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Solution Manual for Trigonometry, 5th Edition

Solution Manual for Trigonometry, 5th Edition makes tackling textbook exercises a breeze, with clear and concise answers to every problem.

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    S OLUTIONS M ANUAL E DGAR R EYES Southeastern Louisiana University T RIGONOMETRY F IFTH E DITION Mark Dugopolski Southeastern Louisiana University

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    Table of Contents Chapter P .................................................................................................1 Chapter 1 ................................................................................................32 Chapter 2 ................................................................................................83 Chapter 3 ..............................................................................................137 Chapter 4 ..............................................................................................195 Chapter 5 ..............................................................................................245 Chapter 6 ..............................................................................................302

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    P.1 The Cartesian Coordinate System 1 For Thought 1. False, the point (2 , − 3) is in Quadrant IV. 2. False, the point (4 , 0) does not belong to any quadrant. 3. False, since the distance is √ ( a − c ) 2 + ( b − d ) 2 . 4. False, since Ax + By = C is a linear equation. 5. True 6. False, since √ 7 2 + 9 2 = √ 130 ≈ 11 . 4 7. True 8. True 9. True 10. False, since the radius is 3. P.1 Exercises 1. ordered 2. Cartesian 3. x -axis 4. origin 5. Pythagorean theorem 6. circle 7. linear equation 8. y -intercept 9. (4 , 1), Quadrant I 10. ( − 3 , 2), Quadrant II 11. (1 , 0), x -axis 12. ( − 1 , − 5), Quadrant III 13. (5 , − 1), Quadrant IV 14. (0 , − 3), y -axis 15. ( − 4 , − 2), Quadrant III 16. ( − 2 , 0), x -axis 17. ( − 2 , 4), Quadrant II 18. (1 , 5), Quadrant I 19. c = √ ( √ 3) 2 + 1 2 = √ 4 = 2 20. Since a 2 + a 2 = √ 2 2 , we get 2 a 2 = 2 or a 2 = 1. Then a = 1. 21. Since b 2 + 2 2 = 3 2 , we get b 2 + 4 = 9 or b 2 = 5. Then b = √ 5. 22. Since b 2 + ( 1 2 ) 2 = 1 2 , we get b 2 + 1 4 = 1 or b 2 = 3 4 . Thus, b = √ 3 2 . 23. Since a 2 + 3 2 = 5 2 , we get a 2 + 9 = 25 or a 2 = 16. Then a = 4. 24. c = √ 3 2 + 2 2 = √ 9 + 4 = √ 13 25. √ 4 · 7 = 2 √ 7 26. √ 25 · 2 = 5 √ 2 27. √ 5 √ 9 = √ 5 3 28. √ 3 √ 16 = √ 3 4 29. √ 2 √ 3 · √ 3 √ 3 = √ 6 3 30. √ 3 √ 5 · √ 5 √ 5 = √ 15 5 31. 2 √ 3 √ 5 · √ 5 √ 5 = 2 √ 15 5 32. 5 √ 3 · √ 3 √ 3 = 5 √ 3 3 33. 1 √ 3 · √ 3 √ 3 = √ 3 3 34. 3 √ 2 · √ 2 √ 2 = 3 √ 2 2 35. √ 2 √ 3 · √ 3 √ 3 = √ 6 3 36. √ 5 √ 2 · √ 2 √ 2 = √ 10 2 37. Distance is √ (4 − 1) 2 + (7 − 3) 2 = √ 9 + 16 = √ 25 = 5, midpoint is (2 . 5 , 5) 38. Distance is √ 144 + 25 = 13, midpoint is (3 , 0 . 5) 39. Distance is √ ( − 1 − 1) 2 + ( − 2 − 0) 2 = √ 4 + 4 = 2 √ 2, midpoint is (0 , − 1)

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    2 Chapter P Algebraic Prerequisites 40. Distance is √ 4 + 4 = 2 √ 2, midpoint is (0 , 1) 41. Distance is √ √ √ √ ( √ 2 2 − 0 ) 2 + ( √ 2 2 − 0 ) 2 = √ 2 4 + 2 4 = √ 1 = 1, midpoint is ( √ 2 / 2 + 0 2 , √ 2 / 2 + 0 2 ) = ( √ 2 4 , √ 2 4 ) 42. Distance is √ ( √ 3 − 0 ) 2 + (1 − 0) 2 = √ 3 + 1 = 2, midpoint is ( √ 3 + 0 2 , 1 + 0 2 ) = ( √ 3 2 , 1 2 ) 43. Distance is √ ( √ 18 − √ 8 ) 2 + ( √ 12 − √ 27 ) 2 = √ (3 √ 2 − 2 √ 2) 2 + (2 √ 3 − 3 √ 3) 2 = √ ( √ 2) 2 + ( − √ 3) 2 = √ 5, midpoint is ( √ 18 + √ 8 2 , √ 12 + √ 27 2 ) = ( 3 √ 2 + 2 √ 2 2 , 2 √ 3 + 3 √ 3 2 ) = ( 5 √ 2 2 , 5 √ 3 2 ) 44. Distance is √ ( √ 72 − √ 50 ) 2 + ( √ 45 − √ 20 ) 2 = √ (6 √ 2 − 5 √ 2) 2 + (3 √ 5 − 2 √ 5) 2 = √ ( √ 2) 2 + ( √ 5) 2 = √ 7, midpoint is ( √ 72 + √ 50 2 , √ 45 + √ 20 2 ) = ( 6 √ 2 + 5 √ 2 2 , 3 √ 5 + 2 √ 5 2 ) = ( 11 √ 2 2 , 5 √ 5 2 ) 45. Distance is √ (1 . 2 + 3 . 8) 2 + (4 . 4 + 2 . 2) 2 = √ 25 + 49 = √ 74, midpoint is ( − 1 . 3 , 1 . 3) 46. Distance is √ 49 + 81 = √ 130, midpoint is (1 . 2 , − 3) 47. Distance is √ π 2 + 4 2 , midpoint is ( 3 π 4 , 1 2 ) 48. Distance is √ π 2 + 4 2 , midpoint is ( π 4 , 1 2 ) 49. Distance is √ (2 π − π ) 2 + (0 − 0) 2 = √ π 2 = π , midpoint is ( 2 π + π 2 , 0 + 0 2 ) = ( 3 π 2 , 0 ) 50. Distance is √ ( π − π 2 ) 2 + (1 − 1) 2 = √ π 2 4 = π 2 , midpoint is ( π + π/ 2 2 , 1 + 1 2 ) = ( 3 π 4 , 1 ) 51. Distance is √ ( π 2 − π 3 ) 2 + ( − 1 3 − 1 2 ) 2 = √ π 2 36 + 25 36 = √ π 2 + 25 6 , midpoint is ⎛ ⎜ ⎜ ⎝ π 3 + π 2 2 , 1 2 − 1 3 2 ⎞ ⎟ ⎟ ⎠ = ( 5 π 12 , 1 12 ) 52. Distance is √ ( π − 2 π 3 ) 2 + ( − 1 + 1 2 ) 2 = √ π 2 9 + 1 4 = √ 4 π 2 + 9 6 , midpoint is ⎛ ⎜ ⎜ ⎝ 2 π 3 + π 2 , − 1 2 − 1 2 ⎞ ⎟ ⎟ ⎠ = ( 5 π 6 , − 3 4 ) 53. Center(0 , 0), radius 4 3 5 x 3 5 y 54. Center (0 , 0), radius 1 2 -2 x 2 -2 y

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    P.1 The Cartesian Coordinate System 3 55. Center ( − 6 , 0), radius 6 -14 -10 x -6 6 y 56. Center (0 , 3), radius 3 3 -3 x 3 7 y 57. Center (2 , − 2), radius 2 √ 2 2 5 x -2 -5 y 58. Center (4 , − 2), radius 2 √ 5 4 10 x -4 -6 y 59. x 2 + y 2 = 7 60. x 2 + y 2 = 12 since (2 √ 3) 2 = 12 61. ( x + 2) 2 + ( y − 5) 2 = 1 / 4 62. ( x + 1) 2 + ( y + 6) 2 = 1 / 9 63. The distance between (3 , 5) and the origin is √ 34 which is the radius. The standard equation is ( x − 3) 2 + ( y − 5) 2 = 34. 64. The distance between ( − 3 , 9) and the origin is √ 90 which is the radius. The standard equation is ( x + 3) 2 + ( y − 9) 2 = 90. 65. Note, the distance between ( √ 2 / 2 , √ 2 / 2) and the origin is 1. Thus, the radius is 1. The standard equation is x 2 + y 2 = 1. 66. Note, the distance between ( √ 3 / 2 , 1 / 2) and the origin is 1. Thus, the radius is 1. The standard equation is x 2 + y 2 = 1. 67. The radius is √ ( − 1 − 0) 2 + (2 − 0) 2 = √ 5. The standard equation is ( x +1) 2 +( y − 2) 2 = 5. 68. Since the center is (0 , 0) and the radius is 2, the standard equation is x 2 + y 2 = 4. 69. Note, the center is (1 , 3) and the radius is 2. The standard equation is ( x − 1) 2 +( y − 3) 2 = 4. 70. The radius is √ (2 − 0) 2 + (2 − 0) 2 = √ 8. The standard equation is ( x − 2) 2 + ( y − 2) 2 = 8. 71. We solve for a . a 2 + ( 3 5 ) 2 = 1 a 2 = 1 − 9 25 a 2 = 16 25 a = ± 4 5 72. We solve for a . a 2 + ( − 1 2 ) 2 = 1 a 2 = 1 − 1 4 a 2 = 3 4 a = ± √ 3 2 73. We solve for a . ( − 2 5 ) 2 + a 2 = 1 a 2 = 1 − 4 25 a 2 = 21 25 a = ± √ 21 5

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    4 Chapter P Algebraic Prerequisites 74. Solve for a : ( 2 3 ) 2 + a 2 = 1 a 2 = 1 − 4 9 a 2 = 5 9 a = ± √ 5 3 75. y = 3 x − 4 goes through (0 , − 4), ( 4 3 , 0 ) . 2 5 x -3 -4 y 76. y = 5 x − 5 goes through (0 , − 5) , (1 , 0). 2 4 x -5 5 y 77. 3 x − y = 6 goes through (0 , − 6), (2 , 0). 1 3 x -6 3 y 78. 5 x − 2 y = 10 goes through (0 , − 5), (2 , 0). 1 3 x -5 5/2 -4 y 79. x + y = 80 goes through (0 , 80), (80 , 0). 80 40 x 20 80 y 80. 2 x + y = − 100 goes through (0 , − 100), ( − 50 , 0).  40 20 x  100  50 y 81. x = 3 y − 90 goes through (0 , 30) , ( − 90 , 0). -90 -45 x 30 15 y 82. x = 80 − 2 y goes through (0 , 40) , (80 , 0). 80 40 x 40 20 y

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    P.1 The Cartesian Coordinate System 5 83. 1 2 x − 1 3 y = 600 goes through (0 , − 1800), (1200 , 0). 1200 600 x -1800 -900 y 84. 2 3 y − 1 2 x = 400 goes through (0 , 600), ( − 800 , 0). -800 -400 x 600 300 y 85. Intercepts are (0 , 0 . 0025) , (0 . 005 , 0). 0.005 x 0.0025 y 86. Intercepts are (0 , − 0 . 3) , (0 . 5 , 0). 0.5 -0.5 x -0.3 0.3 y 87. x = 5 3 7 x -5 5 y 88. y = − 2 5 -5 x -1 -3 y 89. y = 4 4 -4 x 3 5 y 90. x = − 3 -2 -4 x -3 3 y 91. x = − 4 -3 -5 x -4 4 y

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    6 Chapter P Algebraic Prerequisites 92. y = 5 -5 5 x 4 6 y 93. Solving for y , we have y = 1. -1 1 x 2 -1 y 94. Solving for x , we get x = 1 . -1 2 x 1 -1 y 95. y = x − 20 goes through (0 , − 20), (20 , 0).  20 20 x 20  20 y 96. y = 999 x − 100 goes through (0 , − 100), (100 / 999 , 0). 0.5 1 x  100 100 y 97. y = 3000 − 500 x goes through (0 , 3000), (6 , 0). 6  4 x 3000 1000 y 98. y = 1 300 (200 x − 1) goes through (0 , − 1 / 300), (1 / 200 , 0). 0.01  0.01 x  0.01 0.01 y 99. The hypotenuse is √ 6 2 + 8 2 = √ 100 = 10. 100. The other leg is √ 10 2 − 4 2 = √ 84 = 2 √ 21 ft. 101. a) Let r be the radius of the smaller circle. Consider the right triangle with vertices at the origin, another vertex at the center of a smaller circle, and a third vertex at the center of the circle of radius 1. By the Pythagorean Theorem, we obtain 1 + (2 − r ) 2 = (1 + r ) 2 5 − 4 r = 1 + 2 r

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    P.1 The Cartesian Coordinate System 7 4 = 6 r r = 2 3 . The diameter of the smaller circle is 2 r = 4 3 . b) The smallest circles are centered at ( ± r, 0) or ( ± 4 / 3 , 0). The equations of the circles are ( x − 4 3 ) 2 + y 2 = 4 9 and ( x + 4 3 ) 2 + y 2 = 4 9 102. Draw a right triangle with vertices at the cen- ters of the circles, and another vertex at a point of intersection of the two circles. The legs of the right triangle are 5 and 12. By the Pythagorean theorem, the hypotenuse is √ 5 2 + 12 2 = 13. 103. Let C ( h, k ) and r be the center and radius of the smallest circle, respectively. Then k = − r . We consider two right triangles each of which has a vertex at C . The right triangles have sides that are perpen- dicular to the coordinate axes. Also, one side of each right triangle passes through the center of a larger circle. Applying the Pythagorean Theorem, we list a system of equations ( r + 1) 2 = h 2 + (1 − r ) 2 (2 − r ) 2 = h 2 + r 2 . The solutions are r = 1 / 2, h = √ 2, and k = − r = − 1 / 2. The equation of the smallest circle is ( x − √ 2) 2 + ( y + 1 / 2) 2 = 1 / 4 . 104. We apply symmetry to the centers of the remaining three circles. From the answer or equation in Exercise 103, the equations of the remaining circles are ( x − √ 2) 2 + ( y − 1 / 2) 2 = 1 / 4 ( x + √ 2) 2 + ( y − 1 / 2) 2 = 1 / 4 ( x + √ 2) 2 + ( y + 1 / 2) 2 = 1 / 4 . 105. The midpoint of (0 , 20 . 8) and (48 , 27 . 4) is ( 0 + 48 2 , 20 . 8 + 27 . 4 2 ) = (24 , 24 . 1) . In 1994 (= 1970 + 24), the median age at first marriage was 24.1 years. 106. a) If h = 0, then 0 = 0 . 229 n + 5 . 203. Then n = − 5 . 203 / 0 . 229 ≈ − 22 . 72. The n -intercept is ( − 22 . 72 , 0) . There were no unmarried couples in 1977 ( ≈ 2000 − 22 . 7). Nonsense. b) If n = 0, then h = 0 . 229(0) + 5 . 203 = 5 . 203. The h -intercept is (0 , 5 . 203). In 2000, there where 5,203,000 unmarried- couple households. 107. The distance between (10 , 0) and (0 , 0) is 10. The distance between (1 , 3) and the origin is √ 10. If two points have integer coordinates, then the distance between them is of the form √ s 2 + t 2 where s 2 , t 2 ∈ { 0 , 1 , 2 2 , 3 2 , 4 2 , ... } = { 0 , 1 , 4 , 9 , 16 , ... } . Note, there are no numbers s 2 and t 2 in { 0 , 1 , 4 , 9 , 16 , ... } satisfying s 2 + t 2 = 19. Thus, one cannot find two points with integer coordinates whose distance between them is √ 19. 108. One can assume the vertices of the right triangle are A (0 , 0), B (1 , √ 3), and C (1 , 0). The midpoint of the hypotenuse AB is ( 1 2 , √ 3 2 ) . The distance between the midpoint and C is √ ( 1 2 − 1 ) 2 + ( √ 3 2 − 0 ) 2 = 1, which is also the distance from the midpoint to A , and the distance from the midpoint to B . 111. On day 1, break off a 1-dollar piece and pay the gardener. On day 2, break of a 2-dollar and pay the gar- dener. The gardener will give you back your change which is a 1-dollar piece.

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    8 Chapter P Algebraic Prerequisites On day 3, you pay the gardener with the 1- dollar piece you received as change from the previous day. On day 4, pay the gardener with the 4-dollar bar. The gardener will give you back your change which will consist of a 1-dollar piece and a 2-dollar piece. On day 5, you pay the gardener with the 1- dollar piece you received as change from the previous day. On day 6, pay the gardener with the 2-dollar piece you received as change from day 4. The gardener will give you back your change which is a 1-dollar piece. On day 7, pay the gardener with the 1-dollar piece you received as change from day 6. 112. Let  ABC be a right triangle with vertices at A (2 , 7), B (0 , − 3), and C (6 , 1). Notice, the midpoints of the sides of  ABC are (3 , − 1), (4 , 4) and (1 , 2). The area of  ABC is 1 2 AC × BC = 1 2 √ 4 2 + 6 2 √ 6 2 + 4 2 = 1 2 (52) = 26 . P.1 Pop Quiz 1. The distance is √ 16 + 4 = √ 20 = 2 √ 5. 2. Center (3 , − 5), radius 9 3. Completing the square, we find ( x 2 + 4 x + 4) + ( y 2 − 10 y + 25) = − 28 + 4 + 25 ( x + 2) 2 + ( y − 5) 2 = 1 . The center is ( − 2 , 5) and the radius is 1. 4. The distance between (3 , 4) and the origin is 5, which is the radius. The circle is given by ( x − 3) 2 + ( y − 4) 2 = 25. 5. By setting x = 0 and y = 0 in 2 x − 3 y = 12 we find − 3 y = 12 and 2 x = 12, respectively. Since y = − 4 and x = 6 are the solutions of the two equations, the intercepts are (0 , − 4) and (6 , 0). 6. (5 , − 1) For Thought 1 True, since the number of gallons purchased is 20 divided by the price per gallon. 2. False, since a student’s exam grade is a function of the student’s preparation. If two classmates had the same IQ and only one prepared then the one who prepared will most likely achieve a higher grade. 3. False, since { (1 , 2) , (1 , 3) } is not a function. 4. True 5. True 6. True 7. False, the domain is the set of all real numbers. 8. True 9. True, since f (0) = 0 − 2 0 + 2 = − 1. 10. True, since if a − 5 = 0 then a = 5. P.2 Exercises 1. function 2. independent, dependent 3. domain, range 4. parabola 5. function 6. function 7. Note, b = 2 πa is equivalent to a = b 2 π . Then a is a function of b , and b is a function of a . 8. Note, b = 2(5 + a ) is equivalent to a = b − 10 2 . Then a is a function of b , and b is a function of a . 9. a is a function of b since a given denomination has a unique length. Since a dollar bill and a five-dollar bill have the same length, then b is not a function of a .

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    P.2 Functions 9 10. Since different U.S. coins have different diameters, then a is a function of b , and b is a function of a . 11. Since an item has only one price, b is a function of a . Since two items may have the same price, a is not a function of b . 12. a is not a function of b since there may be two students with the same semester grades but different final exams scores. b is not a function of a since there may be identical final exam scores with different semester grades. 13. a is not a function of b since it is possible that two different students can obtain the same final exam score but the times spent on studying are different. b is not a function of a since it is possible that two different students can spend the same time studying but obtain different final exam scores. 14. a is not a function of b since it is possible that two adult males can have the same shoe size but have different ages. b is not a function of a since it is possible for two adults with the same age to have different shoe sizes. 15. Since 1 in ≈ 2.54 cm, a is a function of b and b is a function of a . 16. Since there is only one cost for mailing a first class letter, then a is a function of b . Since two letters with different weights each under 1/2-ounce cost 47 cents to mail first class, b is not a function of a . 17. Since b = a 3 and a = 3 √ b , we get that b is a function of a , and a is a function of b . 18. Since b = a 4 and a = ± 4 √ b , we get that b is a function of a , but a is not a function of b . 19. Since b = | a | , we get b is a function of a . Since a = ± b , we find a is not a function of b . 20. Note, b = √ a since a ≥ 0, and a = b 2 . Thus, b is a function of a , and a is a function of b . 21. A = s 2 22. s = √ A 23. s = √ 2 d 2 24. d = s √ 2 25. P = 4 s 26. s = P/ 4 27. A = P 2 / 16 28. d = √ 2 A 29. y = 2 x − 1 has domain ( −∞ , ∞ ) and range ( −∞ , ∞ ), some points are (0 , − 1) and (1 , 1)  1 1 x 2  3 y 30. y = − x + 3 has domain ( −∞ , ∞ ) and range ( −∞ , ∞ ), some points are (0 , 3) and (3 , 0)  2 3 x 3  2 y 31. y = 5 has domain ( −∞ , ∞ ) and range { 5 } , some points are (0 , 5) and (1 , 5)  2 2 x 4 6 y 32. y = − 4 has domain ( −∞ , ∞ ) and range {− 4 } , some points are (0 , − 4) and (1 , − 4)  3 3 x  3  5 y

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    10 Chapter P Algebraic Prerequisites 33. y = x 2 − 20 has domain ( −∞ , ∞ ) and range [ − 20 , ∞ ), some points are (0 , − 20) and (6 , 16)  6 6 x 30 10  30 y 34. y = x 2 + 50 has domain ( −∞ , ∞ ) and range [50 , ∞ ), some points are (0 , 50) and (5 , 75) 5 10 x 100 150 y 35. y = 40 − x 2 has domain ( −∞ , ∞ ) and range ( −∞ , 40], some points are (0 , 40) and (6 , 4)  6 6 x 30 50 y 36. y = − 10 − x 2 has domain ( −∞ , ∞ ) and range ( −∞ , − 10], some points are (0 , − 10) and (4 , − 26)  4 4 x  20  30 10 y 37. y = x 3 has domain ( −∞ , ∞ ) and range ( −∞ , ∞ ), some points are (0 , 0) and (2 , 8) 1 2 x 8  8 y 38. y = − x 3 has domain ( −∞ , ∞ ) and range ( −∞ , ∞ ), some points are (0 , 0) and (0 , − 8)  1 2 x 8  8 y 39. y = √ x − 10 has domain [10 , ∞ ) and range [0 , ∞ ), some points are (10 , 0) and (14 , 2) 10  10 40 x  2 4 y 40. y = √ x + 30 has domain [ − 30 , ∞ ) and range [0 , ∞ ), some points are ( − 30 , 0) and ( − 26 , 2)  30 10 x  2 6 y

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    P.2 Functions 11 41. y = √ x + 30 has domain [0 , ∞ ) and range [30 , ∞ ), some points are (0 , 30) and (400 , 50) 200 600 x 30 60 y 42. y = √ x − 50 has domain [0 , ∞ ) and range [ − 50 , ∞ ), some points are (0 , − 50) and (900 , − 20) 1000 2000 x  50 10 y 43. y = | x | − 40 has domain ( −∞ , ∞ ) and range [ − 40 , ∞ ), some points are (0 , − 40) and (40 , 0) 20 60 x  40 10 y 44. y = 2 | x | has domain ( −∞ , ∞ ) and range [0 , ∞ ), some points are (0 , 0) and (1 , 2)  1 1 x  2 2 4 y 45. y = | x − 20 | has domain ( −∞ , ∞ ) and range [0 , ∞ ), some points are (0 , 20) and (20 , 0)  10 20 50 x 20 40 y 46. y = | x + 30 | has domain ( −∞ , ∞ ) and range [0 , ∞ ), some points are (0 , 30) and ( − 30 , 0)  30  60 x 30 10 y 47. 3 · 4 − 2 = 10 48. 3(16) + 4 = 52 49. − 4 − 2 = − 6 50. − 8 − 2 = − 10 51. | 8 | = 8 52. | − 1 | = 1 53. 4 + ( − 6) = − 2 54. 24 · 6 = 144 55. 80 − 2 = 78 56. 2 / 2 = 1 57. 3 a 2 − a 58. 4 b − 2 59. f ( − x ) = 3( − x ) 2 − ( − x ) = 3 x 2 + x 60. g ( − x ) = 4( − x ) − 2 = − 4 x − 2 61. Factoring, we get x (3 x − 1) + 0. So x = 0 , 1 / 3 . 62. Since 4 x − 2 = 3, we get x = 5 / 4 . 63. Since | a + 3 | = 4 is equivalent to a + 3 = 4 or a + 3 = − 4, we have a = 1 , − 7 . 64. Since 3 t 2 − t − 10 = ( t − 2)(3 t + 5) = 0, we find t = 2 , − 5 / 3. 65. C = 353 n 66. P = 580 n

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    12 Chapter P Algebraic Prerequisites 67. C = 35 n + 50 68. C = 2 . 50 + 0 . 50 n 69. We find C = 4 B 3 √ D = 4(12 + 11 / 12) 3 √ 22 , 800 ≈ 1 . 822 and a sketch of the graph of C = 4 B 3 √ 22 , 800 is given below. 5 10 20 B 1 2 3 C 70. Solving for B , we get 4 B 3 √ 22 , 800 < 2 B < 3 √ 22 , 800 2 B < 14 ft , 2 in. Then the maximum displacement is 14 ft, 2 in. With D fixed, we get that C becomes larger as the beam B becomes larger. Thus, a boat is more likely to capsize as the beam gets larger. 71. Let N = 2, B = 3 . 498, and S = 4 . 250. Then D = π 4 B 2 · S · N = 81 . 686 in. 3 Then D ≈ 81 . 7 in. 3 . 72. Let N = 2, B = 3 . 518, and S = 4 . 250. Then D = π 4 B 2 · S · N = 82 . 622 in. 3 Using the unrounded answer to Exercise 71, the difference in the displacement is 82 . 622 − 81 . 686 ≈ 0 . 94 in. 3 . 73. Solving for B , D = π 4 B 2 · S · N 4 D πS · N = B 2 B = 2 √ D πS · N . 74. Solving for V , CR = 1 + πB 2 · S 4 V CR − 1 = πB 2 · S 4 V 1 CR − 1 = 4 V πB 2 · S V = πB 2 · S 4( CR − 1) . 75. Pythagorean, legs, hypotenuse 76. circle, radius, center 77 . √ (2 + 3) 2 + ( − 4 + 6) 2 = √ 29 78. ( 4 − 6 2 , − 8 + 16 2 ) = ( − 1 , 4) 79. If we replace x = 0 in 4 x − 6 y = 40, then − 6 y = 40 or y = − 20 / 3. The y -intercept is (0 , − 20 3 ). If we replace y = 0 in 4 x − 6 y = 40, then 4 x = 40 or x = 10. The x -intercept is (10 , 0). 80. The diagonal is √ 3 2 + 7 2 = √ 58 ft. 81. Rewriting the equation, we find 1 3 3 · 3 100 · 1 3 4 · 3 2 x = 1 3 · 3 x 1 3 3 · 3 100 · 1 3 4 · 3 2 x = 3 x − 1 3 2 x +93 = 3 x − 1 2 x + 93 = x − 1 x = − 94 .

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    P.3 Families of Functions, Transformations, and Symmetry 13 82. First, 9 = ( a + b ) 2 = ( a 2 + b 2 ) + 2 ab = 89+ 2 ab . Then ab = − 40. Thus, a 3 + b 3 = ( a + b )( a 2 + b 2 − ab ) = 3(89 + 40) = 387 . P.2 Pop Quiz 1. Yes, since r ≥ 0 and r = √ A/π 2. Since A = s 2 and s ≥ 0, we obtain s = √ A . 3. No, since b = ± a . 4. [1 , ∞ ) 5. [2 , ∞ ) 6. f (3) = 3(3) + 6 = 15 7. We find 2 a − 4 = 10 2 a = 14 a = 7 . For Thought 1. False, it is a reflection in the y-axis. 2. False, the graph of y = x 2 − 4 is shifted down 4 units from the graph of y = x 2 . 3. False, rather it is a left translation. 4. True 5. True 6. False, the down shift should come after the reflection. 7. True 8. False, since the domains are different. 9. True 10. True, since f ( − x ) = − f ( x ) where f ( x ) = x 3 . P.3 Exercises 1. rigid 2. nonrigid 3. reflection 4. upward translation, downward translation 5. right, left 6. stretching, shrinking 7. odd 8. even 9. transformation 10. family 11. f ( x ) = √ x , g ( x ) = − √ x 2 4 x 2 -2 y 12. f ( x ) = x 2 + 1, g ( x ) = − x 2 − 1 2 -2 x 4 -4 y 13. y = x , y = − x 4 -4 x 4 -4 y

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