pc chapter 1 solutions (3)

29
Chapter 1: Packing your Suitcase Lesson 1.1.1 1-1. a. Independent variable = distance from end of tube to the wall. Dependent variable = width of field of view. e. The equation depends on the length and diameter of the tubes used. The students should have a slope between 0.12 and 0.14 with a y-intercept around 3.5 cm if they use a paper towel core. 1-2. Answers will depend on the students’ tube and models. 1-3. Answers will depend on the students’ tube and models. Review and Preview 1.1.1 1-5. a. parabola y = x 2 b. cubic y = x 3 c. hyperbola, inverse variation, d. exponential y = 2 x reciprocal function y = 1 x e. absolute value y = x f. square root y = x

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Page 1: PC Chapter 1 Solutions (3)

Chapter 1: Packing your Suitcase Lesson 1.1.1 1-1. a. Independent variable = distance from end of tube to the wall. Dependent variable = width of field of view. e. The equation depends on the length and diameter of the tubes used. The students should

have a slope between 0.12 and 0.14 with a y-intercept around 3.5 cm if they use a paper towel core.

1-2. Answers will depend on the students’ tube and models. 1-3. Answers will depend on the students’ tube and models. Review and Preview 1.1.1 1-5. a. parabola y = x2 b. cubic y = x3 c. hyperbola, inverse variation, d. exponential y = 2x

reciprocal function y = 1

x

e. absolute value y = x f. square root y = x

Page 2: PC Chapter 1 Solutions (3)

1-6. Examples of non-functions are a circle, x2 + y2 = r2 , and a “sleeping” parabola, x = y2 .

Other answers are acceptable. 1-7. a. slope = 17!8

7!4=94

point ! slope form" y ! 8 = 94(x ! 4)

slope ! intercept form" y ! 8 = 94x ! 9

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!y = 94x !1

b. slope = 17!87!4

=94

point ! slope form"

y ! 20 = 94(x ! (!12))

slope ! intercept form"

y ! 20 = 94x + 27!or !y = 9

4x + 47

1-8. a. 2

2= 2 !2

23= 2 !2 !2

24= 2 !2 !2 !2

b. 25 = 2 !2 !2 !2 !2 = 32

26 = 2 !2 !2 !2 !2 !2 = 64

c. They are half as large each time. Divided by 2, or multiplied by ½ is also acceptable. d. 20

= 2 ! 12= 1,!2"1

= 1 ! 12=

12

, 2"2=

12!

12=

12

,!2"3=

14!

12

= 18

,!2"4=

18!

12=

116

2"n=

1

2n"1!

1

21=

1

2n"1+1=

1

2n

1-9. a. 2!4

"22= 2!2 !!!Check :!

1

16" 4 =

1

4 b. 2!1 "2!2 = 2!3 Check :!

1

2"1

4=1

8

c. 20 !2"3 = 2"3 !!!!!Check :!1 !1

8=1

8

1-10. a. x(2x + 5) b. 3xy3(xy2 ! 3) c. 17x3y(1! 2xy) 1-11. a. (2x)3 = 23 ! x3 = 8x3 b. (3x ! 2)2

= (3x ! 2)(3x ! 2)

= 9x2 ! 6x ! 6x + 4

= 9x2 !12x + 4

c. (3x)4 = 34 ! x4 = 81x4 d. (3x)!3 = 1

(3x)3=

1

33 x3=

1

27x3

1-12. a. a

b b. c

b c. a

c

Page 3: PC Chapter 1 Solutions (3)

Lesson 1.1.2 1-14.

a. b. y = 1.5x looks like the others, but would graph to the right of y = 2x .

c. 0 < b < 1 Example: 1-15. This is the graph of y = x shifted up five units. 1-16. y = x2 ! 4 1-17. This is the graph of y = x3 shifted left three units. 1-18.

y = x ! 2 1-19. Parent graph: y = 1

x

Shifted right four units: y = 1

x!4

Shifted down three units: y = 1

x!4! 3

Page 4: PC Chapter 1 Solutions (3)

1-20. a.

A vertical stretch occurs with each of the y-values doubling.

b. A vertical stretch with stretch factor 2. 1-21. a. y = x3 b. See graph at right. c. Stretched vertically by 1

2, some may

prefer to call this a “compression.” 1-22. y = !2x2

Review and Preview 1.1.2 1-23. a. x + 2 b. 2x !1 c. x

2+ 4 d. 5x

1-24. a. f (4) = 2 ! 42 " 3 = 32 " 3 = 29

b. f (!5) = 2 " (!5)2 ! 3 = 50 ! 3 = 47

c. f (3b) = 2 ! (3b)2 " 3 = 2 !9b2 " 3 = 18b2 " 3

d. f (a +1) = 2 ! (a +1)2 " 3 = 2a2 + 4a + 2 " 3 = 2a2 + 4a "1

x y = x2 y = 2x2 –4 16 32 –3 9 18 –2 4 8

–1 1 2 0 0 0 1 1 2 2 4 8 3 9 18 4 16 32

Page 5: PC Chapter 1 Solutions (3)

1-25. (3x + 2)2 = (3x + 2)(3x + 2) = 9x2 + 6x + 6x + 4

= 9x2 +12x + 4 ! 9x2 + 4

The middle terms must be included.

1-26. a. ab !ac = a(b+c) b. a!b "ac = a(!b+c) = a(c!b)

c. Cannot be simplified. d. a !ab = a1 !ab = a(1+b) e. a0 !ab = a(0+b) = ab f. a(b+c) !a2c = a(b+c+2c) = a(b+3c) 1-27. a. 3x2y ! (27x " 4)

b. (2x +1) 3+ x + 5[ ] = (2x +1)(x + 8)

c. (3x ! 7) 2(3x ! 7) + (x ! 2)[ ] = (3x ! 7) 6x !14 + x ! 2[ ] = (3x ! 7)(7x !16) d. (x + y)(m + x + y) 1-28. a. (5a!2 )2 = 52 "a!2"2 = 25a!4 b. (m!1

n!2 )3 = m!1"3

n!2"3

= m!3n!6

c. (2x!1)2(2x0 ) = 22 " x!1"2 "2 = 8x!2 1-29. a. 2

2!x= 2

2x b. 2!5"4 = 2!20

c. 2

3( )!1

= 3

2( )1

= 3

2 d. 2

3( )!2

= 3

2( )2

= 32

22= 9

4

1-30. a.

cos 26! =x

18

0.899 =x

18

x = 16.18

b.

cos 70! =8

x

x !0.342 = 8

x = 23.39

c. 222 = x2 +102

484 = x2 +100

384 = x2

x = 384 = 19.60

d.

sin 41! =x

12

0.656 =x

12

x = 7.87

Lesson 1.1.3 1-31. a. It multiplies the input by two and then adds 1. b. We hope that they will think that 3 would come out the top and if that is true, then the

machine must “undo” itself—working backwards in a sense. c. Subtract one and then divide by two.

Page 6: PC Chapter 1 Solutions (3)

1-32. a. Subtract 6, then multiply by 2. b. f !1(x) = 2(x ! 6) 1-33. a. f (x) + g(x) = 3x ! 5 + x2 + 2 = x2 + 3x ! 3 b. f (x)g(x) = (3x ! 5)(x2 + 2) = 3x3 ! 5x2 + 6x !10 c. f (g(x)) = 3(x2 + 2) ! 5 = 3x2 + 6 ! 5 = 3x2 +1 d. g( f (x)) = (3x ! 5)2 + 2 = 9x2 ! 30x + 25 + 2 = 9x2 ! 30x + 27 1-34. a. f (x) = x3

! 4x

f (x) = x(x2! 4) = x(x + 2)(x ! 2)

0 = x(x + 2)(x ! 2)

Either x = 0, x + 2 = 0, x ! 2 = 0

x = !2, 0, 2

b.

c. Shifted left two units d. e. g(x) = (x + 2)3 ! 4(x + 2) 1-35. a.

2 hours ! 3

miles

hour= 6 miles

b. c. d. miles

hr!hr = miles

Review and Preview 1.1.3 1-36. a. 3x3 + 7 ! (x2 !1) = 3x3 + 7 ! x2 +1

= 3x3 ! x2 + 8

b. 3x2 +7

x2!1, x " ±1

c. (3x3 + 7)(x2 !1) = 3x5 ! 3x3 + 7x2 ! 7

Page 7: PC Chapter 1 Solutions (3)

1-37. a. y = 3x3 ! 5

x = 3y3 ! 5

x + 5 = 3y3

x+53

= y3

x+53

3 = y"

x+53

3 = f !1(x)

b. y = (2x + 4)1/2

x = (2y + 4)1/2

x2 = 2y + 4

x2 !42

= y"

x2 !42

= g!1(x)

c. y = 12x2 ! 2

x = 12y2 ! 2

x + 2 = 12y2

2x + 4 = y2

2x + 4 = y"

2x + 4 = h!1(x)

1-38. a. g(h(4)) = g 1

2!16 " 2( ) = g(6) = 2 !6 + 4 = 16 = 4

b. h(g(-1)) = h 2 ! (-1) + 4( ) = h 2( ) = 12! 2( )

2" 2 = 1" 2 = "1

c. g(h(!2)) = g 12" 4 ! 2( ) = g(0) = 2 "0 + 4 = 2

d. Part (c) does not have the same output as input. e. A function has one output (y) for every input (x), but y = ± 2x + 4 has two outputs for

every input. 1-39. a.

sin x = 712

sin!1(sin(x)) = sin!1 0.583( )

x = 35.7!

b.

cos y = 515

cos!1(cos(y)) = cos!1 0.333( )

y = 70.5!

1-40. a. cubic function y = x3

flipped over y-axis y = !x3

shifted down 3 units y = !x3! 3

shifted right 2 units y = ! x ! 2( )3! 3

b. exponential function y = 2x

shifted down 3 units y = 2x! 3

shifted right 2 units y = 2x!2! 3

1-41. a. g(x ! 2) = x ! 2 +1 = x !1 b. f (g(8)) = f ( 8 +1) = f (3) = 32 + 2(3) = 15 c. g( f (8)) = g(82 + 2(8)) = g(80) = 80 +1 = 9 d. f ( f (1)) = f (12 + 2(1)) = f (3) = 32 + 2(3) = 15 e. f (x +1) = (x +1)2 + 2(x +1) = x2 + 2x +1+ 2x + 2 = x2 + 4x + 3

f. g( f (x)) = x2 + 2x +1 = (x +1)2 = x +1

Page 8: PC Chapter 1 Solutions (3)

1-42. h( j(x)) = 3(ax + b) ! 2 = 3ax + 3b ! 2

j(h(x)) = a(3x ! 2) + b = 3ax ! 2a + b

3b ! 2 = !2a + b

2a + 2b = 2

a + b = 1

Lesson 1.1.4 1-44. a. Shifts right three units and up two units. b. f (x + 4) ! 2 c. The point is on the x-axis. It does not change since –0 = 0. d. It still does not move. 1-45. Shifted left two units: g(x) = f (x + 2) Shifted down one unit: g(x) = f (x + 2) !1 1-46. a. It is stretched then shifted down 3. b. It is shifted down 3 and then stretched. c. k(x) = 2 f (x) ! 3 , m(x) = 2( f (x) ! 3) = 2 f (x) ! 6. These two functions are not equivalent. 1-48. a. They are the same. b. 2 ! x ! 4 c. Yes, replace x with x – 2 in the inequalities and solve. d. 0 ! x + 3 ! 2 ! 3 ! 3 ! 3

! 3 " x " !1

0 ! x " 2 ! 2

+2 + 2 + 2

2 ! x ! 4

Page 9: PC Chapter 1 Solutions (3)

Review and Preview 1.1.4 1-49. a. f (x + 2) +1 Shifted left two units and up

one unit.

b. 2 f (x) + 2 Vertical stretch by a factor of two, up two.

c. ! f (x + 4) Flipped over x-axis, and shifted left four units. 1-50.

!A = 180! " 90! " 38! = 52!

cos 38! =15

c

0.788 =15

c

c =15

0.788= 19.04 cm

sin 38! =b

19.04

b = 0.6157 #19.04 = 11.72 cm

1-51. a. 50(1.5) + 75(0.5) = 75 + 37.5 = 112.5 miles b. two rectangles c. 50(1.5) + 75(0.5) = 75 + 37.5 = 112.5 miles d. miles

hour!hours = miles

1-52. a. f (g(!2)) = f ((!2)2 !1) = f (3) = 2(3) + 5 = 11 b. g f h(2)( )( ) = g f 2 + 2( )( ) = g f (2)( ) = g 2(2) + 5( ) = g(9) = 92 !1 = 81!1 = 80

c. y = 2x + 5

x = 2y + 5

x ! 5 = 2y

x!52

= y

12x ! 5( ) = f !1(x)

d. f g h(x)( )( ) = f x + 2( )2!1"

#$% = f (x +1) = 2(x +1) + 5 = 2x + 2 + 5 = 2x + 7

Page 10: PC Chapter 1 Solutions (3)

1-53. The graph does not give a full line. The line starts at (!2 , 3) and then follows the line

y = 2x + 7 . Since h(x) is defined for only values of x ! "2 , the composite function is only defined for x ! "2 .

1-54. a. Opposite = !54

Reciprocal = 5!4 = 1

54

b. Opposite = !3!5

Reciprocal = 35

c. Opposite = 11!6

Reciprocal = !116 d. Opposite = !

27

Reciprocal = 72

e. Opposite = !119( )2

Reciprocal = 119( )

!2= 9

11( )2

f. Opposite = !713( )

!5= !

713( )

!5

Reciprocal = 713( )

5

1-55.

xm

xn= x

m! x

"n= x

m+("n)= x

m"n Lesson 1.2.1 1-57. a. (7, 2) b. m =

5!2

7!1=3

6=1

2 c. (x, 2) d. m =

y!2

x!1

e. 12=

y!2

x!1

12(x !1) = y ! 2

y ! 2 = 12(x !1)

f. (x, y1) g. y ! y1, x ! x1 h. y!y1x!x1

1-58. Yes, the point (0, 0) is on the line because f (0) = 2

3!0" 0 = 0 .

y = 23(x ! 2) + 5

y ! 5 = 23(x ! 2)

This is the same as point-slope form. 1-59. original function ! y = mx

right shift h units ! y = m(x " h)

shifted up k units ! y = m(x " h) + k

Page 11: PC Chapter 1 Solutions (3)

1-60. The point-slope form requires a point on the line and the slope of the line. bThe slope-

intercept form requires the slope and the y-intercept. 1-61. a. y = 3

5(x !10) ! 3

b. y ! 7.3 = 2.85(x ! 6.1) c. m =

21!8

15!4=13

11 Point-slope form: y ! 8 =

1311

(x ! 4) or y ! 21 = 1311

(x !15)

d. m =9.78!6.24

5.1-4.3= 4.425

Point-slope form: y ! 6.24 = 4.425(x ! 5.1) or y ! 6.24 = 4.425(x ! 4.3) 1-62. The negative reciprocal. slope =

!

2

5

1-63. The negative reciprocal. The product of a slope and the perpendicular slope should be –1. 1-64. When the slope is zero. 1-65. a. m = 3

y ! 7 = 3(x + 2)

b. m1 =34

! m1 = m! = "43

y " 20 = "43(x "12)

1-66. a. AB = (15 ! 3)2 + (12 ! 3)2 = 144 + 81 = 225 = 15

b. midpoint of AB = 3+152

, 3+122( ) = 18

2, 15

2( ) = 9, 7.5( ) Review and Preview 1.2.1 1-67. a. Parent Graph: y = 1

x

Shifted left 2 units: y = 1x+2

Shifted down 3 units: y = 1x+2

! 3

b. Parent Graph: y = x2

Shifted right 2 units: y = (x ! 2)2

Shifted up 1 unit: y = !(x ! 2)2+1

Page 12: PC Chapter 1 Solutions (3)

1-68. a. 50mph ! 3hours = 150 miles b. It is a rectangle. c. height = 50 mph, base = 3 hours,

50 mileshr( ) ! (3hrs) = 150 miles

1-69. a. 1

275=

1

(33)5=

1

315= 3!15

b. 1

8( )x

= 1

23( )

x

= 2!3( )

x

= 2!3x

c. 16x !1

32( )2"x( )

= 24( )x

!1

25( )

2"x( )= 24x ! 2"5( )

2"x( )= 24x !2"10+5x = 24x"10+5x = 29x"10

1-70. f (x +1) = x+1+1

x+1!2=

x+2x!1

=12

" 2(x + 2) = 1(x !1)

2x + 4 = x !1

x + 4 = !1

x = !5

1-71.

a. (23)(x+3) = 25

23x+9 = 25

! 3x + 9 = 5

3x = "4

x = "43

b. (33)2x = 1

32( )

(x!1)

36x = (3!2 )(x!1)

36x = 3!2x+2

" 6x = !2x + 2

8x = 2

x = 14

c. 1

53( )

(2x!3)

= 1

52

(5!3)(2x!3) = 5!2

" 5!6x+9 = 5!2

!6x + 9 = !2

!6x = !11

x = 116

1-72. a. 3x ! 6 ! 2x !14 = 2x +17

x ! 20 = 2x +17

!x ! 20 = 17

!x = 37" x = !37

b. (x + 5)(x ! 2) = 0

Either x + 5 = 0 or x ! 2 = 0

x = !5, 2

c. x2! 7x +12 = 0

(x ! 4)(x ! 3) = 0

Either x ! 4 = 0 or x ! 3 = 0

x = 3, 4

d. x3+ x

2! 6x = 0

x(x2+ x ! 6) = 0

x(x + 3)(x ! 2) = 0

x = 0, x + 3 = 0!or x ! 2 = 0

x = !3, 0, 2

Page 13: PC Chapter 1 Solutions (3)

1-73. a. g( f (x)) = x2 + x +1 b. f (g(x)) = (x +1)2 + x

= x2 + 2x +1+ x

= x2 + 3x + 2

c. x2+ 3x +1 = 1

x2+ 3x = 0

x(x + 3) = 0

x = 0 or x = !3

1-74.

sin!P = 8

16= 1

2

sin"1 sin!P( ) = sin"1 1

2( )!P = 30!

!R = 180! " 90! " 30! = 60!

sin 60! = r

16

3

2#16 = r

r = 8 3 $ 13.86 cm

Lesson 1.2.2 1-75. a. The coefficients a, b and c.

b.

!b + D

2a= R and

!b ! D

2a= S

c. R and S; the Quadratic Formula has two solutions because of the ± in the formula. 1-78. Sierpinski’s Triangle

a. By choosing a random integer: 0, 1, or 2. b. T = 0 chooses A, T = 2 chooses B. c. :(X+T)/2, :(Y+T)/2Y, and Y/2Y.

Page 14: PC Chapter 1 Solutions (3)

Review and Preview 1.2.2 1-79. The program will “crash” since the program tries to take the square root of a negative

number. 1-80. a. Flip over y axis: ! f (x)

Shifted right 3 units: ! f (x + 3)

Shifted up 2 units: ! f (x + 3) + 2

g(x) = ! f (x) + 3

b. Shifted up 1 unit: f (x) +1

Shifted left 1 unit: f (x +1) +1

Doubled: 2(f (x +1) +1) = 2 f (x +1) + 2

h(x) = 2 f (x +1) + 2

1-81. Slope of line m =

9!(!3)

8!2=126= 2

Midpoint of line = 8+22,9+(!3)

2( ) = 102, 62( ) = 5, 3( )

Slope of perpendicular line = !1

2

Equation of line y ! 3 = !1

2x ! 5( )

1-82. a. p = 6, q = 2 b. not in pq form c. p = x + 3y, q = 2 ! r d. not in pq form 1-83. a. 62 ! (62 )"3 !1 = 62 !6"6 = 6"4

b. (52 )2 !5"3

(53)"2=54 !5"3

5"6=51

5"6= 51"("6) = 57

c. 3 !1997

8 !1994=3

8!1997"94 =

3

8!193

d. 3 !19-97

8 !19"94=3 !1994

8 !1997=3

8!1994"97 =

3

8!19"3

1-84. a. Example: x = 2, y = 3

! (2 + 3)2 " 4 + 9

25 " 13

b. Example: p = 2, q = 3

! 22 + 32 " 2 + 3

13 " 5

Solution continues on next page. →

Page 15: PC Chapter 1 Solutions (3)

1-84. Solution continued from previous page.

c. Example: w = 4

! 3 " 4#2 $ 1

3"42

316

$148

d.

e. Example: x = 5

! 3 "25 # 65

3 " 32 # 7776

96 # 7776

1-85. a. Impossible, different bases. b. Impossible, bases are being added.

c. 22+3 = 25 d. d. 22!3

= 26

e. 22=3

= 2!1

f. Impossible, bases are being subtracted. 1-86. y = 5

2x ! 3

x = 52y ! 3

x + 3 = 52y

2(x+3)

5= y

2(x+3)

5= f !1(x)

1-87. a. x(x + 8) b. 6x(x + 8) 1-88. Circumference of circle = 2! "1 = 2! Length of

AB! =

1

4!2" =

"

2

1-89.

Area = base !height

30 = 12!12 !h

30 = 6 !h

5 inches = h

sin 36! = 5KL

KL =5

0.5878= 8.51 inches

Example: a = 2, b = 3

! (2"1 + 3"1)"1 # 2 + 3

12+ 13( )

"1# 5

56( )

"1# 5

65# 5

Page 16: PC Chapter 1 Solutions (3)

Lesson 1.3.1 1-90. a. 1

2K ah= b. sinC =

h

b ! h = b sinC

c. K =1

2ah ! K =

1

2ab sinC

1-91.

Area = 1

2(6)(4) sin 76! = 12 !0.970 = 11.644 cm2

1-92. K =

1

2bh

sin A =h

c! h = c sin A

K =1

2bc sin A

1-93.

SA of one side =12

(10)(10) sin 40!

= 50 !0.643 = 32.139 ft2

Total surface area of sides = 4(32.139) = 128.558 ft2

1-94. a. sin A =

h

b ! h = b sin A b. sin B =

h

a ! h = a sin B

c. h = b sin A, h = a sin B

b sin A = a sin B

sin Aa

=sin Bb

d. h = b sinC, h = c sin B

b sinC = c sin B

sin Bb

=sin Cc

e. sin A

a=sin B

b=sin C

c

1-95. 1-96. a. !NAT = 180

!"100

!" 38

!= 42

! b.

sin 42!

200=sin 100

!

y

sin 42!

200=sin 38!

x

0.0033x = 0.6157

x = 184.018 ft

0.0033x = 0.985

x = 294.354 ft

sin Pp

=sinQ

q=sin Rr

Page 17: PC Chapter 1 Solutions (3)

1-97. Cannot in (a) and (b) because you will get two unknowns in any form of the equation.

Cannot in (d) for the same reason, and also because the triangle is not determined. Note that (c) is the only diagram in which you’re given exactly one side.

Review and Preview 1.3.1 1-98. a. b.

!G = 180! " 64! " 38! = 78!

8

sin 78!=

OG

sin 64!

8 #0.8988 = OG #0.9781

7.1904

0.9781= OG

7.351 in = OG

8

sin 78!=

DG

sin 38!

8 !0.6157 = DG !0.9781

4.9256

0.9781= DG

5.035 in = DG

c. Area of !DOG =

8"5.036"sin 64!

2==

40.288"0.89882

= 18.102 sq. in. 1-99. f (x + 2) = (x + 2)2

+ 2(x + 2)

= x2+ 4x + 4 + 2x + 4

0 = x2+ 6x + 8

0 = (x + 2)(x + 4)

Either x + 2 = 0 or x + 4 = 0

! x = "2 or x = "4

1-100. x

6= 90

x = 906

1-101. yn = x

y = xn

Page 18: PC Chapter 1 Solutions (3)

1-102. a. 641/3( )

2

= 42 = 16 b. 1251/3( )4

= 54 = 625

c. 813/4 = 81

1/4( )3

= 33 = 27 d. 27

8( )1/3!

"#$

%2= 3

2( )%2

= 2

3( )2

= 4

9

1-103. a. y = 2x !13

x = 2y !13

x3 = 2y !1

x3+12

= y

12(x3 +1) = f !1(x)

b. y = 12(x ! 3) +1

x = 12(y ! 3) +1

2(x !1) = y ! 3

2(x !1) + 3 = y

2(x !1) + 3 = g!1(x)

c. y = 2x3/2

x = 2y3/2

x2= y3/2

x2( )2/3

= y

x2( )2/3

= h!1(x)

1-104. a. Distance for one revolution = Circumference of circle = 2 !" ! r = 2 !" !1 = 2" feet b. 10 = 2 !" ! r

10

2"= r # r =

5

"= 1.592 feet

1-105. a. n

2+ n

2= d

2

2n2= d

2

2n2= d

n 2 = d

b.

1

2!90! = 45!

c. (2n)2 = n2 + k2

4n2 = n2 + k2

3n2 = k2

3n2 = k

n 3 = k

d. y = 60! (equilateral triangle)

Page 19: PC Chapter 1 Solutions (3)

Lesson 1.3.2 1-106. a. e = b ! d b. h

2+ e

2= c

2 ! h

2= c

2" e

2 c. h

2+ d

2= a

2 ! h

2= a

2" d

2 d. e. cosC =

d

a ! d = a cosC

f. d = a cosC

c2 = a2 + b2 ! 2bd

cosC =d

a, d = a cosC

1-107. a.

c2= 1102 +1262 ! 2(110)(126) cos 74!

c2= 12100 +15876 ! 7640.668

c2= 20335.332

c = 20335.332 = 142.6 ft

Yes, 150 feet of fencing is enough.

b.

sin 74!

142.6=sin B126

0.0067 = sin B126

0.849 = sin B

sin!1 0.849 = sin!1(sin B)

58.1! = "B

sin 74!

142.6=sin C110

0.0067 = sin C110

0.7415 = sinC

sin!1 0.7415 = sin!1(sinC)

47.9! = "C

c.

Area =12

(110)(126) sin 74! = 6930 !0.9613 = 6661.54

2300 (area of home)

6661 (area of lot)= 0.3453

Yes, the area of the home would be more than 13

of the lot size.

1-108. It is not possible in (c) or (d) because you will get two unknowns in any form of the

equation. You can solve (c) with the Law of Sines. The triangle for (d) is not determined. 1-109.

cos C = cos90! = 0

!c2!= a2

!+ b2!– 2ab 0( )

c2!= a2

!+ b2

h2= c2

! e2 " h2

= a2! d2

c2! e2

= a2! d2

" c2= a2

! d2+ e2

c2= a2

! d2! d2

+ e2

c2= a2

! d2+ b2

! 2bd + d2

c2= a2

+ b2! 2bd

Page 20: PC Chapter 1 Solutions (3)

1-110. c

2= 102 +142 ! 2(10)(14) cos 60°

c2= 296 ! 280 " 1

2

c2= 156

c = 156 = 12.490

1-111.

BE2= 3.52

+ 2.82! 2(2.8)(3.5) cos 43!

BE2= 20.09 !14.3345

BE2= 5.7555

BE = 2.399 km

1-112.

c2= 102

+ 202! 2(10)(20) cos 30!

c2= 500 ! 346.4102

c2= 153.5898

c = 12.393 cm

1-113.

x2= 282 + 422 ! 2(28)(42) cos X!

x2= 784 +1764 ! 2352 cos X!

x2= 2548 ! 2352 cos X!

x2+ 2352 cos X! = 2548

If X = 0!

x2+ 2352 = 2548

x2= 196

x = 196 = 14

If X = 180!

x2! 2352 = 2548

x2= 4900

x = 4900 = 70

14 < x < 70 inches Review and Preview 1.3.2 1-114. a.

62 = 52 + 82 ! 2 "5 "8 cos A

36 = 89 ! 80 cos A

53

80= cos A

0.6625 = cos A

48.51! = #A

6

sin 48.51!=

5

sin B

6 sin B = 5 !0.7491

sin B =3.7454

6= 0.6242

"B = 38.62!

"C = 180! # 38.62! # 48.51! = 92.87!

b. A =128 !6 ! (sin 38.62) = 8!6!0.6242

2=29.9592

= 14.98 square meters

Page 21: PC Chapter 1 Solutions (3)

1-115. a. x

2+1+ 2x ! 8 = x

2+ 2x ! 7

b. (2x - 8) 12x +1( ) = x2 + 2x ! 4x ! 8 = x2 ! 2x ! 8

c. 2( 12x +1) ! 8 = x + 2 ! 8 = x ! 6

d. (x2 +1)2 +1 = x4 + 2x2 +1+1 = x4 + 2x2 + 2 1-116. a.

1

2! 360

!= 180

! b. 1

2!2" = " meters c.

AB! =

1

6!2" =

"

3meters

1-117. a. 52 ! 52( )

"3

!53 = 52+ "6( )+3 = 5"1

b. 32( )2

! 3"3

33( )

"2=34! 3

"3

3"6

=3

3"6

= 31" "6( ) = 37

c. 5 !1498

8 !1495=5

8!1498"95 =

5

8!143

d. 5 !14-98

8 !14"95=5

8!14

"98" "95( ) =5

8!14"3

1-118. a. 8

1/3( )2

= 22 = 4 b. 1001/2( )

3

= 103 = 1000

c. 1251/3( )2

= 52 = 25 1-119. 3x ! 7y = 42

!7y = !3x + 42

y = 37x ! 6

perpendicular slope = !73

equation of line " y + 8 = !73x + 3( )

1-120. a. 2x + 3y + 6 = 6x ! 30

3y = 4x ! 36

y = 4

3x !12

b. 6x +1 = 6y!!!!!y ! 0

y = 6

6x + 1

6

y = x + 1

6!!x ! "

1

6

1-121. a. (2x ! 3y)(2x + 3y) b. 2x3(4 ! x4 ) = 2x3(2 + x2 )(2 ! x2 )

Page 22: PC Chapter 1 Solutions (3)

1-122. 3! x " 0

!x " !3

x # 3

Lesson 1.4.1 1-123. d. 2π radius lengths = circumference 1-124. Length of AB! = 1 unit . 1-125. C = 2! "1 = 2! 1-126. a. 360° b. 2π radians c. 2! = 6.2832 , nearest whole number = 6 1-127.

1-128. a. Degrees in half a circle: 180! b. π radians = 180˚ Approximate radians in half a circle: 3 Exact radians in half a circle: !

c.

!

3=180!

3= 60

! d.

200! !"

180!=200!"

180!=10"

9

1-129. a.

180!

!= 57.296! b.

!

180!= 0.017

c. Very different. A radian is much larger, almost 60 times as large.

Page 23: PC Chapter 1 Solutions (3)

1-130. a.

180!!

"

180!= " b.

!36! "#

180!=

!36#

180

= !#

5

c.

2! ! "!

180!=2!!!

180!=

!2

90

1-131. a.

3!

2"180!

!=540

2= 270! b.

!7"

6#180!

"= !

1260

6= !210

!

c. 2 !

180!

"= 360

"( )!

Review and Preview 1.4.1 1-132. Radians per minute = 500 !2" = 1000"

Radians per second =1000"

60=

100"6

=50"

3

1-133.

1-134. G R Y 16!6

x: 116!6

: 232!6

96x=

196

x = 9216 teaspoons

x =92166

= 1536 ounces

x =1536128

= 12 gallons

1-135. x2 + b = 7

x2 = 7 ! b

x = ± 7 ! b , b " 7

1-136. a. 9: x2 + 6x + 9 = (x + 3)(x + 3) = (x + 3)2 b. 8: x2 ! 8x +16 = (x ! 4)(x ! 4) = (x ! 4)2

Page 24: PC Chapter 1 Solutions (3)

1-137.

a. d = !4 ! (!2)( )2+ 2 ! (!6)( )

2

d = (!2)2 + 82 = 4 + 64 = 68

b. m =2!(!6)

!4!(!2)=

8!2

= !4 point slope form ! y + 2 = "4(x + 4)

point slope form ! y + 6 = "4(x + 2)

slope intercept form ! y + 2 = "4x "12

y = "4x "14

1-138. a.

tan A = 47

tan!1(tan A) = tan!1 47( )

"A = 29.74!

b.

tan B = 74

tan!1(tan B) = tan!1 74( )

"B = 60.26!

1-139. a. (a ! 3)(a ! 3!1) = (a ! 3)(a ! 4) b. 5x(x ! 3) + 4(x ! 3) = (x ! 3)(5x + 4) 1-140. Slopes: a. !2 b. 2 c. 2 d. 3 e. !2 Parallel Lines (same slope) ! a and e, b and c Lesson 1.4.2 1-141. a. Radian measure for a: b. !

4, 3!4, 5!4, 7!4

Half circle: 12!2" = "

Quarter circle: 14!2" =

"

2

Three fourths of a circle: 34!2" =

3"

2

c. !

6, 2!6=

!

3, 3!6=

!

2, 4!6=2!

3, 5!6

d. 7!

6, 8!6=4!

3, 9!6=3!

2, 10!6

=5!

3, 11!6

Page 25: PC Chapter 1 Solutions (3)

1-142. a. In the center of each quadrant. b. y-axis c. Closest to the x-axis. 1-143. a. !

2"

3+6"

3=4"

3 b. !

5"

4+8"

4=3"

4

c. !11"

6+12"

6=

"

6

1-144. a. 10!

3=9! +!

3= 2! +

4!

3=4!

3 b. 17!

4=16! +!

4= 4! +

!

4=

!

4

c. !25"

6= !

24" +"

6= !4" !

"

6= !

"

6 or

11"

6

1-145. a. The speed does not change. The ratio of the distance and time is constant, or for a set time

interval, an object will travel a set distance. b. Faster on the inside. The CD must go around more times on an inside track to cover the

same distance as a point on the outside of the CD. c. 200(2! "5.25) # 6597 cm d. Distance around innermost track = 2! "2 6597

2! "2= 524.97

525 rotations

1-146. a. 5280 feet = 5280 !12 inches = 63360 inches

63360 in

902 rev= 70.2439 inches in one revolution

C = 70.2439 = "d

d = 22.36 inch diameter

b. 902 !" !26 = 73676.63 inches

73676.63

12= 6140 feet

6140

5280= 1.163 miles

c. Linda could get a speeding ticket. 40

22.36=

x

26

x =40!2622.36

= 46.5 mph

Review and Preview 1.4.2 1-147. a.

120!

1!

"

180!=120"

180=2"

3 b.

!225!

1"

#

180!=

!225#

180= !

5#

4

c.

80!

1!

"

180!=80"

180=4"

9

Page 26: PC Chapter 1 Solutions (3)

1-148. a.

tan x = 2012

= 53

tan-1(tan x) = tan!1 53( )

x = 59.0!

b.

10

sin 35!=

x

sin 80!

10 !0.9848 = x !0.5736

9.848

0.5736= 17.17 = x

c.

x2= 6.52 + 7.12 ! 2 "6.5 " 7.1cos119!

x2= 42.25 + 50.41! 92.3(!0.4848)

x2= 92.66 + 44.75 = 137.41

x = 137.41 = 11.72

d.

60 =1

2!15 ! x ! sin 28!

60

7.5= 0.4695x

8

0.4695= 17.04cm = x

1-149. a. y = x + 23

x = y + 23

x3 = y + 2

x3 ! 2 = y

x3 ! 2 = f !1(x)

b.

c. f !1( f (6)) = f !1 6 + 23( ) = f !1 2( ) = 23 ! 2 = 6

f f !1(2)( ) = f 23 ! 2( ) = f 6( ) = 6 + 23 = 83 = 2

Composing 1and

!ff in either order returns the original number. 1-150. ax2 + bx + c = d(x2 ! 2ex + e2 ) + f

ax2 + bx + c = dx2 ! 2dex + (de2 + f )

a = d, b = !2de, c = de2 + f

1-151. a. Parent Graph ! y = x3

Transformation

Flip over y-axis ! y = "x3

Shifted right two units ! y = " x " 2( )3

Shifted down three units ! y = " x " 2( )3" 3

b. Parent graph ! y = x

Transformation

Shifted left one unit ! y = x +1

Shifted down two units ! y = x +1 " 2

-6 -4 -2 2 4 6

-6

-4

-2

2

4

6

x

y

f(x)

f -1(x)

Page 27: PC Chapter 1 Solutions (3)

1-152. f (g(x)) = (x ! 3)2

! 2(x ! 3) + 5

f (g(x)) = x2! 6x + 9 ! 2x + 6 + 5

f (g(x)) = x2! 8x + 20

x2! 8x + 20 = 8

x2! 8x +12 = 0

(x ! 2)(x ! 6) = 0

x ! 2 = 0 or x ! 6 = 0

x = 2 or x = 6

1-153. a. !!!

906030 !! b.

sin 30!=

PQ

12

1

2!12 = 6 = PQ

c.

cos 30! =QR

12

3

2!12 = 6 3 = QR

d. sin P =6 3

12=

3

2

1-154. a. b. c.

k(x ! 2) + 3 !k(x) ! 2

12k(x) +1

Page 28: PC Chapter 1 Solutions (3)

Closure Problems CL 1-155. a. f (x) ! g(x) = 2x2 ! x ! 3x +1 = 2x2 ! 4x +1 b. g( f (x)) = 3(2x2 ! x) !1 = 6x2 ! 3x !1

c. g(x+2)

f (x+2)=

3(x+2)!1

2(x+2)2 !(x+2)=

3x+6!1

2(x2 +4x+4)!x!2=

3x+5

2x2 +8x+8!x!2=

3x+5

2x2 +7x+6

CL 1-156. a. b. y ! 4 = !

12(x ! 2)

CL 1-157. a.

Third angle = 180! ! 50! ! 45! = 85! b. Law of Sines

sin 85!

7=sin 50!

a

0.1423 =0.766

a

a =0.766

0.1423= 5.383

sin 85!

7=sin 45!

b

0.1423 =0.707

b

b =0.707

0.1423= 4.969

c. ASA

CL 1-158. a.

c2= 42 + 72 ! 2(4)(7) cos 50! b. Law of Cosines

c2= 16 + 49 ! 36

c2= 29

c = 29 = 5.386

c. SAS

CL 1-159.

a. 25x4 = 5 !5 ! x2 ! x2 = 5x2 b. (xy2 )3

(x2y3)1/2=

x3y6

xy3/2= x3!1y6!3/2 = x2y9/2

c. x3+ (x2 )1/2 = x3 + x

CL 1-160. a. 3 !2" = 6" b. 26

2!6" = 78" in/sec # 20.4 ft / sec

Page 29: PC Chapter 1 Solutions (3)

CL 1-161. a.

sin 45! = 4hyp.

hyp =4

1/ 2= 4 2

Isosceles triangle

leg = 4

b.

sin 45! =hyp.

6

hyp =1

2!6 =

6

2=

6 22

= 3 2

Isosceles triangle

leg = 3 2

c.

sin 30! = 2hyp.!!!!!!!!cos 30! =

leg

4

hyp = 21 2

= 4 !!!!!!!!leg = 32! 4 = 2 3

CL 1-162. f (x) = 2 x ! 3 +1 Parent graph: y x= Stretched: 2y x= Shifted right three units: 2 3y x= ! Shifted up one unit: 2 3 1y x= ! + Inverse: x = 2 y ! 3 +1

x !1 = 2 y ! 3

x!12

= y ! 3

x!12( )

2= y ! 3

x!12( )

2+ 3 = 1

4(x !1)2 + 3 = y

f !1(x) = 14(x !1)2 + 3

CL 1-163. a. b. CL 1-164.

a.

Total distance = 30 !2 + 40 !1+ 50 !1

2

= 60 + 40 + 25 = 125 miles

b. 125

2 +1+ 12

=125

3.5= 35.7 mph