14. P1P2   1 2 3 4  4 3  1 2 3 4  60˚  1 2 3 4  240˚  7 6  0    30˚  240˚   6  D   3  6  4 3  210˚  0˚   90˚  120˚  210˚  5 3  5 6  300˚  10. Sample answer: 
2,   2  270˚  180˚  11 6  3 2  2 3  330˚  210˚  2,  9. 30˚  150˚  B  0˚  300˚  150˚  1 2 3 4  4 3  60˚  13.  0  7 6  300˚  240˚   2  5 6  330˚  210˚   6    1 2 3 4   3  5 6  0˚  1 2 3 4   2  2 3  30˚  A  180˚  7.  2 3  30˚  150˚  Pages 557–558  12.  60˚  270˚  300˚  2 3   2   3  6  5 6  F 0    1 2 3 4 11 6  7 6 4 3  3 2  5 3  18.  90˚  120˚  19.  60˚  180˚  0˚  2 4 6 8   6  H  0    1 2 3 4  G 330˚  210˚ 240˚  20.  270˚  2  2 3  7 6  300˚  21. 150˚  90˚  J 4 3  22.  3 2  2  2 3  23.   3  90˚  L 4 3  24.  90˚  120˚  240˚  5 3  3 2  25.  60˚  180˚  330˚  210˚ 240˚  26.   2  2 3  27.  0    180˚  1 2 3 4  Q  7 6 4 3  3 2  30˚ 1 2 3 4    270˚    3  7  3   2(1) → 2,   3  6 0    1 2 3 4 11 6  7 6  35.  6   2  5 6  4 3   3  120˚  5 3  3 2 90˚  60˚ 30˚  150˚ 180˚  0 1 2 3 4  1 2 3 4  4 3  300˚  36.  120˚  90˚  30˚  180˚  1 2 3 4  240˚  275  0˚  330˚  210˚  (
r, v  (2k  1)180°) → (
2, 60°  (1)180°) → (
2, 240°) → (
2, 60°  (3)180°) → (
2, 600°)  37.  60˚  150˚    240˚  5 3  3 2  270˚  300˚  0˚  330˚  210˚  11 6  7 6  → 2, 3  2(0) → 2, 3  → 2,   2  2 3  0˚  7    300˚  270˚    28. Sample answer: 2, 3, 2, 3, (
2, 240°), (
2, 600°) (r, v  2k)   0˚  330˚  2 3  330˚ 240˚  5 3  1 2 3 4  5 6  60˚  210˚  11 6  30˚  240˚  34.  33.  60˚  210˚  5 3  150˚ R  90˚  180˚  1 2 3 4  90˚  10  150˚   3  3 2    120˚  11 6  120˚   6  5 6  32.  0  P  4 3   3  300˚  7 6  300˚  270˚  0˚   6    N  4  31. Sample answer: (4, 675°), (4, 1035°), (
4, 135°), (
4, 495°) (r, v  360k°) → (4, 315  360(1)°) → (4, 675°) → (4, 315  360(2)°) → (4, 1035°) (
r, v  (2k  1)180°) → (
4, 315  (
1)180°) → (
4, 135°) → (
4, 315  (1)180°) → (
4, 495°)  330˚  5 6  0˚  1 2 3 4   2  2 3  30˚  150˚  270˚    → 1, 3  (1) → 1, 3  60˚  M  210˚  11 6  7 6  13  → 1, 3  (3) → 1, 3  1 2 3 4  1 2 3 4    (
r, v  (2k  1))  30˚  180˚  7  → 
1, 3  2(2) → 
1, 3  300˚  150˚ 0    270˚    → 
1, 3  2(1) → 
1, 3  330˚  120˚   6  5 6  (r, v  2k)  0˚  1 2 3 4  240˚  5 3  10  1,  3   60˚  210˚  11 6  7  13 4  , 1, , 30. Sample answer: 
1, 3, 
1,  3   3  30˚  1 2 3 4  7 6  5 3  K  180˚  0    3 2  120˚   6  5 6  11 6 4 3   3  29. Sample answer: (1.5, 540°), (1.5, 900°), (
1.5, 0°), (
1.5, 360°) (r, v  360k°) → (1.5, 180°  360(1)°) → (1.5, 540°) → (1.5, 180°  360(2)°) → (1.5, 900°) (
r, v  (2k  1)180°) → (
1.5, 180°  (
1)180°) → (
1.5, 0°) → (
1.5, 180°  (1)180°) → (
1.5, 360°)   3  5 6  30˚  150˚   2  2 3  2 3  270˚  2  300˚   3  6  5 6  0    1 2 3 4  7 6  11 6 4 3  3 2  5 3  Chapter 9  38.  90˚  120˚  39.  60˚  30˚  150˚ 180˚  1 2 3 4  0˚  150˚  240˚  40.  90˚  120˚  30˚ 1 2 3 4  0˚  50a.  330˚  210˚ 240˚  300˚  270˚  49a. When v  120°, r  17. The maximum speed at 120° is 17 knots. 49b. When v  150°, r  13. The maximum speed at 150° is 13 knots.  60˚  r
0  180˚  330˚  210˚  90˚  120˚  30˚  180˚  1 2 3 4  240˚  270˚  0˚  11 6 4 3  50b.  300˚  1  25 
 10 cos 12    26 
 10 cos 12  2  N  N  120  120  2 2    360 ((300 ) 
 360 ((25) ) 120    360  (90,000 
 625)   93,593 ft2 If each person's seat requires 6 ft2 of space,    7      
 
3  3 or 120°  2 2   A 360 (R ) 
 360 (r )  12  52 
 2(1)(5) cos 4 
 6 3    3  5 3  3 2  Let R  3 100 or 300 and let r  0.25 100 or 25.  41. r  
2  or r  

2  for any v. 2 
 2( 42  6 4)(6) c os (10 5° 
 170°)  42. P1P2  
  
16 36  48 
 cos°) (
65  
52 48 
 cos) (
65  5.63 43. P1P2   0 1 2 3 4  7 6  330˚  210˚   6    60˚  150˚   3  5 6  300˚  270˚   2  2 3  93,593  there are 6 or 15,599 seats.  7  51. The distance formula is symmetric with respect to (r1, v1) and (r2, v2). That is,   5.35  
 r22   r12 
  2r2r1  cos (v1 
 v2)  44. P1P2  (
2.5)2  (
1.75)2 
 2(
2.5)(
1.75)cos   21  6.25  30.0625 
 8.75 cos 
 40    21  9.3125 
 8.75 cos 
 40      2  
5 
 8     
 r12   r22 
  2r1r2 cos[
  (v2 
  v1)]     
 r12   r22 
  2r1r2  cos(v2 
 v1)    52a.    120˚  90˚  60˚ 30˚  150˚   3.16 180˚  45. P1P2  
 1.32  (
3.6 )2 
 2(1.3)(
  3.6) co s (
62 ° 
 (
47°))   
1.69   12.966  9.3(
62° cos °)  47  
14.65 6  9.3(
15° cos )  4.87  46. r   
 (
3)2   42  240˚  5  180° 
 53°  127° Sample answer: (5, 127°) 47. There are 360° in a circle. If the circle is cut into 6 360  equal pieces, each slice measures 6 or 60°. Beginning at the origin, the equation of the first line is v  0°. The equation of the next line, rotating counterclockwise, is v  0  60 or 60°. The equation of the last line is v  60  60 or 120°. Note that lines extend through the origin, so 3 lines create 6 pieces.     3 mph  54. 3, 
2, 4  1, 
4, 0  (3)(1)  (
2)(
4)  (4)(0) 380  11 No, the vectors are not perpendicular because their inner product is not 0.   r1 
 r2 Chapter 9  300˚   8 mph  48. P1P2  
 r12   r22 
  2r1r2  cos (v  
 v)  
 r12   r22 
  2r1r2 2 
 (r1
 r 2)  270˚  52b. P1P2 
 52  62 
 2( 5)(6) c os (34 5° 
 310°)   
25 36  60 
 cos (35°)  
61 60 
 cos (35°)  3.44 No; the planes are 3.44 miles apart. 53. Draw a picture. sin v  38 Boat 1 sin (sin v)  sin138 v  22.0°  4    0˚  330˚  210˚  sin v  5, v  53°  
 r12 
  r22 
  2r1r2  cos 0  2 4 6 8  276  55. Rewrite y  9x 
 3 as 9x 
 y 
 3  0. d     
2 4 
1 4 0 1 
1 0 
4  (
1) 1 
1 0  
2 
1 
3 5 
3 4 5 4 
3 4 5  
2(
5) 
 4(5) 
 1(1)  
11 64. 11 
 (
3)  14 11 
 (
2)  13 11 
 (
1)  12 11 
 0  11 {(
3, 14), (
2, 13), (
1, 12), (0, 11)} For each x-value, there ia a unique v-value. Yes, the relation is a function. 65. Since the two triangles formed are right triangles, the side opposite the right angles, A B , intercept an arc measuring 180°, or half the circle.  AB  is a diameter. C  d 50  d 50  d The correct choice is E.  Ax1  By1  C  
 A2   B2 9(
3)  (
1)(2)  (
3)  
 92  ( 
1)2 
32  
82  
32 
82    
82  
82   63.  
32
82      82  56.   16
82  41  1 
 sin2a  sin2a    Distance is always positive. 1  
 1 sin2a csc2a 
 1    cot2a 
3   57. Arc cos  2  30° 1 In a 30°-60°-90° right triangle, the angle opposite the smallest leg is 30°. 58. y  5 cos 4v Amplitude  5; Period  24 or 2  2   3 
  59. b sin A  18.6 sin 30°  9.3 Since a  b sin A, there is one solution. Find B. Find C. 18.6  sin B  9.3   sin 30°  18.6 sin 30°  9.3   60°   sin B  90°  B Find c. c  sin 60°  Graphs of Polar Equations  Page 565  Check for Understanding  1. Sample answer: r  sin 2v The graph of a polar equation whose form is r  a cos nv or a sin nv, where n is a positive integer, is a rose. 2. 
1  sin v  1 for any value of v. Therefore, the maximum value of r  3  5 sin v is r  3  5(1) or 8. Likewise, the minimum value of r  3  5 sin v is r  3  5(
1) or 
2. 3. The polar coordinates of a point are not unique. A point of intersection may have one representation that satisfies one equation in a system, another representation that satisfies the other equation, but no representation that satisfies both simultaneously. 4. Barbara is correct. The interval 0  v   is not always sufficient. For example, the interval 0  v   only generates two of the four petals for the rose r  sin 2v. r  sin 2v is an example where values of v from 0 to 4 would have to be considered.  C  180° 
 90° 
 30°  18.6 sin 30°  9.3 sin B  9-2  9.3   sin 30°  c sin 30°  9.3 sin 60° 9.3 sin 60°  c sin 30°  c  16.1 60. 3 or 1 positive f(
x)  
x3 
 4x2 
 4x 
 1 0 negative P 1 Q  Since there are only positive real zeros, the only rational real zero is 1. 61. x
 3 x  5x2  2x 
 3 x2 5 x  
3x 
 3 
3x 
 15  10 10  → 0. Therefore, the slant As x →  ,  x5 asymptote is y  x 
 3. 62. y-axis: For x: f(x)  x4 
 3x2  2 For 
x: f(
x)  (
x)4 
 3(
x)2  2  x4 
 3x2  2 So, in general, point (
x, y) is on the graph if and only if (x, y) is on the graph.  5.  90˚  120˚  30˚  150˚ 180˚  1  2  0˚  330˚  210˚ 240˚  270˚  cardioid  277  6.  60˚  300˚  120˚  90˚  60˚ 30˚  150˚ 180˚  2 4 6 8  0˚  330˚  210˚ 240˚  270˚  300˚  limaçon  Chapter 9  7.  2 3   2  8.   3  6  5 6   11 6 4 3  11.  3 2   2  5 3  0  4 3  3 2  If 
   or 
   5  6  equation, r  1. If 
   3  2  2 3  4 3  3 2  330˚  120˚  270˚  300˚  90˚  60˚ 30˚  150˚ 1  0˚  2  330˚  210˚  300˚  240˚  rose 17.  270˚  300˚  rose 2 3   2  0  4 8 12 16  4 3  3 2  0  19.  2 3    4 3  cardioid  278  3 2  5 3  0˚  330˚ 270˚  300˚  2 3   2   3  6  5 6 0  11 6  7 6  4 6 8  limaçon 20.  1 2 3 4  2  240˚   6  5 6  60˚ 30˚  180˚  5 3   3  90˚  210˚  Spiral of Archimedes  2  120˚ 150˚  11 6  7 6  5 3  18.   3  6    14  10b. Sample answer: 0  v   3 Begin at the origin and "spiral" twice around it, or through 4 radians. Move straight up through 4  2 or 92 radians. Now move to the left slightly, through approximately 92  6 or 14  radians. 3  Chapter 9  0˚  is substituted in either  11 6  7 6  270˚  60˚  1 2 3 4  0˚ 180˚  2  330˚  5 6  4 8 12 16  16.  1  90˚  30˚  240˚  30˚  240˚   6    300˚  lemniscate  60˚  210˚   3  5 6  90˚  180˚  1, 6, 1, 56, and 
2, 32. 10a.  270˚  210˚  5 3  150˚  original equation, r  
2. The solutions are  2  3 2  120˚  180˚  0  11 6  120˚  is substituted in either original  330˚  150˚  spiral of Archimedes  
  6 or 
  56 or 
  32  0˚  2 4 6 8  210˚  14.   3  5 10 15 20  15.  30˚  240˚   6  4 3  2 sin 
  2 cos 2
 sin 
  cos 2
 sin 
  1 
 2 sin2 
 2 sin2 
  sin 
 
 1  0 (2 sin 
 
 1)(sin 
  1)  0 2 sin 
 
 1  0 or sin 
  1  0 sin 
  12 sin 
  
1   6  2 3   2  7 6  5 3  60˚  cardioid    11 6  90˚  150˚  300˚  5 6  2  7 6  270˚  circle 13.  1  330˚ 240˚  120˚  0˚ 180˚  1 2 3 4  210˚   6    12.  60˚  30˚  180˚   3  5 6  90˚  150˚  spiral of Archimedes 2 3  Exercises  120˚  0  11 6 4 3  rose 9.  3 6 9 12  7 6  5 3  3 2   6    0  Pages 565–567   3  5 6  1  7 6   2  2 3    1 2 3 4  0  11 6  7 6 4 3  lemniscate  3 2  5 3  21.  90˚  120˚  22.  60˚  120˚  30˚  150˚ 180˚  330˚ 240˚  30˚ 0˚  1 2 3 4  330˚  210˚  300˚  270˚  28. (1, 0.5), (1, 1.0), (1, 2.1), (1, 2.6), (1, 3.7), (1, 4.2), (1, 5.2), (1, 5.8)  60˚  150˚  0˚ 180˚  1  210˚  90˚  240˚  270˚  300˚  rose cardioid 23. Sample answer: r  sin 3v The graph of a polar equation of the form r  a cos 3v or r  a sin 3v is a rose with 3 petals. 24. Sample answer: r  2v   4   a2  r  12
  1  2  a  r  2  
  25. 3  2  cos 
 1  cos 
 
0 The solution is (3, 0)  2 3   2   3  6  5 6   1 2 3 4  4 3  26. 1  cos 
  1 
 cos 
 2 cos 
  0 cos 
  0  2 3  Substituting each angle into either of the original equations gives r  1, so the solutions of the system are 1, 27.  2 3   2   3  6    1  2   and 1, .  0  [
4, 4] scl1 by [
4, 4] scl1  11 6  7 6  3  2   2  5 3  3 2  5 6  
  2 or 
  32  [
6, 6] scl:1 by [
6, 6] scl1 30. (3.6, 0.6), (2.0, 4.7)  0  11 6  7 6    2  [
2, 2] scl1 by [
2, 2] scl1 29. (2, 3.5), (2, 5.9)  4 3  3 2  31a. If the lemniscate is 6 units from end to end, then a  12(6) or 3.  5 3  r2  9 cos 2
 or r2  9 sin 2
 31b. If the lemniscate is 8 units from end to end, then a  12(8) or 4.   3  r2  16 cos 2
 or r2  16 sin 2
   6  5 6  32.   1  2  4 3  3 2  90˚  60˚  0 30˚  150˚  11 6  7 6  120˚  180˚  5 3  2 4 6 8  330˚  210˚  2 sin 
  2 sin 2
 sin 
  sin 2
 sin 
  2 cos 
 sin 
 0  2 cos 
 sin 
 
 sin 
 0  sin 
 (2 cos 
 
 1) sin 
  0 or 2 cos 
 
 1  0  0˚  240˚  270˚  300˚  This microphone will pick up more sounds from behind than the cardioid microphone. 33. 0  v  4: Begin at the origin and curl around once, or through 2 radians. Curl around a second time and go through 2  2 or 4 radians. 34. All screens are [
1, 1] scl1 by [
1, 1] scl1  cos 
  12 
  0 or  or 
  3 or 
  53 If 
  0 or 
   is substituted in either original equation, r  0. If 
  3 or 
  53 is substituted in either original equation, r  
3  or r  

3 , respectively. The solutions are (0, 0), (0, ),  
3, 3, and 

3, 53.  279  Chapter 9  
  
  34a. r  cos 2  r  cos 7  
  
  r  cos 4  r  cos 9  
  When n  11, the innermost loop will be on the left and there will be an additional outer ring. 35. Sample answer: r  
1 
 sin v A heart resembles the shape of a cardioid. The sine function orients the heart so that the axis of symmetry is along the y-axis. If a  
1, the heart points in the right direction. 36a. For a limaçon to go back on itself and have an inner loop, r must change sign. This will happen if b a. 36b. For the other two cases, a  b. Experimentation shows that the dimple disappears when a  2b, so there is a dimple if b  a  2b. 36c. For this remaining case, there is neither an inner loop nor a dimple if a  2b. 37a. Subtracting a from v rotates the graph counterclockwise by an angle of a. 37b. Multiplying v by 
1 reflects the graph about the polar axis or x-axis. 37c. Multiplying the function by 
1 changes r to its opposite, so the graph is reflected about the origin. 37d. Multiplying the function by c results in a dilation by a factor of c. (Points on the graph move closer to the origin if c  1, or farther away from the origin if c 1.) 38. Sample answer: (4, 405°), (4, 765°), (
4, 
135°), (
4, 225°) (r, 
  360k°) → (4, 45°  360(1)°) → (4, 405°) → (4, 45°  360(2)°) → (4, 765°) (r, 
  (2k  1)180°) → (
4, 45°  (
1)180°) → (
4, 
135°) → (
4, 45°  (1)180°) → (
4, 225°)  r  cos 6  
  r  cos 8  When n  10, two more outer rings will appear. 
  34b. r  cos 3  
  r  cos 5  Chapter 9  280  i j k 2 3 0 
1 2 4 3 0  2 0  2 3   i
 j k 2 4 
1 4 
1 2     12i 
 8j  7k  12, 
8, 7 2, 3, 0 12, 
8, 7  24  (
24)  0 or 0 
1, 2, 4 12, 
8, 7  
12  (
16)  28 or 0 40. 3.5 cm, 87°   39. v  x2 r cos 
  2 2  r cos 
 r  2 sec 
 4. To convert from polar coordinates to rectangular coordinates, substitute r and v into the equations x  r cos v and y  r sin v. To convert from rectangular coordinates to polar coordinates, use the equation r  
 x2  y2 to find r. If x 0, v  Arctan yx. If x  0, v  Arctan yx  . If x  0, you can use 2 or any coterminal angle for v. 3.    w  2  sin x 2 41.  cos4 x  cos2 x sin2 x  tan x sin2 x  cos2 x (cos2 x  sin2 x) sin2 x  (cos2 x)(1) sin2 x  cos2 x   tan2 x   tan2  y  x   tan2 x  r  tan2 x  tan2 x   42. Find C. C  180° 
 21°15
 
 49°40
  109°5
 Find b. b  sin 49°40
  O  28.9   sin 109°5
   
4  or 2  28.9 sin 49°40
  sin 109°5
  2, 34  b  23.3  6. r  
 (
2)2   (
5 )2  Find a. a  sin 21°15
  28.9   sin 109°5
  a sin 109°5
  28.9 sin 21°15
 28.9 sin 21°15
   7.  a  sin 109°5
 a  11.1 NY LA Miami $240 $199 $260 $254 $322 $426  43. Bus Train  8.  12 1 3 44. 1  6  1    8  4  8  1   6   8  9.  8  13   4 8 1 3  16 8 So      3 8 3 3   16  16  10.  2 6  3  The correct choice is A.  9-3 Page 571  11.  Polar and Rectangular Coordinates  12.  Check for Understanding  x   34  
 2 

2   
5   v  Arctan  
2     5.39  4.33  
29 (5.39, 4.33) x  
2 cos 43 y  
2 sin 43 1  
3  (1, 
3 ) x  2.5 cos 250° y  2.5 sin 250°  
0.86  
2.35 (
0.86, 
2.35) y2 r sin v  2 2  r sin v r  2 csc v x2  y2  16 (r cos v)2  (r sin v)2  16 r2(cos2 v  sin2 v)  16 r2  16 r  4 or r  
4 r6 
 x2  y2  6 x2  y2  36 r  
sec v r  r  1. Sample answer: (2
2 , 45°) 22  22 r  
  x 
 r cos   5. r  
 (

2  )2  (
 )2 v  Arctan  2  b sin 109°5
  28.9 sin 49°40
 b  y 
 r sin   
1    r cos v  
1  1 x  
  Arctan 22  x  
1   
8   45°  2
2  2. The quadrant that the point lies in determines y y whether v is given by Arctan x or Arctan x  .  281  Chapter 9  13a.  90˚  120˚  22. x  
1 cos 
6  60˚  
3    
1  2   30˚  150˚ 180˚  
3   1 2 3 4  

23 , 12  330˚  210˚ 240˚  270˚  23. x  
2 cos 270° 0 (0, 2) 24. x  4 cos 210°  300˚  13b. No. The given point is on the negative x-axis, directly behind the microphone. The polar pattern indicates that the microphone does not pick up any sound from this direction.  
3  
2   4  25.  Pages 572–573 Exercises 
2 2 2  14. r  
2  ( 
2) v  Arctan 2  26.   
4   
8  or 2
2  Add 2 to obtain v  74.  2
2, 74  15. r  
 02  12  27.   
1  or 1 Since x  0 when y  1, v  2.  1, 2  16. r  
 12  ( 
3 )2  v  Arctan    
4  or 2   3  2, 3    1 2 
4  17. r   28.  
3   1      
3  2 
 4   14 6  24 or 12  12, 43  18. r  
 32  82  
73   8.54 (8.54, 1.21) 42  ( 
7)2 19. r  
  v  Arctan  
3   
 4  
14     29.  
3  Arctan  or 43 1  30. v  Arctan 83  1.21 
7  v  Arctan  4   31.   
65   8.06  
1.05 Add 2 to obtain v  5.23. (8.06, 5.23) 20. x  3 cos 2 0 (0, 3)  y  3 sin 2 3 32.  21. x  12 cos 34  y  12 sin 34   12 
2   12 
2   
4   
2   4  
 2  
2   

42 , 
42   Chapter 9  
 2  33.  282   
1
12  12   
2  0˚  y  
1 sin 
6    y  
2 sin 270° 2 y  4 sin 210°  4
12   
2
3   
2 (
2
3 , 
2) x  14 cos 130° y  14 sin 130°  
9.00  10.72 (
9.00, 10.72) x  
7 r cos v  
7 
7 r   cos v r  
7 sec v y5 r sin v  5 5 r   sin v r  5 csc v x2  y2  25 (r cos v)2  (r sin v)2  25 r2(cos2 v  sin2 v)  25 r2  25 r  5 or r  
5 x2  y2  2y (r cos v)2  (r sin v)2  2r sin v r2 (cos2 v  sin2 v)  2r sin v r2  2r sin v r  2 sin v x2 
 y2  1 (r cos v)2 
 (r sin v)2  1 r2 (cos2 v 
 sin2 v)  1 r2 (cos 2v)  1 1  r2   cos 2v 2 r  sec 2v x2  (y 
 2)2  4 x2  y2 
 4y  4  4 (r cos v)2  (r sin v)2 
 4r sin v  0 r2(cos2 v  sin2 v) 
 4r sin v  0 r2 
 4r sin v  0 r2  4r sin v r  4 sin v r2 
 x2  y2  2 x2  y 2  4 r  
3 2 
 x  y2  
3 x2  y2  9  v  3  34.  y 43. horizontal distance: 25(4  2 cos 120°)  75 m east vertical distance: 25(3  2 sin 120°)  118.30 m north  y  y Arctan x  3 y  
3     x 1  2 6 3  y  
3 x  1  
 3  2  1  2y y2 r  3 cos v r2  3r cos v x2  y2  3x 37. r2sin 2v  8 r22 sin v cos v  8 2r sin v r cos v  8 2yx  8 xy  4 38. yx 36.  Arctan  6.07  5.47 6.07  5.47  5.47  cos 47.98°   tan v   Arctan 1  r  sin v r2  r sin v x2  y2  y 40. x  325 cos 70°  111.16 (111.16, 305.40)  r  8.17  r 8.17∠47.98° 44d. 8.17 sin (3.14t  47.98°) 45. r  2a sin v  2a cos v r2  2ar sin v  ar cos v x2  y2  2ay  2ax 2 2 x 
 2ax  y 
 2ay  0 (x 
 a)2  (y 
 a)2  2a2 The graph of the equation is the circle centered at (a, a) with radius 
2 a.  1  39.    4  r sin v    r cos v  47.98  v; 47.98° 5.47  r cos 47.98°  v  4  5  4  y  325 sin 70°  305.40  46.  120˚  90˚  60˚ 30˚  150˚  — 41. — 6 
 6    5  24 4  24  
    24  180˚  1 2 3 4  0˚  330˚  210˚   0.52 unit 42. Drop a perpendicular from the point with polar coordinates (r, v) to the x-axis. r is the length of the hypotenuse in the resulting right triangle. x is the length of the side adjacent to angle v, so cos v  xr. Solving for x gives x  r cos v. y is the y length of the side opposite angle v, so sin v  r. Solving for y gives y  r sin v. (The figure is drawn for a point in the first quadrant, but the signs work out correctly regardless of where in the plane the point is located.)  240˚  270˚  300˚  47. Sample answer: (
2, 405°), (
2, 765°), (2, 225°), (2, 585°) (r, v  360k°) → (
2, 45°  360(1)°) → (
2, 405°) → (
2, 45°  360(2)°) → (
2, 765°) (
r, v  (2k  1)180°) → (2, 45°  (1)180°) → (2, 225°) → (2, 45°  (3)180°) → (2, 585°) 48. r2  502  4252  2 50 425 cos 30° r  382.52 mph  y (r, )  50  sin v  r  382.52   sin 30°  50 sin 30°  382.52 sin v  r  50 sin 30°  382.52    O  x  44a. x  4 cos 20° y  4 sin 20°  3.76  1.37 3.76, 1.37 x  5 cos 70° y  5 sin 70°  1.71  4.70 1.71, 4.70 44b. 3.76, 1.37  1.71, 4.70  3.76  1.71, 1.37  4.70  5.47, 6.07 44c. 5.47  r cos v; 6.07  r sin v    r sin v  y  x y  x  120˚  O  x  35. r  2 csc v r  r  2   sin v    30˚ 425 mph 50 mph  3°45
  v The direction is 3°45
 west of south.  x  283  Chapter 9  3. The graph of the equation x  k is a vertical line. Since the line is vertical, the x-axis is the normal line through the origin. Therefore, f  0° or f  180°, depending on whether k is positive or negative, respectively. The origin is k units from the given vertical line, so p  k. The polar form of the given line is k  r cos (v 
 0°) if k is positive or 
k  r cos (v 
 180°) if k is negative. Both equations simplify to k  r cos v. 4. You can use the extra ordered pairs as a check on your work. If all the ordered pairs you plot are not collinear, then you have made a mistake.  sin2 A  cos A 
 1 1 
 cos2 A  cos A 
 1 0  cos2 A  cos A 
 2 0  (cos A  2)(cos A 
 1) cos A  2  0 or cos A 
 1  0 cos A  
2 cos A  1 A  0° y 50.  49.  2  y 
 2 cos   1  O 
1  90˚ 180˚ 270˚ 360˚ 
  5. 
 A2   B2  
 32  ( 
4)2  5 Since C is negative, use 5.  
2  4  f  Arctan 
43  x  3,  1 2 2  (   
53° or 307° p  r cos (v 
 f) 2  r cos (v 
 307°)  )  6. 
 A2   B2  
 (
2)2   42  2
5  Since C is negative, use 2
5 .  52. Enter the x-values in L1 and the f(x)-values in L2 of your graphing calculator. Make a scatter plot. The data points are in the shape of parabola. Perform a quadratic regression. a  
0.07, b  0.73, c  
1.36 Sample answer: y  
0.07x2  0.73x 
 1.36 53. 2 1 0 0 
3 0 
20 2 4 8 1 0 2 0  1 2 4 5 10  0 x4  2x3  4x2  5x  10 625 
 145  54. m   25 
 17   60 55. x y and y If x  2  4  9   x   y 
   0 
 2
 2
 2
 5 5 5   9
5  2
5  
5  , sin f  , p   cos f  
 10 5 5  f  Arctan(
2)  
63° Since cos f  0, but sin f the second quadrant. f  180° 
 63° or 117° p  r cos (v 
 f)  (y 
 145)  60(x 
 17)   9
5  10  y  60x 
 875 z, so x  
 45y 
 2  0  cos f  35, sin f  
5, p  2  30˚  
3   cos 210°  
2  3 x 5  y  51. The terminal side is in the third quadrant and the reference angle is 210 
 180 or 30°.  0, the normal lies in   r cos (v 
 117°)  7. 3  r cos (v 
 60°) 0  r cos (v 
 60°) 
 3 0  r (cos v cos 60°  sin v sin 60°) 
 3  z.  z, then 0  xz  1.  The correct choice is C.   
3  r sin v 
 3 0  12r cos v   2  
3 y 
 3 0  12x   2  9-4  y 
 6 or 0  x  
3 x  
3 y 
 6  0  Polar Form of a Linear Equation  Pages 577–578  8.  Check for Understanding  r cos (v  4)  2  1. The polar equation of a line is p  r cos (v 
 f). r and v are the variables. p is the length of the normal segment from the line to the origin and f is the angle the normal makes with the positive x-axis. 2. For r to be equal to p, we must have cos (v 
 f)  1. The first positive value of v for which this is true is v  f.  Chapter 9  r  2 sec v  4  r cos v cos 4 
 sin v sin 4 
 2  0  
2  2   
2  r sin v 
 2  0 r cos v 
  2  
2 x 2   
2 y 
 2  0 
 2  x 
 
2 y 
 4  0 
2  284  9.  2 3   2  10.   3  6  5 6  90˚  30˚  180˚  1 2 3 4  2 4 6 8  7 6  240˚  270˚  4  f  Arctan 
3  
53° Since cos f 0, but sin f  0, the normal lies in the fourth quadrant. f  360° 
 53° or 307° p  r cos (v 
 f) 2.1  r cos (v 
 307°)  300˚  Since the shortest distance is along the normal, the answer is (p, f) or 5, 56. 2 3   2  15. 
 A2   B2  
 32  22  
13  Sinc C is negative, use 
13 .   3  6  5 6  3  x  
13  0    3 2  5 3  16.  4  x  
41   25 Since C is positive, use 
25. cos f  24  25y 
 4  0 7 24 ,  
2 5 , sin f   25  5  10  y 
  0 
  1 
41 
4   
5
41   10
41  4
 41 , p    , sin f   cos f   41 41 41  f  Arctan 
4  
51° Since cos f 0, but sin f  0, the normal lies in the fourth quadrant. f  360° 
 51° or 309° p  r cos (v 
 f) 5  p4  f  Arctan 
274  
74° Since cos f  0, but sin f 0, the normal lies in the second quadrant. f  180° 
 74° or 106° p  r cos (v 
 f) 4  r cos (v 
 106°) 13. 
 A2   B2  
 212  202   29 Since C is negative, use 29. 21 x 29  5
 13   r cos (v 
 34°) 13 A2   B2  
 42  ( 
5)2 
   
14  Since C is negative, use 
41 .  Pages 578–579 Exercises 2B  2  
 12. 
A 72  ( 
24)2 7 
2 5x  5  f  Arctan 23  34° p  r cos (v 
 f)  11 6 4 3  2  y 
  0  13 13 
 
  3
 2
13 5
 13 13  , sin f  , p   cos f   13 13 13  2 4 6 8  7 6  4  cos f  5, sin f  
5, p  2.1  11a. p  r cos (v 
 f) → 5  r cos v 
 56  11b.  1 
 18 y 
 21 0 0 0 3  5 3  3 2  6 x 10  0˚  330˚  210˚  11 6 4 3  14. 
 A2   B2  
 62  ( 
8)2  10 Since C is negative, use 10.  60˚  150˚ 0    120˚   10
41   41   r cos (v 
 309°)  A2   B2  
 (–1)2  32 17. 
  
10  Since C is negative, use 
10 . 1 
x 
 10  cos f    2209y 
 8279  0  3 
10  
 10 
 10  7 
10    y 
   0 3
 10 7
 10 , p   sin f   10 10  f  Arctan (
3)  
72° Since cos f < 0, but sin f > 0, the normal lies in the second quadrant. f  180° 
 72° or 108° p  r cos (v 
 f)  cos f  2219, sin f  2209, p  3 f  Arctan 2201  44° p  r cos (v 
 f) 3  r cos (v 
 44°)  7
 10   10  r cos (v 
 108°)  18. 6  r cos (v 
 120°) 0  r (cos v cos 120°  sin v sin 120°) 
 6  
3  r sin v 
 6 0  
 12 r cos v   2  
3 y 
 6 0  
 12x   2  y 
 12 or 0  
x  
3 
x  
3 y 
 12  0  285  Chapter 9  28.  19. 4 
 r cos 
v  4  0 
 r 
cos v cos 4  sin v sin 4  4  x  2 y  8 or 0 
 2 x  2 y  8 
 0 2 20. 2 
 r cos (v  ) 0 
 r (cos v cos   sin v sin ) 2 0 
 r cos v  2 0 
 x  2 x 
 2 21. 1 
 r cos (v  330°) 0 
 r (cos v cos 330°  sin v sin 330°)  1 0
  r 
cos v cos  r cos 
v   sin v sin  7  6    1 y 2  13 5  
 13  30˚  150˚ 180˚  3 6 9 12  330˚  210˚ 240˚  26. 120˚  270˚ 90˚  300˚  0˚  180˚  1 2 3 4 330˚  210˚ 240˚  270˚  300˚   3  11 6  7 6  120˚ 30˚   2  0  27.  3 2 90˚  5 3 60˚ 30˚  150˚ 180˚  2 4 6 8  0˚  330˚  210˚ 240˚  5  13   
0  3 13 5 13 , p 
  sin f 
  13 13  r cos (v  56°)  Use a graphing calculator and the INTERJECT feature to find solutions to the system at (2.25, 0.31) and (5.39, 0.31). Since p, the length of the normal, must be positive, use f 
 2.25 and p 
 0.31. 0.31 
 r cos (v  2.25) 32a. p 
 r cos (v  f) → 6 
 r cos (v  16°) Since the shortest distance is along the normal, the closest the fly came was p or 6 cm. 32b. (p, f) or (6, 15°) 33. Since both normal segments have length 2, p must be 2 in both equations. Since the two lines intersect at right angles, their normals also intersect at right angles. This can be achieved by having the two f-values differ by 90°. To make sure neither line is vertical, neither f-value should be a multiple of 90°. Therefore, a sample answer is 2 
 r cos (v  45°) and 2 
 r cos (v  135°). 34. m 
 0 (y  4) 
 0(x  5) → y  4 
 0 cos f 
 0, sin f 
 1, p 
 4 Since cos f 
 0 when sin f 
 1, f 
 90°. p 
 r cos (v  f) 4 
 r cos (v  90°)  1 2 3 4  4 3    → p 
 2 cos 
76  f   6    3  13   → p 
 3 cos 
4  f  5 6  60˚  150˚  Chapter 9  0˚  5 3  31. p 
 r cos (v  f)   11 
 0  2 3  3 2  56° p 
 r cos (v  f)    11 
 0  25.    2 13 , cos f 
  13 f 
 Arctan 32   
 11  60˚  4 3  
 13  Since C is negative, use 13 .  y  10 
 0 x  3 90˚  11 6   A2   B2 
  22  32   3 1 r cos v   r sin v  5 
 0 2 2  3 1 x  y  5 
 0 2 2  120˚  7 6  2  x  y  22 
 0 3 23. r 
 5 sec (v  60°) r cos (v  60°) 
 5 r (cos v cos 60°  sin v sin 60°)  5 
 0  24.  0 2 4 6 8  (y  1) 
 3(x  4) → 2x  3y  5 
 0  r 
 11 sec 
v  76  7  6   3  6    5 3  3 2   2  4 2  or  30. m 
  6 3  1  3  r cos v  r sin v  11 
 0  2 2  3 x  2  11 6  2  13   22. 7  6  0  4 3  2 3 5 6  2 4 6 8  7 6  sin v  1  x  y  2 or 0 
 3 3 x  y  2 
 0  29.   3  6    2 2  x  y  4 0
 2 2  0
   2  5 6   2 2  r cos v   r sin v  4 0
 2 2  1 3  r cos v  r 2 2 1 3 x  y  1 2 2  2 3  270˚  300˚  286  35a.  120˚  90˚  40. x  3y 
 6  60˚  x  6  y
 3  30˚  150˚  y 
 13x  2  0˚  x 
 t, y 
 13t  2  12 5 25 0 37 505 0  180˚  N  r2 41. A 
  360 
  330˚  210˚ 240˚  270˚  65  62 
 360 
  300˚  35b. p 
 r cos (v  f) → p 
 125 cos (130  f) → p 
 300 cos (70  f) Use a graphing calculator and the INTERSECT feature to find the solutions to the system at (45, 124.43) and (135, 124.43). Sinc p, the length of the normal, must be positive, use f 
 135° and p 
 124.43. 124.43 
 r cos (v  135°) 36. k 
 r sin (v  a) k 
 r [sin v cos a  cos v sin a] k 
 r sin v cos a  r cos v sin a k 
 y cos a  x sin a This is the equation of a line in rectangular coordinates. Solving the last equation for y yields k . The slope of the line shows y 
 (tan a)x   cos a that a is the angle the line makes with the x-axis. To find the length of the normal segment in the figure, observe that the complementary angle to a in the right triangle is 90°  a, so the v-coordinate of P in polar coordinates is 180°  (90°  a) 
 a  90°. Substitute into the original polar equation to find the r-coordinate of P: k 
 r sin (a  90°  a) k 
 r sin 90° k
r Therefore, k is the length of the normal segment.  y  20.42 ft2 42. Since 360° lies on the x-axis of the unit circle at (1, 0), sin 360° 
 y or 0.  (1, 0)  x  43. 2x3  5x2  12x 
 0 x(2x2  5x  12) 
 0 x(2x  3)(x  4) 
 0 x 
 0 or x 
 32 or x 
 4 44.  c2  d2 
 48 (c  d)(c  d) 
 48 12(c  d) 
 48 cd
4  Page 579 1.  Mid-Chapter Quiz 90˚  120˚  2.  60˚  180˚  2 4 6 8  240˚  120˚  P  270˚ 90˚    37. p 
 r cos (v  f) → p 
 40 cos (0°  f) → p 
 40 cos (72°  f) Use a graphing calculator and the INTERSECT feature to find the solutions of the system at (144, 32.36) and (36, 32.36). Since p, the length of the normal, must be positive, use f 
 36° and p 
 32.36. 32.36 
 r cos (v  36°) 38. r
6  x2  y2 
 6 x2  y2 
 36 39. The graph of a polar equation of the form r 
 a sin nv is a rose.  0˚  330˚ 270˚  300˚  5. r 
  (2  )2  ( 2 )2 
 4  or 2  11 6  7 6  4.  2 4 6 8  240˚  0 1 2 3 4  4 3  60˚  210˚   6    30˚  180˚  x   3  0˚  300˚  150˚   2  5 6  330˚  210˚  
  2 3  30˚  150˚  3.  O  y  2 3  5 3  3 2  2   3  6  5 6   1  0  11 6  7 6 4 3  5 3  3 2  v 
 Arctan 
 4  
  2      Since (2 , 2 ) is in the third quadrant, v 
   4 or 54.  
2, 54  6. r 
  02  ( 4)2 
 16  or 4 Since x 
 0 when y 
 4, v 
 32.  
4, 32  287  Chapter 9  x2  y2 
 36  x2  y2 
 36  r 
 6 or r 
 6 8. r 
 2 csc v r sin v 
 2 y
2   9. A2   B2 
  52  ( 12)2 
 13 Since C is positive, use 13.  4. Sample answer: x2  1 
 0 (x  i)(x  i) 
 0, where the solutions are x 
 x2  xi  xi  i2 
 0 x2  (1) 
 0 x2  1 
 0 6 4 2 5. i 
 (i )  i2 
 12  (1) 
 1 6. i10  i2 
 (i4)2  i2  i2 
 (1)2i2  i2 
 1  (1) or 2 7. (2  3i)  (6  i) 
 (2  (6))  (3i  i) 
 4  4i 8. (2.3  4.1i)  (1.2  6.3i) 
 (2.3  (1.2))  (4.1i  (6.3i)) 
 3.5  10.4i 9. (2  4i)  (1  5i) 
 (2  (1))  (4i  5i) 
 1  9i 10. (2  i)2 
 (2  i)(2  i) 
 4  4i  i2 
 3  4i  7.  5  12 3 y   
 0 1 3x   13 13 5  12 3 , p 
  cos f 
 1 3 , sin f 
  13 13 2 f 
 Arctan 
15   67° Since cos f 0, but sin f  0, the normal lies in the second quadrant. f 
 180°  67° or 113° p 
 r cos ( f) 3  13  10.  
 r cos (v  113°)   A2   B2 
  (2)2   (6 )2  i i 1  2i  
    11.  1  2i 1  2i 1  2i  
 210  Since C is negative, use 210 . 6 2 10  2  i  2i2   
 1  4i2  2 210     y   
 0 210 x cos f 
   10 1 , 0  f 
 Arctan 
  sin f 
  310  10,  p
    i2  
 5  10    10  
 25  15i  3  1  72° Since cos f 0 and sin f the third quadrant. f 
 180°  72° or 252° p 
 r cos (v  f) 10    10  12. (2.5  3.1i)  (6.2  4.3i) 
 (2.5  (6.2))  (3.1i  4.3i) 
 3.7  7.4i N  0, the normal lies in  Pages 583–585  
 r cos (v  252°)  Page 583  Check for Understanding  1. Find the (positive) remainder when the exponent is divided by 4. If the remainder is 0, the answer is 1; if the remainder is 1, the answer is i; if the remainder is 2, the answer is 1; and if the remainder is 3, the answer is i. 2.  Complex Numbers (a  bi ) Reals (b 
 0)  Imaginary (b  0)  19. 
12  i  (2  i) 
 
12  (2)  (i  (i))  Pure Imaginary (a 
 0)  
 32  2i 20. (3  i)  (4  5i) 
 (3  (4))  (i  (5i)) 
 7  4i 21. (2  i)(4  3i) 
 8  10i  3i2 
 5  10i 22. (1  4i)2 
 (1  4i)(1  4i) 
 1  8i  16i2 
 15  8i  3. When you multiply the denominators, you will be multiplying a complex number and its conjugate. This makes the denominator of the product a real number, so you can then write the answer in the form a  bi. Chapter 9  Exercises  13. i6 
 i4  i2 
 1  1 
 1 14. i19 
 (i4)4  i3 
 14  i 
 i 15. i1776 
 (i4)444 
 1444 
1 16. i9  i5 
 (i4)2  i  (i4)2  i3 
 12  i  12  i 
 i  (i) or 0 17. (3  2i)  (4  6i) 
 (3  (4))  (2i  6i) 
 1  8i 18. (7  4i)  (2  3i) 
 (7  2)  (4i  3i) 
 9  7i  Simplifying Complex Numbers  9-5  i.  288  3i 3i  
  34.  (2  i)2 (2  i)(2  i)  23. (1  7 i)(2  5 i) 
 2  5 i  27 i  35 i2 
 (2  35 )  (27   5 )i 24. (2  3 )(1  12 ) 
 (2  3 i)(1  12 i) 
 2  212 i  3 i  36 i2 
 2  43 i  3 i  6 
 8  33 i  3i   
 4  4i  i2  3i  
 3  4i  3  4i 3i    
 3  4i 3  4i 9  9i  4i2   
 9  16i2  13  9i  
 25  1  2i 2i 2i  
    25.  1  2i 1  2i 1  2i 2  3i  2i2  
 1  4i2  
 1235  295i  4  3i  
 5  35.  (1  i)2  (3  2i)2  
 45  35i  
  3  2i i 3  2i  4  
 26.  4  i   4  i 4  i  
  5  12i 2i    
 5  12i 5  12i 10i  24i2   
 25  144i2  11  0 
 11  17i 7  27.  
 
 
 
  5i  5i    24  10i  
 169  5i  5i  24 10   i 
  169 169  25  10i  i2  25  i2  36a. Z 
 R  (XL  XC )j → Z 
 10  (1  2)j → Z 
 10  j ohms → Z 
 3  (1  1)j → Z 
 3  0j ohms 36b. (10  j)  (3  0j) 
 (10  3)  (1j  0j) 
 13  j ohms  24  10i  26 12 5   i 13 13  28. (x  i)(x  i) 
 0 x2  i2 
 0 x2  1 
 0 29. (x  (2  i))(x  (2  i)) 
 0 (x  2  i)(x  2  i) 
 0 x2  2x  xi  2x  4  2i  xi  2i  i2 
 0 x2  4x  4  1 
 0 x2  4x  5 
 0 30. (2  i)(3  2i)(1  4i) 
 (6  i  2i2)(1  4i) 
 (8  i)(1  4i) 
 8  31i  4i2 
 12  31i 31. (1  3i)(2  2i)(1  2i) 
 (2  8i  6i2)(1  2i) 
 (4  8i)(1  2i) 
 4  16i  16i2 
 12  16i 32.  1   3 i 2  1  2 i  
  1  1  36c. S 
 Z → S 
  6  3j 1 6  3j    
 6  3j 6  3j 6  3j   
 36  9j2  6  3j  
 45 0.13  0.07j siemens  37a. x 
  
 3  4i 37b. No 37c. The solutions need not be complex conjugates because the coefficients in the equation are not all real. 37d. (3  4i)2  8i(3  4i)  25 
 0 7  24i  24i  32  25 
 0 0
0 (3  4i)2  8i(3  4i)  25 
 0 7  24i  24i  32  25 
 0 0
0 38. f(x  yi) 
 (x  yi)2 
 x2  2xyi  y2 
 (x2  y2)  2xyi 39a. z0 
 2  i z1 
 i(2  i)  i2 or 1  2i z2 
 i(2i  1) 
 2i2  i or 2  i z3 
 i(2  i) 
 2i  i2 or 1  2i z4 
 i(1  2i) 
 i  2i2 or 2  i z5 
 i(2  i) 
 2i  i2 or 1  2i  1   3 i 2 i 1  2    i 1  2 1  2 i 1 2    i  3i  6i2 2 2 1  2i  
12  6  
22  3i  
  3    6 3 2   
  i 
 
16   3  3 6 2  2 i  3   6i  2  2 i 3  6 i  3  6 i 3  6 i  
    
  6  2 6i  3 2i   12i2  9  6i2  
  (6  2 3)  (2 6  3 2)i  15   23  8i  (8i)2  4(1)( 25)  2(1)  8i 36   
 2  
  2  33.  1  2i  i2  9  12i  4i2  2i  
 5  12i  12  11i  2i2  16  i2  10  11i  
 17 5i  5i  (1  i)(1  i) 
  (3  2i)(3  2i)   26   2   
   i 
 
25   15  15 5  289  Chapter 9  39b. z0 
 1  0i z1 
 (0.5  0.866i)(1  0i) 
 0.5  0.866i z2 
 (0.5  0.866i)(0.5  0.866i) 
 0.25  0.866i  0.75 
 0.500  0.866i z3 
 (0.5  0.866i)(0.500  0.866i) 
 0.250  0.750 
 1.000  0.000i z4 
 (0.5  0.866i)(1.000) 
 0.500  0.866i z5 
 (0.5  0.866i)(0.500  0.866i) 
 0.250  0.866i  0.75 
 0.500  0.866i  46. tan a 
 43 tan2 a  1 
 sec2 a 4 2  3  
   1
 25  9 9  25 3  5  sin2  
 
 
  2  48. h 
 x3   x3   tan 52° 
  x  45  x tan 52°  45 tan 52° 
 x3  x tan 52°  x3  
 45 tan 52° 45 tan 52° tan 52°  3   x 
  x 127.40 h 
 x3  
 127.40(3 ) 221 ft  3 10  p
 20  8˚  52˚ 45 ft  h  60˚ x  49. Enter the x-values in L1 and the f(x)-values in L2 of your graphing calculator. Make a scatter plot. The data points are in the shape of a parabola, so a quadratic function would best model the set of data. 50. Let d 
 depth of the original pool. The second pool's width 
 5d  4, the length 
 10d  6, and the depth 
 d  2. (5d  4)(10d  6)(d  2) 
 3420 (50d2  70d  24)(d  2) 
 3420 50d3  100d2  70d2  140d  24d  48 
 3420 50d3  170d2  164d  3372 
 0 25d3  85d2  82d  1686 
 0 Use a graphing calculator to find the solution d 
 3. The dimensions of the original pool are 15 ft by 30 ft by 3 ft. 51. 80 
 k(5)(8) 2
k y 
 2(16)(2) 
 64  0  11 6 5 3  44. x  (3), y  6 
 t 1, 4 x  3, y  6 
 t 1, 4  45. u 
 1 8, 6, 4  2 2, 6, 3 4  
 2, 32, 1  4, 12, 6 
 6, 227, 5  Chapter 9  30˚  120˚   3  3 2       6  4 3  12  y 
 3.5 cos 
6t  r cos (v  162°)  7 6  4  5   period 
 1 2 or 6  18° Since cos f 0, but sin f  0, the normal lies in the second quadrant. f 
 180°  18° or 162° p 
 r cos (v  f)  7 14 21 28  5  cos B 
 1 3  1  6 210     25   cos2 B 
  169  47. amplitude 
 2(7) or 3.5   x  y   
 0  5 6   cos2 B 
 1  33   62  ( 2)2 
 210  Since C is positive, use 210 .   2  2  
1123  
 65   A2   b2 
  2 3  
 sin B B  cos2 B 
 1  16  25 4  5  3  41. c1(cos 2t  i sin 2t)  c2(cos 2t  i sin 2t) 
 c1 cos 2t  c1i sin 2t  c2 cos 2t  c2i sin 2t 
 (c1  c2)(cos 2t)  (c1  c2)(i sin 2t) 
 (c1  c2)(cos 2t) only if c1 
 c2  43.  
 sin2 B    
 
5
1 3   
 5 
 13   11  3 10  
 20  sin2  
 csc2 B  cos (a  B) 
 cos a cos B  sin a sin B  11  2i   11  2i  
   169  144 144  169 12  13  
 cos2 a  sin a 
  1  (3  4i)(1  2i)  2 3 210  210  310  10    , cos f 
  10 , sin f 
  10 1 f 
 Arctan 3  5  
 sec2 a  sin2 a 
  2  1  
1 
 csc2 B 2  a  3 2  2  i 
  12 5   125  42.  1  cot2 B 
 csc2 B  sin2 a  
5 
 1  1  1  11  2i 1  11  2i 11  2i  125  sec2  
 cos a a  cos2 a 
 1   40. (1  2i)3 
  (1  2i)3  
  cot B 
 152  290  52.  4. The conjugate of a  bi is a  bi. (a  b i)(a  bi) 
  a2  b2, so the friend's method gives the same answer. Sample answer: The absolute value of 2  3i is  22  32 
 13 . Using the friend's method, the absolute value is (2  3 i)(2  3i) 
 4  9
 13 . 5. 2x  y  (x  y)i 
 5  4i 2x  y 
 5 xy
4 2x  (x  4) 
 5 y 
 x  4 x
1 y 
 (1)  4 or 3 6. 7.  y 
 7  x2 x 
 7  y2 x  7 
 y2 x  7 
 y2 x 7 
y f1(x) 
 7  x  y  53.  (6, 8)  (6, 1)  O (1, 1)  x  f(x, y) 
 2x  y f(1, 1) 
 2(1)  1 or 3 f(6, 1) 
 2(6)  1 or 11 f(6, 8) 
 2(6)  8 or 4 The maximum value is 3 and the minimum value is 11. 54. x  2y  7z 
 14 x  3y  5z 
 21 y  2z 
 7 x  3y  5z 
 21 → 5x  15y  25z 
 105 5x  y  2z 
 7 5x  y  2z 
 7 16y  27z 
 112 y  2z 
 7 → 16y  32z 
 112 16y  27z 
 112  6y  27z 
 112 1  59z 
 0 z
0 y  2(0) 
 7 → y
7 x  2(7)  7(0) 
 14 → x 
 0 (0, 7, 0) 55. Since BC 
 BD, m∠BDC 
 m ∠DCB 
 x m ∠DBC 
 180  120 or 60. x  x  60 
 180 2x 
 120 x 
 60 x  40 
 60  40 or 100 The correct choice is A.  i  2  2  1  1  O  2 1 (2, 1)  i  1  1  2  2  1    2  z 
  12  ( 2  )2 
 3  2 v 
 Arctan 
2  2  z 
  (22  (1)2 
 5  2 8. r 
  2  ( 2)2  7   or 22  
 8  
 4  v is in the fourth quadrant. 7 7 2  2i 
 22  
cos 4  i sin 4 5 9. r 
  42  52 v 
 Arctan 
4 
 41  
 0.90 4  5i 
 41 (cos 0.90  i sin 0.90) 0  (2)2   02 10. r 
    v 
 Arctan  2     or 2 
 4 
 is on the x-axis at 2. 2 
 2 (cos   i sin ) 11.  The Complex and Polar Form of Complex Numbers  Pages 589–590  2 1 O  1  2 3   2   3 
 (4, 3 )  i  5 6  9-6    2  (1, 
2 )    11 6 4 3  1. To find the absolute value of a  bi, square a and b, add the squares, then take the square root of the sum.   2. i 
 0  i; cos 2 
 0 and sin 2 
 1   i 
 cos 2  i sin 2 3. Sample answer: z1 
 i, z2 
 i  3 2    5 3   4
cos 3  i sin 3  3 —— 2  
 4
  i
 1  2  i 
 2  23    2 3   6  2 4 6 8 7 6  Check for Understanding  12.  0    
  i   2   3  6  5 6   (2, 3)  0 1 2 3 4  7 6    11 6 4 3  3 2  5 3  2(cos 3  i sin 3) 2(0.99  i(0.14)) 
 1.98  0.28i  z1  z2 
 z1  z2 i  (i) 
 i  i 0 
 i  i 0 2i  291  Chapter 9  13.  2 3   2  i  19.   3  20.  i  i   6  5 6  (2, 3)  3  ( 2 , 2
) 0    1 2 3 4 7 6     O  5 3  3 2  3 (cos 2  i sin 2 3 
 2 (1  i(0)) 3 
 2  14.    O  11 6 4 3  (3, 4)  2) z 
  22  32 
 13   z 
  32  ( 4)2 
 25  or 5  21.  i  22.  i  1  i  0.63 1  0.90  0.36 1 1.30  O  0.38i  O     O (0, 3)  1 (1, 5)  102  152 15a. magnitude 
  
 325  18.03 N  (1)2   (5 )2 z 
  02  ( 3)2 z 
  
 26  
 9  or 3 23. 24.  15  15b. 
 
 Arctan 1 0 56.31°  (1, 
5 )  i  i  2  Pages 590–591  (4, 
2 )  1  Exercises  16. 2x  5yi 
 12  15i 2x 
 12 5y 
 15 x
6 y 
 3 17. 1  (x  y)i 
 y  3xi 1
y x  y 
 3x x  (1) 
 3x 1 
 2x 1  
 x 2 18. 4x  (y  5)i 
 2x  y  (x  7)i y5
x7 4x 
 2x  y y 
 x  12 4x 
 2x  (x  12) 3x 
 12 x 
 4 y 
 (4)  12 or 8  2 1 O  1  2      O  1 2  z 
  (1)2   (5  )2 
 6  25. r 
  (4)2   62 
 52  or 213   z 
  42  ( 2  )2 
 18  or 32   v 
 Arctan 
3 3  26. r 
  32  32    
 18  or 32   
 4     3  3i 
 32  
cos 4  i sin 4 27. r 
  (1)2   (  )2 3   3   v 
 Arctan 
 1    4   or 2 
 4  
 3  v is in the third quadrant. 4  4  i 
 2
cos 3  i sin 3 1  3 28. r 
  62  ( 8)2   or 10 5.36 
 100 v is in the fourth quadrant. 6  8i 
 10(cos 5.36  i sin 5.36)  Chapter 9  292  8  v 
 Arctan 
6  2   v 
 Arctan 
 4    1  29. r 
  (4)2   12   2.90 
 17 v is in the second quadrant. 4  i 
 17 (cos 2.90  i sin 2.90)  39.  5  
 2
 0  
 9  or 3 v is on the x-axis at 3. 3 
 3(cos 0  i sin 0)  41.  
0   42   or 42  
 
 32 v is on the x-axis at 42 . 42  
 42  (cos   i sin ) 2   34. r 
 0  (2)2 
 4  or 2 3 Since x 
 0 when y 
 2, v 
 2.  i   3  2 3  
  (3, 4 )  6  5 6   1 2 3 4    4 3  5 3  3 2       2  32  3 2   3    7 6    11 6 4 3  3 2  5 3  3(cos   i sin ) 
 3(1  0) 
 3  1  O  0.50 0.39i 0.60 0.39i  0.25 i 1  44.    i i  1  1  
 2  2i  32  2 3  i   2  38.   3  2 3   6  5 6  0    1 2 3 4 7 6  0 1 2 3 4  5 3  
 2  2 i 37.   6  0.44 0.44i  5 3     2   3  i  1    cos 
6  i sin 
6  
 3
2  i 
2   2  6  4 3  3
cos 4  i sin 4  i  (3, )    11 6  (1, 
 ) 6 11  7 6  2 3  1  0 1 2 3 4  5 3  0.44 0.5i   6    11 6  7 6  i   3  5 6 0  (5, 0) 0  43.  2  3 2  5 6  5(cos 0  i sin 0) 
 5(1  0) 
5  3  36.  42.   3  2 4 6 8  3 2  11 6  2.5(0.54  i(0.84))   6  4 3  0    
 1.35  2.10i  7 6  2i 
 2
cos 2  i sin 2 2 3  i    v 
 Arctan     3    5 6 0   33. r 
 (4 )2   2 02   2  2 3   2  6  2.5(cos 1  i sin 1)   2 2    2 i 
 2  v 
 Arctan 
3  32  02 32. r 
   (2.5, 1)   4 3  5  i
   3  7 6  2
cos 4  i sin 4 2 2   2  1 2 3 4  5 3  3 2  i      11 6  4   or 25  2.03 
 20 v is in the second quadrant. 2  4i 
 25 (cos 2.03  i sin 2.03)  2 3 5 6  0 1 2 3 4  (2, 5
 ) 4 7 6 4 3   v 
 Arctan 
 2     31. r 
  (2)2   42  40.   3  6    21   or 29 5.47 
 841 v is in the fourth quadrant. 20  21i 
 29(cos 5.47  i sin 5.47)  35.   2  i  5 6  v 
 Arctan 
2 0   2  202  (21 )2 30. r 
   2 3  (2, 4
 ) 3 4 3 3 2 4  11 6 5 3 4  2
cos 3  i sin 3  3  
 2
2  i 
2 1  i 
 1  3    i   2  1   3  1  O   6  5 6  1  0    3 6 9 12  (10, 6) 7 6    45.  4 3  3 2  i 1  11 6  0.5  0.5i  5 3  1  10(cos 6  i sin 6)  O  1    0.5  0.5i 1  10(0.960  i(0.279)) 
 9.60  2.79i  293  Chapter 9  46a. 40∠30° 
 40(cos 30°  j sin 30°)  3  2  
 40
   j
  1  2  51. (6  2i)(2  3i) 
 12  22i  6i2 
 6  22i 52. x 
 3 cos  135° y 
 3 sin  135°    
 34.64  20j 60∠60° 
 60(cos 60°  j sin 60°)  1 3 
 60
2  j
 2   2  49a. 49b. 49c. 49d. 50a.      71.96   v 
 Arctan  64.64  
  
 
  tan 60°  tan 45°  1  tan 60° tan 45° 3 1 1  (3 )(1) 3   1 1   3    1  3  1  3  3   3  1  3    4  23   
 2  
 2  3  v  55. w 
 t 12(2)  
 1 or 24 v 
 rq 
 18(24) or 432 cm/s 432 cm/s 
 4.32 m/s 13.57 m/s  14.11   
 
 Arctan  21.69  12  56. sin A 
 1 8 A 
 sin1 
3 2  A 41.8° 47. 2a   
 3a 1 5  2a  1 
 3a  5 4
a 58. as x → , y → ; as x → , y →  y 
 2x2  2 x 1000 10 0 10 1000  1.41(cos 0.79  i sin 0.79)  y 2  106 202 2 202 2  106  59. In the fourth month, the person will have received 3 pay raises. $500(1.10)3 
 $665.50 The correct choice is D.    
cos 4  i sin 4 
 102  
 14.14(cos 0.79  i sin 0.79). 50d. To multiply two complex numbers in polar form, multiply the moduli and add the amplitudes. (In the sample answer for 50c, note that 5.18  1.89 
 7.07, which is coterminal with 0.79.)  Chapter 9  32  
   z2 
 5(cos 0.93  i sin 0.93) z1z2 
 52  (cos 1.71  i sin 1.71) 
 7.07(cos 1.71  i sin 1.71) 50c. Sample answer: Let z1 
 2  4i and z2 
 1  3i. Then z1 
 25 (cos 5.18  i sin 5.18) 4.47(cos 5.18  i sin 5.18), z2 
 10 (cos 1.89  i sin 1.89) 3.16(cos 1.89  i sin 1.89), and z1z2 
 (2  4i)(1  3i) 
 10  10i   
 2  53. magnitude 
  (3)2   72 
 58  u  7j u 3, 7 
 3i 54. tan 105° 
 tan (60°  45°)  25.88 0.58 21.69  14.11j 
 25.88 (cos 0.58  j sin 0.58) ohms Translate 2 units to the right and down 3 units. Rotate 90° counterclockwise about the origin. Dilate by a factor of 3. Reflect about the real axis. Sample answer: let z1 
 1  i and z2 
 3  4i. z1z2 
 (1  i)(3  4i) 
 1  7i  50b. z1 
 2 
cos 4  i sin 4  
 2  32 32   
 2 , 2   96.73 48° v(t) 
 96.73 sin (250t  48°) 47. The graph of the conjugate of a complex number is obtained by reflecting the original number about the real axis. This reflection does not change the modulus. Since the amplitude is reflected, we can write the amplitude of the conjugate as the opposite of the original amplitude. In other words, the conjugate of r(cos v  i sin v) can be written as r(cos (v)  i sin (v)), or r(cos v  i sin v). 48a. 10(cos 0.7  j sin 0.7) 7.65  6.44j 16(cos 0.5  j sin 0.5) 14.04  7.67j 48b. (7.65  6.44j)  (14.04  7.67j) 
 (7.65  14.04)  (6.44j  7.67j) 
 21.69  14.11j ohms  2 48c. r 
 21.69  14 .112  
 3
2  32  
 30  51.96j 46b. (34.64  20j)  (30  51.96j) 
 (34.64  30)  (20j  51.96j) 
 64.64  71.96j 46c. v(t) 
 r sin (250t  v°) 2  71 r 
  64.64 .962   2  
 3
2  294    3  9-6B Graphing Calculator Exploration:  5. r 
 4  Geometry in the Complex Plane  2  v 
 6  3 3    
 6 or 2  
cos 
2  i sin 
2 
 34(0  (1)i)  3  4  Page 592  3  
 4i  1. They are collinear.  9    v 
 4  
2  4  6. r 
 2 or 2  9  2  11  
 4  4 or 4 11  11   2   2  2
cos 4  i sin 4 
 2
2  i
2 
 2   2 i 7. r 
  1 (6) 2    5  v 
 3  6  or 3  2  5  7  
 6  6 or 6  2. Yes. M is the point obtained when T 
 0, and N is the point obtained when T 
 1. 3. The points are again collinear, but closer together.  7  7   3  3
cos 6  i sin 6 
 3
2  i
2 1  33  3  
 2  2i 22  ( 23 )2 8. r1 
  
 16  or 4 r 
 4(23 ) or 83   r2 
  (3)2   (3  )2 
 12  or 23   23  3  v1 
 Arctan 
2     v2 
 Arctan 
3    5  
 6  
 3   5  v 
 3  6 2  9-7  7  3  1  11  11      
 2
cos 6  j sin 6  3
cos 3  j sin 3  r 
 2(3) or 6  11    11  2  13    v 
 6  3 
 6  6    
 6 or 6    V 
 6
cos 6  j sin 6 volts  Pages 596–598  Exercises  10. r 
 4(7) or 28    2  v 
 3  3 3  
 3 or   Check for Understanding    7  
 12  43 i 9. E 
 IZ  28(cos   i sin ) 
 28(1  i(0)) 
 28  1. The modulus of the quotient is the quotient of the moduli of the two complex numbers. The amplitude of the quotient is the difference of the amplitudes of the two complex numbers. 2. Square the modulus of the given complex number and double its amplitude. 3. Addition and subtraction are easier in rectangular form. Multiplication and division are easier in polar form. See students' work for examples. 4. r 
 2  2 or 4  7  83 
cos 6  i sin 6 
 83 
2  i
2  Products and Quotients of Complex Numbers in Polar Form  Pages 596  5  
 6  6 or 6  4. The points are on the line through M and N. 5. If one of a, b, or c equals 0, then aK  bM  cN is on KMN. If none of a, b, or c equals 0, then aK  bM  cN lies on or inside KMN. 6. M is the point obtained when T 
 0 and N is the point obtained when T 
 1. Thus, a point between M and N is obtained when 0 T 1. 7. The distance between z and 1  i is 5. This defines a circle of radius 5 centered at 1  i. 8. The distance between a point z and a point at 2  3i is 2. z  (2  3i) 
 2  6  3    2    v 
 4  4  11. r 
 2 or 3   
 4 or 2    3
cos 2  i sin 2 
 3(0  i(1)) 
 3i 12. r 
  1  2  3    1    v 
 3  6  or 6  2  3      
 6  6 or 6  v 
 2  2  
cos 6  i sin 6 
 16
23  i
12  1  6  4  
 2 or 2  3  4(cos 2  i sin 2) 
 4(1  i(0)) 
4  1    
 12  12 i  295  Chapter 9  3  13. r 
 5(2) or 10  24. r1 
  22  ( 2)2 r2 
  (3)2   32 
 8  or 22  
 18  or 32  r 
 22 (32 ) or 12  v 
   4 4  3  7  
 4  4 or 4 7  7   2  10
cos 4  i sin 4 
 10
2  i
2  2  v1 
 Arctan 
2  2      7  3  7  10    11  3  3  
 12i 25. r1 
  (2 )2  (2 )2 r2 
 (3   )2  ( 2 32  )2 
 4  or 2 
 36  or 6 r 
 2  6 or 12  11  v1 
 Arctan 
  2 v2 
 Arctan 
  3
cos 6  i sin 6 
 3
2  i 
2 1  33  7  17. r 
  2   2   2  
  22    2  7    12(cos   i sin ) 
 12(1  i(0)) 
 12 (3 )2  ( 1)2 r2 
  22  (23  )2 26. r1 
  
 4  or 2 
 16  or 4  3  v 
 4  4 4  
 4 or   2  20  4  7  
  
cos   2   i sin   2  
 
  4 (0 3  22 
cos   i sin    3  4  2 5  4 or 4   2 2 2  i 2    2           3  
    1  5    v 
 3  3  or 8  4  
 3 4  4   3  8
cos 3  i sin 3 
 8
2  i
2 1  
 4  43 i  Chapter 9  r 2 
  62  62 
 72  or 62      
    v2 
 Arctan 
6 6    
 4    22  
 3 (0  i(1)) 2 2  
 3i  12
cos 6  i sin 6 
 12
2  i
2 
 63   6i  8 62  4  32  42   6  2  2  2 2  
 or 3 4 2 v1 
 Arctan  42  3 
 4 3  v 
 4  4 2  
 4 or 2 2   2  cos   i sin  2 2 3  
  
  1  r 
    i (1))   
 22
      )2   2 (42 )2 27. r1 
 (4 
 64  or 8  v 
 4  2 5  4    
 4  4i  
 6  6 or 6  23. r 
  10   3  v 
 3  
6  22. r 
 2(6) or 12  4  1  2  11  
cos 6  i sin 6 
 12
23  i
12  7 22    6 6 15  6 or 2  
 2  2i    5  1  2  4 3i  
 5  4  11  
 6  6 or 6  11  3  ) or 22  21. r 
 2(2  5  
 3  v 
 6  3  3.10  2.53i 
  4  3  11  v 
 6  3  20. r 
 15 or 3  v2 
 Arctan 
   2  1 3   
 6  v 
 2  3.6 or 5.6  4[cos (5.6)  i sin (5.6)]  23   v1 
 Arctan 
  2  2(cos   i sin ) 
 2(1  i(0)) 
 2 18. r 
 3(0.5) or 1.5 v 
 4  2.5 or 6.5 1.5(cos 6.5  i sin 6.5) 1.46  0.32i 4  1  r 
 4 or 2  or 2  19. r 
 1 or 4  5  12  
 3  33 i 7  
 4  
 4 or    3 2  6(cos 300°  i sin 300°) 
 6
  i
  5  3 2 32   v 
 4  4  v 
 240°  60° 
 300° 1  2  2   2  
 4  3  
 2  2i 16. r 
 2(3) or 6    12
cos 2  i sin 2 
 12(0  i(1))  
 6  6 or 6 11    
 4 or 2    14  3  v 
 4  4  v 
 3  2  15. r 
 1 or 3  3  
 4  
 4   5 v 
 3  6 2 5 
 6  6 3  
 6 or 2  18
cos 2  i sin 2 
 18(0  i(1)) 
 18i  3  7  
 52   52 i 14. r 
 6(3) or 18   v2 
 Arctan 
 3     2  296  E  28. I 
 Z  5  34. 13  5   
 3  2j  r1 
 13 13  13  r cos 
v  6 
 5  r 2 
  32  ( 2)2 
 13   r
cos v cos 6  sin v sin 6  5 
 0  v2 
 Arctan 
3  2x  2y  5 
 0  5  5   3  1  2r cos v  2r sin v  5 
 0  r 
  or 13    3  2  v1 
 0  35.  v 
 0  (0.59) or 0.59 I 
 13 (cos 0.59  j sin 0.59) 29. Z 
 
  3  2j amps  130˚  x lb  x lb 23 lb  x lb  23 lb  50˚  r 2 
  42  ( 3)2 
 25  or 5  r1 
 100  x lb  Prop  Since the triangle is isosceles, the base angles are  100  r 
 5 or 20  180  50  congruent. Each measures 2 or 65°.  3  v2 
 Arctan 
4  v1 
 0  v 
 0  (0.64) or 0.64 z 
 20(cos 0.64  j sin 0.64) 
 16  12j ohms 30. Start at z1 in the complex plane. Since the modulus z1 5 of z2 is 1, z1z2 and z will 6 2 both have the same modulus as z1. Then z1z2  z1 and z can be located by 2  rotating z1 by 6 7 6 counterclockwise and clockwise, respectively.  23  sin 50°  0.64  23 sin 65°  sin 50°  2 3  i  z1z2   2  z1 z2   3  36.  6 0  1 2 3 4    3 2      5  
x  27.21 x; 27.21 lb cos 2x  sin x 
 1 1  2 sin2 x  sin x 
 1 2 sin2 x  sin x 
 0 sin x (2 sin x  1) 
 0 sin x 
 0 or 2 sin x  1 
 0 x 
 0°  11 6 4 3  x   
 sin 65°  23 sin 65° 
 x sin 50°  1  sin x 
 2  x 
 30° y 
 cos x x 
 cos y arccos x 
 y 38. BC 
 ED 
 BE 
 AF 
 CD 
 3 AB 
 FE 
 2 AC 
 AB  BC 
 2  3 or 5 FD 
 FE  ED 
 2  3 or 5 perimeter of rectangle ACDF 
 3  5  3  5 or 16 perimeter of square BCDE 
 4(3) or 12 16  12 
 4 The correct choice is C.  5 3  37.  31a. The point is rotated counterclockwise about the origin by an angle of v. 31b. The point is rotated 60° counterclockwise about the origin. 32. Since a 
 1, the equation will be the form z2  bz  c 
 0. The coefficient c is the product of the 7 7 solutions, which is 6
cos 6  i sin 6, or 33   3i in rectangular form. The coefficient b is the opposite of the sum of the solutions, so convert the solutions to rectangular form to do the addition. 5  b 
 3
cos 3  i sin 3  2
cos 6  i sin 6 3 3  
 
2  2i  (3   i) 3  2 33  
 
2  3   
2i 3  Therefore, the equation is z2  
 2 33  2  
  1  x  y 10 
 0 3  0.59  E  I 100  4  3j  r 
 5 sec
v  6  3 2  iz  (33  3i) 
 0.  52  ( 12)2 33. r 
   9-8   3    Page 602  12  
 
 Arctan 
5  2  Powers and Roots of Complex Numbers Graphing Calculator Exploration  1. Rewrite 1 in polar form as 1(cos 0  i sin 0). Follow the keystrokes to find the roots at 1, 0.5  0.87i, and 0.5  0.87i.   or 13 5.11 
 169 5  12i 
 13(cos 5.11  i sin 5.11)      2. Rewrite i in polar form as 1
cos 2  i sin 2. Follow the keystrokes to find the roots at 0.92  0.38i, 0.38  0.92i, 0.92  0.38i, and 0.38  0.92i.  297  Chapter 9    3. Rewrite 1  i in polar form as 2 
cos 4  i sin    4  .  v 
 2  1  6  
cos 
16
2  i sin 
16
2    
 1
cos 1 2  i sin 12  1  Follow the keystrokes to find the roots at 1.06  0.17i, 0.17  1.06i, 0.95  0.49i, 0.76  0.76i, and 0.49  0.95i. 4. equilateral triangle 5. regular pentagon 6. If a  0 and b 
 0, then a  bi 
 a. The principal roots of a positive real number is a positive real number which would lie on the real axis in a complex plane.  Pages 604-605  0.97  0.26i  1  3   5    4  1  4  3  3  4n  3  4n  3  3  x1 
 cos 8  i sin 8 x2 
 cos x3 
 cos x4 
 cos  0.38  0.92i  7 7   i sin  8 8 11 11   i sin  8 8 15 15   i sin  8 8    4  i  1  0.92  0.38i 0.38  0.92i 0.92  0.38i  0.38  0.92i  0.92  0.38i 1  1  O  0.92  0.38i 0.38  0.92i 1  10. 2x3  4  2i 
 0 → x3 
 2  i Find the third roots of 2  i. (2)2   (1 )2 
 5  r 
   a a  ai  O  a    1   1  3  ) x2 
 (5    4  . By De Moivre's Theorem,   the polar form of (a  ai)2 is 2a2
cos 2  i sin 2.  i sin  ) x3 
 (5      23
cos (3)
6  i sin (3)
6  1 3       
 8 cos 
2  i sin 
2 
 
 8(0  i (1)) 
 8i  Chapter 9  v  2 v  2   i sin  
cos  3 3 v  4 v  4     cos  i sin 
 3 3   O 1 1.29  0.20i 1  5  32  ( 5)2 or 34  v 
 Arctan 
3 6. r 
  34 4 (cos (4)(
)  i sin (4)(v)) 
 644  960i  v  0.47  1.22i  1  v 
 Arctan 
 or 6  (3 )2  ( 1)2 or 2 5. r 
   1  3  v  i    Since cos 2 
 0, this is a pure imaginary number.    v  2n  x1 
 (5 ) 3 
cos 3  i sin 3 
 0.47  1.22i  4. Shembala is correct. The polar form of a  ai is a2 
cos  v  2n  
 (5 ) 3 
cos 3  i sin 3 1     4  1   v 
 Arctan 
 2     3.605240263 1 1   (2  i) 3 
 [5 (cos (v  2i)  i sin (v  2n))] 3  a  ai  a  1   3  
 cos 8  i sin 8  a  ai  a  3  v 
 2  (i) 
 1 (cos 
2  2n  i sin 
2  2n 4  
 4  4i 2. Finding a reciprocal is the same as raising a number to the 1 power, so take the reciprocal of the modulus and multiply the amplitude by 1.  i  
cos 
 (
)  i sin (3)(
)  02  ( 1)2 
 1 r 
   
cos (5)
   i sin (5)
  5 5 
 42 
cos 4  i sin 4   2 2 
 42  
2  i
2  a  ai  2.677945045  1  3  
 0.82  1.02i 9. x4  i 
 0 → x4 
 i Find the fourth roots of i.  Check for Understanding  (2  )5  1   v 
 Arctan 
 2     (2)2   (1 )2 or 5  8. r 
   1. Same results, 4  4i; answers may vary. (1  i)(1  i)(1  i)(1  i)(1  i) 
 (1  2i  i2)(1  2i  i2)(1  i) 
 (2i)(2i)(1  i) 
 4(1  i) 
 4  4i (1  i)5  → r 
 2 , v 
 4  3.    02  12 or 1 7. r 
   1.030376827  298  1    0.82  1.02i  1.29  0.20i 0.81  1.02i  11. For w1, the modulus 
 ( 0.82  (0.7 )2)2 or 1.13. For w2, the modulus 
 1.132 or 1.28. For w3, the modulus 
 1.282 or 1.64. This moduli will approach infinity as the number of iterations increases. Thus, it is an escape set.  21. r 
  (2)2   12 
 5    v 
 Arctan 
 2    1  2.677945045  ) ( 5  1  4  
cos 
 (v)  i sin 
 (v) 1  4  1  4  
 0.96  0.76i  1  v 
 Arctan 
4  42  ( 1)2 
 17  22. r 
   Pages 605–606 12.  33 
  Exercises   cos (3) 6  i sin   27 cos 2  i sin 2  
  
   
    6  (3)
    13.  
 
  ) (17  (22 )  
   
 
     
  
   
 162 
   i sin  2  2   i
  2  2  7    1      v 
 Arctan 
1 
 3  4  
 16 
  i
  3  2  
 8  83 i  
cos 
12
2  i sin 
12
2  2n  2n  
 cos 3  i sin 3    x1 
 cos 0  i sin 0 
 1 6  3  v 
 Arctan 
  32  ( 6)2 
 35  16. r 
     4  4  1   3  1  2  0.9827937232 (13 )2 (cos (2)(v)  i sin (2)(v)) 
 0.03  0.07i  2  O  1  v 
 Arctan 
2 4  1   1  
3 i 2  
 1.107148718 4 (25 ) (cos (4)(v)  i sin (4)(v)) 
 112  384i 2   3   1  
3 i  3  2  1  i  v 
 Arctan 
2  22  42 
 25  18. r 
   2  x3 
 cos 3  i sin 3 
 2  2i  (35 )4 (cos (4)(v)  i sin (4)(v)) 
 567  1944i 22  32 
 13  17. r 
   2  x2 
 cos 3  i sin 3 
 2  2i  1.107148718  1     v 
 2  
 0.71  0.71i 26. x3  1 
 0 → x3 
 1 Find the third roots of 1. r 1
  12  02 
 1 v
0 1   3 1 
 [1 (cos (0  2n)  i sin (0  2n))] 3  
 16 
cos 3  i sin 3 1 2  
cos 
14
34  i sin 
14
34  
 0.91  0.61i 1  2   3  24 
cos (4)
3  i sin (4)
3  1  4  02  12 
 1 25. r 
     12  ( 3  )2 
 2 15. r 
  4  
  ( 2 )  7  4  21  4  
 16  16i   1   v 
 Arctan 
 1    3  v 
 Arctan 
    
cos 
13
4  i sin 
13
4  
 4  7   2  2  2  
 1.37  0.37i  (22 )3
cos (3)
9  i sin (3)
4 
 162  
cos    v 
 Arctan 
2 
 4  (1)2   (1 )2 
  2  24. r 
   
 162   162 i 2 2  14. r 
  (2)  2 
 22  21  4  1  3  0.2449786631  1  3  22  22 
 22  23. r 
     cos (5) 4  i sin (5) 4 5 5 32 cos 4  i sin 4   2 2 32 2  i 2  
  
cos 
 (v)  i sin 
 (v) 1  3  
 1.60  0.13i  
 27(0  i(1)) 
 27i 25    1  3  2  1    19. 32 5 
cos 
5
3  i sin 
5
3 1  
 2
cos  2  15  1   i sin  2  15    1.83  0.81i 20. r 
  (1)2   02 
 1 1  4  1  v
  
cos 
 ()  i sin 
 () 1  4    1  4    
 cos 4  i sin 4 0.71  0.71i  299  Chapter 9  27. x5  1 → x5 
 1 Find the fifth roots of 1. r 
  (1)2   02 
 1  i  
 2  
 2i  v
  1  5  
 2  
 2i  1 1  5  (1) 
 [1 (cos (  2n)  i sin (  2n))]   2n    2n  
 cos 5  i sin 5 x1 
 cos x2 
 cos x3 
 cos x4 
 cos x5 
 cos  O  1      i sin  
 0.81  0.59i 5 5 3 3   i sin  
 0.31  0.95i 5 5 5 5   i sin  
 1 5 5 7 7   i sin  
 0.31  0.95i 5 5 9 9   i sin  
 0.81  0.59i 5 5    1  1 2i 
 2  
  2i 
 2  
  30. x4  (1  i) 
 0 → x4 
 1  i Find the fourth roots of 1  i. 1     v 
 Arctan 
1 
 4 1  r 
  12  12 
 2    1     (1  i) 4 
 2 
cos 
4  2n  i sin 
4  2n 4  i  1  0.31  0.95i  1     8n    8n  
 (2 ) 4 
cos 16  i sin 16  0.81  0.59i  i 1  O  1  1.07  0.21i 0.81  0.59i  0.31  0.95i 1  3  1  v
0  2i 2
    1   9  9  1  ) x3 
 (2 1  x4 
 (2 )  3  2  3  17    4  1   (1  3 i) 4  4  64 
 [64 (cos (0  2n)  i sin (0  2n))] n  n   i sin    2  1   x1 
 2  1   x2 
 2 4   (cos   i sin ) 
 22  x3 
 22  
  1   x3 
 2 4   
 22i  Find the fourth roots of 16. 2  02 
 16  r 
 (16) v
 1  4  1   1   i sin  4  28  0.59  1.03i  1.03  0.59i    2n     i sin  28  1  4  x1 
 2 
cos 4  i sin 4 
 2   2 i  i sin  4  
cos 12  i sin 12 
 0.59  1.03i 10 10  
cos 1 2  i sin 12  
 1.03  0.59i 22 22 
cos 12  i sin 12 
 0.59  1.03i i  
 2 
cos 4  i sin 4 3  4 5  4 7  4  4  6n  x4 
 2 4 
cos 12  i sin 12 
 1.03  0.59i  (16) 
 [16 (cos (  2n)  i sin (  2n))]   2n  4  6n  1  4   
 22i  3 3 x4 
 22  cos 2  i sin 2 3x4  48 
 0 → x4 
 16  1   
 2 4 
cos 12  i sin 12  x1 
 22  (cos 0  i sin 0) 
 22    2  4  
 2 
cos 
3  2n  i sin 
3  2n 4  
 22  
cos 2  i sin 2  3  4 5  4 7  4  17   
cos 1 6  i sin 16  
 1.07  0.21i 25 25 
cos 16  i sin 1 6  
 0.21  1.07i   3  1  4    1  4   
    or  v 
 Arctan 
 3 3 1   2i 2
  1  4  x2 
 22  
cos  1  4  i 
 0 → x4 
 1  3 i. 31. 2x4  2  23 Find the fourth roots of 1  3 i. r 
  (1)2   (  )2 
 2 3  2  Chapter 9    ) 4 
cos 16  i sin 16 
 0.21  1.07i x2 
 (2  3 2 1 O 1  x4 
 2 
cos  1   2  2
 2  x3 
 2 
cos  0.21  1.07i  x1 
 (2 ) 4 
cos 16  i sin 1 6  
 1.07  0.21i  i  2
 2  x2 
 2 
cos  1  O 1 1.07  0.21i  28. 2x4  128 
 0 → x4 
 64 Find the fourth roots of 64. r 
  642  02 
 64  29.  1  0.21  1.07i  1   
 2  2i  
 2  2i  
 2  2i  O  0.59  1.03i 1  300  1 1.03  0.59i  32. Rewrite 10  9i in polar form as 9  10   cos 
tan1 
 181  38a. The point at (2, 2) becomes the point at (0, 2). From the origin, the point at (2, 2) had a length of 22  and the new point at (0, 2) has a length 2  of 2. The dilation factor is 2.  9  10    i sin 
tan1 
 .  Use a graphing calculator to find the fifth roots at 0.75  1.51i, 1.20  1.18i, 1.49  0.78i, 0.28  1.66i, and 1.66  0.25i. 33. Rewrite 2  4i in polar form as 25 [cos (tan1 (2))  i sin (tan1 (2))]. Use a graphing calculator to find the sixth roots at 1.26  0.24i, 0.43  1.21i, 0.83  0.97i, 1.26  0.24i, 0.43  1.21i, and 0.83  0.97i. 34. Rewrite 36  20i in polar form as  2 y (0, 2) 1  x 2   2  2  
  3  4  38b.  For w2, the modulus 
 (0.81)2 or 0.66. For w3, the modulus 
 (0.66)2 or 0.44. This moduli will approach 0 as the number of iterations increase. Thus, it is a prisoner set. 36a. In polar form the 31st roots of 1 are given by 2n  x3 
 cos x4 
 cos x5 
 cos x6 
 cos   i sin  i sin  i sin  i sin  2  3 3  3 4  3 5  3  
  
  1  
 2(cos 90°  i sin 90°)    5    10  v 
 6  3 11   3  11  
 6  6 or 6 1  45°   3  1 2    
   3 i 2  
 1 
  2  43. cos 22.5° 
 cos 
2  x2 
 cos 3  i sin 3 
 2  2i 2  3 3  3 4  3 5  3    
 33   3i 41. (2  5i)  (3  6i)  (6  2i) 
 (2  (3)  (6))  (5i  6i  2i) 
5i 42. x 
 t, y 
 2t  7  n  1    22 (cos 45°  i sin 45°)  11  
 cos 3  i sin 3    2   i sin 
2  6
cos 6  i sin 6 
 6 
2  i 
2  1     
  40. r 
 2(3) or 6  1 6 
 [1 (cos (0  2n)  i sin (0  2n))] 6 x1 
 cos 0  i sin 0 
 1  (cos 45°  i sin 45°)   2  2   2 2  The square is rotated 90° counterclockwise and dilated by a factor of 0.5. 39. The roots are the vertices of a regular polygon. Since one of the roots must be a positive real number, a vertex of the polygon lies on the positive real axis and the polygon is symmetric about the real axis. This means that the non-real complex roots occur in conjugate pairs. Since the imaginary part of the sum of two complex conjugates is 0, the imaginary part of the sum of all the roots must be 0.   cos 3 1  i sin 31 , n 
 0, 1, . . . , 30. Then 2n   a 
 cos 31 . The maximum value of a cosine expression is 1, and it is achieved in this situation when n 
 0. 36b. From the polar form in the solution to part a, we 2n 2n  get b 
 sin 3 1 . b will be maximized when 31 is  as close to 2 as possible. This occurs when n 
 8, 16 so the maximum value of b is sin 3 1 , or about 0.9987. 37. x6  1 
 0 → x6 
 1 Find the sixth roots of 1. r 
  12  02 
 1 v
0 n  2  
 0.5  0.5i  2 
 
122  
 2 or 0.81.  1   1  2  5  35. For w1, the modulus 
  2n  1 O 1  4106  cos 
tan1 
9  i sin 
tan1 
9. Use a graphing calculator to find the eighth roots at 1.59  0.10i, 1.05  1.19i, 0.10  1.59i, 1.19  1.05i, 1.59  0.10i, 1.05  1.19i, 0.10  1.59i, and 1.19  1.05i. 5  (2, 2) 2  2  
   1 3 2  2i  1 3   i 2 2  
 
  1  cos 45°    2    2  1 2  2  2  2     4    2    2  2  44. Find B. B 
 180°  90°  81°15
 
 8°45
 Find a. a  tan 81°15
 
 2 8 28 tan 81°15
 
 a 181.9 
 a  301  Chapter 9  Find c.  15. Sample answer: (4, 585°), (4, 945°), (4, 45°), (4, 405°) (r, v  360k°) → (4, 225°  360(1)°) → (4, 585°) → (4, 225°  360(2)°) → (4, 945°) (r, v  (2k  1)180°) → (4, 225°  (1)180°) → (4, 45°) → (4, 225°  (1)180°) → (4, 405°)  90˚ 16. 17. 2  60 120 2  28  cos 81°15
 
 c 28   c
 cos 81°15
  c 
 184.1 45. Let x 
 the number of large bears produced. Let y 
 the number of small bears produced. x  300 y 1200 y  400 (300, 900) 1000 x  y  1200 800 f(x, y) 
 9x  5y (800, 400) 600 f(300, 400) 
 9(300)  5(400) 400 
 4700 200 (300, 400) x f(300, 900) 
 9(300)  5(900) O 200 600 1000 
 7200 f(800, 400) 
 9(800)  5(400) 
 9200 Producing 800 large bears and 400 small bears yields the maximum profit. 46. 0.20(6) 
 1.2 quarts of alcohol 0.60(4) 
 2.4 quarts of alcohol 1.2  2.4  64  ˚  ˚  3  180˚  1 2 3 4  240˚  270˚  18.  90˚  120˚  3.6  
 1 or 36% alcohol 0  19.  1. 3. 5. 7. 9.  Check for Understanding 2. 4. 6. 8. 10.  absolute value prisoner pure imaginary rectangular Argand  120˚  Polar iteration cardioid spiral of Archimedes modulus  11. 120˚  12.  180˚  0˚  1 2 3 4  13.  2 3  270˚  2  240˚  300˚  14.   3  5 6  C   6  1 2 3 4  7 6  11 6 4 3  Chapter 9  3 2  5 3   2  4 3  3 2  270˚  
22   
 32  (32 , 32 ) 25. x 
 2 cos 330°  
   5 3  
2  3   2  
 3  (3 , 1)  302  8 16 24 32  0  11 6  7 6  2 3  3 2  5 3   2   3  6  5 6   2 4 6 8  0  11 6  7 6 4 3  300˚  
6  11 6  0˚  330˚  limaçon 24. x 
 6 cos 45°  D  7 6    23.  2 4 6 8   3  0   3  6  4 3  30˚  240˚  1 2 3 4  5 3  5 6  0˚  60˚  210˚   6    90˚  300˚  5 6 0    2 3  270˚  3 2  2  Spiral of Archimedes  180˚  330˚  210˚  2 3  300˚  150˚  0˚  1 2 3 4  B  330˚ 240˚  30˚  180˚  A 210˚  120˚  11 6  21.  330˚  22.  60˚  150˚  30˚  150˚  7 6  60˚  210˚ 270˚  1 2 3 4  4 3  2 4 6 8   3  0    circle 90˚  120˚  60˚  0˚  30˚  180˚  Skills and Concepts  90˚  90˚   2  5 3   6  300˚  150˚  240˚  Pages 608–610  270˚  3 2  5 6  330˚  20.  Pages 607  2 3  60˚  210˚ 240˚  11 6 4 3  1 2 3 4  Chapter 9 Study Guide and Assessment  0 1 2 3 4  7 6  30˚  180˚   6    300˚  150˚  The correct choice is A.  0˚  330˚  210˚  3  5 6  30˚  150˚  3 2  rose y 
 6 sin 45°  
22   
6  
 32   y 
 2 sin 330°  
 1  
 2 2 
 1  5 3  
34 2  
 2
2  
34 2  
 2
2  26. x 
 2 cos  
 2  (2 , 2 )  27. x 
 1 cos 
      y 
 1 sin 
  v 
 Arctan 
    y  6 or 0 
 x  3 x  3 y  6 
 0  35. 4 
 r cos 
v  2  4  3  
    3   3    5    
 50  or 52   
 4  
52, 4  36.    v 
 Arctan 
 3  1  30. r 
  (3)2   12  3.16 
 10 (3.16, 2.82)  37.  2.82  38.  v 
 Arctan 
4 2  31. r 
  42  22 
 20  4.47 (4.47, 0.46)   A2   B2 
  0.46  39.   22  12  40.  
 5  Since C is positive, use 5 . 2 5   1 5  25  cos f 
 5, 1 f 
 Arctan 2  3 5     x   y   
 0  41.  5   35   sin f 
 5, p 
 5  5  2i  20  18i  4i2  16  42.   32  12  5i  1   2i  1   2i 1  2 i  5i 1   2i  
    
  5  52 i  i  2 i2  1  2i2  
  (5  2 )  (52   1)i  3 1  52   5  2   
 3  3i  v 
 Arctan 
2 2  22  22 43. r 
   4  10  x   y   
 0    
 8  or 22   10  22  
cos 4  i sin 4   2 10  sin f 
 1 , p 
 5 0  18° Since cos f 0 and sin f the third quadrant. f 
 180°  18° or 198° p 
 r cos (v  f)  18  
 29  29i  0, the normal lies in  
 10  Since C is positive, use 10 .  210   5  4  2i     
 5  2i 5  2i 16  18i  
 r cos (v  207°)  
   4  2i  5  2i  
 29  27° Since cos f 0 and sin f the third quadrant. f 
 180°  27° or 207° p 
 r cos (v  f)  1  10 310  cos f 
 10, 1 f 
 Arctan 3  0 
 0  r sin v  4 0 
 y  4 0 
 y  4 or y4
0 i10  i25 
 (i4)2  i2  (i4)6  i 
 (1)2  (1)  (1)6  i 
 1  i (2  3i)  (4  4i) 
 (2  (4))  (3i  (4i)) 
 2  7i (2  7i)  (3  i) 
 (2  (3))  (7i  (i)) 
 1  6i 3 3 i (4  3i) 
 4i  3i4 
 4(i)  3(1) 
 3  4i (i  7)(i  7) 
 i2  14i  49 
 1  14i  49 
 48  14i  
 25  4i2  
    A2   B2 
    0 
 r cos v cos 2  r sin v sin 2  4  v 
 Arctan 
5  29. r 
  52  52  3  10   3   0 
 2x  2 y  3     (3  )2  ( 3)2   or 23  
 12  33.  3   1  0 
 2r cos v  2 r sin v  3  
1  
23, 43  35   5    0 
 r cos v cos 3  r sin v sin 3  3 1    2  
0 (0, 1)  32.    
 2     2  28. r 
  34. 3 
 r cos 
v  3  y 
 2 sin  
 4    44. r 
  12  ( 3)2  v 
 Arctan 
1  2 3   
 10 10  (cos 5.03  i sin 5.03)  0, the normal lies in  45. r 
  (1)2   (3  )2    v 
 Arctan 
 1  3   2   or 2 
 4 2
cos 3  i sin 3 2  
 r cos (v  198°)  5.03  
 3  2  46. r 
  (6)2   (4 )2  4   v 
 Arctan 
 6      or 213  
 52 213  (cos 3.73  i sin 3.73) 47. r 
  (4)2   (1 )2  1   v 
 Arctan 
 4      
 17 17  (cos 3.39  i sin 3.39)  303  3.73  3.39  Chapter 9  48. r 
  42  02 
 16  or 4 4(cos 0  i sin 0) (2  )2   2 02 49. r 
  
 8  or 22  22 (cos   i sin )  v
0  02  32 50. r 
   v
  60. r 
 ( 3  )2   (1 )2 
 2 27 
cos  v
    2 3 5 6  
 128 
    2   6    2 cos 6  i sin 6 1 3  
 2 2  i 2  
  
  
   ( 3, 5
 3) 4 3    2 2 12 cos 3  i sin 3 1 3  
 12 2  i 2  
   3  
  5  11 6  
 4 5  15  2  
 16  16i 1  4  1    1      0.92  0.38i       2  1  3  3  3  1  1       1.24  0.22i  
 162   162 i 55. r 
 2(5) or 10 10 (cos 2.5  i sin 2.5) 8.01  5.98i  v 
 2  0.5 or 2.5  Page 611  Applications and Problem Solving  65. lemniscate  8  56. r 
 2 or 4  5  7  10      v
   3  
cos  i sin  3 1 3  
 2 
2  i 
2  
    2    3  6    r cos 
v  2  5 
 0   r cos 
 cos 2  r sin v sin 2  5 
 0    6   6  or  r sin v  5 
 0 y5
0 y 
 5    3 E  68. I 
 Z 50  180j  32    
 4  5j  
 4  4i 2.2  50  180j  0.5 (cos 0.9  i sin 0.9) 0.31  0.39i 59. r 
  22  22 
 22    
 
    v 
 Arctan 
2 
 4 2  4  5j    
 4  5j  4  5j  v 
 1.5  0.6 or 0.9   58. r 
  4.4 or 0.5  59.04°   67.  3  2  125  
 21,25 0  145.77 (145.77, 59.04°)    4cos 
2  i sin 
2 
 6  6 or 2 
 4(0  i(1)) 
 4i  v 
 Arctan 
7 5   752  1252 66. r 
   7  v 
 6  3    3    1  
 2 3 
cos 
18  i sin 
18  
 4  4 or 4  2       1  200  470j  900j2  16  25j2 1100  470j2  41  26.83  11.46j amps    (2 )8 
cos (8)
4  i sin (8)
4 
 4096 (cos 2  i sin 2) 
 4096  304     v 
 Arctan 
  3 
 6  2 
cos 
3
6  i sin 
3
6  v 
 4  2  2   Chapter 9    1 
cos 
4
2  i sin 
4
2  
 32
2  i 
2  3  v 
 2  (3 )2  12 
 2 64. r 
   3  or    02  12 
 1 63. r 
  
 cos 8  i sin 8  32
cos 4  i sin 4  3  2   2  
 162  
2  i 
2    54. r 
 8(4) or 32  57. r 
  15  
 162  
cos 4  i sin 4  
 6  63 i  6  4  5  (22 )3 
cos (3)
4  i sin (3)
4     
  2   v 
 Arctan 
 2     (2)2   (2 )2 
 22  62. r 
   5 3  3 2  
  3  
 4(cos 3  i sin 3) 
 4  5 5 3 cos 3  i sin 3 1 3  
 3 2  i 2 3 33  
 2  2i   2 v 
 3  3 or 3  53. r 
 4(3) or 12  
  1  (2 )4 
cos (4)
4  i sin (4)
4  1 2 3 4  7 6  
 3 i  
   v 
 Arctan 
 1    3  0  5 3  3 2  
       
 4   3  6    11 6 4 3  i 2  5 6  1 2 3 4   7 6    2 3  0    
  
 643   64i  52.  ( 2, 
6 )    (1)2   12 
 2  61. r 
    3  i 2  
  
 128 
  
 9  or 3   3
cos 2  i sin 2 51.   1  
  6 3    (7) 6  i sin (7) 6 7 7 cos 6  i sin 6  1 3 2  i 2  v 
 Arctan  Page 611  180  (c  b) 
 180  x x
cb The correct choice is E. 4. Volume 
 
wh 16,500 
 75  w  10 16,500 
 750w 22 
 w  Open-Ended Assessment  1a. Sample answer: 4  6i and 3  2i (4  6i)  (3  2i) 
 (4  3)  (6i  2i) 
 7  4i 1b. No. Sample explanation: 2  3i and 5  i also have this sum. (2  3i)  (5  i) 
 (2  5)  (3i  (i)) 
 7  4i 2a. Sample answer: 4  i z 
  42  12 
 17  2b. No. Sample explanation: 1  4i also has this absolute value. z 
  12  42 
 17   w 5.  8  h  C  The answer choices include sin x. Write an expression for the height, using the sine of x. h  1  sin x 
 8  A 
 2bh 1  8 sin x 
 h  
 2 (10)(8 sin x)  
 40 sin x The correct choice is B. 3. Since PQRS is a parallelogram, sides PQ and SR are parallel and m∠Q 
 m∠S 
 b.  M c˚  P  Q b˚  T x˚  R a˚  S  1  10  The correct choice is A. 6. Consider the three unmarked angles at the intersection point. One of these angles, say the top one, is the supplement of the other two unmarked angles, because of vertical angles. So the sum of the measures of the unmarked angles is 180°. The sum of the measures of the marked angles and the three unmarked angles is 3(180), since these angles are the interior angles of three triangles. m(sum of marked angles)  m(sum of unmarked angles) 
 3(180) m(sum of marked angles)  180 
 3(180) m(sum of marked angles) 
 360 The correct choice is C. 7. Subtract the second equation from the first. 5x2  6x 
 70 5x2  6y 
 10 6x  6y 
 60 x  y 
 10, so 10x  10y 
 100. The correct choice is E. 8. Since ∠B is a right angle, ∠C is a right angle also, because they are alternate interior angles. In the triangle containing ∠C, 90  x  y 
 180 or x  y 
 90. The straight angle at D is made up of 3 angles. 120  x  x 
 180 2x 
 60 or x 
 30 x  y 
 90 (30)  y 
 90 y 
 60 The correct choice is B. 9. In the slope-intercept form of a line, y 
 mx  b, m represents the slope of the line, and b represents the y-intercept. Since the slope is 3 given as 2, the slope-intercept form of the line is 3 y 
 2x  b. Since (–3, 0) is on the line, it satisfies the 3 9 equation. 0 
 2(–3)  b. So b 
 2. The correct choice is D. 10. Note that consecutive interior angles are supplementary. 110  2x 
 180 y  x 
 180 2x 
 70 y  (35) 
 180 x 
 35 y 
 145 The answer is 145.  B  10  1      1099 
 100100  10100 9  SAT and ACT Practice  x˚  1  100100   
 10100  1. ∠a and ∠b form a linear pair, so ∠b is supplementary to ∠a. Since ∠b and ∠d are vertical angles, they are equal in measure. So ∠d is also supplementary to ∠a. Since ∠d and ∠f are alternate interior angles, they are equal. So ∠f is supplementary to ∠a. And since ∠f and ∠h are vertical angles, ∠h is supplementary to ∠a. The angles supplementary to ∠a are angles b, d, f, and h. The correct choice is A. 2. Draw the given triangle and draw the height h from point B.  A  
 
 75 ft  The correct choice is A.  Chapter 9 SAT & ACT Preparation Page 613  h 
 10 ft  O N  In SMO, c  b  a 
 180 or a 
 180  (c  b). Also, x  a 
 180 or a 
 180  x since consecutive interior angles are supplementary.  305  Chapter 9  Chapter 10 Conics 10-1  8. AB 
  (x2   x1)2  (y2  y1)2  Introduction to Analytic Geometry  Pages 619–620  
  (6  3 )2  (2  4)2 
 13  slope of  AB   Check for Understanding  y2  y1  1. negative distances have no meaning 2. Use the distance formula to show that the measure of the distance from the midpoint to either endpoint is the same. a 5 3a. Yes; the distance from B to A is 2 and the a 5 distance from B to C is also 2. 3b. Yes; the distance from B to A is  a2  b2 and the distance from C to A is also  a2  b2. 3c. No; the distance from A to B is  a2  b2, the distance from A to C is a  b, and the distance from B to C is b2 . 4. (1) Show that two pairs of opposite sides are parallel by showing that slopes of the lines through each pair of opposite sides are equal. (2) Show that two pairs of opposite sides are congruent by showing that the distance between the vertices forming each pair of opposite sides are equal. (3) Show that one pair of opposite sides is parallel and congruent by showing that the slopes of the lines through that pair of sides are equal and that the distances between the endpoints of each pair of segments are equal. (4) Show that the diagonals bisect each other by showing that the midpoints of the diagonals coincide. 5. d 
  (x2   x1)2  (y2  y1)2 d 
  (5  5 )2  (1 1  1 )2 2 2 d 
  0  1 0 d 
  100 or 10 x1  x2 y1  y2    2 , 2  
  5  5 1  11 ,  2 2  

  
 (5, 6)   m
 x x 2  24  1  2    
 6  3 or  3  DC 
  (x2   x1)2  (y2  y1)2 
  (8  5 )2  (7  9)2 
 13  slope of  DC  y2  y1   m
 x x 2  79  1  2    
 8  5 or  3  Yes; A  and DC 
 13 , B D C  since AB 
 13 2 and A B D C  since the slope of A B  is 3 and the 2 slope of  DC  is also 3. (x2   x1)2  (y2  y1)2 9. XY 
  
  [1  (3)]2  ( 6  2 )2 
 68  or 217  XZ 
  (x2   x1)2  (y2  y1)2 
  [5  ( 3)]2   (0  2)2 
 68  or 217  Yes; X  and XZ 
 217 , Y X Z , since XY 
 217 therefore XYZ is isosceles. 10a. y D (c, a)  A(0, a) E B (0, 0)    O  C (c, 0)  10b. BD 
  (c  0 )2  (a  0)2 2 2 
  c  a  (x2   x2)2  (y2  y1)2 6. d 
  d 
  (4  0)2  (3  0)2 2 2 d 
  (4)   (3 ) d 
 25  or 5  AC 
  (c  0 )2  (0  a)2 2 2 
  c  a Thus, A C B D .  x1  x2 y1  y2  0  (4) 0  (3) ,  
2, 2 
 
 2 2  10c. The midpoint of  AC  is 
2, 2. The midpoint of c a B D  is 
2, 2. Therefore, the diagonals intersect c a at their common midpoint, E
2, 2. Thus, AE  E C  and B E E D . c  
 (2, 1.5)  [0  ( 2)]2   (4  2)2 7. d 
  2 2 d 
  2  2 d 
 8  or 22  x1  x2 y1  y2  2  0 2  4 ,  
2, 2 
 
 2 2  a  10d. The diagonals of a rectangle are congruent and bisect each other. 11a. Both players are located along a diagonal of the field with endpoints (0, 0) and (80, 120). The kicker's teammate is located at the midpoint of this diagonal.  
 (1, 3)  x1  x2 y1  y2  0  80 0  120   
2, 2 
 
 2 , 2  
 (40, 60)  Chapter 10  x  306  11b. d 
  (x2   x1)2  (y2  y1)2  19. d 
  (x2   x1)2  (y2  y1)2  d 
  (40  0)2  (60  0)2 2 2 d 
  40  60 d 
 5200  d 
 2013  or about 72 yards  Pages 620–622  d 
  (c  2  c)2  (d   1  d)2 2 2 d 
  2  ( 1) d 
 5  x1  x2 y1  y2  cc2 dd1 ,  
2, 2 
 
 2 2 2c  2 2d  1     

 2 , 2  1 
 
c  1, d  2  Exercises  12. d 
  (x2   x1)2  (y2  y1)2  20. d 
  (x2   x1)2  (y2  y1)2  d 
  [4  ( 1)]2   (13  1)2 d 
  52  1 22 d 
 169  or 13  d 
  [w  ( w  2 )]2  ( 4w   w)2 d 
  22  ( 3w)2 d 
  4  9 w2 or  9w2  4  x1  x2 y1  y2  1  4 1  13 ,  
2, 2 
 
 2 2  x1  x2 y1  y2  w  2  w w  4w ,  
2, 2 
 
 2 2 5 
 
w  1, 2w  
 (1.5, 7)  (x2   x1)2  (y2  y1)2 13. d 
  d 
  (1  1)2  (3  3)2 2 2 d 
  (2)   (6 ) d 
 40  or 210   21.  d 
  (x2   x1)2  (y2  y1)2  20 
 (2a   a)2   [7  (9)]2 2 2 20 
 (3a)   16 20 
  9a2  256 400 
 9a2  256 144 
 9a2 a2 
 16 a 
 16  or 4 22. Let D have coordinates (x2, y2).  x1  x2 y1  y2  1  (1) 3  (3) ,  
2, 2 
 
 2 2  
 (0, 0)  (x2   x1)2  (y2  y1)2 14. d 
  d 
  (0  8 )2  (8  0)2 d 
  (8)2   82 d 
 128  or 82   4  x2 1  y2  
2, 2 
 
3, 52  x1  x2 y1  y2  80 80   
2, 2 
 
 2 , 2   4  x2  2  
 (4, 4)  
 3  1  y2  2  5  
 2  4  x2 
 6 1  y2 
 5 x2 
 10 y2 
 6 Then D has coordinates (10, 6). y 23. Let the vertices A(2, 3) of the quadrilateral D (3, 2) be A(2, 3), B(2, 3), C(2, 3), and D(3, 2). x O A quadrilateral is a parallelogram if one pair B (3, 2) of opposite sides are C (2, 3) parallel and congruent. AD   and B C  are one pair of opposite sides. slope of  AD slope of  BC    (x2   x1)2  (y2  y1)2 15. d 
  d 
  [5  ( 1)]2   [3  ( 6)]2 2 2 d 
  6  3 d 
 45  or 35  x1  x2 y1  y2  1  5 6  (3)  ,  
2, 2 
 
 2 2  
 (2, 4.5)  (x2   x1)2  (y2  y1)2 16. d 
  d 
  (72   3  )2   2 [1  (5)]2 2  42 d 
  (42 ) d 
 48  or 43  x1  x2 y1  y2    72  5  (1) 32 ,  
2, 2 
 
   2 2  y2  y1   m
 x x  
 (52 , 3)  2  1  23  (x2   x1)2  (y2  y1)2 17. d 
    
 3  (2)  d 
  (a  a )2  ( 9  7 )2 2 2 d 
  0  ( 16) d 
 256  or 16  1  
 5  y2  y1   m
 x x 2  1  3  (2)   
 2  (3) 1  
 5  Their slopes are equal, therefore A D B C . AD 
  (x2   x1)2  (y2  y1)2  x1  x2 y1  y2  a  a 7  (9)   
2, 2 
 
 2 , 2  
  [3  ( 2)]2   (2  3)2 2 2 
  5  ( 1) 
 26  BC 
  (x2   x1)2  (y2  y1)2  
 (a, 1)  (x2   x1)2  (y2  y1)2 18. d 
  d 
  [r  2  (6   r)]2  (s  s)2 2 2 d 
  (8)  0 d 
 64  or 8  
  [2  ( 3)]2   [3  ( 2)]2 2 2 
  5  ( 1) 
 26  The measures of  AD  and B C  are equal. Therefore A BC D B C . Since A D B C  and A D  , quadrilateral ABCD is a parallelogram; yes.  x1  x2 y1  y2  6rr2 ss ,  
2, 2 
 
 2 2 2r  4 2s     

 2 , 2  
 (r  2, s)  307  Chapter 10  24. Let the vertices of the quadrilateral be A(4, 11), B(8, 14), C(4, 19), and D(0, 15). A quadrilateral is a parallelogram if both pairs of opposite sides are parallel. A B  and D C  are one pair of opposite sides. slope of  AB  y2  y1  m
 x x 2  20  y C (4, 19)  16 D (0, 15) 12  1  E FH  G  since the slope of E F  is 2 and the slope 1   of  HG  is  2 . Thus the points form a parallelogram.  EF ⊥F G  since the product of the 1 2 FG slopes of E F  and  , 2  1, is 1. Therefore, the points form a rectangle. 28. Let A(0, 0), B(b, c), and C(a, 0) be the vertices of a triangle. Let D be the midpoint of  AB  and E be the midpoint of B C .  B (8, 14) A (4, 11)  8 4  x 4 O  4  8  12  y  slope of  DC  y2  y1  m
 x x  1  2  14  11  B (b, c)  1  19  15   
 84   
 40  3  D  4  
 2  DC Since  AB  ⁄  , quadrilateral ABCD is not a parallelogram; no. 25. The slope of the line through (15, 1) and (3, 8) should be equal to the slope of the line through (3, 8) and (3, k) since all three points lie on the same line. slope through (15, 1) slope through (3, 8) and (3, 8) and (3, k) y2  y1  8  1  
 3  15 9 1   
 18 or 2 k8 1  
  ⇒ k 6 2  2  
 
  DE 
  1  k  (8)  3  (3) k8  6  1  
 5  y D (b  a, c)  
  (4  2 )2  (4  5)2 
 5  HG 
  (x2   x1)2  (y2  y1)2  O A(0, 0)  
  (2  0 )2  (0  1)2 
 5  slope of  EF  y2  y1 1  1    
 4  2 or  2  slope of  FG  y2  y1   m
 x x 1  2    
 2  4 or 1  slope of  HG  y2  y1   m
 x x 1  1    
 2  0 or  2  E FH  and HG 
 5 .  G  since EF 
 5 Chapter 10  C (a, c)  B (b, 0)  x  AC 
  (a  0 )2  (c  0)2 2 2 
  a  c BD 
  (b  a  b)2  (c  0)2 
  a2  c2 AC 
  a2  c2 
  a2  c2 
 BD, so the diagonals of an isosceles trapezoid are congruent.   m
 x x  2  a  Since DE 
 2AC, the line segment joining the midpoints of two sides of a triangle is equal in length to one-half the third side. 29. In trapezoid ABCD, let A and B have coordinates (0, 0) and (b, 0), respectively. To make the trapezoid isosceles, let C have coordinates (b  a, c) and let D have coordinates (a, c).  Yes, AB 
 4, BC 
 4, and CA 
 4. Thus, A B B C   CA. Therefore the points A, B, and C form an equilateral triangle. 27. EF 
  (x2   x1)2  (y2  y1)2  01  c  ba b 2 c c 2       
2    
 2  2 2 a2  (3  1)2  (0   0)2 CA 
  
 16  or 4  2  ba   or  
  4 2  [1  ( 1)]2   (0  23 )2 BC 
  
 16  or 4  04  ba c0  c  (a  0 )2  (0  0)2 AC 
  2 
  a or a  2  (2 AB 
  [1  (3)] 3   0)2 
 16  or 4  2  b  The coordinates of E are 
2, 2 or 
2, 2.  (x2   x1)2  (y2  y1)2 26. d 
   45  0b 0c  The coordinates of D are 
2, 2 or 
2, 2.  y2  y1  1  C (a, 0) x  O A (0, 0)   m
 x x   m
 x x 2  E  
 4 or 1  308  30. In ABC, let the vertices be A(0, 0) and B(a, 0). Since  AC  and B C  are congruent sides, let the third a vertex be C
2, b. Let D be the midpoint of  AC  and let E be the midpoint of  BC .  32. Let the vertices of quadrilateral ABCD be A(a, e), B(b, f), C(c, g), and D(d, h). The midpoints of  AB , B CD C ,  , and D A , respectively, cd gh  ad eh  N
2, 2, and P
2, 2.  C ( a2, b)  y  bc fg  ab ef  are L
2, 2, M
2, 2,  y D  B (b, f )  L  A(a, e) E  O  O A(0, 0)  D (d, h)  x  B (a, 0)  The coordinates of D are:  
   or 
  The coordinates of E are:  
   or 
  a  2  0 b0  ,  2 2 a  2  a b0  ,  2 2  0  0 
   b2  
  
   BD 
  b 1 9a   
   b2  
a  a 
0   4   2 2  4   2  2  b  2  2  .  3a b ,  4 2  2  .  x  C (c, g) fg ef     2 2  bc ab    2 2 gh eh    2 2  ad cd     2 2  ge   or  c  a.  eg   or  a  c.  These slopes are equal, so  LM N P . fg gh    2 2 The slope of M N  is  bc cd    2 2 ef eh    2 2 The slope of P L  is  ad ab     2 2  2  Since AE 
 BD, the medians to the congruent sides of an isosceles triangle are congruent. 31. Let A and B have coordinates (0, 0) and (b, 0) respectively. To make a parallelogram, let C have coordinates (b  a, c) and let D have coordinates (a, c). y D (a, c)  N  The slope of N P  is  9a2  4  1  2  M  The slope of L M  is  a b ,  4 2  AE 
  3a  4  O  P  f h   or  bd .  hf   or  d  b.  These slopes are equal, so  MN P L . Since L M N P , and M N P L , PLMN is a parallelogram. 33. Let the vertices of the y rectangle be A(3, 1), B B(1, 3), C(3, 1), and D(1, 3). Since the area A of a rectangle is the length x O times the width, find the  C (a  b, c)  C  measure of two consecutive sides,  AD  and D C .  O A(0, 0)  B (b, 0)  (x2   x1)2  (y2  y1)2 AD 
   x  ab 0c ,  or 2 2 ab0 c0 ,  2 2  The midpoint of  BD  is 
  D    
  [1  ( 3)]2   (3  1)2 2 2 
  4  ( 4) 
 32  or 42  DC 
  (x2   x1)2  (y2  y1)2  ab c ,  . 2 2 ab c    or 2 , 2  
    The midpoint of  AC  is 
  
 . Since the diagonals have the same midpoint, the diagonals bisect each other.  
  (3  1 )2  [ 1  ( 3)]2 2 2 
  2  2 
 8  or 22  Area 
 
w 
 (42 )(22 ) 
 16 The area of the rectangle is 16 units2. 34a. y 20  10 20 10 O  10  20  30  x  10  309  Chapter 10  37a. Find a representation for MA and for MB. MA 
  t2  ( 3t  1 5)2 
  t2  9 t2  9 0t  2 25 
  10t2  90t  225  34b. The two regions are closest between (12, 12) and (31,0). d 
  (x2   x1)2  (y2  y1)2 
  [31  (12 )]2  ( 0  1 2)2 2 2 
  43  (12 ) 
 1993  or about 44,64 The distance between these two points is about 44.64 pixels, which is greater than 40 pixels. therefore, the regions meet the criteria. 35. Let the vertices of the isosceles trapezoid have the coordinates A(0, 0), B(2a, 0), C(2a  2c, 2b), D(2c, 2b). The coordinates of the midpoints are: P(a, 0), Q(2a  c, b), R(a, 2b), S(c, b). y D (2c, 2b)  R  MB 
  (t  9 )2  (3 t  12 )2 2 2 
  t  1 8t  8 1  9 t  7 2t  144  
  10t2  90t  225 By setting these representations equal to each other, you find a value for t that would make the two distances equal. MA 
 MB 2  10t  90t  225 
  10t2  90t  225 Since the above equation is a true statement, t can take on any real values. 37b. A line; this line is the perpendicular bisector of A B . b a2  b2 v 
 Arctan a 38. r 
   C (2a  2c, 2b)  12  
  (5)2   122 S  O A (0, 0)  Q  P   
 Arctan  5  
 169  or 13 1.176005207 (5  12i)2 
 132 [cos 2v  i sin 2v] 
 119  120i u 39. If v 
 (115, 2018, 0), then u 
  v 1152  20182  02 
 40855 49  or about 2021 The magnitude of the force is about 2021 N.  B (2a, 0) x  PQ 
  (2a  c  a )2  (b  0)2 2 2 
  (a  c )  b QR 
  (2a  c  a )2  (b  2b )2 
  (a  c )3  b2 RS 
  (a  c )2  (2 b  b )2 2 2 
  (a  c )  b PS 
  (a  c )2  (0  b)2 
  (a  c )2  b2 So, all of the sides are congruent and quadrilateral PQRS is a rhombus. 36a. distance from fountain to rosebushes: d 
  (x2   x1)2  (y2  y1)2  1  1    40. 2 sec2 x 
  1  sin x  1  sin x  2 sec2 x 
 2 sec2 x 
 2 sec2 x 
 2 sec2 x 
 41. s 
 rv 11.5 
 12v  (1  sin x)  (1  sin x)  (1  sin x)(1  sin x) 2  1  sin2 x 2  
 cos2 x 2 sec2 x  11.5  v 
 1 2 radians 11.5  12  180°    54.9° 42. sin 390° 
 sin (390°  360°) 1 
 sin 30° or 2 43. z2  8z 
 14 2 z  8z  16 
 14  16 (z  4)2 
 2 z  4 
 2  z 
 4 2  44. x2 
 16 x 
 16  or 4 y2 
 4 y 
 4  or 2 Evaluating (x  y)2 when x 
 4 and y 
 2 results in the greatest possible value, [4  (2)]2 or 36.  d 
  [1  ( 3)]2   (3  2)2 d 
 41  or 241  meters distance from rosebushes to bench: d 
  (x2   x1)2  (y2  y1)2 d 
  (3  1 )2  [3  ( 3)]2 d 
 210  or 410  meters distance from bench to fountain: d 
  (x2   x1)2  (y2  y1)2 d 
  (3  3)2  (2   3)2 d 
 37  or 237  meters Yes; the distance from the fountain to the rosebushes is 241  or about 12.81 meters. The distance from the rosebushes to the bench is 410  or about 12.65 meters. The distance from the bench to the fountain is 237  or about 12.17 meters. 36b. The fountain is located at (3, 2) and the rosebushes are located at (1, 3).  10-2  Circles  x1  x2 y1  y2  3  1 2  (3)  ,  
2, 2 
 
 2 2 1 
 
1, 2  Chapter 10  Page 627  Check for Understanding  1. complete the square on each variable  310  2. Sample answer: (x  4)2  (y  9)2 
 1, (x  4)2  (y  9)2 
 2, (x  4)2  (y  9)2 
 3, (x  4)2  (y  9)2 
 4, (x  4)2  (y  9)2 
 5 3. Find the center of the circle, (h, k), by finding the midpoint of the diameter. Next find the radius of the circle, r, by finding the distance from the center to one endpoint. Then write the equation of the circle in standard form as (x  h)2  (y  k)2 
 r2. 4. The equation x2  y2  8x  8y  36 
 0 written in standard form is (x  4)2  (y  4)2 
 4. Since a circle cannot have a negative radius, the graph of the equation is the empty set. 5. Ramon; the square root of a sum does not equal the sum of the square roots. 6. (x  h)2  (y  k)2 
 r2 (x  0)2  (y  0)2 
 92 x2  y2 
 81  9.  y  (5  23, 2)  x2  y2  Dx  Ey  F 
 0 02  02  D(0)  E(0)  F 
 0 42  02  D(4)  E(0)  F 
 0 02  42  D(0)  E(4)  F 
 0 F
0 F
0 4D  F 
 16 ⇒ D 
 4 4E  F 
 16 E 
 4 x2  y2  4x  4y 
 0 x2  4x  4  y2  4y  4 
 0  4  4 (x  2)2  (y  2)2 
 8 center: (h, k) 
 (2, 2) radius: r2 
 8 r 
 8  or 22  11. x2  y2  Dx  Ey  F 
 0 12  32  D(1)  E(3)  F 
 0 ⇒ D  3E  F 
 10 52  52  D(5)  E(5)  F 
 0 ⇒ 5D  5E  F 
 50 52  32  D(5)  E(3)  F 
 0 ⇒ 5D  3E  F 
 34 D  3E  F 
 10 (1)(5D  5E  F) 
 (1)(50) 4D  2E 
 40 5D  5E  F 
 50 (1)(5D  3E  F) 
 (1)(34) 2E 
 16 E 
 8 4D  2(8) 
 40 4D 
 24 (6)  3(8)  F 
 10 D 
 6 F 
 20 x2  y2  6x  8y  20 
 0 x2  6x  9  y2  8y  16 
 20  9  16 (x  3)2  (y  4)2 
 5 center: (h, k) 
 (3, 4) radius: r2 
 5 r 
 5  12. (x  h)2  (y  k)2 
 r2 [x  (2)]2  (y  1)2 
 r2 (x  2)2  (y  1)2 
 r2 (1  2)2  (5  1)2 
 r2 25 
 r2 (x  2)2  (y  1)2 
 25  10.  x  (0, 9)  (x  h)2  (y  k)2 
 r2 [x  (1)]2  (y  4)2 
 [3  (1)]2 (x  1)2  (y  4)2 
 16 (1, 8) y  (1, 4)  (3, 4)  x  O 8.  x2  y2  4x  14y  47 
 0 x2  4x  4  y2  14y  49 
 47  4  49 (x  2)2  (y  7)2 
 100  y  (2, 3)  x  O (8, 7)  x (5, 2)  (9, 0)  7.  (5,2  23)  O  y  O  2x2  2y2  20x  8y  34 
 0 2x2  20x  2y2  8y 
 34 2(x2  10x  25)  2(y3  4y  4) 
 34  2(25)  2(4) 2(x  5)2  2(y  2)2 
 24 (x  5)2  (y  2)2 
 12  (2, 7)  311  Chapter 10  13. midpoint of diameter: x1  x2 y1  y2    2 , 2  
  18.  2  10 6  (10) ,  2 2  

    
 (4, 2)  (x  h)2  (y  k)2 
 r2 9 2 [x  (5)]2  (y  0)2 
 
2 81  (x  5)2  y2 
 4  (x2   x1)2  (y2  y1)2 radius: r 
   y  
  [4  ( 2)]2   [6  (2)]2 2 2 
  6  8 
 100  or 10 (x  h)2  (y  k)2 
 r2 (x  4)2  [y  (2)]2 
 102 (x  4)2  (y  2)2 
 100 14. (x  h)2  (y  k)2 
 r2 (x  0)2  (y  0)2 
 (1740  185)2 x2  y2 
 19252  Pages 627–630  (  5, 9  2  )  (5, 0)  (  19 , 2  0  x  O  )  19. (x  h)2  (y  k)2 
 r2 (x  6)2  (y  1)2 
 62 (x  6)2  (y  1)2 
 36  y  Exercises  15. (x  h)2  (y  k)2 
 r2 (x  0)2  (y  0)2 
 52 x2  y2 
 25  (6, 7)  y (0, 5)  (6, 1)  (0, 1)  x  O  (5, 0)  x  O  20.  16.  (x  h)2  (y  k)2 
 r2 [x  (4)]2  (y  7)2 
 (3  )2 (x  4)2  (y  7)2 
 3  y  x (3, 2)  (4  
 3, 7)  (7, 2)  (4, 7 
3)  17.  (3, 2)  O  y (4, 7)  O  (x  h)2  (y  k)2 
 r2 (x  3)2  [y  (2)]2 
 [2  (2)]2 (x  3)2  (y  2)2 
 16  y  21. 36  x2 
 y2 x2  y2 
 36  x  (0, 6)  (x  h)2  (y  k)2 
 r2 2  2 [x  (1)]2  [y  (3)]2 
 
  2  (x  1)2  (y   y O  (6, 0)  1 3)2 
2  x 22.  (1, 6 2
2 )  2  
2 , 2  (  y  3  x2  y2  y 
 4 1  3  1 2  2  
1  1  x2  y2  y  4 
 4  4  3 )  x2  
y   (1, 3)  Chapter 10  x  O  312    (0, 12 ) (0,  12 )  O  x  (1,  12 )  23.  x2  y2  4x  12y  30 
 0 x2  4x  4  y2  12y  36 
 30  4  36 (x  2)2  (y  6)2 
 10  y O  27.  x2  y2  14x  24y  157 
 0 x2  14x  49  y2  24y  144 
 157  49  144 (x  7)2  (y  12)2 
 36  y O  x (2, 6  
 10 )  (7, 6) (1, 12)  (2, 6) (7, 12)  (2  
 10, 6)  x2  y2  Dx  Ey  F 
 0 02  (1)2  D(0)  E(1)  F 
 0 ⇒ E  F 
 1 (3)2  (2)2  D(3)  E(2)  F 
 0 ⇒ 3D  2E  F 
 13 (6)2  (1)2  D(6)  E(1)  F 
 0 ⇒ 6D  E  F 
 37  E  F 
 1 (1)(3D  2E  F) 
 (1)(13) 3D  E 
 12 3D  2E  F 
 13 (1)(6D  2E  F) 
 (1)(37) 3D  2E 
 24 3D  E 
 12 3(6)  E 
 12 3D  E 
 24 E 
 6 6D 
 36 D
6 (6)  F 
 1 F 
 7 x2  y2  6x  6y  7 
 0 x2  6x  9  y2  6y  9 
 7  9  9 (x  3)2  (y  3)2 
 25 center: (h, k) 
 (3, 3) radius: r2 
 25 r 
 25  or 5 29. x2  y2  Dx  Ey  F 
 0 72  (1)2  D(7)  E(1)  F 
 0 ⇒ 7D  E  F 
 50 112  (5)2  D(11)  E(5)  F 
 0 ⇒ 11D  5E  F 
 146 32  (5)2  D(3)  E(5)  F 
 0 ⇒ 3D  5E  F 
 34 7D  E  F 
 50 (1)(11D  5E  F) 
 1(146) 4D  4E 
 96 11D  5E  F 
 146 (1)(3D  5E  F) 
 1(34) 8D 
 12 D 
 14 4(14)  4E 
 96 4E 
 40 7(14)  (10)  F 
 50 E 
 10 F 
 58 x2  y2  14x  10y  58 
 0 x2  14x  49  y2  10y  25 
 58  49  25 (x  7)2  (y  5)2 
 16 center: (h, k) 
 (7, 5) radius: r2 
 16 r 
 16  or 4  2x2  2y2  2x  4y 
 1 2x2  2x  2y2  4y 
 1  24.  28.  2
x2  1x  4  2(y2  2y  1) 
 1  2
4  2(1) 1  1  1 2 3 x  2  2(y  1)2 
 2 1 2 3 x  2  (y  1)2 
 4  2
    
    y  (1 2 
3, 1) ( 12 , 1) 3 O ( 12 , 1  
 ) 2  25. 6(x2  x  6x2  12x  6y2  36y 
 36  2x  1)  6(y2  6y  9) 
 36  6(1)  6(9) 6(x  1)2  6(y  3)2 
 96 (x  1)2  (y  3)2 
 16  y O  x (1, 3)  (3, 3) (1, 7)  16x2  16y2  8x  32y 
 127 16x2  8x  16y2  32y 
 127  26.  2 2  16
x2  2x  1 6   16(y  2y  1) 
 127  16
 16   16(1) 1  1  1  1 2  16
x  4  16(y  1)2 
 144 2  
x  14  y  x   (y  1)2 
 9  ( 134, 1)  ( 14 , 1) O  x  ( 14 , 2)  313  Chapter 10  x2  y2  Dx  Ey  F 
 0 (2)2  72  D(2)  E(7)  F 
 0 ⇒ 2D  7E  F 
 53 (9)2  02  D(9)  E(0)  F 
 0 ⇒ 9D  F 
 81 (10)2  (5)2  D(10)  E(5)  F 
 10 ⇒ 10D  5E  F 
 125 2D  7E  F 
 53 (1)(9D  F) 
 (1)(81) 7D  7E 
 28 DE
4 10D  5E  F 
 125 (1)(9D  F) 
 (1)(81) D  5E 
 44 D  5E 
 44 D E
4 4E 
 40 E 
 10 D  (10) 
 4 D 
 6 9(6)  F 
 81 F 
 135 x2  y2  6x  10y  135 
 10 x2  6x  9  y2  10y  25 
 135  9  25 (x  3)2  (y  5)2 
 169 center: (h, k) 
 (3, 5) radius: r2 
 169 r 
 169  or 13 31. x2  y2  Dx  Ey  F 
 0 (2)2  32  D(2)  E(3)  F 
 0 ⇒ 2D  3E  F 
 13 62  (5)2  D(6)  E(5)  F 
 0 ⇒ 6D  5E  F 
 61 02  72  D(0)  E(7)  F 
 0 ⇒ 7E  F 
 49 2D  3E  F 
 13 (1)(6D  5E  F) 
 (1)(61) 8D  8E 
 48 D  E 
 6 6D  5E  F 
 61 (1)(7E  F) 
 (1)(49) 6D  12E 
 12 D  2E 
 2 D  E 
 6 D  2E 
 2 E 
 4 E 
 4 D  2(4) 
 2 D 
 10 7(4)  F 
 49 F 
 21 x2  y2  10x  4y  21 
 0 x2  10x  25  y2  4y  4 
 21  25  4 (x  5)2  (y  2)2 
 50 center: (h, k) 
 (5, 2) radius: r2 
 50 r 
 50  or 52   x2  y2  Dx  Ey  F 
 0 42  52  D(4)  E(5)  F 
 0 ⇒ 4D  5E  F 
 41 (2)2  32  D(2)  E(3)  F 
 0 ⇒ 2D  3E  F 
 13 (4)2  (3)2  D(4)  E(3)  F 
 0 ⇒ 4D  3E  F 
 25 4D  5E  F 
 41 (1)(2D  3E  F) 
 (1)(13) 6D  2E 
 28 3D  E 
 14 4D  5E  F 
 41 (1)(4D  3E  F) 
 (1)(25) 8D  8E 
 16 D  E 
 2 3D  E 
 14 (1)(D  E) 
 (1)(2) 2D 
 12 D 
 6 6  E 
 2 E
4 2(6)  3(4)  F 
 13 F 
 37 x2  y2  6x  4y  37 
 0 x2  6x  9  y2  4y  4 
 37  9  4 (x  3)2  (y  2)2 
 50 center: (h, k) 
 (3, 2) radius: r2 
 50 r 
 50  or 52  33. x2  y2  Dx  Ey  F 
 0 12  42  D(1)  E(4)  F 
 0 ⇒ D  4E  F 
 17 22  (1)2  D(2)  E(1)  F 
 0 ⇒ 2D  E  F 
 5 (3)2  02  D(3)  E(0)  F 
 0 ⇒ 3D  F 
 9 D  4E  F 
 17 (1)(2D  E  F) 
 (1)(5) D  5E 
 12 D  4E  F 
 17 (1)(3D  F) 
 (1)(9) 4D  4E 
 8 D  E 
 2 D  5E 
 12 D  E 
 2 6E 
 14 7 E 
 3  30.  Chapter 10  32.  D  
3 
 2 7  1  D 
 3 3
3  F 
 9 1  F 
 8 x2  y2  Dx  Ey  F 
 0 1 7 x2  y2  3x  3y  8 
 0 1  7  1  49  1  49  2     x2  3x  3 6  y  3 y  36 
 8  36  36 2  
x  16  314  7 2   
y  6 
 1 8 169  39. (x  h)2  (y  k)2 
 r2 (x  5)2  (y  1)2 
 r2 x  3y 
 2 x  3y  2 
 0 ⇒ A 
 1, B 
 3, and C 
 2 Ax1  By1  C  r
  A2   B2   center: (h, k) 
 
6, 6 1 7  169  radius: r2 
 1 8  r 
  18 169  13  132    or  
 6 32   x2  y2  Dx  Ey  F 
 0 02  02  D(0)  E(0)  F 
 0 ⇒  34.  (1)(5)  (3)(1)  2  
     12  32  F
0 (2.8)2  02  D(2.8)  E(0)  F 
 0 ⇒ 2.8D  F 
 7.84 (5)2  22  D(5)  E(2)  F 
 0 ⇒ 5D  2E  F 
 29 2.8D  0 
 7.84 5(2.8)  2E  (0) 
 29 2.8D 
 7.84 2E 
 15 D 
 2.8 E 
 7.5 x2  y2  2.8x  7.5y  0 
 0 x2  2.8x  1.96  y2  7.5y  14.0625 
 1.96  14.0625 (x  1.4)2  (y  3.75)2 
 16.0225 or about (x  1.4)2  (y  3.75)2 
 16.02 35. (x  h)2  (y  k)2 
 r2 [x  (4)]2  (y  3)2 
 r2 (x  4)2  (y  3)2 
 r2 (0  4)2  (0  3)2 
 r2 25 
 r2 (x  4)2  (y  3)2 
 25 36. (x  h)2  (y  k)2 
 r2 (x  2)2  (y  3)2 
 r2 (5  2)2  (6  3)2 
 r2 18 
 r2 (x  2)2  (y  3)2 
 18 37. midpoint of diameter:   or 10 
    10 10  2    (x  5)2  (y  1)2 
 10 40. center: (h, 0), radius: r 
 1 (x  h)2  (y  k)2 
 r2 2  
22  h 1  2   
2  0 
 12 2   1  2  2 h  h2  2 
 1 h 
 1  1 h2  2  
     
0 h h  2  h 
 0 or h 
 2 2 2 (x  0)  (y  0) 
 1 x  2   
  x2  y2 
 1  or  41a. (x  h)2  (y  k)2 
 r2  2     (y  0)2 
 1 2  
x  2   y2 
 1  12 2  (x  0)2  (y  0)2 
 
2  x2  y2 
 36 41b. x2  y2 
 36 y2 
 36  x2 y 
  36   x2 dimensions of rectangle: 36   x2 2x by 2y ⇒ 2x by 2  x1  x2 y1  y2  2  (6) 3  (5) ,  
2, 2 
 
 2 2  
  41c. A(x) 
 2x 2 36   x2  
 (2, 1) r 
  (x2   x1)2  (y2  y1)2  
 4x 36   x2  
  (2  2)2  (1  3)2 2 2 
  (4)   (4 ) 
 32  (x  h)2  y  k)2 
 r2 2 [x  (2)]2  [y  (1)]2 
 
32  2 2 (x  2)  (y  1) 
 32 38. midpoint of diameter: x1  x2 y1  y2  
  (x  5)2  (y  1)2 
 10     41d.  3  2 4  1  
2, 2 
 
2, 2 1 5 
 
2, 2  [0, 10] scl:1 by [0, 100] scl:20 41e. Use 4: maximum on the CALC menu of the calculator. The x-coordinate of this point is about 4.2. The maximum area of the rectangle is the corresponding y-value of 72, for an area of 72 units2.  r 
  (x2   x1)2  (y2  y1)2 2 2 2  1  
12  
52   5 2 3 2 
 
 2   
2  
  42a.   
  2 17  (x  h)2  (y  k)2 
 r2 2  2  2  17  x  
12  
y  52 
 
 2  2 2 
x  12  
y  52 
 127  [15.16, 15.16] scl:1 by [5, 5] scl:1 42b. a circle centered at (2, 3) with radius 4 42c. (x  2)2  (y  3)2 
 16  315  Chapter 10  42d. center: (h, k) 
 (4, 2) radius: r2 
 36 r 
 36  or 6 2nd [DRAW] 9:Circle( - 4 , 2 , 6 )  2(13)  (9)  F 
 5 F 
 30 x2  y2  13x  9y  30 
 0 x2  13x  42.25  y2  9y  20.25 
 30  42.25  20.25 (x  6.5)2  (y  4.5)2 
 32.5 y 45a. (x  h)2  (y  k)2 
 r2 y
x (x  k)2  (y  k)2 
 22 (x  k)2  (y  k)2 
 4 45b. (x  1)2  (y  1)2 
 4 O (x  0)2  (y  0)2 
 4 x (x  1)2  (y  1)2 
 4  [15.16, 15.16] scl:1 by [5, 5] scl:1 43a. (x  h)2  (y  k)2 
 r2  45c. All of the circles in this family have a radius of 2 and centers located on the line y 
 x. 46a. (x  h)2  (y  k)2 
 r2 (x  0)2  (y  0)2 
 142 x2  y2 
 196 y2 
 196  x2 y 
  196  x2 2 y 
  196  x 46b. No, if x 
 7, then y 
 147  12.1 ft, so the truck cannot pass. 47. x2  y2  8x  6y  25 
 0 x2  8x  16  y2  6y  9 
 25  16  9 (x  4)2  (y  3)2 
 0 radius: r2 
 0 r 
 0  or 0 center: (h, k) 
 (4, 3) Graph is a point located at (4, 3). 48a. (x  h)2  (y  k)2 
 r2 (x  0)2  (y  0)2 
 r2 x2  y2 
 r2 2 (475)  (1140)2 
 r2 1,525,225 
 r2 x2  y2 
 1,525,225 48b. r2 
 1,525,225 r 
 1,525 ,225  or 1235 A 
 r2 
  (1235)2 or approximately 4,792,000 ft2  24 2  (x  0)2  (y  0)2 
 
2  x2  y2 
 144 43b. x2  y2 
 6.25 ⇒ r12 
 6.25 r1 
 6.25  or 2.5 If the circles are equally spaced apart then radius r2 of the middle circle is found by adding the radius of the smallest circle to the radius of the largest circle and dividing by two. 12   2.5  r2 
 2 or 7.25 area of area of area of 
  region B middle circle smallest circle 
 r22  r12 
  (r22  r12) 
  (7.252  2.52) 
  (46.3125) or about 145.50 The area of region B is about 145.50 in2. 44.  y x  5y  3 
 0 (2, 1)  (12, 3)  x (5, 1) 2x  3y  7 
 0  O 4x  7y 
 27  x2  y2  Dx  Ey  F 
 0 22  12  D(2)  E(1)  F 
 0 ⇒ 2D  E  F 
 5 52  (1)2  D(5)  E(1)  F 
 0 ⇒ 5D  E  F 
 26 122  32  D(12)  E(3)  F 
 0 ⇒ 12D  3E  F 
 153 2D  E  F 
 5 (1)(5D  E  F) 
 (1)(26) 3D  2E 
 21 2D  E  F 
 5 (1)(12D  3E  F) 
 (1)(153) 10D  2E 
 148 3D  2E 
 21 10D  2E 
 148 13D 
 169 D 
 13 3(13)  2E 
 21 2E 
 18 E 
 9  Chapter 10  48c.  25002  4,792,000  25002  
 0.23328  about 23%  y4   49a.  PA  has a slope of  B  has slope of x  3 and P y4 . If P A  x3 y4 y4    x3 x3 y2  16   x2  9 2 y  16  y4  y4    ⊥P B  then  x  3  x  3 
 1.  
 1 
 1  
 x2  9 x2  y2 
 25 49b. If P A ⊥P B , then A, P, and B are on the circle x2  y2 
 25. 50. d 
  (x2   x1)2  (y2  y1)2 d 
  (2  4)2  [6   (3)]2 d 
  (6)2   92 d 
 117   316  51. (2  i)(3  4i)(1  2i) 
 (6  8i  3i  4i2)(1  2i) 
 [6  8i  3i  4(1)](1  2i) 
 (10  5i)(1  2i) 
 10  20i  5i  10i2 
 10  20i  5i  10(1) 
 20  15i u cos v u sin v  1gt2 y 
 tv 52. x 
 tv  (0, 0), (0, 9), (5, 8), (10, 5) f(x, y) 
 320x  500y f(0, 0) 
 320(0)  500(0) 
 0 f(0, 0) 
 320(0)  500(9) 
 4500 f(5, 8) 
 320(5)  500(8) 
 5600 f(10, 5) 
 320(10)  500(5) 
 5700 The maximum profit occurs when 10 cases of drug A and 5 cases of drug B are produced. 57b. When 10 cases of drug A and 5 cases of drug B are produced, the profit is $5700. 58. y  2  1  x 
 t(60) cos 60°  y 
 t(60) sin 60°  2(32)t2  x 
 60t cos 60° y 
 60t sin 60°  16t2 x 
 60(0.5) cos 60° y 
 60(0.5) sin 60°  16(0.5)2 x 
 15 y 21.98076211 15 ft horizontally, about 22 ft vertically  x  5  53. A 
 2 or 2.5 2  20 
 k  
  x  x  x    and k 
 1 0  y 
 A cos (kt)  y 
 2.5 cos 
10t 54. s 
  x  1 (a  b  2 1 (15  25 2  y x  x  x  x  x  y  y 
 5x  2x The correct choice is A  c)  35)  
 37.5 a)(s  b)(s  c) k 
 s(s  
 37.5(3 7.5  7.5 15)(3 7.5 25)(3  3.5) 
 26,36 7.187 5  162 units2 55. v 
  v20  6 4h  10-3  Pages2 637–638 1.   152  64h  95 
 952 
 152  64h 952  152 h 
 6 4  O  4 y
0  8  (y  k)2 (x  h)2   b 
1 2 a2 (y  0)2 [x  (7)]2   3 
1 2 62 y2 (x  7)2    
 1 36 9 a2  b2 foci: c 
   x  5y 
 45  (5, 8) (10, 5) 4 (0, 0) x
0  3x  5y 
 55 12  16  20  Check For Understanding  y x2   a2  b2 
 1 y2 x2   82  52 
 1 y2 x2    
 1 64 25  2. Since the foci lie on the major axis, determine whether the major axis is horizontal or vertical. If the a2 is the denominator of the x terms, the major axis is horizontal. If the a2 is the denominator of the y terms, the major axis is vertical. 3. When the foci and center of an ellipse coincide, c 
 0. c e 
 a c2 
 a2  b2 0 
 a2  b2 0 e 
 a b2 
 a2 e
0 b
a The figure is a circle. c 4. In an ellipse, b2 
 a2  c2 and a 
 e. c  
 e b2 
 a2  c2 a c 
 ae b2 
 a2  a2e2 2 2 2 c 
a e b2 
 a2(1  e2) 5. Shanice; an equation with only one squared term cannot be the equation of an ellipse. 6. center: (h, k) 
 (7, 0) a 
 0  6 or 6 b 
 7  (4) or 3  h 
 137.5 ft 56. y 
 6x4  3x2  1 b 
 6a4  3a2  1 (x, y)  (a, b) x-axis: (x, y) 
 (a, b) b 
 6a4  3a2  1; no y-axis: (x, y) 
 (a, b) b 
 6(a)4  3(a)2  1 b 
 6a4  3a2  1; yes y 
 x: (x, y) 
 (b, a) a 
 6b4  3b2  1; no origin: f(x) 
 f(x) f(x) 
 6(x)4  3(x)2  1 f(x) 
 (6x4  3x2  1) f(x) 
 6x4  3x2  1 f(x) 
 6x4  3x2  1 no The graph is symmetric with respect to the y-axis. 57a. Let x 
 number of cases of drug A. Let y 
 number of cases of drug B. x 
  10 y y9 x 
 10 (0, 9) 12 3x  5y  55 y
9 x  5y  45 8  Ellipses  24 x  c 
  62  32 c 
 33   317  (h, k  c) 
 (7, 0 33 ) 
 (7, 33 )  Chapter 10  7. center: (h, k) 
 (0, 0) a2 
 36 b2 
 4 c 
  a2  b2 a 
 36  or 6 b 
 4  or 2 c 
 36 4   or 42  foci: (h c, k) 
 (0 42 , 0) or ( 42 , 0) major axis vertices: (h a, k) 
 (0 6, 0) or ( 6, 0) minor axis vertices: (h, k b) 
 (0, 0 2) or (0, 2)  10.  center: (h, k) 
 (1, 2) a2 
 9 b2 
 4 a 
 9  or 3 b 
 4  or 2 foci: (h, k c) 
 1, 2 5   y (0, 2)  (6, 0)  O  
  (6, 0)  (0, 2)    c 
  a2  b2 c 
 9   4 or 5   major axis vertices: (h, k a) 
 (1, 2 3) or (1, 1), (1, 5) minor axis vertices: (h  b, k) 
 (1 2, 2) or (3, 2), (1, 2) y  x  (1, 1)  8. center: (h, k) 
 (0, 4) a2 
 81 b2 
 49 c 
  a2  b2 a 
 81  or 9 b 
 49  or 7 c 
 81 9  4 or 42  foci: (h c, k) 
 (0 42 , 4) or ( 42 , 4) major axis vertices: (h a, k) 
 (0 9, 4) or ( 9, 4) minor axis vertices: (h, k b) 
 (0, 4 7) or (0, 11), (0, 3)  y  9x2  4y2  18x  16y 
 11 9(x2  2x  ?)  4(y2  4y  ?) 
 11  ?  ? 9(x2  2x  1)  4(y2  4y  4) 
 11  9(1)  (4) 9(x  1)2  4(y  2)2 
 36 (x  1)2 (y  2)2    
 36 4 9  (1, 2)  O  x (3, 2)  (1, 2)  (1, 5)  11. center: (h, k) 
 (2, 3) 8 a 
 2 or 4  (0, 11)  2  b 
 2 or 1  (y  k)2 (x  h)2   b 2 a2 [y  (3)]2 [x  (2)]2   1 2 42 (y  3)2 (x  2)2    16 1  (0, 4)  (9, 4)  (9, 4)  O  x (0, 3)  9.  
1 
1  12. The major axis contains the foci and it is located on the x-axis.  25x2  9y2  100x  18y 
 116 25(x2  4x  ?)  9(y2  2y  ?) 
 116  ?  ? 25(x2  4x  4)  9(y2  2y  1) 
 116  25(4)  9(1) 25(x  2)2  9(y  1)2 
 225  1  1 0  0  center: (h, k) 
 
2, 2 or (0, 0) c 
 1, a 
 4 c2 
 a2  b2 12 
 42  b2 b2 
 15  (x  2)2 (y  1)2    
 1 9 25  center: (h, k) 
 (2, 1) a2 
 25 b2 
 9 c 
  a2  b2 a 
 25  or 5 b 
 9  or 3 c 
 25 9   or 4 foci: (h, k c) 
 (2, 1 4) or (2, 5), (2, 3) major axis vertices: (h, k a) 
 (2, 1 5) or (2, 6), (2, 4) minor axis vertices: (h b, k) 
 (2 3, 1) or (1, 1), (5, 1) (2, 6)  
1  (x  h)2  a2 (x  0)2  42  (y  k)2   b 
1 2 (y  0)2   15 
 1 x2  16  y2   15 
 1  13. center: (h, k) 
 (1, 2)  y  y  The points at (1, 4) and (5, 2) are vertices of the ellipse.  (1, 4)  (5, 1)  (2, 1)  O  (5, 2) (1, 1)  (1, 2)  x  a 
 4, b 
 2  (2, 4)  Chapter 10  x  O  (x  h)2  a2 (x  1)2  42 (x  1)2  16  318  (y  k)2   b 
1 2 (y  2)2   2 
1 2 (y  2)2   4 
 1  14. center: (h, k) 
 (3, 1) a
6 c e 
 a 1  3  c  
 6   foci: (h, k c) 
 
3, 4 39 19. center: (h, k) 
 (2, 1) a2 
 4 b2 
 1 c 
  a2  b2 a 
 4  or 2 b 
 1  or 1 c 
 4   1 or 3  foci: (h, k c) 
 (2, 1 3 ) major axis vertices: (h, k a) 
 (2, 1 2) or (2, 3), (2, 1) minor axis vertices: (h b, k) 
 (2 1, 1) or (1, 1), (3, 1) y  2
c c2 
 a2  b2 22 
 62  b2 4 
 36  b2  b2 
 32  (y  k)2 (h  h)2   b 2 a2 (y  1)2 (x  3)2   32 62 (y  1)2 (x  3)2    36 32  
1 
1 
1  (2, 3)  15. The major axis contains the foci and is located on the x-axis. center: (h, k) 
 (0, 0) c 
 0.141732 1 a 
 2(3.048) or 1.524  (1, 1)  (3, 1)  O  x  (2, 1)  c2 
 a2  b2 (0.141732)2 
 (1.524)2  b2 0.020 
 2.323  b2 b2 
 2.302 b 1.517 (x  h)2 (y  k)2   b 2 a2 (x  0)2 (y  0)2   1.5242  1.5172 x2 y2   1.5242  1.5172  (2, 1)  20. center: (h, k) 
 (6, 7) a2 
 121 b2 
 100 a 
 121  or 11 b 
 100  or 10 a2  b2 c 
  c 
 121   100 or 21  foci: (h, k c) 
 (6, 7 21 ) major axis vertices: (h, k a) 
 (6, 7 11) or (6, 18), (6,  4) minor axis vertices: (h b, k) 
 (6 10, 7) or y (4, 7) (16, 7),  
1 
1 
1  (6, 18)  Pages 638–641  Exercises  16. center: (h, k) 
 (0, 5) a 
 0  (7) or 7 b 
 5  0 or 5 (x  h)2 (y  k)2   b 2 a2 (x  0)2 (y  (5)]2   5 2 72 (y  5)2 x2    25 49 a2  b2 c 
   (4, 7)  
1  
1  O  x (6, 4)  21. center: (h, k) 
 (4, 6) a2 
 16 b2 
 9 c 
  a2  b2 a 
 16  or 4 b 
 9  or 3 c 
 16 9   or 7  foci: (h c, k) 
 (4 7 , 6) major axis vertices: (h a, k) 
 (4 4, 6) or (8, 6), (0, 6) minor axis vertices: (h, k b) 
 (4, 6 3) or y 3), (4, 9) (4,  
1 
1 
1  x  O  c 
  42  22 or 23  foci: (h c, k) 
 (2 23 , 0) 18. centers: (h, k) 
 (3, 4) a 
 4  12 or 8 b 
 3  2 or 5 (y  k)2 (x  h)2   b 2 a2 (y  4)2 [x  (3)]2   5 2 82 (y  4)2 (x  3)2    64 25 a2  b2 c 
   (16, 7)  
1  c 
  72  52 or 26  foci: (h c, k) 
 (0 26 , 5) 
 ( 26 , 5) 17. center: (h, k) 
 (2, 0) a 
 2  2 or 4 b 
 0  2 or 2 (x  h)2 (y  k)2   b 2 a2 [x  (2)]2 (y  0)2   2 2 42 (x  2)2 y2    16 4 a2  b2 c 
   (6, 7)  (4, 3) (4, 6) (0, 6)  (8, 6)  (4, 9)  
1 
1 
1  c 
  82  52 or 39   319  Chapter 10  y  22. (h, k) 
 (0, 0) a2 
 9 b2 
 4 c 
  a2  b2 a 
 9  or 3 b 
 4  or 2 c 
 9   4 or 5  foci: (h, k c) 
 0, 0 5  or 0, 5   
   
  (3, 8)    (3, 4)  major axis vertices: (h, k a) 
 (0, 0 3) or (0, 3) minor axis vertices: (h b, k) 
 (0 2, 0) or ( 2, 0) y  (8, 4)  (2, 4)  O  x  (3, 0)  (0, 3)  25. (2, 0)  (2, 0)  x  O  (x  3)2  8  (0, 3)  23.  y2  (x  1)2  1  c 
  a2  b2 c 
 24 8   or 4 foci: (h, k c) 
 (3, 1 4) or (3, 5), (3, 3) major axis vertices: (h, k a) 
 (3, 1 26 ) minor axis vertices: (h b, k) 
 (3 22 , 1)  (y  3)2   4 
 1  center: (h, k) 
 (1, 3) b2 
 1 a2 
 4 a 
 4  or 2 b 
 1  or 1 foci: (h, k c) 
 1, 3 3   
    c 
  a2  b2 c 
 4   1 or 3   (3, 1  26) y  major axis vertices: (h, k a) 
 (1, 3 2) or (1, 1), (1, 5) minor axis vertices: (h b, k) 
 (1 1, 3) or (2, 3), (0, 3) y  O (0, 3) (1, 3)  24.  (1, 1)  (3, 1) (3  22, 1) (3  22, 1)  26.  (2, 3) (1, 5)  16x2  25y2  96x  200y 
 144 16(x2  6x  ?)  25(y2  8y  ?) 
 144  ?  ? 16(x2  6x  9)  25(y2  8y  16) 
 144  16(9)  25(16) 16(x  3)2  25(y  4)2 
 400 (y  4)2   16 
 1  center: (h, k) 
 (3, 4) b2 
 16 c 
  a2  b2 a2 
 25 a 
 25  or 5 b 
 16  or 4 c 
 25 16   or 3 foci: (h c, k) 
 (3 3, 4) or (6, 4), (0, 4) major axis vertices: (h a, k) 
 (3 5, 4) or (8, 4), (2, 4) major axis vertices: (h, k b) 
 (3, 4 4) or (3, 8), (3, 0)  6x2  12x  6y  36y 
 36 6(x2  2x  ?)  6(y2  6y  ?) 
 36 6(x2  2x  1)  6(y2  6y  9) 
 36  6(1)  6(9) 6(x  1)2 2 6(y  3)22 
 96 (x  1) ( y  3)    
 1 16 16 center: (h, k) 
 (1, 3) a2 
 16 b2 
 16 c 
  a2  b2 a 
 16  or 4 b 
 16  or 4 c 
 16 16   or 0 foci: (h c, k) or (h, k c) 
 (1, 3) Since a 
 b 
 4, the vertices are (h 4, k) and (h, k 4) or (5, 3), (3, 3), (1, 1), (1, 7) y (1, 1)  x  O  (1, 3) (5, 3) (3, 3)  (1, 7)  Chapter 10  x  O  (3, 1 2 6)  x  (x  3)2  25  (y  1)2   24 
 1  center: (h, k) 
 (3, 1) a2 
 24 b2 
 8 a 
 24  or 26  b 
 8  or 22     8x  6y  9 
 0 4(x2  2x  ?)  (y2  6y  ?) 
 9  ?  ? 4(x2  2x  1)  (y2  6y  9) 
 9  4(1)  9 4(x  1)2  (y  3)2 
 4 4x2  3x2  y2  18x  2y  4 
 0  6x  ?)  (y2  2  ?) 
 4 3(x2  6x  9)  (y2  2y  1) 
 4  3(9)  1 3(x  3)2  (y  1)2 
 24 3(x2  320  27.  18y2  12x2  144y  48x 
 120 18(y2  8y  ?)  12(x2  4x  ?) 
 120  ?  ? 18(y2  8y  16)  12(x2  4x  4) 
 120  18(16)  12(4) 18(y  4)2  12(x  2)2 
 216 (y   12 4)2  center: (h, k) 
 (2, 4) a2 
 18 a 
 18  or 32   (x   29. 49x2  16y2  160y  384 
 0 49x2  16(y2  10y  ?) 
 384  ? 49x2  16(y2  10y  25) 
 384  16(25) 49x2  16(y  5)2 
 784 x2  16  2)2   18 
 1  ( y  5)2   49 
 1  center: (h, k) 
 (0, 5) b2 
 16 c 
  a2  b2 a2 
 49 a 
 49  or 7 b 
 16  or 4 c 
 49 16   or 33  foci: (h, k c) 
 (0, 5 33 ) major axis vertices: (h, k a) 
 (0, 5 7) or (0, 2), (0, 12) minor axis vertices: (h b, k) 
 (0 4, 5) or ( 4, 5)  b2 
 12 b 
 12  or 23   c 
  a2  b2 c 
 18 12   or 6  foci: (h c, k) 
 (2 6 , 4) major axis vertices: (h g, k) 
 (2 32 , 4) minor axis vertices: (h, k b) 
 (2, 4 23 ) y  y (0, 2)  (2, 4  23)  x  O (4, 5)  (2, 4) (2  32, 4)  (0, 5)  (4, 5)  (2  32, 4) (0, 12)  O (2, 4  23)  x 9y2  108y  4x2  56x 
 484  12y  ?)  4(x2  14x  ?) 
 484  ?  ? 9(y2  12y  36)  4(x2  14x  49) 
 484  9(36)  4(49) 9(y  6)2  4(x  7)2 
 36  30. 28.  9(y2  4x2  8y  9x2  54x  49 
 0  2y  ?)  9(x2  6x  ?) 
 49  ?  ? 2 4(y  2y  1)  9(x2  6x  9) 
 49  4(1)  9(9) 4(y  1)2  9(x  3)2 
 36  4(y2  ( y  1)2  9    (x  3)2  4  (y  6)2  4  
1  center: (h, k) 
 (7, 6) b2 
 4 c 
  a2  b2 a2 
 9 a 
 9  or 3 b 
 4  or 2 c 
 9   4 or 5  foci: (h c, k) 
 (7 5 , 6) major axis vertices: (h a, k) 
 (7 3, 6) or (10, 6)(4, 6) minor axis vertices: (h, k b) 
 (7, 6 2) or (7, 4), (7, 8)  center: (h, k) 
 (3, 1) b2 
 4 c 
  a2  b2 a2 
 9 a 
 9  or 3 b 
 4  or 2 c 
 9   4 or 5  foci: ( j, k c) 
 (3, 1 5) major axis vertices: (h, k a) 
 (3, 1 3) or (3, 4), (3, 2) minor axis vertices: (h b, k) 
 (3 2, 1) or (5, 1), (1, 1) y  y  (3, 1)  (7, 4) (5, 1) (4, 6)  x  O  x  O  (3, 4)  (1, 1)  (x  7)2   9 
 1  (3, 2)  (10, 6) (7, 6) (7, 8)  31. a 
 7, b 
 5  (x  h)2 ( y  k)2   b 
1 2 a2 [x  (3)]2 [y  (1)]2   5 
1 2 72 (x  3)2 ( y  1)2    
 1 49 25  321  Chapter 10  32. The major axis contains the foci and it is located on the x-axis. 2  2 0  0 center: (h, k) 
 
0, 2 or (0, 0)  36. The major axis contains the foci and it is the vertical axis of the ellipse. 5  (1)  c 
 2 or 3  c 
 2, a 
 7 c2 
 a2  b2 22 
 72  b2 b2 
 45 (x  h)2  a2 (x  0)2  72  33. b 
 6
  1  1 1  5  center: (h, k) 
 
2, 2 or (1, 2)  (y  k)2  b 2 (y  0)2  45 x2 y2    49 45  ( y  k)2  a2 (2  2)2  a2  
1  (x  h)2  a2 (x  0)2  82  
1 
1  37.  (y  k)2  (y  0)2   6 
1 2 y2   36 
 1  5
c  1  (1) 1  (5)  center: (h, k) 
 
2, 2 or (1, 2) c 
 1  k c 
 1  (2) or 3 c2 
 a2  b2 2 32 
 213   b2  
  
 52  9 b2 
 43   a 
 213  
1 b2 
 a2  c2 b2 
 102  52 b2 
 75  ( y  0)2   75 
 1 x2  100  y2   75 
 1  (4, 0)  
1  4  8  12  x  tangent vertices: (4, 0), (0, 7) a 
  7  0 or 7 b 
 4  0 or 4  (4, 7)  12  
1 
1  39.  (7, 5)  12 8 4 O  
1  ( y  k)2  O  The horizontal axis of the ellipse is the major axis.  12 (2, 9) 8 (11, 5) 4 (2, 1)  9 
 b2  (x  h)2  b2 (x  1)2  9   b 
1 2  4 (0, 7) 8  y  35.  
1  y  38.    (y  k)2 (x  h)2   b 2 a2 [x  (1)]2 [y  (2)]2     43 (21 3)2  (y  2)2 (x  1)2    52 43  ( y  k)2   a2 ( y  2)2   18 1 c  
  2 a a  
 c 2 10  
 c 2 (x  h)2  a2 (x  0)2  102  34. The major axis contains the foci and it is the vertical axis of the ellipse.  b2  32   b2 
 1 9  b2   b 
1 2 x2  64  (y   c2 
 a2  b2 32 
 a2  9 18 
 a2  1)2   b 
1 2 02  a2  3 a 4 3 a 4  8
a  (x  h)2   b 
1 2  4  ( y  k)2 (x  h)2   b 2 a2 [y  (7)] (x  4)2   4 2 72 2 ( y  7) (x  4)2    49 16 2 b 
 a2(1  e2)  
1 
1 
1  3 2  b2 
 221  
4  b2  8x  
  28  16    or 1.75  Case 1: Horizontal axis is major axis. 11  7 5  5 ,  2 2  enter: (h, k) 
 
 ha
7 2  a 
 7 a
9 (x  h)2  a2 [x  (2)]2  92 (x  2)2  81  Chapter 10      (y  k)2  b2 (y  5)2  42 (y  5)2  16  (x  h)2 ( y  k)   b 2 a2 (x  0)2 ( y  0)2     1.75 22 y2 x2     1.75 4   or (2, 5)  kb
9 5b
9 b
4  
1 
1 
1  Case 2: Vertical axis is major axis.  
1  ( y  k)2  a2 ( y  0)2  22  
1 
1  (x  h)2   b 
1 2 (x  0)2    1.75 
 1 y2  4  322  x2    1.75 
 1  44. 4x2  y2  8x  2y 
 1 4(x2  2x  ?)  (y2  2y  ?) 
 1  ?  ? 4(x2  2x  1)  (y2  2y  1) 
 1  4(1)  1 4(x  1)2  (y  1)2 
 4 (y  1)2 
 4  4(x  1)2 y  1 
  4   4(x   1)2 y 
  4  4 (x  1 )2  1 Vertices: (0, 1), (2, 1), (1, 1), (1, 3)  40. The major axis contains the foci and it is the horizontal axis of the ellipse. 31 55  center: (h, k) 
 
2, 2 or (2, 5) foci: (3, 5) 
 (h  c, k) 3
hc 3
2c 1
c c  e 
 a 0.25 
  b2 
 a2(1  e2) b2 
 42(1  0.252) b2 
 15  1  a  a
4  41.  (x  h)2 ( y  k)2   b 2 a2 (x  2)2 ( y  5)2   15 42 (x  2)2 ( y  15)2    16 15 20 a 
 2 or 10 b2 
 a2(1  e2)  
1 
1  2  b2 
 1021  
1 0 7  [4.7, 4.7] scl:1 by [3.1, 3.1] scl:1 45. 4x2  9y2  16x  18y 
 11 2 4(x  4x  ?)  9(y2  2y  ?) 
 11  ?  ? 4(x2  4x  4)  9(y2  2y  1) 
 11  4(4)  9(1) 4(x  2)2  9(y  1)2 
 36 9(y  1)2 
 36  4(x  2)2  
1   or 51  (y  k)2 (x  h)2   b 
1 2 a2 (y  0)2 (x  3)2   51 
 1 102 y2 (x  3)2    
 1 100 51  y1
 y
  c  5   3  b2 
 a2(1  e2)    b2 
 4  [7.28, 7.28] scl:1 by [4.8, 4.8] scl:1 46. 25y2  16x2  150y  32x 
 159 25(y2  6y  ?)  16(x2  2x  ?) 
 159  ?  ? 25(y2  6y  9)  16(x2  2x  1) 
 159  25(9)  16(1) 25(y  3)2  16(x  1)2 
 400 25(y  3)2 
 400  16(x  1)2  (y  k)2 (x  h)2   b 
1 2 a2 [y  (1)]2 (x  1)2   4 
 1 32 2 (y  1) (x  1)2    
 1 9 4 x2  4y2  6x   24y 
 41  6x  ?)   6y  ?) 
 41  ?  ? (x2  6x  9)  4(y2  6y  9) 
 41  9  4(9) (x  3)2  4(y  3)2 
 4 4(y  3)2 
 4 (x  3)2 (x2  2   5 2  b2 
 321  
3  5   
 a  a
3 major axis: vertical axis  43.  2  Vertices: (1, 1), (5, 1), (2, 3), (2, 1)  ) 
 (h, k  c) 42. focus: (1, 1  5 1  5 
kc 1  5  
 1  c 
c 5 e 
 a  36  4(x  2)      9 36  4(x  2)    1 9  (y  3) 
  4(y2  (y  3)2 
 y3
 y
  y
  400  16(x  1)     25  400  16(x  1)     5 3 2  2  Vertices: (4, 3), (6, 3), (1, 7), (1, 1)  4  (x  3)2  4  4  (x  3)     4 4  (x  3)     3 4 2  2  vertices: (5, 3), (1, 3), (3, 2), (3, 4)  [15.16, 15.16] scl:1 by [10, 10] scl:1  [7.28, 7.28] scl:1 by [4.8, 4.8] scl:1  323  Chapter 10  47. The target ball should be placed opposite the pocket, 5  feet from the center along the major axis of the ellipse. The cue ball can be placed anywhere on the side opposite the pocket. The ellipse has a semi-major axis of length 3 feet and a semi-minor axis of length 2 feet. Using the equation c2 
 a2  b2, the focus of the ellipse is found to be 5  feet from the center of the ellipse. Thus the hole is located at one focus of the ellipse. The reflective properties of an ellipse should insure that a ball placed 5  feet from the center of the ellipse and hit so that it rebounds once off the wall should fall into the pocket at the other focus of the ellipse. 48. A horizontal line; see students' work.  8  52a. a 
 2 or 4 b
3 a2  b2 c 
  c 
  42  32 c 
 7  foci: (h c, 0) 
 (0 7 , 0) or ( 7 , 0) , 0) from The thumbtacks should be placed ( 7 the center of the arch. 52b. With the string anchored by thumbtacks at the foci of the arch and held taunt by a pencil, the sum of the distances from each thumbtack to the pencil will remain constant. 53a. GOES 4; its eccentricity is closest to 0.  96  49a. a 
 2 or 48 b
  46  2  c  6955  or 23  (h, k) 
 (0, 0) (x  h)2  a2 (x  0)2  482  (y  k)2  b 2 (y  0)2   232 x2 y2    2304 529  x  
1  a  x2 y2    9 4 a2 
 9  
1  A 
 r2  
1  A
rr A
ab A 
 ab  [figure not drawn to scale] x 
 a  c  Earth's radius x 
 6955  361.66  6357 x 
 959.66 x 960 km 54. x2  y2  Dx  Ey  F 
 0 2 0  (9)2  D(0)  E(9)  F 
 0 ⇒ 9E  F 
 81 72  (2)2  D(7)  E(2)  F 
 0 ⇒ 7D  2E  F 
 53 (5)2  (10)2  D(5)  E(10)  F 
 0 ⇒ 5D  10E  F 
 125  9E  F 
 81 (1)(7D  2E  F) 
 (1)(53)  7D  7E 
 28 DE
4 7D  2E  F 
 53 (1)(5D  10E  F ) 
 (1)(125) 12D  8E 
 72 (8)(D  E) 
 (8)(4) 12D  8E 
 72 4D 
 40 D 
 10 DE
4 9E  F 
 81 10  E 
 4 9(6)  F 
 81 E 
 6 F 
 135 x2  y2  Dx  Ey  F 
 0 x2  y2  10x  6y  135 
 0 (x2  10x  25)  (y2  6y  9) 
 135  25  9 (x  5)2  (y  3)2 
 169  
1  b2 
 4 a
3 b
2 A 
 ab A 
 (3)(2) A 
 6 units2 51. If (x, y) is a point on the ellipse, then show that (x, y) is also on the ellipse. x2  a2 (x)2  a2  y2   b2 
 1 (y)2   b 
1 2 x2  a2  y2   b2 
 1  Thus, (x, y) is also a point on the ellipse and the ellipse is symmetric with respect to the origin.  Chapter 10  x  
1  49b. c 
 c 
 2304   529 c 42.13 He could have stood at a focal point, about 42 feet on either side of the center along the major axis. 49c. The distance between the focal points is 2c. 2c 
 2(42) 
 84 about 84 ft 50a. x2  y2 
 r2 x2 y2   r2 r2 x2 y2   b2 a2  
 0.052  c 
 361.66  O c  
e  
1   a2  b2  50b.  c  a  y  53b.  324  center: (h, k) 
 (5, 3) radius: r2 
 169 r 
 13 y  60. Let h 
 0.1. x  h 
 x  0.1 
 1  0.1 or 1.1 f(x  0.1) 
 f(1.1) 
 (1.1)2  4(1.1)  12 
 15.19 x  h 
 x  0.1 
 1  0.1 or 0.9 f(x  0.1) 
 f(0.9) 
 (0.9)2  4(0.9)  12 
 14.79 f(x) 
 16 f(x) f(x  0.1) and f(x) f(x  0.1), so the point is a location of a minimum. 61. The graph of the parent function g(x) 
 x is translated 2 units right.  (5, 16) (5, 3)  (8, 3)  x  O  55. Graph the quadrilateral with vertices A(1, 2), B(5, 4), C(4, 1), and D(5, 4). A quadrilateral is a D (5, 4) y parallelogram if one pair of opposite sides C (4, 1) are parallel and congruent. x O  g (x )  A(1, 2) B (5, 4)  slope of  DA   slope of  CB   m
   m
 x x  
 
  y2  y1   x2  x1 2  4  1  (5) 6 3  or  4 2  y2  y1 2  
 
  62. Initial location: (2, 0) Rot90 
 0 1 2 
 0 or (0, 2) 1 0 0 2 1 0 2 Rot80 
 
 2 or (2, 0) 0 1 0 0 0 1 2 0 Rot270 
 
 or (0, 2) 1 0 0 2  1  4  1  54 5 1      The slopes are not equal, so D CB A    . The quadrilateral is not a parallelogram; no. 56. cos 2v 
 1  2 sin2 v 7 2  cos 2v 
 1  2
8 34  17  2  k  57. A 
 4 c k c 2  
 180  4  k
2  
 20°  h
0  
 20°  Page 641  c 
 40° y 
 A cos(kx  c)  h y 
 4 cos[2x  (40°)]  0 y 
 4 cos(2x  40°) 58. A 
 180°  (121° 32
  42° 5
) or 16° 23
 a  sin A 4.1  sin 16° 23
 4.1 sin 42° 5
  sin 16° 23
  b   
 sin B b   
 sin 42° 5
  
b  a  sin A 4.1  sin 16° 23
 4.1 sin 121° 32
  sin 16° 23
               63. mQTS  mTSR 
 180 a  b  c  d 
 180 b  b  c  c 
 180 2b  2c 
 180 b  c 
 90 The correct choice is C.   cos 2v 
 6 4 or  32  A
  x  O  p  b  c 
 180 p  90 
 180 p 
 90  Graphing Calculator Exploration  1. Sample answer: The graph will shift 4 units to the right.  c   
 sin C c   
 sin 121° 32
  
c  9.7 b 12.4 c 59. P(x) 
 x4  4x3  2x2  1 P(5) 
 54  4(5)3  2(5)2  1 P(5) 
 74 P(5) 0; no, the binomial is not a factor of the polynomial.  [15.16, 15.16] scl:2 by [10, 10] scl:2  325  Chapter 10  2. Sample answer: The graph will shift 4 units to the left.  2. transverse axis: vertical 2a 
 4 a
2 An equation2in standard form of the hyperbola y y2     must have 22 or 4 as the first term; b. c  3. e 
 a, so ae 
 c and a2e2 
 c2. Since c2 
 a2  b2 we have a2e2 
 a2  b2 2 2 a e  a2 
 b2 a2(e2  1) 
 b2 4. With the equation in standard form, if the first expression contains "x", the transverse axis is horizontal. If the first expression contains "y", the transverse axis is vertical. 5. center: (h, k) 
 (0, 0) a2 
 25 b2 
 4 c 
  a2  b2 a
5 b
2 c 
 25 4   or 29  transverse axis: horizontal foci: (h  c, k) 
 (0 29 , 0) or ( 29 , 0) vertices: (h a, k) 
 (0 5, 0) or ( 5, 0)  [15.16, 15.16] scl:2 by [10, 10] scl:2 3. Sample answer: The graph will shift 4 units up.  [15.16, 15.16] scl:2 by [10, 10] scl:2 4. Sample answer: The graph will shift 4 units down.  b (x a 2 (x 5 2 x 5  asymptotes: y  k 
 y0
 y
 8   0)  y  (5, 0) 4  (5, 0)  8 4 O 4  [18.19, 18.19] scl:2 by [12, 12] scl:2 5. Sample answer: The graph will rotate 90°.   h)  4  8  x  8  6. center: (h, k) 
 (2, 3) a2 
 16 b2 
 4 c 
  a2  b2 a 
 16  or 4 b 
 4  or 2 c 
 16 4   or 25  transverse axis: vertical foci: (h, k c) 
 
2, 3 25  vertices: (h, k a) 
 (2, 3 4) or (2, 7), (2, 1) asymptotes: y  k 
  [15.16, 15.16] scl:2 by [10, 10] scl:2 6. For (x  c), the graph will shift c units to the left. For (x  c), the graph will shift c units to the right. 7. For (y  c), the graph will shift c units down. For (y  c), the graph will shift c units up. 8. The graph will rotate 90°.  y3
 y3
  y (2, 7)  (2, 3)  10-4  Hyperbolas x  O  Pages 649–650  (2, 1)  Check For Understanding  1. The equations of both hyperbolas and ellipses have x2 terms and y2 terms. In an ellipse, the terms are added and in a hyperbola these terms are subtracted.  Chapter 10  326  a (x b 4 (x 2   h)  2)  2(x  2)  7.  11. 2b 
 6 b
3 x1  x2 y1  y2 33 40 center: 
2, 2 
 
2, 2 
 (3, 2) transverse axis: vertical a 
 4  2 or 2  y2  5x2  20x 
 50 y2  5(x2  4x  ?) 
 50  ? y2  5(x2  4x  4) 
 50  (5)(4) y2  5(x  2)2 
 30 y2  30  (x  2)2   6 
 1  xh
x2 yk
y h
2 k
0 center: (h, k) 
 (2, 0) a2 
 30 b2 
 6 c 
  a2  b2 a 
 30  b 
 6  c 
 30 6   or 6 transverse axis: vertical foci: (h, k c) 
 (2, 0 6) or (2, 6) vertices: (h, k a) 
 
2, 0 30  or 
2, 30  asymptotes: y  k 
 y0
 y
  ( y  k)2  a2 ( y  2)2  22 ( y  2)2  4  (x  2) 5  x1  x2 y1  y2  0  0 6  (6)  c2  a2 
 2 (2,  30) (2, 0)  62  a2 
 2 or 18  ( y  k)2  a2 ( y  0)2  18  4 8x 30) (2,   (x  h)2   b 
1 2 (x  0)2   18 
 1 y2 x2    
 1 18 18 x1  x2 y1  y2    2 , 2  8  13. center: 
  8. center: (h, k) 
 (0, 5) transverse axis: horizontal a 
 5, b 
 3 (x  h)2  a2 (x  0)2  52 x2  25  (x  3)2   9 
 1  
 (0, 0) transverse axis: vertical c 
 distance from center to a focus 
 0  6 or 6 b2 
 c2  a2 b2 
 a2 2 2 2 a 
c a b2 
 18 2a2 
 c2  a (x  h) b  30 (x  2)  6  8  4 O 4  (x  3)2   3 
1 2  12. center: 
2, 2 
 
2, 2  y  4  (x  h)2   b 
1 2  10  (10) 0  0   
 
2, 2  ( y  5)2  
 (0, 0) transverse axis: horizontal c 
 distance from center to a focus 
 10  0 or 10 c e 
 a b2 
 c2  a2  ( y  5)2  5  3  ( y  k)2   b 
1 2  3 
1 2  9 
 1  9. c 
 9 quadrants: II and IV transverse axis: y 
 x vertices: xy 
 9 3(3) 
 9 (3, 3)  a
6  (x  h)2  a2 (x  0)2  62  xy 
 9 3(3) 
 9 (3, 3)  y  b2 
 102  62 b2 
64 ( y  k)2   b 
1 2 ( y  0)2   64 
 1 x2  36  y2   64 
 1  14a. The origin is located midway between stations A and B; (h, k) 
 (0, 0). The stations are located at the foci, so 2c 
 130 or c 
 65.  (3, 3)  y  O  x (3, 3)  A(65, 0)  10. center: (h, k) 
 (1, 4) (x  h)2 ( y  k)2   b 2 a2 (x  1)2 [y  (4)]2     52 22 (x  1)2 ( y  4)2    25 4  10  
 a  O  B (65, 0) x  The difference of the distances from the plane to each station is 50 miles. 50 
 2a (Definition of hyperbola) 25 
 a b2 
 c2  a2 b2 
 652  252 b2 
 3600  
1 
1 
1  327  Chapter 10  16. center: (h, k) 
 (0, 5) a2 
 9 b2 
 81 a 
 9  or 3 b 
 81  or 9 c 
  a2  b2 c 
 9 1  8 or 310  transverse axis: horizontal foci: (h c, k) 
 (0 310 , 5) or ( 310 , 5) vertices: (h a, k) 
 (0 3, 5) or ( 3, 5) b asymptotes: y  k 
 a(x  h)  transverse axis: horizontal (x  h)2 (y  k)2   b 2 a2 (x  0)2 (y  0)2    252 3600 x2 y2    625 3600  
1 
1 
1  a, k) 
 (0  14b. Vertices: (h  asymptotes: y  k 
 y0
 y
 Plane located on this branch  60  25, 0) or ( 25, 0)  b (x  h) a 3  600 (x  25 12 x 5  0)  y  9 (x 3  y5
  3x   0)  y  40  (0, 5) Station B (65, 0)  20  Station A (65, 0)  y5
  60 40 20 O 20  20  (3, 5)  (3, 5)  60 x  40  O  x  40  17. center: (h, k) 
 (0, 0) b2 
 49 a2 
 4 a 
 4  or 2 b 
 49  or 7 c 
  a2  b2 c 
 4 9  4 or 53  transverse axis: horizontal foci: (h c, k) 
 (0 53 , 0) or ( 53 , 0) vertices: (h a, k) 
 (0 2, 0) or ( 2, 0)  60  14c. Let y 
 6. x2  625 x2  625 x2  625  y2    3600 
 1 62    3600 
 1 36    3600 
 1 x2  625 x2  625 x2  36   
1 3600  asymptotes: y  k 
  
 1.01  y0
  
 625(1.01) x2 
 631.25 x 
 631.2 5  x 25.1 Since the phase is closer to station A than station B, use the negative value of x to locate the ship at (25.1, 6).  y
 8  b (x  h) a 7 (x  0) 2 7 x 2  y  4 (2, 0)  (2, 0)  4 2 O  2  4x  4  Pages 650–652  8  Exercises  15. center: (h, k) 
 (0, 0) a2 
 100 b2 
 16 a 
 100  or 10 b 
 16  or 4 2 2 c 
  a  b c 
 100   16 or 229  transverse axis: horizontal foci: (h c, k) 
 (0 229 , 0) or ( 229 , 0) vertices: (h a, k) 
 (0 10, 0) or ( 10, 0) asymptotes: y  k 
 y0
 y
  b (x  h) a 4 (x  0) 10 2 x 5  18. center: (h, k) 
 (1, 7) a2 
 64 b2 
 4 a 
 64  or 8 b 
 4  or 2 c 
  a2  b2 c 
 64 4   or 217  transverse axis: vertical foci: (h, k c) 
 (1, 7 217 ) vertices: (h, k a) 
 y (1, 7 8) or (1, 15), (1, 1)  y  4 (10, 0)  (10, 0) 16 8 O 4  8  y7
  4(x  1)   h)  (1, 15)   (1)] (1, 7)  16 x  O (1, 1)  8  Chapter 10  y7
  a (x b 8 [x 2  asymptotes: y  k 
  8  328  x  19.  x2  4y2  6x  8y 
 11 (x2  6x  ?)  4( y2  2y  ?) 
 11  ?  ? (x2  6x  9)  4( y2  2y  1) 
 11  9  (4)(1) (x  3)2  4( y  1)2 
 16 (x  3)2  16  21. 16y2  25x2  96y  100x  356 
 0 16(y2  6y  ?)  25(x2  4x  ?) 
 356 16(y2  6y  9)  25(x2  4x  4) 
 356  16(9)  25(4) 16(y  3)2  25(x  2)2 
 400  ( y  1)2  (y  3)2 (x  2)2    
 1 25 16   4 
 1  center: (h, k) 
 (3, 1) b2 
 4 c 
  a2  b2 a2 
 16 2 a 
 16  or 4 b 
 4  or 2 c 
 16 4   or 25  transverse axis: horizontal foci: (h c, k) 
 (3 25 , 1) vertices: (h a, k) 
 (3 4, 1) or (1, 1), (7, 1) b asymptotes: y  k 
 a(x  h) y  (1) 
 y1
  center: (h, k) 
 (2, 3) b2 
 16 c 
  a2  b2 a2 
 25 a 
 25  b 
 16  c 
 25 16  or 5 or 4 or 41  transverse axis: vertical foci: (h, k c) 
 (2, 3 41 ) vertices: (h, k a) 
 (2, 3 5) or (2, 8), (2, 2) asymptotes: y  k 
  2 [x  (3)] 4 1 (x  3) 2  y3
   h)  2)  y  y  (2, 8)  O (7, 1)  a (x b 5 (x 4  (2, 3)  x (1, 1)  (3, 1)  x  O (2, 2)  20.  4x  9y2  24x  90y  153 
 0 9( y2  10y  ?)  4(x2  6x  ?) 
 153 9(y2  10y  25)  4(x2  6x  9) 
 153  9(25)  4(9) 9( y  5)2  4(x  3)2 
 36 ( y  5)2  4    (x  3)2  9  22.  36(x2x1)49(y2 6y9)
216936(1)49(9) 36(x1)2 49(y3)2 
1764 (x 1)2 (y 3)2    
1 49 36  
1  center: (h, k) 
 (1, 3) b2 
 36 c 
  a2  b2 a2 
 49 a 
 49  b 
 36  c 
 49 36  or 7 or 6 or 85  transverse axis: horizontal foci: (h c, k) 
 (1 85 , 3) vertices: (h a, k) 
 (1 7, 3) or (8, 3), (6, 3)  center: (h, k) 
 (3, 5) b2 
 9 c 
  a2  b2 a2 
 4 a 
 4  or 2 b 
 9  or 3 c 
 4   9 or 13  transverse axis: vertical foci: (h, k c) 
 (3, 5 13 ) vertices: (h, k a) 
 (3, 5 2) or (3, 7), (3, 3) asymptotes: y  k 
 y5
 y5
  36x2 49y2 72x294y
2169  36(x2 2x?)49(y2 6y?)
2169??  a (x  h) b 2 [x  (3)] 3 2 (x  3) 3  asymptotes: y  k 
 y  (3) 
 y3
  y   h)  1)  1)  y  (3, 7) (3, 5)  O  (3, 3)  b (x a 6 (x 7 6 (x 7  x  (6, 3)  O  329  (1, 3)  x (8, 3)  Chapter 10  23.  27. c 
 49 quadrants: I and III transverse axis: y 
 x vertices: xy 
 49 7(7) 
 49 (7, 7)  25y2  9x2  100y  72x  269 
 0 25(y2  4y  ?)  9(x2  8x  ?) 
 269  ?  ? 25(y2  4y  4)  9(x2  8x  16) 
 269  25(4)  9(16) 25(y  2)2  9(x  4)2 
 225 (y  2)2 (x  4)2    
 1 9 25  center: (h, k) 
 (4, 2) b2 
 25 c 
  a2  b2 a2 
 9 a 
 9  b 
 25  c 
 9 5  2 or 3 or 5 or 34  transverse axis: vertical foci: (h, k c) 
 
4, 2 34  vertices: (h, k a) 
 (4, 2 3) or (4, 5), (4, 1) asymptotes: y  k 
 y2
 y2
  a (x b 3 [x 5 3 (x 5  y (7, 7)  (7, 7)   (4)] 28. c 
 36 quadrants: II and IV transverse axis: y 
 x vertices: xy 
 36 6(6) 
 36 (6, 6)   4)  y  (4, 5)  3 y  2 
 (x  4) 5  O  (4, 1)  y  (6, 6)  24. center: (h, k) 
 (4, 3) transverse axis: vertical a 
 4, b 
 3  x  O (6, 6)  (x  h)2   b 
1 2 (x  4)2   3 
1 2  29. 4xy 
 25 25 xy 
 4  (x  4)2   9 
 1  25. center: (h, k) 
 (0, 0) transverse axis: horizontal a 
 3, b 
 3 (x  h)2  a2 (x  0)2  32  25  c 
 4  (y  k)2  quadrants: II and IV transverse axis: y 
 x  (y  0)2  vertices:   b 
1 2  3 
1 2 x2  9  (y  (x    b 2 a2 (y  0)2 [x  (4)]2   1 2 22 y2 (x  4)2    4 1 k)2  b)2  25  xy 
 4  25  25  2
2 
 4  xy 
 4  
   5 5   
 2 2 5 5 ,   2 2  y2   9 
 1  
  26. center: (h, k) 
 (4, 0) transverse axis: vertical a 
 2, b 
 1  5 5  4    
52, 52  y  
1  ( 52, 52 )  
1  x  
1  O  ( 52,  52 )  Chapter 10  xy 
 36 6(6) 
 36 (6, 6)  x  (4, 2   34)  (y  k)2  a2 (y  3)2  42 (y  3)2  16  x  O   h)  (4, 2   34) 3 y  2 
  (x  4) 5 (4, 2)  xy 
 49 7(7) 
 49 (7, 7)  330  25  34. 2b 
 8 b
4  30. 9xy 
 16 xy 
 c
  16  9 16  9  x1  x2 y1  y2  3  (3) 9  (5)  center: 
2, 2 
 
2, 2 
 (3, 2) transverse axis: vertical a 
 distance from center to a vertex 
 2  9 or 7  quadrants: I and III transverse axis: y 
 x 16  xy 
 9  16  4 4 16   
  3 3 9 4 4 ,  3 3  3
3 
 9  xy 
 9  vertices:  
  
   4    4  16  
43, 43  y 35.  x  O  8  (8) 0  0 ,     
 
 2 2  
  
 (0, 0) transverse axis: horizontal c 
 distance from center to a focus 
 0  8 or 8 b2 
 c2  a2 b2 
 a2 a2 
 c2  a2 b2 
 32 2a2 
 c2  ( 43, 43 ) ( 43,  43 )  (y  k)2 (x  h)2   b 
1 2 a2 (y  2)2 [x  (3)]2   4 
1 2 72 (y  2)2 (x  3)2    
 1 49 16 x1  x2 y1  y2 center: 2, 2  c2  a2 
 2  31. center: (h, k) 
 (4, 2) (y  k)2  a2 [y  (2)]2  22 (y  2)2  4  82  (x  h)2  a2 
 2 or 32   b 
1 2  (x  h)2  a2 (x  0)2  32  (x  4)2   3 
1 2 (x  4)2   9 
 1  x1  x2 y1  y2  c  5  4  (x  h)2  a2 [x  (3)]2  42 (x  3)2  16  (x  0)2   72 
 1 x2   72 
 1  5  (5) 2  2  
 (0, 2) transverse axis: horizontal c 
 distance from center to a focus 
 0  5 or 5 b2 
 c2  a2 b2 
 52  32 or 16    (y  1)2   9 
 1  3  (y  k)2    (y  1)2   9 
 1  y  2 
 4(x  4) a  b 3  b   b 
1 2 (y  2)2  16 (y  2)2  16  (y  k)2   b 
1 2  37. center: (h, k) 
 (4, 2) a 
 distance from center to a vertex 
 2  5 or 3 transverse axis: vertical 4y  4 
 3x 4y  4  12 
 3x  12 4y  8 
 3x  12 4(y  2) 
 3(x  4)  center: 
2, 2 
 
2, 2  (x  b)2  a2 (x  0)2  32 x2  9  5  
 a  a
4 transverse axis: horizontal  (x  h)2  x1  x2 y1  y2  b2 
 c2  a2 b2 
 52  42 b2 
 9  e 
 a   b 
1 2  33. 2a 
 6 a
3  y2   32 
 1  36. centers: (h, k) 
 (3, 1) c 
 distance from center to a focus 
 3  2 or 5  
 (0, 0) transverse axis: vertical a 
 distance from center to a vertex 
 0  3] or 3 c 
 distance from center to a focus 
 0  (9) or 9 b 2 
 c 2  a2 b2 
 92  32 or 72  y2  9  (y  0)2   32 
 1 x2  32  0  0 3  (3)  32. center: 
2, 2 
 
2, 2  (y  k)2  a2 (y  0)2  32  (y  k)2   b 
1 2  
1  3  
 4 3  
 4  b
4  
1  (y  k)2  a2 (y  2)2  32 (y  2)2  9  331  (x  h)2   b 
1 2 (x  4)2   4 
1 2 (x  4)2   16 
 1  Chapter 10  38. center: (h, k) 
 (3, 1) a 
 distance from center to a vertex 
 3  5 or 2 transverse axis: horizontal 3x  11 
 2y 3x  11  4 
 2y  4 3x  15 
 2y  4 3(x  5) 
 2(y  2) 3 (x 2  transverse axis: horizontal (x  h)2  a2 (x  0)2 __ 81  2  3  
 2 3  
 2  a  b  b
3  
  0  0 8  (8)   ,  2 2  

  c  4  3  
  8  a  a
6 transverse axis: vertical (y   a2 (y  0)2  62 k)2  40.  (x     b2 
 c2  a2 b2 
 82  62 b2 
 28  a2 
 4
5  6  
  
1  PV 
 505 (101)V 
 505 V 
 5.0 dm3 43c. PV 
 505 (50.5)V 
 505 V 
 10.0 dm3 43d. If the pressure is halved, then the volume is doubled, or V 
 2(original V ). 44. In an equilateral hyperbola, a 
 b and c2 
 a2  b2. c 2 
 a2  a2 a
b c2 
 2a2 c 
 a2  c  9  (9) 0  0 ,   0 2  

  Since e 
 a, we have    c  e 
 a  
 (0, 0) c 
 distance from center to a focus 
 0  9 or 9 b2 
 c 2  a 2 b2 
 a2 a2 
 92  a2 2a2 
 81 81 a2 
 2  
1  43b.  a
5 transverse axis: horizontal (x  h)2 (y  k)2   b 
1 2 a2 (x  4)2 [y  ( 3)]   11 
 1 52 (x  4)2 (y  3)2    
 1 25 11 x1  x2 y1  y2    center: 2 , 2  
1  O 2 4 6 8 10 P  10  (2) 3  (3) ,   
 
 2 2  b2 
 c2  a2 b2 
 62  52 b2 
 11  
 a  64  a2 
 5  V  (x  0)2  c  16  250 200 150 100 50   28 
 1  e 
 a  Chapter 10  a 
 2b c2 
 a2  b2 42 
 (2b)2  b2 16 
 5b2  43a. quadrants: I and II transverse axis: y 
 x  
 (4, 3) c 
 distance from center to a focus 
 4  10 or 6  41.  a2 
 (2b)2   b 
1 2  
  1  1 5  (3)    
 
 2 , 2  2  h)2  y2 x2    
 1 36 28 x1  x2 y1  y2 centers: 2, 2  6  5  
  16  
 b2 5 (y  k)2 (x  h)2   b 2 a2 (x  1)2 (y  1)2    64 16   5 5 5(y  1)2 5(x  1)2    64 16  
 (0, 0) c 
 distance from center to a focus 
 0  8 or 8 e 
 a  
1  
 (1, 1) c 
 distance from center to a focus 
 1  5 or 4 transverse axis: vertical  3  39.  (y  0)2 __ 81  2  42. center: 
   5) 
 y  2  (x  b)2 (y  k)2   b 
1 2 a2 (x  3)2 [y  (1)]2   3 
1 2 22 (x  3)2 (y  1)2    
 1 4 9 x1  x2 y1  y2    center: 2 , 2    2x2 2y2    
 1 81 81 x1  x2 y1  y2  ,  2 2  y  2 
 2(x  5) b  a b  2  (y  k)2   b 
1 2  e
  a2   a   e 
 2 Thus, the eccentricity of any equilateral hyperbola is 2 .  81  b2 
 2  332  y  45a.  The lightning is 2200 feet farther from station B than from station A. The difference of distances equals 2a. 2200 
 2a (Definition of hyperbola) 1100 
 a b 2 
 c 2  a2 b 
  c2  a2 b 
  10,56 02  1 1002 b 10,503 center: (h, k) 
 (0, 0) transverse axis: horizontal  150 ft  x  O  2a 
 150 a 
 75 5  3  c  75  
  (x  h)2  a2 (x  0)2  11002 x2  11002  b2 
 c2  a2 b2 
 1252  752 b2 
 10,000 b 
 100  c  e 
 a  125 
 c transverse axis: horizontal center: (h, k) 
 (0, 0) (x  h)2  a2 (x  0)2  752  (y  k)2  (y  0)2    1002 
 1  45b.  y2    1002 
 1  y (x, 100) (0, 100)  48a.  1002    1002 
 1  x2  752  (x, y) 
 (x, 100)  1
1 x2  752 x2  x2  16  (y  k)2   b 
1 2 (y  0)2   11 
 1 y2   11 
 1  y2   9 
 1  asymptotes: y  k 
  
2  y0
  
 11,250 x 106.07 ft  x2  y
  y2    base:  752  1002 
 1 x2 (350)2   752  1002 x2  752  12.25 x2  752 x2  y2  9  
1  PF1  PF2  2a  center: (h, k) 
 (0, 0) b2 
 9 a2 
 16 a 
 16  or 4 b 
 9  or 3 transverse axis: horizontal vertices: (h a, k) 
 (0 4, 0) or ( 4, 0)  y2    top:  752  1002 
 1 x2  752  y2    10,5032 
 1  x2  25  (x, 350)  (0, 350)  x2  (x  h)2  a2 (x  0)2  52  x  O  (y  0)2    10,5032 
 1  47. center: (h, k) 
 (0, 0) c 
 0  6 or 6 PF1  PF2 
 10 2a 
 10 a
5 b2 
 c2  a2 b2 
 62  52 b2 
 11 transverse axis: horizontal   b 
1 2  x2  752  (y  k)2   b 
1 2  (x, y) 
 (x, 350)    x2  16  b (x a 3 (x 4 3 x 4   h)  h)  
1  center: (h, k) 
 (0, 0) a2 
 9 b2 
 16 a 
 9  or 3 b 
 16  or 4 transverse axis: vertical vertices: (h, k a) 
 (0, 0 3) or (0, 3)  
1 
 13.25  
 74,531.25 x 273.00 ft 46. The origin is located midway between stations A and B. The stations are located at the foci, so 2c 
 4 or c 
 2 miles. c 
 2 mi  asymptotes: y  k 
 y0
 y
  a (x b 3 (x 4 3 x 4   h)  0)  y  5280 ft   c 
 2 mi   1 mi  c 
 10,560 ft d 
 rt d 
 1100(2) or 2200 ft  y O A  (10,560, 0)  x  B  O  (10,560, 0) x  333  Chapter 10  48c. 48d.  (y  2)2  25 (x  3)2  16     (x  3)2  16 (y  2)2  25  y  51.  48b. They are the same lines. 
1  A (1, 3)  center: (h, k) 
 (3, 2) b2 
 25 a2 
 16 a 
 16  or 4 b 
 25  or 5 transverse axis: horizontal vertices: (h a, k) 
 (3 4, 2) or (7, 2), (1, 2) asymptotes: y  k 
 y2
 (y  2)2  25    (x  3)2  16  b (x a 5 (x 4  C (6, 2)  B (2, 1)  AB 
  (2  1 )2  ( 1  3 )2 
 5 2 BC 
  (2  6 )  ( 1  2 )2 
 5 2 2 (6  3 )  (2  6) 
 5 CD 
  AD 
  (3  1 )2  (6  3)2 
 5 Thus, ABCD is a rhombus. The slope of A D 
   h)  3)  
1  asymptotes: y  k 
 y2
  a (x b 5 (x 4  63  31   h)  3)  8 4 8  x  4 x1  x2 y1  y2  1 x 2  2  2 3  (3)  49. center: 
2, 2 
 
2, 2  50.  4   3   2 y  3 
 0  y  6 
 0 x  3 x  55.  
 (2, 0) a
4 c 
 distance from center to a focus 
 0  3 or 3 b2 
 a2  c2 b2 
 42  32 or 7 major axis: vertical (y  k)2  a2 (y  0)2  42 y2  16  31  3    or 4 and the slope of  AB 
 1  2 or  3 .  Thus, A D  is perpendicular to A B  and ABCD is a square. 52. (r, v) 
 (90, 208°) (r, v  360 k°) 
 (90, 208°  360(1)°) 
 (90, 152°) (r, v  (2k  1)(180°)) 
 (90, 208  (2(1)  1)(180°)) 
 (90, 28°) 53. 4(5)  1(2)  8(2) 
 6 No, the inner product of the two vectors is not zero. 54. x cos f  y sin f  p 
 0 x cos 60  y sin 60  3 
 0  y  4  x  O  center: (h, k) 
 (3, 2) b2 
 16 a2 
 25 a 
 25  or 5 b 
 16  or 4 transverse axis: vertical vertices: (h, k a) 
 (3, 2 5) or (3, 7), (3, 3)  4 O  D (3, 6)  
1  9000 m 30˚ 60˚  (x  h)2   b 
1 2  x   tan 30° 
  9000 9000 tan 30° 
 x 5196 x d 
 rt  (x  2)2  7 
 1 (x  2)2  7 
 1 x2  y2  4x   14y  28 
 0 (x2  4x  ?)  (y2  14y  ?) 
 28  ?  ? (x2  4x  4)  (y2  14y  49) 
 28  4  49 (x  2)2  (y  7)2 
 81  d  t 5196  15  
r 
r  346.4 r about 346 m/s  y 56.  x  O (2, 7) (11, 7)  (2, 16)  Chapter 10  Since 0.2506 is closer to zero than 0.6864, the zero is about 1.3.  334  7. y2  4x  2y  5 
 0 y2  2y 
 4x  5 y2  2y  ? 
 4x  5  ? y2  2y  1 
 4x  5  1 (y  1) 
 4(x  1) vertex: (h, k) 
 (1, 1) 4p 
 4 p
1 focus: (h  p, k) 
 (1  1, 1) or (2, 1) directrix: x 
 h  p x
11 x
0 axis of symmetry: y 
 k y 
 1  Since 0.0784 is closer to zero than 0.2446, the zero is about 0.6. 57. Case 1: r is positive and s is negative. Case 2: r is negative and s is positive. I. r3  s3 is false if r is negative. II. r3 
 s2 is false for each case. III. r4 
 s4 is true for each case. The correct choice is C.  y x
0  10-5  Parabolas  Pages 658–659  O  x  (2, 1) (1, 1)  Check for Understanding  1. The equation of a parabola will have only one squared term, while the equation of a hyperbola will have two squared terms. 2. vertex: (h, k) 
 (2, 1) p 
 4 (x  h)2 
 4p(y  k) (x  2)2 
 4(4)( y  1) (x  2)2 
 16( y  1) 3. The vertex and focus both lie on the axis of symmetry. The directrix and axis of symmetry are perpendicular to each other. The focus and the point on the directrix collinear with the focus are equidistant from the vertex. 4. (h, k) 
 (4, 5) p 
 5 (y  k)2 
 4p(x  h) (y  5)2 
 4(5)[x  (4)] (y  5)2 
 20(x  4) 5a. ellipse 5b. parabola 5c. hyperbola 5d. circle 6. vertex: (h, k) 
 (0, 1) 4p 
 12 p
3 focus: (h, k  p) 
 (0, 1  3) or (0, 4) directrix: y 
 k  p y
13 y 
 2 axis of symmetry: x 
 h x
0 y  8. x2  8x  4y  8 
 0 x2  8x 
 4y  8 x2  8x  ? 
 4y  8  ? x2  8x  16 
 4y  8  16 (x  4)2 
 4(y  2) vertex: (h, k) 
 (4, 2) 4p 
 4 p 
 1 focus: (h, k  p) 
 (4, 2  (1)) or (4, 1) directrix: y 
 k  p y 
 2  (1) y
3 axis of symmetry: x 
 h x 
 4  y y
3 (4, 2) (4, 1)  O x  9. vertex: (h, k) 
 (0, 0) opening: downward p 
 4 (x  h)2 
 4p(y  k) (x  0)2 
 4(4)(y  0) x2 
 16y  (0, 4) (0, 1)  y  O  x  x  O y 
 2  335  Chapter 10  10.  (y  k)2 
 4p(x  h) (1  5)2 
 4p[2  (7)] (6)2 
 36p 1
p (y  k)2 
 4p(x  h) (y  5)2 
 4(1)[x  (7)] (y  5)2 
 4(x  7)  12b. The maximum height is s 
 52 ft. 12c. Let s 
 0. s 
 56t  16t2  3 0 
 16t2  56t  3  (h, k)  (7, 5); (x, y)  (2, 1)  y  t
  b  b2  4 ac  2a  t
  56  562  4(1 6)(3)  2(16)  t 3.6 or 0.05 3.6 s  Pages 659–661 (7, 5)  x O 11. vertex: (h, k) 
 (4, 3) opening: upward (x  h)2 
 4p(y  k) (5  4)2 
 4p[2  (3)] 12 
 20p (x   1  20 h)2  Exercises  13. vertex: (h, k) 
 (0, 0) 4p 
 8 p
2 focus: (h  p, k) 
 (0  2, 0) or (2, 0) directrix: x 
 h  p x
02 x 
 2 axis of symmetry: y 
 k y
0 (h, k)  (4, 3); (x, y)  (5, 2)  y  
p 
 4p(y  k)  O  (x  4)2 
 4
2 0 (y  3) 1  (2, 0)  x  x 
 2  1  (x  4)2 
 5( y  3)  y 14. vertex: (h, k) 
 (0, 3) 4p 
 4 p
1 focus: (h, k  p) 
 (0, 3  (1)) or (0, 2) directrix: y 
 k  p y y 
 3  (1) y
4 y
4 (0, 3) axis of symmetry: x 
 h (0, 2) x
0  x O (4, 3)  s 
 v0t  16t2  3 s 
 56t  16t2  3 2 16t  56t 
 s  3  12a.  O  x  16
t2  5t  ? 
 s  3  ? 7  16
t2  2t  16 
 s  3  (16)
16 7  49  
 
     
  opening: downward  
     50 45 40 35 30 25 20 15 10 5  O Chapter 10  15. vertex: (h, k) 
 (0, 6) 4p 
 4 p
1 focus: (h  p, k) 
 (0  1, 6) or (1, 6) directrix: x 
 h  p x
01 x 
 1 y axis of symmetry: y 
 k y
6  7 2 t  4 
 s  52 7 2 1 t  4 
 1 6 (s  52) 7 2 1 t  4 
 4 6 4 (s  52) 7 s k) 
 4, 52 55  16
  vertex: (h,  49  x 
 1  (1, 6)  (0, 6)  1  2  3  4  x  O  336  x  16.  y 2  12x 
 2y  13 y2  2y 
 12x  13 y2  2y  ? 
 12x  13  ? y2  2y  1 
 12x  13  1 ( y  1)2 
 12(x  1) vertex: (h, k) 
 (1, 1) 4p 
 12 p 
 3 focus: (h  p, k) 
 (1  (3), 1) or (4, 1) directrix: x 
 h  p x 
 1  (3) x
2 axis of symmetry: y 
 k y
1  18. x2  10x  25 
 8y  24 (x  5)2 
 8(y  3) vertex: (h, k) 
 (5, 3) 4p 
 8 p 
 2 focus: (h, k  p) 
 (5, 3  (2)) or (5, 1) directrix: y 
 k  p y 
 3  (2) y
5 axis of symmetry: x 
 h x 
 5 y y
5  y (5, 3)  x
2 (1, 1)  O  O  17.  (5, 1)  x  (4, 1)  x 19. y2  2x  14y 
 41 y2  14y 
 2x  41 2 y  14y  ? 
 2x  41  ? y2  14y  49 
 2x  41  49 (y  7)2 
 (2x  4) vertex: (h, k) 
 (4, 7) 4p 
 2  y  2 
 x2  4x x2  4x 
 y  2 x2  4x  ? 
 y  2  ? x2  4x  4 
 y  2  4 (x  2)2 
 y  2 vertex: (h, k) 
 (2, 2) 4p 
 1  2  focus: (h  p, k) 
 
4  2, 7 or 
2, 7 1  1  focus: (h, k  p) 
 
2, 2    or 
2,  1  4  1    x 
 4  2  7 4  9  x 
 2  directrix: y 
 k  p 1  axis of symmetry: y 
 k y 
 7  y 
 2  4 y
  9 4  y  axis of symmetry: x 
 h x
2 y  O  7  directrix: x 
 h  p  p 
 4  1  1  p 
 4 or 2  1  2  3  x
  x  ( 72 , 7)  (2,  74 ) (2, 2)  x  O 9 2  y 
  94  (4, 7)  337  Chapter 10  vertex: (h, k) 
 (3, 2) 4p 
 10  20. y2  2y  12x  13 
 0 y2  2y 
 12x  13 y2  2y  ? 
 12x  13  ? y2  2y  1 
 12x  13  1 ( y  1)2 
 12(x  1) vertex: (h, k) 
 (1, 1) 4p 
 12 p
3 focus: (h  p, k) 
 (1  3, 1) or (4, 1) directrix: x 
 h  p x
13 x 
 2 axis of symmetry: y 
 k y
1  10  5  p 
 4 or 2 focus: (h, k  p) 
 
3, 2  2 or 
3, 2 5  9  directrix: y 
 k  p 5  y 
 2  2 1  y 
 2 axis of symmetry: x 
 h x
3 y  (3, 92 )  y x 
 2  O (1, 1)  (4, 1)  O  x  23. 2y2  16y  16x  64 
 0 2y2  16y 
 16x  64 2(y2  8y  ?) 
 16x  64  ? 2(y2  8y  16) 
 16x  64  2(16) 2(y  4)2 
 16x  32 (y  4)2 
 8x  16 (y  4)2 
 8(x  2)  21. 2x2  12y  16x  20 
 0 2x2  16x 
 12y  20 2(x2  8x  ?) 
 12y  20  ? 2(x2  8x  16) 
 12y  20  2(16) 2(x  4)2 
 12y  12 (x  4)2 
 6(y  1) vertex: (h, k) 
 (4, 1) 4p 
 6 6  y 
  12 x  (3, 2)  vertex: (h, k) 
 (2, 4) 4p 
 8 p 
 2 focus: (h  p, k) 
 (2  (2), 4) or (4, 4) directrix: x 
 h  p y x 
 2  (2) x
0 O x axis of symmetry: y 
 k x
0 y 
 4  3  p 
 4 or 2 focus: (h, k  p) 
 
4, 1  2 or 
4, 2 3  1  directrix: y 
 k  p 3  (4, 4)  y 
 1  2 5  y 
 2 axis of symmetry: x 
 h x
4 y  24. vertex: (h, k) 
 (5, 1) opening: right hp
2 5  p 
 2 p
7 (y  k)2 
 4p(x  h) (y  1)2 
 4(7)[x  (5)] (y  1)2 
 28(x  5)  (4, 12 ) O  x  (4, 1)  y 
  52  y 8  22. 3x2  30x  18x  87 
 0 3x2  18x 
 30y  87 3(x2  6x  ?) 
 30y  87  ? 3(x2  6x  9) 
 30y  87  3(9) 3(x  3)2 
 30y  60 (x  3)2 
 10y  20 (x  3)2 
 10(y  2)  Chapter 10  4 (5, 1) 12 8 4 O 4 8  338  x  (2, 4)  25. opening: left focus: (h  p, k) 
 (0, 6) hp
0 h  (3) 
 0 h
3 (y  k)2 
 4p(x  h) (y  6)2 
 4(3)(x  3) (y  6)2 
 12(x  3)  28. vertex: (h, k) 
 (2, 3) (y  k)2 
 4p(x  h) [1  (3)]2 
 4p[3  (2)] 42 
 4p 4 
 p (y  k)2 
 4p(x  h) (y  3)2 
 4(4)(x  2) (y  3)2 
 16(x  2)  k
6  y 16 12 8 (3, 6)  O 4  8  y O  x  (2, 3)  4 8 4  (h, k)  (2, 3); (x, y)  (3, 1)  x  4  29. opening: upward p
2 focus: (h, k  p) 
 (1, 7) h 
 1 kp
7 k2
7 k
5 2 (x  h) 
 4p(y  k) [x  (1)]2 
 4(2)(y  5) (x  1)2 
 8(y  5)  26. opening: upward vertex: (h, k) 
 
4,  51  2   or (4, 3)  focus: (h, k  p) 
 (4, 1) kp
1 3  p 
 1 p
2 (x  b)2 
 4p(y  k) (x  4)2 
 4(2)[y  (3)] (x  4)2 
 8(y  3) y  O  (1, 5)  O  x 30. opening: downward vertex: (h, k) 
 (5, 3) (x  h)2 
 4p(y  k) (1  5)2 
 4p[7  (3)] (4)2 
 16p 1 
 p (x  h)2 
 4p(y  k) (x  5)2 
 4(1)(y  3) (x  5)2 
 4(y  3)  (4, 3)  27. opening: downward vertex: (h, k) 
 (4, 3) (x  h)2 
 4p(y  k) (5  4)2 
 4p(2  3) 12 
 4p  (h, k)  (4, 3); (x, y)  (5, 2)  1  4 
 p  y O  (x  h)2 
 4p(y  k) (x   4)2  
 4
  (x   4)2  
 (y  3)  y  O  y  (y  3)  1 4  x  (maximum) (h, k)  (5, 3); (x, y)  (1, 7)  x (5, 3)  (4, 3)  x  31. opening: right vertex: (h, k) 
 (1, 2) ( y  k)2 
 4p(x  h) (0  2)2 
 4p[0  (1)] (2)2 
 4p 1
p (y  k)2 
 4p(x  h) (y  2)2 
 4(1)[x  (1)] (y  2)2 
 4(x  1)  (h, k)  (1, 2); (x, y)  (0, 0)  y  (1, 2)  O  339  x  Chapter 10  34a. Let y 
 income per flight. Let x 
 the number of $10 price decreases. Income 
 number of passengers  cost of a ticket y 
 (110  20x)  (140  10x) y
 15,400  1100x  2800x  200x2  32. opening: upward h
 
  x1  x2   2 12 3  or  2 2  vertex: (h, k) 
 
2, 0 3  (x  h)2 
 4p(y  k) 2  
1  32  (x   1  
 4p 4 1  
 p 16 h)2 
 4p(y  3 2 x  2 3 y  2  
 
  y 
 200
x2  2x  15,400 17  (h, k)  
2, 0; 3  
 4p(1  0)  y  15,400 
 200
x2  2x 17  (x, y)  (1, 1)  17 2  1   200   k)   
 4
 (y  0)  
 14y 1  16  1  ( 32 , 0)  2  x  33a. vertex: (h, k) 
 (0, 0) depth: x 
 4 p
2 y (4, y 1)  O  (2, 0)  (4, 0) x  (4, y 2)  ( y  k)2 
 4p(x  h) (y  0)2 
 4(2)(x  0) y2 
 8x y 
 8x  y 
 8(4)  y 
 42  y1 
 42 , y2 
 42   (h, k)  (0, 0); p2  (y  19,012.5) 
 
x     17 2  4    17  3 2  3 2  y  15,400 100
2 
 100 x2  3x  
2 y  15,625 
 100
x  1   100  (y  15,625) 
 
x   3 2  2      3 2  2    The vertex of the parabola is at 
2, 15,625, and because p is negative, it opens downward. So the vertex is a maximum and the number of $10 3 price decreases is 2 or 1.5. number of passengers 
 110  10x 
 110  10(1.5) 
 125 This is less than 180, so the new ticket price can be found using 1.5 $10 price decreases. cost of a ticket 
 140  10x 
 140  10(1.5) 
 $125  diameter 
 y1  y2 
 42   (42 ) 
 82  in. 33b. depth: x 
 1.25(4) x
5 ( y  k)2 
 4p(x  h) (y  0)2 
 4(2)(x  0) (h, k)  (0, 0); y2 
 8x p2 y 
 8x  y 
 8(5)  y 
 210  y1 
 210 , y2 
 210  diameter 
 y1  y2   (210 ) 
 210 
 410  in.  Chapter 10  17  The vertex of the parabola is at 
4, 19,012.5, and because p is negative it opens downward. So the vertex is a maximum and the number of $10 17 price decreases is 4 or 4.25. number of passengers 
 110  20x 
 110  20(4.25) 
 195 However, the flight can transport only 180 people. number of passengers 
 110  20x 180 
 110  20x 3.5 
 x Therefore, there should be 3.5 $10 price decreases. cost of ticket 
 140  10x 
 140  10(3.5) 
 $105 34b. Let y 
 income per flight. Let x 
 the number of $10 price decreases. Income 
 number of passengers  cost of a ticket y 
 (110  10x)  (140  10x) y 
 15,400  300x  100x2 y 
 100(x2  3x)  15,400 y  15,400 
 100(x2  3x)  1  O  17 2  y  15,400  200
4 
 200 x2  2x  
9  y  340  3  35a. Let (h, k) 
 (0, 0). x2 
 4py x2 
 4
8y 1  2x2 
 y x2 
 4py x2 
 1(1)y 1 x 4  38b. 4p 
 16 p 
 4 focus of parabola 
 center of circle vertex: (h, k) 
 (1, 4) focus: (h, k  p) 
 (1, 4  (4)) or (1, 0) diameter 
 latus rectum 
 16  x2 
 4py  1  p  8  x2 
 4
4y 1  1  p  4  x2 
 y  y  p1  2x 2 
 y  x2 
 y  
y  1  radius 
 2(16) 
8 (x  h)2  (y  k)2 
 r2 (x  1)2  (y  0)2 
 82 (x  1)2  y2 
 64 39. center: (h, k) 
 (2, 3) a2 
 25 b2 
 16 a 
 25  or 5 b 
 16  or 4 c 
  a2  b2 c 
 25 16   or 41  transverse axis: vertical foci: (h, k c) 
 (2, 3 41 ) vertices: (h, k a) 
 (2, 3 5) or (2, 8), (2, 2) a asymptotes: y  k 
 b (x  h)  The opening becomes narrower. 1 2 x 4  O  
y  x  35b. The opening becomes wider. 36a. Sample answer: opening: upward vertex: (h, k) 
 (0, 0)  y (2100, 490) (0, 0) 2100 10 ft O  500 ft  y3
  2100 x  y  roadway  (x  h)2 
 4p(y  k) (2100  0)2 
 4p(490  0) (h, k)  (0, 0) 2250 
 p (x, y)  (2100, 490) (x  h)2 
 4p(y  k) (x  0)2 
 4(2250)( y  0) x2 
 9000y 36b. x2 
 9000y (720)2 
 9000y 57.6 
 y 57.6  10 
 67.6 ft 37. (y  k)2 
 4p(x  h) y2  2ky  k2 
 4px  4ph y2  4px  2ky  k2  4ph 
 0 y2  Dx  Ey  F 
 0 (x  h)2 
 4p(y  k) x2  2hx  h2 
 4py  4pk x2  4py  2hx  h2  4pk 
 0 x2  Dx  Ey  F 
 0 38a. 4p 
 8 p 
 2 or p 
 2 opening: right (y  k)2 
 4p(x  h) (y  2)2 
 4(2)[x  (3)] (y  2)2 
 8(x  3) opening: left (y  k)2 
 4p(x  h) (y  2)2 
 4(2)[x  (3)] (y  2)2 
 8(x  3)  5 (x 4   2)  (2, 8)  (2, 3)  O  x  (2, 2)  40. 4x2  25y2  250y  525 
 0 4x2  25(y2  10y  ?) 
 525  ? 4x2  25(y2  10y  25) 
 525  25(25) 4x2  25(y  5)2 
 100 x2  25  ( y  5)2   4 
 1  center: (h, k) 
 (0, 5) a2 
 25 b2 
 4 a 
 25 or 5 b 
 4  or 2 2 2 c 
  a  b c 
 25 4   or 21  foci: (h c, k) 
 (0 21 , 5) or ( 21 , 5) major axis vertices: (h a, k) 
 (0 5, 5) or ( 5, 5) minor axis vertices: (h, k b) 
 (0, 5 2) or (0, 3), (0, 7) y  O  x  (0, 3) (5, 5)  (0, 5) (5, 5) (0, 7)  341  Chapter 10  41.  
 0   6  6    3  6  
6, 3    2  12  
12, 2  2  3  6  
6, 23  5  6  6  
6, 56  12  (12, )  7  6  6  
6, 76  4  3  6  
6, 43  3  2  12  3  2  
12,   5  3  6  
6, 53  6  
6, 116     11  6   2  2 3  45. 19 t 14  19 19 t 33 t 
 33 or 27 perimeter 
 14  19  t 
 14  19  27 
 60 The correct choice is C.  (r, 
) (12, 0)  12 cos 2
 12  
6, 6  Page 661  
  (6  3 )2  (9  3)2 2 2 
  3  6 
 45  BC 
  (x2   x1)2  (y2  y1)2 
  (9  6 )2  (3  9)2 2 2 
  3  ( 6) 
 45  Since AB 
 BC, triangle ABC is isosceles. 1b. AC 
  (x2   x1)2  (y2  y1)2 
  (9  3 )2  (3  3)2 2 2 
  6  0 
6 perimeter 
 AB  BC  AC 
 45   45 6 19.42 units 2. Diagonals of a rectangle intersect at their midpoint. x1  x2 y1  y2 midpoint of A C  
 
2, 2   3  6  5 6   0  3 6 9  11 6  7 6 4 3  3 2  5 3  4  5 9  5  
 
2, 2  42. 2n, where n is any integer 43.  3.  The measure of a is 360° 12 or 30°. a   cos 30° 
  6.4  a  44.    6.4 cm  6.4 cos 30° 
 a 5.5 a; 5.5 cm  
 (0.5, 7) x2  y2  6y  8x 
 16 (x2  8x  ?)  (y2  6y  ?) 
 16  ?  ? (x2  8x  16)  (y2  6y  9) 
 16  16  9 (x  4)2  (y  3)2 
 9 center: (4, 3): radius: 9  or 3 y (4, 6)  (4, 3)  4   g(x)   x2  1  x 10,000 1000 100 10 0 10 100 1000 10,000  (7, 3)  y  g(x) 4  108 4  106 4  104 0.04 4 0.04 4  104 4  106 4  108  O  x  (x  h)2  (y  k)2 
 r2 2 [x  (5)]2  (y  2)2 
 
7  (x  5)2  (y  2)2 
 7 y 5a. Let d1 be the greatest distance from the d2 satellite to Earth. Let d1 O d2 be the least x a c distance from the satellite to Earth. 4.  y → 0 as x → , y → 0 as x →    Chapter 10  Mid-Chapter Quiz  1a. AB 
  (x2   x1)2  (y2  y1)2  342  1  7.  a 
 2(10,440) a 
 5220 c  
 e a c  5220  (y  4)2  2  
 0.16  
 3960 d1 
 a  c  Earth radius d1 
 5220  835.20  3960 d1 
 2095.2 miles d2 
 major axis  d1  Earth diameter d2 
 10,440  2095.2  7920 d2 
 424.8 miles 5b. (h, k) 
 (0,0) a 
 5220 b2 
 a2(1  e2) b2 
 (5220)2 (1  0.162) b2 
 26,550,840.96  vertices: (h, k  y  (4) 
 y4
  y  (4, 5)  1)  x  (1, 4  2) (1, 4) (1, 4  2)  8. To find the center, find the intersection of the asymptotes. y 
 2x  4 2x 
 2x  4 4x 
 4 x
1 y
21 y
2 The center is at (1, 2). Notice that (4, 2) must be a vertex and a equals 4  1 or 3. Point A has an x-coordinate of 4. Since y 
 2x, the y-coordinate is 2  4 or 8. The value of b is 8  2 or 6.  (y  5)2   9 
 1  (x  1)2  (y  2)2  The equation is 9  36 
 1.  y  (1, 5)  (1)]  O  center: (h, k) 
 (4, 5) b2 
 9 c 
  a2  b2 a2 
 25 a
5 b
3 c 
 25 9   or 4 major axis vertices: (h a, k) 
 (4 5, 5) or (9, 5), (1, 5) minor axis vertices: (h, k b) 
 (4, 5 3) or (4, 2), (4, 8) foci: (h c, k) 
 (4 4, 5) or (8, 5), (0, 5)  9. y2  4x  2y  5 
 0 y2  2y 
 4x  5 y2  2y  ? 
 4x  5  ? y2  2y  1 
 4x  5  1 (y  1)2 
 4(x  1) vertex: (h, k) 
 (1, 1) 4p 
 4 p
1 focus: (h  p, k) 
 (1  1, 1) or (2,  1) axis of symmetry: y 
 k y 
 1 y directrix: x 
 h  p x
0 x
11 x
0 (1, 1) O  x  (4, 2)  2   22   a (x  h) b 2   [x  6  3   (x  3  asymptotes: y  k 
   554 
 0 9(x2  8x  ?)  25(y2  10y  ?) 
 544  ?  ? 9(x2  8x  16)  25(y2  10y  25) 
 544  9(16)  25(25) 9(x  4)2  25(y  5)2 
 225  O  c 
  a2  b2 c 
 2   6 or 22   a) 
 
1, 4  c) 
 
1, 4  foci: (h, k  (x  b)2 (y  k)2   b 
1 2 a2 (x  0)2 (y  0)2    
 1 5220 26,550,840.96 x2 y2     
 1 27,248,400 26,550,840.96 9x2  25y2  72x  250y  (x  4)2  25  (x  1)2   6 
 1  center: (h, k) 
 (1, 4) b2 
 6 a2 
 2 a 
 2  b 
 6  transverse axis: vertical  c 
 835.20 1 radius of Earth 
 2(7920)  6.  3y2  24y  x2  2x  41 
 0 3(y2  8y  ?)  (x2  2x  ?) 
 41  ?  ? 3(y2  8y  16)  (x2  2x  1) 
 41  3(16)  1 3(y  4)2  (x  1)2 
 6  (9, 5)  (4, 8)  (2, 1)  343  x  Chapter 10  10. vertex: (h, k)  (5, 
1) (x 
 h)2  4p(y 
 k) (9 
 5)2  4p[
2 
 (
1)] 42  
4p 
4  p (x 
 h)2  4p(y 
 k) (x 
 5)2  4(
4)[y 
 (
1)] (x 
 5)2  
16(y  1)  10-6  3. Sample answer: rectangular equation: y2  
x parametric equations: y  t  (h, k)  (5, 1) (x, y)  (9, 2)  y  t, x  
t2, 
   t   4. A  1, c  9; since A and C have the same sign and are not equal, the conic is an ellipse. x2  9y2  2x 
 18y  1  0 (x2  2x  ?)  9(y2 
 2y  ?)  
1  ?  ? (x2  2x  1)  9(y2 
 2y  1)  
1  1  9(1) (x  1)2  9(y 
 1)2  9  Rectangular and Parametric Forms of Conic Sections  Page 665  y2  
x t2  
x 
t2  x  (x  1)2  9  ( y 
 1)2   1  1  center: (h, k)  (
1, 1) b2  1 a2  9 a3 b1 vertices: (h  a, k)  (
1  3, 1) or (2, 1), (
4, 1) (h, k  b)  (
1, 1  1) or (
1, 2), (
1, 0)  Graphing Calculator Exploration  1.  y  (4, 1)  Tmin: [0, 6.28] step: 0.1 [
7.58, 7.58] scl1 by [
5, 5] scl1 1a. (
1, 0) 1b. clockwise 2.  (1, 2) (2, 1)  (1, 0) O  (1, 1)  x  5. A  1, C  0; since C  0, the conic is a parabola. y2 
 8x  
8 y y2  8x 
 8 y2  8(x 
 1) vertex: (h, k)  (1, 0) (1, 0) opening: right O x Tmin: [0, 6.28] step: 0.1 [
7.58, 7.58] scl:1 by [
5, 5] scl1 2a. (0, 1) 2b. clockwise 3.  6. A  1, C  
1; since A and C have different signs, the conic is a hyperbola. x2 
 4x 
 y2 
 5 
 4y  0 (x2 
 4x  ?) 
 (y2  4y  ?)  5  ?  ? (x2 
 4x  4) 
 (y2  4y  4)  5  4 
 4 (x 
 2)2 
 (y  2)2  5 (x 
 2)2  5  center: (h, k)  (2, 
2) a2  5 a  
5  vertices: (h  a, k)  (2  
5 , 
2) b asymptotes: y 
 k   a(x 
 h)  Tmin: [0, 6.28] step: 0.1 [
7.58, 7.58] scl1 by [
5, 5] scl1 an ellipse 4. The value of a determines the length of the radius of the circle. 5. Each graph is traced out twice.  
5    (x 
 2) y 
 (
2)     
5  y  Page 667  y  2   (x 
 2)  Check for Understanding  1. For the general equation of a conic, A and C have the same sign and A C for an ellipse. A and C have opposite signs for a hyperbola. A  C for a circle. Either A  0 or C  0 for a parabola. 2. 
   t    Chapter 10  ( y  2)2  
 5  1  O  x  (2  
 5, 2)  (2  
 5, 2)  (2, 2)  344  7. A  1, C  1; since A  C, the conic is a circle. x2 
 6x  y2 
 12y  41  0 (x2 
 6x  ?)  (y2 
 12y  ?)  
41  ?  ? (x2 
 6x  9)  (y2 
 12y  36)  
41  9  36 (x 
 3)2  (y 
 6)2  4 center: (h, k)  (3, 6) radius: r2  4 r2 y  10. Sample answer: Let x  t. y  2x2 
 5x y  2t2 
 5t x  t, y  2t2 
 5t, 
   t   11. Sample answer: x2  y2  36 x2 y2    36 36 x 2 y 2   6 6 cos2 t  sin2 t     (3, 8)  x 2  6 x  6     (5, 6)  (3, 6)       cos2  1 1 1  y 2  6  t  y  6   cos t   sin2 t  sin t  x  6 cos t y  6 sin t x  6 cos t, y  6 sin t, 0  t  2 12.  x  O  t2  x  8 0 y2  x  80 80x  y2 y2  80x  
t2  8. y  
 6t  2 y  
x2 
 6x  2 t 
6 
5 
4 
3 
2 
1 0  x 
6 
5 
4 
3 
2 
1 0  (x, y) (
6, 2) (
5, 7) (
4, 10) (
3, 11) (
2, 10) (
1, 7) (0, 2)  y 2 7 10 11 10 7 2  Pages 667–669  Exercises  13. A  1, C  0; since C  0, the conic is a parabola. x2 
 4y 
 6x  9  0 x2 
 6x  ?  4y 
 9  ? x2 
 6x  9  4y 
 9  9 (x 
 3)2  4y vertex: (h, k)  (3, 0) opening: upward  y  y  O  O 9. x  2 cos t x  2  cos2  t  x 2  2  x  y  3  sin2   sin t  y t
  t1  
 2  y 2   3  1  x2  4  14. A  1, C  1; since A  C, the conic is a circle. x2 
 8x  y2  6y  24  0 2 (x 
 8x  ?)  ( y2  6y  ?)  
24  ?  ? (x2 
 8x  16)  ( y2  6y  9)  
24  16  9 (x 
 4)2  (y  3)2  1 center: (h, k)  (4, 
3) radians: r2  1 r1 y  y  3 sin t   cos t  t

  y2   9  1  t
0  O  x  (3, 0)  x O (3, 3)  3
 t
 2  (4, 3)  x  (4, 4)  t 0  x 2  y 0  (x, y) (2, 0)    2  0  3  (0, 3)    
2  0  (
2, 0)  0  
3  (0, 
3)  3  2  345  Chapter 10  17. A  1, C  0; since C  0, the conic is a parabola. x2  y  8x 
 16 x2 
 8x  ?  y 
 16  ? x2 
 8x  16  y 
 16  16 (x 
 4)2  y y vertex: (h, k)  (4, 0) opening: upward  15. A  
1, C  3; since A and C have different signs, the conic is a hyperbola. x2 
 3y2  2x 
 24y 
 41  0 
x2  3y2 
 2x  24y  41  0 3(y2  8y  ?) 
 (x2  2x  ?)  
41  ?  ? 3(y2  8y  16) 
 (x2  2x  1)  
41  3(16) 
 1 3(y  4)2 
 (x  1)2  6 (y  4)2  2  (x  1)2  
 6  1  center: (h, k)  (
1, 
4) a2  2 a  
2  vertices: (h, k  a)  (
1, 
4  
2 )  O  a  asymptotes: y 
 k   b(x 
 h) y 
 (
4)   y4  
2    
6  x  18. A  C  D  E  0; the conic is a hyperbola. 2xy  3  [x 
 (
1)]  
3  (x 3  (4, 0)  3  xy  2   1)  quadrants: I and III transverse axis: y  x  y  vertices:    or  62, 62, 
 32, 
 32 or 

26 , 

26   (1, 4  
2)  O  x  3 , 2  3  2  y  (1, 4)  (
26 , 
26 ) (1, 4  
2)  O  (
26 , 
26 )  16. A  9, C  25; since A and C have the same sign and are not equal, the conic is an ellipse. 9x2  25y2 
 54x 
 50y 
 119  0 9(x2 
 6x  ?)  25(y2 
 2y  ?)  119  ?  ? 9(x2 
 6x  9)  25(y2 
 2y  1)  119  9(9)  25(1) 9(x 
 3)2  25(y 
 1)2  225 (x 
 3)2  25  19. A  5, C  2; since A and C have the same sign and are not equal, the conic is an ellipse. 5x2  2y2 
 40x 
 20y  110  0 2 5(x 
 8x  ?)  2(y2 
 10y  ?)  
110  ?  ? 5(x2 
 8x  16)  2(y2 
 10y  25)  
110  5(16)  2(25) 5(x 
 4)2  2(y 
 5)2  20  ( y 
 1)2   9  1  center: (h, k)  (3, 1) b2  9 a2  25 a5 b3 vertices: (h  a, k)  (3  5, 1) or (8,1), (
2, 1) (h, k  b)  (3, 1  3) or (3, 4), (3, 
2)  (x 
 4)2 ( y 
 5)2    4 10 ( y 
 5)2 (x 
 4)2    10 4  (3, 4)  (2, 1)  O  (4, 5  
 10)  y (2, 5)  (8, 1)  x  (4, 5)  (6, 5)  (3, 2) (4, 5  
 10)  O  Chapter 10  1 1  center: (h, k)  (4, 5) b2  4 a2  10 a  
10  b2 vertices: (h, k  a)  (4, 5  
10 ) (h  b, k)  (4  2, 5) or (6, 5), (2, 5)  y  (3, 1)  x  346  x  20. A  1, C 1; since A  C, the conic is a circle. x2 
 8x  11  
y2 (x2 
 8x  ?)  y2  
11  ? (x2 
 8x  16)  y2  
11  16 (x 
4)2  y2  5 center: (h, k)  (4, 0) radius: r2  5 r  
5   23.  
4y2  10x  16y 
 x2 
 5 x2 
 4y2  10x 
 16y  5  0 A  1, C  
4; since A and C have different signs, the conic is a hyperbola. x2 
 4y2  10x 
 16y  5  0 2 (x  10x  ?) 
4(y2  4y  ?)  
5  ?  ? (x2  10x  25) 
 4(y2  4y  4)  
5  25 
 4(4) (x  5)2 
 4(y  2)2  4 (x  5)2  4  y (4, 
 5)  center: (h, k)  (
5, 
2) a2  4 a2 vertices: (h  a, k)  (
5  2, 
2) or (
3, 
2), (
7, 
2)  (4  
5, 0)  O  x  (4, 0)  ( y  2)2  
 1  1  b  asymptotes: y 
 k   a(x 
 h) 1  y 
 (
2)   2[x 
 (
5)]  21. A  
9, C  8; since A and C have different signs, the conic is a hyperbola. 8y2 
 9x2 
 16y  36x 
 100  0 8(y2 
 2y  ?) 
 9(x2 
 4x  ?)  100  ?  ? 8(y2 
 2y  1) 
 9(x2 
 4x  4)  100  8(1) 
 9(4) 8(y 
 1)2 
 9(x 
 2)2  72 ( y 
 1)2  9  1  y  2   2(x  5)  y (5, 2)  (x 
 2)2  x  O  
 8  1  center: (h, k)  (2, 1) a2  9 a3 vertices: (h, k  a)  (2, 1  3) or (2, 4), (2, 
2)  (7, 2)  (3, 2)  a  asymptotes: y 
 k   b(x 
 h) 24.  3  (x 
 2) y
1 2
 2 3
 2  y 
 1   4(x 
 2)  y (2, 4) (2, 1)  x  O (2, 2)  Ax2  Bxy  Cy2  Dx  Ey  F  0 2x2  0  2y2  (
8)x  12y  6  0 2x2  2y2 
 8x  12y  
6 A  C; circle 2(x2 
 4x  ?)  2(y2  6y  ?)  6  ?  ? 2(x2 
 4x  4)  2(y2  6y  6y  9)  
6  2(4)  2(9) 2(x 
 2)2  2(y  3)2  20 (x 
 2)2  (y  3)2  10 y center: (h, k)  (2, 
3) 2 radius: r  10 (2, 3  
 10) r  
10  x O  22. A  0, C  4; since A  0, the conic is a parabola. 4y2  4y  8x  15 4(y2  y  ?)  
8x  15  ? 4  (2  
 10, 3) (2, 3)    
8x  15  4     
8x  16    
2x  4    
2(x 
 2) 1 vertex: (h, k)  2, 
2 1 y2  y  4 1 2 4 y  2 1 2 y  2 1 2 y  2  opening: left  1  4  y  25. y  2t2 
 4t  1 y  2x2 
 4x  1  O  x  y t 
1 0 1 2  ( 2 ,  12 ) O  x  347  x 
1 0 1 2  y 7 1 
1 7  (x, y) (
1, 7) (0, 1) (1, 
1) (2, 7) Chapter 10  26. x  cos 2t y  sin 2t cos2 2t  sin2 2t  1 x 2  y2  1 t 0   2    3  2  x 1  y 0  (x, y) (1, 0)  0 
1  1 0  (0, 1) (
1, 0)  0  
1  (0, 
1)  29.  x  
 sin 2t  y  2 cos 2t y  2  
x  sin 2t 2t   cos2  y 2  2     sin2    2t  1  (
x)2  y2  4   cos 2t  y t
0  1    x2  1  x2   y2  4  1  O  x  y  t
0    4   2  x  O  x 0  y 2  (x, y) (0, 2)  
1  0  (
1, 0)  0  
2  (0, 
2)  1  0  t 0  3  4  27.  x  
cos t 
x  cos t cos2 t  sin2 t  1 (
x)2  y2  1 x 2  y2  1 t 0   2    3  2  y  sin t  30.  x  2t 
 1 x  1  2t x1  2  y 0  (x, y) (
1, 0)  0 1  1 0  (0, 1) (1, 0)  0  
1  (0, 
1)  t 0 1 2 3 4  t
0  x 31.  28. x  3 sin t  y  2  2  y  x 2   3  1  y2  4 x2  9     x2  9 y2  4  t
0  O  x 2  5 x  5       2    3  2  Chapter 10  x 2  y 0  (x, y) (0, 2)  3 0  0 
2  (3, 0) (0, 
2)  
3  0  (
3, 0)  y  3   sin 2t  2t  1  y 2  3  1 x2 y2     1 9 9 x2  y2  9     x2 y2    25 25 x 2 y 2   5 5 cos2 t  sin2 t  x     t 0  y  3 sin 2t  32. Sample answer: x2  y2  25  1 1  (x, y) (
1, 0) (1, 1) (3, 
2 ) (5, 
3 ) (7, 2)  y 0 1 
2  
3  2  x  
3 cos 2t      cos t  cos2 t  sin2 t  1  x 
1 1 3 5 7  x 2 
3  x  O  x 
3  cos 2t cos2 2t  sin2  y  2 cos t   sin t  2y  t x1  2  y  y  x  3  y  y  
t  x 
1  O  (1, 0)       cos2   cos t  t  1 1 1  2  5y  y  5   sin2 t  sin t  x  5 cos t y  5 sin t x  5 cos t, y  5 sin t, 0  t  2  348  33. Sample answer: x2  y2 
 16  0 x2  y2  16 x2 y2    16 16 y 2 x 2   4 4 cos2 t  sin2 t     x 2  4 x  4          cos2  39b.  t 0 1 2 3  1 1 1  t   cos t  y 2  4  y  4  x 0 1 2 3  (x, y) (0, 0) (1, 1) (2, 4) (3, 9)  y 0 1 4 9  y  sin2 t  sin t  x  4 cos t y  4 sin t x  4 cos t, y  4 sin t, 0  t  2 34. Sample answer: x2  4  2  2x  y2  y 2   5  1  cos2 t  sin2 t  1 2  2x  x  2  x  O   25  1   cos2 t  cos t  y 2  5  y  5   sin2 t  sin t  x  2 cos t y  5 sin t x  2 cos t, y  5 sin t, 0  t  2 35. Sample answer: y2  16   x2  1 y 2  x2  4  1 cos2 t  sin2 t  1 x2  cos2 t x  cos t  39c. 39d. y 2  4  y  4   sin2 t 40a.   sin t  y  4 sin t x  cos t, y  4 sin t, 0  t  2 36. Sample answer: Let x  t. y  x2 
 4x  7 y  t2 
 4t  7 x  t, y  t2 
 4t  7, 
  t   37. Sample answer: Let y  t. x  y2  2y 
 1 x  t2  2t 
 1 x  t2  2t 
 1, y  t, 
  t   38. Sample answer: Let y  t. (y  3)2  4(x 
 2) (t  3)2  4(x 
 2) 0.25(t  3)2  x 
 2 0.25(t  3)2  2  x x  0.25(t  3)2  2, y  t, 
  t   39a. Answers will vary. Sample answers: Let x  t. x  
y t  
y t2  y x  t, y  t2, t  0 Let y  t. x  
y x  
t x  
t, y  t, t  0  40b.  Tmin: [0, 5] step: 0.1 [
7.58, 7.58] scl1 by [
5, 5] scl1 yes There is usually more than one parametric representation for the graph of a rectangular equation. a circle with center (0, 0) and radius 6 feet (x 
 h)2  (y 
 k)2  r2 (x 
 0)2  (y 
 0)2  62 x2  y 2  36 Sample answer: x2  y2  36 x2 y2    36 36 x 2 y 2   6 6 sin2 (qt)  cos2 (qt)        1 1  1 Since the paddlewheel completes a revolution in 2 2 seconds, the period is q  2, so q  . sin2 (t)  cos2(t)  1 2  6x  x  6   sin2(t)  sin(t)  2  6y  y  6   cos2 (t)  cos(t)  x  6 sin(t) y  6 cos (t) x  6 sin (t), y  6 cos (t), 0 t  2 40c. C  2r C  26 C  37.7 ft The paddlewheel makes 1 revolution, or moves 37.7 ft in 2 seconds. 37.7 ft  2s  60 s  1131 ft  The paddlewheel moves about 1131 ft in 1 minute.  349  Chapter 10  41a. A  2, C  5; since A and C have the same sign and A C, the graph is an ellipse. 2x2  5y2  0 5y2  
2x  44. After drawing a vertical line through (x, y) and a horizontal line through the endpoint opposite (x, y), two right triangles are formed. Both triangles contain an angle t, since corresponding angles are congruent when two parallel lines are cut by a transversal. Using the larger triangle, x cos t  a or x  a cos t. Using the smaller triangle,  2  y2  
5x 2  y  
5x  y  sin t  b or y  b sin t. 45. x2 
 12y  10x  
25 x2  10x  ?  12y 
 25  ? 2 x  10x  25  12y 
 25  25 (x  5)2  12y vertex: (h, k)  (
5, 0) 4p  12 p3 focus: (h, k  p)  (
5, 0  3) or (
5, 3) axis of symmetry: x  h x  
5 directrix: y  k 
 p y0
3 y  
3  This equation is true for (x, y)  (0, 0). The graph is a point at (0, 0); the equation is that of a degenerate ellipse. 41b. A  1, C  1; since A  C, the graph is a circle. x2  y2 
 4x 
 6y  13  0 (x2 
 4x  ?)  (y2 
 6y  ?)  
13 (x2 
 4x  4)  (y2 
 6y  9)  
13  4  9 (x 
 2)2  (y 
 3)2  0 center: (h, k)  (2, 3) radius: 0 The graph is a point at (2, 3); the equation is that of a degenerate circle. 41c. A  
9, C  1; since A and C have different signs, the graph is a hyperbola. y2 
 9x2  0 y2  9x2 y  3x The graph is two intersecting lines y  3x; the equation is that of a degenerate hyperbola. 42. The substitution for x must be a function that allows x to take on all of the values stipulated by the domain of the rectangular equation. The domain of y  x2 
 5 is all real numbers, but using a substitution of x  t2 would only allow for values of x such that x  0. 43a. center: (h, k)  (0, 0) (x 
 h)2  (y 
 k)2  r2 (x 
 0)2  (y 
 0)2  6 x2  y2  36 43b. x2  y2  36 y2 x2    36 36 y 2 x 2   6 6 sin2 t  cos2 t     x 2  6 x  6         sin2 t  sin t  y  O x  (5, 0)  46. c  
25 quadrants: II and IV transverse axis: y  
x vertices: xy  
25 5(
5)  
25 (5, 
5)  1  xy  
25 
5(5)  
25 (
5, 5)  y  1 1  (5, 5) y 2  6 y  6      cos2 t  O   cos t  x (5, 5)  x  6 sin t y  6 cos t x  6 sin t, y  6 cos t Since the second hand makes 2 revolutions, 0  t  4.  47.  43c.  3x2  3y2 
 18x  12y  9 3(x2 
 6x  ?)  3(y2  4y  ?)  9  ?  ? 3(x2 
 6x  9)  3(y2  4y  4)  9  3(9) 
 3(4) 3(x 
 3)2  3(y  2)2  48 (x 
 3)2  (y  2)2  16 center: (h, k)  (3, 
2) (3, 2) y radians: r2  16 r4  O  (3, 2)  Tmin: [0, 4] step: 0.1 [
9.10, 9.10] scl1 by [
6, 6] scl1  Chapter 10  (7, 2)  350  x  x  48.  y  kxz y  kxz 16  k(5)(2) y  1.6(8)(3) 1.6  k y  38.4 54. 5 9  5(
3) 
 7(9) 7 
3  
78 Yes, an inverse exists since the determinant of the matrix 0. 53.  60˚  y  30    y  x  cos 60°  30  sin 60°  30  y2 
 y1   55. m   x 
x  x  30 cos 60° y  30 sin 60° x  15 lb y  15
3  lb 49. y  
0.13x  37.8 0.13x  y 
 37.8  0 A  0.13, B  1, C  
37.8 Car 1: (x1, y1)  (135, 19) d1  d1     2     1  7
4  3 
 (
6) 1  3  y 
 y1  m(x 
 x1) 1  1  y 
 4  3(x  6) or y 
 7  3(x 
 3)  Ax1  By1  C  
 A2   B2  y  mx  b 4  0.13(135)  1(19)  (
37.8)  2  12  
 (0.13)  Ax2  By2  C  
 A2   B2  d2   0.13(245)  1(16)  (
37.8)  2  12 
 (0.13)  y  mx  b 1  b  y  3x  6  6b 56. (1 # 4) @ (2 # 3)  1 @ 2 2 The correct choice is B.  d1  
1.24 The point (135, 19) is about 1 unit from the line y  
0.13x  37.8. Car 2: (x2, y2)  (245, 16) d2   1 (
6) 3  Transformation of Conics  10-7  d2  9.97 The point (245, 16) is about 10 units from the line y  
0.13x  37.8. Car 1: the point (135, 19) is about 9 units closer to the line y  
0.13x  37.8 than the point (245, 16).  Pages 674–675  Check for Understanding  1. Sample answers: (h, k)  (0, 0) x  y2  y  1  50. Let v  Sin1 2.  x 
 y2  1  Sin v  2 v  30  1   sin (2 30)  sin 60 
3    2  (h, k)  (
3, 3) (x 
 h)  (y 
 k)2 (x  3)  (y 
 3)2 y  1  51. s  2(a  b  c) 1  s  2(48  32  44) s  62 a)(s 
 b)(s 
 c) K  
s(s 
 K  
62(62 )(62 
 48 
2 32)(64) 
 4 K  
46872 0  K  685 units2 52. 
2y  
 3 
 
2y   3  
1 
2y 
 3  
2y   3
1 2y 
 3  2y  3 
 2
2y 31  
7  
2
2y 3  7  2 49  4 37  4 37  8  x  O  sin 2 Sin1 2  sin (2v)  x  3 
 (y  3)2  O  x  2. Replace x with x
 cos 30°  y
 sin 30° or   
2y  3  
3   x
 2   2y  3  Replace y with 
x
 sin 30°  y
 cos 30° or   2y  
2x
,   1  1   2y
. 
2   y
. 2  y  351  Chapter 10  y  3.  x2 100  8. B2 
 4AC  0 
 4(1)(1)  
4 A  C  1; circle  90° or 270°  y2   25 
 1    2 x
 cos 4  y
 sin 4 
 5  90˚  x2 25  y2   100 
 1  y  7 2 
 5  2 4 49 y 
 5  8 9 y  8 9 y  8 
 k 9 y  8 
 5 31 y 
 8 31 y  8 31 y 
 8                  7 2 x2 
 2x 7 7 2 2 x2 
 2x  4 7 2 2 x 
 4 7 2 2 x 
 4 2 7 2 x 
 4 
 h 2 7 2 x 
 4  4 9 2 2 x  4 18 81 2 x2  4x  1 6 81 2x2  9x  8 2x2  9x 
 y  14             2(x
)2  2(y
)2 
 5
2 x
 
 5
2 y
 
 6  0 
 4AC  42 
 4(9)(4) 9.  
128 A C; ellipse B2  B   tan 2v   A
C 4   tan 2v   9
4  tan 2v  0.8 2v  38.65980825° v  19° 10. B2 
 4AC  52 
 4(8)(
4)  153 hyperbola        B   tan 2v   A
C 5   tan 2v   8 
 (
4)  tan 2v  0.416666667 2v  22.61986495° v  11° 11. 3(x 
 1)2  4(y  4)2  0 3(x2 
 2x  1)  4(y2  8y  16)  0 3x2 
 6x  3  4y2  32y  64  0 4y2  32y  (3x2 
 6x  67)  0  (h, k)  (4, 5)                0 2x2  9x 
 y  14  0 7. B2 
 4AC  0 
 4(1)(
1) 4 hyperbola  ↑ a  1 (x
)2 4  
3   12 x
  
23 y
  ↑ c  b  
 b2 
 4ac   2  
 
2x
  2y
  9 
3   1  3   2x
y
  4(y
)2 
  y  
32  
 322 
 4(4)(3 x2 
 6 x  67)  2(4)  y  
32  
 
48x2  96 x 
 48  8  y  
32  
 
48(x 
 1)2  8  x  1, y  
4; point  3 1 
3  (x
)2 
x
y
  (y
)2  9 4 4 2 1 1 
2(x
)2  
3 x
y
  2(y
)2  9 x
y
 
 (y
)2  
18 (x
)2 
 2
3  y x  O  (x
)2 
 2
3 x
y
 
 (y
)2  18  0  Chapter 10  ↑ b  y  
  2a  x 2 
 y2  9 (x
 cos 60°  y
 sub 60°)2 
 (
x
 sin 60°  y
 cos 60°)2  9 2    1 1 5
2  5
2  (x
)2  x
y
  (y
)2 
 x
 
  y
 2 2 2 2 1 1  2(x
)2 
 x
y
  2(y
)2  3 5
2  5
2  (x
)2  (y
)2 
 2x
 
 2y
  3 x
 
 5
2 y
  6 2(x
)2  2(y
)2 
 5
2  2  4. Ebony; B2 
 4AC  6
3  
 4(7)(13)  0 and A C 5. B2 
 4AC  0 
 4(1)(1)  
4 A  C  1; circle (x 
 h)2  (y 
 k)2  7 (x 
 3)2  (y 
 2)2  7 (h, k)  (3, 2) 2 x 
 6x  9  y2 
 4y  4  7 x2  y2 
 6x 
 4y  6  0 2 6. B 
 4AC  0 
 4(2)(0) 0 parabola y  2x2 
 7x  5 y 
 5  2x2 
 7x y
5  x2 
 5x  y2  3    y
 sin 4    2  
x
sin 4  y
 cos 4  3 2 
22 x
  
22 y
 
 5
22 x
  
22 y
 2 
2  
2   
2x
  2y
  3    x  270˚ O   x
 cos 4  (1, 4)  352  14. B2 
 4AC  0 
 4(4)(5)  
80 A C; ellipse  1  y  6x2  12a.  1  
x
 sin 30°  y
 cos 30°  6(x
 cos 30°  y
 sin 30°)2 2  
2x
  2 y
  62x
  2y
 
3   1  1 
3   1  1  
3   1 3 (x
)2 4   4 x
y
 1  4(y
)2  1  
3   1  
3   
2x
  2y
  6  4x2  5y2  20 4(x 
  5(y 
 k)2  20 4(x 
 5)2  5(y  6)2  20 (h, k)  (5, 6) 4(x2 
 10x  25)  5(y2  12y  36)  20 4x2 
 40x  100  5y2  60y  180  20 4x2  5y2 
 40x  60y  260  0 2 15. B 
 4AC  0 
 4(3)(1)  
12 A C; ellipse 3x2  y2  9 2 3(x 
 h)  (y 
 k)2  9 3(x  1)2 (y 
 3)2  9 (h, k)  (1, 3) 3(x2  2x  1)  y2 
 6y  9  9 3x2  6x  3  y2 
 6y  9  9 3x2  y2  6x 
 6y  3  0 2 16. B 
 4AC  0 
 4(12)(4)  
192 A C; ellipse 4y2  12x2  24 2 4(y 
 k)  12(x 
 h)2  24 4(y 
 4)2  12(x  1)2  24 (h, k)  (1, 4) 4(y2 
 8y  16)  12(x2  2x  1)  24 4y2 
 32y  64  12x2  24x  12  24 y2 
 8y  16  3x2  6x  3  6 3x2  y2  6x 
 8y  13  0 2 17. B 
 4AC  0 
 4(9)(
25)  900 hyperbola 9x2 
 25y2  225 2 9(x 
 h) 
 25(y 
 k)2  225 9(x 
 0)2 
 25(y 
 5)2  225 (h, k)  (0, 5) 9x2 
 25(y2 
 10y  25)  225 9x2 
 25y2  250y 
 850  0 18. (x 
 3)2  4y 2 x  6x  9 
 4y  0 B2 
 4AC  0 
 4(1)(0) 0 parabola (x  3)2  4y (x  3 
 h)2  4(y 
 k) (x  3  7)2  4(y 
 2) (h, k)  (7, 2) (x  10)2  4y 
 8 x2  20x  100  4y 
 8 2 x  20x 
 4y  108  0 19. B2 
 4AC  0 
 4(1)(0) 0 parabola x2 
 8y  0 2 (x
 cos 90°  y
 sin 90°) 
8(
x
 sin 90°  y
 cos 90°)  0 (y
)2 
 8(
x
)  0 (y
)2  8x  0 h)2  2
3   1  2 
2x
  2y
  8(x
)2  12x
y
  2 4 (y
)  ↑ a                      
12x
  12
3 y
  3(x
)2  2
3 x
y
  (y
)2 0  3(x
)2  2
3 x
y
  (y
)2  12x
 
 12
3 y
 3(x
)  2
3 x
y
  (y
)2  12x
 
 12
3 y
  0 12b. 3x2  2
3 xy  y2  12x 
 12
3 y  0 2 1y  (2
3 x 
 12
3 )y  (3x2  12x) 0 ↑ b  ↑ c  y  b2 
 4 ac 
b  
  2a  y  
(2
3 x 
 12
3 )  
 (2
3 x 
 12 
3  )2  
 4(1 )(3x2  12x)  2(1)  y  

3 x  6
3   
 12x2 
 144x   432 
 12 x2 
 4 8x  2  y  

3 x  6
3   

192 x2  43  2  Pages 675–677  1  and y  6x2  Exercises  13. B2 
 4AC  0 
 4(3)(0) 0 parabola y  3x2 
 2x  5 y 
 5  3x2 
 2x y 
 5  3x2 
 3x 2  1 2  1 2  y 
 5  33  3 x2 
 3x  3 2  1 2  y 
 3  3x 
 3 14  2  y 
 3 
 k  3x 
 3 
 h 14  1  2  y 
 3  3  3x 
 3 
 2 14  y
 y
 y
  1  5  3 5  3 5  3  7 2 x 
 3 14 x2 
 3x   3  3    (h, k)  (2, 3)   9 49  49   3x2 
 14x  3  0  3x2 
 14x 
 y  18 3x2 
 14x 
 y  18  0  353  Chapter 10  20. B2 
 4AC  0 
 4(2)(2)  
16 A  C; circle  24. B2 
 4AC  0 
 4(16)(
4)  256 hyperbola 16x2 
 4y2  64 16(x
 cos 60°  y
 sin 60°)2 
 4(
x
 sin 60°  y
 cos 60°)  64  2x2  2y2  8  2(x
 cos 30°  y
 sin 30°)2   2(
x
 sin 30°  y
 cos 30°)2  8          16  
 4AC  1 hyperbola  x
 cos    4  6  y2  8x  0  3    
8   
8    23. B2 
 4AC  0 
 4(1)(1)  
4 A  C; circle  x
 cos   2 
3  y
 2  1 x
 2    1 (x
)2 4   y
 sin    
3    2  3     
x
 sin  
 5    3  4  tan 2v  1 2v  45° v  23° 27. B2 
 4AC  (
1)2 
 4(1)(
4)  17 hyperbola   
8    3   3  
 5x    y
 sin  y
 cos  y2    3  2  3      tan 2v   A
C  3  
1   tan 2v   1 
 (
4) 1  tan 2v  
5 2v  
11.30993247° v  
6° 28. B2 
 4AC  82 
 4(8)(2) 0 parabola B  tan 2v   A
C  3   2 1 
 3  
2x
  2 y
  3  
3  y
 2  5
3   5  
 3  3  1      2(y
)2  (y
)2  
  5 x
 2  
  5
 3 y 2  8   tan 2v   8
2  3  4  tan 2v  3 2v  53.13010235° v  27° 29. B2 
 4AC  92 
 4(2)(14)  
31 A C; ellipse  
 5x
 
 5
3 y
 
 6  0  B   tan 2v   A
C 9   tan 2v   2 
 14 3  tan 2v  
4 2v  
36.86989765° v  
18° Chapter 10  
3   2  1   tan 2v   9
5   4(x
)2 
 2x
y
  4(y
)2  3 2(x
)2  
 3  1   2 x
y
  4(y
)2  B   2x
y
  4(y
)2 
 2x
 
 2y
 (x
)2  3 (x
)2 4  B  x2  
 5x
 cos  1 x
 2  2  1   tan 2v   A
C   16 (x
)2 
 (y
)2 
 16  0    3  1  
 2x
y
  4(y
)2  64  26. 32 
 4AC  42 
 4(9)(5)  
164 A C; ellipse  xy  
8     y
 sin 
x
 sin 4  y
 cos 4 
2  
2  
2  
2  x
  y
 
x
  y
 2 2 2 2 1 1 1 1 
2(x
)2  2x
y
 
 2x
y
  2(y
)2 (x
)2 
 (y
)2    
3   3 (x
)2 4  1 
3  3 (x
)2 
 x
y
  (y
)2  4 2 4 18 63 6 5 5
3  (x
)2   x
y
  (y
)2  (x
)2 
 x
y
 4 2 4 4 2 15  4(y
)2  23 
3  21 (x
)2  x
y
  (y
)2 
 30  4 2 4 23(x
)2  2
3 x
y
  21(y
)2 
 120    4(y
)2  4
3 x
  4y
  0    3  5  3(y
)2  16
3 x
  16y
  0 
 4(0)(0)    4  2   2x
y
  4(y
)2  
3   2  22.  1  62x
  2y
  5
2x
  2y
  30  
x
 sin 6  y
 cos 6  8x
 cos 6  y
 sin 6  0 2 
12x
  
23 y
  8
23 x
  12y
  0 12  
 3  
3   x
y
  12(y
)2 4(x
)2  8
3 
 3(x
)2  2
3 x
y
 
 (y
)2  64 x
y
  11y
)2 
 64  0 (x
)2  10
3 25. 6x2  5y2  30 2 6(x
 cos 30°  y
 sin 30°)  5(
x
 sin 30°  y
 cos 30°)2  30  
 4AC  0 
 4(1)(0) 0 A C; parabola  B2  1 (x
)2 4  
4  B2  1 
3  (x
)2 
  x
y
 2 4 x
y
  (x
)2 
 2
3  
3   1  (x
)2  (y
)2 
 4  0  21.  2  162x
  2y
 
 4
2x
  2y
  64  2 2 1 1 
3  
3  2 2x
  2y
  2 
2x
  2y
  8 3 1 
3  2 4(x
)2  2x
y
  4(y
)2 1 3 
3   2 4(x
)2 
2x
y
  4(y
)2  8 3 1 1 (x
)2  
3 x
y
  2(y
)2  2(x
)2 2 3 
 
3 x
y
  2(y
)2  8 2(x
)2  2(y
)2  8  354  30  30 0 0  30. B2 
 4AC  42 
 4(2)(5)  
24 A C; ellipse  36.  B   tan 2v   A
C 4   tan 2v   2
5 4 
3  tan 2v  2v  
53.13010235° v  
27° 2 31. B2 
 4AC  4
3  
 4(2)(6) 0 parabola  y  0  
 0 
 4 (1)(x2)  2(1)  
 
4x2  y   2 x  0, y  0; point  y  B  A
C  4
3  2
6  (0, 0)  x  O   tan 2v  

3 2v  
60° v  
30° 32. B2 
 4AC  42 
 4(2)(2) 0 parabola  37. (1)y2  x2 
 2xy  y2 
 5x 
 5y  0  (
2x 
 5)y  (x2 
 5x)  0 ↑ b  ↑ c  ↑ a    A  C; v  4 or 45° 33.  
b  
 b2 
 4 ac  2a          tan 2v   y          tan 2v   (x 
 2)2  (y 
 2)2  4(x  y)  8 x2 
 4x  4  y2 
 4y  4  4x  4y  8 (1)y2  0y  x2  0 ↑ ↑ ↑ a b c  (x 
 2)2 
 (x  3)2  5(y  2) 2 x 
 4x  4 
 x2 
 6x 
 9  5(y  2) 
10x 
 5  5(y  2) 
2x 
 1  y  2 y 
2x 
 3  y y  
2x 
 3 line O x  y y y y  
b  
 b2 
 4 ac  2a 
(
2x 
 5)  
 (
2x 
 5)2 
 4(1)(x2 
 5x)  2(1) 2x  5  
 4x2  20x  25 
 4x2  20x  2 2x  5  
40x   25  2  y 
 2x  3            34. 2x2  6y2  8x 
 12y  14  0 x2  3y2  4x 
 6y  7  0 2 3y  (
6)y  (x2  4x  7)  0 ↑ a  ↑ b  ↑ c  38.  [
6.61, 14.6] scl1 by [
2, 12] scl1 2x2  9xy  14y2  5 14y2  (9x)y  (2x2 
 5)  0  
 b  
 b2 
 4 ac  2a  y  
(
6)  
 (
6)2  
 4(3 )(x2  4x   7)  2(3)  y  6  
 
12x2 
 48 x 
 48  6  y  
b  
 b2 
 4 ac  2a  y  6  
 
12(x  2)2  6  y  
9x  
 (9x)2 
 4(14) (2x2 
 5)  2(14)  y  
9x  
 
21x2  28 0  28          y  ↑ a  x  
2, y  1; point y  ↑ b  ↑ c  (2, 1)  O  x  35. y2 
 9x2  0 y2  9x2 y  
 9x2 y   3x intersecting lines  y y 
 3x  O  [
7.58, 7.58] scl1 by [
5, 5] scl1  x  y 
 3x  355  Chapter 10  8x2  5xy 
 4y2  
2 (
4)y2  (5x)y  (8x2  2)  0  42.  ↑ a  ↑ c  ↑ a  ↑ b  ↑ c  y  
b  
 b2 
 4 ac  2a  y  
b  
 b2 
 4 ac  2a  y  
5x  
 (5x)2 
 4(
4 )(8x2  2)   y  
4x  
 (4x)2 
 4(6)(9 x2 
 2 0)  2(6)  y  153x2  32 
5x  
   y  
4x  
 
212 x2  4 80  12  [
7.85, 7.85] scl1 by [
5, 5] scl1 43a. y  [
7.58, 7.58] scl1 by [
5, 5] scl1 2x2  4
3 xy  6y2  3x  y 2 6y  (4
3 x 
 1)y  (2x2  3x)  0  y  ↑ b  (0, 5280)          ↑ a          40.  ↑ b  9x2  4xy  6y2  20 6y2  (4x)y  (9x2 
 20)  0                 39.  ↑ c  (1320, 1320)  
b  
 b2 
 4 ac  2a  O (5280, 0) x  
 (4
3 x 
 1)2 
 4(6 )(2x2  3x)  y  
(4
3 x 
 1)   2(6)  y  
4
3 x  1  
 48x2 
 8
3 x  1 
 48x2  
 72x  12  y  
4
3 x  1  
 
8
3  x 
 72 x1  12  T(1320, 1320) 43b. circle center: (h, k)  (1320, 1320) radius: r  1320 (x 
 h)2  (y 
 k)2  r2 (x 
 1320)2  (y 
 1320)2  13202 (x 
 1320)2  (y 
 1320)2  1,742,400 44a. B2 
 4AC  0 
 4(1)(0) 0 parabola; 360° 44b. B2 
 4AC  0 
 4(8)(6)  
192 A C; ellipse; 180° 44c. B2 
 4AC  42 
 4(0)(0)  16 hyperbola; 180° 44d. 32 
 4AC  0 
 4(15)(15)  
900 A  C; circle; There is no minimum angle of rotation, since any degree of rotation will result in a graph that coincides with the original. 45. Let x  x
 cos v  y
 sin v and y  
x
 sin v  y
 cos v. x 2  y2  r 2 (x
 cos v  y
 sin v)2  (
x
 sin v  y
 cos v)2  r2 (x
)2 cos2 v  x
y
 cos v sin v  (y
)2 sin2 v  (x
)2 sin2 v 
 x
y
 cos v sin v  (y
)2 cos v  r2 [(x
)2  (y
)2] cos2 v  [(x
)2  (y
)2] sin2 v  r2 [(x
)2  (y
)2](cos2 v  sin2 v)  r2 [(x
)2  (y
)2](1)  r2 (x
)2  (y
)2  r2 2 2 46a. B 
 4AC  
10
3  
 4(31)(21)  
2304 A C; elliptical  [
8.31, 2.31] scl1 by [
2, 5] scl1 x 
 2
2 y  
12 2x2  4xy  2y2  2
2  (4x 
 2
2 )y  (2x2  2
2 x  12)  0               2y2          41.  ↑ b  ↑ c  ↑ a  y  
b  b2 
 4ac  2a 
(4x 
 2
2 ) 
 (4x 
  2
2  )2  
 4(2 )(2x2  2
2 x  12)  2(2)  y  
4x  2
2   
 16x2 
 16
2  x8 
 16x2 
 16 
2 x 
 96  4  y  
4x  2
2   
 
32
 x 
 8 2 8  4  y  [
10.58, 4.58] scl1 by [
2, 8] scl1  Chapter 10  356  48a. center: (h, k)  (0, 0) major axis: horizontal a  
81  or 9 b  
36  or 6 c  
 a2 
 b2 c  
81 36 
 c  
45  or 3
5   31x2 
 10
3 xy  21y2  144 21y2  (
10
3 x)y  (31x2 
 144)  0 ↑ a            46b.  ↑ b  ↑ c  y  
b  
 b2 
 4 ac  2a  y  
(
10
3 x)  
 (
10
 x)2 
 3 4(21) (31x2  
 144)  2(21)  y  x  

230  4x2  12,096  10
3  42  y  8 4  y 
8  
4 O  4  8  x  
4  O  
8  x 48b. T(3
5, 0) B   46c. tan 2v   A
C  tan 2v   
10
3   31 
 21  48c.  x2 y2     1 81 36 (x 
 h)2 (y 
 k)2     1 81 36 (x
 
 3
5  )2 (y
)2     1 36 81  49. A  
3, C  5; since A and C have different signs, the conic is a hyperbola. 50. (h, k)  (2, 
3)   tan 2v  

3 2v  
60° v  
30° B  c   47a. tan 2v   A
C  e  a c 2
6     1 5 2
6    c 5 b2  a2  
2
 3   tan 2v   9 
 11  tan 2v  
3  2v  60° v  30° The graph of this equation has been rotated 30°. To transform the graph so the axes are on the x- and y-axes, rotate the graph 
30°.  1  b2  2 5 major axis: horizontal  2  
3   1  2  
 3  (x 
 h)2 (y 
 k)2   b 2 a2 (x 
 2)2 [y 
 (
3)]2    1 1  25 (x 
 2)2  25(y  3)2  2  3 1 
3  (x
)2 
 x
y
  (y
)2 
 4 4 2 3 1 
3  
3  2     2
3  4 (x
)  4 x
y
 
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 4(y
)2 1 3 
 3  11 4(x
)2  2x
y
  4(y
)2 27 9 6 9
3  (x
)2 
 x
y
  (y
)2 
 (x
)2 4 4 4 2 6 6
3  2
 3 2 
 4x
y
  4x
y
  4(y
) 11 33 11
3   4(x
)2  2x
y
  4(y
)2 8(x
)2  12(y
)2  (y 
 k)2  a2 [y 
 (
3)]2  12  (y  3)2   
 24  0 51.  9  (x
)2  3  1 1 1  major axis : vertical  1  12x
  
23 y
  1112x
  
23 y
  2
6  2  b2  12 
 5  47b. B2 
 4AC  
2
3  
 4(9)(11)  
384 A C; the graph is an ellipse. 9x2 
 2
3 xy  11y2 
 24  0 9[x
 cos (
30°)  y
 sin (
30°)]2 
 2
3  [(x
 cos (
30°)  y
 sin (
30°)] [
x
 sin (
30°)  y
 cos (
30°)]  11 [
x
 sin (
30°)  y
 cos (
30°)2] 
 24  0 2x
 
 2y
 92x
 
 2y
 
 2
3  
 c2  v r  
 24  0  0 1 120˚  (x 
 h)2   b 1 2  (x 
 2)2  1 1  25 25(x 
 2)2  1    30° 1.4  60° 3.9  90˚  1 2 3 4  240˚  (y
)2  357  180° 
1  0˚  330˚  210˚   2  1  150° 
1  30˚  180˚  
 24  0  120° 
1.4  60˚  150˚  
 24  0  90° 
3.9  270˚  300˚  Chapter 10  52.  58.  N E  W 5  c  S  53. cos 70°  0.34 cos 170°  
0.98 cos 70° 54.  5  16  5  180°     16   56.25 15   566 0  h (x )   56° 15
  55.  2y  5   y2  3y  2 2y  5   y2  3y  2  2y  5    (y  2)(y  1) A  B     y2  y1  x  2y  5  A(y  1)  B(y  2) 2(
2)  5  A(
2  1)  B(
2  2) 1  
A 
1  A 2y  5  A(y  1)  B(y  2) 2(
1)  5  A(
1  1)  B (
1  2) 3B  56.  A B    y2 y1 x1 x2   y2  y1 12  y2    
1  5a8b5 180a b  1  b and 2  a, the expression is always larger  3     y2  y1  2213 36  3223 36  72 36  expression is always less than    or 2. The correct choice is B.  m  3  4    3 n 8    1 p 2  4m  9n 
 2p  
4  44  8n  2p  9n 
 2p  
4 3  1  3  21 n 2   
7 2 n  
3 6m  12n  5p  
1  8m 
 3n 
 4p  6  8m 
 3
3 
 4p  6 6m  12
3  5p  
1 8m  2 
 4p  6 6m 
 8  5p  
1 8m 
 4p  4 6m  5p  7 2m 
 p  1 2m 
 p  1 → 10m 
 5p  5 6m  5p  7 → () 6m  5p  7 16m  12 2  10-8  Systems of Second-Degree Equations and Inequalities  Page 682  Check for Understanding  1. Possible number of solutions: 0, 1, 2, 3, or 4  3  2n  2p  9n 
 2p  
4  2  3  m  4 8m 
 3n 
 4p  6   
 3
23 
 4p  6  3  4  6  2 
 4p  6 1  p  2  34, 
23, 12  Chapter 10  1 9  than   . Since b  2 and a  3, the  y2  9.6 57. 8m 
 3n 
 4p  6 8m  6  3n  4p  8  a2b3 36  59. The expression  6 2 simplifies to . Since  5  4  3  h(x) 
 [[x]]  3 
6 
5 
4 
3 
2 
1 0 1 2 3  x 
3  x  
 2 
2  x  
1 
1  x  0 0x1 1x2 2x3 3x4 4x5 5x6 6x7  c2  a2  b2 c2  82  52 c  
89  c  9.4 about 9.4 m/s  8  358  2. Sample answer: y  x2, x2  (y  3)2  9  7. x2  y2  4 y2  4 
 x2 9x2 
 4y2  36 2 9x 
 4(4 
 x2)  36 13x2 
 16  36 x2  4 x  2 (2, 0)  y y 
 x2  x  O x 2  (y  3)2 
 9  x2  y 2  4 22  y2  4 y2  0 y0  y 3. The system contains equation(s) that are equivalent. The graphs coincide. 4. Graph each second-degree inequality. The region in which the graphs overlap represents the solution to the system. 5. x 
 y  0 xy (x 
 1)2 (y 
 1)2    20 5 (x 
 1)2 (x 
 1)2    20 5 
 1)2  4(x 
 1)2  (2, 0)  (2, 0)  O  8.  1 1   20 (x x2 
 2x  1  4(x2 
 2x  1)  20 5x2 
 10x  5 
 20  0 5(x2 
 2x 
 3)  0 5(x 
 3)(x  1)  0 x
30 x10 x3 x  
1 (3, 3), (
1, 
1)  x  xy  1 x(x2)  1 x3  1 x1 (1, 1)  x2  y 12  y 1y  y (1, 1)  O  x  y (3, 3)  9.  y  10.  y 4  O x (1, 1)  4  8  12  x  
4 
8  6. x  2y  10 x  10 
 2y x2  y2  16 (10 
 2y)2  y2  16 100 
 40y  5y2  16 5y2 
 40y  84  0  
12  11.  y 8  y  
b  
 b2 
 4 ac  2a  y  2 
 4( 
(
40)  
 (
40) 5)(84)  2(5)  y  40  
 
80  10  4 
8 
4 O 
4  no solution  4  8  x  
8  y  O  O  x  O  x  359  Chapter 10  12a. Let x  side length of flowerbed 1. Let y  side length of flowerbed 2. A1  x x or x2 A2  y y or y2 Total Area  x2  y2 680  x2  y2 x2  y2  680 Difference of Areas  x2 
 y2 288  x2 
 y2 x2 
 y2  288 12b.  40  y  y (2, 1)  O  20  15.  40 x  
40  12c. x2  y2  680 y2  680 
 x2 x2 
 y2  288 2 x 
 (680 
 x2)  288 2x2  968 x2  484 x  22 22 ft and 14 ft  x  (2, 1)  Since side length cannot be negative, an estimated solution is (22, 14) (22, 14).  20 
40 
20 O 
20  (2, 1), (
2, 
1)  x2  y2  680 222  y2  680 y2  196 y  14  
1  2x  y 
1 
 2x  y 4x2  y2  25 2 4x  (
1 
 2x)2  25 4x2  1  4x  4x2  25 8x2  4x 
 24  0 4(2x2  x 
 6)  0 4(2x 
 3)(x  2)  0 2x 
 3  0 x20 x  1.5 x  
2 
1  2x  y 
1  2x  y 
1  2(1.5)  y 
1  2(
2)  y 
4  y 3y (1.5, 
4), (
2, 3)  y (2, 3)  Pages 682–684  Exercises  13. x 
 1  0 x1 y2  49 
 x2 y2  49 
 (
1)2 y2  48 y  6.9 (1, 6.9)  O y  (1, 6.9) (1.5, 4)  O  x  16. x 
 y  2 x2y x2  100 
 y2 (2  y)2  100 
 y2 4  4y  y2  100 
 y2 2 2y  4y 
 96  0 2(y2  2y 
 48)  0 2(y 
 6)(y  8)  0 y
60 y80 y6 y  
8 x
y2 x
y2 x
62 x 
 (
8)  2 x8 x  
6 (8, 6), (
6, 
8)  (1, 6.9)  14. xy  2 2  y  x x2  3  y2  2 2  x2  3  x 4  x2  3  x2 x4  3x2  4 x4 
 3x2 
 4  0 (x2 
 4)(x2  1)  0 x2 
 4  0 x2  1  0 2 x 4 x2  
1 x  2 xy  2 xy  2 2(y)  2 (
2)y  2 y1 y  
1  y (8, 6)  O  (6, 8)  Chapter 10  x  360  x  19. x  y  
1 y  
1 
 x (y 
 1)2  4  x (
1 
 x 
 1)2  4  x (
2 
 x)2  4  x 4  4x  x2  4  x x2  3x  0 x(x  3)  0 x0  17. x 
 y  0 xy  (x 
 1)2  
 y2 9 (y 
 1)2  
 y2 9 
 1)2 
 9y2  1 1  9 (y y2 
 2y  1 
 9y2  9 
8y2 
 2y 
 8  0 4y2  y  4  0 y  
b  
 b2 
 4 ac  2a  y  
1  
 12 
 4(4)(4)   2(1)  y  
1  

63   2  x30 x  
3 x  y  
1 
3  y  
1 y2  x  y  
1 0  y  
1 y  
1 (0, 
1), (
3, 2)  y  no solution y  (3, 2)  O O  x  (0, 1)  x  20. xy  6  0 6  18. x2  2y2  10 x2  10 
 2y2 3x2  9 
 y2 3(10 
 2y2)  9 
 y2 30 
 6y2  9 
 y2 
5y2  
21 y2  4.2 y  2.0 3x2  9 
 y2 3x2  9 
 (2.0)2 x2  1.6 x  1.3 (1.3, 2.0), (1.3, 
2.0)  y  
x x2  y2  13 6 2  x2  
x  13 36  x2  x2  13 x4  36  13x2 x4 
 13x2  36  0 (x2 
 9)(x2 
 4)  0 x2 
 9  0 x2  9 x  3 xy  6  0 3y  6  0 y  
2 xy  6  0 
3y  6  0 y2 (3, 
2), (
3, 2)  3x2  9 
 y2 3x2  9 
 (
2.0)2 x2  1.6 x  1.3  y (1.3, 2.0)  (1.3, 2.0)  O  x  x2 
 4  0 x2  4 x  2 xy  6  0 2y  6  0 y  
3 xy  6  0 
2y  6  0 y3 (2, 
3), (
2, 3)  y  (1.3, 2.0)  (2, 3) (3, 2)  (1.3, 2.0)  O  x (3, 2) (2, 3)  361  Chapter 10  23. xy  
4  21. x2  y 
 3  0 x2  
y  3 x2  4y2  36 
y  3  4y2  36 4y2 
 y 
 33  0 (y 
 3)(4y  11)  0 y
30 y3 x2  y 
 3  0 x2  3 
 3  0 x2  0 x0 (0, 3), (2.4, 
2.8)  4  y  
x x2  25 
 9y2 144  x2  25 
 x 2  4y  11  0 y  
2.8 x2  y 
 3  0 x2  (
2.8) 
 3  0 x2  5.8 x  2.4  x4  25x2 
 144 x4 
 25x2  144  0 (x2 
 9)(x2 
 16)  0 x2 
 9  0 x2 
 16  0 x2  9 x2  16 x  3 x  4 xy  
4 xy  
4 3y  
4 4y  
4 y  
1.3 y  
1 xy  
4 xy  
4 
3y  
4 
4y  
4 y  1.3 y1 (3, 
1.3), (
3, 1.3) (4, 
1), (
4, 1) 24. 25. y y  y (0, 3)  x  O  (2.4, 2.8)  (2.4, 2.8)  4 2  x2  25 
 9
x  8 4  O  22. 2y 
 x  3  0 2y  3  x x2  16 
 y2 (2y  3)2  16 
 y2 4y2  12y  9  16 
 y2 5y2  12y 
 7  0 y  
b  
 b2 
 4ac   2a  y  
12 
 122 
 4(5)(
 7)  2(5)  y  
12  
284   10  y  0.5 or 2y 
 x  3  0 2(0.5) 
 x  3  0 x  4.0 (4.0, 0.5), (
2.8, 
2.9)  
8 
4  4  8  x  
8  y  26.  x  O  y  
2.9 2y 
 x  3  0 2(
2.9) 
 x  3  0 x  
2.8  y  27.  y 8  (4, 0.5)  O  4  x 
8 
4 O 
4  (2.8, 2.9)  Chapter 10  O  
4  
8  362  4  8x  x  28.  37a. 2x  2y  P xy  A 2x  2y  150 xy  800 37b. A system of a line and a hyperbola may have 0, 1, or 2 solutions. 37c. y  29.  y  y 8 4 
8 
4 O 
4  x  O  8x  4  80 40  
8  
80 
40 O  30.  31. 8  40  80  x  
40 
80  y  y  4  37d. xy  800 800  8 4 O  4  8  y  x  x  4  2x  2y  150  x  O  2x  2x  150 800  8  1600  2x  x  150  32.  33. 8  2x2  1600  150x x2 
 75x  800  0  y  y  4  
8 
4 O  8x  4  
8 
4 O 
4  
4  4  x  
8  
8  x  75  
 2425  2  x  12.88 or x  62.12 xy  800 xy  800 12.88y  800 62.12y  800 y  62.11 y  12.88 12.9 m by 62.1 m or 62.1 m by 12.9 m 38a. (h, k)  (0, 4) (x, y)  (6, 0) (x 
 h)2  4p(y 
 k) (6 
 0)2  4p (0, 4) 36  
16p 
2.25  p (x 
 h)2  4p(y 
 k) (x 
 0)2  4(
2.25)(y 
 4) x2  
9(y 
 4) 2 x  
9(y 
 4), y  0 y 38b.  34. parabola: vertex: (1, 
3) (y 
 k)2  4p(x 
 h) (
5  3)2  4p(
1 
 1) 1  
2  p (y 
 3)2  4
2(x 
 1) 1  (y  3)2  
2(x 
 1) line: m  
2, b  
7 y  mx  b y  
2x 
 7 35. circle: center: (0, 0), radius: 2
2  x2  y2  r2 x2  y2  8 hyperbola: (
2)(
2)  4 xy  4 36. large ellipse: a  5, b  4, center  (0, 0) y2  a2 y2  25  x  
b  
 b2 
 4 ac  2a 2 
 4(1)(800 
(
75)  
 (
75) )  2(1)  x  4  x O  x2   b2  1 x2   16  1 (interior is shaded)  small ellipse: a  3, b  2, center  (0, 
1) x2 (y 
 k)2    a2  b2 x2 (y  1)2    9 4  1 1 (exterior is shaded)  363  Chapter 10  38c. (h, k)  (0, 3) (x, y)  (6, 0) (x 
 h)2  4p(y 
 k) (6 
 0)2  4p(0 
 3) 36  
12p 
3  p (x 
 h)2  4p(y 
 k) (x 
 0)2  4(
3)(y 
 3) x2  
12(y 
 3) 2 x  
 12(y 
 3), y  0 39. xy  
12  y  40b.  estimate: (40, 30)  80 40  O 
80 
40  40  80x  
40  40c. x2  y2  2500 x2  2500 
 y2 x2  (y 
 30)2  1600 2500 
 y2  y2 
 60y  900  1600 
60y  1800  0 y  30 (x 
 35)2  (y 
 18)2  169 x2 
 70x  1225  (30 
 18)2  169 x2 
 70x  1200  0 (x 
 30)(x 
 40)  0 x 
 30  0 or x 
 40  0 x  30 x  40 Check (30, 30) and (40, 30): x2  y2  2500 x2  y2  2500 2 2 2 30  30  2500 40  302  2500 1800 2500 2500  2500  (40, 30) 41. x  3y  k 2y2  3y  k 2y2  3y 
 k  0  12  y  
x x  
y  1  x  

x  1 12  x2  12  x 
 x 
 12  0 (x 
 4)(x  3)  0 x
40 x30 x4 x  
3 xy  
12 xy  
12 4y  
12 
3y  
12 y  
3 y4 (4, 
3) (
3, 4) Check that (4, 
3) and (
3, 4) are also solutions of y2  25 
 x2. y2  25 
 x2 y2  25 
 x2 (
3)2  25 
 (4)2 (4)2  25 
 (
3)2 99  16  16  (4, 
3), (
3, 4) x2  3  1  y2  2y 
 2k  0 3 2  y2  2y  4  0 Complete the square.  y  3  2  y  34 1 3 2 
2k  4  (3, 4)  1  9  
2k  1 6  x  O  0  9  k  
8  (4, 3)  42a. 8  y  4  40a. first station: (h, k  (0, 0) x2  y2  r2 x2  y2  502 x2  y2  2500 second station: (h, k)  (0, 30) (x 
 h)2  (y 
 k)2  r2 x2  (y 
 30)2  402 x2  (y 
 30)2  1600 third station: (h, k)  (35, 18) (x 
 h)2  (y 
 k)2  r2 (x 
 35)2  (y 
 18)2  132 (x 
 35)2  (y 
 18)2  169  
8 
4 O 
4  4  8  
8  42b. yes; (6, 2) or (6, 
2) 42c. Earth's surface: x12  y12  40 x 2 1 40  2    
4  0 x1  y12   4 0 1 2    1   
4  0 y1  cos2 t  sin2 t  1 2    
4  0 x1  x1  2
10    cos2 t  cos t   cos t x1  2
10  Chapter 10  x  364  2    
4  0 y1  y1  2
10    sin2 t  sin t  y1  2
10  sin t  49. No; the domain value 4 is mapped to two elements in the range, 0 and 
3. 50. area of rectangle  
q  8(4) or 32 area of circles  2 (r2)  2(4p) or 8p area of shaded region  32 
 8p The correct choice is E.  asteroid: Let y2  
t. x2  0.25y2  5 x2  0.25 t2  5 42d.  10-8B Graphing Calculator Exploration:  Shading Areas on a Graph  Tmin  
8, Tmax  8, Tstep  0.13 [
15.16, 15.16] scl1 by [
10, 10] scl  1  Page 686  x2    y2  1 9  43.  1.  (x
 cos 30°  y
 sin 30°)2   (
x
 sin 30°  y
 cos 30°)2  1 9 2 
 3 1 x
  y
 2 2 2 1 
 3   
2x
  2y
  1 9      3 
 3 1 (x
)2  x
y
  (y
)2 4 2 4  9      
3   1  3   4(x
)2 
 2x
y
  4(y
)2  1  3 1 9 27 
3  9
 3 (x
)2  x
y
  (y
)2  (x
)2 
x
y
   (y
)2  9 4 4 4 4 2 2 x
y
  7(y
)2 
 9  0 3(x
)2 
 4
3  44.  x  4t  1  x
1  4  y  5t 
 7  y7  5  t x
1  4  [
9.1, 9.1] scl1 by [
6, 6] scl1 2.    t  y7  5  (x 
 1)(5)  (4)(y  7) 5x 
 5  4y  28 5x 
 4y 
 33  0    45. 4 csc v cos v tan v  4 sin v (cos v) cos v  1  sin v  [
9.1, 9.1] scl1 by [
6, 6] scl1 3.  4 46. r  10 cm or 0.10 m v  rq v  (0.10)1 5 2p  v  3.14 m/s 47.  48.  r 0 1 2  1 1 1 1  0 0 1 2  
4 
4 
3 4  0 0 1 4  [
9.1, 9.1] scl1 by [
6, 6] scl1 4.  between 1 and 2 y  (x  2)2 
 3 x  (y  2)2 
 3 x  3  (y  2)2 
x  3y2 
x 3
2y y  [
9.1, 9.1] scl1 by [
6, 6] scl1 5a. 3 5b. Find the points of intersection for the boundary equation by using the TRACE function.  y 
 (x  2)2  3  O x y 
  
 x32  365  Chapter 10  14. (x 
 h)2  (y 
 k)2  r2 2 (x 
 0)2  (y 
 0)2  3
3  x2  y2  27 y  5c. SHADE(

((36
X2)/9),
((36
X2)/9),
6,
2,3,4); SHADE(
X22,
((36
X2)/9),
2,2,3,4,); SHADE(

((36
X2)/9), 
((36
X2)/9),2,6,3,4)  (0, 3
3) (3
3, 0)  O  [
9.1, 9.1] scl1 by [
6, 6] scl1 6. See students' work.  x  15. (x 
 h)2  (y 
 k)2  r2 (x 
 2)2  (y 
 1)2  22 (x 
 2)2  (y 
 1)2  4  y  (2, 1)  Chapter 10 Study Guide and Assessment  (4, 1)  O  x (2, 1)  Page 687  Understanding and Using the Vocabulary  1. true 3. false; transverse 5. false; hyperbola  16.  2. false; center 4. true 6. false; axis or axis of symmetry 8. false; parabola 10. false; ellipse  7. true 9. true  x2  y2  6y x2  (y2 
 6y  ?)  0  ? x2  (y2 
 6y  9)  0  9 x2  (y 
 3)2  9  y (0, 6) (3, 3)  (0, 3)  Pages 688–690  Skills and Concepts  11. d  
 (x2 
  x1)2  (y2 
 y1)2 d  
 (
3 
 1)2  [
4 
 (
6)]2 d  
20  or 2
5  x1  x2 y1  y2    2 , 2    1  (
3) 
6  (
4) ,  2 2     (
1, 
5)  O 17.    x  x2  14x  y2  6y  23  14x  ?)  (y2  6y  ?)  23  ?  ? 2 (x  14x  49)  (y2  6y  9)  23  49  9 (x  7)2  (y  3)2  81 (x2  2 x2 
 x (y2 
  y1)2 12. d  
 1)    d  
 (a  3 
 a)2  (b   4 
 b)2 d  
25  or 5  (7, 6)  y  (7, 3)  O  x1  x2 y1  y2  aa3 bb4 ,  2, 2   2 2   (a  1.5, b  2)  (16, 3)  (x2 
  x1)2  (y2 
 y1)2 13. AB  
  
 [3 
 ( 
5)]2   [4 
 (
2)]2  
100  or 10 DC  
 (x2 
  x1)2  (y2 
 y1)2  18.   
 (10 
 2)2  [3 
  (
3)]2  
100  or 10 BC  
 (x2 
  x1)2  (y2 
 y1)2  
 (10 
 3)2  (3 
  4)2  
50  or 5
2  AD  
 (x2 
  x1)2  (y2 
 y1)2  3x2  3y2  6x  12y 
 60  0 x2  y2  2x  4y 
 20  0 (x2  2x  ?)  (y2  4y  ?)  20  ?  ? (x2  2x  1)  (y2  4y  4)  20  1  4 (x  1)2  (y  2)2  25  y (1, 3)   
 [2 
 ( 
5)]2   [
3 
(
2 )]2  
50  or 5
2  Yes; AB  DC  10 and BC  AD  5
2 . Since opposite sides of quadrilateral ABCD are congruent, ABCD is a parallelogram.  Chapter 10  x  O (1, 2)  366  x (4, 2)  minor axis vertices: (h, k  b)  (3, 
1  2) or (3, 1), (3, 
3)  x2  y2  Dx  Ey  F  0 12  12  D(1)  E(1)  F  0 ⇒ D  E  F  
2 (
2)2  22  D(
2)  E(2)  F  0 ⇒ 
2D  2E  F  
8 (
5)2  12  D(
5)  E(1)  F  0 ⇒ 
5D  E  F  
26 D  E  F  
2 (
1)(
5D  E  F)  (
1)(
26) 6D  24 D4 
2D  2E  F  
8 (
1)(D  E  F)  (
1)(
2) 
3D  E  
6 
3(4)  E  
6 E6 D  E  F  
2 4  (6)  F  
2 F  
12 x2  y2  Dx  Ey 
 F  0 x2  y2  4x  6y 
 12  0 2 (x  4x  ?)  (y2  6y  ?)  12  ?  ? (x2  4x  4)  (y2  6y  9)  12  4  9 (x  2)2  (y  3)2  25 center: (h, k)  (
2 
 3) r  
25  or 5 20. center: (h, k)  (5, 2) a2  36 b2  16 a  
36  or 6 b  
16  or 4 a2 
 b2 c  
 c  
6  
 4 or 
2  foci: (h, k  c)  (5, 2  
2 ) major axis vertices: (h, k  a)  (5, 2  6) or (5, 8), (5, 
4) minor axis vertices: (h  b, k)  (5  4, 2) or (9, 2), (1, 2) y (5, 8)  19.  (1, 2)  (5, 2)  O  y (3, 1)  O  (8, 1) x  (2, 1) (3, 3)  22.  6x2  4y2  24x 
 32y  64  0 6(x2  4x  ?)  4(y2 
 8y  ?)  
64  ?  ? 6(x2  4x  4)  4(y2 
8y  16)  
64  6(4)  4(16) 6(x  2)2  4(y 
 4)2  24 (x  2)2 (y 
 4)2     24 4 6  center: (h, k)  (
2, 4) b2  4 a2  6 a  
6  b  
4  or 2 c  
 a2 
 b2 c  
6  
 4 or 
2  foci: (h, k  c)  (
2, 4  
2 ) major axis vertices: (h, k  a)  
2, 4  
6  minor axis vertices: (h  b, k)  (
2  2, 4) or (0, 4), (
4, 4) y (2, 4  
 6) (2, 4) (0, 4)  (4, 4) (2, 4  
 6)  O 23.  x  x2  4y2  124  8x  48y (x2 
 8x  ?)  4(y2 
 12y  ?)  
124  ?  ? (x2 
 8x  16)  4(y2 
 12y  36)  
124  16  4(36) (x 
 4)2  4(y 
 6)2  36 (x 
 4)2 (y 
 6)2     36 36 9  (9, 2)  center: (h, k)  (4, 6) a2  36 b2  9 a  
36  or 6 b  
9  or 3 c  
 a2 
 b2 c  
36  
 9 or 3
3  foci: (h  c, k)  (4  3
3 , 6) major axis vertices: (h  a, k)  (4  6, 6) or (10, 6), (
2, 6) minor axis vertices: (h, k  b)  (4, 6  3) or (4, 9), (4, 3) y  x (5, 4)  21.  (3, 1)  4x2  25y2 
 24x  50y  39  4(x2 
 6x  ?)  25(y2  2y  ?)  39  ?  ?  4(x2 
 6x  9)  25(y2  2y  1)  39  4(9)  25(1) 4(x 
 3)2  25(y  1)2  100 (x 
 3)2 (y  1)2     1 25 4  center: (h, k)  (3, 
1) b2  4 a2  25 a  
25  or 5 b  
4  or 2 c  
 a2 
 b2 c  
25 4 
  or 
21  foci: (h  c, k)  (3  
21 , 
1) major axis vertices: (h  a, k)  (3  5, 
1) or (8, 
1), (
2, 
1)  (4, 9) (2, 6)  (4, 6) (10, 6) (4, 3)  O  367  x  Chapter 10  24. (h, k)  (
4, 1) a9 b6  (y 
 k)2 (x 
 h)2   b 2 a2 (y 
 1)2 [x 
 (
4)]2   6 2 92 (y 
 1)2 (x  4)2    81 36  center: (h, k)  (0, 
2) b2  1 a2  4 a  
4  or 2 b  
1  or 1 c  
 a2  b2 c  
4   1 or 
5  transverse axis: horizontal foci: (h  c, k)  (0  
5 , 
2) or (
5 , 
2) vertices: (h  a, k)  (0  2, 
2) or (
2, 
2), (2, 
2)  1 1 1  25. center: (h, k)  (0, 0) b2  16 a2  25 a  
25  or 5 b  
16  or 4 c  
 a2  b2 c  
25 16   or 
41  transverse axis: horizontal foci: (h  c, k)  (0  
41 , 0) or (
41 , 0) vertices: (h  a, k)  (0  5, 0) or (
5, 0), (5, 0)  b  asymptotes: y 
 k   a(x 
 h) 1  y 
 (
2)   2(x 
 0) 1  y  2   2x  y  b  asymptotes: y 
 k   a (x 
 h)  x  O  4  y 
 0   5(x 
 0)  (2, 2)  (2, 2) (0, 2)  4  y   5x  y 28. (5, 0)  (5, 0)  x  O  9x2 
 16y2 
 36x 
 96y  36  0 9(x2 
 4x  ?) 
 16(y2  6y  ?)  
36  ?  ? 9(x2 
 4x  4) 
 16(y2  6y  9)  
36  9(4) 
 16(9) 9(x 
 2)2 
 16(y  3)2  
144 (y  3)2 (x 
 2)2  
   1 9 16  center: (h, k)  (2, 
3) b2  16 a2  9 a  
9  or 3 b  
16  or 4 c  
 a2  b2 c  
9 6  1 or 5 transverse axis: vertical foci: (h, k  c)  (2, 
3  5) or (2, 2), (2, 
8) vertices: (h, k  a)  (2, 
3  3) or (2, 0), (2, 
6) a asymptotes: y 
 k   b(x 
 h)  26. center: (h, k)  (1, 
5) a2  36 b2  9 a  
36  or 6 b  
9  or 3 c  
 a2  b2 c  
36 9   or 3
5  transverse axis: vertical foci: (h, k  c)  1, 
5  3
5  vertices: (h, k  a)  (1, 
5  6) or (1, 1), (1, 
11) a  y 
 (
3)   4(x 
 2)  3  6  y  3   4(x 
 2)  asymptotes: y 
 k   b(x 
 h)  3  y 
 (
5)   3(x 
 1) y  5   2(x 
 1)  y  y (2, 0)  O  (1, 1)  O  x (1, 5)  (2, 6)  (1, 11)  27.  x2 
 4y2 
 16y  20 x2 
 4(y2  4y  ?)  20  ? x2 
 4(y2  4y  4)  20 
 4(4) x2 
 4(y  2)2  4 x2  4  Chapter 10  (2, 3)  (y  2)2  
 1  1  368  x  29. c  9 quadrants: I and III transverse axis: y  x vertices: xy  9 3(3)  9 (3, 3)  33. vertex: (h, k)  (1, 
2) 4p  
16 p  
4 focus: (h  p, k)  (1  (
4), 
2) or (
3, 
2) directrix: x  h 
 p x  1 
 (
4) x5 axis of symmetry: y  k y  
2  xy  9 (
3)(
3)  9 (
3, 
3)  y (3, 3)  y O  x  (3, 3)  O  (3, 2)  30. 2b  10 b5  x1  x2 y1  y2  1  1 
1  5  center: 2, 2  2, 2  x
5   (1, 2) transverse axis: vertical a  distance from center to a vertex  2 
 (
1) or 3 (y 
 k)2  a2 (y 
 2)2  32 (y 
 2)2  9  34. y2  6y 
 4x  
25 y2  6y  ?  4x 
 25  ? y2  6y  9  4x 
 25  9 (y  3)2  4(x 
 4) vertex: (h, k)  (4, 
3) 4p  4 p1 focus: (h  p, k)  (4  1, 
3) or (5, 
3) directrix: x  h 
 p x4
1 x3 axis of symmetry: y  k y  
3  (x 
 h)2  
 b 1 2 (x 
 1)2  
 5 1 2 (x 
 1)2  
 25  1 x1  x2 y1  y2  
2  6 
3  (
3)  31. center: 2, 2  2, 2  (2, 
3) a  distance from center to a vertex  2 
 (
2) or 4 c  distance from center to a focus  2 
 (
4) or 6 b2  c2 
 a2 b2  62 
 42 b2  20 transverse axis: horizontal  y  (4, 3)  (5, 3)  x
3  35.  32. vertex: (h, k)  (5, 3) 4p  8 p2 focus (h, k  p)  (5, 3  2) or (5, 5) directrix: y  k 
 p y3
2 y1 axis of symmetry: x  h x5 y  x2  4x  y 
 8 x2  4x  4  y 
 8  4 (x  2)2  y 
 4 vertex: (h, k)  (
2, 4) 4p  1 1  p  4 focus: (h, k  p)  
2, 4  4 or (
2, 4.25) 1  directrix: y  k 
 p 1  y  4 
 4 y  3.75 axis of symmetry: x  h x  
2  (5, 5)  O  x  O  (x 
 h)2 (y 
 k)2  
 b 1 2 a2 (x 
 2)2 [y 
 (
3)]2  
 20  1 42 (x 
 2)2 (y  3)2  
   1 16 20  (5, 3)  x  (1, 2)  y (2, 4.25)  y 
 3.75 (2, 4)  y
1  x O  369  x  Chapter 10  44. x  2 sin t  36. vertex: (h, k)  (
1, 3) (y 
 k)2  4p(x 
 h) (7 
 3)2  4p[
3 
 (
1)] 16  8p 2p Since parabola opens left, p  
2. (y 
 k)2  4p(x 
 h) (y 
 3)2  4(
2)[x 
 (
1)] (y 
 3)2  
8(x  1)  37. vertex: (h, k)  5,  38. 39. 40. 41. 42.  2
4  2  x  2  t 
2 
1 0 1 2  x 
2 
1 0 1 2  y 
1 2 3 2 
1  y  3   sin t t  sin2  x 2  2     y 2  3 x2 y2    4 9     1 1  x 0  y 3  (x, y) (0, 3)    2  2  0  (2, 0)    0  
3  (0, 
3)  3  2  
2  0  (
2, 0)  y 
1 1 3.5  (x, y) (0, 
1) (2, 1) (3, 3.5)  y t
0 t

 2 t 
 3
 2  x  O t

  45. x  
t x2  t  (x, y) (
2, 
1) (
1, 2) (0, 3) (1, 2) (2, 
1)  t  y  2 
 1 x2  y  2 
 1 t 0 4 9  y  x 0 2 3  y  t
9  x  O  O 43. x  cos 4t cos2 4t  sin2 4t  1 x2  y 2  1 t 0  y  sin 4t  x 1  y 0  (x, y) (1, 0)    8   4  0  1  (0, 1)  
1  0  (
1, 0)  3  8  0  
1  (0, 
1)  x2 y2    49 49 x 2 y 2   7 7 sin2 t  cos2 t     x 2  7  
  t
8 
  t
4  x  7  t
0  O      sin2 t  sin t  x  7 sin t  x t  t
0  x  46. Sample answer: Let x  t. y  2x2  4 y  2t2  4, 
  t   47. Sample answer: x2  y2  49  y  Chapter 10   cos t  t1  cos2  t 0   or (5, 
1)  focus: (h, k  p)  (5, 2) kp2 
1  p  2 p3 (x 
 h)2  4p(y 
 k) (x 
 5)2  4(3)[y 
 (
1)] (x 
 5)2  12(y  1) A  5, c  2; ellipse A  C  0; equilateral hyperbola A  C  5; circle C  0; parabola y  
t2  3 y  
x2  3  y  3 cos t  3
 
 8  370  1 1 1  y 2  7  y  7   cos2 t  cos t  y  7 cos t, 0  t  2p  54. B2 
 4AC  (
6)2 
 4(1)(9) 0 parabola  48. Sample answer: x2 y2    36 81 x 2 y 2   9 6 cos2 t  sin2 t     x 2  6  x  6     1 1  B   tan 2v   A
C  1  y 2  9   cos2 t  y  9   cos t  x  6 cos t 49. Sample answer: Let y  t. x  
y2 x  
t2, 
  t   50. B2 
 4AC  0 
 4(4)(9)  
144 A C; ellipse  
6   tan 2v   1
9   sin2 t  3  tan 2v  4   sin t  4x2  9y2  36  p 2 p p 2 x
 cos  y
 sin 6  9 
x
 sin 6  y
 cos 6 2 2 1 1 
3  
3  4 2x
  2y
  9 
2x
  2y
 3 1 
3  4 4(x
)2  2x
y
  4(y
)2 1 3 
3   9 4(x
)3 
 2x
y
  4(y
)2 3(x
)2  2
3 x
y
  (y
)2  4    p  6       2v  36.86989765° v  18° 55. (x 
 1)2  4(y 
 1)2  20 (y 
 1)2  4(y 
 1)2  20 5(y 
 1)2  20 (y 
 1)2  4 y 
 1  2 y  3 or 
1 y3 xy x3 y  
1 xy x  
1 (3, 3), (
1, 
1)  y  9 sin t, 0  t  2p       9  9
3   27  5
3   31   36  36  y   4(x
)2 
 2 x
y
  4(y
)2  36 21 (x
)2 4  O  2  56. 2x 
 y  0 2x  y y2  49  x2 (2x)2  49  x2 4x2  49  x2 3x2  49 x  4.04 2x 
 y  0 2(4.04) 
 y  0 y  8.08 (4.0, 8.1), (
4.0, 
8.1)  
 42x
 
 2y
  0 
2   x  (1, 1)  
 2 x
y
  4(y
)2  36 21(x
)2 
 10
3 x
y
  31(y
)2 
 144  0 2 51. B 
 4AC  0 
 4(0)(1) 0 parabola y2 
 4x  0 2 (
x
 sin 45°  y
 cos 45°) 
 4(x
 cos 45°  y
 sin 45°)  0  

22 x
  
22 y
  (3, 3)   36  
2   1 1 (x
)2 
 x
y
  (y
)2 
 2
2 x
  2
2 y
  0 2 2 x
  4
2 y
  0 (x
)2 
 2x
y
  (y
)2
 4
2  52. B2 
 4AC  0 
 4(4)(
16)  256 hyperbola 4x2 
 16(y 
 1)2  64 4(x 
 h)2 
 16(y 
 k 
 1)2  64 4(x 
 1)2 
16(y  2 
 1)2  64 4(x 
 1)2 
 16(y  1)2  64 4(x2 
 2x  1) 
 16(y2  2y  1) 64 4x2 
 8x  4 
 16y2 
 32y 
 16 
 64  0 x2 
 4y2 
 2x 
 8y 
 19  0  2x 
 y  0 2(
4.04) 
 y  0 y  
8.08  y (4.0, 8.1)  (4.0, 8.1)  O  x   
 4(6)(8) 53. B2 
 4AC  2
3 2  A   
180 C; ellipse B   tan 2v   A
C 2
 3   tan 2v   6
8  tan 2v  

3  2v  
60° v  
30°  371  Chapter 10  57. x2 
 4x 
 4y  4 x2 
 4x 
 4  4y (x 
 2)2  4y  0 2 (x 
 2)  x2 
 4x 
 4  0 2 x 
 4x  4  x2 
 4x 
 4  0 2x2 
 8x  0 2x(x 
 4)  0 2x  0 x
40 x0 x4 (x 
 2)2  4y  0 (x 
 2)2  4y  0 (0 
 2)2  4y  0 (4 
 2)2  4y  0 y  
1 y  
1 (0, 
1), (4, 
1)  61.  Page 691  r  
 (12 
 0)2  (16 
 0)2 r  20 x2  y2  r2 x2  y2  202 x2  y2  400 63b. area of watered portion  pr2  p202  1256.6 ft2 area of backyard  
q  50(40)  2000 ft2 area of nonwatered portion  2000 
 1256.6  743.4 ft2 743.4  percent not watered   2000  (4, 1)  58. xy  
4 4  y  
x x2  y2  12 4 2  x2  
x  12 16  x2  x2  12   0.37  x4  16 
 12x2  0 (x2)2 
12(x2)  16  0  x2   12  
 122 
 4(1)(1 6)  2(1)  about 37%  (3.2, 1.2) (1.2, 3.2)  60.  y 4 4 O  4  x  4  Chapter 10  c2  (y 
 k)2   b 1 2 (y 
 0)2  34,560,000 y2  34,560,000  1 1  65. a  3.5 b3 a2 
 b2 c  
 c  
 3.52 
 32 c  1.8 about 1.8 feet from the center  x  x  a2  (x 
 h)2  a2 (x 
 0)2  60002  x2   36,000,000  O  O  c   0.2   6000   
 b2  60002 
 12002 b2  34,560,000  (1.2, 3.2)  y  e  a 1200  c  b2  y  59.  c  64. 2a  2,000 a  6000  or x2  1.528 x2  10.472 x   3.236 x  1.236 xy  
4 xy  
4 3.236y  
4 1.236y  
4 y  
1.236 y  
3.236 xy  
4 xy  
4 
3.236y  
4 
1.236y  
4 y  1.236 y  3.236 (3.2, 
1.2), (
3.2, 1.2), (1.2, 
3.2), (
1.2, 3.2)  (3.2, 1.2)  Applications and Problem Solving  63a. r  
 (x2 
  x1)2  (y2 
 y1)2  x  
b  
 b2 
 4 ac  2a  O  x  O 2  O  x2   y  2  y  (0, 1)  62.  y  372  x  Page 691  Open-Ended Assessment  3. The information in the question confirms the information given in the figure. Recall the formula for the area of a triangle — one half the base times the height. The triangle DCB is obtuse, so the height will lie outside of the triangle. Let D C  be the base. The length of the base is 6. The height will be equal to 7, since it is a line segment parallel to A D  through point B.  1. Sample answer: c  e  a 1  9  c   a  Let a  9. 1  9  c   9  b2  a2 
 c2 b2  92 
 12 b2  80  c1 x2 y2   a2  b2 x2 y2   92  80 x2 y2    81 80  1  1  A  2bh 1   2(6)(7) or 21  1  The correct choice is A. 4. The problem asks how many more girls there are than boys. First find how many girls and how many boys there are in the class. One method is to find the fraction of girls in the whole class and the fraction of boys in the whole class. Since the ratio of girls to boys is 4 to 3, the 4 fraction of girls in the whole class is 7. Find the number of girls in the class by multiplying this fraction by 35. 4 (35)  20 There are 20 girls in the class. 7  1  2. Sample answer: axis of symmetry: x  h x  2, so h  2 focus: (h, k  p)  (2, 5) kp5 Let k  2, p  3. (x 
 h)2  4p(y 
 k) (x 
 2)2  4(3)(y 
 2) (x 
 2)2  12(y 
 2)  3  Using the same process, the fraction of boys is 7. 3 (35) 7  SAT & ACT Preparation Page 693  SAT & ACT Practice  Red 3 60  There are 15 boys in the class.  So there are 5 more girls than boys. The correct choice is D. Another method is to use a "Ratio Box." First enter the given information, shown in the darker cells below. Then enter the number for the total of the first row, 7. To go from the total of 7 to the total of 35, you must multiply by 5. Write a 5 in each cell in the second row.  1. Add the two numbers of parts to get the whole, 8. 3 The fraction of red jelly beans to the whole is 8. The total number of jelly beans is 160. The 3 number of red jelly beans is 8(160) or 60. The correct choice is C. Or you can use a ratio box. Multiply by 20. Green 5   15  Girls 4 5 20  Whole 8 160  Boys 3 5 15  Total 7 5 35  Then multiply the two numbers in the first column to get 20 girls, shown with a dark border. Multiply the second column to get 15 boys. Subtract to find there are 5 more girls than boys. 5. Set A is the set of all positive integers less than 30. Set B is the set of all positive multiples of 5. The intersection of Sets A and B is the set of all elements that are in both Set A and Set B. The intersection consists of all positive multiples of 5 which are also less than 30. The intersection of the two sets is {5, 10, 15, 20, 25}. The correct choice is A.  2. Notice the capitalized word EXCEPT. You might want to try the plug-in method on this problem. Choose a value for b that is an odd integer, say 1. Then substitute that value for b in the equation. a2b  122 2 a (1)  122 a2  122 a  12 Check the answer choices for divisors of this value of a. 12 is divisible by 3, 4, 6, and 12, but not by 9. The correct choice is D.  373  Chapter 10  6. For a quadratic equation in the form y  a(x – h)2  k, the coordinates of the vertex of the graph of the function are given by the ordered pair (h, k). So the vertex of the graph 1 of y  2(x – 3)2 + 4 has coordinates (3, 4). The correct choice is C. 7. On the SAT, if you forget the relationships for 45° right triangles, look at the Reference Information in the gray box at the beginning of each mathematics section of the exam. The measure of each leg of a 45–45–90 triangle is equal to the length of the hypotenuse divided by 
2 . Multiply both numerator and denominator by 
2  and simplify. BC  9. Let d represent the number of dimes in the jar. Since there are 4 more nickels than dimes, there are d  4 nickels in the jar. So, the ratio of dimes d to nickles in the jar is . This ratio is less d4 than 1. The only answer choice that is less than 8 d 8 1 is choice A, . If   , then d  16. So, 10 d4 10 there are 16 dimes and 16  4 or 20 nickels in the 8 16 jar, and   .  total liters  total bottles 8  20  8   
2  8 
2     
2  
2  8
2     2 or 4
2  1 x 7 1 x 7  x   1 2  x  0.4 or 5 The correct answer is .4 or 2/5.  100   x 1   5(100)  20  x  140 The correct choice is E.  Chapter 10  x liters    1 bottle  20x  8  The correct choice is D. You could also use the Pythagorean Theorem and the fact that the two legs must be equal in length, but that method might take more time. 8. Form a ratio using the given fractions as numerator and denominator. Write a proportion, using x as the unknown. Multiply the crossproducts. Solve for x. 1  7  1  5  10  20  The correct choice is A. 10. Set up a proportion.  374  Chapter 11 Exponential and Logarithmic Function 3   Page 695  3   8. 32 5 
 (25) 5  Real Exponents  11-1  15   
 25  
 23 
8 2 9. (3a )3  3a5 
 33  a6  3a5 
 34a1  Graphing Calculator Exploration  1.  81  
 81a1 or a 1   1   10.  m3n2   m4n5 
 (m3n2) 2  (m4n5) 2 3   2   7   7   4   5   
 m2n2  m2n2 2.  11.  
 m 2 n 2 or m3n3mn    
 
 8n  27  4 n  1  2  8n  27  4 n    n   7   82  22  
  n   4 2 n   5.  a m  b  
   
 bm, when b  
  
  2 2  22  2n  3n   3. am  an 
 amn 4. (am)n 
 amn am  7   (23) 2  2 2  n  (22) 2   2  2  0  7   3n  2  7   2n   
 2  22  2 2 5n  7  2  
2  Page 700  n1  2  Check for Understanding  
 22n3  2  1. The quantities are not the same. When the negative is enclosed inside of the parentheses and the base is raised to an even power, the answer is positive. When the negative is not enclosed inside of the parentheses and the base is raised to an even power, the answer is negative. 2. If the base were negative and the denominator were even, then we would be taking an even root of a negative number, which is undefined as a real number. 3. Laura is correct. The negative exponent of 10 represents a fraction with a numerator of 1 and a denominator of a positive power of 10. The product of this fraction and a number between 1 and 10 is between 0 and 1. 1  4. 54 
 54  5.  1   
 625  2  
196   
  
 1  3  6. 216 
 (63) 3  3  
6 
6  1   1   1   1   4   8   
 22  x2  y2 1   
 2 2  x2y4 or x2242  1   13.  169x5 
 (169x5) 2 1   1   
 169 2  (x5) 2 5   
 13x 2  1   14.  a2b3c4 d5 
 (a2b3c4d5) 4 4  1   1   1   1   
 (a2) 4 (b3) 4 (c4) 4 (d5) 4 1   3   5   
 a 2 b 4 c d 4 1   3 1    1   1   15. 6 4 b 4 c 4 
 (6b3c) 4  1   5   3   16. 15x 3 y 5 
 15x 15 y 15  1   
  6b3c 4  
 15 (x5y3) 15 
 15  x5y3 15  17.  p4q6r5 
 (p4q6r5) 3  1   1  2  1  2  1  3 1   1   
 (p4) 3 (q6) 3 (r5) 3  1  2  7. 27   3  
 27  3  4   5   
 p 3 q2r 3  1  2  
 (33)  3 3  2  
 22n3  2n1  12. (2x4y8) 
 2 2  (x4) 2  (y8) 2  
  

  
  1  3  1  9 2  16 16 2  9 162  2 9 256  81  1  2  
 pq2r pr2 3  1  2  
3 3  18.  4   
 32 
 32 
9  4  5  y 
 34 4 5   5 4  
y   5   
 34 4  1   y 
 (345) 4 y  375  82.1 Chapter 11  19. A 
 r2 r 
 3.875  107 m A 
 (3.875  107 m)2 
 (3.875)2  (107)2 m2 
 (15.015625  1014) m2 4.717  1013 m2  33. 
216  
 (216 3 )2  Pages 700–703  34. 81 2  81  20.  (6)4  
 
  21.  22. (5  3)2 
 
 225  23.  2   2   
 (63) 3 
 62 
 36  
   1 64 
  64 1  
  1296 24  
 24(1) 2 1 
 25  3  
78  8  35.  
  

   36.  1   7 (128 )4   
  1  4  
  (128) 7 1  4  1  1  
 y8  y8  1  
  1 1    3 9 
 
 1   27  3  38.  1  4  9 9  4   (4y4) 2  3  2  
1  3   
4 (y4) 2 3   6   1   12   
 22y 2  1   27. 729 3 
(93) 3 
9  
 8y6  1  39.   (27p3q6r1) 3  33  2  
  1  2  3   1  2  1   1   1   1   6   1  1  1   1  1   
 28x8  1   
 28x8 or  
 26  1   1  4  30. 64 2 
 (26) 12  31. 16  1   
 2 2 or 2   1  3   40. [(2x4]2 
  [(2x)4]2  1   
 2  62  
  1  1  
  16 4 1  1  
  (24) 4 1  2  1   1   1   41. (36x6) 2 
 36 2 (x6) 2    
 6x3  1  256x8 b2n  42.  b2n  
  1  2    1   
 (b2n  b2n) 2 1   
 (b4n) 2 
 b2n    43.  37(32)4  6  2n  1 4n   2  
    1 2  2n  n  4 1   (33) 2  2n 2 
  4  315  
 3 9 
 36 
 729  Chapter 11  3   
 3pq2r  1  2  
 22  22  62  
  1   1  r3  
3  (37)(94)  276   1   1   
 33p3q3  29. 2  12 
 2  (2  6)  32.  1   1   33  1   1   
 27 3 (p3) 3 (q6) 3 
r 3  
 (33) 3 (p3) 3 (q6) 3 
r 3   (33) 3  
 32 1  2  3   
 (22) 2 (y4) 2  26. 81 2 
 (92) 2 
9  3    [(2)7] 7 1   
 (2) 4 1 
 1 6 2 3 (3n ) 
 33(n2)3    y8 37. (y2)4  y8 
  (y2)4  1  27  2    
 27n6   25. (31  33)1 
  31  31  28.    
 89  7  8 8 3  7 83  73 512  343  1   1  1  1  1  
  1   
 (92) 2  
 9  9  
  3  
  1  2  (9 2) 2  
 32 24.  1   
 216 3  1   Exercises  1  (6)4 1  1296 152  2  3  
  376  n   2  1   4  1   4  1   4  1   44. 
3m 2  27n 4  
 34
m 2  274
n4  
 34m2(33)4n 45.  1 f 1 6 4  256g4h4  
    
 (f16  2561  g4  h4) 1 4  (2561)  1 4  1 4  (h4)  1  63.  46.   x2
x  x  
 x2
x   3  4  6  3  4  3  4  
 x2  x  1  3  47. 
2x y  1  4  3  3  4  6  4  2  3  
3x y  
 6x  4  3  2  2  4   a5b7c 3  1   1   1   1   1   64.  20x3y6 
 (20) 2 (x3) 2 (y6) 2  1  6    1   
 (a5) 3 (b7) 3 (c) 3 3  
 2xy35x  1  6  
x  1 
 x 
x  1 5  4  1  4  x  1   
 ab2 a2bc  4 f4g  h  or  1   
 defd   4  
 f 4 256  g  h1 f 4gh1  1   62. d3e2f 2 
 (d3) 2 (e2) 2 (f 2) 2  4  1  (g4)   0.33   0.69  1  4  
4  1  5  3  3  
 316m2n 
 (f16)   ) (245   61. x 
  14.2 
 x  65.  3  2  2    (14.2) 3 
 (x 0.17 x  1  6  3  3  66.  3  2  5   2 3  712 
 15a 2  )  712  15 2 712   5 15  y  
   1  2  
 6x y 48.  m  1 1    1    a  
 
a n  m n  
a 
a 
a  4.68  1   49.  m6n 
 (m6) 2 
n 2   1 1   n m  
 m3n  5   5 2    1   x 3.79 68. d 
 6.794  103 km so r 
 3.397  103 km 4 V 
 3r3  1   1   3   4  
 x 2 y 2 1  3  1  3  51.  8x3y6 
 8 (x3) (y6)  
 3(3.397  103 km)3  1  3  1.64  1011 km3 10 1 2 10 5 7 69. y 
 3x; x 
 8, 6, 5, 33, 2, 3, 9, 3, 2  
 2xy2  1   1   1   52. 17 x14y7z12  
 17(x14) 2 (y7) 7 (z12) 7 7  
  12   17x2yz 7 1   1   1   53.  a10b2   c2 
 (a10) 5 (b2) 5 (c2) 4 4  2  5  1  2  
 a2b c 1   1   1   8 27 8 54. 60 r80s56 t27 
 60(r80) 8 (s56) (t ) 27  8 10 
 60r st 8  5  8  4   2  3  3 2    57. p 3 q 2 r 3 
 p 6 q 6 r 6  58.  6  3  21  1  3  2  1   23  
  p4q3r2 1  7  3  8  8  1   
 23 3  
 2   7  21  59. 13a b 
 13a b 1  60.  
 13   (n3m9) 2  3x  y  8  38  6  36  5  35  9  6561 1  729 1  243   a3b7  10   3 33 1  2  3  2  3  3  10  9  3  5  3  3  7  2  3  1.395 1.732 2.080 3.389 14.620 46.765  21  1   69a. If x 0 then y  0. If x 
 0 then y 
 1. Since x 0, y  0 and y 1. So, 0 y 1. 69b. If x  0 then y  1. If x 1 then y 3. So, 1 y 3. 69c. If x  1 then y  3. So, y  3.  1   
 (n3) 2 (m9) 2 3   x  10  33 1  2 2  3 10  9 5  3 7  2  56. (7a) b 
  75a5b3  5  55. 16 
 16  1 1    2   (x 2 ) 5 
 (28) 5  1  mn  50.  xy3 
 (x) 2 (y3) 2  2   a  x 2 
 28  mn  1  5  5 2    
 (a 2 ) 5  1  
 a   5  5   
 a2  x5 
 3.5 67. 8  1  2  1  nm  3  5   724 
 15a 2  12  9   
 n2m2  
 nm4mn   377  Chapter 11  69d. If the exponent is less than 0, the power is greater than 0 and less than 1. If the exponent is greater than 0 and less than 1, the power is greater than 1 and less than the base. If the exponent is greater than 1, the power is greater than the base. Any number to the zero power is 1. Thus, if the exponent is less than zero, the power is less than 1. A power of a positive number is never negative, so the power is greater than 0. Any number to the zero power is 1 and to the first power is itself. Thus, if the exponent is greater than zero and less than 1, the power is between 1 and the base. Any number to the first power is itself. Thus, if the exponent is greater than 1, the power is greater than the base.  n factors                     m factors              m factors        m factors  74b. (am)n 
 a  a  ...  a  a  a  ...  a  ...  a  a  ...  a 
 m  n factors        a  a  ...  a 
 amn m factors              74c.  m factors        m factors  (ab)m  
 ab  ab  ...  ab 
 a  a  ...  a  b  b  ...  b 
 ambm m factors        a m  b  
   74e.  am  an  
  a  am  a   b  ...  b 
 bm        74d.  a  b  a  a  ...  a  m factors         mn 
 n factors a a  a  ...  a  y  75.  1   70. r 
 (1.2  1015)A 3 If r 
 2.75  1015 then 1  2.75  105 
 (1.2  1015)A 3 2.75  1015  1.2  1015  
A  1   2.29 A 3 12.04 A Since 12.04 12, which is the mass number of carbon, the atom is carbon. 71. 32(x 4x) 
 16(x 4x3) (25)(x 4x) 
 (24)(x 4x3) 2(5x 20x) 
 2(4x 16x12) 5x2  20x 
 4x2  16x  12 x2  4x  12 
 0 (x  6)(x  2) 
 0 x6
0 x2
0 x 
 6 x
2 2  76.  2  2  2  2  72a.  2  Wind Speed 5 10 15 20 25 30  G M e t2  42  
  Wind Chill 0.8 12.7 22.6 29.9 35.3 39.2  O 2  4  1   r
   
23   22 2   3  
 Arctan 3   
4 
 6   So, (23   2i) 
 4
cos 6  i sin 6. Use De Moivre's Theorem.  4
cos 6  i sin 6  1  5  1   1   1   
45 
cos 
5  6sin 
5  6 1  5  G 
 6.67  1011    1  5    
 4 cos 30  4 i sin 30 
 1.31  0.14i 78.  2 3   2   3  6  5 6   1 2 3 4  7 6  mn factors                    n factors  2   v 
 Arctan  23  
 16   r  42,250,474.31 m 73b. 42,250,474.31 m 
 42,250.47431 km 42.250.47431  6380 
 35870.47431 35,870 km  3 2  Lemniscate  378  0  11 6 4 3  74a. aman 
 a  a  ...  a  a  a  ...  a 
 a  a  ...  a 
 amn  Chapter 11  x  6  77. (23   2i) 5 Convert to polar form r(cos v  i sin v).  (6.67  1011)(5.98  1024)(86,400)2  42  m factors  y  2 4 6 8  Mt 
 5.98  1024 t 
 1 day 
 86,4000 seconds r3  y2 
 12x (y  0)2 
 4(3)(x  0) Vertex is at (0, 0) and p 
 3. The parabola opens to the right so the focus is at (0  3, 0) or (3, 0). Since the directrix is 3 units to the left of the vertex, the equation of the directrix is x 
 3. 8 6 4 2  72b. A 5-mile per hour increase in the wind speed when the wind is light has more of an effect on perceived temperature than a 5-mile per hour increase in the wind speed when the wind is heavy. 73a. r3 
  x  O  1  3  5 3  u sin v  1qt2  h. 79. Use the equation y 
 tv 2 u v  
 105 g 
 32 h
3 v 
 42  87. The time it takes to paint a building is inversely proportional to the number of painters. k 48 
 8  1  y 
 t(105)(sin 42°)  2(32)t2  3  k 
 384  
 16t2  (105 sin 42°)t  3 Find t when y 
 0 (i.e., the ball is on the ground). t
  384  So t 
 1 6 t 
 24 The correct choice is E.  105 sin 42° (105 in s) 42°16)(  4(3)   2(16)  t 
 0.04, 4.43 So, the ball hits the ground after about 4.43 s. u 80. TC 
 (2  3), (6  (4)), (5  6) 
 1, 10, 11 u TC  
  (2  3 )2  (6  ( 4))2  (5  6)2 
 222  81. Sample answer: 1 tan S cos S 
 2 sin S  cos S  cos S  11-2 Page 705  Exponential Functions Graphing Calculator Exploration  1. positive reals 2. (0, 1) 3. For a 
 0.5 and 0.75, y →  as x → , y → 0 as x → . For a 
 2 and 5, y → 0 as x → , y →  as x → . 4. horizontal asymptote at y 
 0, no vertical asymptotes 5. Yes; the range of an exponential function is all positive reals because the value of any positive real number raised to any power is positive. 6. Any nonzero number raised to the zero power is 1. 7. The graph of y 
 bx is decreasing when 0 b 1 because multiplying by number between 0 and 1 results in a product less than the original number. The graph of y 
 bx is increasing when b  1 because multiplying by a number greater than 1 results in a product greater than the original number. 8. There is a horizontal asymptote at y 
 0 because a power of a positive real number is never 0 or less. As a number between 0 and 1 is raised to greater and greater powers, its value approaches 0. As a number greater than 1 is raised to powers approaching negative infinity, its value approaches 0.  1   1 
 2 1  sin S 
 2 82. cot v 
 0  v 
 2 Period of cot v is  so cot v 
 0  v 
 2  n, where n is an integer. 83. r  6h 
 150 m 150 m   r
 6h  r 
 25 m/h 84. 90°, 270° 85. x3  25x 
 0 Degree of 3 so there are 3 complex roots. x3  25x 
 0 x(x  5)(x  5) 
 0 x
0 x5
0 x5
0 x
5 x
5 3; 5, 0, 5 86.  Page 708  Check for Understanding  1. Power; in a power function the variable is the base, in an exponential function the variable is the exponent. 2. Both graphs are one-to-one, have the domain of all reals, a range of positive reals, a horizontal asymptote of y 
 0, a y-intercept of (0, 1), and no vertical asymptote. The graph of y 
 bx is decreasing when 0 b 1 and increasing when b  1. 3. If the base is greater than 1, the equation represents exponential growth. If base is between 0 and 1, the equation represents exponential decay. 4. The graphs of y 
 4x and y 
 4x  3 are the same except the graph of y 
 4x  3 is shifted down three units from the graph of y 
 4x.  [5, 5] scl:0.5 by [5, 5] scl:0.5  abs. min; (0.75, 1.88)  379  Chapter 11  5.  x  y  1  1  9  0 1 2  1 3 9  y  11. y 
 3x  x 2 1 0 1  y  y 9 3 1  12.  y 
 3x  1  9  x  y  2 1  3 34 1 32  0 1 2  4 2 0  Use (0, 0) as a test point. ?  ?   20  4  0 14 0  3  The statement is true so shade the region containing (0, 0). 14.  y  x  O y  15.  16.   0.45% 9b. Use N 
 N0(1  r)t. N  8,863,052(1  0.0045)20 N 9,695,766  x  y  1  1  2  0 1 2  1 2 4  y 
 2x  y  x y 2 4 1 2 0 1 1 1 2  y 
 2x  x 3 2 1 0  y 
 2x 3  x y 3 1 2 2 1 4 0 8  O  x  y O  x  y 
 2x 3  x  y  1  3 14  0 1 2  1 2 14  y  (0, 0) 0  ?  ?   42  2  0 1  2 0  1; no  O  x 1 0 1  y  y 5 1 y 
 ( 15 ) x  1  5  Exercises O  y 17.  O  y 
 2x  x  x 2 1 0 1  y 4 2 1 1  2  380  x  4x  2  x  y  (0, 0)  
   ? 1 0 0 2 ? 0 1   y  ( 12 ) x  O  Chapter 11  x y  y 1 2 4 8  y  282,167  
 40,309.57 7 40,309.57   0.0045 8,863,052  10.  1 2 4  2x  4  8. Use N 
 N0 (1  r)t where N0 
 3750, r 
 0.25, and t 
 2. N 
 3750 (1  0.25)2 
 3750 (0.75)2  2109.38 9a. 9,145,219  8,863,052 
 282,167  Pages 708–711  0 1 2  x  O  O 13.  0  1  y  x  O 7.  y 1 2  x  O 6.  x  x  18.  y  (0, 0)  3  1  2  ?  204  4 5 6  1 2 4  ?  24  0 0 0  1  16  23b. The graph of y 
 3x is a reflection of the graph of y 
 3x across the x-axis.  y  x  y  2x  4    x  O 19.  x 1 0 1  y  y 100 1  B  (0, 1)  1 (1, 100 )  x  O 20.  x 1 0 1  y 5 1  y (1, 5)  y 
 5x  1  5  C 21.  x 1 0 1 A  (0, 1)  y 49 7 1  [10, 10] scl:1 by [10, 10] scl:1 23c. The graph of y 
 7x is a reflection of the graph of y 
 7x across the y-axis.  y 
 0.01x  1  100  O  (1, 15 )  [10, 10] scl:1 by [1, 9] scl:1  x  1 x  23d. The graph of y 
 
2 is a reflection of the graph of y 
 2x across the y-axis.  y (0, 7)  y 
 71  x (1, 1)  O  x  22. [10, 10] scl:1 by [1, 9] scl:1 y 24a. 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2  [5, 5] scl:1 by [5, 5] scl:1 22a. The graph of y 
 5x is a reflection of y 
 5x across the x-axis. The graph of y 
 5x is a reflection of y 
 5x across the y-axis. 22b. The graph of y 
 5x  2 is shifted up two units, while the graph of y 
 5x  2 is shifted down two units. 22c. The graph of y 
 10x increases more quickly than the graph of y 
 5x. The graphs are not the same because 52x 10x. 23a. The graph of y 
 6x  4 is shifted up four units from the graph of y 
 6x.  O 2 4 6 8 1012 141618 2022x 24b. y 
 9.25(1.06)50 y 170.386427 thousand y $170,400 25a. y 
 (0.85)x  [10, 10] scl:1 by [1, 9] scl:1  381  Chapter 11  25b. y  28d. Sample answer: A borrower might choose the 30-year mortgage in order to have a lower monthly payment. A borrower might choose the 20-year mortgage in order to have a lower interest expense. 29a. P 
 4000, n 
 43, i 
 0.0475  1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1  O  (1  0.0475)43  1  Fn 
 4000   0.0475 535,215.918; $535,215.92 29b. P 
 4000, n 
 43, i 
 0.0525 (1  0.0525)43  1  Fn 
 4000   0.0525  1 2 3 4 5 6 7 8x  611,592.1194 or $611,592.12 $611,592.12  535,215.92 
 $76,376.20 30. The function y 
 ax is undefined when a 0 and the exponent x is a fraction with an even denominator. 31a. Compounded once: I 
 1000[(1  0.05)1  1] 
 50; $50 Compounded twice:  25c. y 
 (0.85)12 0.14 or 14% 25d. No; the graph has an asymptote at y 
 0, so the percent of impurities, y, will never reach 0. 26a. N 
 876,156(1  0.0074)15 978,612.2261 or 978,612 26b. N 
 2,465,326(1  0.0053)15 2,668,760.458 or 2,668,760 26c. 152,307 
 139,510(1  r)10 152,307  139,510  
  0.05 2  I 
 1000[
1  2  1]  
 (1  r)10  152,307  139,510    1  10  50.625; $50.63 Compounded four times:  
1r  0.05 4  I 
 1000[
1  4  1]  N 
 139,510(1  r)25  50.9453; $50.94 Compounded twelve times:  N 
 173,736.7334 or 173,737 26d. 191,701 
 168,767(1  r)10 191,701  168,767  
 (1  r)10  
 168,767  191,701  0.05 12  I 
 1000
1  12  1  10  51.1619; $51.16 Compounded 365 times:  
1r  0.05 365   I 
 1000
1   365   N 
 168,767(1  r)25 N 
 232,075.6889 or 232,076 35   5280 ft  1 hr     27b. s 
  1 hr  1 mi  3600 s  
 0.517x The return is 5.17% 0.05 365  Super Saver: I 
 x
1    1 365   
 6.16 ft/s  36.16   O 
 100(3 5 )  5800.16 5800 units 28a. Pn 121,000, n 
 30  12 or 360, i 
 0.075 12 or 0.00625  
 0.513x The return is 5.13% Money Market Savings 31c.  1  (1  0.00625)360  0.00625  121,000 
 P   P 846.04955; $846.05 28b. Pn 
 121,000, n 
 20  12 or 240, i 
 0.0725 12 or 0.00604 121,000 
 P    0.0725 240  1  
1  1 2  ———  
  0.0725  12    365   (1  0.05) 
 
1   365  x  1  365  (1.05) 1  365  x   
1 365  365[(1.05)  1] 
 x 0.04879 
 x; 4.88% 32. 4x2(4x)2 
 4x2(4)2(x)2 4x2   
 16x2    1  
 4 33. y 
 15 Use y 
 r sin v. y 
 15 So, 15 
 r sin v. 34. 3, 9  2, 1 
 (3)(2)  (9)(1) 
3 3; no because the inner product does not equal 0.  P 956.35494; $956.35 28c. 30 year: I 
 360(846.05)  121,000 
 $183.578 20 year: I 
 240(956.35)  121,000 
 $108,524 Chapter 11   1  51.2675; $51.26 31b. Let x represent the investment. Statement savings: I 
 x[(1  0.051)1  1] 
 0.051x The return is 5.1% 0.0505 12 Money Market Savings: I 
 x
1  1  1 2   27a. O 
 100(3 5 ) 
 100(33) 
 2700 units 4.2 mi   1  382  35.  
cos 78  i sin 78  33cos 
4  i sin 
4 1 7  7  
 
3  33 cos 
8  4  i sin 
8  4 5 5 
 3 
cos 8  i sin 8  1  3  y  y 
 1x  y 
 x 1 3  O  
 0.66  1.60i 36. s 
 72t  16t2  4 s  4 
 16t2  72t s  4  (16)(5.0625) 
 16(t2  4.5  5.0625) (s  85) 
 16(t  2.25)2 Vertex: (2.25, 85) Maximum height: 85 feet. 37a.  y  42.  x  The parent graph is translated 3 units left. The vertical asymptote is now x 
 3. The horizontal asymptote, y 
 0, is unchanged. CAB 
 16 43. CAC 
 32 100.53 50.27 100.53  50.27 
 50.26 or about 50. The correct choice is E.  (12, 8)  (7, 6)  11-3 O  x  Page 714 37b. 38.  39.  (12  (7)) (8  6) ,  2 2  
   
 (2.5, 7)  2    1 rev 
 4800 4800  V 
 9.2 1 138,732.73 or about 139,000 cm/s 40.  21˚ 200 ft  x  tan 69° 
  Check for Understanding  1. C 2. If k is positive, the equation models growth. If k is negative, the equation models decay. 3. Amount in an account with a beginning balance of $3000 and interest compounded continuously at an annual rate of 5.5%. 4. reals, positive reals 5. Sample answer: Continuously compounded interest is a continuous function, but interest compounded monthly is a discrete function. 6a. growth 6b. 33,430 6c. y 
 33,430e0.0397(60) 361,931.0414 or 361,931 7. A 
 12,000e0.064(12) 25,865.412 or $25,865.41  sin4 A  cos2 A 
 cos4 A  sin2 A (sin2 A)2  cos2 A 
 cos4 A  sin2 A (1  cos2 A)2  cos2 A 
 cos4 A  sin2 A 2 (1  2cos A  cos4 A)  cos2 A 
 cos4 A  sin2 A cos4 A  1  cos2 A 
 cos4 A  sin2 A cos4 A  sin2 A 
 cos4 A  sin2 A 2900 rev  1  The Number e  x  200  200 tan 69° 
 x 521.02 
 x about 521 feet  Pages 714–717  Exercises  8. p 
 (100  18)e06(2)  18 
 42.6 or 43% 9a. y 
 84e0.23(15)  76 
 78.66 or 78.7°F 9b. too cold; After 5 minutes, his coffee will be about 90°F.  41.  10a.  [4, 4] scl:1 by [1, 10] scl:1 10b. symmetric about y-axis Sample answer: y 
 948.4x  4960.6  383  Chapter 11  11a. Annually: I 
 100[(1  0.08)1  1] 
 80 $80.00; 8%  14b. 2 decimal places; 4 decimal places; 6 decimal places 14c. always greater 15a. 5 days: P 
 1  e0.047(5)  0.20943 20.9% 20 days: P 
 1  e0.047(26)  0.60937 60.9% 90 days: P 
 1  e0.047(90)  0.98545 98.5% 20.9%; 60.9% 98.5%  0.08 2  Semi-annually: I 
 1000
1  2  1 
 81.6  $81.60; 8.16%  Quarterly: I 
 1000
1   82.4316  Daily: I 
 1000
1      1 $82.43; 8.243%  Monthly: I 
 1000
1   82.9995  0.08 4  4  0.08 12  12     1 $83.00; 8.3%  0.08 365  365     1  15b.   83.2776 $83.28; 8.328% 0.08(1) Continuously: I 
 1000(e  1)  83.2871 $83.98; 8.329% Interest Compounded Annually Semi-annually Quarterly Monthly Daily Continuously  Effective Annual Yield 8% 8.16% 8.243% 8.3% 8.328% 8.329%  Interest $80.00 $81.60 $82.43 $83.00 $83.28 $83.29  about 29 days 15c. Sample answer: The probability that a person who is going to respond has responded approaches 100% as t approaches infinity. New ads may be introduced after a high percentage of those who will respond have responded. The graph appears to level off after about 50 days. So, new ads can be introduced after an ad has run about 50 days. 16a. all reals 16b. 0 f(x) 1 16c. c shifts the graph to the right or left  r n  11b. continuously 11c. E 
 
1  n  1 r 11d. E 
 e  1 12a. y 
 525(1  e0.038(24))  314.097 314 people 12b.  (1  0.035)8  1   17. 120,000 
 P  0.035  after about 61h  120,000  P(9.051687) P  13,257.19725 $13,257.20 8   3   1   18. x 5 y 5 z 5 
 x x3y3z 5  19. y 
 6x2 v 
 45° x sin 45°  y cos 45° 
 6(x cos 45°  y sin 45°)2  [0, 100] scl:10 by [0, 550] scl:50 13a. P 
 1  e6(0.5)  0.95021 95% 13b. 0.02h  1    60 min  1h  
 1.2 6x2   12xy   6y2  2 2  y 2  1 x2 2  1 y2 2   2   2 y 2  
 6
  2 x   2  2 x    2  y 2  
 6
   2 x  2  y 2  
 3x2  6xy  3y2  2   x 
 0.02; about 0.02 h   2   2 x   xy   2 x  2 y 
 6x2  12xy  6y2  2 x  2 y 
 0 1   (5)2   (1 )2 v 
 Arctan  20. r 
  5   
 26   3.34 26 (cos 3.34  i sin 3.34)  [0, 1] scl:0.1 by [0, 1] scl:0.1 2(10)  1 10  2(10)  1  14a. For x 
 10: 
    about 1.2 min 21 10  19  

  2(100)  1 100   For x 
 100: 
 2(100)  1   201 100   
 
 199    2.718304481  2(1000)  1 1000  2(1000)  1  For x 
 1000: 
     2.720551414    2001 1000   
 
 1999    2.718282055 2.720551414; 2.718304481; 2.718282055 Chapter 11  384      y  21.  Page 717  Mid-Chapter Quiz  1. 642 
 8 1  3 2. () 343 2 
 (343 31 )2  
 ((73) 3 )2 2 
7 1 
 4 9  (0, 150)  (x, y)  3.  d
  8x3y6  1  3   
 27w z  6 9  8x3z9 1   y 28˚  O  
  x  x  u u d 
 x, y F 
 0, 150 x   cos 28° 
 10  2xz3  sin 28° 
  4.  x  10    x 
 10 sin 28° x 4.6947  28.  1   3   
 a3b 2 1   5. (125a2b3) 3 
  125a2 b3 3 
  53a2b3 3 
 5b a2 2 6. 1.75  10 
 0.094  A3  3  2 2 1.75  10 
 0.94 A 3  1.75  102  0.94 2  2  3  3   
 A2  3 2    
 (A 2 ) 3  151.34  A 1.51  102 mm2 7. 1,786,691 
 1,637,859 (1  r)8 1,786,691  1,637,859  3x  2  6 3x  8 8 x  3  
 (1  r)8  1  8  1,786,691 8  
 1,637,859  
 [(1  r) ] 1,786,691  
 1,637,859   1 
 r  1  8  1  8    4  3        1   1   r  Store the exact value in your calculator's memory. N 
 1,637,859 (1  0.011)24 Use the stored value for r. 
 2,216,156.979 2,126,157  3 2 2 3 9 6 6 9 25. 3 
 2 6 5 1 6 18 15 3 J
(9, 6), K
(6, 18), L
(6, 15), M
(9, 3); The dilated image has sides that are 3 times the length of the original figure. 4x  y 6 26. 
 x 2y  12 4x  y 
 6 4(2y  12)  y 
 6 x 
 2(6)  12 x 
 2y  12 9y 
 54 x 
 0 y
6 (0, 6) 27. {4, 2, 5}; {5, 7}; yes    2   a6b3 
(a4b3) 2  1.75  10  
 0.94  24. 3x  2  6 3x  2  6 3x  4 4 x  3 x  1   
 (a6) 2 (b3) 2   23. 2x 3
4 2x  3 
 16 2x 
 13 13 x 
 2  x  1   (23) 3 (x3) 3 (z9) 3  1 1 1    (33) 3 (w6) 3 ( y6) 3   or w2xy2z3 
 3 3w2y2  x 
 10 cos 28° x 8.8295 d 
 8.8295, 4.6947 u u W
F d W 
 0, 150  8.8295, 4.6947 
 0  704.205 
 704.2 ft-lb 22. y 
 1.5(10)2  13.3(14)  19.4 
 2.4  8 3  1    3 
 
 27w6y6       0.011  0.052 (4)(3.5)  8. A 
 3500 
1  4 
 3500(1.013)14 
 4193.728 $4193.73 48.1  9a. y 
 6.7e 15  0.271292 271,292 ft3  The correct choice is D.  48.1   9b. y 
 6.7e 50  2.560257 2,560,257 ft3  200   10. 2 years: n 
  1  20e0.35(2)   18.3 200   15 years: n 
  1  20e0.35(15)   181  4  200   60 years: n 
  1  20e0.35(60)   200 18.3; 181; 200  385  Chapter 11  2  14. log7 n 
 3 log7 8  Logarithmic Functions  11-4  2   log7 n 
 log7 8 3 2   Pages 722–723  n 
 83 n
4 15. log6 (4x  4) 
 log6 64 4x  4 
 64 x 
 15  Check for Understanding  y 
 3x and log3 x are similar in that they are y 
 3x both continuous, one-toone, increasing and inverses. x O y 
 3x and log3 x are not y 
 log3 x similar in that they are inverses. The domain of one is the range of another and the range of one is the domain of the other. y 
 3x has a y-intercept and a horizontal asymptote whereas y 
 log3 x has a x-intercept and a vertical asymptote. 2. Let bx 
 m, then logb m 
 x. (bx)p 
 mp bxp 
 mp logb bxp 
 logb mp xp 
 logb mp p logb m 
 logb mp 1.  y  1  16. 2 log6 4  4 log6 16 
 log6 x 1   log6 42  log6 16 2 
 log6 x 42  1 log6  
 log6 x 16 4 42 1 
 x  16 4  x
8  17. x y 1 0 2 1 4 2  y y 
 log 1 x 2  x  O  18. x 1 6  y 0 1  y  y  3.  y 
 log 1 x 5  O  y  log6 x  x  O y 
 log5 x  t   19. 16 
  3.3 log 1024 4  t 
 16(33 log4 1024) t 
 16(33  5) t 
 264 h  Log5 x is an increasing function and log1 x is a 5 decreasing function. 4. Sean is correct. The product property states that logb mn 
 logb m  logb n.  log4 1024 
 x 4x 
 1024 22x 
 210 2x 
 10 x
5  1  5. In half-life applications r 
 2. So, (1  r) becomes 
 N 
 N0(1    
   1 1 1  2 or 2 . Thus, the formula 1 t  r)t becomes N 
 N0 2 .  3   
   
215   1  2  7.  8. log7 y 
 6  9. log8 4 
 3  1  1  2x 
 1 6 2x 
 24 x 
 4 1   12. log7  343 
 x  7x  
  1  343 73  
 x 
 3  7x  Chapter 11  1  3  
5  1  2  21. 16 2 
 4  13. log2 x 
 5 25 
 x 32 
 x  386  5    22. 74 
  2401  23. 4 2 
 32  24. ex 
 65.98  25. 
6  
 36  1  11. log10 0.01 
 x 10x 
 0.01 10x 
 102 x 
 2  Exercises 1   20. 27 
 3 1  6. 9 2 
 27  10. log2 16 
 x  Pages 723–725  4  3  26. log81 9 
 2  27. log36 216 
 2  28. log 18 512 
 3  29. log6 3 6 
 2  1  x  30. log16 1 
 0 32. log8 64 
 x  31. logx 14.36 
 1.238  8x 
 64 8x 
 82 x
2 33. log125 5 
 x  x 
 2 3 47. log1010 
x 3 10x 
 10  10x 
 10  125x 
 5 (53)x 
 51 3x 
 1  1  3  1  x 
 3  1  1  x 
 3  
 32 2x 
 25 x
5 35. log4 128 
 x  34. log2 32 
 x 4x 
 128 22x 
 27 2x 
 7 7 x 
 2 or 3.5  9x 
 96 x
6 37. log49 343 
 x  36. log9 96 
 x 49x 
 343 72x 
 73 2x 
 3  2x  1  1   1   x  2  8 
 84 x  2  
4  x
8 40. 104 log102 
 x 41. logx 49 
 2 x2 
 49 x
7  39. log8 4096 
 x 10log1024 
 x 24 
 x 16 
 x  1   log7 (8x) 2 
 log7 16 1   (8x) 2 
 16 8x 
 256 x 
 32 52. 2 log5 (x  2) 
 log5 36 log5 (x  2)2 
 log5 36 (x  2)2 
 36 x2
6 x
8 53. x y 1 0  42. log3 3x 
 log3 36 3x 
 36 x 
 12 43. log6 x  log6 9 
 log6 54 log6 9x 
 log6 54 9x 
 54 x
6 44. log8 48  log8 w 
 log8 6 48  log8 w 
 log8 6 48  w  
6  6w 
 48 w
8 45. log6 216 
 x 6x 
 216 6x 
 63  1   x 
 9 2  27 3 x
33 x
9 49. log5 (x  4)  log5 8 
 log5 64 log5 (x  4)(8) 
 log5 64 (x  4)(8) 
 64 x4
8 x
4 50. log4 (x  3)  log4 (x  3) 
 2 log4 (x  3)(x  3) 
 2 42 
 (x  3)(x  3) 16 
 x2  9 25 
 x2 5
x 1 51. 2(log7 x  log7 8) 
 log7 16 1 (log 8x) 
 log 16 7 7 2  38. log8 16 
 x x 
8  
 4096  4  1   log12 x 
 log12 9 2  27 3  3  x 
 3  1   log12 x 
 log12 9 2  log12 27 3  x 
 2 or 1.5 8x 
 16 23x 
 24 3x 
 4  1  48. log12 x 
 2 log12 9  3 log12 27  2  1  2  4  1  y y 
 log4 x  O  x  x
3 46. log5 0.04 
 x 5x 
 0.04 5x 
 52  387  Chapter 11  54. x 1 2 4  y 0 3 6  20 15 10 5  O 5 10 15 20 55. x 2 6  y  r 410  62a. 5000  2500
1  4 r 40  27  
1  4  y 
 3 log2 x 5 10 1520 2530 3540 x  r 40  1   2 40 
 
1  4    1  40  r  1.0175  1  4 0.0699  r 6.99%  y  y 0 1  r 40  2 
 
1  4  62b.  r 28  2 
 
1  4  62c.  y 
 log5 (x  1)  1   r  2 28 
 1  4 r  x  1.0251  1  4 0.1004  r 10.04%  56. x 1 2 4  y  log2 x  1  1 —  63a. n 
 log2  y  y 0 1 2  63b. 3 
 log2 p  1  4  1  23 
 p  n 
 log2 4  8 
 p1  2n 
 4 n
2  x  1  8  
p 1  less light; 8  57. x 1 2 4  64. Let y 
 loga x, so x 
 ay. x 
 ay logb x 
 logb ay logb x 
 y logb a  y  y 0 2 4  logb x   y
 log a b  loga x 
  x y  2 log2 x  58. x 0 9  logb x  logb a  P  65a.  16 15.5 P 15 log2.72 (14.7) 
 0.02h 14.5 14 13.5 13 12.5  y  y 0 1  y  log10 (x  1)  x  2 O  4  2  4h  P   65b. log2.72  14.7 
 0.02(1) P  14.7  59. Use N 
 N0(1  r)t; r 
 1 since the rate of growth is 100% every t time periods. 64,000 
 1000 (1  1)t 64 
 2t log2 26 
 log2 2t 6
t t  15 
 90 min 60. All powers of 1 are 1, so the inverse of y 
 1x is not a function. 61. Let logb m 
 x and logb n 
 y. So, bx 
 m and by 
 n. m bx xy  
  n by 
 b m  n m  logb n m logb n  Chapter 11  
 2.720.02  P 
 14.7 (2720.02) 14.4 psi P   65c. log2.72  14.7 
 0.02(6.8)  P 
 14.7(2.720.136) 16.84 psi 1 t  6.8 
 38
2  66.  log log  
 bxy  6.8  38 6.8  38 6.8  38  1 t  
 
2  1  
 t log 2  t
  
xy  1 t  
 log 
2 6.8 log  38 __ 1 log  2  t 
 2.5 2.5  3.82 
 9.55 about 9 days  
 logb m  logb n  388  75. cos (A  B) 
 cos A cos B  sin A sin B 5 35 cos A 
 1 cos B 
 37 3  67. 69.6164 68. 90,000 
 P    
    0.115 12.30 1  1   12 ______ 0.115  12    x2  52 
 132 x2 
 144 x 
 12 12 So, sin A 
 1 3  P  891.262 $891.26 69. ellipse 9x2  18x  4y2  16y  11 
 0 9x2  18x  4y2  16y 
 11 2 9(x  2x  1)  4(y2  4y  4) 
 11  9  16 9(x  1)2  4(y  2)2 
 36 (x   4 1)2  (y   x2  352 
 372 x2 
 144 x 
 12 12 So, sin B 
 3 7 5  35  12  12     cos (A  B) 
 1 3  37  13  37 175  144    
 481  481 31   
 481  2)2  76. y 
 A sin (kt  c)  h   9 
 1  90  64  90  64  h 
 2  A 
 2  y  
 13  2  k  
4   
 77  k 
 2   y 
 13 sin 
2t  c  77   64 
 13 sin 
2(1)  c  77  x  O    13 
 13 sin 
2  c   1 
 sin 
2  c   sin1(1)  2 
 c  70. r 
 3, v 
 2 (3 cos 2t, 3 sin 2t)  3.14  1   (1))2   (3   3)2 71. AB 
 ( 2 2 
  0  ( 6) 
6 BC 
 ( 3  (( 1))2   0  ( 3))2 2 2 
  4  3 
5 AC 
 ( 3  ( 1))2   (0  3)2 2 2 
  4  3 
5 3  3  So, y 
 13 sin 
  2  3  2  3  2  2    
 5  2 cos 
4  3  i sin 
4  3 17 17  cos 1 2  i sin 12 17 17  cos 1 2  10i sin 12  
 10
 
 10     2.59  9.66i 17   3.14  77  77. c2 
 (6.11)2  (5.84)2  2(6.11)(5.84) cos 105.3 c2 
 37.3321  34.1056  71.3648 cos 105.3 c2 90.2689 c 9.5 (6.11)2 (5.84)2  (9.5)2  2(5.84)(9.5) cos A 37.3321 34.1056  90.25  110.96 cos A cos A 0.7843 A 38.34 or 38° 20
 B 180  (105° 18
  38° 20
) 36° 22
 c 
 9.5, A 
 38° 20
, B 
 36° 22
 78. M  72. 5
cos 4  i sin 4  2
cos 3  i sin 3  
  c  k 2  P  c˚ x˚  Q b˚  17   10
cos 1 2  i sin 12 , 2.59  9.66i  73. (3  4j)(12  7j) 
 36  21j  48j  28j2 
 64  27j 64  27j volts y 74.  S  N mSMN 
 mQNM alternate interior angles x° 
 b°  c° Exterior Angle Theorem The correct choice is E.  C  A  c˚  50˚ 50˚  D  O  x  B  Both vectors have the same direction. 50° south of u u east. Therefore, AB and CD are parallel.  389  Chapter 11  Common Logarithms  11-5 Page 730  x1  4    x1  (x  3) log 3  log 2  4 log 4  Check for Understanding  (4x  12)log 3  4 log 2  (x  1) log 4 4x log 3  12 log 3  4 log 2  x log 4  log 4 4x log 3  x log 4  4 log 2  log 4  12 log 3 x(4 log 3  log 4)  4 log 2  log 4  12 log 3 x  4 log 2  log 4  12 log 3  4 log 3  log 4  x  4.84 17.  [10, 10] scl:1 by [20, 100] scl:10 5.5850 200  18a. R 
 log 
 1.6   4.2 
 6.3 18b. 10 times; According to the definition of logarithms, R in the equation R 
 log 
T  B is an exponent of the base of the logarithm, 10. 105 is ten times greater than 104. a  y  10.  Pages 730–732  y  log (x  3)  log 18   11. log12 18 
  log 12   1.1632 log 15   12. log8 15 
  log 8   1.3023 2.2x  5 
 9.32 (x  5) log 2.2 
 log 9.32 (x  5) 
  22. log 0.06 
 log 
0.01  2 12  log 9.32  log 2.2  
 log 0.01  log  x  7.83 x2 14. 6 
 4x (x  2) log 6 
 x log 4 x log 6  2 log 6 
 x log 4 2 log 6 
 x log 4  x log 6 2 log 6 
 x (log 4  log 6) 2 log 6  log 4  log 6  4.3x x log 4.3  Chapter 11  1  1  
 2  1.0792  2(0.6021) 
 1.2218 23. log 36 
 log (4  9) 
 log 4  log 9 
 0.6021  0.9542 
 1.5563 24. log 108,000 
 log (1000  12  9) 
 log 1000  log 12  log 9 
 3  1.0792  0.9542 
 5.0334  8.84 x 76.2 log 76.2 log 76.2  log 4.3  x  2.97  12  1  42  
 log 0.01  log 12  2 log 4  
x  x  Exercises  19. log 4000,000 
 log (100,000  4) 
 log 100,000  log 4 
 5  0.6021 
 5.6021 20. log 0.00009 
 log (0.00001  9) 
 log 0.00001  log 9 
 5  0.9542 
 4.0458 21. log 1.2 
 log (0.1  12) 
 log 0.1  log 12 
 1  1.0792 
 0.0792  x  15.  4  3x3  2
4  1. log 1 
 0 means log10 1 
 0. So, 10° 
 1. log 10 
 1 means log10 10 
 1. So, 101 
 10. 2. Write the number in scientific notation. The exponent of the power of 10 is the characteristic. 3. antilog 2.835 
 102.835 
 683.9116 4. log 15 
 1.1761 log 5 
 0.6990 log 3 
 0.4771 log 5  log 3 
 0.6990  0.4771 
 1.1761 5. log 80,000 
 log (10,000  8) 
 log 104  log 8 
 4  0.9031 
 4.9031 6. log 0.003 
 log (0.001  3) 
 log 103  log 3 
 3  0.4771 
 2.5229 7. log 0.0081 
 log (0.0001  34) 
 log 104  4 log 3 
 4  4(0.4771) 
 2.0915 8. 2.6274 9. 74,816.95  13.  3x3  2  4x1  16.  390  25. log 0.0048 
 log (0.0001  12  4) 
 log 0.0001  log 12  log 4 
 4  1.0792  0.6021 
  2.3188 26. log 4.096 
 log (0.001  46) 
 log 0.001  6 log 4 
 3  6(0.6021) 
 0.6124 1  27. log 1800 
 log 
100  9  4 2   46.  log 3  7 log 2   x log 3  log 2  x 
 9.2571 47. logx 6  1 log 6  log x  1  
 log 100  log 9  2 log 4 
 3.2553  log 6  29. 2.9515 31. 2.001 33. 2.1745 log 8   undefined. When x 1,  log x is negative, which is not greater than 1. So, x must also be greater than 1. Therefore, 1 x 6. log 625   34. log2 8 
  log 2   35. log5 625 
  log 5  
3  48.  
4 log 24  log 4   36. log6 24 
  log 6   37. log7 4 
  log 7  1.7737 38. log6.5 0.0675 
  0.7124  1  2  5 log 4  3 log 3   x 2 log 4  log 3  log 15  1 log  2  40.  2x 
 95 x log 2 
 log 95  49.  log 95   x
 log 2  41.  42x5  3x3 (2x  5) log 4  (x  3) log 3 2x log 4  5 log 4  x log 3  3 log 3 2x log 4  x log 3  5 log 4  3 log 3 x(2 log 4  log 3)  5 log 4  3 log 3  log 0.0675  log 0.5  3.8890 39. log 15 
  1  log 6  log x 6x log 6  When x 
 1, log 1 
 0, which means  log x is  1  
 2  0.9542  2(0.6021) 28. 1.9921 30. 0.871 32. 3.2769  3x1  2x7 (x  1) log 3  (x  7) log 2 x log 3  log 3  x log 2  7 log 2 x log 3  x log 2  log 3  7 log 2 x(log 3  log 2)  log 3  7 log 2  x 6.5699 5x 
 4x  3 x log 5 
 (x  3) log 4 x log 5 
 x log 4  3 log 4 x(log 5  log 4) 
 3 log 4  x  2.1719 0.52x4  0.15x (2x  4) log 0.5  (5  x) log 0.1 2x(log 0.5)  4 log 0.5  5 log 0.1  x log 0.1 2x log 0.5  x log 0.1  5 log 0.1  4 log 0.5 x(2 log 0.5  log 0.1)  5 log 0.1  4 log 0.5 x  Change inequality sign because (2 log 0.5  log 0.1) is negative.  3 log 4   x
 log 5  log 4  x 42.  1  3  50. log2 x 
 3 x 
 23 x 
 0.1250  18.6377  log x 
 log 8 1  3  x 
8 x 512 43.  5 log 0.1  4 log 0.5  2 log 0.5  log 0.1  x  3.8725 51. x  log 52.7  log 3  x  3.6087  52.  0.1643x 
 0.38x (43x) log 0.16 
 (8  x) log 0.3 4 log 0.16  3x log 0.16 
 8 log 0.3  x log 0.3 3x log 0.16  x log 0.3 
 8 log 0.3  4 log 0.16 x(3 log 0.16  log 0.3) 
 8 log 0.3  4 log 0.16 x
  8 log 0.3  4 log 0.16  3 log 0.16  log 0.3  x  0.3434  [10, 10] scl:1 by [3, 3] scl:1  44. 4 log (x  3) 
 9  53.  9  log (x  3) 
 4 9  (x  3) 
 antilog 4 9  x 
 antilog 4 3 x 174.8297 45. 0.25 
 log 16x 0.25 
 x log 16  [1, 10] scl:1 by [1, 3] scl:1  0.25   x
 log 16  x  0.2076  391  Chapter 11  54.  t 1 0.8  0.9535 
 
2  60b.  1  log 0.9535 
 0.8t log 2 log 0.9535  1 log  2    log  x    
 t log 0.8  12.0016 t 12 years 61. Sample answer: x is between 2 and 3 because 372 is between 100 and 1000, and log 100 
 2 and log 1000 
 3.  [10, 1] scl:1 by [2, 10] scl:1 55.  log 0.9535  1 log  2  
 0.8t  0.3210  1   62a. L 
 10 log  1.0  1012  
 10(log 1  log (1.0  1012)) 
 120 dB  62b.  [5, 5] scl:1 by [10, 50] scl:10 56.  x  I   20 
 log  1.0  1012  2 
 log I  log (1.0  1012) 2 
 log I  12 10 
 log I 1  1010 
 I; 1  1010 W/m2  0.1975  1 t  63. Use N 
 N0
2 .  N 
 630 micrograms 
 63  104 gram N0 
 1 milligram 
 1.0  103 gram 1 t  6.3  104 
 (1.0  103)
2  [5, 5] scl:1 by [3, 10] scl:1  6.3  104  1    log  1.0  103 
 t log 2  x
2  57.  0.6666 t 0.6666  5730 3819 yr 64. loga y 
 loga P  loga q  loga r p  loga y 
 loga q  loga r pr  loga y 
 loga q pr  y 
 q  [5, 5] scl:1 by [5, 10] scl:1 100  65. logx 243 
 5 x5 
 243 x
3  10.3   58a. h 
 9 log  14.7  1.7 mi 100  P   4.3 
 9 log  14.7  58b.  66.  0.3870 
 log P  log 14.7 0.3870  log 14.7 
 log P 0.7803 log P 6.03 P; 6 psi 59a. M 
 5.3  5  5 log 0.018 1.58 59b. 5.3 
 8.6  5  5 log P 8.3 
 5 log P 1.66 
 log P 0.0219 P  increasing from  to  1 2    9 1 0.8  60a. q 
 
2  1 0.1342  
 
2  
 0.9112 $91,116  Chapter 11  1   1   1   67. (a4b2) 3 c 3 
 (a4) 3 (b2) 3 (c2) 3 3 
 a  ab2c2 2 2 68. (5)  (0)  D(5)  E(0)  F 
 0 5D  F  25 
 0 2 2 (1)  (2)  D(1)  E(2)  F 
 0 D  2E  F  5 
 0 2 2 (4)  (3)  D(4)  E(3)  F 
 0 4D  3E  F  25 
 0  392  5D  0E  F  25 
 0 () D  2E  F  5 
 0 4D  2E  20 
 0 4D  3E  F  25 
 0 () D  2E  F  5 
 0 3D  E  20 
 0 4D  2E  20 
 0 2(3D  E  20 
 0) 10D  60 
 0 D 
 6 4(6)  2E  20 
 0 2E  4 
 0 E
2 5(6)  0(2)  F  25 
 0 F5
0 F
5 x2  y2  6x  2y  5 
 0 (x2  6x  9)  (y2  2y  1) 
 5  9  1 (x  3)2  (y  1)2 
 5 69. 70.  74. f(x) 
 x3  2x2  11x  12 f(1) 
 1  2(1)  11(1)  12 Test f(1). f(1) 
 0 (x  1) is a factor. 1 1 2 11 12 1 1 12 1 1 12  0 x2  x  12 
 0 (x  4)(x  3) 
 0 So, the factors are (x  4)(x  3)(x  1). y 75.  y 
 5x 3  2x  5  Neither; the graph of the function is not symmetric with respect to either the origin or the y-axis. 76. 7  5  4  1 
 17 17,000,000 The correct choice is D.  65   25  18  4 ,  
 (25 , 11) 
 2 2  r
6 r2 
 36 x2  y2 
 36  71.  90˚  120˚  60˚ 30˚  150˚ 180˚  11-6 Page 735  330˚  210˚ 270˚  Check for Understanding  1. ln e 
 1 is the same as loge e 
 1. And e1 
 e. So, ln e 
 1. 2. The two logarithms have different bases. log 17 ⇒ 10x 
 17 or x 
 1.23 ln 17 ⇒ ex 
 17 or x 
 2.83 3. ln 64 
 4.1589 ln 16 
 2.7726 ln 4 
 1.3863 ln 16  ln 4 
 2.7726  1.3863 
 4.1589 4. The two equations represent the same thing, A 
 Pert is a special case of the equation N 
 N0ekt and is used primarily for computations involving money. 5. 4.7217 6. 1.1394 7. 15.606 8. 0.4570  300˚  u 72. AB 
 (6  5), (5  6) 
 1, 1 u (6  5 )2  ( 5  6 )2 AB  
  
 2  1.414 73.    Natural Logarithms  0˚  2 4 6  240˚  x  O  3.65 cm  a  ln 132   9. log5 132 
  ln 5  b  v 
 36  3.0339 10. log3 64 
  10 
 360°  cos 36° 
  a  3.65  sin 36° 
  a 
 3.65 cos 36° 2.9529 1  Use A 
 2aP, where P A  b  3.65  11.  b 
 3.65 sin 36° 2.1454 10(2.1454)  3.7856 18 
 e3x ln 18 
 3x ln e ln 18  3  21.454.  ln 64  ln 3  
x  0.9635 
 x  1 (2.9529)(21.454) 2  31.6758 or 31.68 cm2  393  Chapter 11  12.  10 
 5e5k 2 
 e5k ln 2 
 5k ln e ln 2  5  ln 0.512   33. log8 0.512 
  ln 8  0.3219 34. log6 323 
  
k  3.2246  0.1386 k 13. 25ex 100 ex 4 x ln e ln 4 x 1.3863 14. 4.5  e0.031t ln 4.5  0.031t ln e ln 4.5  0.031  ln 303  1.6 ln 2  88   
 35. log5 288 ln 5  36.  1.7593 6x 
 72 x ln 6 
 ln 72  37.  2x 
 27 x ln 2 
 ln 27  ln 72  ln 27   x
 ln 2   x
 ln 6  t 38.  48.5186  t 15.  x  13.57  2.3869 9x4 
 7.13 (x  4) ln 9 
 ln 7.13 x ln 9  4 ln 9 
 ln 7.13 x ln 9 
 ln 7.13  4 ln 9  4.7549  ln 7.13  4 ln 9   x
 ln 9  39.  x 4.8940 3x 
 32  x ln 3 
 ln 3  ln 2  ln 3  ln 2    x
 ln 3  [20, 20] scl:2 by [4, 20] scl:2 x  26.90  16.  41.  x 1.3155 60.3 e0.1t ln 60.3 0.1t ln e ln 60.3  0.1  t  40.9933 42.  450  760 450  ln  760 450 ln  760  0.125  
  1  2  e0.125a    4.1926  a; 4.2 km  Pages 736–737 18. 20. 22. 24. 26. 28.  19. 21. 23. 25. 27. 29. ln 56   30. log12 56 
  ln 12  1.6199 32. log4 83 
  ln 25  0.075  0.2705 0.9657 2.2322 1.2134 0.9966 0.2417  
 2x ln e 
x  y  y  42.9183  ln 36   31. log5 36 
  ln 5  2.2266  ln 83  ln 4  3.1875  Chapter 11  
 e2x  0.3466 x 44. 25 e0.075y ln 25 0.075y ln e  Exercise  5.4931 6.8876 0 10.4395 0.0233 146.4963   1 
 e2x 1  2 1  ln 2 1 ln 2  2  
a  
t  0.7613 t 22 
 44 (1  e2x)  43.  
  
 0.125a ln e  
  t 6.2e0.64t 
 3et1 ln 6.2  0.64t ln e 
 ln 3  (t  1) ln e ln 6.2  ln 3  1 
 0.36t ln 6.2  ln 3  1  0.36  [15, 30] scl:5 by [50, 150] scl:10 17a. p 
 760e0.125(3.3) 
 760e0.4125 503.1 torrs 17b. 450 
 760e0.125a  394  40. 25ex 
 1000 ex 
 40 x ln e 
 ln 40 x 3.6889  45.  5x  76  x ln 5  ln 7  ln 6   x  4.72  52.  ln 7  ln 6    x ln 5  46.  x  1.7657 12x4  4x (x  4) ln 12  x ln 4 x ln 12  4 ln 12  x ln 4 x ln 12  x ln 4  4 ln 12 x(ln 12  ln 4)  4 ln 12  [10, 10] scl:1 by [10, 75] scl:5  4 ln 12   x ln 12  ln 4 2   x  0.37  53.  x  9.0474  47. x 3  27.6 3  x  (27.6) 2 x  144.9985 48.  x  3.76 [5, 5] scl:1 by [2, 5] scl: 0.5 0.6 
 1e  54.  ln 0.6 
  t  20,000(4  1011)  ln e  20,000(4  1011) ln 0.6 
 t 4.09  107 t 7 4.09  10 s  [5, 5] scl:1 by [50, 700] scl:50 49.  t  20,000(41011)  x  7.64  1 t  2.8 
 9
2  55.  2.8  1  ln 9 
 t ln 2 2.8 ln 9 __ 1 ln 2  1.6845 t 1.6845  8  24 
 323.4236 324 h 56a. ln 180  72 
 k(0)  c 4.6821 c 56b. ln 150  72 
 k(2)  4.6821  [70, 10] scl:1 by [3, 10] scl:1 50.  t  
t  133.14  ln 78  4.6821  2  
k  0.1627 k 56c. ln 100  72 
 (0.1622)t  4.6821 ln 28  4.6821  0.1627  
t  8.3 t about 6.3 min 57. e2x  4ex  3 
 0 (ex  3)(ex  1) 
 0 ex  3 
 0 ex 
 3 x ln e 
 ln 3 x 1.0986 0 or 1.0986 58a. 2 
 e0.063t ln 2 
 0.063t ln e  [10, 150] scl:10 by [100, 2000] scl:100 51. x 2.14  [6, 6] scl:1 by [4, 24] scl:2  ln 2  0.063  8.3  2 
 6.3  ex  1 
 0 ex 
 1 x ln e 
 ln 1 x
0  
t  11.0023 t about 11 years 58b. See students' work.  395  Chapter 11  1800 
 5000 ln r  59.  1800  5000 1800  5000  68. 2x  5y  3 
 0  A2   B2 
  22  ( 5)2 
 29   
 ln r  2x  60a. ln  1  2 1  2  1 ln  2  1622  229   
 1ek(1622)  329   
 1622 k ln e 
k  1.7  
t  
  tan  
  2   
  29   529   329   3 29   29y  29 
 0; 29  0.56; 112°  Natural Logarithms and Area  Pages 738–739  y 0.9, 1.1 x2 0.9  4 x 4.9  x 2.2, 2.2  1. 0.69314718 2. 0.6931471806; It is the same value as found in Exercise 1 expressed to 10 decimal places. 3a. The result is the opposite of the result in Exercise 1. 3b. Sample answer: a negative value 4a. 0.69314718 4b. 1.0986123 4c. 1.3862944 4d. 0.6931471806, 1.098612289, 1.386294361 4e. The value for each area is the same as the value of each natural logarithm. 5. 0.5108256238; 0.6931471806; 0.9162907319; These values are equal to the value of ln 0.6, ln 0.5, and ln 0.4. 6. If k  1, then the area of the region is equal to ln k. If 0 k 1, then the opposite of the area is equal to ln k. 7. The value of a should be equal to or very close to 1, and the value of b should be very close to e. This prediction is confirmed when you display the actual regression equation. 8. Sample answer: Define ln k for k  0 to be 1 the area between the graph of y 
 x, the x-axis, and the vertical lines x 
 1 and x 
 k if k  1 and to be the opposite of this area if 0 k 1. Define e to be the value of k for which the area of the region is equal to 1.  x2 x x  11  4 2.9  1.7, 1.7  y  x  m3    146 cm3   1003 cm3 
 c  0.00765 c; 0.00765 N  m 66. x 
 0.25 cos  y 
 0.25 sin  
 0.25 
0 (0.25, 0) 67. u a 
 1, 2  3 4, 3 
 1, 2  12, 9 
 13, 7  Chapter 11  tan  cos  11-6B Graphing Calculator Exploration:  1 1 (4)(4  4)  8 1 65   8  52.4 N  m2  
  0.56 units  5   29 529   29 ___ 229   29 5 2  69. y 
 70 cos 4v 70. d 
 800  (10  55) 
 250 The correct answer is 250.  3   65.  329   
 112°  63. 16 4 
 8 64. x2 
 y  4 x2  4y2 
 8 (y  4)  4y2 
 8 4y2  y  4 
 0  O  sin  229  29 x  707.9177 t about 708 yr 61. y is a logarithmic function of x. The pattern in the table can be determined by 3y 
 x which can be expressed as log3 x 
 y. 62. 1.2844  y
  529   p 
 29   ln  2.3 
 0.000427t ln e  y
  3  29 x  29y  29 
 0  0.000427 k 60b. 1.7 
 23e(0.000427)(t) 1.7 ln  2.3  0.000427  5y       
 0   29  29 29  antiln
 
r 0.6977 r; about 70%  396  Modeling Real-World Data with 11-7 Exponential and Logarithmic Functions Page 744  0.415   ln  1.0091 
 0.0197x 0.415 ln  1.0091  0.0197  Check for Understanding  1. Replace N by 4N0 in the equation N 
 N0ekt, where N0 is the amount invested and k is the interest rate. Then solve for t. 2. The data should be modeled with an exponential function. The points in the scatter plot approach a horizontal asymptote. Exponential functions have horizontal asymptotes, but logarithmic functions do not. 3. y 
 2e(ln 4)x or y 
 2e1.3863x; ln y 
 ln 2  (ln 4)x or ln y 
 0.6931  1.3863x ln 2  2631.74   ln  2137.52 
 4r  ln 2  5  5 ln  10.0170  0.0301  0.0301x x  23.08  x; 23.08 min  Pages 745–748  Exercises  7. t 
  ln 2  0.0225  ln 2   8. t 
  0.05  30.81 yr 9. t 
  2631.74  2137.52  4  
r 0.0520 r; 5.2% 17. y 
 40  14.4270 ln x 18a. y 
 –826.4217  520.4168 ln x 18b. The year 1960 would correspond to x 
 0 and ln 0 is undefined. 19. Take the square root of each side. y 
 cx2 y 
  cx2 y 
 cx 20a. 1034.34 
 1000(1  r)1 1.03034 
 1  r 0.03034 r; 3.034% 20b. y 
 1000.0006(1.0303)x 20c. y 
 1000.0006(1.0303)x y 
 1000.0006(eln 1.0303)x y 
 1000.0006e(ln 1.0303)x y 
 1000.0006e0.0299x 20d. 1030.34 
 1000er ln  39.61 yr 8.66 yr 6a. y 
 10.0170(0.9703)x 6b. y 
 10.0170(0.9703)x y 
 10.0170(eln 0.9703)x y 
 10.0170e(ln 0.9703)x y 10.0170e0.0301x 6c. 5 10.0170e0.0301x  ln  10.0170  
x  45.10 x 45.10  10 
 35.10 min 16a. y 
 2137.5192(1.0534)x 16b. y 
 2137.5192(1.0534)x y 
 2137.5192(eln 1.0534)x y 
 2137.5192e(ln 1.0534)x y 
 2137.5192e0.0520x 16c. 2631.74 
 2137.52e4r   5. t 
  0.08   4. t 
  0.0175  0.415 
 1.0091e0.0197x  15c.  13.86 yr  ln 2  0.07125  9.73 10. exponential; the graph has a horizontal asymptote 11. logarithmic; the graph has a vertical asymptote 12. logarithmic; the graph has a vertical asymptote 13. exponential; the graph has a horizontal asymptote 14a. y 
 4.7818(1.7687)x 14b. y 
 4.7818(1.7687)x y 
 4.7818(eln 1.7687)x y 
 4.7818e(ln 1.7687)x y 
 4.7818e0.5702x  1030.34   ln  1000 
 r  0.0299 21a.  x ln y  r; 2.99%  0 1.81  50 2.07  100 3.24  150 3.75  190 4.25  200 4.38  21b. ln y 
 0.0136x  1.6889 21c. ln y 
 0.0136x  1.6889 y 
 e0.0136x1.6889 21d. y 
 e0.0136(225)1.6889 115.4572 115.5 persons per square mile 22a. The graph appears to have a horizontal asymptote at y 
 2, so you must subtract 2 from each y-value before a calculator can perform exponential regression. 22b. y 
 2  1.0003(2.5710)x 23a. ln y is a linear function of ln x. y 
 cxa ln y 
 ln(cxa) ln y 
 ln c  ln xa ln y 
 ln c  a ln x  ln 2  14c. Use t 
 k; k 
 0.5702. ln 2   t
 0.5702  1.215 hr 15a. y 
 1.0091(0.9805)x 15b. y 
 1.0091(0.9805)x y 
 1.0091(eln 0.9805)x y 
 1.0091e(ln 0.9805)x y 
 1.0091e0.0197x  397  Chapter 11  32.  23b. The result of part a indicates that we should take the natural logarithms of both the x- and y-values. ln x ln y  6.21 4.49  6.91 4.84  8.52 5.65  9.21 5.99  28. 5
cos   i sin    6   
 5
  3   2  5 3    50; 22 33. Circle X contains the regions a, b, d, and e. Circle Z contains the regions d, e, f, and g. Six regions are contained in one or both of circles X and Z. The correct choice is C.  Chapter 11 Study Guide and Assessment Page 749 1. 3. 5. 7. 9.  11.    4i11   1 2  4  
   1   12. (64) 2 
 8  
   4   
 16  4   5  b   
 sin 48°  3  4  3   
 (44) 4 
 43 
 64 1   3x2  3  1   3x2  
 3 17.  3  
12x4  4  1 3  
 
2 (x4)3 1  
 8x12 18. (w3)4  (4w2)2 
 w12  42  w4 
 16w16 1   1   3  3  1   1   2  19. 
(2a) 3 (a2b) 3  
 (2a) 3   (a2b) 3  
 (2a)(a2b) 
 2a3b  2i11   
 1 or 5 0  1   1   5   9   20. 
3x 2 y 4 (4x2y2) 
 12x 2 y 4  31. 4 units left and 8 units down  398  3  16. 
6a 3  
 63
a 3  
 216a   
 9x2  42˚  3   14. (256 ) 
 (256) 4   15. 3x2(3x)2 
  (3x)2  x 
 10  Chapter 11  
  exponential growth scientific notation natural logarithm exponential function exponential equation  Skills and Concepts  1  1 2  4  13. (27) 3 
 (33) 3 
 34 
 81  30. 5x2  8x  12 
 0 Discriminant: (8)2  4(5)(12) 
 176 The discriminant is negative, so there are 2 imaginary roots.  8  Understanding the Vocabulary  common logarithm 2. logarithmic function 4. mantissa 6. linearizing data 8. nonlinear regression 10.  Pages 750–752  18˚  176   (2, 1)  1  b  8  
 38 (4, 4): f(x) 
 2(4)  8(4)  10 
 50  x  O  60 sin 48° 
 b sin 24° b 
 109.625 about 109.6 ft  60 ft  (4, 1) (2, 8): f(x) 
 2(2)  8(3)  10  y
1  1 i 2  60  sin 24°  
 26  x  2y 
 4  
 2  2i 29.  x
4 
 22 x
2 (4, 1): f(x) 
 2(4)  8(1)  10 (4, 4)  (2, 3)  9.62 6.19  23c. ln y 
 0.4994 ln x  1.3901 23d. ln y 
 0.4994 ln x  1.3901 eln y 
 e0.4994ln x1.3901 y 
 e0.4994ln x  e1.3901 y 
 (eln x)0.4994  4.0153 y 
 4.0153x0.4994 24. 2 
 ek(85) ln 2 
 85k 0.0082 k 12 
 e0.0082t ln 12 
 0.0082t 303 t 303.04 min or about 5 h 25. 0.01 26. log5 (7x) 
 log5 (5x  16) 7x 
 5x  16 2x 
 16 x
8 27a. y 
 x(400  20(x  3)) y 
 x(460  20x) y 
 20x2  460x y  2645 
 20(x2  23x  132.25) y  2645 
 20(x  11.5) vertex at (11.5, 2645), maximum at x 
 11.5 $11.50 27b. At maximum, y 
 2645. $2645   6  (2, 1): f(x) 
 2(2)  8(1)  10  y  21.  22.  1  45. 2 log6 4  3 log6 8 
 log6 x  y  y  1   log6 42  log6 8 3 
 log6 x log6  42  1   
 log6 x  83  y 
 3x  42  1  y 
 ( 12 ) x    
x  83  x  O 23.  O  8
x  x  1  46. log2 x 
 3 log6 27  24.  1   y  log2 x 
 log2 27 3  y y 
 2x 1  1   x 
 27 3 x
3 47. y y 
 2x  2  O x  O 25.  O  x  y 
 log10 x  x  26.  y  y x  O y  2x  O 27. A 
  48. log 300,000 
 log (100,000  3) 
 log 100,000  log 3 
 5  0.4771 
 5.4771 49. log 0.0003 
 log (0.0001  3) 
 log 0.0001  log 3 
 4  0.4771 
 3.5229 50. log 140 
 log (10  14) 
 log 10  log 14 
 1  1.1461 
 2.1461 51. log 0.014 
 log (0.001  14) 
 log 0.001  log 14 
 3  1.1461 
 1.8539 52. 4x 
 6x2 x log 4 
 (x  2) log 6 x log4 
 x log 6  2 log 6 x log 4  x log 6 
 2 log 6 x(log 4  log 6) 
 2 log 6  y  2x  2  1  x  2500e0.065(10)  4788.8520; $4788.85 28. A 
 6000e0.0725(10) 12,388.3866; $12,388.39 29. A 
 12,000e0.059(10) 21,647.8610, $21,647.86 2   1  30. 8 3 
 4  31. 34 
 8 1  32. log2 16 
 4  33. log5 2 5 
 2  34. 2x 
 32 2x 
 25 x
5  35. 10x 
 0.001 10x 
 103 x 
 3  1  36. 4x 
 1 6  1  37. 2x 
 0.5  4x 
 42 x 
 2  2x 
 21 x 
 1 1  38. 6x 
 216  39. 9x 
 9  
 x
3 40. 4x 
 1024 4x 
 45 x
5  
 x 
 1 41. 8x 
 512 8x 
 83 x
3  42. x4 
 81  43.  6x  2 log 6   x
 log 4  log 6  63  1   x 
 (81) 4 x
3 44. log3 3  log3 x 
 log3 45 log3 3x 
 log3 45 3x 
 45 x 
 15  8.84 53. 120.5x 
 80.1x4 0.5x log 12 
 (0.1x  4) log 8 0.5x log 12 
 0.1x log 8  4 log 8 0.5x log 12  0.1x log 8 
 4 log 8 x(0.5 log 12  0.1 log 8) 
 4 log 8 x  91  9x  4  
12  
x  16 
 x  399  x
  4 log 8  0.5 log 12  0.1 log 8  x  8.04  Chapter 11  3x  
14  54.  1  (x  2) log 6  1  x log 6  2 log 6  3x log 4 3x log 4 1  3x log 4  x log 6 x(3 log  1  4  log 100  64.  4x 
 100 ln 100   x
 ln 4  x  2 log 6  1 3 log   log 6 4  65.  3.3219 6x2 
 30 (x  2) ln 6 
 ln 30  1  ln 30  Change the inequality because 3 log 4  log 6 is negative. x  0.6 55. 0  12x8  7x4 (2x 8) log 0.1  (x  4) log 7 2x log 0.1  8 log 0.1  x log 7  4 log 7 2x log 0.1  x log 7  4 log 7  8 log 0.1 x(2 log 0.1  log 7)  4 log 7  8 log 0.1 x   x2
 ln 6 ln 30   x
 ln 6  2  66.  4 log 7  7 log 0.1  2 log 0.1  log 7  x 3.8982 3x1 
 42x (x  1) ln 3 
 2x ln 4 x ln 3  ln 3 
 2x ln 4 x ln 3  2x ln 4 
 ln 3 x(ln 3  2 ln 4) 
 ln 3  ln 3   x
 ln 3  2 ln 4  x  4 56. log (2x  3) 
 log (3  x) log (2x  3) 
 log (3  x)1 (2x  3) 
 (3  x)1 (2x  3)(3  x) 
 1 2x2  3x  8 
 0 3  x  0.6563 67. 
 5x4 4x ln 9 
 (x  4) ln 5 4x ln 9 
 x ln 5  4 ln 5 4x ln 9  x ln 5 
 4 ln 5 x(4 ln 9  ln 5) 
 4 ln 5 94x  73   x 
 4  4 ln 5   x
 4 ln 9  ln 5  1.39, 2.89 57. y  x  y 
 3 log (x  2)  68.  24 ln 24  e2x 2x  0.8967 69. 15ex  200 200 ex  15  ln 24  x  O  1.7829  x ln 4 
 ln 100   2 log 6  x   63. log15 125 
  log 15  2.0959   2 log 6   log 6)  log 125   62. log4 100 
  log 9  6x2  200  x  2  x  ln 15  x  1.5890  x  2.5903 x 3.333  70.  y  58.  [5, 5] scl:1 by [10, 60] scl:10  y  7x  2  O  71.  x  x  x 
 3.42  59.  [1, 5] scl:1 by [1, 10] scl:1 [5, 5] scl:1 by [5, 5] scl:1 log 15   60. log4 15 
  log 4  1.9534  Chapter 11  log 24   61. log8 24 
  log 8  1.5283  400  2.20  ln 2  ln 2   72. t 
  0.028  24.76 74. 18 
 k
  Chapter SAT & ACT Preparation   73. t 
  0.05125  13.52  ln 2  k ln 2  18  Page 755  0.0385; 3.85%  Page 753 75.  Applications and Problem solving 1 t  0.065 
 
2  1  
t  0.6215 t 0.6215  5730 76a.  
 10 log  To find the quadratic equation, multiply these two factors and let the product equal zero. (2x  1)(3x  1) 
 0 6x2  5x  1 
 0 The correct choice is E. 3. The result of dividing T by 6 is 14 less than the correct average.  3561.13 or 3561 yr.  1.15  1010  10 12  20.6  20.6 dB 9  109   76b.  
 10 log  102  39.5  T  6  39.5 dB 8.95   103  T  6   76c.  
 10 log  1012 99.5 99.5 dB 0.014t 77. 200,000 
 142,000e  ln  100  71 100  71  100 ln  71  0.014  ln  1 ln  2  0.20  3.47  
 correct answer  14   14 
 correct average  The correct average is the total divided by the number of scores, 5.  
 e0.014t  T  correct average 
 5  
 0.014t  T  6  T   14 
 5  The correct choice is E. 4. y B x,y  
t  (2 )  24.4 t 1990  24 
 2014 78a. N 
 65  30e0.20(2) 44.89; 45 words per minute 78b. N 
 65  30e0.20(15) 
 63.50; 64 words per minute 78c. 50 
 65  30e0.20t 1  2 1  2  1  2. Since one root is 2, x 
 2, 2x 
 1, and 2x 
 0. 1 Similarly for the root that is 3, 1 x 
 3, 3x 
 1, and 3x  1 
 0.  1  log 0.65 
 t log 2 log 0.65  1 log  2  SAT and ACT Practice  1. To find the greatest possible value, the other 3 values must be as small as possible. Since they are distinct positive integers, they must be 1, 2, and 3. The sum of all 4 integers is 4(11) or 44. The sum of the 3 smallest is 1  2  3 or 6, so the fourth integer cannot be more than 44  6 or 38. The correct choice is B.  y x 2  A  tan A 
  
 e0.20t  y  x  2  C (x, 0) x 2y  
 x  Find the area of ABC.  
 0.20t  1  1  xy  A 
 2bh 
 2xy 
 2 Simplify the ratio.  
t  area of ABC  tan A  t; 3.5 weeks  
  xy  2  2y  x  
 2y 
2 
 4 x  xy  x2  The correct choice is E.  Page 753  Open-Ended Assessment  5.  1  4  1. Sample answer: (n4) (4m)1 2. Sample answer:  x  y  10  
 2y  2yx 
 10y 2x 
 10 x
5 The correct choice is B.  1  log 2  log(x  2) 
 2 log 36  401  Chapter 11  6. C 
 D 2  3  9. A is the arithmetic mean of three consecutive x  (x  2)  (x  4) 
 positive even integers, so A 
   
 D  D
  2  3  and, therefore, r 
  3 3x  6 
 x  2, where x is a positive even integer.   3  1 . 3  Then A is also a positive even integer. Since A is even, when A is divided by 6, the remainder must also be an even integer. The possible even remainders are 0, 2, and 4. The correct choice is C. 10. First notice that b must be a prime integer. Next notice that 3b is greater than 10. So b could be 5, since 3(5) 
 15. (b cannot be 3.) Check to be sure that 5 fits the rest of the inequality. 5 25 1 3(5) 
 15  6(5) 
 6 
 46  Now use this value for the radius to calculate half of the area. 1 A 2  1 2    
 2r2 
 2
3 
 2
9 
 18 1  1  1  1  The correct choice is A. 7. The average of 8 numbers is 20. 20 
  sum of eight numbers  8  sum of eight numbers 
 160 The average of 5 of the numbers is 14. 14 
  So 5 is one possible answer. You can check to see that 7 and 11 are also valid answers. The correct answer is 5, 7, or 11.  sum of five numbers  5  sum of five numbers 
 70 The sum of the other three numbers must be 160  70 or 90. Calculate the average of these three numbers. average 
  sum of three numbers  3  90  
 3 
 30  The correct choice is D. 8. The sum of the angles in a triangle is 180°. Since ∠B is a right angle, it is 90°. So the sum of the other two angles is 90°. Write and solve an equation using the expressions for the two angles. 2x  3x 
 90 5x 
 90 x 
 18 The question asks for the measure of ∠A. ∠A 
 2x 
 2(18) 
 36 The correct choice is C.  Chapter 11  402  Chapter 12 Sequences and Series 11. an 
 a1  (n  1)d 3 
 a1  (7  1)(2) 3 
 a1  12 15 
 a1 12. an 
 a1  (n  1)d 34 
 100  (12  1)d 66 
 11d 6 
 d 13. an 
 a1  (n  1)d 24 
 9  (4  1)d 15 
 3d 5
d 9  5 
 14, 14  5 
 19 9, 14, 19, 24  Arithmetic Sequences and Series  12-1  Pages 762–763  Check for Understanding  1. a1 
 6  4(1) or 2 a2 
 6  4(2) or 2 a3 
 6  4(3) or 6 a4 
 6  4(4) or 10 a5 
 6  4(5) or 14 2, 2, 6, 10, 14; yes, there is a common difference of 4. an 2a.  n  1  14. Sn 
 2[2a1  (n  1)d]  O 1 2 3 4 5 6  35  S35 
 2[2.7  (35  1)  2]  n  
 1435  1  n  Sn 
 2[2a1  (n  1)d]  15.  2  n  210 
 2[2  30  (n  1)(4)]  3  420 
 60n  4n2  4n 4n2  64n  420 
 0 4(n  21)(n  5) 
 0 n 
 21 or n 
 5 Since n cannot be negative, n 
 21. 16. n 
 19, a19 
 27, d 
 1 an 
 a1  (n  1)d 27 
 a1  (19  1)1 9 
 a1  2b. linear 2c. The common difference is 1. This is the slope of the line through the points of the sequence. 3a. The number of houses sold cannot be negative. n  3b. Sn 
 2[2a,  (n  1)d] 10  S10 
 2[2  3750  (10  1)500] 
 $60,000 4. Negative; let n and n  1 be two consecutive numbers in the sequence. d 
 (n  1)  n or 1 5. Neither student is correct, since neither sequence has a common difference. The difference fluctuates between 1 and 1. The second sequence has a difference that fluctuates between 2 and 2. 6. d 
 11  6 or 5 16  5 
 21, 21  5 
 26, 26  5 
 31, 31  5 
 36 21, 26, 31, 36 7. d 
 7  (15) or 8 1  8 
 9, 9  8 
 17, 17  8 
 25, 25  8 
 33 9, 17, 25, 33 8. d 
 (a  2)  (a  6) 
 a  a  2  6 or 4 a  2  4 
 a  6, a  6  4 
 a  10, a  10  4 
 a  14, a  14  4 
 4 
 a  18 a  6, a  10, a  14, a  18 9. an 
 a1  (n  1)d a17 
 10  (17  1)(3) 
 38 10. an 
 a1  (n  1)d 37 
 13  (n  1)5 50 
 5(n  1) 10 
 n  1 11 
 n  n  S19 
 2(a1  a19 ) 19  
 2(9  27) 
 342 seats  Pages 763–765  Exercises  17. d 
 1  5 or 6 7  (6) 
 13, 13  (6) 
 19, 19  (6) 
 25, 25  (6) 
 31 13, 19, 25, 31 18. d 
 7  (18) 
 11 4  11 
 15, 15  11 
 26, 26  11 
 37, 37  11 
 48  19.  20.  21.  22.  403  15, 26, 37, 48 d 
 4.5  3 or 1.5 6  1.5 
 7.5, 7.5  1.5 
 9, 9  1.5 
 10.5, 10.5  1.5 
 12 7.5, 9, 10.5, 12 d 
 3.8  5.6 or 1.8 2  (1.8) 
 0.2, 0.2  (1.8) 
 1.6, 1.6  (1.8) 
 3.4, 3.4  (1.8) 
 5.2 0.2, 1.6, 3.4, 5.2 d 
 b  4  b or 4 b  8  4 
 b  12, b  12  4 
 b  16, b  16  4 
 b  20, b  20  4 
 b  24 b  12, b  16, b  20, b  24 d 
 0  (x) or x x  x 
 2x, 2x  x 
 3x, 3x  x 
 4x, 4x  x 
 5x 2x, 3x, 4x, 5x Chapter 12  40. 4 
 5  (4  1)d 9 
 3d 3
d 5  3 
 2, 2  3 
 1 5, 2, 1, 4 41. 12 
 3   (4  1)d 12  3  
 3d  23. d 
 n  5n or 6x 7n  (6n) 
 13n  13n  (6n) 
 19n, 19n  (6n) 
 25n, 25n  (6n) 
 31n 13n, 19n, 25n, 31n 24. d 
 5  (5  k) or k 5  k  (k) 
 5  2k, 5  2k  (k) 
 5  3k, 5  3k  (k) 
 5  4k, 5  4k  (k) 
 5  5k 5  2k, 5  3k, 5  4k, 5  5k 25. d 
 (2a  2)  (2a  5) or 7 2a  9  7 
 2a  16, 2a  16  7 
 2a  23, 2a  23  7 
 2a  30, 2a  30  7 
 2a  37 2a  16, 2a  23, 2a  30, 2a  37 26. d 
 5  (3  7 ) or 2 7  7  7   2  7  
 9  27 , 9  27   2  7  
 11  37 , 11  37   2 7  
 13  47  9  27 , 11  37 , 13  47  27. a25 
 8  (25  1)3 
 80 28. a18 
 1.4  (18  1)(0.5) 
 9.9 29. 41 
 19  (n  1)(5) 60 
 5(n  1) 12 
 n  1 13 
 n 30. 138 
 2  (n  1)7 140 
 7(n  1) 20 
 n  1 21 
 n 31. 38 
 a1  (15  1)(3) 38 
 a1  42 80 
 a1 2   12  3  
 d 3   12  3 12  23 3    
  , 3 3   12  2 3 12  3 24  3    
  3 3 3 12  2  3 24   3 3 ,  ,  , 12 3 3  42.  3  1  43. d 
 1  2 or 2 a11 
 2  (11  1)
2 3  1  
 3.5  S11 
 2
2  
32 11 3  1  
 11 44. d 
 4.8  (5) or 0.2 a100 
 5  (100  1)0.2 
 14.8 100  S100 
 2(5  14.8) 
 490 45. d 
 13  (19) or 6 a26 
 19  (26  1)6 
 131  1  32. 103 
 a1  (7  1)3 2  103 
 a1  2 2  83 
 a1  26  S26 
 2(19  131)  33. 58 
 6  (14  1)d 52 
 13d 4
d 34. 26 
 8  (11  1)d 18 
 10d 4 15 
 d  
 1456 n  46. 14 
 2[2(7)  (n  1)1.5] 28 
 14n  1.5n2  1.5n 0 
 1.5n2  15.5n  28 n
  )  (4  5 ) or 3 35. d 
 (1  5 a8 
 4  5   (8  1)3 
 17  5  36. d 
 6  (5  i) 
 1  i a12 
 5  i  (12  1)(1  i) 
 5  i  11  11i 
 16  10i 37. d 
 10.5  12.2 or 1.7 a33 
 12.2  (33  1)(1.7) 
 42.2 38. d 
 4  (7) or 3 a79 
 7  (79  1)3 
 227 39. 21 
 12  (3  1)d 9 
 2d 4.5 
 d 12  4.5 
 16.5 12, 16.5, 21  Chapter 12  5 
 2  (5  1)d 3 
 4d 0.75 
 d 2  0.75 
 2.75, 2.75  0.75 
 3.5, 3.5  0.75 
 4.25 2, 2.75, 3.5, 4.25, 5  15.5  (15. 5)2   4(1.5) (28)  2(1.5) 1  n 
 8 or n 
 23 Since there cannot be a fractional number of terms, n 
 8. n  47. 31.5 
 2[2(3)  (n  1)2.5] 63 
 6n  2.5n2  25n 0 
 2.5n2  8.5n  63 n
  8.5  (8.5 )2 4( 2.5)( 63)  2(2.5)  n 
 7 or n 
 3.6 Since n cannot be negative, n 
 7. 48. d 
 7  5 or 2 an 
 5  (n  1)2 
 2n  3 49. d 
 2  6 or 8 an 
 6  (n  1)(8) 
 8n  14  404  58. Sn 
 a1  a2  (a31  a32)  (a41  a42)  (a51  a52)   
 a1  a2  a2  a1  a3  a2  a4  a3   
 a1  a2  a2  a1  (a2  a1)  a2  a3  a2  a3   
0 59. A 
 Pert 
 100e0.07(15) 
 $285.77 60. 4x2  25y2  250y  525 
 0 4x2  25(y2  10y) 
 525 4x2  25(y  5)2 
 100  50. 9:00, 9:30, 10:00, 10:30, 11:00, 11:30, 12:00 n 
 7, d 
 2, a1 
 3 a7 
 3  (7  1)2 
 15 data items per minute 51. Let d be the common difference. Then, y 
 x  d, z 
 x  2d, and w 
 x  3d. Substitute these values into the expression x  w  y and simplify. x  (x  3d)  (x  d) 
 x  2d or z. 52. a1 
 5, d 
 4, n 
 25 a25 
 5  (25  1)4 
 101 25  S25 
 2(5  101)  150n 
 180n  360 30n 
 360 n 
 12 54a. S4 
 (4  2)180° or 360° S5 
 (5  2)180° or 540° S6 
 (6  2)180° or 720° S7 
 (7  2)180° or 900° 360°, 540°, 720°, 900° 54b. The common difference between each consecutive term in the sequence is 180, therefore the sequence is arithmetic. 54c. a35 
 180  (35  1)180 
 5940° 55a. a1 
 1, d 
 2 S5 
  y  O  (5, 5)  (5, 5)  (0, 5) (0, 7)  6  61. r 
 1 2 or 0.5   (5  1)2]  5      v 
 8  2 or 8     0.5
cos 8  i sin 8  10  55b. S10 
 2[2(1)  (10  1)2]  0.46  0.19i  62. 2, 1, 3  5, 3, 0 
 2(5)  (1)(3)  (3)(0) 
7 63. x cos 30°  y sin 30°  5 
 0  
 100 55c. Conjecture: The sum of the first n terms of the sequence of natural numbers is n2. Proof: Let an 
 2n  1. The first term of the sequence of natural numbers is 1, so a1 
 1. Then, using the formula for the sum of an arithmetic series, n Sn 
 2(a1  an)   3 x 2  1   2y  5 
 0  x  y  10 
 0 3 64.  y O   n  2  n  4  Sn 
 2[1  (2n  1)] 
 2(2n) or n2  2 x  6  56. a1 
 5, d 
 7, n 
 15 S15 
  x (0, 3)  
 25  15 [2(5) 2   4 
 1   h 
 0, k 
 5, a 
 5, b 
 2, c 
 21 center: (0, 5) foci: ( 21 , 5) vertices: major → ( 5, 5) minor → (0, 3) and (0, 7)  n  53. Sn 
 2(128°  172°); Sn 
 (n  2)180°  5 [2(1) 2  (y  5)2  x2  25  
 1325 bricks   (15  1)7]  65. Find A. A 
 90°  19° 32
 
 70° 28
 Find a. a  cos 19° 32
 
  4.5  
 810 feet 57. n 
 10, S10 
 5510, d 
 100 10  5510 
 2[2a1  (10  1)100] 5510 
 10a1  4500 1010 
 10a1 101 
 a1 a10 
 101  (10  1)100 
 1001 least: $101, greatest: $1001  4.2 a Find b. b  sin 19° 32
 
  4.5 1.5 b 66. discriminant 
 (3)2  4(4)(2) 
 23 Since the discriminant is negative, this indicates two imaginary roots.  405  Chapter 12  x1 x  3x2 4x  2 (x2  3x) x  2 (x  3) 1 y
x1 1 1 3 4 0 1 2 0 68.  
 2 0 1 1 0 1 3 4  64b.  67.          Value 27,500 25,000 22,500 20,000 17,500 15,000 12,500    10,000 7500 5000 2500 0  A
(1, 2), B
(3, 0), C
(4, 1) 69. a  4b 
 15 → a 
 15  4b 4a  b 
 15 4(15  4b)  b 
 15 60  16b  b 
 15 15b 
 45 b 
 3 4a  (3) 
 15 4a 
 12 a 
3 a  b 
 3  (3) or 6 The correct choice is C.  0  1  2  3 Years  4  5  6  6c. an exponential function 7. r 
  4 — 2  3  or 6  24(6) 
 144, 144(6) 
 864, 864(6) 
 5184 144, 864, 5184 3  8. r 
 2  12-2  
   Geometric Sequences and Series 9.  Page 771  Check for Understanding  28.8(4) 
 115.2, 115.2(4) 
 460.8, 460.8(4) 
 1843.2 115.2, 460.8, 1843.2  1. Both arithmetic and geometric sequences are recursive. Each term of an arithmetic sequence is the sum of a fixed difference and the previous term. Each term of a geometric sequence is the product of a common ratio and the previous term. 2. an 
 (3)11 or 9 a2 
 (3)21 or 27 a3 
 (3)31 or 81 The expression generates the following sequence: 9, 27, 81, . The common ratio is 3, therefore it is a geometric sequence. 3. If the first term in a geometric sequence were zero, then finding the common ratio would mean dividing by zero. Division by zero is undefined. 4. Sample answer: The first term of the series 5  10  20  is 5 and the sum of the first 6 terms of the sequence is 105, but 105 is not greater than 5. 5a. No; the ratio between the first two terms is 2, but the ratio between the next two terms is 3. 5b. Yes; the common ratio is 3 . 5c. Yes; the common ratio is x. 6a. Beginning of Value of Year Computers 1 27,500.00 2 15,125.00 3 8318.75 4 4575.31 5 2516.42 6 1384.03  Chapter 12  
  
 881 , 881 
32 
 21463  9 3 27 27 3   
 ,   2 2 4 4 2 27 81 243 , ,  4 8 16 7.2   r 
 1.8 or 4  2.1  10. r 
 7 or 0.3 an 
 a1rn1 a7 
 7(0.3)71 
 0.005103 11. an 
 a1rn1 24 
 a1(2)51 24 
 16a1 3  
 a 1 2 2.5  12. a3 
 2 or 1.25 1.25  a2 
 2 or 0.625 0.825  a1 
 2 or 0.3125 0.3125, 0.625, 1.25 13. an 
 a1rn1 27 
 1(r)41 27 
 r3 3
r 1(3) 
 3, 3(3) 
 9 1, 3, 9, 27 1   14. r 
  0.5 or 2 a1  a1rn   Sn 
  1r  0.5  0.5(2)9   S9 
  1  (2)  
 85.5  406  15. r 
 1.035 The value of the car after 10, 20, and 40 years will be the 11th, 21st, and 41st terms of the sequence respectively. a11 
 20,000(1.035)111  $28,211.98 a21 
 20,000(1.035)211  $39,795.78 a42 
 20,000(1.035)411  $79,185.19  26. a5 
 8
2  3 51  81  
 2 27.  3 8 3 — r 
 1 or 4  2 1 3 61 a6 
 2 4  
  243   
  2048 0.4   28. r 
  4.0 or 0.01  Pages 771–773  a7 
 40(0.01)71 
 4  1011  Exercises  2  16. r 
 1 0 or 0.2   10  5  29. r 
  or 2   0.4(0.2) 
 0.08, 0.08(0.2) 
 0.016, 0.016(0.2) 
 0.0032 0.08, 0.016, 0.0032 17. r 
  20  8  
 165   or 2.5  192 
 a1(4)61 192 
 1024a1 0.1875 
 a1 31. 322  
 a1(2 )51 322  
 4a1 82  
 a1 30.  50(2.5) 
 125, 125(2.5) 
 312.5, 312.5(2.5) 
 781.25 125, 312.5, 781.25 18. r 
  2  3 2  9  or 3  32.  19. r 
  486 
 a1  a2 
 486
3 or 162 1  or  a3 
 162
3 or 54 1  24  
  
 61225 , 61225 
25 
  3125  3 2 6 6 2   
 ,   25 5 125 125 5 6 12 24 , ,  125 625 3125 3.5  r
 7 or 0.5  486, 162, 54 33. 0.32 
 a1(0.2)51 0.32 
 0.0016a1 200 
 a1 a2 
 200(0.2) or 40 a3 
 40(0.2) or 8 200, 40, 8 34. 81 
 256r51 81   
 r4 256  1.75(0.5) 
 0.875, 0.875(0.5) 
 0.4375, 0.4375(0.5) 
 0.21875 0.875, 0.4375, 0.21875 6 32    21. r 
 —— or 2 62 
2  
 12, 12
2  
 122 , 122 
2  
 24  3  4  , 24 12, 122 3 3  
 3 144
 4 
   3   3   3   3  3
3 
 3 , 3 
3 
 1, 1
3 
 3  3  3 , 1, 3 1  35.  1  23. r 
 i 
  or i  1  i(i) 
 1, 1(i) 
 i, i(i) 
 1 1, i, 1 t5  24. r 
 t8 or t3  36.  t2(t3) 
 t1, t1(t3) 
 t4, t4(t3) 
 t7 t1, t4, t7  5  2  2  2  2  7  7  3  3  2  5  2  9  1 1 b b2 b3 ,     ab a3 , a5 , a7 , a9  407  
  3  4  144,  108  256, 192, 144, 108, 81 54 
 2r41 27 
 r3 3 
 r 2(3) 
 6, 6(3) 
 18 2, 6, 18, 54 4  7 
 7r31  
   3  192, 192
  3  4  49  
 r2 4 7  
 r 2 4 7   
 7 2 4 , 2, 7 7  
  
 a1b , a1b 
ab 
 a1, a1
ab 
 ab, b b b b b b 
 
 , 
 
  a a a a a a a b   b2 a2  
r  256
   3  22. r 
 9 or 3  25.  1 51  6 
 8 1 a1  2  5  
   20.  6 
 a1
3 1  2(3) 
 6, 6(3) 
 18, 18(3) 
 54 6, 18, 54 3 —— 10 — 3 —— 4  91  a9 
 5 
2   2  Chapter 12  37. r 
  38.  5 — 5  3  5.50  43a. r 
 5 or 1.1  or 3  a10 
 5(1.1)101  $11.79 a20 
 5(1.1)201  $30.58 a10 
 5(1.1)401  $205.72 $11.79, $30.58, $205.72  5 5   (3)5 3 3 S5 
  13 605 
 3 13 r 
 65 or 0.2 65(0.2)6  S6 
 65   1  0.2  5  5(1.1)52   43b. S52 
  1  1.1  
 81.2448   $7052.15 43c. Each payment made is rounded to the nearest penny, so the sum of the payments will actually be more than the sum found in b.  3  39. r 
  2   1  S10 
 
  3  or 2  
   3 10 1  1 2 —— 3 1  2  
   44a.  11,605  512  2 3  40. r 
 2 or 3  S8 
 
 
  2  2(3  )8  1  3  160  1  3   160 1  3     1  3 3  1   44b.  ) 160(1  3   
 2  41a.  41b. 41c.  42a.  42b.  1  4 1  5   
 2 0  z  1  1  1  
 z by definition  a2 1  Then an 
 a1rn1 So, an 
 (2)(3)n1 46. a1 
 1, r 
 2.5, n 
 15 a15 
 1(2.5)151 
 372,529  47a. 25
1  1 2  
 $25.05 0.024  47b. No; at the end of two years, she will have only $615.23 in her account. S24 
  
    0.024 24  25.05  25.05 1   12  0.024  1 1 12  
     $615.23 47c.  3  4 
 3a1  750   
    
    0.024 24  a1  a1 1   12  0.24  1 1 12  1.5  a1
1  
1  12    0.024 24  
 a1  $30.54  a1  a28 
 3
3 281 3  a0
1  1 2   $30.54 0.024  
 26,244  Chapter 12  
  2  r 
 a
 3  4 
 a1
3 3  4  1 1    z 20  1  8  5
z 45. a2 
 3(a1) 
 3(2) 
6  
 80(1  3 ) The population doubles every half-hour, so r 
 2. After 1 hour, the number of bacteria is the third term in the sequence and n  1 
 2. After 2 hours, it is the fifth term and n  1 
 4. After 3 hours, it is the seventh term and n  1 
 6. After t hours it is the 2t  1 term and n  1 
 2t. bt 
 b0  22t bt 
 30  22(5) 
 30,720 Sample answer: It is assumed that favorable conditions are maintained for the growth of the bacteria, such as an adequate food and oxygen supply, appropriate surrounding temperature, and adequate room for growth. a7 
 a4r3 12 
 4r3 3 
 r3 3 3 
r a4 
 a1r41  4  3  1 1    1 5 8  
 x 2 13 1  
  80 x 80  
 x 13  a0  $30.48 The least monthly deposit is $30.48.  408  a2  55. Find the amplitude.  48. r 
 a  86  36  A 
 2 or 25  1  r
  1  27 — 1  81  Find h.  
3  86  36  h 
 2 or 61 Find k.  an 
 a1 rn1  2  k  1  n1 6561 
 8 1 3    k 
 2  (6561)(81) 
 3n1 (38)(34) 
 3n1 312 
 3n1 12 
 n  1 13 
 n 6561 is the 13th term of the sequence. 49.    y 
 25 sin 
2t  c  61   36 
 25 sin 
2  1  c  61   1 
 sin 
2  c    
 c  y 
 25 sin 
2t  3.14  61  n  650 
 2[2(20)  (n  1)5] 1300 
 40n  5n2  5n 0 
 5n2  35n  1300 0 
 5(n  13)(n  20) n 
 13 or n 
 20 Since n cannot be negative, n 
 13 weeks.  56. Since 43° 90°, consider the following. b sin A 
 20 sin 43° 13.64 Since 11 13.64, no triangle exists. 57. (n2)  49  2 n 7 Solution set: {nn  3 or n  3}  log 26 5   50. log11 265 
  log 11  2.3269 y  O  52.   A2   B2 
    2 
 2  c  n  Sn 
 2[2a1  (n  1)d]  51.  
4  n 5 4 3 3 4 5  x  m
n1 5  1 or 4 4  1 or 3 3  1 or 2 3  1 or 4 4  1 or 5 5  1 or 6  nm (5)(4) or 20 (4)(3) or 12 (3)(2) or 6  3(4) or 12  4(5) or 20  5(6) or 30  ← least possible value  The answer is 6.   32  ( 52)  
 34  Since C is positive, use 34 . 3  5  5 34  5  3  5   or    , sin f 
   p
 34 , cos f 
   34 34   34  tan f 
 tan f 
  Pages 780–781  5   34   3   34  5 3  1a.  Check for Understanding  an 1.0  f  59° Since cosine is negative and sine is positive, 
 59°  180° or 121°. p 
 r cos(v  f) 5 34   34  Infinite Sequences and Series  12-3  5  x   y   
0  34 34 34     0.5  
 r cos(v  121°)  O 1 2 3 4 5 6 7 8 9 10n  53. 3x  4y 
 5 3  5  y 
 4x  4 3  1b. The value of an approaches 1 as the value of n increases.  5  x 
 t, y 
 4t  4 54. csc v 
 3 1  sin v 1  3  1c. lim  n→  
3  n1  n  
1  
 sin v  409  Chapter 12  1d. lim  n→  n1  n  n  n→ n  
 lim    10. r 
  6 or  2  
10 
1  Sn 
  
  
   2b. If r  11. r 
  
0  1  
 18  
 no limit  3  3  12. r 
  or 3   
 no limit  The sum does not exist since r 
 3   1.  1, then lim rn 
 0. If r 
 1, then  2  13. a1 
 75, r 
 5  n→  Sn 
  n→  3. Sample answer: 2  4  8  4. Zonta is correct. As n approaches infinity the expression 2n  3 will continue to grow larger and larger. Tyree applied the method of dividing by the highest powered term incorrectly. Both the numerator and the denominator of the expression must be divided by the highest-powered term. It is not appropriate to apply this method here since the denominator of the expression 2n  3 is 1. 5. 0; as n → , 5n becomes increasingly large and 1 thus the value 5n becomes smaller and smaller, approaching zero. So the sequence has a limit of zero. n→  lim  n→  Pages 781–783 14. lim  n→  
  5  2  15.  7.  8.  3n  6  7n  0
0  7  
  n→    6n2  5  3n2   
 lim 
3   3n2  6  5  n→  5 1   lim  2 n→ 3 n→ n    n→  5  17. lim  n→  
 2  3  0 or 2   5n  2  2n3 9n3    
 lim 
2   2n2  n3  9  2   lim  
 lim 
3  n  n2  1  1  n→ n  
 lim (3)  lim n→  1  2 n→ n   lim 4  lim n→  
 3  0  4  0 or 3  126    9. 5.1 26  
5 1000  1,000,000    19. Dividing by the highest powered term, n2, we find  1    a1 
  1000 , r 
 1000  Sn 
 5   4  n→  7  126  1  5 1 1   lim   2  lim  3 n→ 2 n→ n n→ n 5 9   0  0 or  2 2 (3n  4)(1  n) 3n2  n  4  
 lim  lim  n2 n2 n→ n→  n→ 9 
 2   18.  5  9  n→  
 lim  
 9 126  n approaches  
 lim 2  lim  3 6 1 lim 
 lim 7  7  n n→ n→ 3 6 1 
 lim 7  lim 7  lim n n→ n→ n→ 3 6 3 
 7  7  0 or 7 7 7  0.7  
 1 0  100   7 1  a1 
 1 0 , r 
 10 7  10 Sn 
 — 1  1 10  5 2  8  n   n2 —— 3 2 n→   n2  n  lim  126  1000 —— 1  1 1000  which as n approaches infinity  800  8    simplifies to  0  0 
 0 . Since this fraction is undefined, the limit does not exist.  126   
5 999 14  
 511  Chapter 12  2  1  n→  1 7 2   lim   lim  n→ 5 n→ n n→ 5 7 2 2 
 5  0  5 or 5 n3  2 2 lim n 
 lim n  n . 2 n→ n→ 2 1 lim n 
 lim 2  n 
 2  0 or 0, but as n→ n→  16. lim  1  
  Exercises  
 lim 
5  n  5  infinity, n becomes increasingly large, so the sequence has no limit.  As n approaches infinity, 2n becomes increasingly large, so the sequence has no limit. 3 , 7  7  2n  5n  
 lim  1  5  75  2 1   5  
 125 m  
 lim 
2n  2n  n→ 5 1 
 lim 2  n n→  5  2n  1  or 3  Sn 
 — 1 1  3  
1  lim rn 
 1. If r  1 then lim rn does not exist.  5  n2  2n  1  4 — 3  4  
   3  4  
0  n→  6. lim  6 —— 1 1  2  
 4  The limits are equal. 2a. See students' work. Student's should draw the following conclusions: 1 n lim 2 n→ 1 n lim  n→ 4 lim (1)n n→ lim (2)n n→ lim (5)n n→  1  3  1  n→ n   lim  410  20. lim  n→  4  3n  n2   2n3  3n2  5  4 3n n2    n 3  n3  n 3 
 lim  2n3 3n2 5 n→     n3  n3  n 3 4 3 1    n3  n2  n  
 lim 3 5 n→ 2     n n3 1 1 1   lim 4  lim  3  lim 3  lim  2  lim n n→ n→ n n→ n→ n n→  
 1 1  lim 2  lim 3  lim n  lim 5  lim  3 n→ n→ n→ n→ n→ n  259  259  259  1000  Sn 
 6  —— 1  1 1000 259   
6 999 7  
 62 7 15  15  1  15  100  the value 3n becomes smaller and smaller, approaching zero. So the sequence has a limit of zero. 22. Dividing by the highest powered term, n, we find  Sn 
 — 1  1 100 15  23.    2  
 lim  n→  5n  n2   lim  n→  5   lim n→ n n→ (1)n 
 lim n 2 n→  
 lim  63  Sn 
  1  1  5  63  1000   — 1  1 100  29   
 110  (1)n   n2  30. The series is geometric, having a common ratio of 0.1. Since this ratio is less than 1, the sum of the 2 series exists and is 9. 12  16 Sn 
  3 1  4 
 64 7.5  32. r 
 5 or 1.5 This series is geometric with a common ratio of 1.5. Since this ratio is greater than 1, the sum of the series does not exist. 5  4  10  1 Sn 
 — 1  1 0 4  
 9  34. The series is arithmetic, having a general term of 7  n. Since lim 7  n does not equal zero, this n→ series has no sum.  51    25. 0.5 1 
 100  10,000 … 1    a1 
  100 , r 
 100 51  100  35. r 
  Sn 
 — 1  1 100 370  1  4 — 1  8  or 2  This series is geometric with a common ratio of 2. Since this ratio is greater than 1, the sum of the series does not exist.  17  
 99 or 3 3 370    26. 0.3 7 0 
  1000  1,000,000 … 370  1   33. r 
 1 0 or 2 10 Sn 
  1 1  2 
 20  1   a1 
 1 0 , r 
 10  51  3   31. r 
 1 6 or 4  4  51  63   
 5   990  (1)n  n2   24. 0.4  
 1 0  100 … 4  63    a1 
  1000 , r 
 100  As n increases, the value of the numerator alternates between 1 and 1. As n approaches infinity, the value of the denominator becomes increasingly large, causing the value of the fraction to become increasingly small. Thus, the terms of the sequence alternate between smaller and smaller positive and negative values, approaching zero. So the sequence has a limit of zero. 4  63    29. 0.26 3  
 1 0  1000  100,000 …  1  5n  (1)n lim n 2 n→  51  5  
 99 or 3 3  1 
 lim   4 n→   1 n which as n approaches infinity 1 (2)n 
 1, but lim n has no limit since lim  4 n→   1 n→ n o2o  1.  1    a1 
  100 , r 
 100  21. As n → , 3n becomes increasingly large and thus  
  15    28. 0.1 5 
 100  10,000 …  40300  (2)n  n  1    a1 
  1000 , r 
 1000   
 2  3  0  5  0 or 0  (2)n  n lim — 4 n→   1 n  259    27. 6.2 5 9 
6 1000  1,000,000 …  1    a1 
  1000 , r 
 1000 370  1000  Sn 
 —— 1  1 1000 370  10    
 999 or 27  411  Chapter 12  36. r 
  1  9 — 2 3  or  
  n2  n→ 2n  1  41b. lim  1 6  n2(2n  1)  n2(2n  1)      (2n  1)(2n  1) 2n  1  
 lim  n2  n→  2n3  n2  2n3  n2  
 lim  4n2  1 n→  2  3 —— Sn 
 1 1  
6  2n2  2 n→ 4n  1 2n2  n2 — 
 lim 4n2 1 n→    n2 n2  
 lim  4  
 7 37. r 
  4  5 — 6  5  2  or 3  
 lim  n→  6  5  Sn 
 — 2 1  3  1  42a. 12 4 
 3 3 r 
 3  a4 
 a1r41 3 3 4 
 a1
3  4 
 a1(3) 4  
 a 1 3   5  1   or  38. r 
  5 5    5 Sn 
   5 1  5  
   5  42b. a28 
 a1r281  1  5  5 — 
    5  5 1  5 1  5 5 1 5 
  1  2 5  4  
 3
3 27 4  
 4(38) 
 26,244 n 1 43. No; if n is even, lim cos 2 
 2, but if n is odd,  4  4 3   3  lim cos 2 
 2.  n→  8 Sn 
   3 1  
2  1  1 D 4   3  2  exists. After 6 hours and before the second 1  1  1  1  1  44b. a1 
 D, r 
 8   3  1  
18n  Sn 
 D —— 1 1  8   3  
 32
1  2  
 7D1  
8 8  
 32  163   1 n    44c. lim Sn 
 S n→  2  40a. a1 
 35 r 
 5  a1   S
 1r   or 14  2  5  D 
  1 1  8  a3 
 14  a4 
 14
5 or 5.6 2  8  
 7 D  a5 
 5.6 35, 14, 14, 5.6, 5.6 14 14  40b. Sn 
 35   2  2   1 5 1  5  8  44d. 350  7D 306.25  D The largest possible dose is 306.25 mg. 20  45a. A side of the original square measures 4 or  70  
 35  3  3  5  5 feet. Half of 5 feet is 2 feet.  2  
 813 m or about 82 m  
522  
522 
 s2  41a. The limit of a difference equals the difference of the limits only if the two limits exist. Since n2  2n  1  nor lim  n→  n2  2n  1  50  4 5 2  2  exists, this  property of limits does not apply.  Chapter 12  1  dose, 2  2  2D or 8D exists.  8
1  2 
 —— 3 1  4  n→  1  44a. After 2 hours, 2D exists. After 4 hours, 2  2D or  
1   8 
   ——   3 1  2 
1  23   neither lim  n→ 1  n  39. r 
 8 or 2  20  3  
 3(39)  
 5(5   1)  a2 
 35
  1  4  n2  
 2  3  
 35  2   
 s2 
s 5 2   feet. Perimeter 
 4  2 or 102  412  51. vy 
 125 sin 20° 
 42.75 miles vx 
 125 cos 20° 117.46 miles   2  10 2   45b. a1 
 20, r 
 2 0 or 2  20 S
   2 1  2  225°  20 
   ——  2  1  2 
1  22     2  
 
  46d. 46e.  
   2  2  1  1 2  1   180˚  8  2,  8 1 , 2  1 , 4  1  8  Page 784 1. 1.618181818 2. N 
 24 x2 
 x  1 x2  x  1 
 0 (1)  (1)2   4(1 )(1)  2(1) 1   5 x 
  2 1 Since the sum 1   1  1 1    x
   3  negative, the value of 1    6  2 4 6 8  1  x 
 1  x  3.  (y  2)2   1 
 1  7 6  1,  4,  Continued Fractions  50.    p : q  12-3B Graphing Calculator Exploration:  16  5 6  2 2,  8   h 
 6, k 
 2, a 
 2, b 
 1, c 
 5 center: (6, 2) foci: (6 5 , 2) vertices: (8, 2) and (4, 2)  2  1, 1,  55. If b 
 1, then 4b  26 
 30 which is divisible by 2, 5, and 6. If b 
 11, then 4b  26 
 70, which is divisible by 7. 4b  26 is not divisible by 4 since 4 divides 4b evenly, but does not divide 26 evenly. The correct choice is B.  48. a16 
 1.5  (16  1)0.5 
9 49. x2  4y2  12x  16y 
 16 (x  6)2  4(y  2)2 
 4  2 3  360˚  possible rational zeros,  2, 13, 9, 2 7  (x  6)2  4  180˚  54. possible values of p: possible values of q:  16  1  O 1  9
3 
 27 8 2  y  53.  47. 3
3 
 2, 2
3 
 13, 13
3 
 9, 2  2  2   4   2  2   17.3032181(0.864605)n 15.0 12.9 11.2 9.7 8.4 7.2 6.2 5.4 4.7 4.0  15.0, 12.9, 11.2, 9.7, 8.4, 7.2, 6.2, 5.4, 4.7, 4.0 The 2000–2001 school year corresponds to the 9th term of the sequence, 4.7. The model is 0.3 below the actual statistic. The 2006–2007 school year would correspond to the 15th term of the sequence. 17.3032181(0.864605)15  2.0 Yes; as n → , 17.3032181(0.864605)n → 0. No, the number of students per computer must be greater than zero.  46c.  1  2  2    40  202  ft or about 68 ft  46b.  1  cos 225°      2  
  20  102  
  2 1  4  n 1 2 3 4 5 6 7 8 9 10  20˚  52. cos 112.5° 
 cos 
2    
1  22   46a.  125 miles  cannot be  1  1  1 1     1  5  is  2 .  0  11 6 4 3  3 2  5 3  413  Chapter 12  11. Answers will vary. Sample answers: A 
 1, B 
 14; A 
 4, B 
 1. A2  B: If A 
 1 and B 
 14,  12  1 4 
 15 .  1 1 4. Let x 
 3   , then x 
 3  x. 1   3 3  Solve for x. x 
 3  1x  42  (1)  
 15. If A 
 4 and B 
 1,   x2 
 3x  1 x2  3x  1 
 0 x
 
  3  (3)2   4(1 )(1)  2(1) 3  13   2  Pages 790–791  Since the sum of positive numbers must remain positive, x 
  See students' work. See students' work. See students' work. In a given trigonometric series where r  1, each succeeding term is larger than the one preceding it. Therefore, the series approaches  and thus does not converge. 2. As n → , S → 6.  1 1 6. Let x 
 A   , then x 
 A  x. 1   A A  Solve for x. 1 x 
 A  x  
  sn  3a.  x2 
 Ax  1  Ax  1 
 0  x
  1.6 1.4 1.2 1 0.8 0.6 0.4 0.2  A   4(1 )(1)  2(1) (A)2  A2   4 A   2  Since the sum of positive numbers must remain positive, x 
  A   A2   4  2  O 1 2 3 4 5 6 7 8  7. Sample program: Program: CCFRAC :Prompt A :Prompt B :Disp "INPUT TERM" :Disp "NUMBER N, N  3" :Prompt N :1 → K :B  1/(2A) → C :Lbl 1 :B  1/(2A  C) → C :K  1 → K :If K N  1 :Then: Goto 1 :Else: Disp C  A 8. For large values of N, the program output and the decimal approximation of  A2   B are equal. B 9. x  A 
 2A   B  2A   2A  x  A 
 2A   n  3b. convergent 3c.  n2  3n  3d. We can use the ratio test to determine whether the series is indeed convergent. (n  1)2  n2   an 
 3n and an1 
  3n1  r  (n  1)2   3n1 
 lim — n2 n→  3n  r 
 lim  n→  r 
 lim  n→  (n  1)2   3n1  3n   n2  n2  2n  1  3n2 2 1 1  n  n  r 
 lim  3 n→ 1  r 
 3  B  xA  Since r  1, the series is convergent.  4. Consider the infinite series an 1. If the lim an 0, the sum of the series does not  (x  x2  2Ax  A2 x2 x  
 2A(x  A)  B 
 2Ax  2A2  B 
 A2  B 
  A2   B B Since the sum A   cannot be B  2A   2A  A)2  n→  exist, and thus the series is divergent. If the lim an 
 0, the sum may or may not exit and  n→  therefore it cannot be determined from this test if the series is convergent or divergent. 2. If the series is arithmetic then it is divergent. 3. If the series is geometric then the series converges for r 1 and diverges for r  1. 4. Ratio test: the series converges for r 1 and diverges for r  1. If r 
 1, the test fails. This test can only be used if all the terms of the  B negative, the value of A   is B  2A   2A    A2  B .  10. They will be opposites. Chapter 12  Check for Understanding  1a. 1b. 1c. 1d.  3   13  . 2  5. The output and the decimal approximation are equal.  x2  Convergent and Divergent Series  12-4  414  series are positive and if the series can be expressed in general form. 5. Comparison test: may only be used if all the terms in the series are positive.  900  1500  1500
5 S 10 
 ——— 3 1  5 3727 m 1500 12b. S 
  3 1  5 3 10  n1  n   5. an 
 2n , an1 
  2n1  r 
 lim  n→  an1  an  n1   2n1  
 lim n n→ 2 n (n  1)2n  
 lim n2n 1 n→ (n  1) 
 lim 2n n→ 1 1 
 lim 2  2n n→ 1 
 2  
  
 3750 m No, the sum of the infinite series modeling this situation is 3750. Thus, the spill will spread no more than 3750 meters.   Pages 791–793  4n  1  4(n  1)  1   6. an 
 4n, an1 
  4(n  1)  r  4(n  1)  1  4(n  1) 
 lim —— 4n  1 n→  4n (4n  3)(4n)  
 lim (4n  4)( 4n  1) n→ 16n2  12n   
 lim 16n2  12n  4 n→ 16n2 12n   n2  n2  r  test provides no information (n  1)  7. The general term is n. 1   n for all n, so divergent 1   9. The general term is  2  n2 .  10. an 
 r  10  
 5 or 2  for all n, so convergent  1  n  2n ,  2n1  5(n  1) 
 lim — 2n n→  5n 2n  2  5n  
 lim  2n  (5n  5) n→ 10n  
 lim  n→ 5n  5 10n  n  
 lim — 5n 5 n→    n  n  8. The series is arithmetic, so it is divergent.   2n1  2n  16  1  n  4   3n1 — 
 lim 4 n→  3n 4  3n 
 lim   n n→ 4  3  3 1 
 3   14. an 
 5n, an1 
  5(n  1)  
 1 6 or 1  1  2  n2  4  convergent  
 lim —— 4 16n2 12n n→     n2  n2  n2  n1  n  Exercises  4   13. an 
 3n , an 1 
  3n1  convergent  r  3    12a. The series is geometric where r 
  1500 or 5 .  divergent 2n  1   an1 
  (n  1)2n1  2n1   15. an 
 n2 , an1 
  (n  1)2  1  (n  1)2n1 
 —— 1  n2n n2n  
 lim  n1 n→ (n  1)2 n  
 lim  n→ 2(n  1) n  n  r  2n1  (n  1)2 
 lim —— 2n n→  n2 2n  2  n2   n 2 n→ 2 (n  2n  1) 2 2n  
 lim  2 n→ n  2n  1  
 lim  2n2  n2  
 lim —— n 1 n→ 2
n  n  
 lim —— n2 2n 1 n→     n 2  n2  n 2 
2 divergent  1   
 2(1  0) 1  
 2 convergent 3  11. The series is geometric where r 
 4. Since  3  4  1, it is convergent.  415  Chapter 12  2  1   24. The general term is  2n  1 .  2    16. an 
  (n  1)(n  2) , an1 
 (n  2)(n  3)  r  1  2n  1  2  (n  2)(n  3) 
 lim —— n→ 2 (n  1)(n  2)  
 lim  n→  3  25. The series is geometric where r 
 4. 3  Since 4  2(n2  3n  2)   2(n2  5n  6) n2  3n  2n  1  2  2n  1  2n  1  17. an 
  1  1  5  n2  n→  1  1   n  2n  1  r  1  
 lim  n→ (2n  1)(2n) 1 
 lim  2  n→ 4n  2n  
0 convergent  5n1  1  2  ...  (n  1)  5n  r  2  convergent 31a. No, MagicSoft let a1 
 1,000,000 to arrive at their figure. The first term of this series is 1,000,000  0.70 or 700,000. 31b. The series is geometric where a1 
 700,000 and r 
 0.70. 700,000  S
 1  0.70  2(n  1)  2(n  2)  2n1  2(n  1)  2(n  2)  2n1 2n  2(n  1) n→  2n 2(n  1)  2(n  2)  2n 
 lim  2n  2(n  1)  2n  2 n→ n2 
 lim 2n n→ n 2 
 lim 2n  2n n→ 1 
 2  
 $2.3 million  r 
 lim  1  1  1   n for all n, so convergent 1   21. The general term is  n3  1 . 1   n3  1  1   n for all n, so convergent n   22. The general term is  n  1. n  n1  1   n for all n, so divergent 5   23. The general term is  n  2. 5  n2  1   n for all n, so divergent  Chapter 12  1  33a. Culture A: 1400 cells, Culture B; 713 cells Culture A generates an arithmetic sequence where a1 
 1000, d 
 200 and n 
 8 a8 
 1000  (8  1)200 
 2400 Only considering cell growth, there are 2400  1000 or 1400 new cells. Culture B generates a geometric sequence where a1 
 1000 and r 
 1.08 a8 
 1000(1.08)81 
1713 A part of a cell cannot be generated. Only considering cell growth, there are 1713  1000 or 713 new cells. 33b. Culture A: a31 
 1000  (31  1)200 
 7000 7000  1000 
 6000 cells Culture B: a31 
 1000(1.08)311  10,062 10,062  1000 
 9062 Culture B; at the end of one month, culture A will have produced 6000 cells while culture B will have produced 9062 cells.    20. The general term is  (2n)2 or 4n2 . 1  4n2  1  32. the harmonic series: 1  2  3  4    convergent 1  1  
 4 or 2  
0 convergent  2n  2(n  1)  2(n  1)  1   2n2 —— 
 lim 2n  1 n→   2n1 (2n  1)  2n  2 
 lim  (2n  1)  2n  2  2 n→ 2n  1  
 lim  n→ 4n  2 1 2n    n n  
 lim — 4n 2 n→    n  n  5n1  
 lim 1  2  ...n (n  1) 5 n→  1  2  ...  n 5n  5(1  2  ...  n) 
 lim  5n(1  2  ...  (n  1)) n→ 5  
 lim  n→ n  1  19. an 
 2n, an1 
  2(n  1)  1     30. an 
  2n1 , an1 
 2n  2  (2n  1)(1  2  ...  (2n  1))  (2n  1)(1  2  ...  (2n  1))   18. an 
  1  2  ...  n , an1 
  1   n for all n, so divergent  29. The series is arithmetic, so it is divergent  2n  1  1  2  ...  (2n  1)  n→  
 lim  an1 
  1   n for all n, so convergent  . 28. The general term is  n   2(n  1)  1  1  2  ...  [2(n  1)  1]  2(n  1)  1  1  2  ...  [2(n  1)  1]  r 
 lim  1   n for all n, so divergent   27. The general term is  5  n2 .  
1 test provides no information 2n  1  , 1  2  ...  (2n  1)  1, it is convergent.   26. The general term is  2n  1 .   n2  n2 
 lim —— 2 n 5n 6 n→     n 2  n2  n 2  n2  1   n for all n, so convergent  416  34a.  u 41. AB 
 5  8, 1  (3) 
 3, 2  2n  1   2n1 1   n1 n→ 2 n 2(2  1) 
 lim 2 n n→ 2 
 lim 2  2n n→  34b. S 
 lim Sn 
 lim n→  2n  
  42. List all cubes from 1 to 200. There are five. The correct choice is E.  n 1 2 3 4 5    
 2 seconds 35a. When the minute hand is at 4, 20 minutes have  passed. This is 20
6 0  or 3 hour. 1  35b.  35c.  1  1 1  the distance between 4 and 5 is (5) or 3 3 1 5 5 1  minutes. This is   or  hour. 36 3 3 60 1 1 65    (5) 
  minutes 36 3 36 5 65 5    
  minute 36 36 3 5 1 1   
  hour 36 60 432 1 1 1 785      (5) 
  minutes 36 432 3 432 5 785 65    
  minute 432 432 36 5 1 1   
  hour 432 60 5184 1 1 ,  432 5184 1 1 1 1   The sequence 3, 3 6 , 432 , 5184 is geometric 1 where r 
 1 2. a1  lim Sn 
 S 
  1 r n→ 1  3  
 
  
     
   Page 793 20  2. S20 
 2[2(14)  (20  1)6] 
 860 3. 189 
 56r41    27  
 r3 8 3  
 r 2 3 56 2 
 84,  
   
  6  3  3(2)8   S8 
  1  (2)  
 255 5. lim  n→  hour  36. lim  n→  5   n2 
 lim — 2 3 n 2n n→     n2 n2  2  2   137.5  1  1  25  S
 — 1 1  1 0 2  
 4 5 1  2   n  1 2 …  (n  1)  , a n 8. an 
  n1 
 10n 10 1 1  2  …  (n  1)  10n1 r 
 lim 12…n n→  10n 10n[1  2  …  (n  1)] 
 lim  10n  10(1  2  …  n) n→ n1 
 lim 10 n→ n1 As n → , 10 → . Since  1  
 2 r cos v  2 r sin v 1  1   7. a1 
 2 5 , r 
 10  91  
 162  38. 19 
 11  (7  1)d 30 
 6d 5
d 11  5 
 6, 6  5 
 1, 1  5 
 4, 4  5 
 9, 9  5 
 14 11, 6, 1, 4, 9, 14, 19 39. 45.9 
 e0.075t ln 45.9 
 0.075t 51.02 t 40. 6 
 12r cos (v  30°) 1  
 r(cos v cos 30°  sin v sin 30°) 2  3  5  861 ft  or 2    3  2n   n2  n2 
 lim —— n2 1 n→   n 2  n2  137.5  a9 
 2 
2   1  2  n2  n2    Sn 
 250   1  0.55  1  0.55  4  
 3 37. r 
  n2  2n  5   n2  1  
1 6. a1 
 250, r 
 0.55 a2 
 250(0.55) or 137.5 a3 
 137.5 a4 
 137.5(0.55) or 75.625 a5 
 75.625   4  4n2  n2  3  4. r 
 3 or 2  The hands will coincide at 4  11 o'clock, approximately 21 min 49 s after 4:00. 4n2  5   3n2  2n  84
2 
 126  56, 84, 126, 189  
 — 1 1  1 2 4  11  Mid-Chapter Quiz  1. a12 
 11  (19  1)(2) 
 25  
   35d.  n3 1 8 27 64 125  r  1, the series is  divergent.  1  0 
 2x  2 y  2  1  9. The series is geometric where r 
 3.  0 
 3 x  y  1  Since r  417  1, it is convergent.  Chapter 12  60  10. 500
1  4 
 515 0.12  13a.  a1 
 515, r 
 1.03 n 
 4 S4 
   389(0.63)n1 n
1 S60 
  515  515(1.03)4  1  1.03  389  389(0.63)60  1  0.63  1051 ft  
 $2154.57  13b. S 
  389  1  0.63  1051 ft  Sigma Notation and the nth term.  12-5  Pages 798–800  Exercises  4  Pages 797–798  14.  Check for Understanding   (2  4  7) 
 (5)  (3)  (1)  1 
 8 5 15.  5a 
 5(2)  5(3)  5(4)  5(5) a
2 
 10  15  20  25 
 70 8 16.  (6  4b) 
 (6  4  3)  (6  4  4)  (6  4  5) b
3  (6  4  6)  (6  4  7)  (6  4  8) 
 (6)  (10)  (14)  (18)  (22)  (26) 
 96 6 17.  (k  k2) 
 (2  22)  (3  32)  (4  42)  k
2 (5  52)  (6  62) 
 6  12  20  30  42 
 110  1. The series 4  6  8  10  12 can be 4  5  n
0  n
1  represented by  2n  4 or by  2n  2. Sample answer: (1)n1 Sample answer: (1)n 9; 2, 3, 4, 5, 6, 7, 8, 9, 10 t
ba1 t
ba1 
 3  (2)  1 
6  2a. 2b. 3a. 3b. 3c.  3  3d.         k  3 
 2  3  1  3  0  5  1  3  2  3 k
2 1  1  1  1  1  1  1    33 1  1  1  1  1  1  
 1  2  3  4  5  6 
1  1  2    1  3    1  4    1  5     (2n  7) 
 (2  1  7)  (2  2  7)  (2  3  7)  n
1  8  18.  1  5        n4 
 54  64  74  84 n
5 n    a
0    7.    p
0  1  1  
 123  8  1  1  
1 
 3 p  15 11 6    1  4  3 0    1  8    8.  1  21.   (8  6n) n
1     5
  9.   (3k  1) k
0  22.  1  1  1   (0.5)i 
 (0.5)3  (0.5)4  (0.5)5  
 8  16  32 
 56 23.  k! 
 3!  4!  5!  6!  7! k
3 
 6  24  120  720  5040 
 5910 10 24.  4(0.75)p 
 4(0.75)0  4(0.75)1  4(0.75)2  p
0 4(0.75)3    4(0.75) 
 4  3  2.25  1.6875    4(0.75) 4  S
 1  0.75    7  3    1  
 852  5  3   4    11.  1  1  3 3   3
2n n
2 1  
 16   (3)n  n
1  Chapter 12  1  1    12.   
2  4r 
 
2  41  
2  42  
2  43 r
1 
 42  162  642  135  64  
 20   5n n
1  
 133  i
3  4  10.  1  3  3 2  
    3m1 
 301  311  321  331 m
0 
 3  1  3  9  5
4 
 5
4  5
4  5
4  5
4   3 10  5 4 15 45 
 5  4  16  
 16  32  64  128  256 
 496  3  20.  1  16  3 1   2j 
 24  25  26  27  28  j
4  1  5  Sn 
 3 1  4 5  19.  
 20  21  22  23  24 1  2  8  1   (n  3) 
 (1  3)  (2  3)  (3  3)  n
1 (4  3)  (5  3)  (6  3) 
 (2)  (1)  0  1  2  3 
3 5 5.  4k 
 4  2  4  3  4  4  4  5 k
2 
 8  12  16  20 
 56 4.  1  2a  7  7  There are 6 terms.  4  6  
 5  3  3  2  6  6.  5  418    25.  2 n   4
5 n
1  2 1  2 2  2   2 3  47b. 0  1(1  x)  2(2  x)  3(3  x)  4(4  x)  5(5  x) 
 25 1  x  4  2x  9  3x  16  4x  25  5x 
 25 55  15x 
 25 15x 
 30 x
2  
 4
5  4
5  4
5    4
5 
  8  5    16  25    32  125  2   5     4
    8  5  S
 — 2 1  5 
  5  26.    8  3  or  n  in  n
2  4  27.  7  48a. false;  3k 
 33  34    37  2 23  
 (2   (3   (4   (5  
 (1)  (3  i)  (5)  (5  i) 
 14 i3)   (3k  3) k
1   28. 30.  2k  10  32.    5k  1 k
2 1  34.    35.   (1)kk2  36.     (1)n1 2 n
0 32 n    39.    k   2k  3    k 22  k
1  41.    k
11  42.    38.    40.    k
2  2k  7  48c. true; 2 7  5  k!    (k  1)!  44.  
 18  32    98 
 270  2n2 
 18  32    98 
 270  9  7  7  n
3  n
3  k
1   (4  p) 
 4  5    13 
 85  p
0  10  Since 85  105,  (5  n) k
1  9   (4  p). p
0  49a. 49b. 49c. 50a.  (a  1)(a)(a  1)(a  2)!  (a  2)!  
 a(a  1)(a  1) 
  n
3  48d. false;  (5  n) 
 6  7    15 
 105  1 k 3 k!  1  (a  b)!  (a  b  1)!  n
2  n
3  10  k2   
 a(a  1)  
  n2  Since 270 
 270, 2 n2 
  2n2.  (a  2)!  (a  1)!  (a  2)!  9  n
3   
 a(a  1)(a  2)!  43.  8  Since 49 
 49,  (2n  3) 
  (2m  5).   3k!  (a  2)!  a!  (2m  5) 
 1  3    13 
 49  m
3   k k
1   b
7  n
2  9    2k  1 k
1    k
3  8   (13  4k) k
0  k
0  9  48b. true;  (2n  3) 
 1  3    13 
 49   (2)3k    k
2  37.  9   3a. a
3   4k k
0    7  Since there are two 37 terms,  3k   3b  3   2  5k k
1  3b 
 37  38  39  b
7  k
0  4  33.    i5)  3  k
4  31.  i4)  4  12  29.  k
3  9  i2)  (a  b)(a  b  1)!  (a  b  1)!  
 a  45. 43.64  6! 5! or 120 4! or 24, "LISTEN" On an 8  8 chessboard, there is 1-8  8 square. On an 8  8 chessboard, there are 4-7  7 squares, one in each of the four corners. 50b. For the 6  6 squares, begin in one corner. For different configurations, you can move it over, up to 2 more spaces, or down, up to 2 more spaces. Thus, there are 3  3 or 9-6  6 squares. Continue this procedure for the other sizes of squares. 9-6  6, 16-5  5, 25-4  4, 36-3  3, 49-2  2, and 64-1  1 8  50c.  
 1  4  9  16  25  36  49  64 
 204    46a.   500,000(0.35)n n
1  3  175,000   51. The general term is  n  2.   46b. S 
  1  0.35  3  n2  
 269,239 people 46c.  269,230  500,000   n2 
 12  22  32  42  52  62  72  82 n
1  1   n for all n, so divergent  52. 0.42(21) 
 8.82 liters If with each stroke 20% is removed, then 80% remains. a1 
 21(0.80) 
 16.8 a2 
 16.8(0.80) 
 13.44 a3 
 13.44(0.80) 
 10.752 a4 
 10.752(0.80) 
 8.6016 It will take 4 strokes for 42% of the air to remain.  53.8%  46d. The ad agency assumes that the people who buy the tennis shoes will be satisfied with their purchase. 47a. (x  3)  (x  6)  (x  9)  (x  12)  (x  15)  (x  18) 
 3 6x  63 
 3 6x 
 60 x 
 60  419  Chapter 12  51  2c. Even indexed terms are negative and odd indexed terms are positive. 3. The sum of the exponents of each term is n. 4. The exponents must add to 12, so the exponent of y is 12  7 or 5. To find the coefficient of the term use the formula  53. 322  
 a1
2  322  
 4a1 82  
 a1  82 
2  
 16, 16
2  
 162 ,  54. 55.  56. 57.  
2  
 32 162 82 , 16, 162 , 32 log10 0.001 
 3 x2  y2  Dx  Ey  F 
 0 (0, 9) → 81  9E  F 
 0 (7, 2) → 49  4  7D  2E  F 
 0 (0, 5) → 25  5E  F 
 0 9E  F 
 81 9(4)  F 
 81 5E  F 
 25 F 
 45 14E 
 56 53  7D  2(4)  45 
 0 E 
 4 7D 
 0 D
0 x2  y2  4y  45 
 0 x2  (y  2)2 
 49 (2   i)(42   i) 
 8  52 i  i2 
 7  5i2  vy 
 59 sin 63° 52.57 ft/s 59 ft/s vx 
 59 cos 63° 63˚ 26.79 ft/s 2x1  3y1  9  n  (x  y)n 
   r
0  Evaluate the general term for n 
 12 and r 
 5. 5.  35  6  5  4  3  2  a 12345 36  6  5  4  3  2  1   123456 
 a6  18a5  135a4     1458a  729    3  2  5  (y)2  3  2  1  50 (y)3  321  
 125  75y  15y2  y3 8. (3p  2q)4 
 (3p)4 (2q)0  4(3p)3 (2q)1 4  3(3p)2 (2q)2  4  3  2(3p)(2q)3      321 21  x1 4y1 4    17  4  3  2  1(3p)0(2q)4  4321 81p4  216p3q  216p2q2   
 9.  2   16q4   96pq3  7! (a)7r(b)r r!(7  r)! 7! 75(b)5  
 5!(7  5)! (a) 7654321  2 5 
   5432121 a b  
 21a2b5 10.  The Binomial Theorem  r 9! (x)9r 3  r!(9  r)! 3 9! 93 3  
  3!(9  3)! (x)  
  Check for Understanding  
   
    987654321  321654321  x6  33   
 2523  x6  1a. n 
 0: 1 n 
 1: 1  1 or 2 n 
 2: 1  2  1 or 4 n 
 3: 1  3  3  1 or 8 n 
 4: 1  4  6  4  1 or 16 n 
 5: 1  5  10  10  5  1 or 32 1, 2, 4, 8, 16, 32 1b. 2n 2a. The second term of (x  y)3 is 3x2y. It is negative. The second term of (x  y)4 is 4x3y. It is negative. The second term of (x  y)5 is 5x4y. It is negative. 2b. The third term of (x  y)3 is 3xy2. It is positive. The third term of (x  y)4 is 6x2y2. It is positive. The third term of (y  y)5 is 10x3y2. It is positive. Chapter 12  540a3  1215a2    7. (5  y)3 
 53(y)0  3  52(y)   21  
 3  Pages 803–804  32  6  5   4  12 a  34  6  5  4  3  3(5  m) 
 2(9  m) 15  3m 
 18  2m m
3 The correct choice is D.  12-6  3  10c2d3  5cd4  d5  6a5   2  1234 a    13 x  
413  
217 317 y  917   413 
0 5m  9m  
  a6  33  6  5  4  217 x1  317 y1  917 
 13 x1  413 y1  413   59.  3)6   3  123 a  x1  4y1  4  
  12!  x7y5 
 792x7y5. 5!7! c5  5c4d  10c3d2   6. (a   , d 
  58. d1 
  2  13  17 2x1 3y1 9   1 3  n! xnryr. r!(n  r)!  11. (H  T)5 
 H5  5H4T  10H3T2  10H2T3  5HT2  T5 11a. 1 11b. 10 11c. 1  5 or 6 11d. 10  10  5  1 or 26  Pages 804–805  Exercises  12. (a  b)8 
 a8  8a7b  28a6b2  56a5b3  70a4b4  56a3b5  28a2b6  8ab7  b8 6 13. (n  4) 
 n6  24n5  240n4  1280n3  3840n2  6144n  4096 14. (3c  d)4 
 81c4  108c3d  54c2d2  12cd3  d4 15. (2  a)9 
 512  2304a  4608a2 5376a3  4032a4  2016a5  672a6  144a7  18a8  a9  420  7  6  d5  22  8  7(p2)6(q)2   16. (d  2)7 
 d7  20  7  d6  21   21 7  6  5  d4  23    321   24. (p2  q)8 
 (p2)8(q)0  8(p2)7(q)1   21  7  6  5  4  d3  24  4321    7  6  5  4  3  d2  25  54321    7  6  5  4  3  2  d1  26  654321    7  6  5  4  3  2  1  d0  27  7654321  8  7  6(p2)5(q)3  8  7  6  5  4(p2)3(q)5  54321 8  7  6  5  4  3(p2)2(q)6   654321 8  7  6  5  4  3  2(p2)1(q)7   7654321 8  7  6  5  4  3  2  1(p2)0(q)8   87654321    
 d7  14d6  84d5  280d4  560d3  672d2  448d  128 17. (3   x)5  
  35(x)0  5    5  4  3  2  1  30(x)5   54321 243  405x  270x2  
  
 p16  8p14q  28p12q2  56p10q3  70p8q4  56p6q5  28p4q6  8p2q7  q8 3 6 25. (xy  2z ) 
 (xy)6(2x3)0  6(xy)5(2z3)1  5  4  33(x)2   21 5  4  3  2  31(x)4  4321  34(x1)1  5  4  3  32(x)3   321   90x3  15x4  x5 4    3(4a)2(b)2  43 4  3  2  1(4a)0(b)4     4321 321 
 256a4  256a3b  96a2b2  16ab3    2(4a)1(b)3    b4  
 x6y6  12x5y5z3  60x4y4z6  160x3y3z9  240x2y2z12  192xyz15  64z18  19. (2x  3y)3 
 (2x)3(3y)0  3(2x)2(3y)1 3  2(2x)(3y)2  
    8x3  4  36x2y    3  2  1(2x)0(3y)3  321  54xy2  0    26.  27y3  27.  4  3  2(3m)1(2  )3  321  m3  108m2  
 81m4  1082 242 m  4 6  0  6  5(c)4(1)2  6  5  4(c)3(1)3  321 6  5  4  3(c)2(1)4  4321 6  5  4  3  2(c)1(1)5  54321 6  5  4  3  2  1(c)0(1)6  654321    21     5     1  3  4  5  23. 
3a  3b 
 (3a)4
3b  4(3a)3
3b 2  0  2  
   1  
   2 2 2 3 4  3(3a)2 3b 4  3  2(3a)1 3b    21 321 2 4 0   4  3  2  1(3a) 3 b   4321 
 81a4  72a3b  24a2b2     
   32  9  9! (3c)96(2d)6 6!(9  6)!  30.  1 10! (x)107(y)7 7!(10  7)! 2  a5  22   
 35  16a4  (b3) 
 560a4b3  
 84  27c3  64d6 
 145,152c3d6 1  
 120  8x3  (y7)  11! (2p)115(3q)5 5!(11  5)!  84  4  
y  
 70x2y2  33. (M  W)8 
 M 8  8M 7W  28M 6W 2  56M 5W 3  70M 4W 4  56M 3W 5  28M 2W 6  8MW 7  W 8 70  56  28  8  1 or 163 34. Sample answer: Treat a  b as a single term and expand [a  b)  c]12 using the Binomial Theorem. Then evaluate each (a  b)n term in the expansion using the Binomial Theorem. 35. (T  F)12 
 T12  12T11F  6610F2  220T9F3  495T8F4  792T7F5  924T6F6  792T5F7  495T4F8  220T3F9  66T2F10  12TF11  F12 35a. 495 35b. 924  792  495  220  66  12  1 or 2510 36. Find the term for which both x's have the same exponent. This will occur for the middle term of the expansion, the 4th term when n 
 6. Use the  
   4  29.  
x  5 3 2  4 
 3 2 n  8 n  5n  20n  40n  32 2  7! (2a)73(b)3 3!(7  3)!  8!  4!(8  4)!     321 4321 1 0 5   5  4  3  2  1 2 n (2)  54321  1    
 462  64p6  (243q5) 
 7,185,024p6q5 32. The middle term is the fifth term.  
   4  2  
  28.  31.  5  4 2n (2)2  21  
12n  2 
 
12n (2)0  5
12n (2)1  1 1 5  4  3
2n (2) 5  4  3  2
2n (2)  x5y4  
 15x3y7  
 c3  6c2c  15c2  20cc  15c  6c  1 1 3 22.  3 8! (a)83 2  3!(8  3)! 87654321 
  32154321 
 1122 a5  5  21. 
c  1 
 
c (1)  6
c (1)1    987654321  432154321  
  
 126x5y4   )4 4  3  2  1(3m)0(2  4321    6    9! (x)94(y)4 4!(9  4)!  1  20. 
3m  2  
 (3m)4
2   4(3m)3
2   )2 4  3(3m)2(2  21  6  5  4(xy)3(2z3)3  321 6  5  4  3(xy)2(2z3)4  4321 6  5  4  3  2(xy)1(2z3)5  54321 6  5  4  3  2  1(xy)0(2z3)6  654321  6  5(xy)4(2z3)2    21   18. (4a  b)4 
 (4a)4(b)0  4(4a)3(b)1   21    21  8  7  6  5(p2)4(q)4      4321 321  16  ab3  81b4  421  Chapter 12  Binomial Theorem to find the 4th term for the 1 6 expansion of 3x  4x .  
    3  
41x   6! (3x)3 3!3!  12-7  135  
 16  Page 809  37a. Sample answer: 1  0.01 37b. Sample answer: 1.04060401 (1  0.01)4 
 14  4  13  0.011  6  12  0.012  4 11  0.013  0.014 
 1.04060401 7   5  2k 
 (5  2  2)  (5  2  3)  (5  2  4) x
2   (5  2  5)  (5  2  6)  (5  2  7) 
 1  (1)  (3)  (5)  (7)  (9) 
 24  2n  x11   6.  11!  2n1   39. an 
 n! an1 
  (n  1)!  r  Pages 811–812  2n1  (n  1)! 
 lim —— 2n n→  n!  
 lim  n→  2n  2  n!   2n  (n  1)  n!  2  n→ n  1  
0 convergent 1  40. This is a geometric series where r 
 2.  
 41.  2  3 — 1 1  2 1 13  Pn 
 P  n  1  (1  i)   i  150,000 
 P    
    0.08 12(30)  1 1 12 ——— 0.08  12    (0.8)2  (1.36)2  (1.36)3  (1.36)4  1  1.36  0.925  0.419  0.143 3.85 9. sin x  r  0.3183098862 mi r  0.3183098862(5280) 1681 feet 44. Test all answer choices that are prime integers. You can eliminate answer choice C. ? ? 5 A: 3(2)  10  6 (2)  sin   x3  x5  x7  x9  x  3!  5!  7!  9! sin 3.1416 (3.1416)3 (3.1416)5 (3.1416)7 3.1416  3  5  7 ! ! ! (3.1416)9   9 ! 3.1416  5.1677  2.5502  0.5993  0.0821  0.0069 3  3   3  
cos 4  i sin 4 
 2  ei 4 10. 2  ?   Chapter 12  (0.8)4   3  4 8. e1.36 
 1  1.36  2 ! ! !  
4  ? 5 6 10  3 ; false ? ? 5 B: 3(3)  10  6 (3) ? ? 5 10  9 2 ; false ? ? 5 D: 3(11)  10  6 (11) 55 ? ?   10  33  6 ; true  (0.8)3    7. e0.8 
 1  0.8  2 !  3!  4! 1  0.8  0.32  0.08  0.017 2.22  150,000 
 P(136.283491) $1100.65 
 P u  (8  6)j u  (2  3)k u  
 [4  (2)]i 42. MK u u u 
 6i  2j  5k 43. s 
 rv 0.25 
 r  Check for Understanding  1. The approximation given in Example 1 only used the first five terms of the exponential series. Using more terms of the exponential series would give an approximation closer to that given by the calculator. 2. Sample answer: 2  x  1.5 3. The problem seems to imply that siblings mate. Genetically, this can lead to problems. Another problem is the assumption that each birth produces only two offspring, one male and one female. Rabbits are more likely to give birth to more than two offspring and the ratio of male to female births is not guaranteed to be 1 to 1. 4. an1 
 an  an1 for n  2 5. ln (7) 
 ln (1)  ln (7) i  1.9459 6. ln (0.379) 
 ln (1)  ln (0.379) i  0.9702  
 lim  S
  Graphing Calculator Exploration  1. Sample answer (without zooming): 2.4  x  2.4 2. Sample answer for greatest difference: about 0.08; The least difference is 0. 3. Sample answer: 3.4  x  3.4; sample answer: about 0.15; 0 4. Sample answer: 3.8  x  3.8; sample answer: about 0.05; 0 5. larger  37c. 1.04060401; the two values are equal. 38.  Special Sequences and Series   3  11. 1  3 i 
 2
2  2i 1  2  2  
 2
cos 3  i sin 3 
 2e  The correct choice is D.  422  2 i 3  12a.  A 
 Pert 2P 
 Pe(0.06)5 P 
 e0.3    cos 6  (0.5236)2  0.2742    (1.1)3  3!    (0.2)2    (4.2)3    cos   x6  33. r 
  34. r 
    sin 4  7  7  7   35. r 
  32  32 or 32    v 
 Arctan 
3 or 4 3  (2.73)4         3  3i 
 32 
cos 4  i sin 4 
 32  ei 4 36. Sample answer: A transcendental number is one that cannot be the root of an algebraic equation with rational coefficients. Examples are  and e. (3.1416)8  eix  eix  2i  
 
  cos x  i sin x  (cos x  i sin x)  2i 2i sin x  2i  
 sin x  cos x  i sin x  cos x  i sin x eix  eix  
  2 2 2 cos x 
 2  x9  x  3!  5!  7!  9!  
 cos x 38. See students' work.  sin 0.7854 (0.7854)3  7  43   4i 
 8
cos 6  i sin 6 
 8ei 6  37.  x7   2  3 (4)2 or 8 
4 4  1  4.9348  4.0588  1.3353  0.2353 0.9760 actual value: cos  
 1 x5  3   3     or  v 
 Arctan 
 6 4 3  cos 3.1416  x3    3   4  6  8 1  2 ! ! ! !  26. sin x  2 2 
 2  2  or 2   
  2   2 i 
 2
cos 4  i sin 4 
 2ei 4  (3.5)4  4!  (3.1416)6        2  x8  (3.1416)4     or 3 v 
 Arctan 
 4 2   1  2!  4!  6!  8! (3.1416)2       3   i 
 2
cos 6  i sin 6 
 2ei 6  1  2.73  3.726  3.391  2.314 13.16 x4         (0.2)4   3  4 24. e2.73 
 1  2.73  2 ! ! !  x2  5   5  32. r 
  (3 )2  12 or 2 1  v 
 Arctan  or 6 3  (0.55)4  (2.73)3  sin 1.5708    1  3.5  6.125  7.146  6.253 24.02 (2.73)2  x9  1  i 
 2 
cos 4  i sin 4 
 2  ei 4  1  0.55  0.151  0.028  0.004 1.73   0.8660  x7  31. r 
  12  12 or 2  1  Q 
 Arctan 1 or 4   3  4 22. e0.55 
 1  0.55  2 ! ! !  (3.5)3  3!  0.0056  1.5708  0.6460  0.0797  0.0047  0.0002    (4.2)4  (0.55)3  0.0206  30. i 
 cos 2  i sin 2 
 ei 2  1  4.2  8.82  12.348  12.965 39.33 (0.55)2  x5  5   3  4 21. e4.2 
 1  4.2  2 ! ! !  25. cos x  (0.5236)8  29. 5
cos 3  i sin 3 
 5ei 3  1  0.2  0.02  0.0013  0.00007 0.82 (4.2)2  0.7516  1.0000  actual value: sin 2 
 1   3  4 20. e0.2 
 1  (0.2)  2 ! ! !  23. e3.5 
 1  3.5   (0.5236)6  x  3!  5!  7!  9!  sin 2  (1.1)4  4!  (0.2)3  x3  28. sin x  1  1.1  0.605  0.222  0.06 2.99  (3.5)2  2!  (0.5236)4  0.8660   3 actual value: cos 6 
 2  ln (4) 
 ln (1)  ln 4 i  1.3863 ln (3.1) 
 ln (1)  ln (3.1) i  1.1314 ln (0.25) 
 ln (1)  ln (0.25) i  1.3863 ln (0.033) 
 ln (1)  ln (0.033) i  3.4112 ln (238) 
 ln (1)  ln (238) i  5.4723 ln (1207) 
 ln (1)  ln (1207) i  7.0959  19. e1.1 
 1  1.1   x8    1  2  24   720  40,320  Exercises  (1.1)2  2!  x6  1  2  4  6  8 ! ! ! !   1.345 approximately 1.345P 12b. No, in five years she will have increased her savings by about 34.5%, not 100%. 12c. The approximation is accurate to two decimal places.  13. 14. 15. 16. 17. 18.  x4  cos 0.5236  (0.3)2  2!  
 1  0.3   Pages 812–814  x2  1  2!  4!  6!  8!  27. cos x  (0.7854)5  (0.7854)7  0.7854  3  5  7 ! ! ! (0.7854)9   9 ! 0.4845  0.2989  0.1843  0.1137     0.7854  6   120  5040  362, 880  0.7071   2 actual value: sin 4 
 2  0.7071  423  Chapter 12  39. If you add the numbers on the diagonal lines as shown, the sums are the terms of the Fibonacci sequence.  0.01   45a. a1 
 0.005, r 
  0.005 or 2  a3 
 0.005(2)31 
 0.020 cm a4 
 0.020(2) 
 0.040 cm 45b. 0.005(2)n1 45c. a10 
 0.005(2)101 
 2.56 cm a100 
 0.005(2)1001 3.169  1027 cm  1 1  1 1  2 3  1  5  1  2  1  8 13  1  3  1 1 1  4 5  3 6  4  10  6  15  1  10 20  . . .  1 5  2   46.  1  15  6  Fn  1  2 1.75 1.5 1.25 1 0.75 0.5 0.25  14  48. r 
 6 or 4     (0.65)2    (0.65)3  (0.65)4  42a. 42b.  
  75.5  sin 140°    50a.  6  44.        12xy5    y6  v  
 t 5(2) radians  
 10 radians per second v  50b. v 
 r t 
 
2 ft(10 radians/s) 1  15.7 ft/s 51. Let m 
 multiple choice. Let e 
 essay. m  e  30 1m  12e  96 e f(m, e) 
 5m  20e 30 f(24, 6) 
 5(24)  20(6) m  e 
 30 
 240 m  12e 
 96 20 f(0, 8) 
 5(0)  20(8) 
 160 10 (0, 8) (24, 6) f(30, 0) 
 5(30)  20(0) m
0 
 150 m 10 20 (0, 0) (30, 0) e
0 f(0, 0) 
 0 To receive the highest score, answer 24 multiple choice and 6 essay.   160x3y3  2k  k
1  Chapter 12  30 sin 140°  
 75.5   
 1 second  6  5  4  3  2(2x)1(y)5  54321  60x2y4  30   
 sin v  v 
 Arcsin 
 14°48
     
    30 sin 140°  6  5  4  3(2x)2(y)4  4321  6  5  4  3  2  1(2x)0(y)6  654321 64x6  192x5y  240x4y2  15   sin v 
  75.5  6  5(2x)4(y)2  6  5  4(2x)3(y)3  15  4 cos 8  i sin 8 49. u r 2 
 302  502  2(30)(50) cos 140° u r 
 75.5 N   43. (2x  y)6 
 (2x)6(y)0  6(2x)5(y)1   21   321  15  
 8 or 8  5000 1  0.65  2  3  4 ! ! ! $9572.29 No, the account will be short by more than $30,000. about 42 years; 47 years old 40,000 
 Pe0.65 $20,882 
 P Every third Fibonacci number is an even number. Every fourth Fibonacci number is a multiple of 3.  41c. 41d.    v 
 8  4  n  1 2 3 4 5 6 7 8 9 10  40d. yes; 1.618 40e. The two ratios are equivalent to three decimal places. 40f. See students' work. 41a. A 
 Pert 
 5000e0.05(13) 
 5000e0.65  41b.  1   
 8 3 or 2  8  40b. neither 40c. Fn  O  1  3  47. y2  Dx  Ey  F 
 0 (0, 0): F 
 0 (2, 1): 1  2D  E  F 
 0 (4, 4): 16  4D  4E  F 
 0 2D  E 
 1 4D  2E 
 2 4D  4E 
 16 → 4D  4E 
 16 2E 
 14 2D  E 
 1 E
7 2 2D  7 
 1 y  3x  7y 
 0 2D 
 6 D
3  1  1 2 3 5 8 13 21 34 55 89 , , , , , , , , ,  1 1 2 3 5 8 13 21 34 55  40a.  83   424  2  52. (DC)2  22 
 
18  (DC)2 
 14 DC 
 14   24  10. pn 
 40 
 0.60 p1 
 0.60  1.75(0.60)(1  0.60) 
 1.02 (1.02)(40) 41 p2 
 1.02  1.75(1.02)(1  1.02) 
 0.9843 (0.9843)(40) 39 p3 
 0.9843  1.75(0.9843)(1  0.9843) 
 1.0113 (1.0113)(40) 40 p4 
 1.0113  1.75(1.0113)(1  1.0113) 
 0.9913 (0.9913)(40) 40 p5 
 0.9913  1.75(0.9913)(1  0.9913) 
 1.0064 (1.0064)(40) 40 p6 
 1.0064  1.75(1.0064)(1  1.0064) 0.9951 (0.9951)(40) 40 p7 
 0.9951  1.75(0.9951)(1  0.9951) 1.0036 (1.0036)(40) 40 p8 
 1.0036  1.75(1.0036)(1  1.0036) 0.9973 (0.9973)(40) 40 p9 
 0.9973  1.75(0.9973)(1  0.9973) 1.002 (1.002)(40) 40 p10 
 1.002  1.75(1.002)(1  1.002) 0.9985 (0.9985)(40) 40 41, 39, 40, 40, 40, 40, 40, 40, 40, 40  2   
14  12-8 Page 819   52 
 (BC)2 39 
 (BC)2 39  
 BC The correct choice is C.  Sequences and Iteration Check for Understanding  1. Iteration is the repeated composition of a function upon itself. 2. It is the sequence of iterates produced when a complex number is iterated for a function f(z). 3. If the prisoner set is connected, then the Julia set is the boundary between the prisoner set and the escape set. If the prisoner set is disconnected, then the Julia set is the prisoner set. 4. f(1) 
 (1)2 
 1 f(1) 
 12 
 1 f(1) 
 12 
 1 f(1) 
 12 
 1 1, 1, 1, 1 5. f(2) 
 2  2  5 
 1 f(1) 
 2  (1)  5 
 7 f(7) 
 2  (7)  5 
 19 f(19) 
 2  (19)  5 
 43 1, 7, 19, 43 6. z0 
 6i z1 
 0.6(6i)  2i 
 5.6i z2 
 0.6(5.6i)  2i 
 5.36i z3 
 0.6(5.36i)  2i 
 5.216i 7. z0 
 25  40i z1 
 0.6(25  40i)  2i 
 15  26i z2 
 0.6(15  26i)  2i 
 9  17.6i z3 
 0.6(9  17.6i)  2i 
 5.4  12.56i 8. z0 
 0, f(z) 
 z2  (1  2i) z1 
 02  (1  2i) 
 1  2i z2 
 (1  2i)2  (1  2i) 
 1  4i  4i2  1  2i 
 2  6i z3 
 (2  6i)2  (1  2i) 
 4  24i  35i2  1  2i 
 31  22i 9. z0 
 1  2i, f(z) 
 z2  (2  3i) z1 
 (1  2i)2  (2  3i) 
 1  4i  4i2  2  3i 
 1  i z2 
 (1  i)2  (2  3i) 
 1  2i  i2  2  3i 
 2  5i z3 
 (2  5i)2  (2  3i) 
 4  20i  25i2  2  3i 
 19  23i  Pages 820–821  Exercises  11. f(x0) 
 f(4) 
 3(4)  7 
5 f(x1) 
 f(5) 
 3(5)  7 
8 f(x2) 
 f(8) 
 3(8)  7 
 17 f(x3) 
 f(17) 
 3(17)  7 
 44 12. f(2) 
 (2)2 
 4 f(4) 
 42 
 16 f(16) 
 162 
 256 f(256) 
 2562 
 65.536 13. f(4) 
 (4  5)2 
 1 f(1) 
 (1  5)2 
 16 f(16) 
 (16  5)2 
 121 f(121) 
 (121  5)2 
 13,456 14. f(1) 
 (1)2  1 
 0 f(0) 
 02  1 
 1 f(1) 
 (1)2  1 
 0 f(0) 
 02  1 
 1  425  Chapter 12  19. z0 
 1  2i z1 
 2(1  2i)  (3  2i) 
 2  4i  3  2i 
 5  2i z2 
 2(5  2i)  (3  2i) 
 10  4i  3  2i 
 13  2i z3 
 2(13  2i)  (3  2i) 
 26  4i  3  2i 
 29  2i 20. z0 
 1  2i z1 
 2(1  2i)  (3  2i) 
 2  4i  3  2i 
 1  6i z2 
 2(1  6i)  (3  2i) 
 2  12i  3  2i 
 5  14i z3 
 2(5  14i)  (3  2i) 
 10  28i  3  2i 
 13  30i 21. z0 
 6  2i z1 
 2(6  2i)  (3  2i) 
 12  4i  3  2i 
 15  2i z2 
 2(15  2i)  (3  2i) 
 30  4i  3  2i 
 33  2i z3 
 2(33  2i)  (3  2i) 
 66  4i  3  2i 
 69  2i 22. z0 
 0.3  i z1 
 2(0.3  i)  (3  2i) 
 0.6  2i  3  2i 
 3.6  4i z2 
 2(3.6  4i)  (3  2i) 
 7.2  8i  3  2i 
 10.2  10i z3 
 2(10.2  10i)  (3  2i) 
 20.4  20i  3  2i 
 23.4  22i  15. f(0.1) 
 2(0.1)2  0.1 
 0.08 f(0.08) 
 2(0.08)2  (0.08) 0.09 f(0.09) 
 2(0.09)2  0.09 0.07 f(0.07) 
 2(0.07)2  (0.07) 0.08 2  16a. t1 
 1 
 2 2  t2 
 2 
 1 2  t3 
 1 
 2 2  t4 
 2 
 1  2 t10 
 2 
 1 2  1  16b. t1 
 4 
 2 t2 
 t3 
 t4 
  2  
4 1  2 2 1  
  4 2 2  
4 1  2   t10 
  2  1  2  
4  2  16c. t1 
 7 t2 
 t3 
 t4 
  2  2  7 2  7 2  2  7   t10 
  2  2  7  
7  
7  
7 2  16d. The values of the iterates alternate between x 0 and x0.  1  z1 
 3
3  3i  2i 1  17. z0 
 5i z1 
 2(5i)  (3  2i) 
 3  8i z2 
 2(3  8i)  (3  2i) 
 6  16i  3  2i 
 9  14i z3 
 2(9  14i)  (3  2i) 
 18  28i  3  2i 
 21  26i 18. z0 
 4 z1 
 2(4)  (3  2i) 
 11  2i z2 
 2(11  2i)  (3  2i) 
 22  4i  3  2i 
 25  6i z3 
 2(25  6i)  (3  2i) 
 50  12i  3  2i 
 53  14i  Chapter 12  2  23. z0 
 3  3i 2  
 1  2i  2i 
1 z2 
 3(1)  2i 
 3  2i z3 
 3(3  2i)  2i 
 9  6i  2i 
 9  8i 24. z0 
 0  i, f(z) 
 z2  1 z1 
 (i)2  1 
 2 z2 
 (2)2  1 
3 z3 
 32  1 
8  426  25. z0 
 i, f(z) 
 z2  1  3i z1 
 i2  1  3i 
 3i z2 
 (3i)2  1  3i 
 8  3i z3 
 (8  3i)2  1  3i 
 64  48i  9  1  3i 
 56  45i 26. z0 
 1, f(z) 
 z2  3  2i z1 
 12  3  2i 
 4  2i z2 
 (4  2i)2  3  2i 
 16  16i  4  3  2i 
 15  18i z3 
 (15  18i)2  3  2i 
 225  540i  324  3  2i 
 96  542i 27. z0 
 1  i, f(z) 
 z2  4i z1 
 (1  i)2  4i 
 1  2i  1  4i 
 2i z2 
 (2i)2  4i 
 4  4i z3 
 (4  4i)2  4i 
 16  32i  16  4i 
 28i  2  31.   2   2  2  1  x 
 percent 0.10 0.325 0.8734 1.1498 0.7192 1.2241 0.5383 1.1596 0.6969 1.225 0.5359 1.1577 0.7013 1.225 0.5359 1.1577 0.7013 1.225  x  2.5x(1  x) 0.325 0.8734 1.1498 0.7192 1.2241 0.5383 1.1596 0.6969 1.225 0.5359 1.1577 0.7013 1.225 0.5359 1.1577 0.7013 1.225 0.5359  2002  1984 
 18; After 18 years, about 54% of the maximum sustainable population is present. 32. f(z) 
 z2  c 1  15i 
 (2  3i)2  c 1  15i 
 4  12i  9i2  c 4  3i 
 c  33. 2 2 x1 
  
 2  2  28. z0 
 2  2i, f(z) 
 z2  
 2  Iteration 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18  1  z1 
 2  2i 
 2  i  2i2 
 1 z2 
 (i)2 
 1 z3 
 12 
 1 29. z0 
 1  i, f(z) 
 z2  2  3i z1 
 (1  i)2  2  3i 
 1  2i  1  2  3i 
2i z2 
 (2  i)2  2  3i 
 4  4i  1  2  3i 
 5  7i z3 
 (5  7i)2  2  3i 
 25  70i  49  2  3i 
 22  73i 30. p1 
 p0  rp0 p1 
 2000  (0.052)(2000) 
 $2104 p2 
 2104  (0.052)(2104) 
 $2213.41 p3 
 2213.41  (0.052)(2213.14) 
 $2328.51 p4 
 2328.51  (0.052)(2328.51) 
 $2449.59 p5 
 2449.59  (0.052)(2449.59) 
 $2576.97  x1 
   2   2  
 2    34. See students' work. Sample topics for discussion are judging soil quality and detection of heat stress in cows. 35a. 1.414213562, 1.189207115, 1.090507733, 1.044273782 35b. f(z) 
 z, z0 
 2 35c. 1 35d. 1        36. 2
cos 3  i sin 3 
 2ei 3 37.  8! (2a)84(3b)4 4!(8  4)!  
 70  16a4  81b4 
 90,720a4b4  38. Convergent; the series is geometric with 1  r 
 4  1.  39. The distance between the vertices is 130 ft. 2a 
 130, so a 
 75. c  7  91  e 
 a 
 5 
 65 b2 
 c2  a2 b2 
 912  652 
 4056 x2  a2  427  y2  x2  y2     b2 
 1 ⇒  4225  4056 
 1  Chapter 12  40.  n1 sin I 
 n2 sin r 1.00 sin 42° 
 2.42 sin r r 
 Arcsin r  12-9      Page 826  16°  40 ft  56˚ 42˚  41b. Let h 
 height of the building. Let x 
 distance from the point of elevation to the center of the base of the building. h  40  h  tan 42° 
 x  tan 56° 
 x h  40  h   x
 tan 42°  tan 56° 
 h (h  40) tan 42°  tan 56° 
 h 1.6466  40  1  h 40  0.6466 
 h h  62 feet  No, the height of the building is about 62 feet for a total of about 102 feet with the tower. 42. x 2 for all x, so infinite discontinuity 43. y x  y 
 14 (5, 9)  y
9 (8, 6)  y
5 (3, 5)  (8, 5)  x
3  x
8 x  O  f(x, y) 
 2x  8y  10 f(3, 9) 
 2(3)  8(9)  10 
 88 f(5, 9) 
 2(5)  8(9)  10 
 92 f(8, 6) 
 2(8)  8(6)  10 
 74 f(8, 5) 
 2(8)  8(5)  10 
 66 f(3, 5) 
 2(3)  8(5)  10 
 56 max: 92, min: 56 44.  HL  H  L 
 2 2H  2L 
 H  L H 
 3L H  
 3 The correct choice is D. L  Chapter 12  Check for Understanding  1. The n  1 case shows that the premise is true for an infinite number of cases. 2. Provide a counterexample. 3a. n(n  2) 3b. Since 3 is the first term in the sequence of partial sums and 1(1  2) 
 3, the formula is valid for n 
 1. Since 8 is the second term in the sequence of partial sums and 2(2  2) 
 8, the formula is valid for n 
 2. Since 15 is the third term in the sequence of partial sums and 3(3  2) 
 15, the formula is valid for n 
 3. 3c. Sk ⇒ k(k  2); Sk1 ⇒ (k  1)(k  3) 4. 8n  1 
 7r for some integer r. 5. Sample answer: If we wish to prove that we can climb a ladder with an indefinite number of steps, we must prove the following. First, we must show that we can climb off the ground to rung 1. Next, we must show that if we can climb to rung k, then we can climb to rung k  1. 6. Step 1: Verify that the formula is valid for n 
 1. Since 3 is the first term in the sequence and 1(1  2) 
 3, the formula is valid for n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. Sk ⇒ 3  5  7 … (2k  1) 
 k(k  2) Sk1 ⇒ 3  5  7 …  (2k  1)  (2k  3) 
 k(k  2)  (2k  3) 
 k2  4k  3 
 (k  1)(k  3) Apply the original formula for n 
 k  1. (k  1)[(k  1)  2] 
 (k  1)(k  3) The formula gives the same result as adding the (k  1) term directly. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n. 7. Step 1: Verify that the formula is valid for n 
 1. Since 2 is the first term in the sequence and 2(21  1) 
 2, the formula is valid for n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. Sk ⇒ 2  22  23 … 2k 
 2(2k  1) Sk1 ⇒ 2  22  23 …  2k  2k1 
 2(2k  1)  2k1 
 2  2k1  2 
 2(2k1  1) When the original formula is applied for n 
 k  1, the same result is obtained. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n.  41a.  (3, 9)  Mathematical Induction  1.00 sin 42°  2.42  428  8. Step 1: Verify that the formula is valid for n 
 1. 1 Since 2 is the first term in the sequence and 1  The formula gives the same result as adding the (k  1) term directly. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n.  1  1  21 
 2, the formula is valid for n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. 1 1 1 1 1 Sk ⇒ 2  22  23 … 2k 
 1  2k 1  1  1  1  1  1  1     Sk1 ⇒ 2  22  23 … 2k   2k1 
 1  2k  2k1 2  Pages 826–828  1    
1 2  2k  2k1 2  1    
1 2k1  2k1 1   
1 2k1  When the original formula is applied for n 
 k  1, the same result is obtained. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n. 9. Sn: 3n  1 
 2r for some integer r Step 1: Verify that Sn is valid for n 
 1. S1 ⇒ 31  1 or 2. Since 2 
 2  1, Sn is valid for n 
 1. Step 2: Assume that Sn is valid for n 
 k and show that it is also valid for n 
 k  1. Sk ⇒ 3k  1 
 2r for some integer r Sk1 ⇒ 3k1  1 
 2t for some integer t 3k  1 
 2r 3(3k  1) 
 3  2r 3k1  3 
 6r 3k1  1 
 6r  2 3k1  1 
 2(3r  1) Thus, 3k1  1 
 2t, where t 
 3r  1 is an integer, and we have shown that if Sn is valid, then Sk1 is also valid. Since Sn is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Hence, 3n  1 is divisible by 2 for all integral values of n. n(n  1) 10a. 6  4 
 10 10b. an 
 2 10  5 
 15 15  6 
 21 21  7 
 28 28  8 
 36 10, 15, 21, 28, 36 10c. Step 1: Verify that the formula is valid for n 
 1. Since 1 is the first term in the sequence and 1(1  1)(1  2)  6  (1)[3(1)  1]  2  
 1, the formula is valid for n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. k(3k  1)  Sk ⇒ 1  4  7 … (3k  2) 
 2 Sk1 ⇒ 1  4  7 … (3k  2)  (3k  1) k(3k  1)  
 2  (3k  1) k(3k  1)  3k2  5k  2  
 2 (k  1)(3k  2)  
 2 Apply the original formula for n 
 k  1. (k  1)[3(k  1)  1]  2  k(k  1)  (k  1)(k  2)  (k  1)(3k  2)  
 2  The formula gives the same result as adding the (k  1) term directly. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n.  Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. (k  1)(k  2)  2(3k  1)  
 2  2  
 1, the formula is valid for n 
 1.  k(k  1)  Exercises  11. Step 1: Verify that the formula is valid for n 
 1. Since 1 is the first term in the sequence and (1)[2(1)  1] 
 1, the formula is valid for n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. Sk ⇒ 1  5  9 … (4k  3) 
 k(2k  1) Sk1 ⇒ 1  5  9 … (4k  3)  (4k  1) 
 k(2k  1)  (4k  1) 
 2k2  3k  7 
 (k  1)(2k  1) Apply the original formula for n 
 k  1. (k  1)[2(k  1)  1] 
 (k  1)(2k  1) The formula gives the same result as adding the (k  1) term directly. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2. Since it is valid for n 
 2, it is also valid for n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n. 12. Step 1: Verify that the formula is valid for n 
 1. Since 1 is the first term in the sequence and  Sk ⇒ 1  3  6 … 2 
 6 Sk ⇒ 1  3  6 … 2  2 k(k  1)(k  2)  (k  1)(k  2)  
 6  2 
 
  k(k  1)(k  2)  3(k  1)(k  2)  6 (k  1)(k  2)(k  3)  6  Apply the original formula for n 
 k  1. (k  1)[(k  1)  1][(k  1)  2]  6  
  (k  1)(k  2)(k  3)  6  429  Chapter 12  13. Step 1: Verify that the formula is valid for n 
 1. 1 Since 2 is the first term in the sequence and 1 1   21  1 
  2 , the formula is valid for n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. 1 1 1 1 1 Sk ⇒ 2  4  8 … 2k 
 2k  1 1  1 1 1    …  2k 4 8 1 1   2k  1  2k1 2 1   2  2k  1  2k1 2 1 k1   1  k1  2 2 1 k1 1 2  Sk1 ⇒ 2  
 
 
 
    15. Step 1: Verify that the formula is valid for n 
 1. Since 1 is the first term in the sequence and 1[2(1)  1][2(1)  1]  3  1   2k1  Sk1 ⇒ 12  32  52 … (2k  1)2  (2k  1)2 
 
 
 
  When the original formula is applied for n 
 k  1, the same result is obtained. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n. 14. Step 1: Verify that the formula is valid for n 
 1. Since 1 is the first term in the sequence and  
  (k  1)[2(k  1)  1][2(k  1)  1]  3  k2(k  1)2  Sk ⇒ 1  8  27 … k3 
 4 Sk1 ⇒ 1  8  27 … k3  (k  1)3 k2(k  1)2  
 4  (k  1)3  
 
  k2(k  1)2  4(k  1)3  4 (k  1)2[k2  4(k  1)]  4 (k  1)2(k2  4k  4)  4 (k  1)2(k  2)2  4  Apply the original formula for n 
 k  1. (k  1)2[(k  1)  1]2  4  (k  1)2(k  2)2  
 4  The formula gives the same result as adding the (k  1) term directly. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n.  Chapter 12  
  (k  1)(2k  1)(2k  3)  3  The formula gives the same result as adding the (k  1) term directly. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n. 16. Step 1: Verify that the formula is valid for n 
 1. Since S1 ⇒ 1 and 21  1 
 1, the formula is valid for n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. Sk ⇒ 1  2  4 … 2k1 
 2k  1 Sk1 ⇒ 1  2  4 … 2k1  2k 
 2k  1  2k 
 2(2k)  1 
 2k1  1 When the original formula is applied for n 
 k  1, the same result is obtained. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so indefinitely. Thus, the formula is valid for all positive integral values of n. 17. Sn ⇒ 7n  5 
 6r for some integer r Step 1: Verify that Sn is valid for n 
 1. S1 ⇒ 71  5 or 12. Since 12 
 6  2, Sn is valid for n 
 1. Step 2: Assume that Sn is valid for n 
 k and show that it is also valid for n 
 k  1. Sk ⇒ 7k  5 
 6r for some integer r Sk1 ⇒ 7k1  5 
 6t for some integer t 7k  5 
 6r 7(7k  5) 
 7  6r 7k1  35 
 42r 7k1  5 
 42r  30 7k1  5 
 6(7r  5) k1 Thus, 7  5 
 6t, where t 
 7r  5 is an integer, and we have shown that if Sn is valid, then Sk1 is also valid. Since Sn is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Hence, 7n  5 is divisible by 6 for all integral values of n.  
 1, the formula is valid for n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1.  
  k(2k  1)(2k  1)   (2k  1)2 3 k(2k  1)(2k  1)  3(2k  1)2  3 [k(2k  1)  3(2k  1)](2k  1)  3 (2k2  5k  3)(2k  1)  3 (2k  3)(k  1)(2k  1)  3  Apply the original formula for n 
 k  1.  12(1  1)2  4  
  
 1, the formula is valid for n 
 1.  Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. k(2k  1)(2k  1) Sk ⇒ 12  32  52 … (2k  1)2 
  3  430  18. Sn ⇒ 8n  1 
 7r for some integer r Step 1: Verify that Sn is valid for n 
 1. S1 ⇒ 81  1 or 7. Since 7 
 7  1, Sn is valid for n 
 1. Step 2: Assume that Sn is valid for n 
 k and show that it is also valid for n 
 k  1. Sk ⇒ 8k  1 
 7r for some integer r Sk1 ⇒ 8k1  1 
 7t for some integer t 8k  1 
 7r 8(8k  1) 
 8  7r 8k1  8 
 56r 8k1  1 
 56r  7 8k1  1 
 7(8r  1) Thus, 8k1  1 
 7t, where t 
 8r  1 is an integer, and we have shown that if Sn is valid, then Sk1 is also valid. Since Sn is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. hence, 8n  1 is divisible by 7 for all integral values of n. 19. Sn ⇒ 5n  2n 
 3r for some integer r Step 1: Verify that Sn is valid for n 
 1. S1 ⇒ 51  21 or 3. Since 3 
 3  1, Sn is valid for n 
 1. Step 2: Assume that Sn is valid for n 
 k and show that it is also valid for n 
 k  1. Sk ⇒ 5k  2k 
 3r for some integer r Sk1 ⇒ 5k1  2k1 
 3t for some integer t 5k  2k 
 3r 5k 
 2k  3r 5k  5 
 (2k  3r)(2  3) 5k1 
 2k1  3(2k)  6r  9r k1 5  2k1 
 2k1  3(2k)  6r  9r  2k1 
 3(2k)  15r 
 3(2k  5r) Thus, 5k1  2k1 
 3t, where t 
 2k  5r is an integer, and we have shown that if Sn is valid, then Sk1 is also valid. Since Sn is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Hence, 5n  2n is divisible by 3 for all integral values of n. 20. Step 1: Verify that the formula is valid for n 
 1. Since a is the first term in the sequence and 1 [2a  (1  1)d] 
 a, the formula is valid for 2 n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. Sk ⇒ a  (a  d)  (a  2d) … [a  (k  1)d]  Apply the original formula for n 
 k  1. (k  1) {2a 2  (k  1)   [(k  1)  1]d} 
 2(2a  kd)  The formula gives the same result as adding the (k  1) term directly. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
3, and so on indefinitely. Thus , the formula is valid for all positive integral values of n. 21. Step 1: Verify that the formula is valid for n 
 1. 1 Since 2 is the first term in the sequence and 1  11  1  
 2, the formula is valid for n 
 1.  Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. 1  1  1  k      Sk ⇒   2  3  3  4 … k(k  1) 
 k  1 1  1  1  1  1       Sk1 ⇒  1  2  2  3  3  4 … k(k  1)  (k  1)(k  2) k  1    
 k  1  (k  1)(k  2) k(k  2)  1   
 (k  1)(k  2) k2  2k  1   
 (k  1)(k  2) (k  1)2   
 (k  1)(k  2) k1   
 k2  Apply the original formula for n 
 k  1. (k  1)  (k  1)  1  k1   
 k2  The formula gives the same result as adding the (k  1) term directly. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n. 22. Sn ⇒ 22n1  32n1 
 5r for some integer r Step 1: Verify that Sn is valid for n 
 1. S1 ⇒ 22(1)1  32(1)1 or 35. Since 35 
 5  7, Sn is valid for n 
 1. Step 2: Assume that Sn is valid for n 
 k and show that it is also valid for n 
 k  1. Sk ⇒ 22k1  3k1 
 5r for some integer r Sk1 ⇒ 2k3  32k3 
 5t for some integer t 22k1  32k1 
 5r 22k1 
 5r  32k1 22k1  22 
 (5r  32k1)(32  5) 22k3 
 45r  25r  32k3  5(32k1) 2k3 2  32k3 
 45r  25r  32k3  5(32k1)  32k3 
 20r  5(32k1) 
 5(4  32k1) Thus, 22k3  32k3 
 5t, where t 
 4  32k1 is an integer; and we have shown that if Sn is valid, then Sk1 is also valid. Since Sn is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Hence, 22n1  32k1 is divisible by 5 for all integral values of n.  k  
 2[2a  (k  1)d] Sk1 ⇒ a  (a  d)  (a  2d) … [a  (k  1)d] k   (a  kd) 
 2[2a  (k  1)d]  (a  kd) k[2a  (k  1)d]  2(a  kd)  
  2 2ak  k(k  1)d  2a  2kd  
  2 (k  1)2a  [k(k  1)  2k]d  
  2 (k  1)2a  (k2  k)d  
  2 (k  1)2a  k(k  1)d  
  2 (k  1)  
 2(2a  kd)  431  Chapter 12  23. Step 1: Verify that the formula is valid for n 
 1. Since S1 ⇒ [r(cos v  i sin v]1 or r(cos v  i sin v) and r1[cos (1)v  i sin (1)v] 
 r(cos v  i sin v), the formula is valid for n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. That is, assume that [r(cos v  i sin v)]k 
 rk(cos kv  i sin kv). Multiply each side of the equation by r(cos v  i sin v). [r(cos v  i sin v)]k1 
 [rk(cos kv  i sin kv]  [r(cos v  i sin v)] 
 rk1[cos kv cos v  (cos kv(i sin v)  i sin kv cos v  i2 sin kv sin v] 
 rk1[(cos kv cos v  sin kv sin v)  i(sin kv cos v  cos kv sin v)] 
 rk1[cos (k  1)v  i sin (k  1)v] When the original formula is applied for n 
 k  1, the same result is obtained. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n. 24a.  (k  1)2  5(k  1) 
 k2  2k  1  5k  5 
 (k2  5k)  (2k  6) 
 2r  2(k  3) 
 2(r  k  3) Thus, k2  5k 
 2t, where t 
 r  k  3 is an integer, and we have shown that if Sn is valid, then Sk1 is also valid. Since Sn is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Hence, n2  5n is divisible by 2 for all positive integral values of n. 26a. Number of people Number of Interactions 2 1 2
1 3  1 2  3
3  4  1 2 3  4
6      n(n  1)  2  n  26b. Step 1: Verify that Sn ⇒ 0  1  2  3 … n(n  1)   (n  1) 
 2 is valid for n 
 1. Since 0 is the first term in the sequence and 1(1  1)  
 0, the formula is valid for n 
 1. 2 Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. k(k  1)  Sk ⇒ 0  1  2 … (k  1) 
 2  24b. 4  1 
 3 94
5 16  9 
 7 1, 3, 5, 7, … 24c. 2n  1 24d. n2 24e. Step 1: Verify that the formula is valid for n 
 1. Since 1 is the first term in the sequence and 12 
 1, the formula is valid for n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. Sk ⇒ 1  3  5  7 … (2k  1) 
 k2 Sk1 ⇒ 1  3  5  7 … (2k  1)  (2k  1) 
 k2  (2k  1) 
 k2  2k  1 
 (k  1)2 When the original formula is applied for n 
 k  1, the same result is obtained. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n. 25. S1 ⇒ n2  5n 
 2r for some positive integer r Step 1: Verify that S1 is valid for n 
 1. S1 ⇒ 12  5  1 or 6. Since 6 
 2  3, S1 is valid for n 
 1. Step 2: Assume that Sn is valid for n 
 k and show that it is valid for n 
 k  1. Sk ⇒ k2  5k 
 2r for some positive integer r Sk1 ⇒ (k  1)2  5(k  1) 
 2t for some positive integer t  Chapter 12  k(k  1)  Sk1 ⇒ 0  1  2 … (k  1)  k 
 2  k k(k  1)  2k  
 2 k2  k  2k  
 2 k(k  1)  
 2 Apply the original formula for n 
 k  1. (k  1)[(k  1)  1]  2  k(k  1)  
 2  The formula gives the same result as adding the (k  1) term directly. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n. 15(14)  26c. Yes; 15 people will require 2 or 105 interactions and last approximately 105(0.5) or 52.5 minutes.  432  27. Step 1: Verify that Sn ⇒ (x  y)n 
 xn  nxn1y  n(n  1) xn2y2 2!    n(n  1)(n  2)  3!  When the original formula is applied for n 
 k  1, the same result is obtained. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on in indefinitely. Thus, the formula is valid for all positive integral values of n.  xn3y3 … yn is  valid for n 
 1. Since S1 ⇒ (x  y)1 
 x1  y1 or x  y, Sn is valid for n 
 1. Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. k(k  1)  Sk ⇒ (x  y)k 
 xk  kxk1y  2 xk2y2  ! k(k  1)(k  2)  3!  28e. lim  n→  k(k  1)(k  2)  k(k  1)  xk2y2  (x  y)k1 
 x(xk  kxk1y  2 !  k(k1)  2!  … yk)  y(xk  kxk1y   k(k  1)(k  2)  xk2y2  3 xk3y3 … yk) ! k(k  1)  xk1y2 … 
 xk1  kxky  2 ! k(k  1) k2 3 k k k1 2 xy  x y  kx y  2 x y ! k1 … y k(k  1) 
 xk1  (k  1)xky  kxk1y2  2 ! xk1y2 … yk1  31. 25x2  4y2  100x  40y  100 
 0 25(x2  4x)  4(y2  10y) 
 100 25(x  2)2  4(y  5)2 
  100  100  100 25(x  2)2 2 4(y  5)22 
 100 (x  2) (y  5)    
 1 4 25  k(k  1)  
 xk1  (k  1)xky  2 xk1y2 ! k1 … y When the original formula is applied for n 
 k  1, the same result is obtained. Thus if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n. 28a. 0.9  0.09  0.009 … 28b.  9  10n  28c. S 
  ellipse 32. 
 
 250, k 
 4 (250)2    A
 4(4)  12,271.85 m2 33. 34.  a1  a1 rn   1r 9 9 1 n     10 10 10  
    1 10  
  n    
  9 
 
  2    
 2; y 
   10 1  1  101 10n  1  10n  
  
 30° Since 23 is  H  9    Sk1 ⇒ 0.9  0.09  0.009 …   10k  10k1 9    
 10k  10k1  60˚ 1  C  F D  B  G  perimeter is 2(2   42) 
 2 (52) 
  10(10k  1)  9   
 10k1  102  units.  10k1  10  9   
 10k1  
   3  Since A BF E  and B E  measure 1 unit, A H  and   each measure 4 units.  AB  is the hypotenuse of an isosceles right triangle with legs that measure 1 unit. So A 2 units.  CA B  measures   is the hypotenuse of an isosceles right triangle with legs that measure 4 units. So  CA  measures 4 2 units. Since ABCD is a rectangle, the  Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. 10k  1 9   Sk ⇒ 0.9  0.09  0.009 …  10k 
 10k 9  30˚  2  A  Since 0.9 is the first term in the sequence and 101  1  101 
 0.9, the formula is valid for n 
 1.  1  sin 2x  3  cos1 2  28d. Step 1: Verify that Sn: 0.9  0.9  0.009 … 10n  1 9   10n 
 10n is valid for n 
 1.  10k  3  4  negative, x is in the second or third quadrants. x 
 180°  30° or 150° x 
 180°  30° or 210° 35. Since the area E of the square is 25, each side measures 5.  
  1 9 1  1   10 10  n→  Thus, 0.999… 
 1. 29. z1 
 2(4  i)  i 
8i z2 
 2(8  i)  i 
 16  i z3 
 2(16  i)  i 
 32  i 30. 64 
 6  (29  1)d 70 
 28d 5  
 d 2   3 xk3y3 … yk) !  k(k  1)(k  2) xk3y3 3!  1   
 lim 1   10n  
 1  0 or 1  xk3y3 … yk  Sk1 ⇒ (x  y)k(x  y) 
 (x  y)(xk  kxk1y  k(k1) xk2y2 2!  10k  1  10n  The correct choice is B.  10k1  1 k 10 1  433  Chapter 12  4 2  21. r 
 4 or 2   Chapter 12 Study Guide and Assessment  )8 4  4(2  S8 
   Page 829  1  2   Understanding the Vocabulary  1. d 5. k 9. b  2. i 6. f 10. h  Pages 830–831  3. m 7. c  60  2 1    ) 60(1  2   
 12  
 60
1  3n  n  22. lim 
 lim 4n 1 n→ n→    n n 3   
4 6n  3 6n 3   23. lim n 
 lim n  lim n n→ n→ n→  Skills and Concepts  
60 
6  24. Does not exist; lim  n→  25. lim  n→  4n3  3n   n4  4n3  123  1  Sn 
 5  
5 41  504   27. r 
  1260 or 0.4 1260   Sn 
  1  0.4  1  
 2100   16. r 
 4 9 or 7    
    17. a15 
 2.2(2)151 
 36,044.8 18. 8 
 a1(0.2)71 8 
 0.000064a1 125,000 
 a1 19. 125 
 0.2r51 625 
 r4 5
r 0.2( 5) 
 1, 1( 5) 
 0.2, 1, 5, 25, 125 20.  1  7  
  n2  1  343  r  (n  1)2   5n1 
 lim  n2 n→  5n  
 lim  n→  5n(n2  2n  1)   5n  5  n2  n2 2n   lim  2 2 n→ 5n n→ n 1 
 5  0  0 1 
 5  
 lim  5,  5( 5) 
  (n  1)2   28. an 
 5n , an1 
  5n1  25  1  2 n→ n   lim  convergent n5  2.4  r
 1.2 or 2 1.2  1.2(2)9  S9 
  1  (2) 1.2  614.4 
 3  5  29. The general term is n or 1  n. 5  1  1  n  n for all n, so divergent 2  30. The general term is n. 2  n  
 205.2  Chapter 12  123  1000  1  1 1000 123  999   
 5 333  n 
 18 or n 
 19.86 Since n is a positive whole number, n 
 18. 1 1 ,  49 49  123    a1 
  1000 , r 
 1000  1.3  (1.3)3   4(0 .7)(2 50.2)  2(0.7) 1.3 26.5  1.4  1  7  since lim  4n3 3n   n4  n 4   
 lim n4 4n3 n→    n4 n4 0 
 1 123  250.2 
 2n  0.7n(n  1) 0 
 0.7n2  1.3n  250.2  1 1 1 1  7 
 7, 7 1 1 1 , ,  7 49 343  n→  2n ; 3    26. 5.1 2 3 
5 1000  1,000,000 …  n  7  
 lim  
0  250.2 
 2[2(2)  (n  1)(1.4)]  
  2nn3  3n3  n→  2n  3  becomes increasingly large as n approaches infinity, the sequence has no limit.  
 217 n Sn 
 2[2a1  (n  1)d]  n
  2   3n  4n  1  11. d 
 4.3  3 or 1.3 5.6  1.3 
 6.9, 6.9  1.3 
 8.2, 8.2  1.3 
 9.5, 9.5  1.3 
 10.8 6.9, 8.2, 9.5, 10.8 12. a20 
 5  (20  1)(3) 
 52 13. 4 
 6  (5  1)d 10 
 4d 2.5 
 d 6  (2.5) 
 3.5, 3.5  (2.5) 
 1, 1  (2.5) 
 1.5 6, 3.5, 1, 1.5, 4 14. d 
 23  (30) or 7 a14 
 30  (14  1) 7 
 61 14 S14 
 2(30  61) 15.  1   2   
     1  2  4. j 8. e  434  1   n for all n, so divergent  9  31.  46. f(3) 
 (3)2  4 
 13 f(13) 
 132  4 
 173 f(173) 
 1732  4 
 29,933 f(29,933) 
 29,9332  4 
 895,984,493 13; 173; 29.933; 895, 984, 493 47. z0 
 4i z1 
 0.5(4i)  (4  2i) 
 2i  4  2i 
 4 z2 
 0.5(4)  (4  2i) 
 2  4  2i 
 6  2i z3 
 0.5(6  2i)  (4  2i) 
 3  i  4  2i 
 7  3i 48. z0 
 8 z1 
 0.5(8)  (4  2i) 
 4  4  2i 
 2i z2 
 0.5(2i)  (4  2i) 
 i  4  2i 
 4  3i z3 
 0.5(4  3i)  (4  2i) 
 2  1.5i  4  2i 
 6  3.5i 49. z0 
 4  6i z1 
 0.5(4  6i)  (4  2i) 
 2  3i  4  2i 
 2  i z2 
 0.5(2  i)  (4  2i) 
 1  0.5i  4  2i 
 5  1.5i z3 
 0.5(5  1.5i)  (4  2i) 
 2.5  0.5i  4  2i 
 6.5  2.75i 50. z0 
 12  8i z 
 0.5(12  8i)  (4  2i) 
 6  4i  4  2i 
 10  6i z2 
 0.5(10  6i)  (4  2i) 
 5  3i  4  2i 
 9  5i z3 
 0.5(9  5i)  (4  2i) 
 4.5  2.5i  4  2i 
 8.5  4.5i 51. Step 1: Verify that the formula is valid for n 
 1. Since the first term in the sequence is 1 and   (3a  3) 
 (3 5  3)  (3  6  3)  (3  7  3) a
 5  (3.8  3)  (3.9  3)  
 12  15  18  21  24 
 90    32.   (0.4)k 
 (0.4)1  (0.4)2  (0.4)3 … (0.4) k
1 0.4   S
 1  0.4 2  
 3    33.  9   (2n  1) a
0  34.   
1(n2  1) a
1  6!  6!  5 1 4   35. (a  4)6 
 a6   1!(6  1)!  a  (4)  2!(6  2)!  a 6!  6!  3 3    (4)2   3!(6  3)!  a  (4)  4!(6  4)! 6!  1 5   a2  (4)4   5!(6  5)!  a  (4)  6!  6!(6  6)!   a0  (4)6  
 a6  24a5  240a4  1280a3  3840a2  6144a  4096 4!  3 1  36. (2r  3s)4 
 (2r)4   1!(4  1)! (2r) (3s) 4!  2 2   2!(4  2)! (2r) (3s) 4!  3   3!(4  3)! (2r)(3s) 4!  0 4   4!0! (2r) (3s)  
 16r4  96r3s  216r2s2  216rs3  81s4 37.  10!  4!(10  4)!  38.  8!  2!(8  2)!  39.  10!  7!(10  7)!   x107  (3y)7 
 120  x3  2187y7 
 262,440x3y7  40.  12!  5!(12  5)!   (2c)125  (d)4 
 792  128c7  (d)5 
 101,376c7d5   x104  (2)4 
 210  x6  16 
 3360x6  41. 2
cos  3  4   4m82  12 
 28  4096m6  1 
 114,688m6  3  4   
 2e   42. 4i 
 4
cos 2  i sin 2  i sin  1(1  1)  2  Step 2: Assume that the formula is valid for n 
 k and derive a formula for n 
 k  1. k(k  1)  Sk ⇒ 1  2  3 … k 
 2 Sk1 ⇒ 1  2  3 … k  (k  1)  3 i4  k(k  1) k2  k2  k  2k  2 k2  3k  2  
 2 (k  1)(k  2)  7  
 2 7  2  2i 
 22 
cos 4  i sin 4 
 22 e  Apply the original formula for n 
 k  1. (k  1)[(k  1)  1]  2  7 i 4    3    (k  1)(k  2)  
 2 The formula gives the same result as adding the (k  1) term directly. Thus, if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n.  44. r 
  (33 )2  32 or 6  or  v 
 Arctan 
 6 3 3  2k  2  
 2  2  (2) 43. r 
 2 2 or 22   7  k  
 2  2     2  2(k  1)  
 2  2  
 4ei 2  v 
 Arctan 
2 or 4  
 1, the formula is valid for n 
 1.    33   3i 
 6
cos 6  i sin 6 i6  
 6e 45. f(2) 
 6  3  2 
 0 f(0) 
 6  3  0 
 6 f(6) 
 6  3  6 
 12 f(12) 
 6  3  (12) 
 42 0, 6, 12, 42  435  Chapter 12  52. Step 1: Verify that the formula is valid for n 
 1. Since the first term in the sequence is 3 and 1(1  1)(2  1  7)  6  12  54b. S12 
 2(16  368) 
 2304 ft 55. If the budget is cut 3% each year, 97% remains after each year. a1 
 160,000,000, r 
 0.97 a11 
 160,000,000(0.97)111 
 $117,987,860.30 6 56a. One side of the original triangle measures 3 or 2 units. Half of 2 units is 1 unit. Each side of the new triangle measures 1 unit, so its perimeter is 1  1  1 or 3 units.  
 3, the formula is valid for n 
 1.  Step 2: Assume that the formula is valid for n 
 k and derive formula for n 
 k  1. k(k  1)(2k  7)  Sk ⇒ 3  8  15 … k(k  2) 
 6 Sk1 ⇒ 3  8  15 … k(k  2)  (k  1)(k  3) k(k  1)(2k  7)  
 6  (k  1)(k  3) k(k  1)(2k  7)  6(k  1)(k  3)  
 6  6  3  
  k(k  1)(2k  7)  6(k  1)(k  3)  6  
  (k  1)[k(2k  7)  6(k  3)]  6  
  (k  1)(2k2  7k  6k  18)  6  
  (k  1)(2k2  13k  18)  6  S
  Page 833  
  Apply the original formula for n 
 k  1. 
  (k  1)(k  2)(2k  9)  6  Open-Ended Assessment  2. Sample answer: 3n; lim lim  n→  2  n  
 0, but since  n→ 5n lim 3 n→  6  5n2  3n  
  2  n→ n  
 lim   3 5n  becomes increasingly  large as n approaches infinity, the sequence has no limit.  Chapter 12 SAT & ACT Preparation Page 835  SAT and ACT Practice  1. If 40% of the tapes are jazz then 60% of the tapes must be blues. There are 80 tapes. Find 60% of 80. 0.60(80) 
 48 The correct choice is D. 2. Because of alternate interior angles, 150 
 130  unmarked angle of right  20 
 unmarked angle of right  In the right triangle, x  20 
 90 or x 
 70. The correct choice is C. d gallons  d    3. fraction pumped 
  k gallons 
 k  Change this fraction into a percent by multiplying by 100. d 100d percent pumped 
 k  100 or k% The correct choice is A. 4. First calculate the number of caps Andrei has now. He starts with 48 and gives away 13, so he has 48  13 
 35 left. Then he buys 17, so he has 35  17 
 52. Then he trades 6 caps for 8 caps. this leaves him with 52  6  8 or 54. His total is now 54.  Applications and Problem Solving  54a. a1 
 16, d 
 48  16 or 32 A12 
 16  (12  1)32 
 368 ft  Chapter 12  
1  6  5n2  The formula gives the same result as adding the (k  1) term directly. Thus, if the formula is valid for n 
 k, it is also valid for n 
 k  1. Since the formula is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Thus, the formula is valid for all positive integral values of n. 53. Sn ⇒ 9n  4n 
 5r for some integer r Step 1: Verify that Sn is valid for n 
 1. S1 ⇒ 91  41 or 5. Since 5 
 5  1, Sn is valid for n 
 1. Step 2: Assume that Sn is valid for n 
 k and show that it is also valid for n 
 k  1. Sk ⇒ 9k  4k 
 5r for some integer r Sk1 ⇒ 9k1  4k1 
 5t for some integer t 9k  4k 
 5r 9k 
 4k  5r 9(9k) 
 (4k  5r)(4  5) 9k1 
 4k1  5(4k)  20r  25r k1 9  4k1 
 4k1  5(4k)  20r  25r  4k1 
 5(4k)  45r 
 5(4k  9r) k1 k1 Thus, 9 4 
 5t, where t 
 4k  9r is an integer, and we have shown that if Sn is valid, then Sk1 is also valid. Since Sn is valid for n 
 1, it is also valid for n 
 2, n 
 3, and so on indefinitely. Hence, 9n  4n is divisible by 5 for all integral values of n.  Page 833  6  1 1  2  1a. Arithmetic; arithmetic sequences have common differences, while geometric sequences have common ratios. 1b. Sample answer: 1, 4, 7, 10, …; an 
 1  3(n  1)  (k  1)(k  2)(2k  9)  6 (k  1)[(k  1)  1][(2(k  1)  7]  6  1  56b. a1 
 6, r 
 6 or 2  436  54  48  9. The increase from 99 to 100 is 1. So the percent 1 increase from 99 to 100 is .  Percent increase 
 4  100 8 
 12.5% The correct choice is B. 5. Since the figure is not drawn to scale, do not assume that the two lines are parallel, even though they may appear parallel. Since AB 
 AC, ABC is isosceles. So m∠B 
 m∠ACB. m∠B  m∠ACB  80 
 180 2  m∠ACB 
 100 m∠ACB 
 50 Since AD is a line segment, x  70  50 
 180. So, x 
 60. Consider right triangle CDE. x  y 
 90 60  y 
 90 y 
 30 x  y 
 60  30 or 30 The correct choice is C. 6. The population of Rockville is now 20,000 and will double every 8 years. So in 8 years the population will be 40,000 and in 16 years will be 80,000. So f(8) 
 40,000 and f(16) 
 80,000. Choice A is incorrect since f(8) 
 20,000. Choice B is incorrect since f(16) 
 2(20,000)2 or 800,000,000. Choice C  99 1 1  is greater than , or 1%. 99 100  The correct choice is A. 10. Choose a number for the total number of cars in the parking lot. Since the fractions have denominators of 2, 4, and 5, choose a number that is divisible by 2, 4, and 5. Let the number of cars in the parking lot equal 40. 1  5 1  2 1  4   8 blue cars 
 4 blue convertibles  the number of convertibles 
 4 the number of convertibles 
 16  Not Blue and Not Convertible  8 Blue  4 Convertible 16  The number of cars that are neither blue nor convertible is the total number minus the blue cars minus the convertibles plus the number that are both blue and convertible. Neither blue nor convertible 
 40  8  16  4 
 20 percent that are neither blue nor convertible  20,000  is incorrect since f(8) 
  64 . Choice E is incorrect since f(8) 
 20,000. For Choice D, f(8) = 40,000 and f(16) = 80,000. The correct choice is D. 7. Notice that the figure is not drawn to scale. ∠A could be a right angle. To be sure it is, find the slope of AB and compare it to the slope of AC. 10  4   40 
 8 blue cars  20  
 40  100  6    Slope of AB 
  65 
 1  
 50% The answer is 50.  1  Slope of AC 
 6 Since the slopes are negative reciprocals, the line segments are perpendicular, so m∠A 
 90. Therefore, ABC is a 30°-60°-90° right triangle. The hypotenuse, AC, is twice the length of the leg opposite the 30° angle, AB. 2  (1 AB 
  (6  5) 0  4)2 
 1 6  3 or 37  AC 
 237  The correct choice is D. 8. Method 1: Substitute each answer choice for x to test both inequalities. A: (6)  6  0 and 1  2(6)  1. 0  0 and 13  1; false Method 2: Solve each inequality for x. x60 and 1  2x  1 x  6 2x  2 x 1 The solution is 6 x 1. All of the answer choices except A are in this range. The correct choice is A.  437  Chapter 12  Chapter 13 Combinatorics and Probability 4!  13-1  Pages 842–843  4321  
 4  10 or 40 13. Using the Basic Counting Principle, 10  9  8  7  6  5  4  3  2  1 
 3,628,800.  Check for Understanding  15!   14. C(15, 9) 
  (15  9)! 9!  
  Pages 543–545 7!   17. P(7, 7) 
  (7  7)!  
  18a. Using the Basic Counting Principle, 9  10  10  10  10  10  10 
 9,000,000. 18b. Using the Basic Counting Principle, 5  5  5  5  5  5  5 
 78,125. 18c. Using the Basic Counting Principle, 10  10  10  10  10  10  1 
 1,000,000. 18d. Using the Basic Counting Principle, 1  1  1  10  10  10  10 
 10,000. 19. dependent 20. independent 21. dependent  654321  1 5!   8. P(5, 3) 
  (5  3)!  54321   
 21  8!   22. P(8, 8) 
  (8  8)!  
 60  
 
  
  12!  (12  8)! —— 6!  (6  4)! 12! 2!  6! 4! 12  11  10  9  8  7  6  5  4  3  2  1  2  1  6  5  4  3  2  1  4  3  2 1  
  6!   23. P(6, 4) 
  (6  4)!  
  5!   24. P(5, 3) 
  (5  3)!  7!  (7  4)! 4! 7654321  3214321  54321   
 21  
 60  7!   25. P(7, 4) 
  (7  4)!  11. C(20, 15)  
  20!   
 (20  15!) 15!  7654321  321  
 840  20  19  18  17  16  15  14  13  12  11  10  9  8  7  6  5  4  3  2  1  5  4  3  2  1  15  14  13  12  11  10  9  8  7  6  5  4  3  2  1  
 15,504  Chapter 13  654321  21  
 360  
 35  
  87654321  1  
 40,320  
 55,440 10. C(7, 4) 
  7654321  1  
 5040  
 720  
  Exercises  16. Using the Basic Counting Principle, 2  6  4 
 48.  6!  P(12, 8)  P(6, 4)  15  14  13  12  11  10  9  8  7  6  5  4  3  2  1  654321987654321  
 5005 15a. Using the Basic Counting Principle, 10  10  10  10  10 
 100,000. 15b. Using the Basic Counting Principle, 1  9  9  9  10 
 7290 15c. Using the Basic Counting Principle, (10  10  10  10  10  10  10  10  10)  (10  10  10  10  10) 
 999,900,000   7. P(6, 6) 
  (6  6)!  9.  54321    
 1321  32121  1. Sample answer: Both are used to determine the number of arrangements of a group of objects. However, order of the objects is important in permutations. When order of the objects is not important, combinations are computed. 2. Select 2 jacks out of 4—C(4, 2) Select 3 queens out of 4—C(4, 3) number of hands—C(4, 2)  C(4, 3) 3. Sam is correct. The room assignments are an ordered selection of 5 rooms from the 7 rooms. A permutation should be used. 4. blue ————— S-blue S green ———— S-green gray ————— S-gray blue ———— M-blue M green———— M-green gray ———— M-gray blue ————— L-blue L green ———— L-green gray ————— L-gray blue ———— XL-blue XL green ——— XL-green gray ———— XL-gray 5. Using the Basic Counting Principle, 4  3  5  5 
 300. 6. independent 
  5!    12. C(4, 3)  C(5, 2) 
  (4  3)!3!  (5  2)! 2!  Permutations and Combinations  438  38. C(7, 3)  C(8, 5)  9!   26. P(9, 5) 
  (9  5)!  
  7!  7654321  
  4321321   
 15,120 10  9  8  7  6  5  4  3  2  1  321  
 604,800 28.  P(6, 3)  P(4, 2)  
 
 
  54321  P(6, 4)  P(5, 3)  
 
 
  14!   40. C(14, 4) 
  (14  4)! 4!  
  14  13  12  11  10  9  8  7  6  5  4  3  2  1  10  9  8  7  6  5  4  3  2  1  4  3  2  1  
 1001 14!   41. C(14, 5) 
  (14  5)! 5!  6!  (6  4)! — 5!  (5  3)! 6! 2!  5! 2! 654321  54321  P(6, 3)  P(7, 5)  P(9, 6)  14  13  12  11  10  9  8  7  6  5  4  3  2  1  
  98765432154321 
 2002 42. C(18, 12) 18!   
 (18  12)! 12!  
  
  
  18  17  16  15  14  13  12  11  10  9  8  7  6  5  4  3  2  1  6  5  4  3  2  1  12  11  10  9  8  7  6  5  4  3  2  1  
 18,564 43. C(3, 2)  C(5, 1)  C(8, 2) 3! 5! 8!    
 (3  2)! 2!  (5  1)! 1!  (8  2)! 2!  6! 7!     (6  3)! (7  5)! ——— 9!  (9  6)! 6! 7! 3!  9! 3! 2! 6543217654321  98765432121  
  321  54321    
 121  43211   11!   44. P(11, 11) 
  (11  1)!  5!  11  10  9  8  7  6  5  4  3  2  1   31. C(5, 3) 
  (5  3)! 3!  
  1  54321  
 39,916,800   
 21321  
 10  13!  10!  13  12  11  
 286  78 or 22,308 45b. C(4, 1)  C(4, 2)  C(4, 2) 4!  4!  4!     
 (4  1)! 1!  (4  2)! 2!  (4  2)! 2!  4!   33. C(4, 2) 
  (4  2)! 2!  4321  4321  4 321  4321     
 3211  2121  2121   
 2121  
 4  6  6 or 144  
6  12!  12!  (12  4)! 4! 12  11  10  9  8  7  6  5  4  3  2  1  876543214321   45c. C(12, 5) 
  (12  5)! 5!  
  12  11  10  9  8  7  6  5  4  3  2  1  765432154321  
 792 46a. Using the Basic Counting Principle, 10  10  10  10  10 
 100,000 46b. Using the Basic Counting Principle, 10  9  8  7  6 
 30,240 46c. Using the Basic Counting Principle, 5  5  4  4  4 
 1600.  
 495 35. C(9, 9) 
  13  12    
 321  21  10  9  8  7  6  5  4  3  2  1  5432154321  
 252  
  13!    45a. C(13, 3)  C(13, 2) 
  (13  3)! 3!  (13  2)! 2!   32. C(10, 5) 
  (10  5)! 5!  34. C(12, 4) 
  87654321  65432121  
 3  5  28 or 420  
5  
  87654321  65432121  
 5  6  28 or 840  
6 30.  4321    
 43211  2121   6!  (6  3)! — 4!  (4  2)! 6! 2!  4! 3! 65432121  4321321  
 10 29.  87654321  32154321  
 35  56 or 1960 39. C(5, 1)  C(4, 2)  C(8, 2) 5! 4! 8!    
 (5  1)! 1!  (4  2)! 2!  (8  2)! 2!  10!   27. P(10, 7) 
  (10  7)!  
  8!    
 (7  3)! 3!  (8  5)! 5!  987654321  4321  9!  (9  9)! 9!  
1  14!   36. C(14, 7) 
  (14  7)! 7! 14  13  12  11  10  9  8  7  6  5  4  3  2  1  
  76543217654321  5!  
 3432  4!    P(5, 2)  P(4, 3) 
  (5  2)!  (4  3)! 3!  54321  8!    37. C(3, 2)  C(8, 3) 
  (3  2)! 2!  (8  3)! 3! 321   
 121   4321     
 321 1  87654321  54321321  
 20  24 or 480  
 3  56 or 168  439  Chapter 13  47. C(3, 1)  C(4, 1)  C(6, 1)  C(14, 6)  53b. Yes; let h, t, and u be the digits. 100h  10t  u 100h  10u  t 100t  10h  u 100t  10u  h 100u  10t  h  100u  10h  t 200(h  t  u)  20(h  t  u)  2(h  t  u) 
 222(h  t  u)  3! 4! 6! 14!     
 (3  1)! 1!  (4  1)! 1!  (6  1)! 1!  (14  6)! 6! 654321 321 4321    
 2  1  1  3  2  1  1  5  4  3  2  1 1 14  13  12  11  10  9  8  7  6  5  4  3  2  1   87654321654321  
 3  4  6  3003 or 216,216 42!   48a. P(42, 42) 
  (42  42)!  222(h  t  u)  6  
 42! 1.4  1051  54. 2140  (1.058) 
 $2264.12 2264.12(1.058) 
 $2395.44 2395.44(1.058) 
 $2534.38  42!   48b. C(42, 30) 
  (42  30)! 30! 42!   
 12! 30!  10  55.  1.1  48c. C(5, 3)  C(12, 6)  C(10, 6)  C(15, 5) 1010  5!  12!  10!  56. 15!      
 (5  3)! 3!  (12  6)! 6!  (10  6)! 6!  (15  5)! 5!  n! n!   
  [n  (n  1)]! (n  n)! n! n!  
  0! 1! 0!   n3 
 13  2 3  . n
1 7.1x 
 83.1 x ln 7.1 
 ln 83.1 ln 83.1  x 57. x 
  e0.346  2.26 Use a calculator.  1.4 58.  1  n! 
 n!  y 
 4x2 x
 sin 45°  y
 cos 45° 
 4(x
 cos 45°  y
 sin 45°)2 2   2!  2!  21  59.  90˚  120˚  2!  30˚  180˚  1 2 3 4  330˚  210˚ 240˚  270˚  11!   51a. C(11, 4) 
  (11  4)! 4!  6!  5!    51b. C(6, 2)  C(5, 2) 
  (6  2)! 2!  (5  2)! 2! 654321  432121  54321    32121  
 28
2  
 15  10 or 150  2   10!   52. C(10, 2) 
  (10  2)! 2!  
 28
2  61.  or 37(2  5  9) 
 592  360°  period 
 1 or 360° phase shift 
 30°  Chapter 13  2   19.80 ft/s 19.80 ft/s sin 2x  2 sin x 
 0 2 sin x cos x  2 sin x 
 0 2 sin x (cos x  1) 
 0 2 sin x 
 0 or cos x  1 
 0 sin x 
 0 cos x 
 1 x 
 0°, x 
 180°, x 
 360°, or x 
 180° So, x 
 0°, 180°, 360° 62. y 
 8 cos (v  30°) amplitude 
 8  10  9  8  7  6  5  4  3  2  1  8765432121  
 45 3552  6  300˚  r 
 2, r 
 2 cos 2v 2 
 2 cos 2v 1 
 cos 2v 2v 
 0° or 2v 
 360° v 
 0° v 
 180° (2, 180°), (2, 0°) 60. vx 
 28 cos 45°, vy 
 28 sin 45°  11  10  9  8  7  6  5  4  3  2  1  76543214321  
 330  53a.  0˚  21  
 6  2  2  2 or 48  
  60˚  150˚  
 1  1  1  1  
  2   1  0 
 4(x
)2  8x
y
  4(y
)2  2 x
  2 y
        
 (3  3)!  (2  2)!  (2  2)!  (2  2)! 21  2  1  
 720 50b. There are 3 ways to arrange the 3 couples, and 2 ways to arrange each of the two members within a couple. P(3, 3)  P(2, 2)  P(2, 2)  P(2, 2) 3!  2   2 x
  2 y
 
 8
2x
2  x
y
  2y
2  654321  1  
  2  2x
  2y
 
 4
2x
  2y
  6!   50a. P(6, 6) 
  (6  6)!  321  . .  103 
 3025   x
 ln 7.1  
 10  924  210  3003 5.8  109 49. P(n, n  1) 
 P(n, n)  
  
 37(h  t  u)  440  26!  63. Find B. B 
 180°  90°  27° or 63° Find a.   22. linear;  6! 3! 7! 10!  a  15.2  tan 27° 
 15.2 tan 27° 
 a 7.7 a Find c. 15.2  cos 27° 
 c 15.2   c
 cos 27°  c 17.1 360 64. Each hour, an hour hand moves through 1 2 or 1 30°. Since 12 minutes is 5 of an hour, the hour 1 hand moves through an additional 5(30) or 6°. 2(30°)  6° 
 66° The correct choice is A.  5.1  1012  23. 24. 25. 26. 27. 28.  circular; (9  1)! 40,320 circular; (5  1)! 
 24 circular; (8  1)! 
 5040 linear; 6! 
 720 linear; 10! 
 3,628,800 circular; (9  1)! 
 40,320  29. 30. 31. 32.  circular;  
 3,113,510,400 2 circular; (20  1)! 1.22  1017 circular; (32  1)! 8.22  1033 linear; 25! 1.55  1025  33.  8!  2!2!2!2!  (14  1)!  
 2520  34a. (7  1)! 
 720 34b. 7! 
 5040 11!   
 46,200 3!4 !3! 11!  36a.  2! 2 ! 2! 
 4,989,600  35.  Permutations with Repetitions and Circular Permutations  13-2  36b. integral calculus 37a.  Pages 848–849  Check for Understanding  4. 5. 6.  n(n  1)(n  2)(n  3)!  (n  3)! n3  3n2  2n  210  6!   39. C(6, 3) 
  (6  3)! 3!  
  5! 2  
 756  Pages 849–851  Exercises  12.  8!  2! 2!  
 10,080  13.  10!  2! 2!  
 907,200  14.  8!  2!  15.  10!  3! 2!  16.  12!  2! 2! 2!  
 59,875,200  17.  9!  4! 2! 2!  
 3780  18.  7!  2! 2! 2!  
 630  19.  9!  5! 4!  20.  654321  321321  
 20 40. (5x  1)3 
 C(3, 0)  (5x)3  (1)0  C(3, 1)  (5x)2  (1)1  C(3, 2)  (5x)1  (1)2  C(3, 3)  (5x)0  (1)3 
 125x3  75x2  15x  1 41. x log2 413  10. linear;  or 60 9!  5! 2! 2!  
 210  
0 Use a graphing calculator to find the solution at n 
 7.  2  7. circular; (11  1)! or 3,628,800 8. circular; (8  1)! or 5040 9. circular; (12  1)! or 39,916,800  11.  7.85  1017  37b. (43  1)! 1.41  1051 37c. 43! 6.04  1052 38. Let n 
 total number of symbols. Let n  3 
 number of dashes. n!  
 35 3! (n  3)!  1. The circular permutation has no beginning or end. Therefore, the number of different arrangements 1 is always n of a revolution. 2. Sample answer: house or phone numbers where some of the digits repeat 3. Sample answer: The number of permutations of n n! charms on a bracelet with a clasp is . 8!  
 10,080 2! 2! 9!  
 22,680 2! 2! 2! 2! 14!  
 7,567,560 2! 4! 5! 2!  43!  20! 14! 9!  x  log10 413   log10 2  x 8.69 42. h 
 6, k 
 1 3
hp 3
6p 3 
 p (y  k)2 
 4p(x  h) (y  1)2 
 12(x  6) 43. 2(4  3i)(7  2i) 
 2(28  29i  6i2) 
 56  58i  12(1) 
 44  58i  
 20,160 
 302,400  
 126   50! 50!  4! 3! 4! 2! 8! 8! 3! 2! 2! 3! 2! 4!  2.35  1048  21. circular; (12  1)! 
 39,916,800  441  Chapter 13  u u j k 0 3 5 0 3u 2 3u 2 0u i  j  k 0 2 0 2 5 u u  6j u  10k 
 15i  u i 44. u v u w 
 2 2 0 
 5  C(4, 1)  C(3, 2)  
 15, 6, 10 since 2, 0, 3  15, 6, 10 
 30  0  30 or 0 and 2, 5, 0  15, 6, 10 
 30  30  0 or 0, then the resulting vector is perpendicular to u v and u w. 45. x cos 45°  y sin 45°  8 
 0 2  x 2  11.  2    2y  8 
 0  x  2 y  16 
 0 2 y 45˚  12.  8 4 4 O  8  4  P(f) 
 1  P(s)   10. P(s) 
  C(7, 3)  8 x  4  43 
 3 5 12 
 35 12  35 12 — odds 
 23 or 23  35 C(3, 1)  C(4, 2)  P(s) 
  C(7, 3) 36 
 3 5 18   
 35 18  35 18 odds 
 — or 1 7 17  35 4 80   P(rain) 
  100 or 5 4 P(not rain) 
 1  5 1 
 5 1  5 1  odds 
 — 4 or 4  5  12  
 1  35 23  
 35  P(f) 
 1  P(s) 18  
 1  35 17  
 35  8  Pages 856–858  The sparks will be highest at the y-intercept, 82  inches above the center of the wheel. This is 82   8 or about 3.31 inches above the wheel. 46. If x2 
 36, then x  6 or x 
 6. 2x1 
 261 or 32  Exercises  13. P(face card) 
 
  444  52 12 3  or  52 13  6666  14. P(a card of 6 or less) 
 5 2 24  1   2x1 
 261 or  128  6  
 52 or 1 3 10  10  15. P(a black, non-face card) 
 5 2  The correct choice is E.  20  5   
 5 2 or 13  13-3 855–856  17.  Check for Understanding  18.  1. The probability of the event happening is 5050. 2. Answers will vary; see students' work. 3. Sample answer: The probability of the successful outcome of an event is the ratio of the number of successful outcomes to the total number of outcomes possible. The odds of the successful outcome of an event is the ratio of the probability of its success to the probability of its failure. 3  19. 20. 21.  22.  3    4. Geraldo is correct. P(win) 
  2  3 
 5 or 60%. 7  7  1     5. P(softball) 
  3  7  11 
 21 or 3 37  23. 10    6. P(not a baseball) 
  3  7  11 or 21 0   7. P(golf ball) 
  3  7  11 or 0 7  7    8. P(woman) 
  7  4 or 11 C(4, 3)   9. P(s) 
  C(7, 3) 4  odds 
 Chapter 13  4  35 — 31  35  24. P(f) 
 1  P(s) 4  31   
 1  3 5 or 35  
 3 5  10  10  10  10  52 40 10     
 52 or 13 5 5 1    P(red) 
 5  2  3 
 1 0 or 2 1 2 2    P(white) 
  5  2  3 
 10 or 5 52 7   P(not pink) 
  5  2  3 or 10 53 8 4    P(red or pink) 
  5  2  3 
 10 or 5 C(4, 2)  P(2 pop) 
  C(40, 2) 6 1   
 780 or 130 C(8, 2)  P(2 country) 
  C(40, 2) 28 7   
 780 or 195 C(10, 1)  C(18, 1) P(1 rap and 1 rock) 
  C(40, 2) 10  18  
 735 3 180  or  
 13 780 C(22, 2)  P(not rock) 
  C(40, 2) 231 77   
 780 or 260  16. P(not a face card) 
  Probability and Odds  25. Using the Basic Counting Principle, there are 1  1 or 1 way to roll both fives. Using the Basic 1 Counting Principle, P(both fives) 
 3 6.  4  or 3 1  442  C(3, 2)   26. P(s) 
  C(6, 2)  27.  28.  29.  C(3, 1)  C(24, 2)  P(f) 
 1  P(s)   33. P(s) 
  C(27, 3)  3 1 
 1  5 
 1 5 1 4 
 5 
 5 1  5 1  odds 
 — 4 or 4  5 C(4, 2)  P(s) 
 C(6 P(f) 
 1  P(s) , 2) 6 2   
 1  5 
 15 2 3 
 5 
 5 2  5 2 — odds 
 3 or 3  5 C(1, 1)  C(3, 1)  P(s) 
  P(f) 
 1  P(s) C(6, 2) 13 1 
 1  5 
 15 3 1 4  
 1 
 5 5 or 5 1  5 1 — odds 
 4 or 4  5 C(1, 1)  C(2, 1) C(1, 1)  C(3, 1) C(2, 1)  C(3, 1)      P(s) 
  C(6, 2) C(6, 2) C(6, 2) 2 3 6 11    or  
 1  5 15 15 15  34. 35.  30.  31.  32.  233   
 325  P(f) 
 1  P(s) 4  36.  1  4  5 4 — odds 
 1 or 1  5 C(13, 3)  C(13, 2) P(s) 
  C(52, 5) 286  78   12 
 2,598,960 267,696  
 2,598 ,960 429  
 4165   P(4, 2)  P(f) 
 1  P(s) 3736  429    
1 4165 or 4165  37.  4  
 1  15 or 1 5 11  15 11 — odds 
 4 or 4  15 C(11, 3)  P(s) 
  C(27 , 3) 165  
 2925 11  
 19 5 11  19 5 11  odds 
 — 184 or 18 4  195 C(13, 2)  C(11, 1)  P(s) 
 C(27, 3) 78  11  
 2925 22 858   
 2925 or 75 22  75 22 — odds 
 53 or 53  75 C(14, 3)  P(s) 
  C(27, 3) 364  
 2925 28  
 225 28  225 28 —  odds 
 197 or  197  225  92   
1 325  
 1  5 or 5  P(f) 
 1  P(s) 11  P(f) 
 1  P(s)  3  276  
 2925 828 92   
 2925 or 325 92  325 92  odds 
 — 233 or 233  325 1 1   P(s) 
 1   249 or 250 4 P(s) 
 5  429  4165 429  odds 
 — 3736 or 3736  4165 1 1   P(s) 
 1  4 or 5  38. P(s) 
 0.325  P(f) 
 1  P(s) 
 1  0.325 or 0.675  0.325  13    odds 
  0.675 or 27  P(f) 
 1  P(s)  1  1  1  1  1  1     39a. P(s) 
 1 0  9  8 or 720  11   
1 19 5  1  1     39b. P(f) 
 1 0  10  10 or 1000  184  P(s) 
 1  P(f) 1 999   
1 1000 or 1000   
 195  odds 
 P(f) 
 1  P(s)  999  1000 — 1  1000  999  or 1 1  1  1  40a. P(both males) 
 2  2 or 4  22  
 1  75  1  40b. P(s) 
 1  2  53  
 75  
  P(f) 
 1  P(s) 41a.  28   
1 225 197   
 225  41b.  443  1  2  1  1 2 _ odds 
 1 or 1  2 C(15, 10)  C(5, 0) P(s) 
  C(20, 10) 3003  1 21  or  
 184,756 1292 C(15, 8)  C(5, 2)  P(s) 
  C(20, 10) 6435  10  
 184,756 64,350  
 184,756 225  
 646 225  646 225  odds 
 — 421 or 421  646  P(f) 
 1  P(s) 1  1  
 1  2 or 2  P(f) 
 1  P(s) 225   
1 646 421   
 646  Chapter 13  42a. P 
 
  51. Drawing the altitude from one vertex to the opposite side forms a 30°-60°-90° right triangle with hypotenuse 2s. The short side of this right triangle measures s. So the altitude drawn has length s 3. This is the height of the equilateral triangle. The base measures 2s. So the area of the 1 equilateral triangle is   2s  s 3 
  3s2. 2 The correct choice is B.  179,820  151,322  84,475  3273  179,820  151,322 331,142  418,890  0.791 151,322  P(s) 
 1  P(f)   42b. P(f) 
  418,890  151,322   
1 418,890 267,568   
 418,890  odds 
 
  267,568  418,890 —— 151,322  418,890 267,568 133,784  or  151,322 75,661  43. P A  2s 3 s   Q  x  2s  s  s  8x  u Given a pipe PQ and a random cut point, A, APAQ 
 18. If AP is x inches long, then AQ is 8x u inches long. Now, the cut must be made along AP so that the longer piece will be 8 or more times as long as the shorter piece. Thus, the probability 1 x u   that the cut is on AP is  x  8x 
 9 . Since the cut  Page 858  15!  
  20!   2. C(20, 9) 
  (20  9)! 9!  
 167,960 3. Using the Basic Counting Principle, 26  26  26  10  10  10  10 
 175,760,000.  44. This is a circular permutation. (6  1)! 
 5! or 120  12!   4. P(12, 5) 
  (12  5)!  10!   45. C(10, 4) 
  (10  4)! 4!  
  10  9  8  7  6  5  4  3  2  1  6543214321  n  46. Sn 
 2[2a1  (n  1)d] 14  S14 
 2[2(3.2)  (14  1)1.5]  6.  
 7(25.9) 
 181.3 47. Let y 
 7log7 2x log7 y 
 log7 2x y 
 2x So, 7log7 2x 
 2x. 48. Center: (7, 2) r2 
 (10  7)2  (8  2)2 
 32  (10)2 or 109 (x  7)2  (y  2)2 
 109 49. r 
 3  2 or 6   5  7.  
  9!  
 181,440 2! 10!  
 4200 3! 4! 3!  8. This is a circular permutation. (8  1)! 
 5040 C(13, 2)   9. P(both hearts) 
  C(52, 2) 78  1    
 1326 or 17 C(3, 1)  C(3, 1)   10. P(s) 
  C(12, 2) 33 
 6 6 9 3   
 66 or 2 2 3  22 3  odds 
 — 19 or 19  22  5  5  12  11  10  9  8  7  6  5  4  3  2  1  7654321  
 95,040 5. Using the Basic Counting Principle, 18  18  3  6 
 5832.  
 210  v 
 (  4) or 4  15  14  13  12  11  10  9  8  7  6  5  4  3  2  1  10  9  8  7  6  5  4  3  2  1  
 360,360  can be made on either end of the pipe, the actual 2 probability is 9.  
  Mid-Chapter Quiz   1. P(15, 5) 
  (15  5)!    6(cos 4  i sin 4) 
 6 2  
2i 2   2   
 32   32 i u 50. u 
 3, 5  4, 2 
 3  (4), 5  2 
 1, 3  13-4  P(f) 
 1  P(s) 3  
 1  22 19  
 22  Probabilities of Compound Events  Pages 863–864  Check for Understanding  1. The occurrence of one event does not affect another for independent events. The occurrence of the first event affects the occurrence of a second for dependent events.  Chapter 13  444  2a.  diamonds  spades, clubs, hearts  aces  4  4  7  8  26  25  24     20. dependent, 1 5  14  13 
 195 23  22  21  20  19  18  21. dependent, 52  51  50  49  48  47  46  45  44  17  43  16  15  14  19      4 2  41  40 
 1,160,054 12  8  8  32    22. dependent, 28  2 7  26 
 819 5  2b. No, one of the aces can be an ace of diamonds. 2c. P(ace or diamond) 
 P(ace)  P(diamond)  P(ace and diamond) 3. Answers will vary; see students' work. 4. independent, 5. 6. 7. 8. 9.  6  36    3  36  4  1  1  1  11  4  26  2  7  26  12  6  8     26. inclusive, 5 2  52  52 
 13  27. inclusive, 52  52  5 2 
 13  36  35  34  28. exclusive P(at least 3 males) 
 P(3 males)  P(4 males)  P(5 males)  33  C(5, 3)  C(4, 2)  C(5, 4)  C(4, 1)  60  20  81  9  1     
 126  126  126   
 126 or 14  0.518 12. P(5 odd numbers or 5 multiples of 4) 
 P(5 odd numbers)  P(5 multiples of 4)  29. exclusive P(sum of 6 or sum of 9) 
 P(sum of 6)  P(sum of 9)  17  16  15  14  13  35  34  33  29  28  27  26  C(7, 3)  C(7, 3)  25  13  12  94  93  92  435,643  1 6  34  39  or  1 6  15  6  22  11  1  
 64 or 32  C(8, 2)  C(5, 0)  32. inclusive  
 78  78 
  1 6   
 64  6 4  64     
 C(13, 2) C(13, 2)  68  78  302  7  
 C(6, 4)
2  C(6, 5)
2  C(6, 6)
2  15. P(at least 1 right handed pitcher 
 P(1 right-handed pitcher)  P(2 right-handed pitchers) 28  2114  147  31. exclusive P(at least 4 tails) 
 P(4 tails)  P(5 tails)  P(6 tails)  91   
 560,175  40  735    
 300 3 or 429        
 100  99  98  97  96  95  C(8, 1)  C(5, 1)  1225      
 3003  3003  3003  3003  14. P(none if 6 clocks are damaged) 95  C(7, 5)  C(7, 1)    C(14, 6)  0.025  0.007  0.001 0.032 96  C(7, 4)  C(7, 2)  C(7, 6)  C(7, 0)  10  11  1       
 C(14, 6) C(14, 6) C(14, 6)        
7 5  74  73  72  71  14  9  30. exclusive P(at least three women) 
 P(3 women)  P(4 women)  P(5 women)  P(6 women)    
 
75  74  73  72  71  
75  74  7 3  72  71  36  4   
 3 6 or 4  0.029  0.0004 0.029 13. P(5 even numbers or 5 numbers less than 30) 
 P(5 even numbers)  P(5 numbers less than 30)  P(5 even numbers and 5 numbers less than 30) 37  5   
 3 6  36           
 
7 5  74  73  72  71   
 75  74  73  72  71  34  C(5, 5)  C(5, 0)       
 C(9, 5) C(9, 5) C(9, 5)  66 65 64 63 62 P(selecting 5 two digit numbers) 
 75  74  73  72  71  35  4   25. inclusive, 6  6  3 6 
 36  0.025  36  4  256  37  37  4   
 2401  4 3 2   dependent, 1 0  9 
 15 4 3 2 dependent, 6  5 
 5 4 6 10   exclusive, 1 3  13 
 13 15 11 6 20  inclusive, 27  27  2 7 
 27 4 4 8 2    exclusive, 5 2  52 
 52 or 13  38  35  P(winning next four games) 
 7  7  7  7  10. P(selecting 5 even numbers) 
 75  74  73  72  71 11.  7  24. independent 4 P(winning) 
 7  1  72  
  4     23. independent, 1 6  16  16 
 1024  
542  531   
2562  2551  
522  511 12  650  660  55  2     
 2652  2652  2652   
 2652 or 221  Pages 864–867 16. dependent,  Exercises 5  9    4  8  33. exclusive P(at least 2 rock) 
 P(2 rock)  P(3 rock)  5  
 1 8  C(6, 2)  C(5, 1)  5  25     
 C(11, 3) C(11, 3)  1  1  1    
 165  165  18. independent, 6  6 
 3 6 4  C(6, 3)  C(5, 0)  5  17. independent, 9  9 
 8 1 3  75  20  95  19    
 165 or 33  2  19. dependent, 7  6 
 7  445  Chapter 13  26  25  24  23  46. P(word processing or playing games) 
 P(word processing)  P(playing games)  P(both)  22  34. P(all red cards) 
 52  51  50  49  48 253   
 9996  2  35. P(both kings or both aces) 
 P(both kings)  P(both aces) 3  4  12  12  24  2652  2  221  29  47. P(rain or lightning) 
 P(rain)  P(lightning)  P(both)  3    
 2652  2652  
  or  3 4    48. P(even sum) 
 P(3 even cards)  P(2 odd cards and 1 even card) C(5, 3)  C(4, 0)  650  12  660  2652  55  221  3  2  1  or  2  38. P(2 pennies) 
 
  10  30  40  10  
 84  84     
 2652  2652  2652  
  
 84 or 2 1  5  21 1  21    49. P(at least 3 women) 
 P(3 women)  P(4 women)  P(5 women)  4  20  C(6, 3)  C(7, 2)  240  16  15  7  40. P(at least 1 nickel) 
 1  P(no nickels) 
1
    
 
    59  93    3  958  97  7  93  97  9021    
 10,000  10,000  17  30  9979   
 10,000  51. P(supplies or money) 
 P(supplies)  P(money)  P(both)  C(5, 1)  C(7, 1)    
 C(21, 2)  C(21, 2) 36  210 71  210  531  6        
 
 100  100  100  100   
 100  100   13  20  41. P(2 dimes or 1 penny and 1 nickel) 
 P(2 dimes)  P(1 penny and 1 nickel) C(9, 2)  105  50. P(at least 1 doctor) 
 P(1 doctor)  P(both doctors)  4  14  21  420    
 1287 or 143  6    
 420 or 7  
  C(6, 5)  C(7, 0)     
 1287  1287  1287       
 
2 1  20   
 21  20   
 21  20  6  C(6, 4)  C(7, 1)       
 C(13, 5) C(13, 5) C(13, 5)  39. P(2 nickels or 2 silver-colored coins) 7  C(4, 2)  C(5, 1)     
 C(9, 3) C(9, 3)       
 
5 2  51   
 52  51   
 52  51  4  1  
 5  13 10  52  37. P(both red or both queens) 
 P(both red)  P(both queens)  P(both red queens) 25  2  
 5  5  5  36. P(all diamonds) 
 
  26  1  
 60     
 
5 2  51   
 52  51  4  1  
 5  3  4  812  625  1062  531  375     
 2500  2500  2500  35  210    
 2500 or 1250 5  4  3  52a.  2     42. P(all female) 
 1 0  9  8  7 1  P (A )  
 4 2  P (B )  43. P(all female or all male) 
 P(all female)  P(all male) 4  5  3  2  5  4  3  2         
 1 0  9  8  7  10  9  8  7 1  1  2  42  1  21  P (C )   
 4 2  42  
  or  52b. P(A or B or C) 
 P(A)  P(B)  P(C)  P(A and B)  P(A and C)  P(B and C)  P(A and B and C). You must add the intersection of all three sets which have not been accounted for. 53. P(action video or pop/rock CD or romance DVD)  44. P(at least 3 females) 
 P(3 females)  P(4 female) C(5, 3)  C(5, 1)  C(5, 4)  C(5, 0)     
 C(10, 4) C(10, 4) 50  5  55  11    
 210  210   
 210 or 42  4  7  45. P(at least 2 females and at least 1 male) 
 P(2 females and 2 males)  P(3 females and 1 male) C(5, 2)  C(5, 2)  C(5, 3)  C(5, 1)     
 C(10, 4) C(10, 4) 100  50  150  5    
 210  210   
 210 or 7  Chapter 13  446  1  5  2  1  1  1   2  11  9  7  4  4 4  0.93  54a. First consider the probability that no 2 students have the same birthday. The first person in the class can have any birthday; there are 366 choices out of 366 days. The second person has only 365 choices out of 366 days, and so on. So, P(2 students with the same birthday) 
 1  P(no 2 students have the same birthday).  60.  x
  365  
1  366!  348!  36618  
1  366!  (366  348)!  36618  364  349  P(366, 18)   
1 36618  0.346 P(366, n)  4 log 3  2 log 12  log 12  log 3  x 6.7549 61. Let y 
 income and x 
 number of $1.00 increases. income 
 (number of customers)  (cost of a ticket) y 
 (400  20x)(3  x) y 
 1200  340x  20x2 y  1200 
 20(x2  17x) y  1200  1445 
 20(x2  17x  72.25) (y  2645) 
 20(x  8.5)2 The vertex of the parabola is (8.5, 2645). An increase of $8.50 will give a maximum profit of $2645. The price of each ticket should be 3  8.5 or $11.50.      
 1   366  366  366  . . .  366  366  12x  2 
 3x  4 (x  2) log 12 
 (x  4) log 3 x log 12  2 log 12 
 x log 3  4 log 3 x log 12  x log 3 
 4 log 3  2 log 12 x (log 12  log 3) 
 4 log 3  2 log 12  1     54b. 1   366n 2  54c. In part a, there is only a 0.346 probability that 2 students have the same birthday. This is too small. Substitute numbers greater than 18 for n in the inequality of part b. When n is 23, P is about 0.51. So, 23  18 or 5 more students are needed in the class. 55a. inclusive  
2  A 
 4 k  62.  
2    8270 
  4(7) 231,560  
2 
  
 271.5 yards x 
 x1  ta1 y 
 y1  ta2 x 
 1  t(2) y 
 5  t(4) x  1 
 2t y  5 
 4t x  1, y  5 
 t 2, 4 64. 2 tan x  4 
 0 tan x 
 2 tan1(tan x) 
 tan1 2 x 
 63° 26
 65. Since ADB  CBD and they are alternate interior angles,  AD BC L . Simply because a 
 45 does not mean b 
 45, so you cannot conclude that 
3 bisects ABC. The correct choice is B.  55b.  63.  Dead Battery  Flooded Engine  55c. P(flooded engine or dead battery) 
 P(flooded engine)  P(dead battery)  P(both) 1  2  1  
 2  5  10 4  
 5 1  56. P(two threes given a sum of six) 
 5  13-5  57. (7  1)! 
 720 58. Let b 
 basket. Let m 
 miss. 20  Expand (b  m)20 
   r
0  20!  r!(20  r)!  b20  r mr  Pages 870–871  Check for Understanding  1. Sample answer: If A and B are independent events, then P(AB) 
 P(A). Thus, the formula for conditional probability becomes P(A) 
  Find the coefficient of the b15m5 term where r 
 5. 20!  5!(20  5)!  Conditional Probability  b205m5 
 15,504b15m5  P(A and B)  P(B)  15,504 59. No, the spill will spread no more than 2000 meters away.  or P(A)  P(B) 
 P(A and B). This is the  formula for the probability of independent events. 2. S 
 {J spades, Q spades, K spades, J clubs, Q clubs, K clubs} 3. Answers will vary; see students' work.  480   a1 
 1200; r 
  1200 or 0.4 a1   s
 1r 1200   
 1  0.04  4. P(cubes match  sum greater than 5) 
  
 2000  
  447  4  36 — 26  36 2  13  Chapter 13  5. P(queen  face card) 
 
 6. P(all heads   7. P(all heads   8. P(all heads   4  52 — 12  52 1  3  13c. P(not rejected  counterfeit) 
 
  1  8 — first coin is a head) 
 4  8 1 
4 1  8 at least 1 head) 
 — 7  8 1   
7 1  8 at least 2 heads) 
 — 4  8 1   
4  Pages 872–874  
 
  Exercises  14. P(1 head  at least 1 tail) 
 
 15. P(Democrat  man) 
 
  2  4 — 3  4 2  3  4  12 — 8  12 1  2  16. P(first bag  first chip is blue) 
  9. P(numbers match  sum greater than or equal to 9) 2  36 — 10  36 1  5  1  100 — 25  100 1  25  
  4  16 — 10  16 2  5  17. P(girls are separated  girl at an end) 
  10. P(sum is even  sum greater than or equal to 9) 
 
  
  4  36 — 10  36 2  5  18. P(number end in 52  number is even) 
  11. P(numbers match or sum is even  sum greater than or equal to 9) 
 
  
  4  36 — 10  36 2  5  19. P(2 odd numbers  sum is even) 
 
  68  62   12a. P(disease prevented) 
  100  100 13  
 20  20.  12b. P(disease prevented  vaccine) 
 
  68  200 — 100  200 17  25  21.  12c. P(disease prevented  conventional treatment) 
 
 13a.  13b.  62  200 — 100  200 31  50  2  52 — P(ace  black) 
 26  52 1  
 1 3 2  52 P(4  black) 
 — 26  52 1 
 1 3  22. P(face card  black) 
 
  69  100 — P(legal  accepted) 
 70  100 69 
 70 6  100 — P(rejected  legal) 
 75  100 2  
 2 5  Chapter 13  321  5! 43214321  5! 1  8  6  52 — 26  52 3  13  23. P(queen of hearts  black) 
  0  52 — 26  52  
0 24. P(6 of clubs  black) 
 
  448  1  52 — 26  52 1  26  20  72 — 32  72 5  8  12  24 — 20  24 3  5  25. P(jack or ten  black) 
 
  36. A 
 the sum of the cards is 7 or less B 
 at least one card is an ace B
 
 both cards not an ace  4  52 — 26  52 2  13  C(48, 2)  188  26. P(second marble is green  first marble was green) 
 
  
  3 2    8 7 — 3  8 2  7  P(A  B) 
 
  3 5    8 7 — 3  8 5  7  
  37. P(sum greater than 18  queen of hearts) 1  19     52 51 
 — 51 1    52 51  5 4    8 7 — 5  8 4  7  
  C(4, 2)  C(4, 1)  C(20, 1)  C(52, 2) 43   
 663 P(A and B)  P(B) 43  663 — 33  221 43  99  P(A and B) 
  28. P(second marble is yellow  first marble is yellow) 
  19  
 51 38a. C(1, 1)  C(1, 1)  C(4, 1)  C(6, 3)  S U  C  29. P(salmon  bass) 
  
  15  C(1, 1)  C(5, 2)  C(6, 3) 2 4  or  5 10  25  5 155  30. P(not walleye  trout and perch) C(1, 1)  C(1, 1)  C(3, 1)  C(6, 3)  
 
  38b. P(cancer  smokes) 
  C(1, 1)  C(1, 1)  (4, 1)  C(6, 3) 3  4  
  C(1, 1)  C(1, 1)  C(3, 1)  C(6, 3)  P(A  B) 
  C(5, 3)  C(6, 3) 3  
 1 0 C(1, 1) C(1, 1)  (2, 1)  C(6, 3)  2  
 4 or 33.  C(4, 3)  C(6, 3) 1  2  Brown Hair  4  or 5  41. P(passes  studied) 
  Brown Eyes  50% 10%  120  500 — 150  500  Four out of five people who ask questions will make a purchase. Therefore, they are more likely to buy something if they ask questions. 40. Sample answers: The rolls are independent. The number cubes do not have a memory, whether they are fair or biased. Probability does not guarantee an outcome.  32. P(perch and trout  neither bass nor walleye) 
  4  5  20%  
  P(passes and studied)  P(studied) 2  3  P(studied) 2  66  33  42b. P(chip from 3-D Images  defective)  0.10   P(brown eyes  brown hair) 
  0.60  35. P(no brown  5    42a. P(defective) 
  1000 or 500  Exercises 33–35  34. P(no brown  5  P(studied) 
 3  4 or 6  20%  
  25  200 — 30  200 5  6  39. A 
 person buys something B 
 person asks questions  31. P(bass and perch  not catfish) 
  33    P(B) 
 1  P(B
) 
 1   221 
 221  27. P(second marble is yellow  first marble was green) 
  188    P(B
) 
  C(52, 2) 
 221  1  6  
  0.20  hair  brown eyes) 
  0.30 2 
 3 0.20  eyes  no brown hair) 
  0.40 1 
 2  
  21  1000 — 66  1000 7  22 934  467    42c. P(functioning) 
  1000 or 500  449  Chapter 13  42d. Sample answer: A chip from CyberChip Corp. has the least probability of being defective.  53.  6 ft2  
 0.05 21  P(defective from 3-D Images) 
  300  x x  4 ft If x 
 3 ft then the photo would have a negative  
 0.07 20  P(defective from MegaView Designs) 
  200  1  length and width. So, x 
 2 ft or 6 in.  
 0.10  P(A and B  5   by definition. 43. P(A  B) 
  P(B)   54. f(x) 
  x2  4  So, if P(A) 
 P(A  B) then by substitution P(A) 
  5   
 (x  2)(x  2)  or P(A and B) 
 P(A)  P(B). Therefore,  f(x) is discontinuous when x 
 2; f(x) is undefined when x 
 2.  the events are independent. 44. P(at least 3 women) 
 P(3 women)  P(4 women)  P(5 women) C(6, 3)  C(7, 2)  C(6, 4)  C(7, 1)  55. Drop an altitude from B to  AE , from u C to ED , and from E to B C . Label the diagram as shown.  C(6, 5)  C(7, 0)       
 C(13, 5) C(13, 5) C(13, 5) 420  105  531  59  6     
 1287  1287  1287   
 1287 or 143    3(0.5)b 
 3(0.5)1  3(0.5)2  3(0.5)3  .  S
 
  a1  r 1.5  0.5  h  h  E 1  D  1 (BC)h 2 1  area of shaded region 
 2(AE)h  2(ED)h 1  a1 
 1.5, r 
 0.5  
 2h(AE  ED) BC 
 AE  ED since opposite sides of a parallelogram are equal. So, the ratio of the areas 1 1 is 2h(BC)2h(BC) or 11.  or 3  47. They are reflections of each other over the x-axis. y  48.  C  h  area of unshaded region 
  . .  
 1.5  0.75  0.375  . . .  b
1  B  A  45. C(9, 4) 
 126 46.  x  x  25   P(defective from CyberChip) 
  500  P(A and B)  P(B)  (4  2x)(3  2x) 
 6 12  14x  4x2 
 6 x 4x2  14x  6 
 0 2(2x  1)(x  3) 
 0 x 1 x 
 2 or x 
 3 x  x  The correct choice is B. 6 4 2 64 O  13-6  x  2 4 6  4 6  Page 877  The Binomial Theorem and Probability Graphing Calculator Exploration  1. S 
 {0, 1, 2} 2  r 
cos v   49.      2  2. P (Bobby wins) 
 3  5
0  3. Answers will vary. In 40 repetitions, it may be around 0.22. This means that there were exactly 5 wins for 8 or 9 of the 40 repetitions.    r
cos v cos 2  sin v sin 2 
 5 0  r sin v 
 5 y 
 5 y 
 3  2t  50. x 
 4t x  4 x  4  y3  2  
t  
t  y3  
 2 x  y 
 2  3  2 5  y  0.26 5. There is not a large enough sample of trials. 6. Increase the number of repetitions.  (4, 5)  (0, 3) (4, 1)  O  Page 878  x  1   
 2(8)2
1   180  98  71  54.7 ft2 x  52. tan 27° 
 25 12.7  Chapter 13  Check for Understanding  1a. Yes, it meets all three conditions. 1b. No, there are more than 2 possible outcomes. 1c. No, the events are not independent. 2. Sample answer: the probabilities derived from a simulation rather than an actual event  51. A 
 2r2 v 1  1 1  4. P(winning 5 games) 
 C(6, 5)  
3  
3  x; 12.7 m  450  3. First, determine P(right) and P(wrong). Second, set up the binomial expansion (pr  pw)5. Third, determine the term of the expansion. Fourth, substitute the probability values for pr and pw. Last, compute the probability of getting exactly 2 correct answers.  15. P(no more than 3 times correct) 
 1  P(correct 4 times) 2 4 1 0  3  
 1  C(4, 4)
3 
11 
1  1 1 5 4  6  4. P(only one 4) 
 C(5, 1) 
6 
  
   
 
 
  
6  1 1 5 4   6 6  17. P(7 correct) 
  1250  7776  
 
  6. P(at least three 4s) 
 P(three 4s)  P(four 4s)  P(5 fours) 
 
 
  2 2 1 2  3  16. P(correct exactly 2 times) 
 C(4, 2)
3  
  
   C(5, 1)
  
  3 
  
56  1 3 5 C(5, 3) 6 6 1 5  C(5, 5) 6 250 25     7776 7776 23  648  2  1  5  193  
11 1  20. P(at least half correct)  7203  1  
170  2835 243   
 1   100,000  100,000   5   C(5, 5)  
1 0 3  0  
170    48,461  21. P(4 4 4 1 2 C(6, 4) 5 5 3840 768  or  15,625 3125  
 
   5 10 3 3  8  12. P(10 stocks make money) 
 C(13, 10) 
8  
   22. P(3  0.1372  Exercises  13. P(never the correct color) 
 
 
  8  
 4  2 7   
   to 18  
 
   
 
   23. P(at least 2 heads)  2 3 1 1    C(4, 3 3 1 16   1    1 3 81  
 
   
 
 
  2 4 1 0   3 3  4)
  1 2 2 2   answer 3 8 1 1 4  6  9  9  8 1  81 24 9    81 81 11  27  
 C(4, 2)
3  2 0 1 4 C(4, 0) 3 3 1 1  1  8 1 1  81  14. P(correct at least 3 times)  
   1 5 1 5   answer 2 1 1 386       
 252  32  32  1024 252 386   
 1024  1024 319  
 512 1 4 2 0 heads) 
 C(4, 4) 3 3 1 
 1  8 1 1 1 
 8 1 1 3 2 1 heads) 
 C(4, 3) 3 3 1 2  
 4  2 7  3 8 
 8 1  
 C(10, 5)
2  10. P(having rain no more than three days) 
 1  [P(rain on 4 days)  P(rain on 5 days)  
 C(4, 3)
  
   1  1024   
 1024  4  
170     
 100,000 or 20,000  Pages 878–880  1 0 1 10  2  19. P(all incorrect) 
 C(10, 0)
2  9. P(having rain on exactly one day)  
  1  1  1   
 512  0  
130   16,807  11. P(4 do not collapse) 
  1  1  386   
 100,000  96,922  1   
 1024 7    
 100,000 or 50,000  1  1  8. P(not having rain on any day) 
 C(5, 5)
1 0  4  3      10   512  2  1  1024  1  
   1  3  7        
 210  6 4  16  120  128  8  45  256  4 1 5 5 0  6  
 1  C(5, 4)  
1 0  4  
   C(10, 7)
12 
12 1 2 1 9 1 1  C(10, 
  
2  C(10, 9)
2 
2 1 10 1 0  C(10, 10)
2 
2   
 7776  36,015  
 
   1 6 1  2 1 8   8) 2  1  7776  1    
   1  9  8 
 27 1 7 1 3 C(10, 7) 2 2 1 1   120   128  8 15  128  
 C(10, 6)
2  1 4 5 1   6 6  7. P(exactly five 4s) 
 C(5, 5)
6  3  4  9  18. P(at least 6 correct)  
  
   C(5, 4)
  
  0 
  
56  
 C(5, 1)
1 0  1  16  81  65  5. P(no more than two 4s) 
 P(no 4s)  P(one 4)  P(two 4s) 5  
   
 81  3125  7776  1 0 5 C(5, 0) 6 6 1 2  C(5, 2) 6 3125 3125     7776 7776 625  648  16  81  
   
   to 22  answer to 21  1 6 3 4  4 1 81    4096 256  24. P(6 correct answers) 
 C(10, 6)
4 
 210   16  
 27  
   0.016  451  Chapter 13  1 5 3 5  4 1 243    1024 1024  25. P(half answers correct) 
 C(10, 5)
4 
 252   35. Enter 16 nCr X on the Y-menu of your calculator. 16 nCr X represents the coefficients of the binomial expansion where X is the number of games won. P(winning at least 12 games)  
   0.058 26. P(from 3 to 5 correct answers) 
 
  1 3 3 7 C(10, 3) 4 4  C(10, 4) 1 5 3 5  C(10, 5) 4 4 1 2187 1   120  6 4  16,384  210  256  1 243    252   1024  1024  
 
  
 
   729  4096  1   
   36b. P(between 2 and 6 missiles hit the target) 
 1  [P(0 missiles hit the garget)  P(1 missile hits the target)]  1  1.049  104 
 120   
   
 1  
1  1   
 252   2 5 3 5  5 32 243    3125 3125  
   37. P(all men or all women) 4  7  3  
 1  1020 38b. P(exactly half have the disease) 1  1  10  0  75  0.807 40. P(less than or equal to 3 policies) 
 1  P(4 policies)  
   1 4 1 0  2  
 1   C(4, 4)
2 
 1  
1   32. P(at least 2 heads) 1 2 1 1 1 3 1 0 
 C(3, 2) 2 2  C(3, 3) 2 2 1 1 1 
 3  4  2  1  8  1 4 
 8 1 
 2 1 2 1 1 P(exactly 2 tails) 
 C(3, 2) 2 2 1 1 
 3  4  2 3 
 8  
 
   
  15  16  1  16   1  
   or about 0.94  41a. If Trina walks 100 meters, then she has flipped the coin 10 times. To end up where she began, she walked north and south 5 times each. 1 5 1 5  2 1 1    32 32  P(back at her starting point) 
 C(10, 5)
2  
 
   
 252   
   0.246 41b. The closest Trina can come to her starting point is if she flips 6 heads and 4 tails or 4 heads and 6 tails. However, this places her 20 meters from her starting point. The answer for part b is the same as that for part a, 0.246.  34a. The values of the function for 0  x  6 are the coefficients of the binomial expansion. 34b. Change 6 nCr X to 8 nCr X on the Y-menu.  Chapter 13  0  
190   
19060  4 1 96 74    C(75, 1)
 100  
 100   4  1  33.  20  
190    
 1  C(75, 0)
 100   
 4  
 
   0  
140  6.4  106 39. P(at least 2 people do not show up) 
 1  [P(0 people do not show up)  P(1 person does not show up)]  1 3 1 0  2  3)
2  10  
 C(20, 10) 
1 0  0.166 31. P(3 heads or 3 tails) 
 1  8   6  38a. P(all carry the disease) 
 C(20, 20)
1 0  64 81 128 27      
 210   15,625  625  120  78,125  125 3 256 9 512      45   390,625  25  10  1,953,125  5 1024  1 9,765,625  1  
   10   C(10, 10)
1 0 1  4  1 3 1 0   C(3, 2 1 1  1  8  1  0  
160   0.0062  
   C(10, 7)
25 
35 3 2 2 9 3 1  C(10, 
  
5  C(10, 9)
5 
5 2 10 3 0  C(10, 10)
5 
5  
 C(3, 3)
2  10  
 C(10, 10)
1 0  30. P(at least 6 point up) 2 6 3  5 2 8 8) 5  
 
    1077  0.201 
 C(10, 6)
5  
    
 3125  0.215 29. P(exactly 5 point up) 
 C(10, 5)
5  1 0 4 6 1 1 4 5   C(6, 1) 5 5 5 4096 1 1024   6     15,625 5 3125  
 1  C(6, 0)
5  2 3 3 7  5 8 2187     125 78,125  28. P(exactly 3 point up) 
 C(10, 3)
5  3  on each trial is 20% or 5.  2 10 3 0   5 5  
1  13  0.45 or 45% 36a. A success means that a missile hits its target. There are 6 trials and the probability of success  0.25  0.15  0.06 0.46 1024  9,765,625  4  
130   560  
170  
130  7 14 3 2   120  
1 0  
 10  7 15 3 1 7 16 3 1     16  
1 0  
 10   1  
 10  
 10  7  
 
   27. P(all point up) 
 C(10, 10)
  12  
 1820  
1 0  1 4 3 6   4 4  452  50. 4  12 21 62 72 4 32 44 72 1 8 11 18 0 Since the remainder is 0, x  4 is a factor of the polynomial.  41c. P(exactly 20 meters from the starting point) 1 6 1 4 1 4 1 6   C(10, 4) 2 2 2 1 1 1 1     210     64 16 16 64  
 C(10, 6)
2 
 210   
   
 
   0.41 42. P(sum less than 9  both cubes are the same) 
 
  51.  4  36 — 6  36 2  3  1  Area of page for text  Area of entire page  
 
 
  (9  1  1)(12  1.5  1.5)  9  12 63  108 7  12  The answer is 7/12.  43. P(letter is contained in house or phone) 5 5 3   
 2 6  26  26 7  
 2 6  Chapter 13 Study Guide and Assessment  44a. 80, 75, 70, . . . 44b. T 
 80  5n 44c. T 
 125 40,000  n
 1000 or 40  Page 881  125 
 g 5(40) 75 
 g; 75° F 45. 3x  1 
 6x (x  1)log 3 
 x log 6 x log 3  log 3 
 x log 6 x log 3  x log 6 
 log 3 x(log 3  log 6) 
 log 3 log 3  x
 log 3  log 6  7. mutually exclusive 9. conditional  Pages 882–884  
cos  47. 2   i sin     2  6!  
   
 2  0  2  i  654321  3  2 1  
 120 8!   15. P(8, 6) 
  (8  6)!  
  87654321  21  
 20,160  
  (2)2   82  5!   16. C(5, 3) 
  (5  3)! 3!  
 68  or 217   54321   
 21321  55 cm  
 10  106˚  71 cm  Skills and Concepts   14. P(6, 3) 
  (6  3)!  i 
 0  2 u 48. WX 
 6  8, 5  (3) or 2, 8 u (6  8 )2  (5  ( 3))2 WX  
   49.  2. failure 4. probability 6. permutation with repetitions 8. sample space 10. combinatorics  11. Using the Basic Counting Principle, 3  2  1 or 6. 12. Using the Basic Counting Principle, 5  4  3  2  1 or 120. 13. Using the Basic Counting Principle, 6  5  4  3  2 1 or 720.  x 0.38 46. h 
 0, k 
 3, a 
 7, b 
 5, c 
 26  Center: (0, 3) Foci: ( 26 , 3) Vertices: major axis → (7, 3) and (7, 3) minor axis → (0, 2) and (0, 8)   2  Understanding the Vocabulary  1. independent 3. 1 5. permutation  11!   17. C(11, 8) 
  (11  8)! 8!  d  
  11  10  9  8  7  6  5  4  3  2  1  32187654321  
 165 d2 
 712  552  2(71)(55) cos 106° d 101.1 cm  18.  
 
  106˚  71 cm  P(6, 3)  P(5, 3)  6!  (6  3)!  5!  (5  3)! 65432121  54321321  
2  d  5!  3!    19. C(5, 5)  C(3, 2) 
  (5  5)! 5!  (3  2)! 2!  74˚  1  321   
 1   121  55 cm  
3  d2 
 712  552  2(71)(55) cos 74° d 76.9 cm  453  Chapter 13  1  20. There are P(5, 5) ways to arrange the other books if the dictionary is on the left end. The same is true if the dictionary is on the right end. 5!  2  P(5, 5) 
 2   (5  5)!  odds 
  54321  
 2  1  
  
 240 3!  321  7654321  5432121  5   
 1296  
 3  21 or 63  36. dependent, P(two yellow markets) 4 3  
 1 0  9  54321  22.  5!  2! 2!  23.  10!  2! 3! 3!  1  14 — 13  14 1  13  35. independent, P(sum of 2)  P(sum of 6) 1 5  
 3 6  36  7!    21. C(3, 2)  C(7, 2) 
  (3  2)! 2!  (7  2)! 2!  
 121   13   34. P(s) 
 1 4 ; P(f) 
 14   
 2121  
 30 
  2  
 1 5  10  9  8  7  6  5  4  3  2  1  21321321  37. P(selecting a prime number or a multiple of 4) 
 P(prime number)  P(a multiple of 4)  
 50,400  24.  8!  2!  25.  6!  3! 2!  26.  9!  3! 2!  
  87654321  21  6 9  
 20,160 
  
 1 4  654321  32121  38. P(selecting a multiple of 2 or a multiple of 3) 
 P(multiple of 2)  P(multiple of 3)  P(multiple of 2 and 3)  
 60 
  987654321  32121  7  30.  31.  32.  33.  Chapter 13  2  9  
 1 4  C(7, 3)  C(4, 0)  C(5, 0)  C(16, 3) 35  1  1  
 560 1 
 1 6 C(7, 2)  C(4, 1)  C(5, 0) P(2 pennies and 1 nickel) 
  C(16, 3) 21  4  1   
 560 3 
 2 0 C(7, 0)  C(4, 3)  C(5, 0) P(3 nickels) 
  C(16, 3) 141   
 560 1  
 140 C(7, 0)  C(4, 1)  C(5, 2) P(1 nickel and 2 dimes) 
  C(16, 3) 1  4  10  
 560 1   
 14 1 15     P(s) 
 16 ; P(f) 
 16 1  16 — odds 
 15  16 1 
 1 5 3 17  P( f) 
  P(s) 
  20; 20 3  20 odds 
 — 17  20 3  
 1 7 1 139    P(s) 
 140 ; P(f) 
  140 1  140 — odds 
 139  140 1  
 139  27. P(3 pennies) 
  29.  4    
 1 4  14  14  
 30,240  28.  3   
 1 4  14  39. P(selecting a 3 or a 4) 
 P(3) or P(4) 1  1   
 1 4  14 1  
 7 40. P(selecting an 8 or a number less than 8) 
 P(8)  P(less than 8) 1  7   
 1 4  14 4  
 7 41. P(sum less than 5  exactly one cube shows 1) 
 
  4  36 — 10  36 2  5  42. P(different numbers  sum is 8) 
 
  4  36 — 5  36 4  5  43. P(numbers match  sum is greater than or equal to 5) 
 
  4  36 — 30  36 2  15  1 1 1 3  2  44. P(exactly 1 head) 
 C(4, 1)
2 
4 
  1  4  1  2  1 0 1 4  2  45. P(no heads) 
 C(4, 0)
2 
11 1  
 1 6  454    1  8  1  16  
   
   1 2 1 2  2  46. P(2 heads and 2 tails) 
 C(4, 2)
2 1  2. 4x – 3  8 
 32 4x  – 3 
 24 x  – 3
6 (x–3 )2 
 62 x – 3 
 36 x – 3 
 39 The correct choice is E. 3. Find the probability of selecting a green marble from the jar now.  
   1  
 6  4  4 3  
 8 47. P(at least 3 tails) 
 P(3 tails)  P(4 tails)  1 3 1 1   C(4, 2 1 1 1 4  8  2  1  1 6 1 1 1 5    or  4 16 16  
 C(4, 3)
2  
 
  Page 885  
   1 4 1 0  2  4)
2  
   number of green marbles  total marbles  3  1   
 1 5 or 5  Let x represent the number of green marbles 1 2 added so the probability equals 2  5 or 5.  Applications and Problem Solving  3x  48. C(1, 1)  C(6, 4) 
 1  15 or 15  number of green marbles  total marbles  7! 49.  
 2520  5(3  x) 
 2(15  x) 15  5x 
 30  2x 3x 
 15 x
5 The correct choice is C.  2  50. P(at least 1 good chip) 
 1  P(both defective chips) C(3, 2)   
1 C(15, 2) 1  sum of terms   Average 
  number of terms  4.  
 1  3 5 34  20 
  
 35 51b. P(Reba's name, then a male name) 
 
  Page 885    7  14  Open-Ended Assessment 1  1  sum of five terms  5  sum of five terms 
 100 Since one of the numbers is 18, the sum of the other four is 100  18 or 82. The correct choice is C. 5. Select specific numbers for the problem. Let x  y 
 8. Let x  z 
 12. Let z 
 7. x  7 
 12, so x 
 5, an odd number. 5  y 
 8, so y 
 3, an odd number. Statement I is false, and statement III is true. Since y  z or 3  7 
 10, an even number, statement II is true also. The correct choice is E. 6. Since the probability of selecting a blue marble is  7  51a. P(female name excluding Reba) 
 1 5 1  15 1  30  2    
 15  x 
 5  1  1. Yes; sample answer: x  2 
 12, so x 
 6; Two marbles are chosen from a box containing 6 red, 4 blue, and 2 green marbles. What is the probability of choosing a red and a green marble? 2. Sample answer: In a permutation, the order of objects is very important. In a combination, the order of objects is not important.  1  5  and the total number of blue and white marbles  is 200, the number of blue marbles must be 40. So the number of white marbles must be 160. After 100 white marbles are added, the total number of white marbles is 260 and the total of all marbles is 300. The probability of selecting a white marble  Chapter 13 SAT & ACT Preparation  260  13    is  300 or 15 . The correct choice is E.  Page 887  7. ∠A and ∠C must be equal because they are corresponding angles. The correct choice is A. 8. The sample space, or total possible outcomes, is the 52 cards in a deck. The outcome "drawing a diamond" consists of the 13 cards that are diamonds. 13 1 P(diamond) 
 52 or 4  SAT and ACT Practice  1. You might want to draw a diagram of the 20 coins. The first and last coins are heads. The total number of heads is 10. There could be 9 consecutive heads followed by 10 tails and then the final head. The correct choice is D. H  H  H  H  H  H  H  H  H  The correct choice is C. 9. 7 different entrees are offered. 3 are chosen. The number of combinations that can be chosen 7! 3!(7  3)!  7! 3!4!  is C(7, 3) 
  
  
 35. The correct choice is B. H  455  Chapter 13  10.  1 2  The probability that a dart thrown randomly at the target will land in the shaded region is equal to the ratio of the area of the shaded region to the area of the entire target. area of shaded region  P 
  area of target  (1)2  P
  (2)2   P
 4 or  1  4 1  The answer is  4.  Chapter 13  456  Chapter 14 Statistics and Data Analysis 14-1  6f. Sample answer:  The Frequency Distribution  Pages 892–893  Ages of Presidents 12  Check for Understanding  Frequency  1. A line plot, a bar graph, a histogram, and a frequency polygon all show data visually. A line plot shows the frequency of specific quantities by using symbols and a bar graph shows the frequency of specific quantities by using bars. A histogram is a special bar graph in which the width of each bar represents a class interval. A frequency polygon shows the frequency of a class interval using a broken line graph. 2. Choose an appropriate class interval. Use tally marks to determine the number of elements in each class interval. 3a. No; there would be too many classes. 3b. Yes; there would be 9 classes. 3c. Yes; there would be 5 classes. 3d. No; there would only be 3 classes. 3e. No; there would only be 2 classes. 4. See students' work. Age 5a.  4 0  60-69  4  40  50  Pages 893–896 7a.    43026     43212                     43214  43221 45414 43220 43229 ZIP Codes  7b. 43220 7c. Sample answer: to determine where most of their customers live so they can target their advertising accordingly  2000  8a.  Men  Age 65+  30-39  50-64  20-29  35-49  10-19  20-34  0-9  16-19 4  70  Exercises  40-49  0 0 Percent  60  6g. Sample answer: 50–60  50-59  16 12 8  0  Age  70+  1900  8  Women  80 60 40 20 0 0 20 40 60 80 Minutes Behind the Wheel  8 12 16  8b. Sample answer: Men spend more hours driving than women.  5b. In 1999, there are larger percents of older citizens than in 1990. 6a. range 
 69  42 or 27 6b. Sample answer: 5 6c. Sample answer: 40, 45, 50, 55, 60, 65, 70 6d. Sample answer: 42.5, 47.5, 52.5, 57.5, 62.5, 67.5 6e. Sample answer:  9a. Rental Revenue Year Sales Revenue 2003 2000 1997 1990  Ages 40–45 45–50 50–55 55–60 60–65 65–70  Frequency 2 6 12 12 7 3  1985 10 8 6 4 2 0 0 2 4 6 8 10 12 14 Dollars (in billions)  9b. Sales; the sales revenue is increasing, and the rental revenue has started to decrease. 10a. range 
 53  4 or 49 10b. Sample answer: 10 10c. Sample answer: 0, 10, 20, 30, 40, 50, 60 10d. Sample answer: 5, 15, 25, 35, 45, 55  457  Chapter 14  11g. Sample answer:  10e. Sample answer: Grams of Fat 0–10 10–20 20–30 30–40 40–50 50–60  Olympic Winter Games  Frequency 7 11 10 7 2 1  8 6 Frequency 4 2 0  10f. Sample answer:  0 10 20 30 40 50 60 70 80 Number of Nations  12a. range 
 1023  404 or 619 12b. Sample answer: 100 12c. Sample answer:  Grams of Fat in Fast-Food Sandwiches 10  Height (feet) 400–500 500–600 600–700 700–800 800–900 900–1000 1000–1100  8 Frequency 6 4 2 0  10g. 11a. 11b. 11c. 11d. 11e.  0 10 20 30 40 50 60 Grams of Fat  Sample answer: 10–20 range 
 72  16 or 56 Sample answer: 10 Sample answer: 10, 20, 30, 40, 50, 60, 70, 80 Sample answer: 15, 25, 35, 45, 55, 65, 76 Sample answer: Number of Nations 10–20 20–30 30–40 40–50 50–60 60–70 70–80  Frequency 5 4 2 3 1 2 3  12d. Sample answer: 5 4 Frequency 3 2  Frequency 2 3 8 1 1 2 1  1 0  0 400 600 800 1000 Height (feet)  12e. Sample answer: 400–500 13a. American League  Year 2003 2002 2001  11f. Sample answer:  2000  Olympic Winter Games  1999  8  1998  6  1997  Frequency 4  1996  2  1995  0  National League  1994  0 10 20 30 40 50 60 70 80 Number of Nations  80 60 40 20 0  0 20 40 60 80  Greatest Number of Stolen Bases for a Single Player  Chapter 14  458  13b. Sample answer:  16b. $1,200,000  Stolen Bases 30–40 40–50 50–60 60–70 70–80  Frequency 1 6 6 4 3  Sales $800,000 $400,000 $0  1999 2000 Year  16c. See students' work. 17. See students' work. 18. This is a biannual experiment, where 8 mums are involved and there are only two possible outcomes, survival S and not survival N. P(exactly 6 surviving) 
 C(8, 6)(S)6(N)2 
 28(0.8)6(0.2)2 0.29 or 29%  13c. Sample answer:  6 Frequency  1998  4  7   7r(2d)r 19. (c  2d)7 
   r!(7  r)! (c)  2  7!  r
0  0  0  To find the second term, evaluate the general term for r 
 1.  30 40 50 60 70 80 Stolen Bases  7! (c)7r(2d)r r!(7  r)!  13d. 3 players 13e. 7 players 14. Sample answer: 0.1, 0.2, 0.3, 0.4, 0.6, 0.7, 0.8, 0.9, 1.1, 1.2, 1.3, 1.4, 1.6, 1.7, 1.8, 1.9, 2.1, 2.2, 2.3, 2.4  
 7c6  2d 
 14c6d 20.  15. Millions of Tons  300 250  7!   71(2d)1 
 1!(7  1)! (c)  3.6x 
 58.9 x ln 3.6 
 ln 58.9 x 3.18  21.  y 9xy 
 36  x O  200 150 100 50 0  Wheat  Rice  Corn  22. (x  y)2 
 x2  2xy  y2 
 (x2  y2)  2(xy) 
 16  2(8) 
 32  China India United States  16a. The first interval on the vertical axis represents $800,000, but the other intervals represent only $200,000. Therefore, the sales for 1999 appear to be twice the sales of 1998, but in reality they are not.  14-2 Page 903  Measures of Central Tendency Check for Understanding  1. Mean, median, mode; to find the mean, add the values in a set of data and divide the sum by the number of values in the set. To find the median, arrange the values in a set of data from least to greatest. If there is an odd number of values in the set, the median is the middle value. If there is an even number of values in the set, the median is the mean of the two middle values. To find the mode, find the item of data that appears more frequently than any other in the set. 2. Sample answer: {1, 2, 2, 3, 4, 4, 5} The modes are 2 and 4.  459  Chapter 14  1  3. Write the stems 9, 10, 11, 12, 13, and 14 on the left. Write the tens digits as leaves to the right of the appropriate stems. Be sure to order the leaves. 4. Tia; the median 2.5 and the mode 2 do not represent the greater numbers. The mean 8.5 is more representative of all 8 items in the data.  9b. X  
 4 0 (6  3(7)  9  2(13)  2(14)  15  16  17  18  2(19)  3(20)  5(21)  2(23)  28  30  3(31)  32  2(34)  36  38  3(41)  42  47) 
 23.55 9c. Md 
 21 9d. Mode 
 21 9e. Since the mean 23.55, the median 21, and the mode 21 are all representative values, any of them could be used as an average.  1  5. X  
 4(10  10  45  58) or 30.75 10  45  Md 
 2 or 27.5 Mode 
 10 1  6. X  
 1 0 (21  22  23  24  28  29  31  31  34  37) 
 28  Pages 904–907  28  29  Md 
 2 or 28.5  10.  Mode 
 31 1  7. X  
 3(91  94  95  98  99  105  105  107  107  107  111  111  112)  100 103.23  100 10,323 Md 
 10,500 Mode 
 10,700 8a. 2  8  15  6  38  31  13  7 
 120 120 members  1  Md 
 3 Mode 
 3 1  12. X  
 4(17  19  19  21) or 19 Md 
 19 Mode 
 19   fi 
 120  1  13. X  
 8(3  5  5  8  14  15  18  18) 
 10.75   (fi  Xi) 
 2(3)  8(7)  15(11)  6(15)   8  14  Md 
 2 or 11  38(19)  31(23)  13(27)  7(31) 
 2320 2320  X 
 120 or about 19.3  i
1  Visits 1–5 5–9 9–13 13–27 17–21 21–25 25–29 29–33  Members 2 8 15 6 38 31 13 7  Mode 
 5 and 18 1  14. X  
 1 2 (54  58  62  63  64  70  76  76  87  87  98) 
 73.5  Cumulative Members 2 10 25 31 69 100 113 120  70  76  Md 
 2 or 73 Mode 
 87 1  15. X  
 1 2 (5  6  6  6  7  8  9  10  11  11  11  12) 
 8.5 89  Md 
 2 or 8.5 Mode 
 6 and 11 1  16a. X  
 9(117  124  139  142  145  151  Half of the data has been gathered in the 17–21 class. This is the median class. 8d. 69  31 
 38 21  17 
 4 60  31 
 29 Md  17 
 x 38  4   155  160  172) 
 145 lb 16b. Md 
 145 lb  29  
 x  1  16c.  X 
 9(122  129  144  147  150  156  x 3.052631579 Md  17 
 x Md  17 3.1 Md 20.1 9a. stem  160  165  177) 
 150 lb Md 
 150 lb Each will increase by 5 lb.  leaf  0 6 1 3 2 0 3 0 4 1 1|3 
 13 Chapter 14  155  11.  X 
 5(3  3  3  6  12) or 5.4  i
1 8  8c.   160  170) or 155  Mode: none  8  8b.  Exercises  1 X  
 4(140  150 150  160 Md 
 2 or  1  7 3 0 1 1  7 4 0 1 1  7 4 1 1 2  17. X  
 2(35  2(38)  39  44  3(45)  48  2(57)  9 5 6 7 8 9 9 1 1 1 1 3 3 8 2 4 4 6 8 7   59) 45.8 Md 
 45 Mode 
 45  460  1  24. Order the values from least to greatest. The median lies between the fourth and fifth terms. 2, 3, 3.2, 8, x, 11, 13, 14  18. X  
 1 4 (5.2  5.4  5.6  6.0  6.1  6.7  6.8  6.9  7.1  7.6  8.0  8.2  8.6  9.1) 
 6.95  88  x 
 8, since Md 
 2 or 8.  6.8  6.9  Md 
 2 or 6.85  1   25a.  X
 179 [9(245)  14(275)  23(325)  30(375)  Mode: none   33(425)  28(475)  18(525)  12(575)  7(625)  3(675)  1(725)  1(775)] 425.6  1  19. X  
 1 5 (90  91  97  98  99  105  106  109  113  3(118)  120  2(125))  10 
 1088 Md 
 1090 Mode 
 1180 20. stem leaf 1 0 5 5 5 5 7 7 2 0 0 0 5 5 5 5 7 8 3 0 0 5 5 5 4 6 5 5 1|0  10 21a. 135(11) 
 $1485; 145(24) 
 $3480; 155(30) 
 $4650; 165(10) 
 $1650; 175(13) 
 $2275; 185(8) 
 $1480; 195(4) 
 $780 21b. 1485  3480  4650  1650  2275  1480  780 
 $15,800 21c. 11  24  30  10  13  8  4 
 100 100 employees  25b.  Weekly Wages $130–$140 $140–$150 $150–$160 $160–$170 $170–$180 $180–$190 $190–$200  Frequency 11 24 30 10 13 8 4  Cumulative Frequency 11 35 65 75 88 96 100  9 14 23 30 33 28 18 12 7 3 1 1  x 
 20.4 5  Md  400 
 x Md  400 20.5 Md 420.5 1  26a. X  
 5(3.6  3.6  3.7  3.9  4.8) 
 3.92 Md 
 3.7 Mode 
 3.6 26b. Only the mean would change. It would increase to 4.6. 1 X  
 5(3.6  3.6  3.7  3.9  8.2) 
 4.6 26c. The mean increases slightly; the median increases slightly; the mode stays the same. 1 X  
 4(3.6  3.6  3.7  3.9)  x
5 Md  150 x Md  150 5 Md 155 21g. Both values represent central values of the data. 22. 7.5 
  200–250 250–300 300–350 350–400 400–450 450–500 500–550 550–600 600–650 650–700 700–750 750–800  Cumulative Number of Students 9 23 46 76 109 137 155 167 174 177 178 179  
 89.5; Half of the data has been gathered in the 400–450 class. This is the median class. 25c. 109  76 
 33 450–400 
 50 89.5  76 
 13.5 Md  400 
 x 33 13.5  
  50 x  Half of the data has been gathered in the $150–$160 class. This is the median class. 21f. 65  35 
 30 160  150 
 10 50  35 
 15 Md  150 
 x 30 15  
  10 x  1 (2 5  Number of Students  179  2  15,800   21d. X 
 100 or about $158  21e.  Scores  
 3.7 3.6  3.7 Md 
 2 or 3.65   4  5  8  x)  37.5 
 19  x 18.5 
 x 1 23. 6 
 4(x  2x  1  2x  3x  1)  27a. 27b. 27c. 27d.  24 
 8x 3
x  461  Mode 
 3.6 Sample answer: {1, 2, 2, 2, 3} Sample answer: {4, 5, 9} Sample answer: {2, 10, 10, 12} Sample answer: {3, 4, 5, 6, 9, 9}  Chapter 14  28a. stem 0  leaf 1 1 3 3 8 8 0 1 5 9 2  1 2 3 4 5 3 5|3  53  30a. Let  X 
 50 9  1 4 8 3  1 4 9 3  1 4 9 3  1 5 9 5   (X  fi) 
 0 
 (50  5)  (50  20)  (50 i
1  1 2 2 2 2 2 3 3 3 5 5 5 6 6 7 7 7 8 9 8 9 9   37)  (50  44)  (50  52)  (50  68)  (50  71)  (50  85)  (50  x) 0 
 68  x x 
 68 The weight should be hung 68 cm from the end.  
 50 30b. Let X 9   (X  fi) 
 0 
 (50  5)  (50  20)  (50 i
1   37)  (50  44)  (50  52)  (50  68)  (50  71)  (50  85)  (50  x)  (50  x)  1  28b.  X 
 5 0 [7(1)  5(2)  5(3)  3(4)  4(5)  2(6)  3(7)  4(8)  4(9)  10  11  3(13)  15  18  2(19) 25  29  32  53] 
 8.7 28c. Md 
 6 28d. Mode 
 1 28e. The mean 8.7 and the median 6 are representative of the data, but the mode 1 is not representative of the data. 1 [1(170) 27  29a. X   0 
 118  2x 118 
 2x 59 
 x The weight should be hung 59 cm from the end. 1  31a. X  
 1 0 (54  55  59  59  61  62  65  75  162  226)  1000 
 $87,800   6(190)  10(210)  6(230)   3(250)  1(270)] 215.2 29b. Goals 160–180 180–200 200–220 220–240 240–260 260–280  Number of Teams 1 6 10 6 3 1  61,000  62,000  31b. Md 
 2 or $61,500 31c. Mode 
 $59,000 31d. The mean, since it is the greatest measure of central tendency. 31e. The mode, since it is the least measure of central tendency. 31f. Median; the mean is affected by the extreme values of $162,000 and $226,000, and only two people make less than the mode. 31g. Sample answer: I have been with the company for many years, and I am still making less than the mean salary.  Cumulative Number of Teams 1 7 17 23 26 27  27  2  
 13.5; Half of the data has been gathered in the 200–220 class. This is the median class. 29c. 17  7 
 10 220  200 
 20 13.5  7 
 6.5 Md  200 
 x 10 6.5  
  20 x  32a. X    15(2.50)  31(3.00)   37(3.50)  5(4.00) 30.4 32b.  x 
 13 Md  200 
 x Md  200 
 13 Md 213 1  29d.  X 
 2 7 (268  248  245  242  239  239  237  236  231  230  217  215  214  211  210  210  207  205  202  200  196  194  192  190  189  184  179) 215.9 Md 
 211 29e. The mean calculated using the frequency distribution is very close to the one calculated with the actual data. The median calculated with the actual data is less than the one calculated with the frequency distribution.  Chapter 14  1 [12(2.00) 100  Grade Point Averages 1.75–2.25 2.25–2.75 2.75–3.25 3.25–3.75 3.75–4.25  Frequency 12 15 31 37 5  Cumulative Frequency 12 27 58 95 100  The median class is 2.75  3.25. 58  27 
 31 3.25  2.75 
 0.50 50  27 
 13 Md  2.75 
 x 13 31  
  x 0.50 x 0.2096774194 Md  2.75 
 x Md  2.75 0.21 Md 2.96  462  33. He is shorter than the mean (5
 11.6) and the median (5
 11.5).  Measures of Variability  14-3  1  X 
 1  0 (67  68  69  69  71  72  73  74  75  78) 
 71.6 or 5
 11.6  Page 914  71  72  Md 
 2 
 71.5 or 5
 11.5 34. 22 20 18  Frequency  16 14 12 10 8 6 4 2 0  0 60 65 70 75 Speed Limit 3  2  Check for Understanding  1. The median of the data is 70, Q1 is 60, and Q3 is 100. The interquartile range is 40 and the semiinterquartile range is 20. The outliers are 170 and 180. The data in the first two quartiles are close together in range. The last two quartiles are more diverse. 2. square the standard deviation 3. Both the mean deviation and the standard deviation are measures of the average amount by which individual items of data deviate from the mean of all the data. The mean deviation uses the absolute values of the deviations. Standard deviation uses the squares of the deviations. 4. See students' work. 5. interquartile range 
 Q3  Q1 
 41  25 
 16 16 Semi-interquartile range 
 2 
8  3   35. dependent; 11  1 0 
 55 n1  n   36. an 
 3n ; an1 
  3n1  r  n1   3n1 — 
 lim n n→ 3 n  15 6. X 
  37. Fn 
 
  45   5.50  5.50  6.30  7.80  11.00  $4.11 1   7. X 
 200 [15(5000)  30(15,000)  50(25,000)   60(35,000)  30(45,000)  15(55,000)] 
 30,250      864 2 4 6 8  40  1  1, the series is convergent.  8 6 4 2  35  2  (3.24)2 … 8.462  8 (4.29       
  
 $40,305.56 38.  30  $3.54  (1  i)n  1 P i (1  0.03)20  1  1500  0.03    25   12.20  17.20) 
 8.74 1 MD 8(4.29  3.24 … 8.46)  3n(n  1)   n1 n→ 3 (n) n 3 (n  1) 
 lim  n  n→ 3  3n n1 
 lim 3n n→ n 1 
 lim 3n  3n n→ 1 1 
 3  0 or 3  
 lim  Since r  20 1 (4.45 8  (25,250)  15  (15,250)  30 … 24,750  15   200 2  2  2  13,226.39 1  y  8a. X  
 1 2 (65.7  65.9 … 65.9) 
 70.375  69.0  70.3  Md 
 2 or 69.65  O 2 4 6 8x  
  (4.875)2  (4.475)2 … 6.2252  12  4.25  8b. X 
  39. To find the area of the triangle use Hero's formula:  1 (57.3 12   63.3 … 57.5)  80.48  s(s   a)(s  b)(s  c), where Area 
   77.5  82.1  Md 
 2 or 79.8  1 s 
  (a  b  c) and a 
 10, b 
 7, and c 
 5. 2 1 1 So, s 
  (10  7  5) 
 (22) 
 11. 2 2  (23.18)  (22.98) … 25.32   12 2  2  2  17.06   10) (11  7)(11  5) Area 
 11(11  
  11  1  4  6 or 264  The correct choice is A.    16.25.  463  Chapter 14  8c. Los Angeles  1  14. X  
 1 0 (5.7  5.7 … 3.8) 
 4.89 1  MD 
 1 0 (0.81  0.81 … 1.09)  55 60 65 70 75 80 85 90 95 100 105  
 0.672  Las Vegas  
  0.81  0.81 … (1.09)   10 2  2  2  0.73 1  15. X  
 1 2 (369  376 … 454)  55 60 65 70 75 80 85 90 95 100 105  
 403.5  8d. Los Angeles 8e. Los Angeles is near an ocean; Las Vegas is in a desert.  1  MD 
 1 2 (34.5  27.5 … 50.5) 
 20.25 
  Pages 915–917  Exercises  (34.5)  (27.5) … 50.5   12 2  2  2  25.31  9. interquartile range 
 Q3  Q1 
 24  17 
7 7 semi-interquartile range 
 2 or 3.5  1  16. X  
 5(13  22  34  55  91) 
 43 Variation 
  (30)2  (21)2  (9)2  122  482  5  
 774 1   17. X 
 120 [2(3)  8(7) … 7(31)]  19.33  14 16 18 20 22 24 26 28 30  (16.33)  2  (12.33)  8 … 11.67  7   120 2  10. interquartile range 
 Q3  Q1 
 21.5  12 
 9.5 9.5 semi-interquartile range 
 2 or 4.75  2  6.48 1  18. X  
 9 0 [3(57)  7(65) … 12(97)] 
 81.8 
  0  2  (24.8)  3  (16.8)  7 … 15.2  12   90 2  2  2  9.69  10 20 30 40 50 60 70 80  1  19. X  
 8 5 [2(80)  11(100) … 7(180)]  11. interquartile range 
 Q3  Q1 
 10.5  7.6 
 2.9 2.9 semi-interquartile range 
 2 or 1.45  129.65   (49.65)  2  (29.65)  11 … 50.35  7   85 2  2  2  23.29 20a. Md 
 259 mi 20b. Q1 
 129 mi; Q3 
 360 mi 20c. interquartile range 
 Q3  Q1 
 360  129 
 231 mi  5 6 7 8 9 10 11 12 13 14 15 16 12.  231  20d. semi-interquartile range 
 2 
 115.5 mi 20e. An outlier would lie 231  115.5 or 346.5 mi outside of Q1 or Q3. There are no such points. 20f. 1  13. X  
 6(152  158 … 721) 
 381  0 100 200 300 400 500 600 700  1  MD 
 6(229  223 … 340)  20g. The data in the upper quartile is more diverse than the other quartiles. 21. Sample answer: {15, 15, 15, 16, 17, 20, 24, 26, 30, 35, 45} 22a. Md 
 282 22b. Q1 
 42; Q3 
 770  
 211 
  (229)  (223) … 340   6 2  2  2  223.14  Chapter 14  464  22c. interquartile range 
 Q3  Q1 
 770  42 
 728 semi-interquartile range 
  1  25a.  X 
 5 0 [26(9)  12(11) … 2(21)] 
 11  728  2  25b.  
 364 22d. An outlier would lie 728  364 or 1092 points outside of Q1 or Q3. There are no such points.  
  (9  11)  26  (11  11)  12 … (21  11)  2   50 2  2  2  2.94 26. yes; when the standard deviation is less than 1; when both equal 0 or 1 27. See students' work.  22e.  1  28a. X  
 3 5 [2(4.4)  4.9  5.4  5.5  2(6.2)  6.4 0 22f. X 
  200 1 (22 19  400  600  800  1000   23 … 966)  404.42 1  22g. MD  1 9 (382.4  381.4 … 561.58) 316.97  28b. 28c. 29a. 29b. 29c. 29d.   …  19 382.42  22h. Variance   381.42  561.582  118,712.56 22i. 22j. 23a. 23b.  23c.  2.56   118,71 344.55 There is a great variability among the number of teams in women's sports. Q1 
 $3616, Md 
 $4125, Q3 
 $5664 interquartile range 
 Q3  Q1 
 5664  3616 
 2048 An outlier would lie 2048  1024 or $3072 outside of Q1 or Q3. There are two such values, $26,954 and $27,394.   6.5  6.9  7.1  7.4  7.5  7.6  7.7  7.8  7.9  8.0  8.2  8.4  8.5  8.6  8.7  8.8  3(8.9)  9.0  9.2  2(9.3)  9.5  9.6  9.8  9.9] 7.75 Md 
 8.0 Mode 
 8.9 range 
 68  23 or 45 Sample answer: 10 Sample answer: 20, 30, 40, 50, 60, 70 Sample answer: Programs Sold 20–30 30–40 40–50 50–60 60–70  Frequency 2 1 2 5 2  29e. Sample answer:  23d. 4  0  5000 10,000 15,000 20,000 25,000 30,000  Frequency  1  23e. X  
 1 9 (2684  2929 … 27,394)  2  6775.95 MD  0  1 (4091.95  3846.95 … 20,618.05) 19  4463.39 (4091.95)  (3846.95) … 20,618.05   19 2  23f.  2  2  1  30. 9(9!) 
 40,302 ways 31. x1 
 0.5(8) 1 or 3 x2 
 0.5(3)  1 or 0.5 x3 
 0.5(0.5)  1 or 0.75 32. 7 ft 
 7(12) or 84 in. 84  9 
 93 in. 93  
 31 in. 3  7103.45 23g. The data in the upper quartile is diverse. 24a. 0  100 200 300 400 500 600 1  24b. X  
 4 2 (0  0 … 635)  31  31 in. 
 1 2 or 2 ft 7 in.  60.40 MD  0 10 20 30 40 50 60 70 Programs Sold  The correct choice is C.  1 (60.40 42   60.40 … 574.60)  67.87 24c. Variance   (60.40)2  (60.40)2 … (514.60)2  42  14,065.48 5.48   14,06 118.60 24e. The data in the upper quartile is diverse. 24d.  465  Chapter 14  Page 917  Mid-Chapter Quiz  3.  1. Sample answer: 10 2. Sample answer: Exam Scores 50–60 60–70 70–80 80–90 90–100  Frequency 2 4 6 10 8  45 55 65 75 85 95 105  45 55 65 75 85 95 105  The second curve is less variable. 4. Sample answer:  3. Sample answer: Physics Exam 10 8 Frequency 6 4 2 0  5. 50th percentile; it contains half of the data. 6a. 0 50 60 70 80 90 100 Exam Score  4. stem leaf 5 4 5 6 2 2 7 1 5 8 0 2 9 0 2 54 
 54  445 480 515 550 585 620 655  4 6 4 3  5 7 8 9 5 6 7 8 9 9 9 3 5 6 8 9  6b. Since 515 and 585 are within are standard deviation of the mean, it contains 68.3% of the data. 6c. 99.7% of the data lie within 3 standard deviations of the mean. 550 3(35) 
 445  655 6d. 550  480 
 70, 620  550 
 70 tj 
 70 t(35) 
 70 t 
 2 → 95.5% 0.995(200) 
 191 values 7a. Since 22 and 26 are within one standard deviation of the mean, it contains 68.3% of the data. 7b. 24  20.5 
 3.5, 27.5  24 
 3.5 tj 
 3.5 t(2) 
 3.5 t 
 1.75 → 92.9% 7c. 24  0.7(2) 
 22.6 and 24  0.7(2) 
 25.4 22.6  25.4 7d. 24  1.96(2) 
 20.08 and 24  1.96(2) 
 27.92 20.08  27.92  1  5. X  
 3 0 (54  55 … 99) 
 81.1  84  85  6. Md 
 2 or 84.5 7. Mode 
 89 8. 50  60  9. MD 
  70  80  1 (27.1 30  90  100   26.1 … 17.9)  
 10.42 10. Sample answer: The data that are less than the median are more spread out than the data greater than the median.  14-4  The Normal Distribution  Pages 922–923  8a.  Check for Understanding  1. The median, mean, and mode are the same. 2. X  1.5  64 67 70 73 76 79 82  8b.  65 70 75 80 85 90 95  Chapter 14  466  8c. Chemistry; the chemistry grade is 3 standard deviations above the class mean, while the speech grade is only 2 standard deviations above the class mean.  13b.  X  tj 
 180  140  t(20) 
 180 20t 
 40 t 
2 95.5%  2  Pages 923–925  X   tj 
 150 140  t(20) 
 150 20t 
 10 t 
 0.5 38.3%  2  
 47.75%  
 19.15%  47.75  19.15 
 28.6%  Exercises  13c.  9a.  50% 25%  25%  7.5 9 10.5 12 13.5 15 16.5  9b. 12 1(1.5) 
 10.513.5 9c. 12  7.5 
 4.5, 16.5  12 
 4.5 tj 
 4.5 t(1.5) 
 4.5 t 
 3 → 88.7% 9d. 12  9 
 3, 15  12 
 3 tj 
 3 t(1.5) 
 3 t 
 2 → 95.5% 10a. 0.683(200) 
 136.6; about 137 10b. 0.955(200) 
 191 10c.  0.683  2  80  100  120  140  160  180 200  t 
 0.7 corresponds with 50% of the data centered about the mean. The upper limit results in 75% of the data. 140  0.7(20) 
 154 14a. X X   tj 
 7   tj 
 6.5 6  t(35) 
 7 6  t(3.5) 
 6.5 3.5t 
 1 3.5t 
 0.5 t 
 0.29 t 
 0.14 23.6% 8%  
 11.8%  
 4% 2 2  (200) 
 68.3; about 68  11a. 45% corresponds to t 
 0.6. 82 0.6(4) 
 79.6  84.4 11b. 80% corresponds to t 
 1.3 82 1.3(4) 
 76.8  87.2 11c. 82  76 
 6, 88  82 
 6 tj 
 6 t(4) 
 6 t 
 1.5 → 86.6% 11d. 82  80.5 
 1.5, 83.5  82 
 1.5 tj 
 1.5 t(4) 
 1.5 t 
 0.375 → 31.1% 12a. 25% corresponds to t 
 0.3. 402 0.3(36) 
 391.2412.8 12b. 402  387 
 15, 417  402 
 15 tj 
 15 t(36) 
 15 t 
 0.416  → 31.1% 12c. 402  362 
 40, 442  402 
 40 tj 
 40 t(36) 
 40 t 
 1.1  → 72.9% 12d. 45% corresponds to t 
 0.6. 402 0.6(36) 
 380.4  423.6 13a. X X   tj 
 150   tj 
 100 140  t(20) 
 150 140  t(20) 
 100 20t 
 10 20t 
 40 t 
 0.5 t 
2 38.3% 95.5%  
 19.15%  
 47.75% 2 2  14b.  11.8  4 
 7.8% X   tj 
 6.2 6  t(0.35) 
 6.2 0.35t 
 0.2 t 
 0.57 45.1%  
 22.55% 2  X   tj 
 5.5 6  t(0.35) 
 5.5 0.35t 
 0.5 t 
 1.43 83.8%  
 41.9% 2  22.55  41.9 
 64.45% 14c.  80%  10%  10%  4.95 5.30 5.65 6.00 6.35 6.70 7.05  t 
 1.3 corresponds with 80% of the data centered about the mean. The lower limit results in the value above which 90% of the data lies. 6  1.3(0.35) 
 5.545 1 6  15a. P(no tails) 
 
2 or 64 1  1 5  P(one tail) 
 6
2
2 or 3 2 1  3  1 2 1 4 15  or  2 64 1 3 1 3 5     P(three tails) 
 20 2 2 or 1 6 1 4 1 2 5 P(four tails) 
 15 2 2 or 64 1 5 1 3 P(five tails) 
 6 2 2 or 3 2 1 6 1     P(six tails) 
 2 or 64  P(two tails) 
 15
2  
  
 
  
 
  
 
  
   19.15  47.75 
 66.9%  467  Chapter 14  15b.  18.  96%  20 15 Frequency  10 5 0  0  1 2 3 4 5 Number of Tails  6  2% 46  1  15c.  X 
 6 4 [0(1)  1(6)  2(15)  3(20)  4(15)  (0  3)  (1  3) … (6  3)   64 2  2  2  56  61  66  71  76  20%  20%  1.2 15e. They are similar. 16a.  51  96% corresponds to t 
 2.1. 61  2.1(5) 
 50.5 months 19a. 30%   5(6)  6(1)] 
3 15d. j 
  2%  15%  15%  84%  44 8%  3  2  8%  1  0  1  2  19b. 19c.  3 t  The 92nd percentile is the upper limit to 84% of the data that is centered about the mean. 84% corresponds to t 
 1.4. The 92nd percentile is 1.4 standard deviations above the mean. 16b.  20a.  57.6%  20b.  21.2%  21.2%  20c.  20d. 3  2  1  0  1  2  21a.  3 t  t 
 0.8 corresponds to 57.6% of the data centered 100  57.6  about the mean. 2 
 21.2 21.2  57.6 
 78.8 percentile 17a. X   tj 
 22.3 20.4  t(0.8) 
 22.3 0.8t 
 1.9 t 
 2.38 → 98.4% 100  98.4  
 0.8% 2  21b.  17b. 100  0.8 
 99.2%  51  58  65  72  79  86  70% corresponds to t 
 1.0. 65  1.0(7) 
 72 65  1.0(7) 
 58 30% corresponds to t 
 0.4 65  0.4(7) 
 67.8 68 The lowest score for an A is 72, so the highest score for a B is 71. The interval for B's is 6871. a normal distribution with a small standard deviation a normal distribution with a large standard deviation a distribution where values greater than the mean are more spread out than values less than the mean a distribution where all values occur with the same frequency 95% corresponds to t 
 1.96 X X   tj 
 260   tj 
 250 255  1.96j 
 260 255  1.96j 
 250 1.96j 
 5 1.96j 
 5 j 
 2.55 j 2.55 about 2.55 mL X X   tj 
 357   tj 
 353 355  t(2.55) 
 357 355  t(2.55)
 353 2.55t 
 2 2.55t 
 2 t 
 0.78 t 
 0.78 57.6% 147  150  22a. Md 
 2 or 148.5 22b. Q1 
 110, Q3 
 200 22c. interquartile range 
 Q3  Q1 
 200  110 
 90 90  22d. semi-interquartile range 
 2 or 45 22e. 100  Chapter 14  468  200  300  400  500  1  3. About 68.3%; the answer for Exercise 2 can be rounded to 0.683, which equals 68.3%.  23. X  
 9(19  33  42  42  45  48  55  71  79) 48.2 Md 
 45 Mode 
 42 24. y 
 sec(k
  c)  h 2  k: k 
 2 k
4 c c: k 
   4.  0.95449974  c  4  
  c 
 4 h
3 y 
 sec(4
  4)  3 25a. Sample answer: Use a graphing calculator to enter the year data as L1 and the Enrollment data as L2. Then make a scatter plot. The scatter plot indicates that a cubic function would best fit the data. Perform a cubic regression to find the equation y 
 0.05x3  2.22x2  29.72x  366.92. 25b. Sample answer: 2015  1965 
 50 f(50) 
 0.05  503  2.22  502  29.72  50  366.92 
 2553 students 26.  60˚  x˚  3  5.  0.9973002 6. The answer for Exercise 3 can be rounded to 0.954, which is about 95.5%. The answer for Exercise 4 can be rounded to 0.997, which equals 99.7%. 7. 0.9999; t 
 4 corresponds to P 
 0.999. 8.  4 60˚ 1 2  5  30˚  1; no; since the curve is approaching the x-axis asymptotically, the area is probably not exactly equal to 1.  Since vertical angles are equal, m2 
 60. m1  60  30 
 180 m1 
 90 Since an exterior angle of a triangle is equal to the measure of the sum of the two remote interior angles, x  60 
 m1. So, x  60 
 90 and x 
 30. The correct choice is E.  14-5  Sample Sets of Data  Page 930  14-4B Graphing Calculator Exploration: The Standard Normal Curve  Check for Understanding  1. A sample is a subset of a population. However, a sample must be similar in every way to the population. 2. Divide the standard deviation of the sample by the square root of the number of values in the sample. 3. Use a larger sample. 4. Tyler; every twentieth student to enter the school should produce a representative sample. The senior English class will not represent underclassmen. The track team will not represent students who prefer other types of activities.  Page 926 1. The domain of the function is the set of all real numbers. The range is the set of all real numbers 1 . The graph is symmetric y such that 0 y    2 with respect to the y-axis and has the x-axis as a horizontal asymptote. the value of f(x) approaches 0 as x approaches . 2.  73   or 7.3 5. jX 
  00  1 3.4   6. jX 
  50  2  0.22  0.68268949  469  Chapter 14  7. A 1% confidence level is given when P 
 99% and t 
 2.58. 5  or 0.83 jX 
 5   36  internal: X   11.12   20. jX 
  1  000  0.3516452758 interval: X  tjX 
 110 2.58jX 
 109.09110.91 21. P 
 90% corresponds to t 
 1.65. 4  or 0.4 jX 
  1  00  tjX 
 45 2.58(0.83 ) 42.8547.15  8. A 5% confidence interval is given when P 
 95% and t 
 1.96.  interval: X   5.6   jX 
   300  2.4   or 0.24 22. jX 
   100  0.323316156 internal: X   interval: X   tjX 
 55 1.96jX   54.3755.63  17.1  0.9140334473 interval: X  tjX 
 4526 1.96jX 
 4524.214527.79  0.29 9b. P 
 50% corresponds to t 
 0.7. interval: X  tjX 
 27.5 0.7jX 27.3027.70 min 9c. A 1% confidence level is given when P 
 99% and t 
 2.58. interval: X  tjX 
 27.5 2.58jX 27.7628.24 min  Pages 930–932  28   24. jX 
  3  70  1.455650686 interval: X  tjX 
 678 1.96jX 675.15680.85 0.67   25. jX 
  8  0  0.0749082772 interval: X  tjX 
 5.38 1.96jX 5.235.53 1 26a.  X 
 6 [1(4)  3(6) … 2(20)] 4 
 12.375  Exercises  1.8   or 0.2 10. jX 
  8  1 5.8   11. jX 
   250  0.37  26b. j 
  7.8   12. jX 
   140  2  2  3.37  0.42 26d. P 
 0.95 corresponds to t 
 1.96. interval: X  tjX 
 12.375 1.96(0.42) 11.5513.20 min 26e. tjX 
 1 t(0.42) 
 1 t 2.38 → about 98.4%  0.53 2.7   14. jX 
  1  30  0.24 13.5   15. jX 
   375  0.70  12  5.6   27. jX 
   45   16. 0.056 
  N   1.788854382 tjX 
 3 t 
 1.68 → 91.1% 100  91.1 8.9%  N  
 100 N 
 100,000 5.3   17. jX 
  5  0  0.7495331881 interval: X  tjX 
 335 2.58jX 333.07336.93 40  or 5 18. jX 
  6  4  1.4   28a. jX 
   50  0.1979898987 or about 0.20 28b. A 5% confidence interval is given when P 
 95% and t 
 1.96. interval: X  tjX 
 16.2 1.96j X  15.8116.59 mm 28c. P 
 99% corresponds to t 
 2.58. interval: X  tjX 
 16.2 2.58jX 15.6916.71 mm 28d. P 
 0.80 corresponds to t 
 1.3 interval: X  tjX 
 16.2 1.3jX 15.9416.46 mm  tjX 
 200 2.58(5) 
 187.1212.9   19. jX 
   200  0.8485281374 interval: X  tjX 
 80 2.58jX 77.8182.91  45   or 4.5 29a. jX 
   100  Chapter 14  2   26c. jX    64  14  7  00  12  (4  12.375)  (6  12.375) … (20  12.375)   64  3.37  0.66  interval: X   tjX 
 24 1.96(0.24) 23.5324.47   23. jX 
  3  50  3.5   9a. jX 
   150  13. jX 
  tjX 
 68 1.65(0.4) 
 67.3468.66 in.  470  29b. A 1% confidence level is given when P 
 99% and t 
 2.58. interval: X  tjX 
 350 2.58(4.5) 
 338.39361.61 hours 29c. Sample answer: 338 hours, there is only 0.5% chance the mean is less than this number. 30. 10.2064  9.7936 
 0.4128  34c. 50% of 10,000 
 0.50(10,000) 
 5000 tires 34d. X   tj 
 50,000 40,000  t(5000) 
 50,000 5000t 
 10,000 t 
 2 → 95.5% 100%  95.5%  
 2.25% 2  0.4128  2  
 0.2064 tjX 
 0.2064 2.58tjX 
 0.2064 jX 
 0.08  34e.  j   jX 
  N  0.8   0.08 
  N   0.0015(10,000) 
 15 tires  N  
 10 N 
 100 packages  1  35. X  
 8(44  49  55  58  61  68  71  72) 
 59.75 1 MD 
 8(15.75  10.75 … 12.25)  1.8   31a. jX 
   10  0.57 31b. interval: X   
 8.25  tjX 
 4.1 1.96(0.57) 
 2.985.22 hours With a 5% level of confidence, the average family in the town will have their televisions on from 2.98 to 5.22 hours. 31c. Sample answer: None; the sample is too small to generalize to the population of the city.  j
  2  2  2  a1  a1rn   36. Sn 
  1r 1  n 
 10, a1 
 1 6, r 
 4 1  1    (4)10 16 16  S10 
 14  3.2  0.45 32b. P 
 50% corresponds to t 
 0.7. interval: X  tjX 
 42.7 0.7jX 42.3843.02 crackers 32c. Sample answer: No; there is a 50% chance that the true mean is in the interval. However, since 43 is near one end of the interval, they may want to take another sample in the near future. 33a. A 5% confidence interval gives a P 
 95% and t 
 1.96.  349,525  
 16 or 21,845.3125 37.  x
y r cos v 
 r sin v sin v   1
 cos v  1 
 tan v tan1(1) 
 v 45° 
 v 38. tan x  cot x 
 2 1  tan x   tan x 
 2  
 3.136  1.960jX 
 3.136 jX 
 1.6 X  tjX 
 753.136 X   1.96(1.6) 
 753.136 X  
 750 h  (15.75)  (10.75) … (12.25)   8  9.59   32a. jX 
   50  753.136  746.864   0.0225(10,000) 
 225 tires X   tj 
 25,000 40,000  t(5000) 
 25,000 5000t 
 15,000 t 
 3 → 99.7% 100%  99.7%  
 0.15% 2  tan2 x  1  tan x tan2 x  1  tjX  
2  
 2 tan x tan2 x  2tan x  1 
 0 (tan x  1)2 
 0 tan x  1 
 0 tan x 
 1 x 
 45° 39. *2 
 22  2(2) or 0 *1 
 12  2(1) or 1 *2  *1 
 0  (1) or 1 The correct choice is C.  j   33b. jX 
  N  j   1.6 
  1  600  64 
 j; 64 h 34a. 40,000  35,000 
 5000, 45,000  40,000 
 5000 tj 
 5000 t(500) 
 5000 t 
 1 → 68.3% 0.683(10,000) 
 6830 tires 34b. X   tj 
 30,000 40,000  t(5000) 
 30,000 5000t 
 10,000 t 
 2 → 95.5% 95.5%  
 47.75% 2  Chapter 14 Study Guide and Assessment Page 933  Understanding the Vocabulary  1. box-and-whisker plot 2. median 3. standard error of the mean  0.4775(10,000) 
 4775 tires  471  Chapter 14  4. 5. 6. 7. 8. 9. 10.  2  ( 2.4)2  … 2 .62 22. j 
 2.4) 1.74 23. 88  78 
 10 98  88 
 10 tj 
 10 t(5) 
 10 t 
 2 → 95.5% 24. 88  86 
 2 90  88 
 2 tj 
 2 t(5) 
 2 t 
 0.4 → 0.311 25. 90% corresponds to t 
 1.65. interval: X  tj 
 88 1.65(5) 
 79.7596.25 26. 0.683(150) 
 102.45 27. 0.955(150) 
 143.25  range measure of central tendency population bimodal inferential statistics histogram standard deviation  Pages 934–936  Skills and Concepts  11. range 
 14.0  9.0 or 5 12. 9.5, 10.5, 11.5, 12.5, 13.5 13.  Women's Tennis Shoes 18 16 14 12 Frequency 10 8 6 4 2 0  28. 29.  30. jX 
  4.9   120  0.45 25  0   or 1.25 31. jX 
   400  9 10 11 12 13 14 Weight (ounces)  18   or 3.6 32. jX 
   25  1  15   33. jX 
   50  
5 Md 
 5 Mode 
 4  2.121320344 interval: X  tjX 
 100 2.58jX 94.53105.47  1  15. X  
 5(160  200  200  240  250)  30   34. jX 
  1  5  
 210 Md 
 200 Mode 
 200  7.745966692 interval: X  tjX 
 90 2.58jX 70.02109.98  1  16. X  
 5(11  13  15  16  19)  24   35. jX 
   200  
 14.8 Md 
 15 Mode: none  1.697056275 interval: X  tjX 
 40 2.58jX 35.6244.38  1  17. X  
 8(5.9  6.3  6.3  6.4  6.6  6.6  6.7  6.8)  0.5   36. jX 
  2  00  
 6.45  0.035 37. P 
 0.90 corresponds to t 
 1.65. range: X  tjX 
 1.8 1.65(0.035) 1.741.86 h 38. A 5% confidence level is given when P 
 95% and t 
 1.96. range: X  tjX 
 1.8 1.96(0.035) 1.731.87 h 39. A 1% confidence level is given when P 
 99% and t 
 2.58. range: X  tjX 
 1.8 2.58(0.035) 1.711.89 h 40. P 
 0.90 corresponds to t 
 1.65.  6.4  6.6  Md 
 2 or 6.5 Mode 
 6.3 and 6.6 1  18. X  
 8(122  128  130  131  133  135  141  146) 
 133.25 131  133  Md 
 2 or 132 Mode: none 19. interquartile range 
 Q3  Q1 
52 
3 3  20. semi-interquartile range 
 2 or 1.5  1.4   or 0.14 jX 
  1  00  1  21. X  
 1 0 (1  1 … 6)  range: X   
 3.4 1 (2.4 10   2.4 … 2.6)  
 1.6  Chapter 14  
 51.225  0.16  14.  X 
 9(2  4  4  4  5  5  6  7  8)  MD 
  0.683 (150) 2 1.5  jX 
   90  472  tjX 
 4.6 1.65(0.14) 4.374.83 h  Page 937  m 
 x%(10n)  Applications and Problem Solving  x  41a. stem leaf 1 0 3 5 6 7 9 2 1 3 4 5 3 9 9 10 
 10   2000 
  100 (10  20)  1000 
 x The correct choice is D. 4. The numbers in S are positive numbers that are less than 100 and the square root of each number is an integer. So the set S contains perfect squares between 0 and 100. Make a list of the n n  numbers, n, in set S. 1 1 From your list, you can 4 2 see that the median, or 9 3 middle value for n, is 25. 16 4 The correct choice is C. 25 5 36 6 49 7 64 8 81 9  1  41b.  X 
 2(10  13  15  16  17  19  21  23  24  25  39  39) 
 21.75 19  21  41c. Md 
 2 or 20 41d. Mode 
 39 42. X   tj 
 80 75  t(2) 
 80 2t 
 5 t 
 2.5 → 98.8% 100%  98.8%  2  Page 937  
 0.6%  Open-Ended Assessment  5. m∠DBA 
 90  30 
 60 m∠EBC 
 90  40 
 50 m∠ABC 
 180  m∠DBA  m∠EBC 
 180  (60)  (50) or 70 The correct choice is E.  1a. Sample answer: {2, 3, 10, 20, 40} 1b. Sample answer: 15 2. See students' work.  1  6. X  
 6(10  20  30  35  35  50) 
 30 There are 3 numbers larger than 30: 35, 35, and 50. The correct choice is D.  Chapter 14 SAT & ACT Preparation Page 939  SAT and ACT Practice  7.  y  1. The percent increase is the ratio of the number increase to the original amount. Amy Brad Cara Dan Elsa  10  80 30  70 20  80 30  60 20  90  1  
 8 3  
 7 1  O  
 4  The line of best fit has a rise of 2 and a run of 5. So 2 the slope of the line of best fit is 5. The closest 2 1 answer choice to 5 is 2. The correct choice is D. 8. Each year the number of employees increases by 300. The last year of data is 2005. The expected employment in 2007, two years later, will be 2  300 more employees than in 2005. 3100  600 
 3700 The correct choice is D.  1  
 2 2  
 9 1  The largest fraction is 2. Dan has the greatest percent increase. The correct choice is D. 2. a 
 b  bc b  a  b   
 (b  bc) b   
 b(1  c) 1   
 1c  The correct choice is B. 3. Method 1: 0.1%(m) 
 10%(n) 0.001m 
 0.1n  x  m 
 x%(10n) x  m
 100  10n  m 
 100n m 
 0.1xn 100n 
 0.1xn 100 
 x Method 2 Let m represent a large number such as 2000. 0.1% of m 
 0.001(2000) or 2 10% of n 
 2, so 0.1n 
 2 or n 
 20.  473  Chapter 14  9.To find the median of Set A, first rewrite the elements of Set A in order: 4, 1, 2, 3, 7, 11. Since the number of elements is even, the median is the average of the middle two elements: 2 and 3. So the median of Set A is 2.5. To find the mean of Set B, add all of the elements together and divide by the number of elements in the set. The sum of the elements is 15, and there are 6 terms. So the  1  10. X  
 1 0 [820  (65)  (32)  0  1  2  3  32  64  820] 1  
 1 0 (1  2  3) 6  
 1 0 The answer is 0.6, 6/10, or 3/5.  15 6  mean is  or 2.5. The difference between the median of Set A and the mean of Set B is 2.5  2.5 or 0. The correct choice is C.  Chapter 14  474  Chapter 15 Introduction to Calculus 6. lim (1  x  2x  cos x) 
 1  0  20  cos 0 x→0 
111 
1  Limits  15-1 Page 945  x2  x2   
 lim  7. lim  (x  2)(x  2) x2  4  Graphing Calculator Exploration  x→2  x→2  1   
 lim  x2  1.  lim  x→0  ex  1  x  x→2 1 1   
 2  2 or  4 2 x(x  3) x  3x  
 lim   8. lim  2 3 x→0 x  4x x→0 x(x  4) x3  
 lim  2 x→0 x  4 03 3  or  
 02  4 4 x2  3x  10 (x  5)(x  2)    9. lim x2  5x  6 
 lim (x  3)( x  2) x→3 x→3 (3  5)(3  2)   
 (3  3)(3  2) 8(1) 4   
 6(5) or 15 2 2x  5x  2 (2x  1)(x  2)  
 lim  10. lim  2 x→2 x  x  2 x→2 (x  1)(x  2) 2x  1  
 lim  x→2 x  1  
1  2.  2(2)  1   
 2  1 or 1  11a. v (r ) x2  4  lim  2 x→2 x  3x  2  0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02  
4  3. y is undefined when x 
 1. 4.  x2  4  lim  2  3x  2 x x→2  
 lim  x→2  (x  2)(x  2)  (x  2)(x  1) x2   
 lim  x1  x→2 22  
 21  v (r ) 
 0.65(0.52  r 2)  O  or 4  Yes, the limit is the same. 5. No; if the exact answer is a complicated fraction or an irrational number, you may not be able to tell what it is from the decimals displayed by a calculator.  0.2  0.4 0.6  11b. As the molecules get farther from the center and closer to the pipe, r is increasing. As r increases, v(r) gets closer and closer to 0 in./s.  Pages 946–948 Page 946  0.8 r  Exercises  12. The closer x is to 2, the closer y is to 1. So, lim f(x) 
 1. Also f(2) 
 1.  Check for Understanding  x→2  1. Sample answer: The limit of f(x) as x approaches a is the number that the values of f(x) get closer to as the values of x gets closer and closer to a. 2. Sample answer: lim f(x) is the number that the  13. The closer x is to 0, the closer y is to 0. So, lim f(x) 
 0. However, there is point discontinuity x→0  when x 
 0. So f(0) is undefined. 14. The closer x is to 3, the closer y is to 4. So, lim f(x) 
 4. However, there is a point at (3, 2).  x→1  values of f(x) approach as x approaches 1. f(1) is the number that you get if you actually plug 1 into the function. They are the same if f(x) is continuous at x 
 1. 3. Sample answer: If f(x) is continuous at x 
 a you can plug a into the function. If the function is not continuous, you may be able to simplify it and then plug in a. If neither of these methods work, you can use a calculator. Examples will vary. 4. The closer x is to 0, the closer y is to 3. So, lim f(x)  x→3  So f(3) 
 2. 15. lim (4x2  3x  6) 
 4(2)3  3(2)  6 x→2 
 16  6  6 
 16 16. lim (x3  3x2  4) 
 (1)3  3(1)2  4 x→1 
134 
0 17. lim  x→0  x→  
 3. However, there is a point at (0, 1), so f(0) 
 1. 5. lim (4x2  2x  5) 
 4(2)2  2(2)  5 x→2 
 16  4  5 
 17  sin x  x  sin   
  0  
  or 0 18. lim (x  cos x) 
 0  cos 0 x→0 
 0  1 or 1  475  Chapter 15  x2  25  (x  5)(x  5)  x→5  1x  1  x    19. lim  x  5 
 lim x5   1 x  31. lim  
 lim x  1 x  1 x→1 x→1  x→5  
 lim (x  5) x→5  1x  
 5  5 or 10  x→1  20. lim n 
 lim 2n x→0  
 2(0) or 0 21.  x(x  3)   
 lim  (x  5)(x  3)  x→3  32.  x   
 lim  x5  x→3 3  3    
 3  5 or 8  22. lim  x→1  x3  3x2  4x  8  x6  
  23. lim  h→2  1348   4h  4  h2  
 lim  h→2  x→1 1 
 1 or 1 x  2 x4 x4    lim x  2 
 lim x  2  x  2 x→4 x→4 (x  4)(x  2) 
 lim  x4 x→4  
 lim 
x  2  13  3(1)2  4(1)  8  16  x→4  
 4   2 or 4  8  
 7 or 7 h2  33. lim  h→0  (h  2)(h  2)  h2  2h3  h2  5h  h  h→0  h→2   3x  x→3  x  x→0  26.  27.     
 cos () or   1    
 27  18  9 or 2  25.  0    34. lim  cos (x  ) 
 cos (0  )  2(3)2  3(3)  33  2(3)2  3  6 18  9  h→0  35.  x3  x2  2x  lim  3  4x2  2x x x→0  x(x2  x  2)  
 lim  2 x x→0 (x  4x  2) x2  x  2  
 lim  2 x→0 x  4x  2 2 0 02  or 1 
 02  4(0)  2 x cos x x cos x  lim  2  
 lim  x→0 x  x x→0 x(x  1) cos x  
 lim  x→0 x  1 cos 0  
 0  1 or 1 2 (x  2)  4 x2  4x  4  4 lim 4 
 lim x x→0 x→0 x(x  4) 
 lim x x→0  tan 2x  lim x 
 2  x→0  36.  
 lim (x  4) x→0  
 0  4 or 4 28. lim  x→2  (x  1)2  1  x2  x2  2x  1  1  x2 x→2 x(x  2)  
 lim  x→2 x  2  
 lim  ln x lim  x→1 ln (2x  1)  
 lim x  x→2  x3  8   x2  4  
 lim  (x  2)(x2  2x  4)  (x  2)(x  2)  
 lim  x2  2x  4  x2  x→2  
 0.5  37.  x→2  
2 29. lim  x→2 (2)2  2(2)  4 
  2  2 444   or 3 
 4  30.  2x  8  lim  3 x→4 x  64  Chapter 15  h(2h2  h  5)  h  
 2(0)2  0  5 or 5  
 2  2 or 0 
 24. lim  x3  2x2  x  6  
lim  
 lim (2h2  h  5)  
 lim h  2 2x2  1  
 lim  x  x→0  x2  3x  lim  2  2x  15 x x→3  1   
 lim x   (1  x)  2n2  1  8  lim  x→1 x  1  2(x  4) 
 lim  (x  4)(x2  4x  16) x→4 2  
 lim  2 x→4 x  4x  16 2  
 42  4(4)  16 2 1  
 4 8 or 24  476  
 0.5  45. When P 
 0.99, t 
 2.58. Find X.  38.  X   1.4  50  
   0.20 Find the range. X  tj X 
 16.2 2.58(0.20)  
 15.68416.716 mm 9  46. P(not getting a 7) 
 1 0  3x  sin 3x   
 4.5 lim  x2 sin x  0.1  0.06908  0.1  0.06956  1  0.07177  9  59,049  t→0  2 10  1  t  47. (x  3y)5 
   r
0  5!  r!(5  r)!  
 10(x3)(9y2) 
 90x3y2 48. (16y8) 4 
 
16   y8
 4  3   5  5 5  1  49. center 
 
2, 2  x  1  0.5  0.1  0.01  0.001  0.540302  0.877583  0.995004  0.999950  1.000000  x2 1  2  0.5  0.875  0.995  0.99995  1.000000  
 (5, 2) The foci are on the x-axis. 91  a 
 2 or 4 1  (5)  b 
 2 or 3 (x  5)2  42  50.  120˚  (x  5)2  90˚  60˚  1 2 3 4  240˚  16(t  2)(t  2)   
 lim  t2  0˚  330˚  210˚  16t2  64   
 lim  t2  (y  2)2  30˚  180˚  agree with those of cos x.  t→2  (y  2)2   3 
 1 → 16  9 
 1 2  150˚  42b. yes; in the last three columns, all the decimal  t→2  3   3  4  
 23  y6 
 8y6  cos x  43. lim  5!  52(3y)2  (x)5r(3y)r 
  2!(5  2)! (x) 5!  1  d(t)  d(2)  t2  (x)5r (3y)r  2  3 
 2! 3! (x )(9y )  41. No; the graph of f(x) 
 sin
x oscillates infinitely many times between 1 and 1 as x approaches 0, so the values of the function do not approach a unique number. 42a.  places of 1   5!  r!(5  r)!  To find the third term, evaluate the general term for r 
 2.  
 0.07 or 7%  x2  2   
 100,000  5  x   lim  5  P(never getting a 7 in five spins) 
 
1 0  x→0  39. lim a a2  c2 
 a a2  02 c→0 2 
 a Letting c approach 0 moves the foci together, so the ellipse becomes a circle. a2 is the area of a circle of radius a. x 40.  2 10  1 f(x) 
  x x y 1 0.06697  270˚  300˚  u 51. WX 
 3  4, 6  0 
 7, 6 u WX  
  (3  4)2  (6  0)2 
 49 36  
 85   t→2  
 lim 16(t  2) t→2  
 16(2  2) or 64 ft/s 44a. As x approaches 0, the decimals for the values of 1   f(x) 
 (1  x) x approach 2.71828 ... , which is the decimal expansion of e. 44b. He ignored the exponent. As x approaches 0 from  25 in.  1 yd  1 mi    52. C 
 1   36 in.  1760 yd  0.0012395804 mi  65 v 
 C 52437.09741 v v  w 
 t 
  3600 s  1  the positive side, x approaches infinity. A number close to 1 raised to a large power need not be close to 1. If x approaches 0 from the 1 negative side, x approaches negative infinity. A number close to 1 raised to a large negative power need not be close to 1, either.  14.6 rps  477  Chapter 15  y  1   53. csc 270° 
  sin 270°  3. 0.5  1  
 y  270˚  1   
 1 or 1  x  54. Use a graphing calculator to find the rational roots at 1 4  and . 4 3 55.  (0, 1)  4. 1  y 
 4x5  2x2  4 x y 100 10 1  4  100 4  105 6  10  399804  100  4  1010  5. 1  y →  as x → , y →   as x →  1 2 
 (1)(6)  (2)(3) 3 6 
 6  (6) 
 12 57. yes; opposite sides have the same slope 1 y m 
  56.  AB  mDC 
  8 1  8  (8, 4)  1  mBC 
 4  A (0, 3)  1  mAD 
 4 58. If 
 8, then n 
 3. 3n2 
 332 
 35 or 243  6. 3 B  x  2n  D (2, 5)  C (10, 4)  7. The graph of a linear function is a line and the methods in this function will result in the calculation of the slope of that line. 8. If you zoom in on a maximum or minimum point, the graph appears flat. The slope is 0.  15-2A Graphing Calculator Exploration:  dy  9. At (0, 1), dx 1. For other points on the curve, the dy values for y and dx are approximately the same.  The Slope of a Curve  Page 950 1-6. Exact answers are given. Accept all reasonable approximations. 1. 4  15-2  2. 1  Derivatives and Antiderivatives  Pages 957–958  Check for Understanding  1. 4x3 is the derivative of x4. x4 is an antiderivative of 4x3. 1   n1 2. Letting n 
 1 in the expression  n1 x x0  results in 0, which is undefined. Chapter 15  478  3. f(x  h) means substitute the quantity x  h into the function. On the other hand, f(x)  h means substitute x into the function, then add h to the result. Using f(x)  h instead of f(x  h) in the definition of the derivative results in: lim  h→0  f(x)  h  f(x)  h  13.  h  h→0 h  
 lim  C(x) 
 1000  10x  0.001x2 C(x) 
 0  10  1x11  0.001(2)x21 C(x) 
 10  0.002x C(1000) 
 10  0.002(1000) 
8 The marginal cost is $8.  
 lim 1 h→0  Pages 958–960  
1 You would always get 1. 4. f (x) 
 lim  h→0  f(x  h)  f(x)  h  
 lim  3(x  h)  2  (3x  2)  h  
 lim  3x  3h  2  3x  2  h  
 lim  3h  h  h→0 h→0 h→0  
2 15. f(x) 
 lim  h→0  
3 5. f(x) 
 lim  h→0  f(x  h)  f(x)  h  f(x  h)  f(x)  h  
 lim  7(x  h)  4  (7x  4)  h  
 lim  7x  7h  4  7x  4  h  h→0  
 lim  (x  h)2  x  h  (x2  x)  h  
 lim  x2  2xh  h2  x  h  x2  x  h  
 lim  2xh  h2  h  h  16. f
(x) 
 lim  
 lim  h(2x  h  1)  h  
 lim   3(x  h)  (3x)  h  
 lim  3x  3h  3x  h  
 lim  3h  h  h→0 h→0 h→0 h→0  h→0  7h  h→0 h  
 lim 
7  h→0 h→0  
 lim 2x  h  1 h→0  h→0  
 2x  0  1 or 2x  1 6. f(x) 
 2x2  3x  5 f 
(x) 
 2  2x21  3  1x11  0 
 4x  3 7. f(x) 
 x3  2x2  3x  6 f(x) 
 1  3x31  2  2x21  3  1x11  0 
 3x2  4x  3 8. f(x) 
 3x4  2x3  3x  2 f (x) 
 3  4x41  2  3x31  3  1x11  0 
 12x3  6x2  3 9. y 
 x2  2x  3 dy  dx dy  dx  Exercises  f(x  h)  f(x)  h h→0 2(x  h)  2x   
 lim h h→0 2x  2h  3x 
 lim h h→0 2h 
 lim h h→0  14. f (x) 
 lim  h→0  
 3 17. f
(x) 
 lim  h→0  4h  h  h→0  2(x  h)2  5(x  h)  (2x2  5x)  h  
 lim  2x2  4xh  2h2  5x  5h  2x2  5x  h  
 lim  4xh  2h2  5h  h  
 lim  h(4x  2h  5)  h  h→0 h→0  1  h→0  
 4x  2(0)  5 or 4x  5  11. f(x) 
 x3  4x2  x  3  19. f
(x) 
 lim   31  4
x21  x11 F(x) 
  31 x 21 11 1  1  1  h→0   3x  C 12. f(x) 
 5x5  2x3  x2  4  (x  h)3  5(x  h)2  6  (x3  5x2  6)  h  
 lim  x3  3x2h  3xh2  h3  5x2  10xh  5h2  6  x3  5x2  6 ––––––––––––––– h  
 lim  h(3x2  3xh  h2  10x  5h)  h  h→0   51  2
x31  x21 F(x) 
 5
 5  1 x 31 21 1  1  h→0   4x  C  f(x  h)  f(x)  h  
 lim  h→0  1  
 4x4  3x3  2x2  3x  C 1  f(x  h)  f(x)  h  
 lim  h→0  
 3x3  C  1  
 lim 18. f
(x) 
 lim  1  5  4x  4h  9  4x  9  h  
 4   21  C F(x) 
  2  1x  4  
 lim  h→0  f (1) 
 2(1)  2 
4 10. f(x) 
 x2  1  4(x  h)  9  (4x  9)  h  h→0  
 2x  2  f(x  h)  f(x)  h  
 lim  h→0  
 2  1x21  2  lx11  0  f(x  h)  f(x)  h  
 3x2  3x(0)  02  10x  5(0) 
 3x2  10x  1  
 6x6  2x4  3x3  4x  C  479  Chapter 15  20. f(x) 
 8x f
(x) 
 8  1x11 
8 21. f(x) 
 2x  6 f
(x) 
 2  1x11  0 
2 1  y 
 (5x2  7)2 
 25x4  70x2  49 dy  
 100x3  140x dx f 
(1) 
 100(1)3  140(1) 
 240 35. f(x) 
 x6 1  61  C F(x) 
  61 x  34.  4  22. f(x) 
 3x  5 1  f 
(x) 
 3  1x11  0  1  
 7x7  C  1  
 3  36. f(x) 
 3x  4 1  3x2  23. f(x) 
  2x  9 f
(x) 
 3  2x21  2  1x11  0 
 6x  2   11  4x  C F(x) 
 3   1  1x 3  
 2x2  4x  C 37. f(x) 
 4x2  6x  7  1  24. f(x) 
 2x2  x  2  1  4  
 3x3  3x2  7x  C  
x1 25. f(x) 
 x3  2x2  5x  6 f
(x) 
 3x31  2  2x21  5  1x11  0 
 3x2  4x  5 26. f(x) 
 3x4  7x3  2x2  7x  12 f
(x) 
 3  4x41  7  3x31  2  2x21  7  1x11  0 
12x3  21x2  4x  7 27. f(x) 
 (x2  3)(2x  7) 
 2x3  7x2  6x  21 f 
(x) 
 2  3x31  7  2x21  6  1x11  0 
 6x2  14x  6 28. f(x) 
 (2x  4)2 
 4x2  16x  16 f 
(x) 
 4  2x21  16  1x11  0 
 8x  16 29. f(x) 
 (3x  4)3 
 27x3  108x2  144x  64 f
(x) 
 27  3x31  108  2x21  144  1x11  0 
 81x2  216x  144 30. f(x) 
 f
(x) 
 
 31.  y
 dy  dx  2 1 x3  x2  x  9 3 3 2 1   3x31    2x21 3 3 2 2x2  3x  1 x3  38. f(x) 
 12x2  6x  1 1  
 4x3  3x2  x  C 39. f(x) 
 8x3  5x2  9x  3 F(x) 
 8  1  1   21  9  x11  3x  C 5 2  1x 11 5  9  1  2  40. f(x) 
 4x4  3x2  4 1  2  1  1   41    x21  4x  C F(x) 
 4   3 21 4  1x 2  1  5  3 
 2 0 x  9 x  4x  C  41. f(x) 
 (2x  3)(3x  7) 
 6x2  5x  21 1  1   21  5  x11  21x  C F(x) 
 6   2  1x 11 5  
 2x3  2x2  21x  C 42. f(x) 
 x4(x  2)2 
 x6  4x5  4x4 1  1  1   61  4  x51  4  x41  C F(x) 
  6  1x 51 41   1  1x11  0  1  2  4  
 7x7  3x6  5x5  C x3  4x2  x  43. f(x) 
 x  
 3x2  
 x2  4x  1 1 1  21  4  x11  x  C F(x) 
  2  1x 11 1  
 3x3  2x2  x  C 2x2  5x  3   44. f(x) 
  x3  
 3x2  14x  4  
 2x  1 1  11  x  C F(x) 
 2   1  1x 
 x2  x  C 1  1  1  45. Any function of the form F(x) 
 6x6  4x4  3x3  x  C, where C is a constant.  
 2x  1  1  f
(1) 
 2(1)  1 
1  Chapter 15  1 x31 31  
 2x4  3x3  2x2  3x  C  f
(1) 
 3(1)2  14(1)  4 
 7 33. y 
 (x  1)(x  2) 
 x2  x  2 dy  dx  1   21  6   x11  x  C 
 12   21x 11  f 
(1) 
 3(1)2 
3 32. y 
 x3  7x2  4x  9 dy  dx  1   21  6  x11  7x  C F(x) 
 4   2  1x 11  1  f
(x) 
 2  2x21  1  1x11  0  46a. v(12) 
 15  4(12)  8(12)2 
 81 ft/s  480  49a. h(t) 
 3  80t  16t2 h
(t) 
 v(t) 
 0  80  1t11  16  2t21 
 80  32t 49b. v(1) 
 80  32(1) 
 48 ft/s 49c. At the ball's maximum height, the velocity is 0. 0 
 80  32t 80 
 32t 2.5 
 t; 2.5 s 49d. h(2.5) 
 3  80(2.5)  16(2.5)2 
 103 ft  1  46b. v(t) 
 15  4t  8t2 1  v
(t) 
 0  4  1t11  8  2t21 1  
 4  4t 1  v
(12) 
 4  4(12) 
 7 ft/s2 46c. When t 
 12 the car's velocity is increasing at a rate of 7 ft/s per second. 1  46d. v(t) 
 15  4t  8t2 1  1  1   11    t21  C s(t) 
 15t  4   8 21 1  1t  
 15t   2t2    1 t3 24  50. f
(x) 
 lim  h→0  C  When t 
 0, s(t) should equal 0, so C 
 0.  
 lim  exh  ex  h  
 lim  ex  e h  e x  h  
 lim  ex  (eh  1)  h  h→0  1  3 s(t) 
 15t  2t2  2 4t  h→0  1  3 46e. s(12) 
 15(12)  2(12)2  2 4 (12)  h→0  
 540 ft 47. f 
(x) 
 lim  h→0  
 ex  lim  f(x  h)  f(x)  h 1  xh  h→0  h→0  f
(x) 
  h  x(x  h)  
 lim  h h→0 1  h→0 x(x  h) 1  
 x(x  0)  
 lim  1 
  x2 48a. When y 
 2010, I(2010) 2.75. The total amount spent in 2010 on health care will be about $2.75 trillion. 48b. T(y) is approximately linear near (2010, 2.75).  i. e.  x2  3x  3  ex  
 1, so  is its own derivative.  (x  3)(x  1)  x→3  x→3  
 lim x  1 x→3  
 3  1 or 4 53a.  $3.5  170 180 190 200 210 220 230 240 250 260 53b. See students' work. 54. List all pairs of matching numbers and their sums. 1  1 
 2; 2  2 
 4; 3  3 
 6; 4  4 
 8; 5  5 
 10; 6  6 
 12 There are 3 sums out of 6 that are greater than seven. 3 P(sum  7 given that the numbers match) 
 6  Projections  2.5 2.0 1.5 1.0 0.5  1  2001  2004  2007  2010  
 2  2013  55. an 
 a1rn1  Source: Centers for Medicare & Medicaid Services  1 61  a6 
 9
3  Find the slope of the tangent line at (2010, 2.75). T 
(2010)  1
  ex.   
 lim  52. lim  x3 x3  Health Care Spending (Trillions of Dollars)  0  ex  eh  1  h  51a. total revenue 
 cost per cup  number of cups r(p) 
 p(100  2p) 51b. When r(p) is at a maximum, the derivative equals zero. r(p) 
 p(100  2p) 
 100p  2p2 r
(p) 
 100  1p11  2  2p21 
 100  4p 0 
 100  4p 100 
 4p 25 
 p; 25 cents  xh  x(x  h)   
 lim  h h→0  3.0  eh  1  h  A calculator indicates that lim  1  x   
 lim  h h→0 x  x(x  h)  f(x  h)  f(x)  h  2.75 – 2.56  2010 – 2009  9  1    
  243 or  27  56. y 
 136e0.06(30)  74 96°  0.19 In 2010 the amount spent on health care will be increasing at a rate of about $190 billion per year.  481  Chapter 15  57. x2  y2  Dx  Ey  F 
 0 22  (1)2  2D  E  F 
 0 → 2D  E  F  5 
 0 (3)2  02  3D  F 
 0 → 3D  F  9 
 0 12  42  D  4E  F 
 0 → D4E F  17 
 0 2D  E  F  5 
 0 3D F 9 
0 (3D  E  F 9) 
 0 ( D  4E  F  17) 
 0 5D  E  4 
 0 4D  4E 8 
0  x→3  
 18  12  6 
 36 x2  9x  14  x→2  x→2  x7   
 lim  2x  3  x→2 27  
 2(2)  3  28  7  3
  1  3  E 
 3  1  3  F9
0  1  7  The solution of the system is D 
 3, E 
 3, and F 
 8 7  x2  y2  3x  3y  8 
 0 2  
x  16  
y  76 
 11689 5 5 1  3 58. 5
cos 6  i sin 6 
 5
2  i 
2 53   1 0.1 0.01 0.01 0.1 1  0.9093 1.9867 1.9999 1.9999 1.9867 0.9093  h→0  5  
 2  2i 59. x 
 x1  ta1 → x 
 8t  3 y 
 y1  ta2 → y 
 3t  2  
 lim  x2  2xh  h2  3  x2  3  h  
 lim  2xh  h2  h  
 lim  h(2x  h)  h  h→0    O  61.  90˚  180˚  d  H1  270˚  360˚  h→0  450˚  
 2x  0 or 2x 5. f(x) 
  f
(x) 
 0  is a constant. 6. f(x) 
 3x2  5x  2 f
(x) 
 3  2x21  5  1x11  0 
 6x  5  H2 319 m  253 m  7. R(M) 
 M 2
2  3 C  42˚12  C  p : q  C  2532  1 , 2  1  2  3  2  1  2  2  3192  1,  3 , 2  3 2 5 6  1  1   21  7   x11  6x  C F(x) 
 1   21x 11  3 8 5 3 3  1  
 CM  M 2 8. f(x) 
 x2  7x  6 1  7  
 3x3  2x2  6x  C  3 3 0  9. f(x) 
 2x3  x2  8 1  1   31   x21  8x  C F(x) 
 2   31x 21 1  1  
 2 x4  3 x3  8x  C  1 0 1 2 0 1 The rational roots are 3, 2, and 1.  10. f(x) 
 2x4  6x3  2x  5 F(x) 
 1  1  1   41  6  x31  2  x11  5x  C 2   4  1x 31 11  63. 90  x  y  z 
 360 x  y  z 
 270 The correct choice is D.  Chapter 15  1  R
(M) 
 2  2M 21  3  3M 32  
   2(253)(319) cos 42° 12
 d 
 165,7 77 161,4  14  cos2
 42° 1 d 214.9 m  62.  M  
 2M 2  3M 3  A d2  f(x  h)  f(x)  h (x  h)2  3  (x2  3)  h  h→0  y 
 3 sin(  45˚)  x→0  
 lim  h→0  60. y  sin 2x  sin x  x  4. f
(x) 
 lim  or 5 lim x 
 2  x  3.  F 
 8  2  (x  7)(x  2)   
 lim  2. lim  2x2  7x  6 (2x  3)(x  2)  4E 
 3  8 
0  1   4x  6) 
 2(3)2  4(3)  6  1  (4D  4E  8) 
 0 D
  1. lim  Mid-Chapter Quiz  (2x2  4
3  4E  8 
 0  20D  4E  16 
 0 24D  Page 960  2  3  
 5x5  2x4  x2  5x  C  482    3  8.  Area Under a Curve  15-3  0  n  3i 3 3   
 
n n→ i 
 1 n  x3dx 
 lim  n (n  1)  
 4 2  81  4 n→ n 81 
 lim 4 n→ 81 
 lim 4 n→ 81 
 4  
 lim  Page 966  Check for Understanding  1. Sample answer: y 
 2. Sample answer: Subdivide the interval from a to b into n equal subintervals, draw a rectangle on each subinterval that touches the graph at its upper right corner, add up the areas of the rectangles, and then find the limit of the total area of the rectangles as n approaches infinity. 3. Lorena is correct. If the function is decreasing, then the graph will always be above the tops of the rectangles, so the total area of the rectangles will be less than the area under the graph. x4    2  4.  0  n  6  0  
 
 
  1  2  3  1  0      9b.  0   lim  n→ 9 
 lim 2 n→  1  n→  
 1600 ft Yes; integration shows that the ball would fall 1600 ft in 10 seconds of free-fall. Since this exceeds the height of the building, the ball must hit the ground in less than 10 seconds.  1    1  6.  0    6  0    2  10.  0  n   
n  1 
n n→ i 
 1  2  
  
    
 lim  1  4  n  2n  1  
 n  
 lim  1  4  
1   2  2  n→ 1 
 4 unit2   n2  1  0  n  2  
  
  n        3i 2 3  n(n  1)(2n  1)  
 6  27  3 n→ n 9 
 lim 2 n→  2n  3n  n  
 n    9  n→ 2    
 lim 
9  
 lim 36 
2  n  n2  3  ...   
 
n n→ i 
 1 n  x2dx 
 lim 
 lim  216 n(n  1)(2n  1)   3 6 n→ n 2n3  3n2  n 
 lim 36 n 3 n→  
 lim    3  11.  6i 6  
 
n n→ i 
 1 n  x2dx 
 lim   
     i 3 1   n n→ i 
 1 n n2(n  1)2 1 
 lim n4 4 n→  2  n  2  
  
  26  n  x3dx 
 lim  2i  2 2(1) 2(2)    1    1 n n n→ n 2n  n  1 2 2 
 lim n n  n(1  2  ...  n) n→ 2 2 n(n  1) 
 lim n n  n  2 n→ 2 
 lim n(2n  1) n→ 4n 2 
 lim n  n n→ 
 4 units2  
 lim  3 1 1 lim 6 
2  n  n   n→  1  n2  Exercises  (x  1)dx 
 lim  
 9  3 or 3 unit2  n→    Pages 966-968    
  7.    2n3  3n2  n  n3  
2   n(n  1)  
 2  n→    3  n  3200  n2  10  
 lim 1600 
1  n    
  10i  n2  n   
 
   
  n   32 
n
n n→ i 
 1  32tdt 
 lim  
 lim 1600
n 2   n(n  1)(2n  1) 1   3 6 n→ n 9 2n3  3n2  n 
 lim 2 n 3 n→ 1  6  
 576 ft  n→  n  
  6  6i  1152 n(n  1)   2 n2  
 lim  
 
    lim  1  2  2  
   n2   2  i 2 1 3i 2 3    lim   n n n→ i 
 1 n n→ i 
 1 n 27 n(n  1)(2n  1) 
 lim n3 6 n→  
 lim  
1   
  n n 
 lim 576 
n n→ 1 
 lim 576
1  n n→ 10  0  n  2  2  n  n  n→     2  2   32 
n
n n→ i 
 1  
 lim   x dx 
  x dx   x dx 3  n  2n  1  
 n  32tdt 
 lim  2i 2 2   
 
n n→ i 
 1 n  x2dx 
 lim  8 n(n  1)(2n  1) 
 lim n3 6 n→ 4 2n3  3n2  n 
 lim 3 n 3 n→ 3 1 4 
 lim 3 2  n  n2 n→ 8 
 3 units2  5.    9a.  2  1  3  2  3  
2  n3  n1 2  units2  n→  
 72  483  Chapter 15    2  12.  1   
  0  x2dx 
   x dx x dx   x dx  1 1  0  2  x2dx   0 2  2    4  2  17.  0  2  n  n  n  i 2 1  1 n(n  1)(2n  1)   3 6 n→ n 8 n(n  1)(2n  1)      lim n3 6 n→ 3 2 1 2n  3n  n 4 2n3  3n2 n 
 lim 6 n  lim 3 n 3 3 n→ n→  
    
    
  
 lim  n→ 1 8 
 3  3    3  13.  1  6  1  xdx 
    3  0    
2   3  n    
      lim 
2    n→ 4  3  1  n2  3  n    n      1  n2  14.  n  3  i  1    
    0    
    
    
    
 lim  125  6  5  1  0  1  3  n  1  2x3 n  
    
 
  16.    
      
6   
 lim  
 625  units2  n→  Chapter 15  0    10  n2             n  
    
      n   8 
n
n n→ i 
 1 2i  8xdx 
 lim  2  32 n(n  1)   2 2 n→ n n2  n 
 lim 16 n 2 n→ 1 
 lim 16 1  n n→  
 16    15  n      
 lim    
    2  20.   
 
   625  6  3  3i   
  
  
  n    
  48
1  n   n  1 
n     19. lim    
     
sin in  n n→ i 
 1  i 3 1   n n  250 n2(n  1)2 n2(n  1)2 2    lim   4 4 4 n 4 n n→ n→ 625 n2  2n  1 1 n2  2n  1 
 lim 2 n  lim 2 n 2 2 n→ n→ 1 1 625 1 2 2 
 lim 2 1  n  n2  lim 2 1  n  n2 n→ n→ 625 1 
 2  2 or 312 units2 n 5 5i 4 5   x4dx 
 lim n n→ i 
 1 n 0 6n5  15n4  10n3  n 3125  
 lim n 5 30 n→  
 lim  2  2  3 9 3 n(n  1) n(n  1)(2n  1)       2 n 6 2 n→ n n 27 2n2  3n2  n n2  n 9         
 lim n3  n2 3 6 2 n→ 9 9 3 1 1 
 lim 2 2  n  n2  2 1  n  3 n→ 9 
 9  2  3 9 15 
 12  2 or 2 units2  
 lim  lim  2 
  
   2
  
   n→ n→ i 
 1 i
1  
 lim  2  3  
2  n  n2   5i 3 5   n n  2   n (1  2  . . .  n)  n    0  (1  2  . . .  n)  4  n    2  5  3i 2  

  

 

    3     31 2 31 3    n  1 n n→ n 32 2 32  n  n  1  . . . 3n 2 3n  n  n  1 3 9 
 lim n n2 (12  22  . . .  n2) n→  n→ 125 
 3 units2   2x dx 
  2x    n   
n n→ i 
 1  
  
    (x2  x  1)dx  
 lim  
  15.    3  18.   
 
   
 lim  
 
  3 1 32 (2     ) n n2 n→ 3 64 208 
 3  48 or 3 units2  
 lim  xdx  9 1 n(n  1) n(n  1) 
 lim n2 2  lim n2 2 n→ n→ 9 n2  n 1 n2  n 
 lim 2 n  lim 2 n 2 2 n→ n→ 9 1 1 1 
 lim 2 1  n  lim 2 1  n n→ n→ 9 1 
 2  2 or 4 units2 n 5 5i 2 5   x2dx 
 lim n n→ i 
 1 n 0 125 n(n  1)(2n  1)  
 lim n 3 6 n→ 125 2n3  3n2  n 
 lim 6 n 3 n→       
  3  0  
    3  1  3i  4  4i  16 n(n  1)(2n  1)  
  6 n n→ 24 n(n  1)      n 
 2  64 2n  3n  n 96 n  n 
 lim n 
6  n 
2 n→  lim  
n
n  

n  n→ n→ i 
 1 n i
1  
 lim    24  n  
 lim  or 3 units2  xdx   

  

 

  2i 2 2  lim  
n 
n  
 
n  n→ n→ i 
 1 n i
1  
 lim   6 
n
n  4 41 2 41    6 n n n→ n 42 2 42  n  6 n  . . . 2 4n 4n  n  6 n 4 16 
 lim n n2 (12  22  . . .  n2) n→  
 lim  By symmetry  0  
 lim  4i 2   
n n→ i 
 1  (x2  6x)dx 
 lim    1  n4  484  
    
 
     
  1   (x  2)dx 
 lim 4  21.  n   
1  n  2
n n→ i 
 1  1  3  3i  n  
    3 3i  3  n n→ n i 
 1 3 31 
 lim n 3  n  n→  
 lim    3n   
3  n 
 lim  3  n  3n  n3(1  2  . . .  n)  
 lim  3  n  n(n  1)  3n  n3 
 2  n→ n→  n(n  1)(2n  1) 64 n(n  1)    
  8 6n4
 6 2 n n(n  1)(2n  1) n(n  1) 1 4  lim n 
6  n 
2  2 n→ 32 2n  3n  n n n 
 lim 3 
n  32 
n  8 n→ 1 2n  3n  n n n  lim 6 
n  2 
n  2 n→ 32 3 1 1 
 lim 3 
2  n  n  32 
1  n  8 n→ 1 3 1 1  lim 6
2  n  n  2
1  n  2 n→  22.    n    5  23.  3  8x3dx 
    5  0  64  1  
 3  32  8  3  2  2  
  
 24    (x 2  25.    0  63  3  or 45 n  5    
 lim  
2  n3  n1  
 lim  
  n  
 lim  n→  8x3dx n  5i 3 5  5000 n2(n  1)2   4 n4  
 lim 1250 
1  n  n2  1  24.  2  n  1  n2   (x  4x  2)dx 
  (x  4x  2)dx   (x    0  2  n  0  4i 2   
n n→ i 
 1  
 lim  n  2   4x  2)dx  4i  26.  
n4   4n 2  0  n→  2  

n1  x3dx 
 lim  4  8  3  or  2 2  40  3  5i 3 5   n n   
 
   n→  i 
 1 625 n2(n  1)2  
 lim n 4 4 n→     
 lim  625  4   2n  1  
 n  
 lim  625  4  
1  n2  n1  n→  625 
 4   4
n  2  

n n 2    n  n→   4n  1  
  32  3  
   4 
n  2  . . .  

n  4
n 1  n    5  2   2  lim  6  2  2   4
n  2  42   5n4  n2  2  i i 1  
n  4 
n  2
n n→ i 
 1 41 2 41 42 2 4 
 lim n 

n  4 
n  2  

n n→  lim    6n5  3  2  1    2n6  3  1  4    16 
   n n→ 3 4 2n  3n  n  3 
n 16 6 1 4 5 
 lim 3
2  n  n  n  3 n→ 
2  n3  n1  
 1250  162 or 1088  4    
 lim  n→     ...    
  n2  2n  1   lim 162 
n  2 2    22 2  n 2n 2  n   
  n2  2n  1  n→  n→  5  21 2   
n    2  4  
 lim 1250 
n  2   lim 162 
1   5  2i 2   
n   
 5  
  
 lim  648 n (n  1) lim n 
4   n→  n→  2i 2 2  2 32 5 5 5   5 (1  2  . . .  n ) n→ n n 4  n2 (12  22  . . .  n2) 2n6  6n5  5n4  n2 64 
 lim n6  12 n→ n(n  1)(2n  1) 8  n3 6  3i 3 3  2  2i 5  21  

 n   

 

  0  
  n   
n i
1  2  n→ n 22  n 2n  n  lim  8 
n 
n  8 
n 
n  n→ n→ i 
 1 i
1  
 lim  2  n  n→  2    2i 5 2   
 
n  
n 
n n→ i 
 1 n   x2)dx 
 lim  2n3  3n2  n  n3  8x3dx   2  2   22  32  . . .  n2)  3  2  2  27  2  n(n  1)(2n  1)  6  64 
 lim n3 n→ 32 
 lim 3 n→ 32 
 lim 3 n→ 64 
 3  2  2  3  1   
 
   x2dx 
 lim  2  3  4i 2 4   n n  n→ i 
 1 64 
 lim n3 (12 n→  0  or  2  2  3  n→  
9  2  3  n→  4  3  3  
 lim 9  2
1  n 9  2  (1  2  . . .  n)  2n  n→  n2  n  9  4  n  
 lim  
 lim 9  n2 
2 9      32   
3   n ...  
    4 16 2 2 2   2 (1  2  . . .  n ) n→ n n 16  n (1  2  . . .  n)  2n 1 1  lim n n2 (12  22  . . .  n2) n→  
 lim  n2  2  2   4 
n  2  . . .  

n  4 
n  2 2  n  485  Chapter 15    3  27.  0  n  3i 3 3    
 
n n→  i 
 1 2 n  1 x3dx 2  1  
 lim  31a. r (t ) 160  81 n2(n  1)2   
 lim  4 4 n→  2n 81 n2  n  1 
 lim 8 n 2 n→  1 1 81 
 lim 8 1  n  n2 n→   
    
  120    
  80    40  81  28a. f(20)
80  2(20) 
 $40    40  20  0  n  20i 20  
80  2 
20  n 
n n→  i 
 1  n→   n  40i  41 42   
40   n   
40  n   . . . 4n    
 600  2592  1728 
 $1464 31c.  (6  0.06x2)dx n  
 lim  $1464  12  
 $122  32a. v (t )  2  10i 10  
6  0.06 
n 
n  n→  i 
 1 10 
 lim n n→   2  3  2  1  n  
 800  400 or $400 0  3  2  2  10  3  2  2    2  12  n  2  2  n→   12  2  40n  4n0 (1  2  . . .  n) 800 n(n  1) 
 lim 800  n 
2 n→  n n 
 lim 800 400 
n n→  20  n→  n  
 lim  
 lim 800  400
1   12i 2  12i  12  1 12  1 50  36
n  3
n  
 n→  12  2 12  2 2  
50  36
n  3
n   . . . 12  n 12  n 2  
50  36
n  3
n  12 432 
 lim n 50n  n(1  2  . . .  n) n→  432  n(12  22  . . .  n2) 5184 n(n  1) 5184 n(n  1)(2n  1) 
 lim 600  n
2  n
6 n→  n n 2n  3n  n 
 lim 600  2592
n  864
n n→  1 3 1 
 lim 600  2592
1  n  864
2  n  n n→    
40  n  29.  n   50  36
n  3
n 
n n→  i
1  
 lim   
40  n i
1  20  n→  n  
 lim  (50  36t  3t2)dt  
 lim  
 lim  20  n    12  31b.  (80  2x)dx  
 lim  t  O  
 8 or 10.125 ft2  28b.  r (t ) 
 50  36t  3t 2  15  2  10  1  
6  0.06 
 n   2 10  2  
6  0.06 
n   . . . 10  n 2  
6  0.06 
n  6 10 
 lim n6n  n (12  22  . . .  n2) n→  60 n(n  1)(2n  1) 
 lim 60  n 
6 n→  2n  3n  n 
 lim 60  10 
n n→  3 1 
 lim 60  10 
2  n  n n→   v (t ) 
 3.5t  0.25t 2 10  2  5  3  3  v (t ) 
 1.2t  0.03t 2  2  3  2  O  
 60  20 or 40    10  10  6  0.06x2dx 
 2(40) or 80  32b.  By symmetry  0  To make a tunnel 100 ft long, multiply 80 by 100. 80(100) 
 8000 ft3 30. Setting the two functions equal 2 y to each other and solving for x, y 
 x2 we find that the curves cross when x 
 0 and x 
 1. x  x2 y
x 1 for 0  x  1, so we can find the desired area by subtracting the area between the graph of x y 
 x2 and the x-axis from the O 1 area between the graph of 1 y 
 x and the x-axis. These areas are 3 unit2 and 1  2 1  6  1  1 2 3 4 5 6 7 8 9 10t  (3.5t  0.25t2)dt n  10i 2   3.5
n  0.25
n 
n n→  i
1 10i  
 lim  10  2  10  1 10  1   
3.5
 n   0.25
 n   n→  10  2 10  2 2  
3.5
n  0.25
n   . . . 10  n 10  n 2  
3.5
n  0.25
n  10 35 25 
 lim nn(1  2  . . .  n)  n(12  22 n→   . . .  n2) 350 n(n  1) 250 n(n  1)(2n  1) 
 lim n
2  n
6 n→  n n 125 2n  3n  n 
 lim 175
n  3
n n→  125 1 3 1 
 lim 175
1  n  3
2  n  n n→   
 lim  10  n  2  2  3  2  3  2  1  unit2, respectively, so the answer is 2  3 
  2  3  2  unit2.  Chapter 15    10  
 175   486  250  3  or  275  3  m    10  0  40. Use a graphing calculator to find the maximum width of x 
 5.4 cm.  (1.2t  0.03t2)dt n  2  10i 10i 10  1.2
n  0.03
n 
n 10  1 10  1 2   
 
1.2
 n   0.03
 n   10  2 10  2 2  
1.2
n  0.03
n   . . . 10  n 10  n 2  
1.2
n  0.03
n  10 12 
 lim nn(1  2  . . .  n) n→  3  n(12  22  . . .  n2) 120 n(n  1) 30 n(n  1)(2n  1) 
 lim n
2  n
6 n→  n n 2n  3n  n 
 lim 60
n  5
n n→  1 3 1 
 lim 60
1  n  5
2  n  n n→   
 lim  n→  i
1 10 
 lim n n→   [5, 10] scl1 by [30, 90] scl1  2  2  41. 62  82 
 102 36  64 
 100 100 
 100 ABC is a right triangle with a base of 6 inches and a height of 8 inches. The area of  3  2  3  2  2  3  2  
 60  10 or 70 m  1  ABC 
 2(6)(8) or 24 square inches. If the  The first one results in a greater distance covered. 33. The equation y 
  r2  x2 can be rearranged to obtain x2  y2 
 r2, which is the circle centered at the origin of radius r. In the equation y 
  r2  x2, y must be nonnegative, so the graph is only the top half of the circle. Therefore, the value of the 1 integral is 2 the area of a circle of radius r, 1 or 2r2.  rectangle has a width of 3 inches, then it has a 24  length of 3 or 8 inches, since A
w. The perimeter of the rectangle 
 2(3)  2(8) or 22 inches. The correct choice is C.  Page 969  34. f(x) 
 3x2  x2  7x f
(x) 
 3  3x31  1  2x21  7  1x11 
 9x2  2x  7 x2  x→2 x  2  35. lim  22   
 22 0  
 4 or 0 36. log1x 
 3 3  1 3  x 
 
3  15-4  x 
 32 or 27 u 37. u 
 2, 5, 3  3, 4, 7 
 2  (3),  5  4, 3  (7) 
 1, 1,  10 u u u 
 i j  10k 38. cos 2v 
 1  2 sin2 v  The Fundamental Theorem of Calculus  Pages 972–973 1.  Check for Understanding   f(x)dx represents all of bthe function that have f(x) as their derivative. a f(x)dx is a number; it  gives the area under the graph of y 
 f(x) from x 
 a to x 
 b. 2. Sample answer: Let a 
 0, b 
 1, f(x) 
 x, and  3 2  
 1  2
5  g(x) 
 x. Then  f(x)g(x)dx 
  x2dx 
 3, but a 0 b  18  
 1  25  b  1  1  1  a f(x)dx  a g(x)dx 
 0  7  
 2 5  b  1  xdx  0 xdx 
 2  2 
 4. 1  1  1  1  3. If the "C " is included in the antiderivative, it will appear as a term in both F(b) and F(a) and will be eliminated when they are subtracted. 4. Rose is correct; the order does matter. Interchanging the order multiplies the result by 1. In symbols, F(a)  F(b) 
 (F(b)  F(a)). So unless the answer is 0, interchanging the order will give the opposite of the right answer.  39. y 
 2 sin 10v 1  Amplitude 
 2 2  History of Mathematics  1. See students' work. The difference in area should decrease as the number of sides of the polygon increases. 2. The roots of the resulting equation are the zeros of the derivative of the original function. 3. See students' work.    Period 
 10 or 5  5.  (2x2  4x  3)dx 
 2  13x3  4  12x2  3x  C 2  
 3x3  2x2  3x  C 6.  (x3  3x  1)dx 
 14x4  3  12x2  x  C 1  3  
 4x4  2x2  x  C  487  Chapter 15    0  7.  1  2  (4  x2)dx 
 4x  3x3  0  Pages 973–976  2   
 C 1 8 7 16. 6x dx 
 6  8 x  C 15.  
 
40 3 03  
4(2) 3 (2)3 1  1  
 0  
8  3 8    2  8.  0  
 x4dx 
  
    1  3  17.  (x2  2x  4)dx 
 13x3  2  12x2  4x  C 1   
  1   25 5 32   0 5  
 3x3  x2  4x  C    1  5  05 32 or 5 units2  (x2  4x  4)dx 1  18.  1  19.  1  1  (x4  2x2  3)dx 
 15x5  2  13x3  3x  C 1  20.  1  (4x5  6x3  7x2  8)dx 1  
 
3  13  2  12  4  1  2   
3  (1)3  2  (1)2  4  (1)  
 3  
3 or 3 units2   2x dx 
 2  3  3  1  1  

    4  11.  1  
  26  21.  22.    0  1  
  1  2    
 3x3  3x2  3x  C    
 
  33  
3  03    1  2    42    3  1  23.   6  4  2  1  (x2  2)dx 
 3x3  2x 

  1  3   
3  13  2  12  6  1 1  1  112  35  224  35  12.    0  (2x2  3x  2)dx 
 2   24.  2  0  3  1 x2 2  3  
 3x3  2x2  2x   2x 2  2  
    4  13.  
 (x3  2  1 x4 4   x  6)dx 
    0   0 or  1 x2 2  3  2  1  0    4    0.1  0  25.   0  2  0  0  4    1  2  26.  1  2 1 x4 4 0  0  1  4 0  
 
4  44  
4  04 1  
 64  0, or 64 units2 1  3x6dx 
 3  7x7 
  1  1 1  1 3 x7 7 1  
 
7  17  
7  (1)7 3  3  
 7  
7 or 7 unit2 3    0  0.1 1 x2 2 0  27.  2  3  1  1 x3 3  
  0  

 
  488  6  1  (x2  2x)dx 
 3x3  2  2x2  0.1  
 (250  (0.1)2)  (250  02) 
 2.5  0 or 2.5 J  Chapter 15  1  x3dx 
 4x4  1m  
 250x2    1    100 cm 
 0.1 m  500xdx 
 500   2x2  
 4  0 or 4 units2  
 48  (6) or 54 10 cm  1  2  1   
4  24  2  22  6  2  14.  1  (4x  x3)dx 
 4  2x2  1  4x4  
 
2  22  4  24  
2  02  4  04  1  1  13  
  
 
4  44  2  42  6  4 1  
3  
3 or 3 units2  14  3   6x  1  0  3  2 3  2   2  3  
3  23  2  2  33  2  
 
3  23  2  22  2  2 14  3    2  63  
 6  6 or 2 1 x3 3  3  4  
 3  6 2  2  
 18  0 or 18 units2  4  1  

   0  2  3  (x2  x  6)dx 
 3x3  2x2  6x 43  0  3  2  
 3x3  or 40  1  3  3  1  2x2dx 
 2  3x3  14  1  7  (x2  6x  3)dx 
 13x3  6  12x2  3x  C 3  3  1   34 2 81 1    2 2  3  1  3 1 x4 4 1  
 2x4  1  
3x6  2x4  3x3  8x  C  1  10.  1  
 4  6x6  6  4x4  7  3x3  8x  C  1  7  2  
 5x5  3x3  3x  C  1  1  
 3x3  2x2  4x  (3x2  x  6)dx 
 3  13x3  1  12x2  6x  C 
 x3  2x2  6x  C  1  
 3x3  4  2x2  4x  19  Exercises  1 x6 6  
 4x8  C  units2  2 1 x5 5 0  

  1  9.  16  3  x5dx    x2  0 2  0 2   
    1 1   03  02    (2)3  (2)2 3 3 20 20     0   3 or 3 units2  
     (x 3  28.  2  1  1  1   2x  3)dx 
 1  3x3  2  2x2  3x 1  
 3x3  x2  3x  3  3    0  37.  1  1  1  1  1  1  1  29.    0  (x3  x)dx 
    16    0  30.  1    1  1  3  1  39.  1  
  1 x4 4    3   10x  4x2    3  2  1  1  1  1  0  
    4  32.  2  2x3  7  3x4dx 
 3   
 3  1  1    1  35.    1    1 x2 2   4x  0  2  1  
   
  
  
    135  12  1  or  1  1 x3 3  2  
 5x5  3x3  x  
  28  15  2   0 or  1 x4 4    x3    3 x2 2  x  3  2  1  3  3 2  2   32  3 3  15    1  42.  0  15  
 4  0 or 4  1  1  1   
4  24  23  2  22  2  1  5  1  x2  x  2 dx x2   
 
  1  0 1  (x  2)(x  1) dx x2  (x  1)dx  1  
 2x2  x  1 0  
 
2  12  1  
2  02  0 1  34  3  1  3  43.  1    0  
 2 x(4x2  1)dx 
    2  0  (4x3  x)dx 1  1    2 1 x2 2 0  1  
 4  4x4  2x2  0  
  0  
 
5  15  3  13  1  
5  05  3  03  0 1  1  
 
  34  33   5  2  x  (x3  3x2  3x  1)dx  1  4  1    3  2  0  1  1  12    
  20  3 1 x3 3 1    1  1  (x4  2x2  1)dx 2  1  
 4x4  3  3x3  3  2x2  x  
 
4  34  3  33  
4  14  3  13  
  1  3  32  (x  1)3dx 
  1   (1)2  4  1 2  
 (53  52  5)  (13  12  1) 
 105  1 or 104 1 x4 4    3  41.  1  1 x5 5  3  
 9  
3 or 3 5  3  1   32  4  3 2 33 7    or 2 2  45    1  1  
 4  
1 2 36.  1  1  
 
3  33  32  3  3  
3  13  12  3  1  1  1  (x2  2x  3)dx  3  2  (3x2  2x  1)dx 
 3  3x3  2  2x2  x     1   
  x2)dx  (x  3)(x  1)dx  
 3x3  x2  3x  

  (x3  413  4  
 x3  x2  x  3  148  
 3x3  2  2x2  3x  3 3   45    25 5 5 3072 96 2976    or  5 5 5  (x  4)dx 
  74  
 6  6 or 6  
  4 1 x5 5 2  
 34.    3  40.  

  5  0  
 (2  73)  (2  03) 
 686  0 or 686  3    265  7  0  
 5x5  33.  265  23  1  3  
 6  
3  
 4  
4 or 92 units2 6x2dx 
 6  3x3  2  3   
3  (2)3  2  (2)2  8  2  1  7  2  5  3  
 
3  53  2  52  8  5   
4  (1)4  4  (1)2  10  1    5  1  1  1  1  (x2  3x  8)dx  
 3x3  2x2  8x  
 
  34  4  32  10  3  31.  1  
 3x3  3  2x2  8x  1  4  345  0  1  
 8  0 or 8  5  (x3  8x  10)dx 
 4x4  8  2x2  10x  2  1  
 
4  24  2  22  2  
4  04  2  02  0  1  
 4  0 or 4 unit2  3  (x3  x  1)dx 1  1  1  9  
 4x4  2x2  x  
 
4  14  2  12  
4  04  2  02 3  1   
 0  
2 0  or 20  2  38.  1 1 x2 2 0  1  1  9  
 9  3 or 3 units2 1 x4 4  1  
 
5  05  4  04  
5  (1)5  4  (1)4   
3  13  12  3  1  1  0  1  
 5x5  4x4  
 
3  33  32  3  3 11  (x4  x3)dx  1  2  x4  2 0  
 
24  2  22  
04  2  02 1  28  15  1  
18  0 or 18  489  Chapter 15    1  44.  1    1  
  
  x3  

  110 100 90 80 70 60 50 40 30 20 10  (3x2  5x  2)dx  1  
3  1 x3 3    5  5 x2 2    13  5  2  1 x2 2   2x    
  (1)3   2x  1 1  1 1  12   2  1    5  2  2   (1)  2  1    20.5  0  1  1  x3dx 
 4x4  0  1  4   i3 
 
    100.5  45b.  0  
  1    i2 
  i
1  
    2  46.  1  
 6(558  0) 
 $93 1  12  6  49c.  2 1 x2 2 0  

 
  2 4 6  1  
 $105    R  50.  1  R  (R2  x2)dx 
 R2x    3x3 
 
R2  R  3  R3 1   
R2  R  3  (R)3 1  
 3R3  
3R3 2  2  4  
 3R3 1000 
 k(6.4  106)2  51a.  1   1000 
 k   40.96  1012   2  
  1  3  value is  03  4.1  1016   5  0  1 (3)(9) 2  
    3.8  108  51b.  6.4  106  k; 4.1  1016Nm2  4.1  1016x2dx  
 4.1  1016  (1)x1 4.1   4.1  1016   3.8  108  

  1.1   27 . 2  490  6.4  106  6.4  106  4.1  10    
 6.4  10   108  
 6.3  109 J  1 2 3 4 5x y 
 3x  6  3.8  108  3.8  108  1016  
 x  O  Chapter 15  1  
 6(1188  558)  0  47c. Since the function is negative, the integral in part b gives the opposite of the area of the 22 region. The area is 3. y 48. The integral represents the 10 area of a right triangle. The 8 6 4 2  6  6  1  (x2  5)dx 
  12  1  12   
75  6  4  62  6  63  2  negative. Therefore, the sum is negative. since b a f(x)dx is a limit of negative sums, it is also negative. 2 1 x3  5x 3 0 1   23  5 3 22 3  1  1  i
1  0  1  
 6
75  12  4  122  6  122  positive, so each term in the sum  f(xi)x is  2  1  1  n    6  
75  8x  12x2dx  
 6
75x  4x 2  6x3  
 (245,000  22)  (245,000  02) 
 980,000  0 or 980,000 J 47a. Since the graph is below the x-axis, f(x) is negative. Each f(xi) is negative and x is  47b.  12  1  or 338,350  
 245,000 x2    
 6
75x  8  2x2  2  3x3  100(100  1)(2  100  1)  6  490,000 xdx 
 490,000   0  0  1  1    
13  03  100(101)(201)  6  0  6   
75  0  4  02  6  03  338,358.38  0 or 338,358.38 100  1  6  
 6
75  6  4  62  6  63  100.5 1 x3 3 0 1   100.53 3  

  0  1  202(20  1)2  4 400(441)  or 44,100 40  x2dx    
 6
75x  4x2  6x3  44,152.52  0 or 44,152.52 i
1  x 6  
75  8x  12x2dx 1 1 1 1 
 6
75x  8  2x2  2  3x3  
 
  20.54  
  04 20  f (x ) 
 75  8x  12 x 2  1  60  49b.  20.5  1  4  f (x )  O  
 2  
2 or 6 11  45a.  49a.  (x  1)(3x  2)dx  16 6   64.1   108  R R    2  52.  0  1 x2dx 2  n  1 2i 2 2   
 
n n→  i
1 2 n  
 lim  Pages 978–980  4  2 n(n  1)(2n  1)   3 6 n→  2n 3 4 2n  3n2  n 
 lim 6 n 3 n→  2 3 1 
 lim 3 2  n  n2 n→  4 
 3 2x6  3x2  2  
  
 lim  
 
       x→2  12. lim (x3  x2  5x  6) 
 (2)3  (2)2  5(2)  6 x→2 
 8  4  10  6 
4 13. lim (2x  cos x) 
 2(0)  cos 0 x→0 
01 
 1  53. f(x) 
 f
(x) 
 2  6x61  3  2x21  0 
 12x5  6x 54. In a normal distribution, 68.3% of the data lie within 1 standard deviation of the mean. So, 100  68.3 or 31.7% of the data lie outside 1 standard deviation of the mean. Thus, 31.7% of the test-takers scored more than 100 points above or below the mean of 500. 55. To begin with 25 out of the 50 numbers are odd. With each consecutive draw, there is 1 fewer odd numbers and 1 fewer tickets. P(four odd numbers) 
 
  25 24    50 49 253  4606    23  48    14. lim  x→1  15. lim  x→0  5x2  2x  45 ft/s  Vy  5  
  x→0 5 (0) 2  
0  16.  22  47  x2  2x  lim  2  3x  10 x x→4  x→4  x  x→4 x  5 4  
 4  5 or 4  17. lim (x  sin x) 
 0  sin 0 x→0 
0 18. lim  x→0  x2  x cos x  2x  
 lim  x→0  x(x  cos x)  2x  x  cos x  2 x→0 0  cos 0 
 2 1 
 2 x3  2x2  4x  8 (x  2)(x2  4)  lim  
 lim  x2  4 x2  4 x→2 x→2  
 lim  1  2   i
   3  2   
 1  3i  
 lim x  2 x→2  
 2  2 or 4  sin 52° 
 4 5 45 sin 52° 
 vy 35.46 vy; 35.46 ft/s  60. mAE + 
 m∠AXE, so ∠AXE is the smallest angle. If the circle was divided into 5 equal parts, each angle would measure 72°. Since the circle is not divided evenly, the smallest angle AXE is less than 72°. The correct choice is A.  21.  22.  Chapter 15 Study Guide and Assessment Understanding the Vocabulary 2. 4. 6. 8. 10.  x2  6x  9  9  2x x→0 x→0 x(x  6) 
 lim 2 x x→0 x6 
 lim 2 x→0 06 
 2 or 3 x2  9x  20 (x  5)(x  4)  
 lim  lim  x2  5x x(x  5) x→5 x→5 x4 
 lim x x→5 54 1 
 5 or 5 f(x  h)  f(x) f 
(x) 
 lim h h→0 2(x  h)  1  (2x  1) 
 lim  h h→0 2x  2h  1  2x  1 
 lim  h h→0 2h 
 lim h h→0  20. lim  Vx  false; sometimes false; indefinite false; secant false; derivative false; rate of change  x(x  2)   
 lim4  (x  5)(x  2)  
 lim  52˚  Page 977  (x  6)(x  6)  x1  
 lim 2x  vy  59.  1. 3. 5. 7. 9.  x→1  x→1  19.  ; 2
  
 lim  
 1  6 or 5  58. r 
  or 2  
  x2  36  x6  
 lim x  6  56. A 
 Pert 
 600e0.06(15) 
 $1475.76 57. h 
 6; k 
 1 hp
3 6  p 
 3 or p 
 3 (y  k)2 
 4p(x  h) (y  1)2 
 12(x  6) 22   2 2   v 
 3  3 or 3   2 cos 3  i sin 3  Skills and Concepts  11. There is a point at (2, 1) so f(2) 
 1. However, the closer x is to 2, the closer y is to 3. So, lim f(x) 
 3.  true true false; tangent true false; derivative  (x  3)2  9  2x  
 lim  
2  491  Chapter 15  23. f 
(x) 
 lim  h→0  
 lim  h→0  
 
 
  f(x  h)  f(x)  h  1  35. f(x) 
 2x3  2x2  3x  2 1  4(x  h)2  3(x  h)  5  (4x2  3x  5)  h  1  1   6x  C 1 5 
 5x5  4x4  x2  6x  C  
 8x  3  
 lim  x3  3x2h  3xh2  h3  3x  3h  x3  3x  h  
 lim  3x2h  3xh2  h3  3h  h  h→0  h→0  
lim  h→0  37. f(x) 
 (x  4)(x  2) 
 x2  2x  8 1 1  21  2  x11  8x  C F(x) 
  2  1x 11  f(x  h)  f(x)  h (x  h)3  3(x  h)  (x3  3x)  h  h→0  1  
 3x3  x2  8x  C x2  x  38. f(x) 
 x 
x1 1  11  x  C F(x) 
  1  1x   3xh   3)  h h(3x2  h2  1  
 lim 3x2  3xh  h2  3  h→0 
 3x2     2  39.  3  0  25. f(x) 
 2x6 f 
(x) 
 2  6x61 
 12x5 26. f(x) 
 3x  7 f 
(x) 
 3  1x11  0 
 3 27. f(x) 
 3x2  5x f
(x) 
 3  2x21  5  1x11 
 6x  5    1  40.  0  1  f
(x) 
 4  2x21  1  1x11  0 1  
 2x  1 1  29. f(x) 
 2x4  2x3  3x  4 f
(x) 
    4  1  41.   4x41  2  3x31  3  1x11  0  3  1  2i  n(n  1) 8  2 2 n→ n n2  n 
 lim 4 n 2 n→ 1 
 lim 4 1  n n→ 
 4 units2  
 
 
  2       n  i 3 1   
 
n n→ i 
 1 n  x3dx 
 lim  n2(n  1)2 1  4 4 n→ n 1 n2  2n  1 
 lim 4 n 2 n→ 1 2 1 
 lim 4 1  n  n2 n→ 1 
 4 unit2  
 
 
  x2dx 
    4  0       3  x2dx  n  
 2x3  6x2  3    x2dx  0 n  4i 2 4  3i 2 3  lim  
n 
n  
 
n  n→ n→ i 
 1 n i
1  
 lim  30. f(x) 
 (x  3)(x  4) 
 x2  7x  12 f
(x) 
 2x21  7  1x11  0 
 2x  7 31. f(x) 
 5x3(x4  3x2) 
 5x7  15x5 f 
(x) 
 5  7x71  15  5x51 
 35x6  75x4 32. f(x) 
 (x  2)3 
 x3  6x2  12x  8 f
(x) 
 3x31  6  2x21  12  1x11  0 
 3x2  12x  12 33. f(x) 
 8x  64 n(n  1)(2n  1)   3 6 n→ n 27 n(n  1)(2n  1)  lim n3 6 n→ 64 2n3  3n2  n 
 lim 6 n 3 n→  
  
 lim    
  
   lim  
 
  1  
 4x2  C 34. f(x) 
 3x2  2 1   21  2x  C F(x) 
 3   2  1x  
 x3  2x  C  492  64  3 37  3        2n3  3n2  n  n3  
  27  6  n→ 64 
 lim 6(2 n→   11  C F(x) 
 8   1  1x  Chapter 15  n   2 
n
n n→ i 
 1  2x dx 
 lim  
 lim  1  1  
 2x2  x  C  
 lim  28. f(x) 
 4x2  x  4  1  2  3  36. f(x) 
 x4  5x3  2x  6 1 1 1  41  5  x31  2  x11 F(x) 
  41 x 31 11  h→0  
 lim  2  
 8x4  3x3  2x2  2x  C  
 lim 8x  4h  3  h→0  1   11  2x  C 3 1  1x  4x2  8xh  4h2  3x  3h  5  4x2  3x  5 lim  h h→0 8xh  4h2  3h lim h h→0 h(8x  4h  3) lim h h→0  24. f
(x) 
 lim  1   31  2  x21 F(x) 
 2   3  1x 21    27  3  units2    3  n    1  n2 )  27  n→ 6   lim  
2  n3  n1 2    2  42.  1  6x2dx 
 n  
 lim    2    1  6x2dx   0  n  2i 2 2   n n  6
    Page 981  6.x2dx   
   lim  6  
  
   
  
    
 lim 8(2  n→  3  n  1  n2 )      4  2  n→  2  3x2dx 
 3  3x3 
 x3  1h 60  mi 5280 ft r      
 3600 s 1 hr 1 mi 88 ft/s 2 a 
 5  s 
 17.6 ft/s  53a.  53b. v(t) 
  2  1  3    2  
 (3   (3  
 48  12 or 36      1  n2  4  42)  44.  3  n  2  22)  45.    2  2  3  53c. d(t) 
  
  1 x3 3  1  1 x2 2  0  0  t  0   3x  2 2 1 2 3   x  2 x  3x 2 1 (23  2  22  3  2)  1 ( 2)3  2  (2)2  3  (2)  (x  2)(2x  3)dx 
  Page 981  4   7x  6)dx  2  0  
2  1 x3 3  2  1   7  2x2  6x 7  
 3x3  2x2  6x  4    4  1  7  0  Open-Ended Assessment  0  0  16x3dx 
 4x4  1 0  
 4  14  4  04 
4  0  
 
3  43  2  42  6  4  2  t  1. Sample answer: f(x) 
 x2  2x  2; lim(x2  2x  2) 
 12  2(1)  2 x→1 
5 2. Sample answer: g(x) 
 16x3;     (2x  t 0  
 8.8t2  8.8(0)2 
 8.8t2  2  
 12  (16) or 28  4  t   17.6xdx  
 8.8x2  
    0  1  
  46.  t   11 
 17.6   1  1x  3   x  3)dx 
 3  88 ft/s  
 17.6t  17.6(0) 
 17.6t  
 8  (27) or 35 (3x2  2   17.6dx  
 17.6x  
 23  (3)2 2  2  2  0.0000125 m c(x) 
 9x5  135x3  10,000 c
(x) 
 9  5x51  135  3x31  0 
 45x4  405x2 c
(2.6) 
 45(2.6)4  405(2.6)2 or $681.41  4  1  6xdx 
 6  2x2 
 3x2    52.     lim 1 
2   
 16  2 or 14 units2 43.  
  2  51. lim  i 2 1   n n  n→i 
 1 n→ i 
 1 6 48 n(n  1)(2n  1) n(n  1)(2n  1) 
 lim n3 6  lim n3 6 n→ n→ 48 2n3  3n2  n 6 2n3  3n2  n 
 lim 6 n  lim 6 n 3 3 n→ n→  
  Applications and Problem Solving  50 50 1    1m  
 1  t   
 2 m  
 1  100  t→100 2  0  
23  03  72  02  6  0  368  368  Chapter 15 SAT & ACT Preparation  
 3  0 or 3 47.    6x4dx  
6 
  1 x5 5  6 x5 5  C  Page 983  C  48.  (3x2  2x)dx 
 3   49.    1 x3 3  2  1 x2 2  C  1.  
 x3  x2  C (x2  1  5  
 3x3  2x2  2x  C 50.  4  SAT and ACT Practice 1 
  1 1    2 3  1 
  1  (3x5  4x4  7x)dx 
 3  16x6  4  15x5  7  12x2  C 1    1  3  1 
  3 2    6 6  1   5x  2)dx 
 3x3  5  2x2  2x  C 1  1  2   6  7  
 2x6  5x5  2x2  C  
6 The correct choice is A.  493  Chapter 15  2. Let x represent the length of the second side of the triangle. Then the first side has length 2x.  x  2x  Since 16 is composite, 5  ? Clearly the perimeter must be greater than 3x, so eliminate answer choices A and B. Use the Triangle Inequality Theorem. x  2x  ? P x  2x  3x 3x  ? P 6x ?  x  2x P  x  2x  x ?x P  4x Since the perimeter cannot equal 6x or 4x, eliminate answer choices C and E as well. The only possible answer choice is D. 3. Draw a figure. Begin with the 3 parallel lines. Draw the 3 nonparallel lines in positions that are as general as possible. For example, do not draw perpendicular lines or concurrent lines. Draw the first nonparallel line, and mark the intersections. Then draw the second line, making sure it intersects each of the other lines. Then draw the third line, making sure it intersects each of the others.  5.  23  
 69  46  
 23  69  
 34.5  31  
 63  The correct choice is B. 7. The nth term of an arithmetic sequence with first term a1 and common difference d is given by an 
 a1  (n – 1)d. a1 
 4, and d 
 3, so a37 
 4  (37 – 1)(3) 
 4  (36)(3) or 112 The correct choice is C. 8. Draw a figure.  Of the 16 cubes on the front, only the 4 in the center have just one blue face. It is the same on each of the faces. There are 6 faces, so there are 6  4 or 24 cubes with just one blue face. The correct choice is A. 9. To find the value of {{x}}, first find the value of {x}. {x} 
 x2  1. So {{x}} 
 {x2  1}. And {x2  1} 
 (x2  1)2  1 
 (x4  2x2  1)  1 
 x4  2x2 The correct choice is D. 10. There are two distinct prime factors of 20. They are 2 and 5 and their product is 10. There is only one distinct prime factor of 16. It is 2.  2   triangles is 2  9 or 18. The correct choice is B.  16    x2 x  6 x6 0 x2  x  6 0 (x  3)(x  2) 0 At x 
 3 and x 
 2, x y the inequality equals 3 6 0. Test values that are 2 0 greater than and less 0 6 than 3 and 2 to 3 0 determine for which values the inequality 4 6 is less than 0. For values of x that are between 2 and 3, the inequality is true. The correct choice is C.  The answer is 5.  x2  Chapter 15  1  
 2(16) or 8.  
 15  8 or 23.  
 10.5  20  6  2  16  16  21  triangle is 2

 or 9. The area of both 6 2     
 3(5) or 15.  Now, you need to determine which of the choices is equal to 23. Calculate each one.  There are 12 intersections. The correct choice is D. 4. The triangles are right triangles. The vertical angles formed by the two triangles each measure 45°, since 180  135 
 45. The hypotenuse of the triangles is the radius of the circle, which is 6, since the diameter is 12. Using the relationships of 454590 right triangles, the length of each 6 height and base must be . The area of one 1  5  6. Since 5 is prime,  494  Extra Practice 4. [f  g](x) 
 f(g(x)) 
 f(2x3  x2  x  1) 
 2(2x3  x2  x  1) 
 4x3  2x2  2x  2 [g  f](x) 
 g(f(x)) 
 g(2x) 
 2(2x)3  (2x)2  (2x)  1 
 16x3  4x2  2x  1  Lesson 1-1 Page A26 D 
 {2, 1, 2}; R 
 {4, 2, 4}; no D 
 {3, 0.5, 0.5, 3}; R 
 {0.5, 3}; yes D 
 {1, 0, 2, 5, 7}; R 
 {1, 2, 3, 5, 7}; yes D 
 {2, 2.3, 3.2}; R 
 {4, 1, 3, 4}; no f(4) 
 4(4)  2 
16  2 or 14 6. g(3) 
 2(3)2  (3)  5 
 2(9)  3  5 
 18  3  5 or 26 1. 2. 3. 4. 5.  Lesson 1-3 Page A26  3   7. h(1.5) 
  2(1.5)  1. x  2 
 0 x
2  3  
 3 or 1 8. k(5m) 
 3(5m)2  3 
 3(25m2)  3 
 75m2  3  f (x )  x  O  Lesson 1-2  f (x ) 
 x  2  Page A26  2. 3x  4 
 0 3x 
 4  1. f(x)  g(x) 
 2x  1  x2  3x  1 
 x2  5x  2 f(x)  g(x) 
 2x  1  x2 3x  1 
x2  x f(x)  g(x) 
 (2x  1)(x2  3x  1) 
 2x3  5x2  5x  1  4  x 
 3  f (x )  
g(x) 
 g(x) f  f(x)  2x  1   
 x2  3x  1  2. [f  g](x) 
 f(g(x)) 
 f(4x2) 
 3  4x2 [g  f ](x) 
 f(f(x)) 
 g(3  x) 
 4(3  x)2 
 36  24x  4x2 3. [f  g](x) 
 f(g(x)) 
f(x  9)  O  f (x ) 
 3x  4  x  3. 1 
 0, false none  4. 4x 
 0 x
0  f (x )  f (x )  f (x ) 
 4x  1  
 3(x  9)  1  O  1  
 3x  2  x  O  x  f (x ) 
 1  [g  f ](x) 
 g (f(x))  
 g 
3x  1 1  1  
 3x  1  9 1  
 3x  8  495  Extra Practice  5. 2x  1 
 0 2x 
 1  1  6. x  5 
 0 x 
 5 x 
 5  1  x 
 2    5. m 
 
 3  or  3 1  y  6 
 3(x  (2)) 3y  18 
 x  2 x  3y  16 
 0 6. x 
 10 is a horizontal line; perpendicular slope is undefined. y 
 15 or y  15 
 0  f (x )  f (x ) f (x ) 
 2x  1  O x  O  1  x  f (x ) 
 x  5  2    7. m 
 
 5  or  5 2  2  y  (7) 
 5(x  3) 5y  35 
 2x  6 2x  5y  29 
 0  Lesson 1-4 Lesson 1-6 Page A26 1. y 
 mx  b → y 
 2x  1 2. y  2 
 1(x  1) y  2 
 x  1 y 
 x  3  Page A27 1.  y 48  1  3. y 
 mx  b → y 
 4x  3  47  4. y  (4) 
 0(x  (2)) y4
0 y 
 4  46 45  31 2  Enrollment (millions)   5. m 
  2  2 1    
 4 or  2 1  y  1 
 2(x  2)  42  1  41  y 
 2x  2 60  40   6. m 
  0  (1)  
  0  or 6  y 
 mx  b → y 
 6x  6 7. y 
 0 8. y  0 
 1.5(x  10) y 
 1.5x  15  x 0 1990 19911992 199319941995199619971998 1999 2000 2001 Year  2. Sample answer: y 
 0.6091x  1171.6 47.2 – 40.5   m
 2001 – 1990  0.6091 y  40.5 
 0.6091(x  1990) y 
 0.6091x  1171.6 3. Sample answer: y 
 0.6125x  1178; r 0.99 Enter the School Year data as List 1. Enter the Enrollment data as List 2. Perform a linear regression on the graphing calculator. 4. Sample answer: 53.7 million; yes; the correlation coefficient shows a strong correlation. f(2011) 
 0.6125(2011)  1178 53.7 million  Lesson 1-5 Page A27 1. None of these; the slopes are neither the same nor opposite reciprocals. 2. y  (2) 
 1(x  0) y2
x xy2
0 3. y  3 
 2(x  1) y  3 
 2x  2 2x  y  1 
 0 4. y 
 1 is a vertical line; parallel slope is undefined. y 
 12 or y  12 
 0  Extra Practice  43  1  y  1 
 2x  1  6  1  44  496  Lesson 1-7  y  5.  y  6.  1  x  y  4  Page A27 f (x )  1.  x  g (x )  2.  O  g (x ) 
 |x  2|  x  O  O h (x )  3.  x  y  |x | O  4.  Lesson 2-1  x  f (x )  Page A28 y  1.  2x  y  1 
 0  x  O  x  O O  5.  6.  g (x )  x  3y 
 6x  3  k (x )  y 
 2x  1; consistent and dependent  y  2.  g (x ) 
 |5  2x |  O  x  (1, 5)  xy
4  y  3x 
 8  x  O  x O (1, 5); consistent and independent  Lesson 1-8  3.  3x  y 
 1  Page A27 y  1.  2y  6x 
 4  4.  y  y  x1  x no solution; inconsistent 4. 5x  2y 
 1 x  2y 
 5 x  2y 
 5 1  2y 
 5 4x 
 4 2y 
 6 x 
 1 y
3 (1, 3) 5. 2x  4y 
 8 2x  3y 
 8 2x  3y 
 8 2x  3(0) 
 8 y
0 2x 
 8 x
4 (4, 0)  xy3  x  O  x  2  O O  O  y  2. y  3.  y  y  x O  x  4x  2y  6  497  Extra Practice  6. 8x  2y 
 2 3x  4y 
 23  →  8x  2y 
 2 8(1)  2y 
 2 2y 
 10 y
5 (1, 5)  16x  4y 
 4 3x  4y 
 23 19x 
 19 x 
 1  Lesson 2-3 Page A28 1. 2x  y 
 1 y 
 1  2x  y 
 1  2(1) 
1 (1, 1) 2. x  2y 
 5 2x  2y 
 2 3x 
 3 x
1 (1, 2) 3. 3x 
 y  7 3x  7 
 y  Lesson 2-2 Page A28 1. x  y 
 6 x  z 
 2 yz
 8 xy
6 x5
6 x
1 (1, 5, 3) 2. 2x  2y  z 
 6 x  y  2z 
 6  yz
 8 yz
 2 2y 
 10 y
5 x  z 
 2 1  z 
 2 z 
 3 →  x  2y  z 
 7 2x  2y  z 
 6 x  4y 
 13 3x  5y 
 18 → x  4y 
 13 x  4y 
 13 x  4(3) 
 13 x
1 (1, 3, 2) 3. 2x  3y  z 
 1 x  y  z 
 4 3x  2y 
 3 x y z 
 4 3x  2y  2z 
 3 3x  2y 
 3 3(1)  2y 
 3 2y 
 0 y
0 (1, 0, 3)  x  2y 
 1 x  2(1  2x) 
 1 3x  2 
 1 3x 
 3 x 
 1  x  2y 
 5 1  2y 
 5 2y 
 4 y
2 4y 
 5x 4(3x  7) 
 5x 12x  28 
 5x 7x 
 28 x 
 4  y 
 3(4)  7 
 5 (4, 5) 4  7 1  (5) 4. A  C 
 1  0 5 1 28 6 4  4x  4y  2z 
 12 x  y  2z 
 6 3x  5y 
 18       11 6 1 6 10 10 5. D  E 
 4  1 1  (5) 2  (3) 3  2 5 6 
 5 1 4(2) 4(0) 4(3) 6. 4B 
 4(4) 4(3) 4(2) 0 12 
 8 16 12 8 
  3x  5y 
 18 3x  12y 
 39 7y 
 21 y 
 3 x  y  2z 
 6 1  (3)  2z 
 6 2z 
 4 z
2                    7. impossible         2(7) 8. 2C  3A 
 2(0) 2(8)  → 2x  2y  2z 
 8 3x  2y  2z 
 3 5x 
 5 x
1 x  y  z 
 4 1  0  z 
 4 z 
 3 z
3    2(5) 3(4) 3(1) 2(1)  3(1) 3(5) 2(4) 3(2) 3(6)  14  12 10  (3) 03 2  15 16  6 8  18 26 13 
 3 17 22 26 9. impossible 
      5 4 1  2 2 3 1(4)  (5)(2) 1(1)  (5)(3) 
 3(4)  2(2) 3(1)  2(3)  10. ED 
  31          
 14 14 16 3    Extra Practice  498         7 0 3  0 3 2 8 2(7)  0(0)  (3)(8) 
 4(7)  (3)(0)  2(8) 10 22 
 44 15 BC  D 
 10  (4) 44  2 6 23 
 42 18  11. BC 
  24    5 1 4 2(5)  0(1)  (3)(4) 4(5)  (3)(1)  2(4)      0 4. 1  3 6 5 2 0 0 1 3 2 1 3 6 5 2 
 3 2 1 0 W
(3, 3), X
(6, 2), Y
(5, 1), Z
(2, 0)                     y 8  W (3, 3) Z (2, 0) 4 X (6, 2) Y (5, 1) Z (2, 0) 8 4 O 4 8x Y (5, 1) X (6, 2) 4  22  1 15  3      W (3, 3)  8  Lesson 2-4  5. Rot90  Ry 
 x 
 0 1  0 1 1 0 0 
 1 1 0 0 1      Page A28 5 2 2 
 4 6  3  1. 0.5  
  1    1    1 22 1 12    1 1 2    A
 22, 12 , B
(1, 2), C
(1, 3)      10 01  42  3     2 6 
 3 7    42  6 2 3 7    F
(4, 2), G
(2, 3), H
(6, 7)  y  y  B (2, 4)  8  B (1, 2)  F (4, 2) 4  x  O C (1, 3)  2 12,  A (  F (4, 2)  O 8 4 G (2, 3)4  112  )  A (5, 3)  H (6, 7) 8  4  8x  G (2, 3) H (6, 7)  C (2, 6)  2.    Lesson 2-5  6 3 4 1 2 2 2 2  2 5 7 5 3 3 3 3       Page A29  
 4 5 6 1 5 8 4 2 J
(4, 5), K
(5, 8), L
(6, 4), M
(1, 2)      L (4, 7) y 8 L(6, 4) 4  3 7 
 3(2)  (11)(7) 11 2 
 83 3 5 2. 
 3(2)  (7)(5) 7 2 
 29 5 0 1 
 5(6)  2 (0) 3. 1 2 6 1.  M (1, 5) M (1, 2)  8 4 O 4 K (3, 5) 8 K (5, 8)  4  
   8x  J (6, 2) J (4, 5)  
 30 0 2 1 4. 3 1 2 5 1 3 1 2 1 
 1 0 3 2 2 3 1 3 5 3 5 1 
 1(5)  0(19)  2(2) 
1 3 2 1 5. 1 4 2 3 3 4 134 1  2 4 2 
 1 2 3 4 3 3 3 4  0 5 3 0 3. 1 0  5 3 
 2 0 1 2 8 4 8 4            N
(5, 2), P
(3, 8), Q
(0, 4) P (3, 8) N (5, 2)  y 8 4  P (3, 8) N (5, 2)  O  4 8x 4 Q (0, 4) Q (0, 4)  8 4  
 1(11)  3(19)  2(6) 
 34  499  Extra Practice  6.  7.  4 0 1 5 3 6 2 5 2 3 6 5 6 5 3 
4 0  (1) 5 2 2 2 2 5 
 4(36)  0(22) 1(19) 
 163    1 5      9.  54  1 10 5  0 4      2  5      (0, 1)  1  2.  3  2  5  2  y (4, 7)    3  (0, 3)  61 53 
 61 53 1  33  5 1  
  31    1  33 2  11  5  33 1  11  (0, 2)    2 2    1 8 2 2 1 3    31      1 2 x  
 8 2 2 2 y 1 3 x 4 
 y 5       22  6   (1, 1)     O  4 2 x  
 6 3 y 1 7 1 1 2 1 2 
 1 0 3 4 3 4    12.  1  10             2 1  (2, 1)  24  13. 1 2 4  1 3  1 3 1 1 0 4 2     24    1  Extra Practice  vertices: (1, 1), (1, 6), (1, 2), (1, 1) f(x, y) 
 x  y f(1, 1) 
 (1)  1 or 0 → minimum f(1, 6) 
 (1)  (6) or 7 → maximum f(1, 2) 
 1  (2) or 1 f(1, 1) 
 1  (1) or 0 → minimum  1  10  xy 
 2 1  x 4 1  
 y 13 3 3 1 
 1 3 1 10 4 4 2 2              xy 
  10 3 4 1  1 3  xy 
 
  x  (1, 6)   xy 
 31 24 6  7  31 24 43  5 , 2  (1, 1) (1, 2)  (4, 5)  1 4 2 3 1  x  vertices: (0, 3), (4, 7), (4, 2), (0, 2) f(x, y) 
 2x  y f(0, 3) 
 2(0)  3 or 3 → minimum f(4, 7) 
 2(4)  7 or 1 f(4, 2) 
 2(4)  2 or 6 → maximum f(0, 2) 
 2(0)  2 or 2 y 3.         1  (4, 2)  O  2 x  
 22 2 y 6 2 2 
 1 2 2 8 1 1 3 3  11.  x  vertices: (3, 1), (0, 1), (3, 7) f(x, y) 
 4x  3y f(3, 1) 
 4(3)  3(1) or 15 f(0, 1) 
 4(0)  3(1) or 3 → minimum f(3, 7) 
 4(3)  3(7) or 33 → maximum    2  (3, 1)  O    43 65 
 2 43 65  1  3 1  1  5    1 5 6 3 4  3 6  y (3, 7)  1 0 4 0 
 4 0 5 10 10 1 0  
 101 1  8 4  
 10.  1.    3  5 1 5  
  8.  Page A29  3 1 
 1 3 1 5 1 1 2 2  1 2 1  Lesson 2-6    1 2   134    5  2  1  500  4. Let x 
 the number of video store hours. Let y 
 the number of landscaping company hours. y y0 16 x
4 x4 x  y  15 (4, 11) 12  Lesson 2-7 Page A29 1. Let x 
 the number of cars. Let y 
 the number of buses. y 6x  30y  600 x  y  60 60 x  y 
 60 x0 6x  30y 
 600 40 y0 20 x
0 (0, 0)  O  (0, 20)  8 4  O  (50, 10)  1. f(x) 
 4x f(x) 
 4(x) f(x) 
 4x yes 2. f(x) 
 x2  3 f(x) 
 (x)2  3 f(x) 
 x2  3 no  x  y 
 45 20  30  40x  C(x, y) 
 2.95x  2.95y C(10, 20) 
 2.95(10)  2.95(20) or 88.50 C(15, 30) 
 2.95(15)  2.95(30) or 132.75 C(25, 20) 
 2.95(25)  2.95(20) or 132.75 alternate optimal solutions 3. Let a 
 the number of company A's cards. Let b 
 the number of company B's cards. a  b  90 b 80 a 
 40 a  40 b  25 60  40  60  f(x) 
 (x2  3) f(x) 
 x2  3   f(x) 
 
 3x3   1  1   f(x) 
  3(x)3 1  1   f(x) 
  3x3  yes 4. xy 
 2 x-axis  (40, 50) a  b 
 90 (65, 25) (40, 25) 20 b 
 25 20  f(x) 
 (4x) f(x) 
 4x  1   3. f(x) 
  3x3   f(x) 
  3x3  →  y-axis  40  O  16x  Page A30  (10, 20)  10  8 12 y
0  Lesson 3-1  (25, 20)  10  4  C(x, y) 
 5x  7y C(4, 0) 
 5(4)  7(0) or 20 C(4, 11) 
 5(4)  7(11) or 97 C(15, 0) 
 5(15)  7(0) or 75 The maximum earnings is $97.  C(x, y) 
 3x  8y C(0, 0) 
 3(0)  8(0) or 0 C(0, 20) 
 3(0)  8(20) or 160 C(50, 10) 
 3(50)  8(10) or 230 C(60, 0) 
 3(60)  8(0) or 180 The maximum income is with 50 cars and 5 buses. 2. Let x 
 number of gallons of black walnut. Let y 
 number of gallons of chocolate mint. y  2x y 40 y 
 2x y  20 thousand x  y  45 thousand (15, 30) 30 20  (15, 0)  (4, 0)  40 60 x (60, 0)  20 y
0  x  y 
 15  y
x  80a  y 
 x  C(a, b) 
 0.30a  0.32b C(40, 25) 
 0.30(40)  0.32(25) or 20 C(40, 50) 
 0.30(40)  0.32(50) or 28 C(65, 25) 
 0.30(65)  0.32(25) or 27.5 The maximum profit is with 40 cards from company A and 50 cards from company B.  y 
 x, y 
 x 5. y  x2 
 3 → x-axis y-axis y
x y 
 x  ab 
 2 a(b) 
 2 ab 
 2 ab 
 2; no (a)b 
 2 ab 
 2 ab 
 2; no (b)(a) 
 2 ab 
 2; yes (b)(a) 
 2 ab 
 2; yes b  a2 
 3 (b)  a2 
 3 b  a2 
 3; no b  (a)2 
 3 b  a2 
 3; yes (a)  (b)2 
 3 a  b2 
 3; no (a)  (b)2 
 3 a  b2 
 3; no  y-axis  501  Extra Practice  2x2  6. y2 
 7  1  2a2  →  2a2  (b)2 
 7  1  x-axis  b2 
 b2 
  y-axis  b2 
 y
x  (a)2 
 a2 
  y 
 x  (a)2 
 a2 
  x-axis, y-axis 7. x 
 4y x-axis  →  y-axis y
x y 
 x y-axis 8. y 
 3x x-axis  →  y-axis y
x y 
 x  none of these 9. y 
  x2  1 x-axis  y-axis y
x y 
 x  4. g(x) is a translation of f(x) right 1 unit and compressed vertically by a factor of 2.  b2 
 7  1  →  5. y 
 2x  5 1  2a2   1; yes 7 2(a)2   1 7 2a2   1; yes 7 2(b)2   1 7 2b2   1; no 7 2(b)2   1 7 2b2   1; no 7  Lesson 3-3 Page A30 y  2. y  a 
 4b a 
 4(b) a 
 4b; no (a) 
 4b a 
 4b; yes (b) 
 4(a) b 
 4a; no (b) 
 4(a) b 
 4a; no  |x  4|  y  x2  1  O 3.  x  y  O  x  y  4.  x  O  y  2|x  1|  b 
 3a (b) 
 3a b 
 3a; no b 
 3(a) b 
 3a; no (a) 
 3(b) a 
 3b; no (a) 
 3(b) a 
 3b a 
 3b; no  y  
 x2  5.  y  x  O y  6. y  (x  2)2  1  y  |x |  2  O  b 
  a2  1 (b) 
  a2  1 b 
  a2  1 b 
  a2  1; yes b 
  (a)2 1  7. Case 1 x  2  3 (x  2)  3 x  2  3 x  1 {x1  x  5} 8. Case 1 4x  2  18 (4x  2)  18 4x  2  18 4x  16 x  4 {xx  4 or x  5} 9. Case 1 5  2x 9 (5  2x) 9 5  2x  9 2x  14 x 7 {x2 x 7}   a2  1; yes 2  (b)  1 2 a 
 (b)   1; no 2   1 (a) 
 (b) 2 a 
 b  1 2  1; no a 
 b  b
 (a) 
  x-axis, y-axis  Lesson 3-2 Page A30 1. g(x) is a translation of f(x) up 2 units. 2. g(x) is the graph of f(x) expanded vertically by a factor of 3. 3. g(x) is a translation of f(x) left 4 units and down 3 units. Extra Practice  y  1.  502  O  x Case 2 x  2  3 x23 x5  Case 2 4x  2  18 4x  2  18 4x  20 x5  Case 2 5  2x 9 5  2x 9 2x 4 x  2  x  10. Case 1 x  1  3  1 (x  1)  3  1 (x  1)  4 x  1 4 x 5 {xx  5 or x  3} 11. Case 1 2x  3 27 (2x  3) 27 2x  3  27 2x  30 x  15 {x15 x 12} 12. Case 1 3x  4  3x  0 (3x  4)  3x  0 6x  4  0 6x  4 2 x  3 all real numbers  f(x) 
 x2  6 y 
 x2  6 x 
 y2  6 x  6 
 y2 y 
 x  6 f 1(x) 
 x  6; No, it is not a function. 7. f(x) 
 (x  2)2 y 
 (x  2)2 x 
 (y  2)2 x 
y2 y 
 2 x f 1(x) 
 2 x; No, it is not a function. x 8. f(x) 
 2  Case 2 x  1  3  1 (x  1)  3  1 x3  Case 2 2x  3 2x  3 2x x  6.  27 27 24 12  x  y 
 2  Case 2 3x  4  3x  0 (3x  4)  3x  0 4  0; true  y  x 
 2 2x 
 y y 
 2x f 1(x) 
 2x; Yes, it is a function. 9.  1   f(x) 
  x4 1   y
 x4 1   x
 y4  Lesson 3-4  1  y  4 
 x 1  y 
 x  4  Page A30 1.  1  f 1(x) 
 x  4; Yes, it is a function.  f (x )  2. f 1(x )  f 1(x )  x O  3.  10. f (x )  f (x )  O  f(x) 
 x2  8x  2 y 
 x2  8x  2 x 
 y2  8y  2 x  2 
 y2  8y x  2  16 
 (y  4)2 x   18 
 y  4 4 x  18 
 y f 1(x) 
 4 x;  18 No, it is not a function. 11. f(x) 
 x3  4 y 
 x3  4 x 
 y3  4 x  4 
 y3 3  x4
y 3 f 1(x) 
 ; x  4 Yes, it is a function.  f (x )  x  f(x ) f 1(x )  O  x  12.  f (x )  3   f(x) 
  (x  1)2 3   y 
  (x  1)2 3  4.   x 
  (y  1)2  f(x) 
 4x  5 y 
 4x  5 x 
 4y  5 x  5 
 4y  3  (y  1)2 
 x   x 3 y 
 1  x 3 f 1(x) 
 1  x; No, it is not a function. y1
  x5  y 
 4 x5  f 1(x) 
 4; Yes, it is a function. 5.  3  f(x) 
 2x  2 y 
 2x  2 x 
 2y  2 x  2 
 2y x2   y
 2 x2   f 1(x) 
  2 ; Yes, it is a function.  503  Extra Practice  4. yes y 
 x  5  Lesson 3-5  x5 x   3x2  2x  1  x2  3x  Page A31 1. Yes; the function approaches 1 as x approaches 2 from both sides. 2. No; the function is undefined when x 
 3. 3. No; the function is undefined when x 
 1. 4. Yes; the function approaches 1 as x approaches 3 from both sides. 5. jump discontinuity 6. y →  as x → , y →  as x →  7. y →  as x → , y →  as x →  8. y → 0 as x → , y → 0 as x →   5x  1 5x  15 16  Page A31 1. y 
 kx 8 
 k(2) 4
k 2.  g 
 kw 10 
 k(3) 10   3 
 k  Page A31 3.  1. abs. max.: (1, 2) 2. rel. min.: (3, 0), rel max.: (1, 3), abs. min.: (2, 1)  5.  6.  x  x  3  2 1  x   84  rt 
 84  84 
 k  r(7) 
 84 r 
 12 y 
 3xz y 
 3(5)(10) y 
 150  k  y 
 x2 243 
 k  x 2 y
 —    6   5 
 w  k  k  3x  x  10  4 
 3 w t 
 r   27 
  (3)2  3x   y
 x2  10  g 
 3 w  k  4. y 
 kxz 60 
 k(5)(4) 3
k  Page A31  y 
 4x y 
 4(9) y 
 36  t 
 r 6 
 1 4  Lesson 3-7  y
  16   As x → ,  x  3 → 0. So, the graph of f(x) will approach that of y 
 x  5.  Lesson 3-8  Lesson 3-6  1. x 
 2  16   → y
x5 x3  a 
 kbc3 36 
 k(3)(2)3 1.5 
 k  243  y 
 x 2 yx2 
 243 y(5)2 
 243 y 
 9.72 a 
 1.5bc3 a 
 1.5(5)(3)3 a 
 202.5  as x → , y → 3; y 
 3 2x2  2. x 
 3   y
 x3  y
  2x2  x2 —— x 3   x2  x2  y
  2  3 1    x2 x  Lesson 4-1 Page A32 1. yes; f(x) 
 x3  7x2  2x  40 f(2) 
 (2)3  7(2)2  2(2)  40 
 8  28  4  40 
0 2. no; f(x) 
 x3  7x2  2x  40 f(1) 
 (1)3  7(1)2  2(1)  40 
 1  7  2  40 
 36 3. no; f(x) 
 x3  7x2  2x  40 f(2) 
 (2)3  7(2)2  2(2)  40 
 8  28  4  40 
 24 4. yes; f(x) 
 x3  7x2  2x  40 f(5) 
 (5)3  7(5)2  2(5)  40 
 125  175  10  40 
0  no horizontal asymptotes since as x → , y is undefined x5   3. h(x) 
  x2  6x  5  
  x5  (x  5)(x  1)  x 
 5, x 
 1  x5   y
 x2  6x  5  y
  y
  x 5   x2  x2 —— x2 6x 5      x2 x2 x2 1 5    x x2 —— 6 5 1     x x2  as x → , y → 0; y 
 0  Extra Practice  504  5. (x  3)(x  4) 
 0 x2  7x  12 
 0; even; 2 6. (x  (2))(x  (1))(x  2) 
 0 (x  2)(x  1)(x  2) 
 0 (x  2)(x2  x  2) 
 0 x3  x2  4x  4 
 0; odd; 3 7. (x  (1.5))(x  (1))(x  1) 
 0 (x  1.5)(x  1)(x  1) 
 0 (x  1.5)(x2  1) 
 0 x3  1.5x2  x  1.5 
 0; odd; 3 8. (x  (2))(x  (i))(x  i) 
 0 (x  2)(x  i)(x  i) 
 0 (x  2)(x2  1) 
 0 3 x  2x2  x  2 
 0; odd; 1 9. (x  (3i))(x  3i)(x  (i))(x  i) 
 0 (x  3i)(x  3i)(x  i)(x  i) 
 0 (x2  9)(x2  1) 
 0 x4  10x2  9 
 0; even; 0 10. (x  (1))(x  1)(x  2)(x  3) 
 0 (x  1)(x  1)(x  2)(x  3) 
 0 (x2  1)(x2  5x  6) 
 0 4 x  5x3  5x2  5x  6 
 0; even; 4  h2  12h 
 4 h2  12h  36 
 4  36 (h  6)2 
 40 h  6 
 210  h 
 6 210  6. x2  9x  1 
 0 x2  9x 
 1 5.  81  2  
x  92  7. b2  4ac 
  8.  (3)  121  x
 2(4) 3 11 x 
 8 7 x 
 4 or x 
 1 b2  4ac 
 (2)2   w
  or 121; 2 real  4(1)(10) or 44; 2 real  2  44  2(1)  w 
 1 11  9. b2  4ac 
 (5)2  4(12)(6) or 263; 2 imaginary t
   (5) 263   2(12) 5  i 263   t 
 24 10. b2  4ac 
 (6)2  4(1)(13) or 88; 2 real  1. x2  4x  5 
 0 x2  4x 
 5 x2  4x  4 
 5  4 (x  2)2 
 9 x2
 3 x2
3 x
5 2. x2  6x  8 
 0 x2  6x 
 8 x2  6x  9 
 8  9 (x  3)2 
 1 x3
 1 x3
1 x 
 2 3. m2  3m  2 
 0 m2  3m 
 2 9  3m  4 3 2 m  2 3 m  2    
2  6  x 
  3 22  11. b2  4ac 
 (4)2  4(4)(1) or 0; 1 real  (4)  0    n
 2(4)  x  2 
 3 x 
 1  1  n 
 2 12. b2  4ac 
 (6)2  4(4)(15) or 276; 2 real 6   276  6  2 69   3  69    x
 2(4)  x 
 8 x 
 4  x  3 
 1 x 
 4  Lesson 4-3  9  4  17  
   88   x
 2(1)  
 4  m
 4.   77  2 9 77  x 
 2 2 (3)2  4(4)(7)  9  Page A32  
  77  
 4  x  2 
  Lesson 4-2  m2  81  x2  9x  4 
 1  4  Page A32 1. 2 1 10 8 2 16 1 8  8 x  8, R8 2. 1 1 3 4 1 1 2 2 1 2 2 1 x2  2x  2, R1 3. 1 1 0 3 5 1 1 2 1 1 2 3 x2  x  2, R3  17   2 3 17    2 2   8a  6 
 0 a2  4a  3 
 0 a2  4a 
 3 2 a  4a  4 
 3  4 (a  2)2 
 7 a  2 
 7  a 
 2 7   2a2  505  Extra Practice  4. 4 1 2 7 3 4 4 8 4 4 1 2 1 1 0 x3  2x2  x  1 5. f(x) 
 x2  2x  8 f(4) 
 (4)2  2(4)  8 
 16  8  8 
 0; yes 6. f(x) 
 x3  12 f(1) 
 (1)3  12 
 1  12 or 13; no 7. f(x) 
 4x3  2x2  6x  1 f(1) 
 4(1)3  2(1)2  6(1)  1 
 4  2  6  1 
 7; no 8. f(x) 
 x4  4x2  16 f(4) 
 (4)4  4(4)2  16 
 256  64  16 
208; no  3. p: q:  1, 1  2  p : q  1,  2  r 1 1 2 2  p : q  1,  2,  4,  1 , 2  8,  1 , 3  2 , 3  4 , 3  8 , 3  6 6 6   1 7 5   22 29 27   4 33 23   8 25 15   3  6  3  24  12  0  1  2  6  4  24  16  0  r 1 1   Lesson 4-4  2 1  rational roots: 3, 2  Page A32 1. p: q:  1, 1  2,  3,  6  p : q  1,  2,  3,  6  r 1 1 2 2 3 3  1 1 1 1 1 1 1  2 3 1 4 0 5 1  5. 2 or 0 positive f(x) 
 x3  4x2  x  4 1 negative 5 2 6 3 5 10 2  r 1  6 8 0 0 16 24 0  r 1  10  1  2  r 1 1  2 2 2  1 1 3  2 3 5  3 0 8  1 1 9  1  2  2  0  2  2  0  2  2  3   2  9  13  4  1  2  1 1  4 5  1 4  1 1  1  1 2  9  3 5  93  5 0  935  rational zeros: none 7. 2 or 0 positive f(x) 
 4x3  7x  3 1 negative  1  rational root: 2  r  4  0  7  3  3 2  4  6  2  0  4x2  6x  2 
 0 (4x  2)(x  1) 
 0 1  x 
 2, x 
 1 3 1  rational zeros: 2, 2, 1  Extra Practice  4 0  x2  5x  4 
 0 (x  4)(x  1) 
 0 x 
 4, x 
 1 rational zeros: 1, 1, 4 6. 2 or 0 positive f(x) 
 x4  x3  3x2  5x  10 2 or 0 negative  rational roots: 3, 1, 2 2. p: 1 q: 1, 2 1,  2 0 2 10 6  0 2 0 6 2  rational root: 1 4. p: 1, 2, 4, 8 q: 1, 2, 3, 6  2  p : q  1 2 0 3 1  1 1 1 1 1  506  10 10  9360  1  6  Lesson 4-6  8. 3 or 1 positive f(x) 
 x4  x3  4x  4 1 negative r 1  4  1 1  1  1 0  5  4 4  76  0 0  20  Page A33  4 0  300  6  x  1.  rational zero: 1  7  y1  2.  Page A33 r 2 1 0 1 2 3  2 2 2 2 2 2 2  4 8 6 4 2 0 2  5 11 1 5 7 5 1  r 2 1 0 1 2  1 1 1 1 1 1  0 2 1 0 1 2  0 4 1 0 1 4  y    y 
  y1  5  r1  3.    5 13 6 5 4 3  4  1  r1  r 1     
  2  1  4  4  1    2
 2t  t2  4.    2
 t2  t2  2(t  2) 
 4  1 2t  4 
 3 2t 
 7 t 
 3.5 1 4 1    
  5.  3w 5w 15 5  12 
 w 17 
 w; w    Test w 
 1:  r 2 1 0 1 2  0  1  3(1)  1 and 2 3.  y1
0 y 
 1  5(r  1)  4(r  1) 
 1 5r  5  4r  4 
 1 r 
 10  1 and 0, 2 and 3 2.  4  x3
0 x
3  7y  4(y  1) 
 y2 3y  4 
 y2 y2  3y  4 
 0 (y  4)(y  1) 
 0 y4
0 or y
4  Lesson 4-5  1.  x
5  6  x2 
 5x x2  5x  6 
 0 (x  2)(x  3) 
 0 x2
0 or x
2  4  1  4   3  5 1 1 1 1 1 1  0 2 1 0 1 2  1 3 0 1 0 3  4 2 4 4 4 10  2 6 2 2 6 22  17   1 5     Test w 
 1:  w  507  true  1  3(1)  Test w 
 18:  2 and 1, 1 and 0 4-6. Use the TABLE feature of a graphing calculator. 4. 0.3, 1.3 5. 2.2, 0.3, 1.2 6. 1.3, 1.3  ? 1 15 ? 1  15 ? 1  1 5;    5(1)   1 ? 1    5(1)  15 1 1 ? 1      3 5 15 8 ? 1   ; false 15 15 1 1 ? 1     3(18) 5(18) 15 1 1 ? 1      54 90 15 4 ? 1   ; true 135 15  0 or w  17  Extra Practice  x2  x  6.  x4  5. 2x 32  2x  3  4 2x  1   
 x6  (x  6)(x  2) 
 x(x  4) x2  8x  12 
 x2  4x 12 
 4x 3 
 x; x 0 or 6  1  x  2 2x  3  0 2x  3 3 x  2  (1)  2 ? (1)  4  (1)  6 ? 5 3 7 ; false 12 ? 14   1 16 3 ? 1  5; true 42 44  ?  4 46 1  ? 0; false 2 72 ? 74   7 76 5  ? 3; true 7   Test x 
 1:  1  Test x 
 1: Test x 
 4: Test x 
 7: 0  x  ?  Test x 
 2: 2(2)  32 ? 1   2; meaningless ? Test x 
 0: 2(0)   32 ? 3   2; true ? Test x 
 1: 2(1)   32 ?    2; false 5 3  4  6. 2 6a    4 6a  2  256  3 or x  6  6a  258 a  43 6a  2  0 6a  2 1 a  3  Lesson 4-7 Page A33 1. 2 t  3 
 4 2  3t 
 16 3t 
 14 14 t 
 3  1  Solution: 2  x  2  ? Test a 
 0:  6(0)  2 4 4 ? 2   4; meaningless 4  Check:    14 2  3
 3  
4  
4 16 4
4  ? 6(1)  2 4 Test a 
 1:  4 ? 4   4; false 4  ?   2 4 Test a 
 44: 6(44) 4 ? 262   4; true 4  2. 4  x 2
1 Check: 4  11  2
1 x  2 
 3  43
1 x2
9 1
1 x 
 11 3 3 3.  y  7  10 
 2 Check: 505   7  10 
 2 3  y  7 
 8 8  10 
 2 y  7 
 512 2
2 y 
 505 4. a  1  5 
 a   6 a  1  10a   1  25 
 a  6 10a  1 
 30  a 1
3  a1
9 a 
 10 Check: 10    5 
 10 1 6  35
2 2 2 no real solution  Solution: a  43  Lesson 4-8 Page A33 1. f(x) 
 0.75x  2 2. f(x) 
 x3  x2  x  2 3. Sample answer: f(x) 
 0.51x2  0.02x  0.79 4a. Sample answer: y 
 1.632x  99.275 4b. Sample answer: about 122.123 thousand f(x) 
 1.632x  99.275 f(15) 
 1.632(14)  99.275 
 122.123  Lesson 5-1 Page A34 1. 13.75° 
 13°  (0.75  60)
 
 13°  45
 
 13° 45
  Extra Practice  508  2. 75.72° 
 75°  (0.72  60)
 
 75°  43.2
 
75°  43
  (0.2  60) 
 75°  43
  12 
 75° 43
 12 3. 29.44° 
 29°  (0.44  60)
 
 29°  26.4
 
 29°  26
  (0.4  60) 
 29°  26
  24 
 29° 26
 24  1   2. sin v 
  csc v 1   sin v 
  2.5 or 0.4  3. (AC )2  (BC )2 
 (AB)2 142  162 
 (AB)2 452 
 (AB)2 452  
 AB; 452  or 2113  side opposite   sin A 
  hypotenuse  
  6. 38° 15
 10 
  38°  15

    10
  1°  3600  
  38.253°  
  8. 51° 14
 32 
 51°  14

 
 51.242° 9. 850°  360° 
 490° 490°  360° 
 130° 130°; II 10. 65°  360° 
 295° 295°; IV 11. 1012°  360° 
 652° 652°  360° 
 292° 292°; IV 12. 578°  360° 
 218° 218°; III 13. 180°  126° 
 54° 14. 480°  360° 
 120° 120°  360° 
 240° 240°  180° 
 60° 60° 15. 642°  360° 
 282° 360° 282° 
 78° 78° 16. 1154°  360° 
 794° 794°  360° 
 434° 434°  360° 
 74° 74°    32
  1°  60
  1°  3600  21  13  hypotenuse  21  13  1  13  sec A 
 14 or 7  1  13  side adjacent   cot A 
  side opposite 14  7   cot A 
 1 6 or 8  4. (AC )2  (BC )2 
 (AB)2 252  (BC )2 
 282 BC 2 
 159 BC 
 159       side opposite   cos A 
  hypotenuse   159  cos A 
 28  side opposite   csc A 
  side opposite   159   or  csc A 
  159  159  hypotenuse   cot A 
  side opposite   sin A 
  hypotenuse    side adjacent 25  sin A 
 2 8  tan A 
  side adjacent  hypotenuse  281  59  28  tan A 
 2 5  sec A 
  side adjacent 28  side adjacent  251  59  25  sec A 
 25   or  cot A 
   159 159  5. (AC)2  (BC )2 
 (AB)2 92  62 
 (AB)2 117 
 (AB)2 117  
 AB; 117  or 313  side opposite   sin A 
  hypotenuse 213   6   or  sin A 
  313  13  side adjacent   cos A 
  hypotenuse 313   9   or  cos A 
  313  13  side opposite   csc A 
  side opposite  6  csc A 
 6 or 2   tan A 
  side adjacent  hypotenuse   sec A 
  side adjacent 313   hypotenuse  313   2  tan A 
 9 or 3  13  sec A 
 9 or 3  13   side adjacent   cot A 
  side opposite 9  3  cot A 
 6 or 2  Lesson 5-3 Page A34  y  1. tan v 
 x Since tan v 
 0, y 
 0. x  x  cot v 
 y 
 0 cot v is undefined. 2. Sample answers: 90°, 270°  Page A34 1   1. cot v 
  tan v 1 — 5  6  csc A 
 16 or 8  8  hypotenuse  Lesson 5-2  cot v 
  16   sec A 
  side adjacent  1°  
 107.213°   csc A 
  side opposite   tan A 
 1 4 or 7    7. 107° 12
 45 
  107°  12

 60
   45
 3600  1°  71  13  14   or  cos A 
  113 21  13  side opposite   tan A 
  side adjacent  4. 87.81° 
 87°  (0.81  60)
 
 87°  48.6
 
 87°  48
  (0.6  60) 
 87°  48
  36 
 87° 48
 36 1° 1°   5. 144° 12
 30 
 144°  12

 60
   30
 3600  
 144.208° 1°  60
  81  13  16   or  sin A 
  113 21  13  side adjacent   cos A 
  hypotenuse  6  or 5  x  cos v 
 r Since cos v 
 0, x 
 0. On the unit circle, x 
 0 when v 
 90° or v 
 270°.  509  Extra Practice  3. r 
  x2  y2 
  (1)2   (2 )2 
 5  y  2.  a  cos 87° 
 19 19 cos 87° 
 a 1.0  a  x  sin v 
 r  5 2  2  sin v 
  or  5  y tan v 
 x 2  tan v 
  1 r sec v 
 x  5  sec v 
  1  5  cos v 
 r 1  csc v 
 cot v 
  or  5   5   cos v 
  or   csc v 
  or 2  cot v 
  5  r  y 5   2 x  y 1 1  or  2 2  3.  5  y  16.5  16.5   c
 cos 65.4°  c  39.6 4.  12  12   a
 tan 42.5°  a  13.1  x  2   2  22  2 y tan v 
 x 2  tan v 
  2 or 1 r sec v 
 x 22   sec v 
   2 or 2  2   2  cos v 
  or   csc v 
 csc v 
 cot v 
 cot v 
  y  b   sin 75° 
  5.8  2 2 2 r  y 22   or 2  2 x  y 2  or 1 2  5.8 sin 75° 
 b 5.6  b 20  b  20  b  b  tan 42° 
 x tan 42° 
  or  5  cos v 
  29  r csc v 
 y 29  csc v 
 2 x cot v 
 y 5 cot v 
 2  or  b  48˚   x
 tan 48°  42˚  b –– 20  b  tan 48°  x  20 tan 42°  b tan 42° 
 b tan 48° 20 tan 42° 
 b(tan 48°  tan 42°)  x  229   29  20 feet  6. tan 48° 
 x  cos v 
 r  2  sin v 
  29  y tan v 
 x 2 tan v 
 5 r sec v 
 x 29  sec v 
 5  b  sin B 
 c  5.  x2  y2 5. r 
  
  (5)2  (2)2 
 29  sin v 
 r  b  tan B 
 a tan 42.5° 
 a  cos v 
 r  sin v 
  or   a  cos B 
 c cos 65.4° 
 c  4. r 
  x2  y2 
  (2)2   (2)2 
 22  sin v 
 r  a  cos B 
 c  529   29  20 tan 42°  (tan 48°  tan 42°)  85.7  
b b; 85.7 ft  Lesson 5-5   x2  y2  6. r 
 
  (4)2   (3)2 
5 y sin v 
 r 3 sin v 
 5 y tan v 
 x  tan v 
  sec v 
  Page A35 3  cos v 
  cos v 
   
  csc v 
 cot v 
  5    3  1  1   sec A 
  cos A  5  3 x  y  sec A 
  1  1  2  or 2  
  4  sec v 
  4  1  2. Let A 
 cos1 2. Then cos A 
 2.  r  r  x  3  sin arcsin 4 
 4  4  5  csc v 
 y 3  4  3  1. Let A 
 arcsin 4. Then sin A 
 4.  x  r  cot v 
  3  sec cos tan1  1. Then tan A 
 1. 3. Let A 
 tan(tan1 1) 
 1 a  b  5. sin B 
 c  4. tan A 
 b  Lesson 5-4  38  17  tan A 
 2 5 38  Page A34 1.  a  tan A 
 b a  15  tan 38° 
 15 tan 38° 
 a 11.7  a Extra Practice  sin B 
 1 9  510  17  A 
 tan1 25  B 
 sin1 1 9  A  B  56.7°  63.5°  a  a  cos B 
  7. A 
 180°  (60°  75°) or 45°  7. cos B 
 c  6. cos B 
 c 24  30  cos B 
  1  sin B sin C   K 
 2a2  sin A  9.2  12.6  1  sin 60° sin 75°  B 
 cos1 30   B 
 cos1  12.6   K 
 2(8)2  sin 45°  B  B  K  9.2  24  36.9°  8. tan A 
 tan A 
  43.1°  1 1  K 
 2(16)(12) sin 43° K  28.4   A 
 tan1  36.5  A  37.9 units2  8. K 
 2bc sin A  a  b 28.4  36.5  65.5 units2  37.9°  Lesson 5-7 Lesson 5-6 Page A35 1. Since 145°  90°, consider Case II. 5  10; no solution 2. Since 25° 90°, consider Case I. b sin A 
 10 sin 25° 4.226182617 9  4.226182617; 2 solutions a b  
  sin A sin B  Page A35 1. C 
 180°  (75°  50°) or 55° a  sin A 7  sin 75°  b   
 sin B  
 b sin 50°  a  sin A 7  sin 75°  7 sin 50°   c
 sin 75°  b 5.551472956 C 
 55°, b 
 5.6, c 
 5.9 2. B 
 180°  (97°  42°) or 41° b   
 sin B b   
 sin 41° 12 sin 41°   b
 sin 42°  
 
  b
  b  sin B b  sin 49° 10 sin 49°  sin 99°  a   
 sin A a   
 sin 22° 25 sin 22°   a
 sin 117°  6  sin 25°  5.936340197  B 
 sin1  a  c  c  
6 10 sin 25°  B 44.77816685 B 180°  44.8° or 135.2° Solution 1 C  180°  (25°  44.8°) or 110.2°  17.80004338  a  sin A 6  sin 25°  c  c   
 sin C c  sin 110.2° 6 sin 110.2°  sin 2 5°  c 13.32398206 B 
 44.8°, C 
 110.2°, c 
 13.3 Solution 2 C 180°  (25°  135.2°) or 19.8°  5.365247745  a  sin A 6  sin 25°  b c  
  sin B sin C 25 c   
  sin 117° sin 41° 25 sin 41°  c
 sin 117°  a 10.51077021 B 
 117°, a 
 10.5, c 
 18.4  10   
 sin B  6 sin B 
 10 sin 25°  a c  
  sin A sin C 10 c  
  sin 99° sin 32° 10 sin 32°  c
 sin 99°  b 7.641171301 A 
 99°, b 
 7.6, c 
 5.4 4. B 
 180°  (22°  41°) or 117° b  sin B 25  sin 117°  c  c a  
  sin C sin A 12 a  
  sin 42° sin 97° 12 sin 97°  a
 sin 42°  b 11.76557801 B 
 41°, a 
 17.8, b 
 11.8 3. A 
 180°  (49°  32°) or 99° a  sin A 10  sin 99°  c   
 sin 55° 7 sin 55°   b
 sin 75°  c  sin C 12  sin 42°  c   
 sin C  c  c   
 sin C c  sin 19.8° 6 sin 19.8°  sin 25°  c 4.809133219 B 
 135.2°, C 
 19.8°, c 
 4.8 3. Since 56° 90°, consider Case I. C sin B 
 50 sin 56° 41.45187863 34 41.5; no solution  18.40780654  1  5. K 
 2bc sin A 1  K 
 2(12)(6) sin 34° K 20.1 units2 6. C 
 180°  (87°  56.8°) or 36.2° 1  sin A sin B   K 
 2c2  sin C 1  sin 87° sin 56.8°   K 
 2(6.8)2  sin 36.2°  K  32.7 units2  511  Extra Practice  4. C 
 180°  (45°  85°) or 50° a  sin A a  sin 45°  a
  c  sin C 15  sin 50° 15 sin 45°  sin 50°  a  13.8459352  b  sin B b  sin 85°  
 
  
 
  b
  3. b2 
 a2  c2  2ac cos B b2 
 142  182  2(14)(18) cos 48° b2 182.7581744 b 13.51880817 a b  
  sin A sin B 14  sin A  c  sin C 15  sin 50° 15 sin 85°  sin 50°  14 sin 48°  sin A 
 b A 
 sin1  Lesson 5-8  14.2  sin A  Page A35 1. a2 
 b2  c 2  2bc cos A a2 
 62  82  2(6)(8) cos 62° a2 54.93072997 a 7.411526831 a b  
  sin A sin B  B
  A 
 sin1  A  1 1  
  s 
 2(4  7  10)    6 sin 62°  a  s 
 10.5 a)(s  b)(s  c) K 
 s(s  K 
 10.5(1 0.5  .5 4)(10)(10.5  7)  10 K 
 119.43 75  K 10.9 units2 1  6. s 
 2(a  b  c)  112  1   
 16 8   s 
 2(4  6  5) s 
 7.5 a)(s  b)(s  c) K 
 s(s  K 
 7.5(7. 5 )(7.5  4  6)(7.5  5) K 
 98.43 75  K 9.9 unit2  48.1896851 
 
  sin B 
  b  sin B 7  sin B 7 sin A  9  B 
 sin1  14.2 sin 85.3°  5. s 
 2(a  b  c)  B 45.62599479 C 180°  (62°  45.6°) or 72.4° a 
 7.4, B 
 45.6°, C 
 72.4° 2. a2 
 b2  c2  2bc cos A 92 
 72  122  2(7)(12) cos A 112 
 168 cos A  a  sin A 9  sin A  
c  A 31.23444201 B 180°  (31.2°  85.3°) or 63.5° c 
 27.3, A 
 31.2°, B 
 63.5°  6 sin 62°  a  A 
 cos1  c   
 sin 85.3° 14.2 sin 85.3°  6  sin1  14 sin 48°  sin A 
 c   
 sin B  sin B 
  
b  A 50.31729382 C 180°  (50.3°  48°) or 81.7° b 
 13.5, A 
 50.3°, C 
 81.7° 4. c2 
 a2  b2  2ab cos C c2 
 (14.2)2  (24.5)2  2(14.2)(24.5) cos 85.3° c2 744.8771857 c 27.29243825 a c  
  sin A sin C  b 19.50659731 C 
 50°, a 
 13.8, b 
 19.5  a  sin 62°  b   
 sin 48°  1  
9 7 sin A  7. s 
 2(a  b  c) 1  s 
 2(12.4  8.6  14.2)  B 35.43094469 C 180°  (48.2°  35.4°) or 96.4° A 
 48.2°, B 
 35.4°, C 
 96.4°  s 
 17.6 K 
 s(s a)(s   b)(s  c) K 
 17.6(1 7.6  (17.6 12.4) )(17.6 8.6.2)  14 K 
 2800. 512  K 52.9 units2 1  8. s 
 2(a  b  c) 1  s 
 2(150  124  190) s 
 232 a)(s  b)(s  c) K 
 s(s  K 
 232(2 32 150)(2  32 124)(2  32 190)  K 
 86,29 2,864  K 9289.4 units2  Extra Practice  512  Lesson 6-2  9a. 28 in.  Page A36  110˚  1. 2. 3. 4.  35 in.  d2 
 282  352  2(28)(35) cos 110° d2 2679.359481 d 51.8 in. 9b. Area 
    1 2 2(28)(35) 920.9 in2  sin 110°  5  2 
 10 or about 31.4 radians 3.8  2 
 7.6 or about 23.9 radians 14.2  2 
 28.4 or about 89.2 radians 2.1  2 
 4.2 v  q 
 t    4.2  q 
 5 q 2.6 radians/s 5. 1.5  2 
 3 v  q 
 t  Lesson 6-1  3  q 
 2  Page A36  q 4.7 radians/min 6. 15.8  2 
 31.6     1. 120° 
 120°   180°  
  v  q 
 t  2  3  31.6  q 
 1 8    180°  2. 280° 
 280°   q 5.5 radians/s 7. 140  2 
 280  14  
 9  v  q 
 t     3. 440° 
 440°   180°  280  q 
 2 0  22  
 9   q   4. 150° 
 150°   180°  v  5  
 6 5. 6.  8  3 5  12  8  2  q 
 30  180°  
 3    q  
 480° 5  44.0 radians/min  8. q 
 t about 0.2 radian/s  180°  
 12   
 75°  Lesson 6-3  180°  7. 2 
 2   
 114.6°  Page A36  180°  8. 10.5 
 10.5   
 601.6°  5    5  1  1. 1 2. 0 3. 1 4.  9. reference angle:   6 
 6; Quadrant II sin 6 
 2 10. reference angle:  4  3  
 4   ; 3  y y 
 sin x  Quadrant III  1  3   sin 3 
 2 11.  9  4    4  is coterminal with 4; Quadrant I  3  2  2   9  cos 4 
 2 12.  3  2  is coterminal with  
  3  1    y  5.  13. If the diameter 
 10 in., the radius 
 5 in.  y 
 cos x    180  4  
 9 s 
 rv s
5 s  x    2  cos 2 
 0 80° 
 80°   O  4  3  1  O  x  1  4  
9  7.0 in.  513  Extra Practice  Lesson 6-4  Lesson 6-5  Page A36  Page A37  1. 2 
 2;  2  1  
   c  
 2  y  y  2    1. k 
  2 or 2 y 
 sin(2  
)  1  y 
 2 cos   1  O    1  2  3  O    2  2  c   2. 3 
 3;  0.5 
 4  2. k 
 1 or 2  y  y 3 2 1  y 
 3 sin 0.5   O    1 2 3  2  3    O  3.  4  c k  O  2  4  6    8  O  (2  
2 )    2  3    1 2  k  4. A 
 0.5 A
  0.5  y
  0.5 sin 3  
 6  k
  2  6  or  4. A 
 2  1  3  v  2  
    3    c  
 2  1 
   A
  2  y
  2 sin (v  )  1  5. A 
 0.5  2  or 6 k
   3  2 sin 6v 2  k  3  5 3  5 3 v  cos  5 2  h 
 1  2 or 1 c 
   k
    2  4  or  1  2  2  k    c  
 4  k
  8 
 0  2  4  A
  0.5  y
  0.5 sin 8v  3  6. A 
 20  
 4  k
  2  k  c
0  or 8    c  
 2  h
3  4 
 2  A
  20  y
  20 cos (4v  8)  4  h
4  2 or 4 c 
 8  k
   2  2  k  7. A 
 0.25 A
  0.25  y
  0.25 cos 4 v    
8  k
  2  8  or  3  7. A 
 4    4  A
 y
  Extra Practice  2  k  2  2  k  5. A 
 2  y
  or   y 
 sin  1  1  A
  3   2  y y 
 12 cos 4  6. A 
    2  1  2  
   
  1  y
    2  y  A
  y 
 2 cos(  2
)  2  1 2 
 2;  1 
 8   1  2  3   2  1  2  3.    514  2  k  c  
0    
 10  3 2  k 
 10 4 3  1  cos  v   4 5 2    or 5  5  c
0  1  h 
 2  3  Lesson 6-6  4. If y 
 tan 4, then y 
 1. 3  Cos1
tan 4 
 Cos1 y  
 Cos1 (1) 
  Page A37 1a. 12.1  2.7 
 9.4 h 1b. 12.1  2.7 
 14.8 h 1c. m 
 10 represents the middle of October. d 
 2.7 sin (0.5(10)  1.4)  12.1 d 10.9 h 2. A 
  6  2  y
3  2 or 3; 14  cos 7t  
  1  5. Let a 
 Cos1 2 and b 
 Sin1 0. 1  Cos a 
 2  Sin b 
 0    3  b
0  a
    7  sin 
   Sin1 0 
 sin (a  b)  1 Cos1 2  
     
 sin 
3  0   
 sin 3 3   
 2  Lesson 6-7  3   6. Let v 
sin1 2.  3   sin v 
 2  Page A37    2  v 
 3  y 
 y 
 cot   4  8  (    
 cos 
2  3   3  1.  cos 
2 Sin1 2  
 cos (2v) 
 cos 3 1  
 2  )  4   O  2  3    Lesson 7-1  4 8  2.  Page A38 1   1. csc v 
  sin v  y  1  
  2  1  co s v  4    O  3.  1  
  1 2 1  4  y 
 sec   2  2  
    2  3    
  1  15  16    1 
 15   y   4  4 2   O  4   15   15   15  
     y 
 csc(2  2
) 2 3   
  2  4 15  15  1  2. tan v 
  cot v  4  1 
  6  
  Lesson 6-8    v 
 2 3. Let v 
 Tan1 1. Tan v 
 1   6  6      6   
  2 13 13   3. cos 
 cos 6  2 6  
 cos 6  
  Page A37 1. Let v 
 Cos1 0. Cos v 
 0   3 3    6   2. Let v 
 Arcsin 0. Sin v 
 0 v
0  sin (315°)   4. tan (315°) 
  cos (315°)   
  cos(Tan1 1) 
 cos 4 
    2  
  2  sin (45°  (360°))  cos (45°  (360°)) sin  45°  cos 45°  
 tan 45°  v 
 4  515  Extra Practice  5. Sample answer: cos x 
 1  1   5. csc (930°) 
  sin (930°)  
 
 
 
  cot x  csc x cos x  sin x  1  sin x  1  sin (360°(2)  210°) 1 sin (210°) 1 sin (30°) 1 sin 30°  
1  cos x 
 1 6. Sample answer: cot x 
 2 2 tan x sin x  2 cos x 
 csc x  
 csc 30° 6.  
1  sin v  tan v  c os v 
  sin v sin v 1 
 cos v  2    
 cos x  sin x  cos x 
 sin x 1 sin x  cos x  
  2 
  sin x cos x 1 1   2 
 cos x  
 sin x sin x  1  2  
 sec v cos v  sin v  1  2  cos x      sin v   7. cot v tan v  sin v sec v 
  sin v cos v cos v   2
 sin x  sin v  2 
 cot x   
1 cos v  
 1  tan v 8. (1  sin x)(1  sin x) 
 1  sin2 x 
 cos2 x 9.  cot x  sin x  csc x cos x  
 
  Lesson 7-3  cos x   sin x sin x  1   cos x sin x cos x  cos x  sin x  Page A38 1. cos 75° 
 cos (45°  30°) 
 cos 45° cos 30°  sin 45° sin 30° 2   3   2   1  
 2  2  2  2  
 sin x  6  2   
 4 2. sin 105° 
 sin (60°  45°) 
 sin 60° sin 45°  cos 60° cos 45°  Lesson 7-2   3  2   1  2   
 2  2  2  2 6  2   
 4  Page A38 csc2 v 
 cot2 v  sin v   1  sin v  
  csc2 v 
 cot2 v  1 csc2 v 
 csc2 v 2.  sec v  csc v  csc v sec v sec v csc v    csc v sec v csc v sec v 1 1    csc v sec v  
  
 sin v  cos v   p tan 4  tan 6    1  tan  tan  4 6 1  3   3  1  1  3  3  
 sin v  cos v  
  1  3   3   3    1  3  4. tan  7  12   
 2  3   
 tan A sin A 
 tan A sin A 
 tan A sin A 
 tan A sin A  1  3   3  1  3   
 tan 
3  4 
  
 tan A sin A    tan   tan  3 4    1  tan  tan  3 4  3  1  
  1   
  1  1  1 3 1 1  3       3   
 2  3  tan A sin A 
 tan A sin A  Extra Practice     3  
 sin v  cos v  sin v  cos v 
 sin v  cos v 3. sin2 x  cos2 x 
 sec2 x  tan2 x 1 
 tan2 x  1  tan2 x 1
1 4. sec A  cos A 
 tan A sin A 1   cos A cos A cos2 A 1    cos A cos A 1  cos2 A  cos A sin2 A  cos A sin A  sin A cos A      3. tan 12 
 tan 
4  6  1. csc2 v 
 cot2 v  sin v csc v  516  29  5  4  5. sec 12  sec 1 2 1     cos 4  6  sec2 y  1  tan2 y 2  54  1           cos 4 cos 6 
 sin 4 sin 6  9  16 3  4  1   
 2 
3  
2  1   
   2 2 2 2   1  tan2 y  tan2 y  tan y tan x  tan y   tan (x  y)   1 
 tan x tan y 3 3    4 4  1   
6  
 
2   4 4 
6   
 2     
6  
 
2    
2  
6  
6   
2    —— 3 3 1 
 4 4   6. cot 375°  cot 15°  24  1  6
 85  8  
 82 
 52  
39   8  5 5  8  cos y  8  csc y  5  1  tan 45° tan 30°  1  sin y  1    sin y   
3  1 
 3  
3  1  1 3    1  
3  1   3 —— 
3  1 
  3    
3  1   3 —— 
3  1   3  2 2  1 
 5    cos y   21  25    7  16    21  5    7  4  
7   4  2   5  3  4  
 122 
 112  
23   
1  19  12  sin y  11  12   
 280
85  48
3315  179  30°    
 23  12  cos (x 
 y)  cos x cos y  sin x sin y 5    6
3315  
 35
85   6
3315  
 35
85   1. sin 15°  sin 2  8. 
 122 
 52  
119    1 2  680  6
3315   35
85   Page A39  6  7
3    20 sin x   1  6
85  
39  7
85  5      85 8 85 8  Lesson 7-4  sin (x  y)  sin x cos y  cos x sin y 
21   1  cos x cos y  sin x sin y   3 2  1 
 4     5  
39    sec (x 
 y)   cos (x 
 y)   2  
3  7. sin x   7
 85  10. 
 72  62  
85 ; cos x  85; sin x  85  1  tan 45° 
 tan 30°    6  4 — 7  16   7    tan (45° 
 30°)    3  9. If cot x  3, then tan x  4.  
1  19   12    
 23  12  1 
 cos 30°  2 
3  1 
  2  2  
  2 
 
3   55  
2737     2    144  
  2 
 
3   Since 15° is in Quadrant I, sin 15°  2. 150°  2. cos 75°  cos 2   1  cos 150°  2    
3  1 
  2  2  
  2 
 
3     2  
  2 
 
3   Since 75° is in Quadrant I, cos 75°  2.  517  Extra Practice  3. tan    12   tan    6 — 2  3
5   sin 2v  2 sin v cos v   1 
 cos —— 6   1  cos —6—     277 3
5   12
5   cos 2v  cos2 v 
 sin2 v 2 2  41     (2 
 
3  )2  1   
 4 9 2 tan v   tan 2v   1 
 tan2 v        45°  1  cos 45°  2  
5    2 3 3 
5   2  4
5    9 cos 2v  cos2 v 
 sin2 v  
  2  
2   
5  2  2 2   3 
 3  
 2  
  2  Since 22.5° is in Quadrant I, cos 22.5°  .  1   9  2  5  6 — 2  2 tan v   tan 2v   1 
 tan2 v  2 5 2
5   5  1 
 cos 6  2    1 
 5 2
5   2   4
5   
3   1  2  9.    2  
 (3)2  (
1)2  3
 10   2 10 3
10   5  5p  
  2  
3   
 10  
  10   3   . Since 1 2 is in Quadrant I, sin 12  2   
 5 cos 2v  cos2 v 
 sin2 v  6. tan 112.5°  tan 225°   1 
 cos 225°  1  cos 225°    
 2 1   2  
2  1 
 2  
 10 2  4   
 5 2 tan v  2 (
 3)    1 
 (
 3)2  2  
2  2  
2     2  
2  2 
 
2    (2  
2  )2  2  3
10  2  
 10  
 1 0    tan 2v   1 
 tan2 v    3   4  or 1  
2   Since 112.5° is in Quadrant II, tan 112.5°  
1 
 
2 .  518  
 10   
10 ; sin v  10; cos v  
1 0  sin 2v  2 sin v cos v  
  2  
3     2  Extra Practice  2
5   sin 2v  2 sin v cos v  1  2  2    2  8. 
 32 
 22  
5 ; cos v  3, tan v  5    2    3
5    
 4 1  
2   5  1 
 2 12
 5  4. cos 22.5°  cos 2  5. sin 12  sin  2 2 3
5   or 2 
 
3   Since 12 is in Quadrant I, tan 12  2 
 
3 .    3
 5 2   7 
 7  2 
 
3  2 
 
3    2  
3  2 
 
3     2   49  
 3 1 
  2  
 3 1   2    3
5   7. 
 72 
 22  3
5 ; sin v  7; tan v  2  3  
 32 
 ( 
 2)2  
5   3  2 2  3  
5  cos v  3 2
 5 tan v  
 5  10. csc v  
 2 1  sin v  
  sin v  
  Lesson 7-6 Page A39 1. x cos 30°  y sin 30° 
 12  0  sin 2v  2 sin v cos v  2 
 3  
3   2x  
35   2  4
 5   
 9  x  y 
 24  0 
3      2. x cos 3  y sin 3 
 2  0  cos 2v  cos2 v 
 sin2 v 
5  2  1  2  2 2   3 
 
 3  
 3  x  2 y 
 2  0 y 
 4  0 x  
3  1   9  1  3. x cos 150°  y sin 150° 
 2  0  2 tan v   tan 2v   1 
 tan2 v    1   2 y 
 12  0  2 
  
3   1  1  
 2 x  2 y 
 2  0   2
5  2 1 
 
 5 2
5   5  x  y 
 1  0 
 
3 4. 
  
 A2   B2  4x  
 2
 29   
 4
5    
 
 42  1 02  
2
29   10y  10      0  
 2
29  
 2
 29  2
29   5
29   5
29   
 29x 
 29y 
 29  0 5
29   p  29  Lesson 7-5  5
 29  2
29   sin f  
 29, cos f  
 29 5  tan f  2  Page A39  f  Arctan 2  180° f  248° 5  1. 4 cos2 x 
 2  0 1  cos2 x  2 cos x    5. 
 A2   B2  
 12  ( 
1)2  
2   
2   2  x  
2   
 2  x 2  x  45°, 135°, 225°, 315° 2. sin2 x csc x 
 1  0 sin2 x  1  sin  x  90°   
3  cos x  sin x  
2   cos x    2 cos x  cos x  0 x  90°, 270°  
 2  0 or  
2   tan f  
1 f  Arctan (
1)  360° f  315° 6. 
 A2   B2  
 22  32  
13   
 2 cos x  0  
3  sin x  4.  cos x  sin x  
2   
 2 y 
 
2 0  sin f  
 2, cos f  2   cos x  2 cos x 
3  
3  2   p  
2  x
10 sin x  1  3.  y   
   0 
 
2  
 2  2  
 13 
3   sin x  12  2
13  3
13  12
13   x   y 
   13 13 13 12
13  p  13 3
13  2
13  sin f  13, cos f  13 3 tan f  2 3 f  Arctan 2  
20  
3  sin x 
3   2  3   y 
   0 x 
13  
 13  2  sin x  x  60°, 120° 3 cos2 x  6 cos x 
 3 3 cos2 x 
 6 cos x  3  0 3(cos2 x 
 2 cos x  1)  0 3(cos x 
 1)(cos x 
 1)  0 cos x 
 1  0 cos x  1 x  0°  0  f  56°  519  Extra Practice  Lesson 7-7  10.  1.7 cos 70°  h 0.58  h  Page A39 1.  3(2) 
 2(1) 
 2  2  (
 2)2 
3    11.  2    
13    2.  2(3)  4(0) 
 2  2  42 
2    h   cos 76°   2.1  2.1 cos 76°  h 0.51  h  2
 13  13 4  12.    
20   h   cos 30°   3.6  3.6 cos 30°  h 3.12  h  4    2
5   3.  h   cos 70°   1.7  2
5   5 
3(
1)  (
4) 
 1 
2    
 
 (
3)2   12 10 
10   x   sin 70°   1.7  1.7 sin 70°  v 1.60  v v   sin 76°   2.1  2.1 sin 76°  v 2.04  v v   sin 30°   3.6  3.6 sin 30°  v 1.8  v  Lesson 8-2  
10    
 5; d  5 2  4. y  3 x 
 2 → 2x 
 3y 
 6  0 2(4) 
 3(
2) 
 6  2  ( 
2  
3)2     Page A40  u 1. AB  4 
 3, 1 
 6  1, 
5 u AB   
 12  ( 
5)2  
26  u 2. AB  
2 
 (
1), 2 
 3  
1, 
1 u AB   
 (
1)2   (
1 )2  
2  u 3. AB  
1 
 0, 
8 
 (
4)  
1, 
4 u AB   
 (
1)2   (
4 )2  
17  u 4. AB  3 
 1, 
9 
 10  2, 
19 u AB   
 22  ( 
19)2  
365  u 5. AB  
3 
 (
6), 
6 
 0  3, 
6 u 32  ( 
6)2 AB   
  
45   3
5  u 6. AB  0 
 4, 7 
 (
5)  
4, 12 u AB   
 (
4)2   122  
160   4
10  7.  5, 6   
 52  62  
61  u  6j u 5i 8.  
2, 4   
 (
2)2   42  
20   2
5  u  4j u 
2i 9.  
10, 
5   
 (
10)2  (
 5)2  
125   5
5  u 
 5j u 
10i 10.  2.5, 6   
 (2.5)2   62  
42.25   6.5 u  6j u 2.5i  8  13 
 8
 13  13  5. (0, 
1) 0  2(
1) 
 4  d  2  2  2 
1   
6      
5  6
 5 
 5;  6
5   d  5  6. (0, 3) 3(3) 
 2(0) 
 7   d 2 2  
3   (
  2) 2    
13   2
 13   13 7. (2, 
1) 2(2)  5(
1) 
 4  d  2  2 2  5 
  
5    
 29   
0.9; d  0.9 unit  Lesson 8-1 Page A40 1.  2. 1.7 cm 2.1 cm  70˚  104˚  3. 330˚  3.6 cm  4. 3.6 cm; 89° 6. 3.7 cm; 357° 8. 7.2 cm; 330°  Extra Practice  5. 2.6 cm; 23° 7. 1.2 cm; 342° 9. 8.8 cm; 340°  520  11.  2, 
6   
 22  ( 
6)2  
40   2
10  u 
 6j u 2i 2  (
 12.  
15, 
12   
 (
15) 12)2  
369   3
41  u 
 12j u 
15i  u 
1, 5, 2  u i j u k 3 0 4 
1 5 2 u 
 10j u  15k u  
20i   
20, 
10, 15 3, 0, 4  
20 3  (
10) 0  15 4  0; yes 
1, 5, 2  
20 (
1)  (
10) 5  15 2  0; yes u 2, 1, 3  u i j u k  
20, 
10, 15 
20, 
10, 15  Lesson 8-3 1. u p  2 1, 2, 
1  3(
4, 
3, 0  2, 4, 
2  
12, 
9, 0  
10, 
5, 
2 1 u 2. p  1, 2, 
1 
 2 
2, 2, 4  
4, 
3, 0  1, 2, 
1  1, 
1, 
2  
4, 
3, 0  
2, 
2, 
3 3. u p  
2 
2, 2, 4  
4, 
3, 0  4, 
4, 
8  
4, 
3, 0  0, 
7, 
8 u  3 
4, 
3, 0  2 1, 2, 
1 4. p 4    9   
3, 
 4, 0  2, 4, 
2  8, 10, 19  3   
1, 14, 
2  8, 10, 19 0, 0, 0 0, 0, 0 0 
 7, 0 
 0, 0 
 4 
7, 0, 
4  4. 8, 6 5. 3, 4, 0 6. 4, 5, 1 7. 1, 0, 3   8, 10, 19 
1, 
3, 2  8 (
1)  10 (
3)  19 2  0; yes 6, 
1, 
2  8 6  10 (
1)  19 (
2)  0; yes  Page A41 1. magnitude  
 2002  2202  297.32 N  Page A41  3. 
5, 3    Lesson 8-5  Lesson 8-4  2. 
3, 2    
1 1 0 2 1 3 u  3j u 
 3k u  3i  3, 3, 
3 3, 3, 
3 
1, 1, 0  3 (
1)  3 1  1 (
3) 0  0; yes 3, 3, 
3 (2, 1, 3  3 2  3 1  (
3) 3  0; yes u u 10. 
1, 
3, 2 6, 
1, 
2  u i j k 2 
1 
3 6 
1 
2 u  10j u  19k u  8i  Page A40  1. 3, 4    9. 
1, 1, 0  5. 2, 
4, 1  5, 4, 3 u v  7, 0, 4 u v  u v  u v       8. 3, 0, 4  200   direction: tan v   220  2, 5  3 2  4 5  26; no 4, 6  
3 4  2 6  0; yes 2, 
3  
5 2  3 (
3)  
19; no 
2, 
3  8 (
2)  6 (
3)  
34; no 4, 
3, 6  3 4  4 (
3)  0 6  0; yes 
1, 
2, 3  4 (
1)  5 (
2)  1 3  
11; no 1, 1, 2  u i u j u k    v  42.3° u2  502  502 
 2(50)(50) cos 120° 2. r u2  7500 r u  86.60 mph; 30° r u 3. r 2  3502  2802 
 2(350)(280) cos 135° u  339,492.9291 r u  582.66 N r 582.66  sin 135°  280 sin 135°   sin v   582.66  sin v  0.34 v  19.9° u 902  1102 4. r   
 u  142.13 N r    1 0 3 1 1 2 u u  
3i j u k  
3, 1, 1 
3, 1, 1  280    sin v   
3, 1, 1 (1, 0, 3  
3 1  1 0  1 3  0; yes 1, 1, 2  
3 1  1 1  1 2  0; yes  90 N  110 N  521  Extra Practice  2c. y  5  75t sin 25° 
 16t2  Lesson 8-6  2.13   20.6 ft  Page A41 1. x 
 2, y 
 3  t 1, 0 x
2t y
30 x2t y3 2. x 
 (
1), y 
 (
4)  t 5, 2 x  1, y  4  t 5, 2 x  1  5t y  4  2t x  
1  5t y  
4  2t 3. x 
 (
3), y 
 6  t 
2, 4 x  3, y 
 6  t 
2, 4 x  3  
2t y 
 6  4t x  
3 
 2t y  6  4t 4. x 
 3, y 
 0  t 0, 
1 x
30 y 
 0  
t x3 y  
t 5. x  3t y2t y
2 y  x  3 1 x 3      1. The figure is 4 times the original size and reflected over the xy-plane. 2. The figure is half the original size. 3. The figure is 1.5 times the original size and reflected over the yz-plane.  Lesson 9-1  x  →  t  3  →  ty
2 1.  →  t  →  t  x1  2 y  4  →  yt
1 → 1 10 y  1  3x  3  180˚    t  2 3  K  1 2 3 4  0˚  330˚  x  10  3  240˚  300˚  270˚  3.  90˚  120˚  7  3  N  150˚  330˚ 240˚  5.  2 3  300˚  270˚  Page A41 1. vy  70 sin 34°  39.14 yd/s 70 yd/s vx  70 cos 34° 34˚  58.03 yd/s u 2a. x  t v  cos v  75t cos 25° 1 u y  t v  sin v  2gt2  h   2  0   5  75t sin 25° 
 16t2 2b. y  0 when t  ?  7.  2 
 4( 
75 sin 25°  
 (75 si n 25°)  16)(5 )  2(
16)  2 3  3 2  2  t  0  11 6 4 3  522  3 2  P  5 3  11 6  120˚  3 2 90˚  5 3  60˚ 30˚ 0˚  1 2 3 4  330˚  210˚ 240˚  2 3  270˚  2  300˚   3  6  5 6  1 2 3 4  7 6  0 1 2 3 4  7 6  8.   3  6    
0.1468595989 or t  2.127882701 t  2.13 s x  75t cos 25°  75(2.13) cos 25°  145 ft   3  6    5 3  5 6   2  5 6  180˚  11 6 4 3  5 3  3 2  150˚  1 2 3 4  7 6  1  2 3  6.  6    11 6  4 3   3  5 6   75t sin 25°  2(
32)t2  5  0˚  0  1 2 3 4  7 6  30˚  210˚  M    4.  1 2 3 4   3  6  4 3  60˚   2  5 6  ty1  Lesson 8-7  Extra Practice  2. 30˚  180˚  t  60˚  210˚  7. x  3t 
 10  1 x 3  90˚  120˚ 150˚  1  2  y  2x  2  y  Page A41  2  y  4t x  2  Lesson 8-8  Page A42  6. x  
1  2t y  4  2.13 2   5  752 sin 25° 
 162  0    1 2 3 4  7 6  11 6 4 3  3 2  5 3  9. r  
5  or r  

5   6. x  5 cos 240°    Lesson 9-2  90˚  120˚  2.  60˚  30˚  150˚ 180˚  1 2 3 4  0˚  330˚  210˚ 240˚  270˚  300˚  90˚  0    5 10 15 20 11 6  7 6 3 2  5 3  30˚ 1 2 3 4  240˚  270˚  Lesson 9-4  300˚  cardioid  Page A42 1. 

 A2   B2  

 62  ( 
5)2  

61   Lesson 9-3  6 
61   1. r  
 12  ( 
1)2   
2  
1  v  Arctan 1  2   7  4  
2,   v  Arctan 3 0  22  ( 
2  )2 3. r  
  
6  or about 2.45 (2.45, 0.62)  v  Arctan 2  0.62   2 2    5. x  4 cos 2 1   3
10   
 3   4  1  
 3  1  0  2 x  2 y 
 8    x  2y 
 16 0  
3 4. 1  r cos (v  ) 0  r (cos v cos  
 sin v sin ) 
 1 0  
r cos v 
 0 
 1 0  
x 
 1 x1  1  0  3
 10  0  2 r cos v  2 r sin v 
 8  y  4 sin 2  4 (1)  90  
 10   
2    4 (0)  9   y 
   0  3
10  3
10   cos f  1 , sin f  10, p  3
10  0 f  Arctan 3  72° 3
10   r cos (v 
 72°) 3. 8  r cos (v 
 30°) 0  r(cos v cos 30°  sin v sin 30°) 
 8  
2   1  6
61  61  6
61    r cos (v 
 140°) 61 
 A2   B2  
 32  92 3  x 3
10      2 2   
2  
2, 
2  2.  
2   y  2 sin 4  5 
61   sin f  , p   5  0  4. x  2 cos 4 
2   6 
 61  f  Arctan 
6  180°  140°  7  4  2. r  
 32  02  
9  or 3 (3, 0)    5 
61  6 
, 
61   
x  y 
   0 cos f   Page A42  1  6   r sin v  0˚  330˚  210˚  8. y  6 r sin v  6  r  
2 sec v r  6 csc v 2 2 9. x  y  36 (r cos v)2  (r sin v)2  36 r2 (cos2 v  sin2 v)  36 r2  36 r  6 or r  
6 10. x2  y2  3y 2 (r cos v)  (r sin v)2  3r sin v r2 (cos2 v  sin2 v)  3r sin v r2  3r sin v r  3 sin v 11. r  4 12. r  4 cos v r2  16 r2  4r cos v x2  y 2  16 x2  y2  4x   6  5 6  4. Sample answer: r  sin 5v  60˚  150˚ 180˚  3 5
   
2 or about 
4.33   r  
 cos v   3  spiral of Archimedes 120˚   5 
2  2   2  2 3  4 3  circle 3.  
 3     (
2.5, 
4.33) 7. x  
2 r cos v  
2  Page A42 1.  y  5 sin 240°  1 5 
2 5 
2 or 
2.5  1  0, 14  523  Extra Practice  5.  90˚  120˚  Lesson 9-6  60˚ 30˚  150˚ 180˚  1 2 3 4  Page A43  0˚  1. 4x 
 6yi  14  12i 4x  14  330˚  210˚ 240˚  270˚  x  300˚  3.  3  3. i1000  (i4)250  1250 1  (0, 5)  42  12 z  
  
17   4. i12  i4  (i4)3  (i4)
1  13  11 2  4.  11.  (i 
 2)2  4  2i     i sin 4  1  42  ( 

2 )2 v  Arctan 
4  2 7. r  
  
18  or 3
2   5.94 4 
 
2 i  3
2  (cos 5.94  i sin 5.94) 8a. 5(cos 0.9  i sin 0.9)  3.11  3.92j 8(cos 0.4  j sin 0.4)  7.37  3.12j 8b. (3.11  3.92j)  (7.37  3.12j)  (3.11  7.37)  (3.92j  3.12j)  10.48  7.04j ohms 
2   
2 
 i  
2 
 i 
12 
 10i 
 2i2  4 
 i2 
10 
 10i  5   8c. r  
(10.48  )2  (7 .04)2 v  Arctan  10.48   12.63  0.59 10.48  7.04j  12.63 (cos 0.59  j sin 0.59) ohms  i2 
 4i  4  7.04    4  2i 3 
 4i  4 
 2i  4 
 2i 12 
 22i  8i2  16 
 4i2 4 
 22i  20 1 11  
  5 10 i    4  2i       4    4   6. r  
 (
2)2   12 v  Arctan  
2     
5   2.68 
2  i  
5 (cos 2.68  i sin 2.68)  6  2i    4  4  4i  4
2 cos    
2  i    v  Arctan 4   
32  or 4
2     
2 
 2i 12.    5. r  
 42  42  1i  1i 4  5i  i2  1 
 i2 3  5i  2 3 5   i 2 2    z  
 22  ( 

3 )2  
7   i  (2, 
3)  4i    z  
 02  ( 
5)2  
25  or 5  O    1
i  6  2i  
2  i    4  2. i17  (i4) i  14 i i  5. (4 
 i)  (
3  5i)  (4  (
3))  (
i  5i)  1  4i 6. (6  6i) 
 (2  4i)  (6  (
2))  (6i  (
4i))  4  2i 7. (3  i)(5 
 3i)  15 
 4i 
 3i2  18 
 4i 8. (2  5i)2  (2  5i)(2  5i)  4  20i  25i2  
21  20i 9. (1 
 
2 i)(
3 
 
8 i)  
3 
 
8 i  3
2 i  
16 i2  
3 
 2
2 i  3
2 i  4i2  
7  
2 i      O  O    i  (4, 1)  1. i10  (i4) i2 3 1 (
1)  
1  4i  1
i  y  
2  i  Lesson 9-5  10.  12  y  
6  x  3.5 2.  Page A43  
6y  12  14  4  Lesson 9-7 Page A43    1. r  6 4 or 24    v  2  4 3  3  3     4    24cos 4  i sin 4  24 
2  i2 
2   
2    
12
2   12
2 i Extra Practice  524  3 2. r   1 or 6  2  6(cos  7  4  3. d  
 (x2 
  x1)2  (y2 
 y1)2  3  v  4    d  
 (r 
 r )2  (
 2 
 6 )2  7   4  i sin  7 ) 4    6   i  
2   2  
2  
2  d  
 02  ( 
8)2 d  
64  or 8    x1  x2 y1  y2  r  r 6  (
2)   2, 2   2 , 2   3
2  
 3
2 i 3. r  5 2 or 10 v  135°  45°  180° 10(cos 180°  i sin 180°)  10(
1  i(0))  
10  4.      1. 44 cos (4)2  i sin (4)2  256(cos 2  i sin 2)  256(1  i(0))  256  2. r   v  Arctan   12  
5  
 122  (
5)2    v  Arctan    
2  
2   1  3      4  6a.    1  1    1       (5, 2)  x  0  50 0  40 ,  2 2      (25, 20)  v 1    (2, 5)  d  
 252  202 d  
1025  d  5
41  or about 32 ft  Lesson 10-2    1 5 cos 5()  i sin 5() 1  (5, 10) (2, 8)  d  
 (25 
 0)2  (20 
 0)2   
2  3 cos 12  i sin 12  1.08  0.29i (
1)2   02 4. r  
 1  y  6b. d  
 (x2 
  x1)2  (y2 
 y1)2  cos 134  i sin 1312  1   2  y2  2  O   13  1.965587446 133(cos (3)(v)  i sin (3)(v))  2035 
 828i 12  12 3. r  
    (
5, 8)  5. A quadrilateral is a parallelogram if both pairs of opposite sides are parallel. Since only one pair of opposite sides are parallel, the quadrilateral is not a parallelogram.        8 2  y2  16 6  x2  
10 x2  
16 y2  14 Then A has coordinates (
16, 14).  Lesson 9-8 Page A43   (r, 2)  6  x2 2  y2   2 , 2 6  x2   
5 2     1cos 5  i sin 5  Page A44 1.   0.81  0.59i  Lesson 10-1  (x 
 h)2  (y 
 k)2  r2 2 [x 
 (
2)]2  (y 
 2)2  
2  (x  2)2  (y 
 2)2  2 y (x  2)2  (y  2)2 
 2  Page A44 1. d  
 (x2 
  x1)2  (y2 
 y1)2 d  
 (4 
 ( 
2))2   (5 
 2)2  O  x  
 62  32  d d  
45  or 3
5   2.  x1  x2 y1  y2  
2  4 2  5 ,  2, 2   2 2   (1, 3.5) 2. d  d  
 (x2 
  x1)2  (y2 
 y1)2 
 (8 
 ( 
3))2   (
1 
 6)2  d  
 112  (
7)2 d  
170   (x 
 h)2  (y 
 k)2  r2 (x 
 0)2  (y 
 (
4))2  42 x2  (y  4)2  16 y O x x 2  (y  4)2 
 16  x1  x2 y1  y2  
3  8 6  (
1)  ,  2, 2   2 2   (2.5, 2.5)  525  Extra Practice  3. x2  49 
 y2 → x2  y2  49 y2 2 8  7. r  
 (4 
 2 )2  (0 
 (
 5))2 r  
 22  52  x  y 
 49  r  
29 ; r2  29 (x 
 4)2  (y 
 0)2  29 (x 
 4)2  y2  29  4 
8 
4 O 
4  4  8x  Lesson 10-3  
8  4.    6x 
 8y  18  0 (x2  6x  9)  (y2 
 8y  16)  
18  9  16 (x  3)2  (y 
 4)2  7 y 2 2 x2  y2  Page A44 1. center: (h, k)  (0, 0) 10  (x  3)  (y  4) 
 7  a  2 or 5 6  b  2 or 3 (x 
 h)2  a2 (x 
 0)2  52  O  (y 
 0)2   3 1 2 x2  25  x  y2   9  1  c2  a2 
 b2 c2  25 
 9 c2  16 c4 foci: (4, 0) 2. center: (h, k)  (
2, 1)  5. x2  y2  Dx  Ey  F  0 22  (
2)2  2D 
 2E  F  0  → 2D 
 2E  F  
8 02  (
4)2  0(D) 
 4E  F  0 → 
4E  F  
16 (
2)2  (
2)2 
 2D 
 2E  F  0 → 
2D 
 2E  F  
8 2D 
 2E  F  
8 2(2 0 
 2E  F )  2(
8) 
2D 
 2E  F  
8 
 4E  F )  
16 4D  
0 F0 D0 
4E  (0)  
16 
4E  
16 E4 x2  y2  4y  0 center: (0, 
2) x2  (y2  4y  4)  0  4 radius: 2 x2  (y  2)2  4 2 6. x  y2  Dx  Ey  F  0 (
1)2  32 
 D  3E  F  0 → 
D  3E  F  
10 (
4)2  62 
 4D  6E  F  0 → 
4D  6E  F  
52 (
7)2  32 
 7D  3E  F  0 → 
7D  3E  F  
58 2(
D  3E  F )  2(
10) 
4D  6E  F )  2(
52 
4D  6E  F )  2(
52 2(
7D  3E  F )  2(
58) 2D  F )  2(
32 10D 
 F )  2(
64 2D  F  32 
(8)  3E  F  
10 10D 
 F  64 
4(8)  6E  F  
52 12D  96 24) 
 3E  
42 D8 
3E  18 E  
6 
(8) 3(
6)  F  
10 
26  F  
10 F  16 x2  y2  8x 
 6y  16  0 (x2  8x  16)  (y2 
 6y  9)  
16  16  9 (x  4)2  (y 
 3)2  9 center: (
4, 3) radius: 3 Extra Practice  (y 
 k)2   b 1 2  8  a  2 or 4 4  b  2 or 2 (y 
 k)2 (x 
 h)2   b 2 a2 (y 
 1)2 [x 
 (
2)]2     42 22 (y 
 1)2 (x  2)2    16 4 2 c  a2 
 b2  1 1 1  c2  16 
 4 c2  12 c  
12  or 2
3  foci: (
2, 1  2
3 ) 3. The major axis contains the foci and it is located on the x-axis. y center: (h, k)  (4, 1) 8 (x  4)2 (y  1)2  
1 c2  a2  b2 64 16 c2  64  16 4 c2  80 c  
80  or 4
5  4 8 12x 
4 O 
4  foci: (h  c, k)  (4  4
5 , 1) major axis vertices: (h  a, k)  (4  8, 1)  (12, 1) and (
4, 1) minor axis vertices: (h, k  b)  (4, 1  4)  (4, 5) and (4, 
3)  526  Lesson 10-4  Lesson 10-5  Page A45  Page A45 1. vertex  (h, k)  (0, 0) 4p  4 p1 focus: (h  p, k)  (0  1, 0) or (1, 0) directrix: x  h 
 p x0
1 x  
1 axis of symmetry: y  k y0  y  1.  (1, 2  
 58)  8 y  2 
  73 (x  1) 4  y  2 
 73 (x  1) (1, 2)  4 8 x (1, 2  
 58) 
4  
4 O  y  y  2.  8  y 2 
 4x  4 
8 
4 O 
4  x  O  8x  4  xy 
 16  
8  3. center: (h, k)  (
4, 3) (x 
 h)2  a2 [x 
 (
4)]2  32 (x  4)2  9  
 
 
  (y 
 k)2  b2 (y 
 3)2  22 (y 
 3)2  4  2. x2 
 4x  4  12y 
 12 (x 
 2)2  12(y 
 1) vertex  (h, k)  (2, 1) 4p  12 p3 focus: (h, k  p)  (2, 1  3) or (2, 4) directrix: y  k 
 p y1
3 y  
2 axis of symmetry: x  h x2 y  1 1 1  4. transverse axis: x  h  2 foci: (h, k  c)  (2, 7) k  c  
7 (h, k 
 c)  (2, 
3) k 
 c  
3 2k  
4 k  2; c  5 vertices: (h, k  a)  (2, 5) 2a5 (h, k 
 a)  (2, 
1) a3 a2  b2  c2 32  b2  52 b2  16 b4 (y 
 k)2  a2 (y 
 2)2  32 (y 
 2)2  9  x 2  4x  4 
 12y  12  (x 
 h)2  
 b 1 2 
 
  (x 
 2)2  42 (x 
 2)2  16  O  x  1 3. vertex: (h, k)  (
2, 3) focus: (h  p, k)  (0, 3) h  p  0, k  3 
2  p  0 p2 (y 
 k)2  4p(x 
 h) (y 
 3)2  4(2)[x 
 (
2)] (y 
 3)2  8(x  2) 4. directrix: y  k 
 p  
3 focus: (h, k  p)  (0, 
2) k 
 p  
3 
2.5  p  
2 k  p  
2 p  0.5 2k  
5 k  
2.5 (x 
 h)2  4p(y 
 k) (x 
 0)2  4(0.5)[y 
 (
2.5)] x2  2(y  2.5)  1  527  Extra Practice  4. B2 
 4AC  02 
4(4)(
25)  0  400 or 400 Since 400 0, the graph is a hyperbola. 4x2 
 25y2  64 
 
 2 4(x cos 90°  y sin 90°) 
 25(
x
 sin 90°  y
 cos 90°)2  64 4(0  y
)2 
 25(
x
  0)2  64 4(y
)2 
 25(x
)2 
 64  0 2 2  
 4(1)(2) 5. B 
 4AC  
2
2  8 
 8 or 0 parabola  Lesson 10-6 Page A45 1. A  1, C  1; since A  C, the conic is a circle. x2  y2 
 8x  2y  13  0 2 (x 
 8x  16)  (y2  2y  1)  
13  16  1 (x 
 4)2  (y  1)2  4 2. A  1, C  
 4; since A and C have different signs, the conic is a hyperbola. x2 
 4y2  10x 
 16y  
5 2 (x  10x  25) 
 4(y2  4y  4)  
5  25 
 16 (x  5)2 
 4(y  2)2  4 (x  5)2  4  B   tan 2v   A
C 
2
2   (y  2)2   tan 2v   1
2  
 1  1 3. A  0, C  1; since A  0, the conic is a parabola. y2 
 5x 
 6y  9  0 (y2 
 6y  9)  5x 
 9  9 (y 
 3)2  5x 4. A  1, C  2; since A and C have the same sign x2  2y2  2x  8y  15 (x2  2x  1)  2(y2  4y  4)  15  1  8 (x  1)2  2(y  2)2  24 (x  1)2  24  tan 2v  2
2  2v  70.52877937° v  35, 
35° 6. B2 
 4AC  52 
 4(15)(5)  25 
 300 or 
275 Since 
275  0 and A C, the graph is an ellipse. B   tan 2v   A
C 5   tan 2v   15 
 5  (y  2)2   12  1  1  tan 2v  2 2v  26.56505118° v  13°  Lesson 10-7 Page A45 1. B2 
 4AC  02 
 4(1)(1)  0 
 4 or 
4 Since 
4  0 and A  C, the graph is a circle. x 2  y2  9 2 (x 
 1)  [y 
 (
1)]2  9 (x 
 1)2  (y  1)2  9 x2 
 2x  1  y2  2y  1  9 x2  y2 
 2x  2y 
 7  0 2 2. B 
 4AC  02 
 4(4)(1)  0 
 16 or 
16 Since 
16  0 and A C, the graph is an ellipse. 4x2  y2  16 4[x 
 (
3)]2  [y 
 (
2)]2  16 4(x  3)2  (y  2)2  16 2 4(x  6x  9)  (y2  4y  4)  16 4x2  y2  24x  4y  24  0 2 3. B 
 4AC  02 
 4(49)(
16)  0  3136 or 3136 Since 3136 0, the graph is a hyperbola. 49x2 
 16y2  784  2    Lesson 10-8 Page A45 1. xy  3 y  x 2 
 y2  8  3  x  x2  3 2  
 x  8 9  x2 
 x2  8 x4 
 9  8x2 
 8x2 
 9  0 2 (x 
 9)(x2  1)  0 x2 
 9  0 or x2  1  0 x2  9 x2  
1 x  3 x  

1  or i x4  3  If x  3, then y  (3) or 1. 3   If x  
3, then y   (
3) or 
1.   is an imaginary number, disregard Since x  

1 this solution. y (3, 1), (
3, 
1)    49 x
 cos 4  y
 sin 4 
 16 
x
 sin 4   2  2 
2  
2  49 2x
  2y
 
 16 1 1 49 2(x
)2  x
y
  2(y
)2    49 (x
)2 2     49x
y
    y
 cos 4  784  2 
2  
2  
2x
  2y
  1 
 16 2(x
)2 
 x
 y
 1  2(y
)2  49 (y
)2 
 8(x
)2  16x
y
 2 
 8(y
)2       (3, 1)  O  784 (3, 1)  784  784 49(x
)2  98x
y
  49(y
)2 
 16(x
)2  32x
y
 
16(y
)2  1568 33(x
)2  130x
y
  33(y
)2 
 1568  0 Extra Practice  528  x  2. x 
 y  4 xy4  3  11. ((3f)2)  (3f)6  x2  10y2  10 (y  4)2  10y2  10 y2  8y  16  10y2  10 9y2 
 8y 
 6  0  1    (3f)6 1    36 f 6 1  
(
8)  
 (
8)2  
 4(9 )(
6)   y    729f 6  8  2
70   12.  y  1 8  
3a  2  cc  4a  4  
70   y  9 4  
70   c 6 a   c8 a  c
14a 1    c14a  4 
 
70   y  9 or y  9  1   1   6   
1  h   216h  6  14.  y  3  3  
1     216  h9  3  
9  8 4  6   13. (2n 3 3n 2 )6  26n 3 36n 2  64n2 799n3  46,656n5  y  1.4 y  
0.5 If y  1.4, then x  (1.4)  4 or 5.4. If y  
0.5, then x  (
0.5)  4 or 3.5. (5.4, 1.4), (3.5, 
 0.5)  h 3  1 
 216 3  (5.4, 1.4)  1   216 3   h3  
8 
4 O 4 8x 
4 (3.5, 0.5)  3 
  216  h 3  
8  6   h3  Lesson 11-1  15.  3  1   z4(z4) 2  
 z4 z2 3   
 z6 3  Page A46 1.  6   (
12)2     1  (
12)2 1  144  
122  2.  3. (4 6)3  43 63 64 216  13,824  4  23  4.   z3  z2  1   
 122 1  
 144 24   34 16   81  3   3   4m3n2 12 2    19. 15
 r12t2  15r 3 t 3 3  1  2  1  2  1  2  1  2  (32)  (22 5) 1  2  1   (22) 2  3  2   15r4t 3  1  2  5  1   2   1   1   16   20. 
 256x2 y16  256 8 x 3 y 8 8  1  2  2  6   1   20  (32 3)   3 2 15  6
15    (28) 8 x 4 y2 1   2   7. 
625   625 4 4  9    (43) 3 m3n2  16    1   18. 
 64m9n6  64 3 m 3 n 3   4 or 4 6. 27  1 4 8     5   17. 
 a3b5  a 2 b 2  16 16 5. 1  1   16 2 (42) 2  1  2  1   16. (4r2t5)(16r4t8) 4  4r2t5(16 4 r 4 t 4 )  (4r2t5)(2rt2)  8r3t7   2x 4 y2  1    625 2  
625  or 25 1 1 8.   6 3  
 (15)6 15 3  Lesson 11-2  1    152   9.  (2a4)2  10. (x4)3  Page A46  1  225  1.  y    4a8 x5  x12 x5  x17 22  y  2.  (a4)2  y 
 3x  O  529  y 
 3x  x  O  x  Extra Practice  y  3.  9. log36 6  x 36x  6 (62)x  6 62x  61 2x  1 1 x  2 10. log3 y  4 34  y 81  y 11. log5 r  log5 8 r8 12. log5 35 
 log5 d  log5 5  x  O  y 
 3x  1  Lesson 11-3  35  log5 d  log5 5  Page A46  35  d  5 7d  1. p  (100 
 a)e
bt  a p  (100 
 18)e
0.6(2)  18  42.7% 2. y  ae
kt  c y  140e
0.01(10)  70  197° F 3a. y  6.7e  13. log4 
4 x 1   log4 4 2  x 1  log 4 2  
48.1  t 
48.1   y  6.7e 15 y  0.271292 millions of cubic feet y  271,292 ft3 3b. y  6.7e  
48.1  t  1  
48.1   15. 4 log8 2  3 log8 27  log8 a    1   log8 24  log8 27 3  log8 a log8 16  log8 3  log8 a log8 48  log8 a 48  a  r nt    A  Pert  A  P 1  n  A  5000e0.058(20)  A  5000 1  2    0.058 2(20)    A  $15,949.67 A  $15,688.63 The account that compounds continuously would earn $261.04 more than the account compounded semiannually.  Lesson 11-5 Page A47 1. log 5000  log (5 1000)  log 5  log 103  log 5  3 log 10  0.6990  3  3.6990 2. log 0.0008  log (0.0001 8)  log 104  log 8  
4 log 10  log 8  
4  0.9031  
 3.0969 3. log 0.14  log (0.01 14)  log 102  log 14  
2 log 10  log 14  
2  1.1461  
0.8539  Lesson 11-4 Page A47 1   1. 16 4  2 1 3  2  8 1  4  41   3. 4. log8 x  
2 5. logx 32  5 6. log1 16  
2 4  1  7. log5 5  x  log 81   4. log3 81   log 3  1  5x  5 5x  51 x  
1 8. log3 27  x 3x  27 3x  33 x3 Extra Practice  x  14. log4 (2x  3)  log4 15 2x  3  15 2x  12 x6  y  6.7e 50 y  2.560257 millions of cubic feet y  2,256,275 ft3 4. Continuously Semiannually  2.  4x 1  2  4 log 12   5. log6 12   log 6   1.3869 log 29   6. log5 29   log 5   2.0922  530  7.  3x  45 x log 3  log 45  12.  log 45   x log 3  8.  x  3.4650 6x  2x 
 1 x log 6  (x 
 1) log 2 x log 6  x log 2 
 log 2 x log 6 
 x log 2  
log 2 x (log 6 
 log 2)  
log 2  16  10(1  ex) 1.6  1  ex 0.6  ex ln 0.6  x ln e 
0.5108  x  Lesson 11-7  
log 2   x log 6 
 log 2  Page A47  x  
0.6309 9. 5 log y  log 32 log y5  log 32 y5  32 y5  25 y2  ln 2   1. t   0.045   15.40 yr ln 2   2. t   0.06   11.55 yr ln 2   3. t   0.08125  Lesson 11-6   8.53 yr 4a. y  5.2449(1.5524)x 4b. y  5.2449(eln 1.5524)x y  5.2449e0.4398x  Page A47  4c. Use t  k; k  0.4398  ln 2  1. 3.5553 2. 
0.5763 3. 3.4398  ln 2   t 0.4398   1.58 hr  4. log15 10   ln 10  ln 15   0.8503  Lesson 12-1  ln 14   5. log3 14   ln 3   2.4022 ln 350  Page A48   6. log8 350   ln 8  1. d  3 
 7 or 
4 
1  (
4)  
5, 
5  (
4)  
9, 
9  (
4)  
13, 
13  (
4)  
17 
5, 
9, 
13, 
17 2. d  
1 
 0.5 or 
1.5 
2.5  (
1.5)  
4, 
4  (
1.5)  
5.5, 
5.5  (
1.5)  
7, 
7  (
1.5)  
8.5 
4, 
5.5, 
7, 
8.5 3. d  
8 
 (
14) or 6 
2  6  4, 4  6  10, 10  6  16, 16  6  22 4, 10, 16, 22 4. d  2.8 
 3 or 
0.2 2.6  (
0.2)  2.4, 2.4  (
0.2)  2.2, 2.2  (
0.2)  2, 2  (
0.2)  1.8 2.4, 2.2, 2, 1.8 5. d  
x 
 4x or 
5x 
6x  (
5x)  
11x, 
11x  (
5x)  
16x, 
16x  (
5x)  
21x, 
21x  (
5x)  
26x 
11x, 
16x, 
21x, 
26x 6. d  (2y 
 2) 
 (2y 
 4)  2y 
 2y  (
2) 
 (
4) 2 2y  2, 2y  2  2  2y  4, 2y  4  2  2y  6, 2y  6  2  2y  8 2y  2, 2y  4, 2y  6, 2y  8   2.8171 7. 5x  90 x ln 5  ln 90 ln 90   x ln5  8.  x  2.7959 7x  2  5.25 (x  2) ln 7  ln 5.25 x ln 7  2 ln 7  ln 5.25 x ln 7  ln 5.25 
 2 ln 7 ln 5.25 
 2 ln 7   x ln 7  9.  x  
1.1478 4x  4
3  x ln 4  ln 4
3  ln 4
3    x ln 4  x  1.3962 6ex  48 ex  8 x ln e  ln 8 x  2.0794 11. 50.2  e0.2x ln 50.2  0.2x ln e  10.  ln 50.2  0.2  x  x 19.5801  531  Extra Practice  a8  7. an  a1  (n 
 1)d a16  2  (16 
 1)5  77 8. 
20  6  (n 
 1)(
2) 
26  
2n  2 
28  
2n 14  n 9. 42  a1  (12 
 1)4 42  a1  44 
2  a1 10. 30  7  (13 
 1)d 23  12d 11 1 12  d  1    6. r   a10 or a 2  a6 a2   a4, a4a2   a2, a2 a2   1 1  1  1  a4, a2, 1 7. an  a1rn
1 a6  9(2)6
1  288 8. 100  a1(4)8
1 100  16,384 a1 25  4096   a1 1 5
1  9. 10  a1
2 1  10  1 6 a1  11. d  10 
 10.5 or 
0.5 a24  10.5  (24 
 1)(
0.5) a24  
1  160  a1  a2  160 
2 or 
80 1  a3  
80 
2 or 40 1  n  12. Sn  2[2a1  (n 
 1)d]; d  2.8 
 2 or 0.8  160, 
80, 40 10. 256  4r4
1 64  r3 4r 4(4)  16, 16(4)  64 4, 16, 64, 256  12  S12  2[2 2  (12 
 1)(0.8)]  6(4  8.8)  76.8 n  13. Sn  2[2a1  (n 
 1)d] n  80  2[2 (
4)  (n 
 1)4]  a1 
 a1rn  9    11. Sn   1
 r ; r  3 or 3  160  n(
8  4n 
 4) 160  
12n  4n2 0  4(n2 
 3n 
 40) 0  4(n 
 8)(n  5) n  8 or n  
5 Since n cannot be negative, n  8.  3 
 3(3)6   S6   1
3 3 
 2187    
2   1092 12a. There are four 15-minute periods in an hour, so r  24 or 16. bt  b0 16t 12b. b4  12 164  786,432  Lesson 12-2 Page A48 7  1. r  1 4 or 0.5  Lesson 12-3  3.5(0.5)  1.75, 1.75(0.5)  0.875, 0.875(0.5)  0.4375 1.75, 0.875, 0.4375  Page A49  4   2. r   
2 or 
2  1. lim  n→  
8(
2)  16, 16(
2)  
32, 
32(
2)  64 16, 
32, 64 3. r   3  8  1  2  3  or 4  2n  n→ 3n   lim  n4 
 3n  n3 n→   lim n 
 n2  3  n→   lim n. But as n approaches infinity, n becomes     392 , 392 34  12278 , 12278 34  58112  n→  increasingly large, so there is no limit.  9 27 81 , ,  32 128 512 
5 r  10 or 
0.5  3. lim  n→  2.5(
0.5)  
1.25, 
1.25(
0.5)  0.625, 0.625(
0.5)  
0.3125 
1.25, 0.625, 
0.3125  4.  8
2   5. r  8 or 
2  16
2   16
2 , 16
2 
2   32, 32
2   32
2   8n2  6n 
 2  4n2  4n2 
 2n  1  lim  n2  2 n→  8n3  2 n→ 4n   lim  4n2  n2  2n  1  
2  2 n→ 4n   lim  
 n2  n2 ––––  lim n2 2 n→   n2 n2 4
00  532  6n  2 n→ 4n   lim  200 2    10 4  16
2 , 32, 32
2   Extra Practice  4  n→ 3n 2  0  3 2  3   lim  2. Limit does not exist. lim  3 3   8 4  4.  4  2n  3n  5.  n 3 
 n2  4  lim  3 n→ 5  2n  n3  n3  n2  2. an  3, an1  3 r  1
00    02 1   2 6. Limit does not exist.  lim (2n  1   3. an   3n(n  1)3 , an1   n→  becomes increasingly large, so there is no limit. 9  9  — —— 7. 0.0 9 — 100  10,000  . . . 9  r  1  — —— a1  — 100 , r  100 9   100 Sn  — 1 — 1
— 100 1  3  3    8. 0.13   1 0  100  1000  . . . 3  1    a1   100 , r  10  1  convergent  1 — Sn  
 1 1 0  1 
 1 0 2  407  r  407    9. 7.4 0 7 7 1000  1,000,000  . . . 1    a1   1000 , r  1000 407   1000 Sn  7  —— 1  1
 1000  7   5. The general term is  6n  1 .  11  10. a1   r  7  6n  1  or  or  for all n, so divergent 1    6. The general term is  (2n)2 or 4n2 . 1  4n2  1   n for all n, so convergent. 1  1    7. an   2n  1 , an1  2n1  1  Sn  — 1 1 
 2   1  n  1  1  2  1  20  2  20  4   2n1  lim — 4 n→  2n 2n  lim  n n→ 2 2 1  2  convergent   727 1 , 20  4  4   4. an  2n , an1   2n1   1 5  1  40 — 1  20  1   3n1(n  2)3 ––––  lim 1 n→   3n(n  1)3 n 3 (n3  3n2  3n  1)  lim  3n 3(n3  6n2  12n  8) n→   3  3  100  407  1  3n1(n  1  1)3  n3 3n2 3n 1     n 3  n 3  n3  n 3 ———  lim 3n3 6n2 12n 8     n→  n 3  n3  n 3  n 3 1000   3000   —11— 1  n1  3 —  lim n  n→ 3 n1  lim n n→ 1 n  lim n  lim n n→ n→  10 1 test provides no information  1 n n   —  lim 2n —  2 2  n   lim 2 n→ n→ ——  1 n 1) or lim 2n As n approaches infinity, 2n  2n n lim —— n→  2  n  n→  n1  n  4  
 n3  n3  lim  5 2n3 n→    n3 n3  1  10  r  11. Sn does not exist. This series is geometric with a common ratio of 2. Since this ratio is greater than 1, the sum of the series does not exist.  Lesson 12-4 Page A49  1   2n1  1 ——  lim n→ 1 2n  1 2n  1   lim 2n 2 1 n→ 2n 1    2n 2n ——  lim 2n 2 1 n→    2n 2n 10   20 1  2  convergent  1. an  (2n)2, an1  (2n1)2  22n  22n2 22n2 2 n n→ 2 2 n 2 22   lim  22n n→  r  lim  4 divergent  533  Extra Practice  2n1  2n  Lesson 12-6    8. an   n  2 , an1  (n  1)  2  r  2n1  n3  lim — n n→ 2 n2 2n 2(n  2)   lim  n n→ 2 (n  3) 2n  2   lim  n→ n  3 2n 2    n n  lim — n→ n  3 n n 20   10  Page A50 1. (2  x)4  16  32x  24x2  8x3  x4 2. (n  m)5  n5  5n4m  10n3m2  10n2m4  5nm4  m5 3. (4a 
 b)3  64a3 
 48a2b  12ab2 
 b3 6 5 m4(
3)2   4. (m 
 3)6  m6(
3)0  6 m5 (
3)1   2 1  2 divergent    6 5 4 m3 (
3)3  3 2 1    6 5 4 3 2 m1(
3)5  5 4 3 2 1    6 5 4 3 2 1 m0(
3)6  6 5 4 3 2 1    6 5 4 3 m2 (
3)4  4 3 2 1   m6 
 18m5  135m4 
 540m3  1215m2 
 1458m  729 4 3(2r)2(s)2   5. (2r  s)4  (2r)4(s)0  4(2r)3(s)1   2 1  Lesson 12-5  4 3 2 (2r)1(s)3  4 3 2 1(2r)0(s)4  4 3 2 1 24r2s2  8rs3  s4    3 2 1   16r4  32r3s   Page A49  6. (5x 
 4y)3  (5x)3(
4y)0  3(5x)2 (
4y)1  5  1.    n1  3 2(5x)1(
4y)2  (3n 
 1)  (3 1 
 1)  (3 2 
 1)  (3 3 
 1)  (3 4 
 1)  (3 5 
 1)  2  5  8  11  14  40  3 2 1(5x)0(
4y)3  3 2 1  240xy2 
 64y3    2 1   125x3 
 300x2y  8 7 6 5 4 3 2 1  5 4 3 2 1 3 2 1  56x3y5  7.  8!  5!(8 
 5)!  x8
5y5   34 44 54 6  12  16  20  24  72  8.  7!  4!(7 
 4)!  3 9 b7
4 
 34   4 3 2 1 3 2 1 b 3  315b   (k2 
 2)  (32 
2 2)  (42 
2 2)  (52 
 2) k3  9.  10!  2!(10 
 2)!  6   4a  4 k3  2.  7  3.   (6 
 2)  (7 
 2)  7  14  23  34  47  125  8  4.  j  5  8   8 3 4 5 6  7  8  9 2683   3 9240  4  6  7 6 5 4 3 2 1  (4z)10
2 (
w)2 10 9 8 7 6 5 4 3 2 1  8 2   2 1 8 7 6 5 4 3 2 1 (4z) (
w) 8 2  45 65,536z w  2,949,120z8 w2        j3  43  53  63  73 j4 4  7  10.  12!  7!(12 
 7)!  (2h)12
7(
k)7     7  10    x3y5  8  11  12 11 10 9 8 7 6 5 4 3 2 1  7 6 5 3 2 1 5 4 3 2 1 792 32h5 (
k)7  (2h)5(
k)7    
25,344h5k7   3p  30  31  32  33  34 p0  5.   1  3  9  27  81  121    6.    2  n1  3 n  4        S    n1          Lesson 12-7 2  3 3  4     . .2  . . .  Page A50 1. ln (
3)  ln (
1)  ln (3)  i  1.0986 2. ln (
4.6)  ln (
1)  ln (4.6)  i  1.5261 3. ln (
0,75)  ln (
1)  ln (0.75)  i 
 0.2877  3   4     1 12 — 3 1 
 4  6  4  7.  3 1 3 2 2 4  2 4 3   2 4 1 1 7 12  18  23 2 .  (3n  2)  9   4m
1 m0  9.  Extra Practice  (1.2)2  10  8.    r2  (1.2)3  (1.2)4    4. e1.2  1  1.2  2 !  3!  4!  1  1.2  0.72  0.288  0.0864  3.29  
4r     w! w1  10.  (
0.7)2  (
0.7)3  (
0.7)4   3  4 5. e
0.7  1  (
0.7)  2 ! ! !  1 
 0.7  0.245 
 0.057  0.010  0.50  534  (3.65)2  (3.65)3  4. z0  4  4i z1  0.5(4  4i)  i  2  2i  i 2  3i z2  0.5(2  3i)  i  1  1.5i  i  1  2.5i z3  0.5(1  2.5i)  i  0.5  1.25i  i  0.5  2.25i 2  3i, 1  2.5i, 0.5  2.25i 5. p1  p0  rp0 p1  4000  (0.054)4000  4216 p2  4216  (0.054)4216  4443.66 p3  4443.66  (0.054)4443.66  4683.62 p4  4683.62  (0.054)4683.62  4936.54 p5  4936.54  (0.054)4936.54  5203.11 $4216, $4443.66, $4683.62, $4936.54, $5203.11  (3.65)4  6. e3.65  1  3.65  2  3  4 ! ! !  1  3.65  6.661  8.105  7.395  26.81 x2  x4  x6  x8  7. cos x  1 
 2!  4! 
 6!  8!   cos 4  cos 0.7854 (0.7854)2  (0.7854)4  (0.7854)6  (0.7854)8   4 
 6  8  1 
 2 ! ! ! ! 0.6169  0.3805  0.2347  0.1448     1 
 2  24 
  720  40,320   0.7071    
 2  x5  x7  actual value: cos 4  2  0.7071 x3  x9  8. sin x  x 
 3!  5! 
 7!  9!   sin 6  sin 0.5236 (0.5236)3  (0.5236)5  (0.5236)7   5 
 7  0.5236 
 3 ! ! ! (0.5236)9   9 ! 0.1435  0.0394  0.0108  0.0030      0.5236 
 6   120 
 5040  362,880   0.5000    actual value: sin 6  0.5   x2  x4  x6  x8  9. cos 3  1 
 2!  4! 
 6!  8!   cos 3  cos 1.0472 (1.0472)2  (1.0472)4  (1.0472)6  Lesson 12-9  (1.0472)8   4 
 6  8  1 
 2 ! ! ! ! 1.0966  1.2026  1.3188  1.4462     1 
 2  24 
  720  40,320   0.5000 actual value: cos    3  Page A50 1. Step 1: Verify that the formula is valid for n  1. Since S1  2 and 1(1  1)  2, the formula is valid for n  1. Step 2: Assume the formula is valid for n  k and show that it is also valid for n  k  1. 2  4  6  . . .  2k  k(k 
 1) Derive the formula for n  k  1 by adding 2(k  1) to each side. 2  4  6  . . .  2k  2(k  1)  k(k  1)  2(k  1) 2  4  6  . . .  2k  2(k  1)  (k  1)(k  2) Apply the original formula for n  k  1. Sk1  (k  1)((k  1)  1) or (k  1)(k  2) The formula gives the same result as adding the (k  1) term directly. Thus, if the formula is valid for n  k, it is also valid for n  k  1. Since Sn is valid for n  1, it is also valid for n  2, n  3, and so on indefinitely. Hence the formula is valid for all positive integral values of n.   0.5  Lesson 12-8 Page A50 1. f(
2)  2 (
2) or 
4 f(
4)  2 (
4) or 
8 f(
8)  2 (
8) or 
16 f(
16)  2 (
16) or 
32 
4, 
8, 
16, 
32 2. f(4)  42 or 16 f(16)  162 or 256 f(256)  2562 or 65,536 f(65,536)  65,5362 or 4,294,967,296 16, 256, 65,536, 4,294,967,296 3. z0  2i z1  0.5(2i)  i  2i z2  0.5(2i)  i  2i z3  0.5(2i)  i  2i 2i, 2i, 2i  535  Extra Practice  2. Step 1: Verify that the formula is valid for n  1.  4. dependent 5. dependent  1(1  1)(1  2)  6   1, the formula is Since S1  1 and valid for n  1. Step 2: Assume the formula is valid for n  k and show that it is also valid for n  k  1. k(k  1)  5!   6. P(5, 5)   (5 
 5)! 5 4 3 2 1   1  120  k(k  1)(k  2)  1  3  6  . . .  2  6  8!   7. P(8, 3)   (8 
 3)!  Derive the formula for n  k  1 by adding (k  1)(k  2)  2    to each side. k(k  1)   336  (k  1)(k  2)  1  3  6  . . .  2  2   4!   8. P(4, 1)   (4 
 1)!  k(k  1)(k  2)  6  136. . .  (k  1)(k  2)  2 k(k  1) (k  1)(k  2)  2  2  4 3 2 1    3 2 1 4  k(k  1)(k  2)  3(k  1)(k  2)  6 k(k  1) (k  1)(k  2)  6  . . .  2  2   13    10!   9. P(10, 9)   (10 
 9)!    (k  1)(k  2)(k  3)  6  (k  1)((k  1)  1)((k  1)  2)  6  or  (k  1)(k  2)(k  3)  6  10 9 8 7 6 5 4 3 2 1  1   3,628,800 9!   10. P(9, 6)   (9 
 6)!  Apply the original formula for n  k  1. Sk1   8 7 6 5 4 3 2 1  5 4 3 2 1    9 8 7 6 5 4 3 2 1  3 2 1   60,480 7!   11. P(7, 3)   (7 
 3)!  The formula gives the same result as adding the k  1 term directly. Thus, if the formula is valid for n  k, it is also valid for n  k  1. Since the formula is valid for n  1, it is also valid for n  2, n  3, and so on indefinitely. Hence, the formula is valid for all positive integral values of n. 3. Sn: 5n 
 1  2r for some integer r. Step 1: Verify that Sn is valid for n  1. S1  51 
 1 or 4. Since 4  2 2, Sn is valid for n  1. Step 2: Assume that Sn is valid for n  k and show that it is also valid for n  k  1. Sk → 5k 
 1  2r for some integer t Sk1 → 5k1 
 1  2t for some integer t 5k 
 1  2r 5(5k 
 1)  5(2r) 5k1 
 5  10r 5k1 
 1  10r  4 5k1 
 1  2(5r  2) Thus, 5k1 
 1  2t, where t  (5r  2) is an integer. Thus if Sk is valid, then Sk1 is also valid. Since Sn is valid for n  1, it is also valid for n  2, n  3, and so on indefinitely. Hence, 5n 
 1 is even for all positive integral values of n.    7 6 5 4 3 2 1  4 3 2 1   210 12.  P(5, 2)  P(2, 1)      5!  (5 
 2)! — 2!  (2 
 1)! 5! 1!  3! 2! 5 4 3 2 1 1  3 2 1 2 1   10 13.  P(8, 6)  P(7, 4)      8!  (8 
 6)! — 7!  (7 
 4)! 8! 3!  7! 2! 8 7 6 5 4 3 2 1 3 2 1  7 6 5 4 3 2 1 2 1   24 14.    P(5, 2) P(8, 4)  P(10, 1)  5! 8!   (5 
 2)! (8 
 4)! ——— 10!  (10 
 1)!  9! 8! 5!    10! 4! 3!    Lesson 13-1  9 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 5 4 3 2 1  10 9 8 7 6 5 4 3 2 1 4 3 2 1 3 2 1   3360 4!   15. C(4, 2)   (4 
 2)! 2!  Page A51  4 3 2 1    2 1 2 1 6  1. Using the Basic Counting Principle, 6 6 6  216.  10!   16. C(10, 7)   (10 
 7)! 7!  8!   2. P(8, 8)   (8 
 8)!      8 7 6 5 4 3 2 1  1   120   40,320 3. independent Extra Practice  10 9 8 7 6 5 4 3 2 1  3 2 1 7 6 5 4 3 2 1  536  6!  4   17. C(6, 5)   (6 
 5)! 5!    42  4!  4 3 2 1  1 3 2 1  4  9. P(s)  1 6  7 6 5 4 3 2 1  4 3 2 1 3 2 1  1  2 0 1  20 odds  — 19  20   140 3!  8!  (8 
 7)! 7!   19. C(3, 1) C(8, 7)   (3 
 1)! 1! 3 2 1  2 1 1  8 7 6 5 4 3 2 1  1 7 6 5 4 3 2 1  6  10. P(s)  1 6   24 9!  4!  (4 
 3)! 3!   20. C(9, 5) C(4, 3)   (9 
 5)! 5!  9 8 7 6 5 4 3 2 1  4 3 2 1 5 4 3 2 1   4 3 2 1  1 2 1  3  15  1  1  8 — 7  8  2 4   120 1  —1— 5 1  15 odds  — 14  15  Lesson 13-2 Page A51  2.  8!  2! 2!  3.  7!  3!  4.  10!  2!  5.  4!  2!  7 6 5 4 3 2 1  1   5040  1  or 1 9 5  15  P(f )  1 
 P(s) 1   10,080  1  or 7  1   1 
 1 5 14   1 5 1  or 1 4  C(2, 1) C(4, 1)  C(4, 1) C(10, 1)  C(2, 1) C(10, 1)  2 4  4 10  2 10      120  120  120   840   P(f )  1 
 P(s)        C(16, 2) C(16, 2) C(16, 2)  7 6 5 4 3 2 1  3 2 1    7   1 
 8 or 8  12. P(s)  P(1 white, 1 yellow)  P(1 yellow, 1 red)  P(1 white, 1 red)  8 7 6 5 4 3 2 1  2 1 2 1    19    1 
 2 0 or 20  C(2, 1) C(4, 1)    6  P(f )  1 
 P(s)   11. P(s)   C(16, 2)  1.  12  1  8  odds    504  7!  0!  3     8. P(red or green)   8  4 2  14 or 7  7!  (7 
 3)! 3!   18. C(4, 3) C(7, 3)   (4 
 3)! 3!    6  84  6    2     7. P(not red)   8  4  2  14 or 7  6 5 4 3 2 1  1 5 4 3 2 1    4     6. P(green)   8  4  2  14 or 7  1  1  1     1 5  3  6  10 9 8 7 6 5 4 3 2 1  2 1  17   30   1,814,400  P(f)  1 
 P(s)  4 3 2 1    2 1  17  13   1 
 30 or 30   12  odds   6. circular; (4 
 1)! or 6 7. circular; since the bracelet can be turned over (9 
 1)! there are 2 or 20,160 permutations 8. linear; 5! or 120  17  30 — 13  30  17  or 1 3  Lesson 13-4 Lesson 13-3  Page A52 6  2 3    7 14 10 10 25   independent; 1 8 18  81 1 6 3  independent; 2 1 0  10 1 1 1 11  inclusive; 6  6 
 3 6  36 4 4 8 2    exclusive; 5 2  52  52 or 13  1. dependent; 8  Page A51 4  2.  1   1. P(ace)  5 2 or 13  3.  44444  52 20 5     52 or 1 3 6 3  or  52 26  2. P(a card of 5 or less)  3. P(a red face card)   4. 5.  4. P(not a queen)  1 
 P(a queen) 4   1 
 5 2 48  12   52 or 1 3 2  2  1     5. P(blue)   8  4  2  14 or 7  537  Extra Practice  Lesson 13-5  200  1  5  5  Page A52    256    3   5. P(exactly 3 hits)  C(5, 3) 1000  200  8 4    12 11 — 8  12 4  11  1  125   10  2  800   1000   16  25  32    625 or 0.0512  6. P(at least 4 hits)  P(4 hits)  P(5 hits)  2. P(second clip is bluefirst clip was blue)   4  800   1000   256  625    625 or 0.4096  1. P(second clip is bluefirst clip was red)   1   4. P(exactly 1 hit)  C(5, 1) 1000   4    C(5, 4) 1000   4 3    12 11 — 4  12 3  11  200  5   1  625  21  3125  4  5  1  800   1000   1  1  3125  200  1  or 0.00672  3. P(numbers on dice matchsum is greater than 7)    3  36 — 15  36 1 3  or  5 15  Lesson 14-1 Page A53 1. range  70 
 22 or 48  4. P(sum is greater than 7numbers match)    3  36 — 6  36 3 1  or  6 2  2. Sample answer: 10 3. Sample answer: 20, 30, 40, 50, 60, 70, 80 4. Sample answer: 25, 35, 45, 55, 65, 75 5.  5. P(ball is from second boxball is white)   P(2nd box and white)  P(white) 1 3   2 8   ——   1 4 1 3      2 7 2 8 21  53  6.  Lesson 13-6  Grams of Fat Consumed by Adults  6 1 3 1 0   2 2  1. P(all heads)  C(3, 3) 1   1  8      Frequency 4  1  2  1  8  1 2  2. P(exactly 2 tails)  C(3, 2)2 3  1  4  0  1  12  1  2  3   8  1 2 1 1  2 1 1   1  2 8   C(3, 2)2 1  4  3 3  8 4  8     1 3 1 0  2   C(3, 3)2     1  1  8 1 or 2  Extra Practice  0 10 20 30 40 50 60 70 80 Grams of Fat  7. Sample answer: 40–50  3. P(at least 2 heads)  P(2 heads)  P(3 heads)    Frequency 6 7 8 3 5 1  8  Page A52    Grams of Fat 20-30 30-40 40-50 50-60 60-70 70-80  538  5    C(5, 5) 1000   0  800   1000   Lesson 14-2  11. Order the values from least to greatest. The median lies between the fourth and fifth terms. 3, 4, 4, 7, x, 12, 16, 19 7x  Page A53 1. X   1 (130 4  7.5  2   150  180  190)  15  7  x 8x   162.5  150  180  Md  2 or 165 Mode: none  Lesson 14-3  1  2.  X  6(15  16  17  18  18  19)  17.2  17  18  Md  2 or 17.5  Page A54 1. interquartile range  Q3 
 Q1  58 
 39  19  Mode  18 1  3. X   6(25  28  30  36  38  42)  33.2  30  36  2  Md   19  semi-interquartile range  2 or 9.5 or 33  Mode: none 1  4.  X  1 0 (1  2  3  4  5  5  6  9  9  10)  5.4 Md  5 Mode  5 and 9  40  50  60  70  80  2. interquartile range  Q3 
 Q1  7.65 
 2.75  4.9  1  5. X   1 2 (2.3  2.5  4  2(5.6)  6  6.4  6.5  2(7)  8  10)  4.9  semi-interquartile range  2 or 2.45   5.9  6  6.4  Md  2 or 6.2  1  Mode  5.6 and 7  3. X   2  4  5  6  7  8  9 10   175  180  180  195  200  212  220  250)  195.7  1   MD  9 (
45.8  
20.8  
15.8  . . .  54.2)  21.98  1  6. X   1 2 (14  2(15)  16  20  21  24  27  28  36  2(39))  24.5  21  24  Md  2 or 22.5 Mode  15 and 39 1  7. X   1 8 (3.0  3.4  3.6  5.2  2(5.4)  2(5.6) 5.7  6.2  6.3  6.8  7.0  7.1  7.6  7.7  8.2)  5.9 Md  5.7 Mode  5.4 and 5.6  (
45.8)2  (
20.8)2  . . .  54.22  9  j   27.56 1  4. X   11(1.4  2  2.4  2.9  3  3.5  3.7  4.2  4.6  5.3  5.5)  3.5 1 MD  11(
2.1  
1.5  
1.1  . . .  2)  1.05  1  8. X   1 4 (800  820  830  890  960  970  980  1040  2(1050)  1080  1110  1170  1180)  j   995  (
2.1)2  (
1.5)2  (
1.1)2  . . .  22  11   1.26 5a. Md  18  980  1040  Md  2 or 1010 Mode  1050 9. stem  1  2  3  4  5  12  12  3  1 (150 9  16  18  5b. Q1  2 or 17 leaf 2 7 3 4 4 4 0 2 5  Q3  20 5c. interquartile range  Q3 
 Q1  20 
 17 3  9 4 6 8 9 9 5 6 7 9 4  3  5d. semi-interquartile range  2 or 1.5 5e. Any points less than 17 
 1.5(3) or 12.5 and any points greater than 20  1.5(3) or 24.5 are considered outliers. There are no such points.  1  10. 8  6(4  5  6  9  10  x) 48  34  x 14  x  539  Extra Practice  Lesson 15-1  5f. 15 16 17 18 19 20 21 22 23 24  Page A55 1. lim (x2  2x 
 2)  42  2(4) 
 2 x→4  22  Lesson 14-4  2. lim (
x4  x3 
 2x  1)  
(
1)4  (
1)3 
 2(
1)  1 x→
1  Page A54 1a. 25% corresponds to t  0.3. 10  0.3(2)  9.4 
 10.6 1b. 10 
 8  2, 14 
 10  4 tj  2 tj  4 t(2)  2 t(2)  4 t1 t2 68.3%  2   34.15%  95.5%  2  4. lim  x→
4  x→
4  (x  4)(x 
 4)  x4  x→
4  x2  5x  6   2 x→
2 x  x 
 2   47.75%  (x  3)(x  2)  x→
2 (x 
 1)(x  2) x3   lim  x→
2 x 
 1 
2  3   
2 
 1 1  
3 3x  9 3(x  3)   lim  lim  2 x→2 x 
 5x 
 24 x→2 (x 
 8)(x  3) 3   lim  x→2 x 
 8 3   2
8 3 1  
6 or 
2  5. lim  6.   43.3%  2a. 0.683(400)  273.2 2b. 0.955(400)  382 0.683 (400) 2   lim   
4 
 4  
8  1d. 80% corresponds to t  1.3. 10  1.3(2)  7.4 
 12.6  2c.  x2 
 16  x4   lim (x 
 4)  34.15%  47.75%  81.9% 1c. 10 
 7  3, 10 
 10  0 tj  3 tj  0 t(2)  3 t(2)  0 t  1.5 t0 86.6%  2  1  3. lim (x  sin x)  0  sin 0 x→0 0   lim   136.6  Lesson 15-2 Lesson 14-5 Page A55  f(x  h) 
 f(x)  h h→0 5(x  h) 
 5x    lim h h→0 5x  5h 
 5x  lim h h→0 5h  lim h h→0  1. f
(x)  lim  Page A54 1.2  1. jX   or about 0.13  
90 3.4 2. jX   or 0.34 00  
1 12.4 3. jX   or about 0.80 40  
2 4. A 1% confidence level is given when P  99% and t  2.58.  5 2. f
(x)  lim  h→0  4.2  jX    40  
  0.6640783086 interval: X   tjX  150  2.58jX    148.29 
 151.71   lim  9(x  h) 
 2 
 (9x 
 2)  h   lim  9x  9h 
 2 
 9x  2  h   lim  9h  h  h→0 h→0  5. A 1% confidence level is given when P  99% and t  2.58.  f(x  h) 
 f(x)  h  h→0  9  10  jX   78  
  1.132277034 interval: X   tjX  320  2.58jX    317.08 
 322.92  1  2  3. f(x)  2x  3 1  f
(x)  2 1x1
1  0 1   2 4. f(x)  x2  4x  8 f
(x)  2x2
1  4 1x1
1  0  2x  4  Extra Practice  540  5. f(x)  x5 F(x)   6. f(x)   1  21  F(x)  2  7. f(x)  F(x)   8. f(x)    1 x3 3  n  1  11   lim  1  11  n→ 2 2  x11  2x  C  0  x11 
 x  C   lim  x11  x  C  n→ 8  3  56  3      5  2.  1  n    
lim  n→ 5  lim n n→  
 lim  n→ 5  lim n n→    1  n  1  n  0    5 7  n n 2   lim    35  2    25  2n  n→ 35 3  2  0 
 2 32  2 or 16    n  1  3n 
 n(1  2  . . .  n) n(n  1)   1500 
 250 or $1250    1  (x  1)dx 
  0  n  (x  1)dx  Lesson 15-4  5 n  Page A55  x6dx  17x7  C 1 2. 5x4dx  5 5x5  C  n  1.   x5  C   2  . . .  n)  n(n  1)  2  
 lim  1  n    1  2  n→ 1 3 
 lim n 2n  n→ 3 1 
 lim 2  2n n→   2 5  . .  3 
 n  1  2  5  3 
 n1  3 
 n2  .  n→   1  n  1  . . .  n  1  n  n  i   lim 1500 
 2501  n  5 2  n  n  5n00   3 
 n i1  500   1  n  1  . . .  n  1  1 (1 n  500i  n→  5   lim  n→  5  500  n  6 
 0.0021500  n     lim 1500 
 n 2  2  n  n (1  2  . . .  n) 1  n  n  500  n→ n 500  lim n n→  5 i 1 lim  n  1 n     1 n 
 n→ i1  5i  n n→ i1 5 1 5  lim n n n→   lim    2n   8  2   lim  0  (x  1)dx     4n  n2  (6 
 0.002x)dx  n→      88   lim         8 n(n  1)   n 2  2  n3  n1  81  n1  8   n→ i1  3i 3 5 n n n→ i1 45 n(n  1)  lim n2 2 n→ 45 2   lim  2 (n  n) n→ 2n 2 45n 45n    lim  2  lim 2n2 n→ 2n n→      4  3   lim  45  2 45  2   4n      2 3  . .  n2)  2 4 n(n  1)(2n  1)     2 6 n→ n n 8 2n3  3n2  n 16  lim n3 6  n2 n→  n    2 3 2   lim  1500    2 1  2 4 2 2   2 (1  2  . n→ n n 8  n(1  2  . . .  n)  4.  5x dx  lim  2 2 2   4n  4  n    lim  
 x2  x  C  Page A55    2  2 1   n   2  n  2ni   4 n2  4  4n  4  . . .  n  4n  4  Lesson 15-3  1.  2  2i  n n→ i1  2000  3   4x  4)dx   lim  2 x3 
 4x2  2x  C 3 1 3 x3 
 x 
 1 5 4 1 3 1 1   x31 
   5 31 4 11 3 1  x4 
  x2 
 x  C 8 20 x3 
 2x2  x  x 2 x 
 2x  1 1  2  0  x21 
 8   21 
 2 F(x)   21 x     (x 2  3.  1 x51  C 51 1 x6  C 6 2x2 
 8x  2    (x2 
 x  5)dx  13x3 
 12x2  5x  C 1 1 4. (
4x4  x2 
 6)dx  
4 5x5  3x3 
 6x  C  3. 1  n  n  n(n  1)  2  4      2  5.    2  14x6dx  14  2x7  
0  1   
5x5  3x3 
 6x  C 2 1 x7 7 2 2 2   (2 27) 
 (2 (
2)7)  256 
 (
256)  512  541  Extra Practice    6  6.  0  1  (x  2)dx  2x2  2x  6 0  7.    2  4   2 62  2 6 
 2 02  2 0 1  4  1  1    Extra Practice  5  4  (x2 
 2x 
 8)dx  1  1 x2 2  1  
 8x   3x3 
 x2 
 8x  4    1  3  2  1  43 
 4 3 16 16  
 
 3 3 32  3       3x3 
 2     (x 
 4)(x  2)dx    30 
 0  30 (x2 
 4)dx  3x3 
 4x    5  8.  4 
   1  3  23  
 4 2  53  
  52  70  10   3  542  80  4  4  
 8 5 
 3 43 
 42 
 8 4   
3 
 
3    5  5  1  Chapter Tests Chapter 1 Test  12. 16 14 12 10 y  4x  12 8 6 4 2  Page A56 1. D  [
1, 0, 2, 3}; R  {2, 4, 5}; yes 2. D  {
5, 4, 6, 7}; R  {
3, 
2, 0, 2, 7}; no 3. f(4)  4 
 3 42  4 
 48 or 
44 4. f(
7)  
7 
 3(
7)2  
7 
 147 or 
154 5. f(a  2)  a  2 
 3(a  2)2  a  2 
 3(a2  4a  4)  a  2 
 3a2 
 12a 
 12  
3a2 
 11a 
 10  O  
2 
 4  5   15. m   8
0 
6  5  y 
 3  3(x  1) 5  5  y 
 4  
4(x 
 0)  3  5  14  y 
 4  
4x  3  y  3x  3  1200  3   117.89 Im/m2 6b. d cannot be negative or zero. 7. f(x)  g(x)  x2 
 7  x  3  x2  x 
 4 f(x) 
 g(x)  x2 
 7 
 (x  3)  x2 
 x 
 10 f(x) g(x)  (x2 
 7)(x  3)  x3  3x2 
 7x 
 21  y  
4x  4 16. parallel: y 
 2  4(x 
 0) y 
 2  4x 4x 
 y  2  0 1  perpendicular: y 
 2  
4(x 
 0) 1  y 
 2  
4x 4y 
 8  
x x  4y 
 8  0  x2 
 7    x3  8. [f 
 g](x)  f(g(x))  f(4x 
 5)  4x 
 5  1  4x 
 4 [g 
 f ](x)  g(f(x))  g(x  1)  4(x  1) 
 5  4x  4 
 5  4x 
 1 9. [f 
 g](x)  f(g(x))  f(2x2  6)  5(2x2  6)  10x2  30 [g 
 f ](x)  g(f(x))  g(5x)  2(5x)2  6  50x2  6 y 10. 11.  3   8 or 
4  y 
 3  3x  3    4(0.9)2  x  O  1 2 3 4 5x  14. y 
 3  3(x 
 (
1))  P  O  y |x |  2  y  
5
4
3
2
1 
2 
4   6a. E   4d2  f(x)  g(x)  13.  y  1  17. x 
 5y  3 ⇒ m  5 1  parallel: y 
 2  5(x 
 (
1)) 1  y 
 2  5(x  1) 5y 
 10  x  1 x 
 5y  11  0 5  perpendicular: y 
 2  
1(x 
 (
1)) y 
 2  
5(x  1) y 
 2  
5x 
 5 5x  y  3  0 18.  19.  g (x )  f (x )  g (x ) 
 x  1  O  x  O  x  y  x O  x 2x  y 
 1  y 
 3x  6  543  Chapter Tests  20a. 100 80 Economics 60 Grade 40 20 0  20  →  x 
 5y  z  
15 
7x  y  2z  1  →  7x 
 9y  
25 9x 
 11y  
31  →  7x 
 9y  
25 7x 
 9(2)  
25 7x  
7 x  
1 ( 
1, 2, 
4) 5. y  12 
 2x  40 60 80 100 Statistics Grade  20b. Sample answer: Using (47, 67) and (95, 88), y  0.44x  46.44 88 
 67   m 95 
 47   0.4375 y 
 67  0.4375(x 
 47) y 
 67  0.4375x 
 20.5625 y  0.4375x  46.4375 y  0.44x  46.44  4x  6y 
 3z  20 3x 
 15y  3z  
45 7x 
 9y  
25 2x 
 10y  2z  
30 
7x  y  2z  1 9x 
 11y  
31 63x 
 81y  
225 63x 
 77y  
217 
4y  
 8 y2 x 
 5y  z  
15 
1 
 5(2)  z  
15 
11  z  
15 z  
4 3x 
 4  2y 3x 
 4  2(12 
 2x) 3x 
 4  24 
 4x 7x  28 x4  y  12 
 2(4) 4 (4, 4) 6. A  B  
5  (
2) 1  3 4  (
3) 2  0 
7 4  1 2 7. 
3C  
3(
1) 
3(0) 
3 (3) 
3 (3) 
3(2) 
3(
4) 3 0 
9  
9 
6 12  Chapter 2 Test Page A57 1.  4. 4x  6y 
 3z  20 x 
 5y  z  
15  y (1, 3)  2(
2) 2(3) 2(
3) 2(0) 
4 6  
6 0  2x  y 
 1  8. 2B   x  O  xy
4  2B 
 A  
4 
 (
5) 6 
 1 
6 
 4 0 
 2 2. 3x  y  7 5x  2y  12  →  3x  y  7 3(2)  y  7 y1 (2, 1) 3. 4x 
 5y  
2 3x  2y  
13 4x 
 5y  
2 4x 
 5(
2)  
2 4x  
12 x  
3 (
3, 
2)  →  6x  2y  14 5x  2y  12 x  2   9. BC   1 5 
10 
2  
2 3 
3 0  
1 0 3 3 2 
4   
2(
1)  3(3) 
2(0)  3(2) 
2(3)  3(
4) 
3(
1)  0(3) 
3(0)  0(2) 
3(3)  0(
4) 11 6 
18  3 0 
9  12x 
 15y  
 6 12x  8y  
52 
23y  46 y  
2  1 
4 6  1.5 
6 9 
3 3 2 
4.5 4.5 3 A
(1.5, 
4.5), B
(
6, 4.5), and C
(9, 3) y  10. 1.5  B B  8 6 4 2  C  O C  
6
4
2 A 2 4 6 8 10 x 
4 
6 A  Chapter Tests  544  11.  19.  3 1 
6 
3 3 1 
6 
3  2 
5 1 5 
2 5 
1 
5  1 0 0 
1  y 
 12 x  3  F
(3, 
2), G
(1, 5), H
(
6, 
1), J
(
3, 
5)  y  J  y (2, 4) (2, 2) ( 43, 0)  (6, 0) y 
 0 O y 
 3x  4  G  x x
2  F H  f(x, y)  2x 
 y f(
6, 0)  2(
6) 
 0 or 
12 c minimum f(2, 4)  2(2) 
 4 or 0 f(2, 2)  2(2) 
 2 or 2  x  O H  F  f 3, 0  23 
 0 or 3 c maximum 4  G  J  2 4  2(9) 
 (
10)(4) 
10 9  58  y
4  15.  2 1 1 3  16.  1 3 
4 4 
2  x
3  3  5 1  
 5  f(x, y)  12,000x  16,000y f(3, 5)  12,000(3)  16,000(5)  116,000 f(3, 4)  12,000(3)  16,000(4)  100,000 f(5, 4)  12,000(5)  16,000(4)  124,000 The company reaches the most people with 5 ads and 4 commercial minutes.  1  
5 2  5  1 
2 4 
2 4  1 0 
4 3 
4 3 1    
5 2  
5  2  5 3  10  Chapter 3 Test  
5 4 17. The inverse does not exist since  0. 
15 12 2 
1 x 7  18. 3 1 y 3 1 1 1 1 1 1  5 2 
1 
3 2 
3 2 3  1  5  Page A58 1. y  2x  1 x-axis  1  1 1 2 
1 
3 2 3 1  x  O  3 
1  1 3 
1 5 
1 
1 2 2   (5, 4)  (3, 4)  1 2 
1 0 1 3 1 3 0 
2  (
1) 14. 3 0 1 1 1 
2 4 
2 4 1 4 1 
2  1(
1) 
 2(
10) 
 1(3)  16 1  8  20. Let x  number of ads. Let y  number of commercial minutes. x3 y 100x  200y 
 1300 y4 (3, 5) 100x  200y  1300  12. 5 7  5(6) 
 3(7) 3 6 9 13.  4  y-axis  1 x 1 1 7  5 y 
3 2 3  yx  x 2  y 
3  y  
x  →  b  2a  1 
b  2a  1 b  
2a 
 1 no b  2(
a)  1 b  
2a  1 no a  2b  1 a
1 b  2 no 
a  2(
b)  1 a  2b 
 1 a1 b  2 no  none of these  545  Chapter Tests  2  2  →  2. y  
x2  
b   x-axis  b b  y-axis  b yx  a b  y  
x  
a  
a  b →  3. x  y2  3 x-axis  2 
a2 2  no a2 2   
 (a)2 2 
a2 yes 2 
b2 2 
a no 2  
 (
b)2 2 
b2 2  no; y-axis a b2  3  yx y  
x  →  y-axis yx y  
x  f 1(x)   10.  
5
4
3
2 O  Chapter Tests  f (x ) 
 5x  4  3   f(x)   x
2  3   x y
2 3  y 
 2  x 3  y  x  2 3  f 1(x)  x  2; Yes, it is a function. 22  4    11. f(2)   2
2  0  No; the function is undefined when x  2. 12. Yes; the function is defined when x  0, the function approaches 1 as x approaches 0 from both sides, and f(0)  1. 13. y →  as x → , y →  as x → 
 14.  15.  x 3 4 5  y 
11 
12 
11  x 0 1 2  y 
1 
2 
3  16. x  1  At x  4, y is at a minimum value, since f(4)  f(3) and f(4)  f(5).  At x  1, y is at a point of inflection, since f(0) f(1) f(2).  4x   y x
1  y  as x → , y → 4; y  4  x  50 45 40 35 30 25 20 15 10 5  x  O  3  17.  8.  f 1(x ) 
 15 x  45   y x
2  |x  4|  O  x4  5 1 4 x   5 5  f (x )  Yes, it is a function.  y  x, y  
x 5. The graph of g(x) is the graph of f(x) translated left 3 units and reflected over the x-axis. 6. The graph of g(x) is the graph of f(x) stretched vertically by a factor of 4 and then translated down 2 units. y 7.  y  f(x)  5x 
 4 y  5x 
 4 x  5y 
 4 x  4  5y y  a a  (
b)2  3 a  b2  3 yes 
a  b2  3 a  
b2 
 3 no b  a2  3 a  
b 
 3 no  
b  (
a)2  3 
b  a2  3 a  

b 
 3 no; x-axis ab  5 a(
b)  5 
ab  5 no (
a)b  5 
ab  5 no ba  5 ab  5 yes (
b)(
a)  5 ab  5 yes  y-axis  4. xy  5 x-axis  9.  b  
a2  x2 
 4  0 (x  2)(x 
 2)  0 x  2  y as x → , y → 0; y  0  y  2x 2  3 1 2 3 4 5x  546  4x  x  x 1  
  x x  4 y  1 1 
 x x   y x2 
 4  y  x  x2  4 x2  
  x2 x2 1  x  y  4 1 
 x2  0.25  0.004x  6. 
1  
5 
13 53 60 
1 6 7 
60 1 
6 
7 60  0 x3 
 6x2 
 7x  60 7. f(x)  x3  8x2  2x 
 11 f(
2)  (
2)3  8(
2)2  2(
2) 
 11  
8  32 
 4 
 11  9; no 8. f(x)  4x4 
 2x2  x 
 3 f(1)  4(1)4 
 2(1)2  1 
 3 4
2
2  0; yes 9. 1 positive f(
x)  
6x3  11x2  3x 
 2 2 or 0 negative   18a. y   0.25  0.001x 0.25 0.004x     x x  0.25 0.001x     x x  y  y as x  0.25   0.004 x  0.25   0.001 x 0.004  → , y →  0.001  or 4; y  4  18b. As the amount of 4 molar solution added increases, the molarity of the mixture approaches 4. 19. y  kx y  0.25x 0.5  k(2) y  0.25(10) 0.25  k y  2.5 20.  72  k  y  x2 8  y  x2  k  (3)2  18   72  k  1  r  72  x2  1  2 1 
3  4 x  2  x2  
2  6  11  
3  
2  6  14  4  0  6  9  
6  0  6  
1  
1  0  1 1  rational zeros: 
2, 
3, 2  Chapter 4 Test  10. 1 positive f(
x)  x4 
 x3 
 9x2  17x 
 8 3 or 1 negative  Page A59 1. (x 
 4)(x 
 i)(x 
 (
i)  (x 
 4)(x 
 i)(x  i)  (x 
 4)(x2 
 i2)  (x 
 4)(x2  1)  x3 
 4x2  x 
 4 2 2 2. b 
 4ac  (
5) 
 4(1)(4) 9 Since b2 
 4ac 0, there are 2 real roots. n n n n  x  
17 487 
393  
8 3888 3136  1 
1  1 1  2 0  
7 
9  
24 
8  
32 0    r 1 2          1 1 1  
5 
1 5  3 4 5  
9 
10 1  upper bound: 2 f(
x)  
x3  3x2  5x 
 9 r 1 2 3 4 5  4(2)(4)   
7 Since b2 
 4ac  0, there are 2 imaginary roots.  x  
9 63 47  rational zero: 
1  z  x  1 9 
7  11. Use the TABLE function of a graphing calculator; 
0.8, 3.8. 12. Use the TABLE function of a graphing calculator: 
1.3. 13. Sample answer: 2; 
5  n4 n1 3. b2 
 4ac  (
7)2 
 4(1)(
3)  61 Since b2 
 4ac 0, there are 2 real roots.  4.  1 1 1    b2 
 4 ac 
b  
  2a 
(
5)  
 9  2(1) 53  2 53 5
3  or n   2 2  b2 
 4 ac 
b  
  2a 
(
7)  
61   z 2(1) 7  
61  z  2 b2 
 4ac  (
5)2 
  r 8 
8  
b  
 b2 
 4 ac  2a 
(
5)  

7   2(2) 
7 i 5  4  
1 
1 
1 
1 
1 
1  3 2 1 0 
1 
2  5 7 7 5 1 
5  
9 
2 5 6 
5 
34  lower bound: 
5  
3 3 
4 4 2 10 2 1 5 6 2x2  x  5, R6  5. 2  2  547  Chapter Tests  14. Sample answer: 1; 
2 r 1  2 2  
1 4  3 5  1 5  19. 
2x   2  
3x  
5 2x  2  3x 
 5 7x Check: 
2(7)   2  
3(7)  
 5 
16   
16  20. 
11 
 1 9 0m 11 
 10m 81 
10m 70 m  
7 11 
 10m  0 
10m  
11 11 m  10  1 6  upper bound: 1 f(
x)  2x4 
 3x3 
 x2 
 x  1 r 1 2  
3 
1 1  2 2 2  
1 
2 1  
1 
3 1  1 
2 3  lower bound: 
2 15.  1  80  1  1  Test m  
8: 
11 0(
8) 
 1 ? 
91  ? 9.54 ? Test m  0: 
11 
 1 ? 9 0(0) 
11  ? 9 3.32 ? 9 Solution: m  
7   a  1 0 1  a    a 16.  7  80 80  7 4  x
2 4  x
2  3  1   
   x2 
 4 4 3  1     (x  2)(x 
 2) 
 4 (x  2)(x 
 2)  21.  4(x  2)  3 
 4 16(x  2)  12 
 (x  2)(x 
 2) 16x  32  12 
 x2  4 x2  16x  16  0 x x  5  x2  5  Test x   Test x   3  7  
 3  B2  2 3  7  
1  B Let z  
2. 5(
2) 
 11  A(2 (
2) 
 3)  B(
2  2) 
21  
7A 3A  5z 
 11   2z2  z 
 6  22.  3  5  ? 
3  2  1   
   z2 2z 
 3  7x2  18x 
 1   (x2 
 1)(x  2)    7x2  18x 
 1   (x2 
 1)(x  2) 2 7x  18x 
 1  
0.3148 true  5 2     
3: 
3 3(
3) 5 2 
5 ? 
3 
 9 8 
5 ? 
19 false 5 5 2  ?     
1:  
1  2 
1 3(
1) 2 5 ? 
5 
 3 2 5 ? 
53 true 5 5 2  ?    1:  12 1 3(1) 5 2  ? 5   3 3 2 ? 2 13 53 false  Chapter Tests  3  2  
2  2B  5 5 2  ?    
18  2 
18 3(
18) 5 ? 5 1  
1 
1 6 8 
 27 ?  x  
17, 
2  x  0 18. 
y 
2
30 
y 
23 y
29 y  11  B  52 
 11  A2  15x  15(x  2)  2(x  2) 15x  15x  30  2x  4 
2x  34 x  
17; x 
2 or 0  Test x   A      z2 2z 
 3  3  2  
0.3125  5z 
 11    (z  2)(2z 
 3)  Let z  2.   x  3x  Test x  
18:  false  5z 
 11  A(2z 
 3)  B(z  2)  
16  
 162 
 4(1)(16)   2(1) 
16  8
3   2  x  
8  4
3  17.  5z 
 11   2z2  z 
 6 5z 
 11   2z2  z 
 6  9 9 9 true  7x2  18x 
 1  (x  1)(x 
 1)(x  2) A  B  C        x1 x2 x
1   A(x 
 1)(x  2)  B(x  1)(x 
 1)  C(x  1)(x  2)  Let x  1. 7(1)2  18(1) 
 1  A(1 
 1)(1  2)  B(1  1)(1 
 1)  C(1  1)(1  2) 24  6C 4C Let x  
2. 7(
2)2  18(
2) 
 1  A(
2 
 1)(
2  2)  B(
2  1)(
2 
 1)  C(
2  1)(
2  2) 
9  3B 
3  B Let x  
1. 7(
1)2  18(
1) 
 1  A(
1 
 1)(
1  2)  B(
1  1)(
1 
 1)  C(
1  1)(
1  2) 
12  
2A 6A  Check: 
11  
2
30 
9 
30 3
30 00  7x2  18x 
 1   (x2 
 1)(x  2)  548  6  3  4   
     x1 x2 x
1  5. (RS)2  (ST )2  (RT )2 (RS )2  122  152 RS2  225 
 144 RS  
81  or 9 in.  23. quadratic 24. Let x  the height. Then 6x  the length and 6x 
 7  the width. V  x(6x)(6x 
 7) 120  36x3 
 42x2 0  6x3 
 7x2 
 20 Use a graphing calculator to graph the related function. V(x)  6x3 
 7x2 
 20 It appears as if the zero is at x  2. Use the Factor Theorem to check V(2)  6(2)3 
 7(2)2 
 20  0. The height is 2 cm, the length is 6(2) or 12 cm and the width is 12 
 7 or 5 cm. 12 cm by 5 cm by 2 cm 25. Let s  the speed of the freight train. Then 30  s  the speed of the car. 500  Car: 500  (30  s)t ⇒ t   30  s Train: 350  st ⇒ t  500  30  s  side opposite   cos R   hypotenuse  12   cos R  1 5 or 5  4  9   sin R  1 5 or 5   csc R   side opposite  12   csc R  1 2 or 4  4  15  15  5  9  y  
3   tan 60°  1 or 
3  r  7. sec 270°  x 1  sec 270°  0; undefined  350  s  8. sin (
405°)  sin (
45°) sin (
45°)  
y  350   s  
2   sin (
45°)  
2 9. r  
 x2  y2 r  
 32  52 r  
34   cos v   
34   5
34   cos v  34  r  sec v  x  
34   sec v  3  csc v  y csc v  5  3  5  tan v  3  3
34   x  r  cot v  y  
34   cot v  5  3  10. r  
 x2  y2 r  
 (
4)2   22 r  
20  or 2
5    
0.65  y  x  sin v  r 2  
4   cos v   2
5  2
5   2   tan v   
4  cos v  
5  r  sec v  x  cot v  y  csc v  2  2
5   sec v   cot v  2  csc v  
5   sec v  
2  csc v  y   
3.46  tan v  x  
 5  sin v  5  a  360° (
4)  
1245° a 
 1440°  
1245° a  195°; III  y  cos v  r   sin v   2
5    1.14  
1245°  360°  tan v  x  5  sin v  34   2.76  y  cos v  r   sin v   
 34  a  360°(1)  410° a  360°  410° a  50°; I 4.  x  sin v  r  a  360°(
1)  
234° a 
 360°  
234° a  126°; II 3.  3   cos R  1 2 or 4  6. tan 60°  x  a  360°(2)  995° a  720°  995° x  275°; IV  410°  360°   cot R   side opposite  sec R  9 or 3  Page A60  
234°  360°  5  side adjacent  hypotenuse   sec R   side adjacent  Chapter 5 Test  2.  hypotenuse  tan R  9 or 3  y  995°  360°  3  side opposite   tan R   side adjacent  500 s  10,500  350s 150s  10,500 s  70 km/h  1.  side adjacent   sin R   hypotenuse  1  tan v  
2 x  r  2
5   
4 
5   
4  cot v  
2  11. r  
 x2  y2 r  
 02  ( 
3)2 r  
9  or 3 y  sin v  r 
3  tan v  x  y  0  
3  sin v  3  cos v  3  tan v  0  sin v  
1  cos v  0  undefined  r  y 3  
3  r  x 3  0  csc v  csc v   csc v 
1  549  x  cos v  r  sec v  sec v   undefined  x  cot v  y 0   cot v   
3  cot v  0  Chapter Tests  21. Since 98°  90°, consider Case II. c a; one solution  b  12. cos A  c 42  cos °  c  sin 98°  90  42  cos 77°  c  64 sin 98°  90 sin A  c  186.7 13.  64 sin 98°  A  sin1 90  b  a b  13  tan B  tan 27°   A  44.8° B  180° 
 98° 
 44.8° or 37.2° sin 37.2  b  b  13 tan 27° b  6.6 14. sin  90 sin 37.2°   b sin 98°  b  54.9 A  44.8°, B  37.2°, b  54.9 22. Since 31°  90°, consider Case I. a  b sin A 9  20 sin 31° 9  10.3; none 23. a2  b2  c2 
 2bc cos A 132  72  152 
 2(7)(15) cos A 169  274 
 210 cos A 
105  
210 cos A  a  14 sin 32° 17° a  7.5 a  15. cos B  c 23  cos B  37 23  B  cos1 37 B  51.6° a  16. tan A  b   A  cos1  210  105  3  tan A  11  A  60°  3  A  tan1 11  sin 60°  13  A  15.3° 17. Let h  the height.  7 sin 60°  B  sin1 13 65 m  B  27.8° C  180° 
 60° 
 27.8° or 92.2° A  60°, B  27.8°, C  92.2° 24. b2  a2  c2 
 2ac cos B b  
 202  242 
 2(20)( 24) co s 47° b  17.92432912  h  70˚  18. B  180° 
 36° 
 87° or 57°  sin 47°  b  sin B sin C   K  2a2  sin A sin 57° sin 87°  K  410.4  24 sin 47°  C  sin1 b  units2  C  78.3° A  180° 
 47° 
78.3° or 54.7° b  17.9, C  78.3°, A  54.7° 25. Let x  the distance between the transmitters.  1  19. K  2bc sin A 1  K  2(56.4)(92.5) sin 58.4° K  2221.7 units2  20.  180° 
 36°  2 sin 36°  22   72°   s  22 sin 72°  sin 36°  x  35.6 cm Perimeter  22  35.6  35.6  93.2 cm  Chapter Tests  70 miles  sin 72°  x  x sin 36°  22 sin 72°  sin C   24  24 sin 47°  b sin C   K  2(24)2 sin 36° 1  sin B   7  7 sin 60°  13 sin B  h  65  h  65 sin 70° h  61.1 m  1  sin 98°   90  90 sin 37.2°  b sin 98°  a sin A  c a 32° 17
  14  sin 70°   sin A   64  130 miles  130˚  36˚  x  x  x   70°  
 2(70)(130) cos 130° x  
21,80 0 8,200 
 130° cos 1 x  183.0 miles  x2  72˚ 22 cm  550  1302  Chapter 6 Test Page A61  14.  y  Arccsc x x  Arccsc y Csc x  y or y  Csc x  y    180°  1. 225°  225°    5   4   180°  2. 480°  480°  y 
 Arccsc x 
  8   3 5  5  5    4. reference angle: 4 
   4; Quadrant 3  5.  15.  1  7.  1   2  x  y   2  y 
 cos x  
 
  O 2 
1   2  O  
 
 2 2  
2  x    8. 3  3;  9. 
2  2;  or  2  3  2    2  16. sin Arccos 2  sin 60°  4  1  or 3 2  k  10. A  4  
 3  2  k  A  0.5    2    2  c  
3   or   
 3 3  18.  v  v  r t 2  2300  240,000 t 240,000(2)   t 2300  t  656 hours or 27.3 days  y  5 y 
 tan (2  
 ) 4 4 3 2 1 O   2  
3   1    c  
  or 4  y 
 3 cos 2   7   tan 6  
4  4  y  0.5 cos (4v 
 )  1 12. 13. y  
2 
 O  1   2  k    17. tan   sin1 2  tan   6  k  4 or 0.5  y  4 sin 0.5v 
 2  3   2   4 2  A  4 11. A  0.5  x  y 
 tan x  
 2  4  x    y 
 sin x  
1    y 
 arctan x  y  v  85.2 cm/s y 6. 1  O   2  y  tan x x  tan y arctan x  y or y  arctan x  71.1  v  12  
 
  2  
2 
  1  sin 6  2 5 tan 4  v v  r t  O    3. reference angle:  
 6  6; Quadrant 2  y 
 Csc x   2  
2  
4 
1 
2 
3 
4 
5   4   2  73 
 21  73  21  19. A  2  2  k  h  2  A  26  2  h  47  k  1 2   A  26     12  k  6   y  26 sin 6t  c  47   21  26 sin 6 1  c  47   
26  26 sin 6  c   
1  sin 6  c   sin1 (
1)  6  c     
2  6  c 4  
6  c 2  
3  c   2  Sample answer: y  26 sin 6t 
 3  47  551  Chapter Tests  20.  65 miles  1 hour  tan v  v  5280 feet 1 hour   1 mile 3600 seconds 2 v  rg (95.3  )2  tan1  1200(32)   95.3  ft/s  8.  sec x  sin x  sin x   
 cos x  cot x  sec x cos x 
 sin2 x  sin x cos x 1 
 sin2 x  sin x cos x cos x  sin x  v  0.23 radians   cot x  cot x  cot x  cot x  cot x 9.  Chapter 7 Test Page A62 1. sin2 v  cos2 v  1 1 2  3   cos2 v  1 8  cos x cos x    1  sin x 1 
 sin x cos x 
 cos x sin x  cos x  sin x cos x  1 
 sin2 x 2 cos x  1 
 sin2 x 2 cos x  cos2 x 2  cos x  cos2 v  9 cos v   2  10. csc (A 
 B)  csc (A 
 B)  csc (A 
 B)   sec v   1  sin2  cos v  cos2 v 
 sin2 v  or  cos 2v   cot 2v   sin 2v  cot 2v  cot 2v 12. sin 255°  sin (225°  30°)  sin 225° cos 30°  cos 225° sin 30°  3  or 
5   
2 2  
2 

6 
 
2  4 
2   v  v1    cos2  v1  cos2  v  25     5    sin (
420°)    sin (
360° 
 60°)  cos (
360° 
 60°) 
sin 60°    cos 60°     
tan 60° 6. tan v(cot v  tan v)  sec2 v tan v cot v  tan2 v  sec2 v 1  tan2 v  sec2 v sec2 v  sec2 v 2 7. sin A cos2 A  (1 
 sin A)(1  sin A) cos2 A 2 sin A cos2 A    13. tan 12  tan 6  4   5. tan (
420°)   cos (
420°)  sin2 A  
2   
2   
6   4    
3    
4  16  cos v  5       (1 
 sin A)(1  sin A)   (1 
 sin A)(1  sin A) 1 
 sin2 A  (1 
 sin A)(1  sin A) (1 
 sin A)(1  sin A)  (1 
 sin A)(1  sin A)  Chapter Tests  sin v   cot 2v   2 sin v cos v  5 
3  cos2  3 2 
5  1    cos 2v   2 sin v 
 2 cos v  1  1  5 
3  sec B  sin A 
 cos A tan B 1  cos B  sin B  sin A 
 cos A  cos B 1  sin A cos B 
 cos A sin B  11. cot 2v  2 cot v 
 2 tan v  3 
5   4. sin v   csc v  sin v    2 sec x  csc (A 
 B)  csc (A 
 B)  9  sec v    2 sec x  1  cos2 v  2 5 cos v    2 sec x   csc (A 
 B)   sin (A 
 B)   cos2 v  1  1  cos v 1  3 
5   2 sec x  2 sec x  2 sec x  2
2   3  2. tan2 v  1  sec2 v tan2 v  1  (
2)2 tan2 v  3 tan v  

3  3. sin2 v  cos2 v  1  
45   2 sec x      tan 6  tan 4    1 
 tan 6 tan 4  1   1 
 3  1  1 1 
 
3  1 1    1  
3 1 
3    1 1   1
 1  
3  
3  1 2     1  
3  3  1 1 
 3 3 4 2
 3     2 3 3     2  
3   552    1  2  sin v  sin2  3 2  4  v    1 7  sin2 v  1 6 sin v   20.   tan v   cos v  14. sin2 v  cos2 v  1  
7  
4  sin 2v  2 sin v cos v    
7  
4  3  4  
 A2   B2  
 (
1)2   12 or 
2  1  
  
2   
 7  tan f   1    1 6 
 16   1 6 or 8          
7  1 
 
 3  2
7  
 3  2  9 45°  1  cos 45°  2    
2  1  2  2    2  
2   4    2  
2   2  0 
5   p  
5; Quadrant IV     15. cos (22.5°)  cos 2   0  
5  2
5  sin f  
5, cos f  5, 
5  
5 1  tan f  or 
2 2
5   5 1 f  tan1 
2  2   
3
7    333° 22. 2x  y  6 → 2x  y 
 6  0 Ax1  By1  C  d  2  2 
A  B   d d 8
5   5  2(
5)  1(8) 
 6  2  1 2 
2  8 8
5   or 
 
 
5  5 Ax1  By1  C  23. d  2 2  

A  B  x  
3  tan x x 

3  tan x  0 tan x(tan x 
 
3 )  0 tan x  0 or tan x 
 
3 0 x  0° tan x  
3  x  60° 17. cos 2x 
 cos x  0 2 cos2 x 
 cos x 
 1  0 (2 cos x  1)(cos x 
 1)  0 2 cos x  1  0 or cos x 
 1  0 1 cos x  
2 cos x  1 x  120° x  0° 18. sin x 
 cos x  0 sin x  cos x tan2  16.  
2   2  or 
1 
2  
 2 tan1 (
1)   10 5 5 x 
 y 
  5
5  5
 5 5
 5 2
5  
5  
 5 x 
   y 
   5 5 5    2    3
2   
 A2   B2  
 102  (
5)2 or 5
5   2 tan v   tan 2v   1 
 tan2 v 
7  2 
 3  
2    f  135° 21. 5  5y  10x 0  10x 
 5y 
 5  
7  2   4 
 
4 2  3
2   
 2  cos 2v  cos2 v 
 sin2 v 7  
2   sin f  2, cos f  
2, p  2; Quadrant II  3  3 2  3  
2x  2y 
 2  0   
3  3
 7  8  9  1  x  y 
   0 
 
2  
2  
 2   2
44 
7   yx3 
x  y 
 3  0  tan2  d d  3(
6)  4(8)  2  2  42   

3 16   
5  16  5  24. 5x  2y  7 → 5x  2y 
 7  0 3  y  
4x  1 → 
3x 
 4y  4  0 5x1  2y1 
 7  
3x1 
 4y1  4  d1  2  2  d2  2  2  
5   2  5x1  2y1 
 7   
 29    
3x1 
 4y1  4   5  
 3  4  25x1  10y1 
 35  
3
29 x1 
 4
29 y1  4
29  (25  3
29 )x (10  4
29 )y 
 35 
 4
29 0  sin x  cos x  1 tan x  1 x  45° or x  225° 19. 2 cos2 x  3 sin x  3 2(1 
 sin2 x)  3 sin x 
 3  0 2 sin2 x 
 3 sin x  1  0 (2 sin x 
 1)(sin x 
 1)  0 2 sin x 
 1  0 or sin x 
 1  0  v0 2  25. R  g sin 2v v0 2  R  g 2 sin v cos v R  32 255 882  3  4  R  232.32 ft  1  sin x  2 sin x  1 x  30°, 150° x  90°  553  Chapter Tests  16. u r u s  
1(4)  3(3)  4(
6)  
19 4u 4u 17. u r u s  3 i 
 
1 j  
1 3 u k 3 
6 4 
6 4 3  
30u i  10u j 
15u k  Chapter 8 Test Page A63 1. 2.5 cm, 60°  2. 1.6 cm, 25°   
30, 10, 
15 18. No; sinceu r u s  
19 and not 0. 19.  a  b  3. 3.9 cm  125˚  a 46˚ b  125 lb  165˚  3.9 cm, 46°  
  60 lb  2b   4.  40˚  x2  1252  602 
 2(125)(60) cos 55° x  
19,22 5 5,000 
 15° cos 5 x  103.06 lb  334˚  sin 55°  103.06  1.8 cm  sin v   60  103.06 sin v  60 sin 55°  2b   a  60 sin 55°   v  sin1  103.06  v  28.48° v  40°  28.48°  40° or 68.48° 20. x  x1  ta1 y  y1  ta2 x  3  t(2) y  11  t(
5) x  2t  3 y  
5t  11 21. x  2t  3 yt1  a   1.8 cm, 334° u 5. AB  
1 
 3, 9 
 6  
4, 3 u  
 AB (
4)2   (3)2  
25  or 5 6. u AB  3 
 (
2), 10 
 7  5, 3 u  
 AB 52  32  
34  7. u AB  9 
 2, 
3 
 (
4), 7 
 5  7, 1, 2 u 72  12  22 AB   
  
54  or 3
6  u 8. AB  
8 
 (
4), 
10 
 (
8), 2 
 (
2)  
4, 
2, 4 u AB   
 (
4)2   (
2 )2  42  
36  or 6 9. u r 
u s  
1 
 4, 3 
 3, 4 
 (
6)  
5, 0, 10 10. 3u s  3 4, 3 3, 3 
6  12, 9, 
18 2u r  2 (
1), 2 3, 2 4  
2, 6, 8 3u s 
 2u r  12 
 (
2), 9 
 6, 
18 
 8  14, 3, 
26 11. u r  3u s  
1  12, 3  9, 4  (
18)  11, 12, 
14 12. u r   
 (
1)2   32  42  
26  13. u s   
 42  32  (
 6)2  
61  14. u r  
u i  3u j  4u k u u 15. u s  4 i  3 j 
 6u k Chapter Tests  55˚  x  x
3  2  t  y
1 y  y
1t x
3  2 1 1 x 
  2 2  22. vx  100 cos 2° vy  100 sin 2° vx  99.94 mph vy  3.49 mph 23. The figure is four times the original size and reflected over the yz-plane. u 24. AB  1.5 cos 60°, 0, 1.5 sin 60°  0.75, 0, 0.75
3  u 110 lb F  0, 0, 
110  1.5 ft 60˚  u u T  AB u i  0.75 0  u F u u j k 0 0.75
3  0 
110 u u  0 0.75
3 i 
 0.75 0.75
3 j  0.75 0 u k 0 0 0 0 
110 
110 u  82.5j u  0k u  0i  u T   
 02  8 2.52  02  82.5 lb-ft  554  u sin v 
 1gt2 25. y  tv 2  u cos v x  tv  10. r  
 (
2)2   (
3 )2  y  28t sin 35° 
 2(32)t2  x  28t cos 35°  0  4t(7 sin 35° 
 4t) 4t  0 or 7 sin 35° 
 4t  0  x  23.02 feet    
13  3.61 (3.61, 4.12) 11. x  r cos v  1  
7 sin 35°  t0   t 
4   4.12  5 
4   3 cos   t  1.003758764  
3   v  Arctan  
2     y  r sin v     3
2   32  
2   
2   3
2   3
2    
2   2  3
2  3
2  ,  
 2 2  Chapter 9 Test  12. x  r cos v 90˚  120˚ 150˚  2.  60˚  1 2 3 4  0˚  330˚  210˚ 240˚  2 3  270˚  2  300˚  0  5.  2 3  7.  120˚  90˚  x  
2 y 
 10 0  
2  0˚  5
34   300˚  1 2 3 4  240˚  270˚  8. r  
 22  22  
8  or 2
2  2
2, 4  3
34   
3   f  Arctan  
5   180°   211° p  r cos (v 
 f) 3
34   34  0˚   r cos (v 
 211°)  A2   B2  
 22  ( 
4)2 19. 
  
20  or 2
5  2 4 1 x 
 y 
   0 2
5  2
 5 2
 5  330˚  210˚  3
34   cos f  
34, sin f  
34, p  34  30˚  180˚  
2   A2   B2  

 52  32 18. 

  

34  5 3 3 x 
 y 
   0 
 
34  
34  
 34  60˚  150˚  15. x2  y2  3x r2  3r cos v r  3 cos v  
2   
 2  330˚ 270˚  y  r sin v  
4 sin 1.4  
3.94  0  2x  2y 
 5  2 4 6 8  240˚  5 3  3 2   
1  y  
3 r sin v  
3 
3  r sin v  
 2  60˚  210˚  11 6  1  0  2r cos v  2r sin v 
 5 30˚  180˚  0  4 3  90˚   2
2  r  
3 csc v r7 r2  49 x2  y2  49 17. 5  r cos (v 
 45°) 0  r cos (v 
 45°) 
 5 0  r (cos v cos 45°  sin v sin 45°) 
 5  5 3  150˚  1 2 3 4  7 6  120˚  3 2   2
2  16.  11 6  6.  6     3  0  4 3   3  14.  1 2 3 4  5 3  5 6   2  7 6  6   2  5 3   6    C 11 3 2  11 6  5 6  1 2 3 4  4 3  7 6  2 3   2 sin 6   

3  (

3 , 
1) 13. x  r cos v  
4 cos 1.4  
0.68 (
0.68, 
3.94)   6  1 2 3 4  3 2  7   2 cos 6 
3    3  0    4.  6  7 6   2  B  4 3   3  5 6   2 3 5 6  30˚  A  180˚  3.  y  r sin v  7  Page A64 1.  5   3 sin 
4  300˚  
5   v  Arctan   4  2
5   
5   cos f  5, sin f  
5, p  10  2  2  
2  f  Arctan 1  360°  297° p  r cos (v 
 f)  9. r  
 (
6)2   02  
36  or 6 Since x  0 and y  0, v  . (6, )  20.  
 5   r cos (v 
 10 i93  (i4)23 i  297°)   123 i i  555  Chapter Tests  32. x3 
 i  0 → x3  i Find the cube roots of i.  21. (2 
 5i)  (
2  4i)  (2  (
2))  (
5i  4i)  0  (
i)  
i 22. 
6i 
 (
3  2i)  
6i  3 
 2i  8i  3 23. (3  5i)(3 
 2i)  9  9i 
 10i2  19  9i 24. (1 
 3i)(2 
 i)(1  2i)  (2 
 7i  3i2)(1  2i)  (
1 
 7i)(1  2i)  
1 
 9i 
 14i2  13 
 9i 25.  6 
 2i  2i    r  
 02  12  
1  or 1 1   v  2      1     2n      4n    4n      
3   1  x1  cos 6  i sin 6  2  2i  2
i  2
i 12 
 10i  2i2  4 
 i2 10 
 10i  5  5  5  9  9  
3   1  x2  cos 6  i sin 6  
2  2i    x3  cos 6  i sin 6  
i 1  v  Arctan   (
4)2   42 26. r  
  or 4
2   
32 4
2 cos  3  4     i sin  3  4    3  4  3 
 2      
1   v  Arctan  
5     i 
3 2  1 i 2  O   12 i  1  0  27. r  
 (
5)2   02  or 5  
25 5(cos   i sin )  
1 i   3  28. r  4 3  33. E  I Z  8 (cos 307°  j sin 307°) 20 (cos 115°  j sin 115°)  8 20 [cos (307°  115°)  j sin (307°  115°)]  160 (cos 422°  j sin 422°)  160 (cos 62°  j sin 62°)    v  2  4 6     12  7   4 or 4 7  7  12cos 4  i sin 4  122  i
2 
 2  
2    6
2  
 6
2 i 2  2
3  
 3    v  3  6  29. r    3  2   6 or     2  Chapter 10 Test    2cos 2  i sin 2  2(0  i)  2i 12  ( 
1)2 30. r  
  Page A65  
1  v  Arctan 1  2  1. d  
 (x2 
  y1)2  (y2 
 y1)2  7    
2   4 7  d  
 3 
 (
 1))2  (1 
  2)2  7  (1 
 i)8  (
2 )8 cos (8)4  i sin (8)4  d  
 42  ( 
1)2 d  
17    16 (cos 14  i sin 14)  16 (1  0)  16  x1  x2 y1  y2  
1  3 2  1 ,  2, 2   2 2 3  1, 2    31. r  
 02  ( 
27)2 v  
2  
729  or 27 1  3 

27i   (0 
 27i) 3    1  3   
2  2. d  
 (x2 
  y1)2  (y2 
 y1)2  
2      i sin      i sin 
6 1 
3    i 
 2 2  1 27 cos 3  3 cos 
6  3  3
3   1  3  d  
 (2k 
 3k)2  (k 
  1 
 (k  1))2    d  
 (
k)2   (
2 )2 2 d  
 k 4 x1  x2 y1  y2  3k  2k k  1  k 
 1 ,  2, 2   2 2 5     2 k, k  3   2 
 2i  3. r  
 (
8 
 (
6)2  (3  
 (
4 ))2 r  
 (
2)2   (7)2 r  
53  (x 
 h)2  (y 
 k)2  r2 2 (x 
 (
8)2  (y 
 3)2  
53  (x  8)2  (y 
 3)2  53  Chapter Tests  2n   cos 6  i sin 6  6 
 2i  4  
4  1  3   1 5 cos 6  3  i sin 6  3    2i    2 
 2i      (0  i) 3  1 cos 2  2n  i sin 2  2n  556  4. center: (h, k)  (0, 0) a2  10 b2  6 c  
 a2 
 b2 a  
10  b  
6  c  
10 6 
  or 2 foci: (h, k  c)  (0, 2) major axis vertices: (h, k  a)  0, 
10  minor axis vertices: (h  b, k)  (
6 , 0) 5. e   c  a    (x 
 h)2  a2  ( y 
 k)2  b  2 ( y 
 0)2 (x 
 0)2     3 1  4 4y2 x2  3   1  2  1  If c  2, then a  1.  
 a2 
 b2  c  
 12 
 b2  1  2 1  4  10. A  1, C  0; since C  0, the conic is a parabola. x2  6x 
 8y 
 7  0 x2  6x  9  8y  7  9 (x  3)2  8y  16 (x  3)2  8(y  2)  y  1 1  O  1  x   1 
 b2 3  b2  4 6. center: (h, k)  (
4, 2) a2  16 b2  7 b2  c2 
 a2 a4 b  
7  7  c2 
 16 → c  
23  foci: (h, k  c)  
4, 2  
23  vertices: (h, k  a)  (
4, 2  4)  (
4, 6) and (
4, 
2) a asymptotes: y 
 k  b(x 
 h) y y 7. foci: (h, k  c)  11. A  4, C  
1; since A and C have opposite signs, the conic is a hyperbola. 4x2 
 y2  1 x2  1  4  y  4 (x 
 (
4)) 
2 
7  4
 7 
 2   7(x  4) 3 c ⇒ h  
5 e  2  a b2  c2 
 a2  kc 4 k 
 c  
2 2k  2 k  1; c  3  (y 
 k)2 (x 
 h)2  
 b 2 a2 (y 
 1)2 (x 
 (
5))2    
 22 5 (y 
 1)2 (x  5)2  
  4 5  y2  
 1  1  O  x  b2  32 
 22 b2  5  1  12. A  1, C  1; since A  C, the conic is a circle. x2  y2  4x 
 12y  36  0 x2  4x  4  y2 
 12y  36  4 (x  2)2  (y 
 6)2  4  1 1  8. vertex: (h, k)  (0, 
3) 4p  8 ⇒ p  2 focus: (h  p, k)  (0  2, 
3)  (2, 
3) directrix: x  h 
 p x0
2 x  
2 axis of symmetry: y  k y  
3 9. 2k  p  
5 foci: (h, k  p)  (3, 
5) 2k 
 p  
2 directrix: y  k 
 p  
2 2k  p  
7 7  y  O  x  3  k  
2, p  
2 (x 
 h)2  4p(y 
 k)  
32y 
 
72 7 (x 
 3)2  
6y  2  (x 
 3)2  4  557  Chapter Tests  13. A  3, C  
16; since A and C have opposite signs, the conic is a hyperbola. 3x2 
 16y2 
 18x  128y 
 37  0 3(x2 
 6x  9) 
 16(y2 
 8y  16)  37  27 
 256 3(x 
 3)2 
 16( y 
 4)2  
192 ( y 
 4)2  12  14 12 10 8 6 4 2  O 
6
4
2 
2 
4 
6  16. A  0, C  1; since A  0, the conic is a parabola. y2 
 2x  10y  27  0 y2  10y  25  2x 
 27  25 (y  5)2  2x 
 2 (y  5)2  2(x 
 1)  (x 
 3)2  O y  
 64  1  x  y  x 2 4 6 8 1012 14  17. A  1, C  2; since A and C have the same sign and A C, the conic is an ellipse. x2  2y2  2x 
 12y  11  0 2 (x  2x  1)  2(y2 
 6y  9)  
11  1  18 (x  1)2  2(y 
 3)2  8  14. A  9, C  
1; since A and C have opposite signs, the conic is a hyperbola. 9x2 
 y2 
 90x  8y  200  0 9(x2 
 10x  25) 
 (y2 
 8y  16)  
200  225 
 16 9(x 
 5)2 
 ( y 
 4)2  9 (x 
 5)2  1  (x  1)2  8  (y 
 3)2   4  1  y  ( y 
 4)2  
 9  1  y  x  O  18. y  2x2  x 19. x  2 cos t  x  O  15. A  2, C  
13; since A and C have opposite signs, the conic is a hyperbola. 2x2 
 13y2  5  0 2x2 
 13y2  
5 13y2  5 y2  5  13  x  2  x2  5  2  Chapter Tests  sin2   sin t  t1  y 2   2  1  x2  4 x2  y2   4  1   y2  4 20. B2 
 4AC  02 
 4(4)(1)  
16 A C; ellipse 4(x  1)2  (y 
 3)2  36 4(x  1 
 3)2  (y 
 3 
 (
5))2  36 4(x 
 2)2  (y  2)2  36 4x2 
 16x  16  y2  4y  4  36 4x2  y2 
 16x  4y 
 16  0  1  y  O  t  x 2  2  2x2  
   cos t  cos2  
 5  1  y  2 sin t y  2  x  558  21. B2 
 4AC  02 
 4(2)(
1) 8 hyperbola  25.  1  
 3  Replace x with x
 cos 60°  y
 sin 60° or 2x
  2y
. Replace y with 
x
 sin 60°  y
 cos 60° or 
3  
2x
        2 2    8  8  39  y  2(
3) 
 3 or 
9  Chapter 11 Test  8  Page A66 2   2  1  
3  1. 343 3  
343  2 7  49 3  2. 64  1 3 
64  1  4      3  2  5   4  12   20   4. x y2z 4   x 2 y8z 4  x6y8z5  3. ((2a)3)2  (2a)6 1    (2a)6 1    26a6  If x  3, y   
3  23   0.9. 2  1    64a6  
(2)2   2(2)  0. If x  2, y  
 (2, 0), (0.7, 0.9) 23.  27  If a person is walking northeast, the person will first hit the motion detector at (
3, 
9).  0 (x
)2 
 6
3 x
y
 
 5( y
)2  32  0 x2  4y2  4 22. (x 
 1)2  y2  1 x2 
 2x  1  y2  1 x2  4(
x2  2x)  4 x2  y2 
 2x  0 x2 
 4x2  8x  4 2 2 y  
x  2x 3x2 
 8x  4  0 (3x 
 2)(x 
 2)  0 2 x  3 or x  2 2  y  25 
 3 or 5  27  2 2 1 
3  
 3  2y
 
 
2x
  2y
  2 1 3 
3  2 4(x
)2  2x
y
  4(y
)2 3 1 
3  
 4(x
)2 
 2x
y
  4(y
)2  1 3 3 1 
3  (x
)2  
3 x
y
  2(y
)2 
 4(x
)2  2x
y
 
 4(y
)2  2 x
y
  5( y
)2 
 32  
(x
)2  6
3    y  2x 
 3  x 5 or x  
3  1 y
. 2  1 x
 2  x2  y2  90 x2  (2x 
 3)2  9 x2  4x2 
 12x  9  90 5x2 
 12x 
 81  0 (5x 
 27)(x  3)  0  1   6   12   12  (33) 3 a 3 b 3 5. 
 27a6b  3a2b4 3  y  1   2   3   4   6. m 2 n 3  m 6 n 6 6  
 m3n4 y 7.  O  x  24. x2  y2  Dx  Ey  F  0 02  12  D(0)  E(1)  F  0  ⇒ E  F  
1 (
2)2  32  D(
2)  E(3)  F  0 ⇒ 
2D  3E  F  
13 42  52  D(4)  E(5)  F  0 ⇒ 4D  5E  F  
41 
4D  6E  2E  
26 3E  3F  6
3 4D  5E  F  
41 11E  3F  
67 11E  3F  
67 
8E  64 E  
8 E  F  
1 
2D  3E  F  
13 
8  F  
1 
2D  3(
8)  7  13 F7 
2D  4 D  
2 2 2 x  y 
 2x 
 8 y  7  0 x2 
 2x  1  y2 
 8y  16  
7  1  16 (x 
 1)2  (y 
 4)2  10 center: (h, k)  (1, 4) radius  
10   8.  O y  x  O  x  1  2  9. 4  2  10.  11. log5 625  4 13. logx 32  
5 x5  32 1  x5 1  32 1 5  2 1  2    559  3  16   216  12. log8 m  5   32  x5  x5 x Chapter Tests  14. log5 (2x)  log5 (3x 
 4) 2x  3x 
 4 4x 15. 3.6x  72.4 x log 3.6  log 72.4 x 16.  Chapter 12 Test Page A67 1. d  4.5 
 2 or 2.5 7  2.5  9.5, 9.5  2.5  12, 12  2.5  14.5, 14.5  2.5  17 9.5, 12, 14.5, 17 2. d  
1 
 (
6) or 5 a24  
6  (24 
 1)5 a24  109 3. 8  
4  (5 
 1)d 12  4d 3d 
4  3  
1, 
1  3  2, 2  3  5 
4, 
1, 2, 5, 8  log 72.4  log 3.6  x  3.3430 6x
1  82
x (x 
 1) log 6  (2 
 x) log 8 x log 6 
 log 6  2 log 8 
 x log 8 x log 6  x log 8  2 log 8  log 6 x(log 6  log 8)  2 log 8  log 6 2 log 8  log 6   x log 6  log 8 log 15   17. log4 15   log 4   1.9534  n  4. 345  2(2(12)  (n 
 1)5)  log 0.9375   18. log3 0.9375   log 3  690  24n  5n2 
 5n 0  5n2  19n 
 690 0  (5n  69)(n 
 10)   
0.0587 log 3   19. log81 3   log 81  69  1  n  
5 or n  10   4 20. log 542  2.7340 21. ln 0.248  
1.3943 22. antiln (
1.9101)  0.1481  Since there cannot be a fractional number of terms, n  10.  ln 2  23. t  k  5.  ln 2   t 0.054     t  12.84 yr 24. Let x  the original number of bacteria. 3x  xek(6) 3  e6k ln 3  6k k  6.  1  16 1 2  ln 3  6  k  0.1831020481 8x  xe0.1831t 8  e0.1831t ln 8  0.1831t t  1  106 1  5   (5 e  ln 0.2   r 1  1  16, 8, 4, 2, 1 7. r   5  5  2  S10   t  
BC    t  
 106)e 3(2 106)    t  
 6 106  or 2 5 5  
 (2)10 2 2  1
2 5 5120  
  2 2  
1 5115  2  8. does not exist;  t 
 6 106  n3  3  lim  2 3 n→ n  1  t  ln 0.2(
6 106) t  9.66 106 s  3 n   n2  lim  , lim n→ 3  1 n→ n2  1  2  0, n→ n  3  3, and lim  so the denominator approaches 3. As n approaches infinity, the n term in the numerator makes the whole numerator approach infinity, so the entire fraction has no limit. 9.  Chapter Tests   r4 1  ln 8  0.1831  Q(t)  Qe  8     6425, 6425 25   3125  162  8, 82  4, 42   2  t  11.3568626 hours 0.3568626(60)  21 minutes 11 hours, 21 minutes 25.  1  10 2 r  or 5 1  4 1 2 2 2 2    ,   25 5 125 125 5 2 4 8 , ,  125 625 3125 1  16r5
1  560  n3  4  lim  3 n→ 2n  3n  n3 4    n3 n3   lim 2n3 3n n→    n3 n3 10   20 1  2  Evaluate the original formula for n  k  1.  1  . 10. The general term is  3n2 1  3n2    1  n  (k  1)[(k  1)  1][4(k  1)  5]  3  11. The series is arithmetic, so it is divergent. 19  12. Sample answer:   1 5k k   3 k
1  13. Sample answer:  62 k1  14. (2a 
 3b)5 5 4(2a)3(
3h)2   (2a)5(
3b)0  5(2a)4(
3b)1   2 1   15. 16.  5 4 3 2 (2a)1(
3b)4 5 4 3(2a)2(
3b)3     3 2 1 4 3 2 1 5 4 3 2 1(2a)0(
3b)5  5 4 3 2 1 32a5 
 240a4b  720a3b2 
 1080a2b3   810ab4 
 243b5  10 !  a10
5 5!(10 
 5)!  10 9 8 7 6 5 4 3 2 1 25   5 4 3 2 1 5 4 3 2 1  8064a5  8! (3x)8
4(
y)4 4!(8 
 4)! 8 7 6 5 4 3 2 1   4 3 2 1 4 3 2 1  70(81x4)(y4)  (k  1)(k  2)(4k  9)    3 The formula gives the same result as adding the (k  1) term directly. Thus, if the formula is valid for n  k, it is also valid for n  k  1. Since the formula is valid for n  1, it is also valid for n  2. Since it is valid for n  2, it is also valid for n  3, and so on, indefinitely. Thus, the formula is valid for all positive integral values of n. 20. There are 4 groups of three months in a year. There are 4 10 or 40 groups of three months in ten years. So, n  40. Since the interest is 0.08 compounded quarterly, the rate per period is 4 or 0.02. The common ratio r is 1.02. a1  200(1.02)  204  for all n, so convergent  a1 
 a1rn   Sn   1
r  a5 32  S40   204 
 204(1.02)40  1 
 1.02  S40  $12,322.00  (3x)4 (
y)4  Chapter 13 Test   5670x4y4  17. r  
 (
2)2   22 or 2
2   3    v  Arctan  
2    or 4 2  
2  2i  2
2 cos  3  4  Page A68   i sin  3  4    2
2 e  6!   1. P(6, 2)   (6 
 2)!  3 i4    18. z0  2i z1  3(2i)  (2 
 i)  6i  2 
 i  2  5i z2  3(2  5i)  (2 
 i)  6  15i  2 
 i  8  14i z3  3(8  14i)  (2 
 i)  24  42i  2 
 i  26  41i 19. Step 1: Verify that the formula is valid for n  1.   30 7!   2. P(7, 5)   (7 
 5)!    7 6 5 4 3 2 1  2 1   2520 8!   3. C(8, 3)   (8 
 3)! 3!    8 7 6 5 4 3 2 1  5 4 3 2 1 3 2 1   56 5!   4. C(5, 4)   (5 
 4)! 4!  1(1  1)(4 1  5)  Since 2 1(2 1  1)  6 and   6, 3 the formula is valid for n  1. Step 2: Assume that the formula is valid for n  k and prove that it is valid for n  k  1. 2 3  4 5  6 7  . . .  2k(2k  1)  5 4 3 2 1    1 4 3 2 1 5 5. Using the Basic Counting Principle, 5 4 3 2 1  120.  k(k  1)(4k  5)  5!   3 2 3  4 5  6 7  . . .  2k(2k  1)  2(k  1)(2k  3)   6. P(5, 3)   (5 
 3)! 5 4 3 2 1    2 1  60 7. (7 
 1)!  6!  720  k(k  1)(4k  5)   3  2(k  1)(2k  3) k(k  1)(4k  5)  6 5 4 3 2 1  4 3 2 1  6(k  1)(2k  3)   3  3  8. C(3, 1) C(12, 8) 3!    (3 
 1)! 1!  k(k  1)(4k  5)  6(k  1)(2k  3)    3  3 2 1    2 1 1  1485  (k  1)(k(4k  5)  6(2k  3)   3  12!  (12 
 8)! 8! 12 11 10 9 8 7 6 5 4 3 2 1  4 3 2 1 8 7 6 5 4 3 2 1 4!  6!  (6 
 3)! 3! 6 5 4 3 2 1 4 3 2 1   3 2 1 3 2 1 2 1 2 1   9. C(4, 2) C(6, 3)   (4 
 2)! 2!  (k  1)(4k2  17k  18)    3    120  (k  1)(k  2)(4k  9)    3  561  Chapter Tests  4. Sample answer: 1.5, 4.5, 7.5, 10.5, 13.5, 16.5, 19.5 5. Sample answer:  5  9  6  10. P(2 white)  1 0 1   3 1  11. P(s)  4  1  odds   1  4  3  4  Number of Absences 0–3 3–6 6–9 9–12 12–15 15–18 18–21  P(f )  1 
 P(s) 3   1 
 4 or 4 1  or 3  12. P(5 clubs or 5 hearts or 5 spades or 5 diamonds)  P(5 clubs)  P(5 hearts)  P(5 spades)  P(5 diamonds)  452 13    12  51  10  49  11  50    9  48  33  16,660 5  6.  3  36  13. P(sum of 8) P(sum of 4)  3 6 15  5     1296 or 432 1  1  2  14. P(3 odd digits)  2  1  2  1   8 15. P(all red or all blue)  P(all red)  P(all blue) 3  2  11   1 2    1  220 1  20    1  10  5   1 2  4  11  3  10  1  22  26  28  7  2  15  1  5  18. P(both eveneven product) P(both even and even product)  P(even product) 3    2 6 or 13  0.006  semi-interquartile range  2  19. P(no more than two heads)  P(0 heads)  P(1 head)  P(2 heads)   0.003  2 0 1 5 2 1 1 4  C(5, 0) 3 3  C(5, 1) 3 3 2 2 1 3  C(5, 2) 3 3 1 10 40     243  243  243 51 17    243 or 81                12. 2.338 2.340 2.342 2.344 2.346 2.348 2.350 1  13. X   1 5 (2.334  2.338  . . .  2.350)  2.34393333 1  MD  1 5 ( 2.34393333 
 2.334  . . .  2.34393333 
 2.350 )  0.0025  1792    15,625 or 0.114688  14. j   Chapter 14 Test  (2.34393333 
 2.334)2  . . .  (2.34393333 
 2.350)2  15   0.0031 15. 68.3% of the data lie within 1 standard deviation of the mean. 24  2.8  21.2 
 26.8 16. 90% corresponds to t  1.65. 24  1.65(2.8)  19.38
28.62  Page A69 1. range  20 
 1 or 19 2. Sample answer: 3 3. Sample answer: 0, 3, 6, 9, 12, 15, 18, 21 Chapter Tests  6 9 12 15 18 21 Number of Absences  there are 15 terms, the median is the 2 or 8th term. Md  2.344 10. Q1  2.341; Q3  2.347 11. interquartile range  Q3 
 Q1  2.347 
 2.341  0.006   1 8  4 4 1 3  5  3  9. Order the values from least to greatest. Since  P(3's together and odd number)  P(odd number)  20. C(7, 4)5  0  10  10  17. P(3'stogetheroddnumber)  6  3 6 16 25 17 8 5  Md  2 or 10   52 or 1 3    ||| ||||| ||||||||||||| |||||||||||||||||||| |||||||||||||| ||||||| ||||  1     5 2  52 
 52    Frequency  7. X   8 0 (6  16  12  7  . . .  8  9  7  9)  10.44 8. Order the data from least to greatest. Since there are 80 terms, the median is the average of the 40th term and the 41st term.  16. P(ace or black card)  P(ace)  P(black card) 
 P(black ace) 4  27 24 21 18 Frequency 15 12 9 6 3 0  Tallies  562  17. 24 
 18.4  5.6 tj  5.6 t(2.8)  5.6 t2 95.5%  2   47.75% 47.75%  49.85%  97.6% 18. 29.6 
 24  5.6 19. tj  5.6 t(2.8)  5.6 t2 95.5%  2  10. f(x)  6x4 
 2x2 
 30 f
(x)  6 4x4
1 
 2 2x2
1 
 0  24x3 
 4x  32.4 
 24  8.4 tj  8.4 t(2.8)  8.4 t3   47.75%  99.7%  2  2  11. f(x)  2x5 
 4x3  5x2 
 6   49.85%  2  f
(x)  2 5x5
1 
 4 3x3
1  5 2x2
1 
 0 4   10x4 
 12x2  5x  j   jX     
N 3.6   jX  
4 00  jX  0.18   12. f(x)  2x4(x3  3x2)  2x7  6x6 f
(x)  2 7x7
1  6 6x6
1  14x6  36x5 13. f(x)  (x  3)2  x2  6x  9 f
(x)  2x2
1  6 1x1
1  0  2x  6 14. f(x)  
2x  6  20. A 5% level of confidence is given when P  95%. 95% corresponds to t  1.96. X   tjX  57  1.96(0.18)   56.65
57.35  
2   11  6x  C F(x)   1  1x   
x2  6x  C 15. f(x)  
x3  4x2 
 x  4  Chapter 15 Test  
1  Page A70  1  1    x→2  4. lim  x→1     lim (x  3)  lim  x→1  (x 
 2)(x 
 1)  x
1  1  1  x→1  6.  7.   lim  x→3  4   21 
 x11  x  C F(x)   2  1x 11   3x3 
 2x2  x  C   
1 x2 
 9  3 x→3 x 
 27   5x  C  17. f(x)  x  x2 
 4x  1   lim (x 
 2) 5. lim  2 1 1 x31 
  x11 7 11 31 1 2 x4 
 x2  5x  C 8 14 1 1 x4 
 x2  5x  C 8 7 x3 
 4x2  x  x→0  x2 
 3x  2  x
1  2  F(x)  2  So f(2)  
1. 3  1  1  2. The closer x is to 2, the closer y is to 4. So, lim f(x)  4. However, there is a point at (2, 
1). x2  3x  x  4  16. f(x)  2 x3 
 7x  5  x→
1  x→0  1   
4x4  3x3 
 2x2  4x  C  1. The closer x is to 
1, the closer y is to 1. So, lim f(x)  1. Also, f(
1)  1.  3. lim  4   31  x21 
 x11  4x  C F(x)   3  1x 21 11   x dx  lim   2  (x  3)(x 
 3)  (x 
 3)(x2  3x  9)  18.  0  x3   lim  2 x→3 x  3x  9 33   32  3(3)  9 2 6   2 7 or 9 x2 
 2x  3 12 
 2(1)  3  lim 2    3(1)2 
 5 x→1 3x 
 5 2   
2 or 
1 f(x  h) 
 f(x) f 
(x)  lim h h→0 (x  h)2 
 2(x  h) 
 (x2 
 2x)  lim  h h→0 x2  2xh  h2 
 2x 
 2h 
 x2  2x  lim  h h→0 2xh  h2 
 2h  lim h h→0  n  3  n→ i  1  2i 3 2   n n     16 n2 (n  1)2   4 4 n→ n n2  2n  1  lim 4 n 2 n→     lim         lim 41  n  n2  2  1  n→    3  19.  1   4 units2 3x2dx  3  x3    1  20.  0  3 1 x3 3 1  3  1   33 
 13  26 units2 (2x  3)dx  2  1 x2 2   3x   x2  3x   lim (2x  h 
 2) h→0  1  1 0  0   [12  3(1)] 
 [02  3(0)] 4   2x 
 2 1  8. f(x)  2x2 
 7x  1 1  f 
(x)  2 2x21 
 7 1x11  0 x
7 9. f(x)  4x3 
 4 f 
(x)  4 3x3
1 
 0  12x2  563  Chapter Tests    3  21.  1  1  1  (
x2  x  6) dx  
3x3  2x2  6x 1  1  1  1  3  24.  1   
3(3)3  2(3)2  6(3) 
 
3(1)3  2(1)2  6(1) 27  37  25. V  2   2 
 6 22   3 22.  x
 23.    (3x2   4x  7)    Chapter Tests    r  0  
brx2  hxdx h  1 x3 3   2 
r  (1 
 2x)dx  x 
 22x2  C x2  h(t)  3  95t 
 16t2 h
(t)  v(t)  0  95 1t1
1 
 16 2t2
1  95 
 32t h
(2)  v(2)  95 
 32(2)  31 ft/s  h  
3hr r3  h2 hr hr  2 
3  2  2  C  2  3 4 x3  x2  7x  C 3 2 x3  2x2  7x  C  r 1 x2 2 0  r2 
 
3r 03  2 02 h  2  V(h, r)  V(2, 3)  2 
3  2 2 32   6 units3  564  2 32  h        

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Solution of the System 10x+12y1 25x-8y+8

Source: https://usermanual.wiki/Document/glencosolutionmanual.63298667/help

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