Cpm Precalculus Chapter 01 Solutions

8/16/2019 Cpm Precalculus Chapter 01 Solutions

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CPM Educational Program ! 2012 Chapter 1: Pre-Calculus with Trigonometry

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 papertowel 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 = x3

c. hyperbola, inverse variation, d. exponential y = 2 x

reciprocal function y = 1

x

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




1-6.Examples of non-functions are a circle, x

2+ y

2= r

2 , and a “sleeping” parabola, x = y2 .

Other answers are acceptable.

1-7.

a. slope = 17!

87 ! 4 = 94

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

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

y = 94 x ! 1

b. slope = 17!

87 ! 4 = 94

point ! slope form "

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

slope ! intercept form "

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

4 x + 47

1-8.a. 2 2

= 2 !2

23

= 2 !2 !2

2 4= 2 !2 !2 !2

b. 2 5= 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 ! is also acceptable.d. 2 0 = 2 ! 1

2 = 1, 2 " 1 = 1 ! 1

2 = 1

2 , 2 " 2 = 12 ! 1

2 = 1

2 , 2 " 3 = 14 ! 1

2 = 18 , 2 " 4 = 1

8 ! 1

2 = 1

16

2 " n = 1

2 n " 1 ! 1

21 = 1

2 n " 1+ 1 = 1

2 n

1-9.a. 2 ! 4 "2 2

= 2 ! 2 Check : 116

"4 =14 b. 2 ! 1 "2 ! 2

= 2 ! 3 Check : 12 " 1

4 =

18

c. 2 0 !2 " 3= 2 " 3 Check : 1 ! 1

8

= 1

8

1-10.a. x (2 x + 5) b. 3 xy 3 ( xy 2 ! 3) c. 17 x 3 y (1 ! 2 xy )

1-11.a. (2 x )3 = 2 3 ! x 3 = 8 x 3 b. (3 x ! 2) 2

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

= 9 x 2 ! 6 x ! 6 x + 4

= 9 x 2 ! 12 x + 4

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

= 133 x 3

= 127 x 3

1-12.

a. ab b. c

b c. a

c




Lesson 1.1.2

1-14.a. b. y = 1.5 x looks like the others, but would

graph to the right of y = 2 x .

c. 0 < b < 1 Example:

1-15.This is the graph of y = x shifted up five units.

1-16. y = x

2 ! 4

1-17.This is the graph of y = x

3 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




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 = x

3 b. See graph at right.c. Stretched vertically by 1

2 , some mayprefer to call this a “compression.”

1-22. y = ! 2 x

2


1-23.a. x + 2 b. 2 x ! 1 c. x

2+ 4 d. 5 x

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

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

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

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

x y = x2 y = 2 x

2 –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




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

= 9 x 2 + 12 x + 4 ! 9 x 2 + 4

The middle terms must be included.

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

c. Cannot be simplified. d. a !a b= a 1 !a b

= a (1+ b )

e. a 0 !a b = a (0 + b ) = a b f. a (b + c ) !a 2 c = a (b + c + 2 c ) = a (b + 3c )

1-27.a. 3 x 2 y !(27 x " 4)

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

c. (3 x ! 7) 2(3 x ! 7) + ( x ! 2)[ ]= (3 x ! 7) 6 x ! 14 + x ! 2[ ]= (3 x ! 7)(7 x ! 16)

d. ( x + y )( m + x + y )

1-28.a. (5 a

! 2 )2 = 5 2 "a! 2"2 = 25 a

! 4 b. (m! 1n

! 2 )3 = m! 1"3n

! 2"3 = m! 3n

! 6

c. (2 x ! 1 )2 (2 x 0 ) = 2 2 " x

! 1"2 "2 = 8 x ! 2

1-29.a. 2 2 ! x

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

c. 23( )

! 1=

32( )

1=

32 d. 2

3( )! 2

= 3

2( )2

= 32

2 2 =

94

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. 22 2= x

2+ 10 2

484 = x 2 + 100

384 = x 2

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. 1-32.a. Subtract 6, then multiply by 2. b. f ! 1 ( x) = 2( x ! 6)

1-33.

a. f ( x) + g( x) = 3 x ! 5 + x2 + 2 = x2 + 3 x ! 3 b. f ( x)g( x) = (3 x ! 5)( x2 + 2) = 3 x3 ! 5 x2 + 6 x ! 10 c. f (g( x)) = 3( x2 + 2) ! 5 = 3 x2 + 6 ! 5 = 3 x2 + 1 d. g( f ( x)) = (3 x ! 5)2 + 2 = 9 x2 ! 30 x + 25 + 2 = 9 x

2 ! 30 x + 27

1-34.a. f ( x) = x3 ! 4 x

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. mileshr

!hr = miles


1-36.

a. 3 x 3 + 7 ! ( x 2 ! 1) = 3 x 3 + 7 ! x 2 + 1

= 3 x 3 ! x 2 + 8

b. 3 x 2 + 7 x 2 ! 1

, x " ± 1

c. (3 x 3 + 7)( x 2 ! 1) = 3 x 5 ! 3 x 3 + 7 x 2 ! 7




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

x = 3 y3 ! 5

x + 5 = 3 y3

x+

53 = y3

x+ 53

3 = y "

x+ 53

3 = f ! 1 ( x)

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

x = (2 y + 4)1/2

x 2 = 2 y + 4

x 2 ! 42

= y "

x 2 ! 42

= g! 1 ( x )

c. y = 12 x2 ! 2

x = 12 y2 ! 2

x + 2 = 12 y2

2 x + 4 = y2

2 x + 4 = y "

2 x + 4 = h ! 1 ( x)

1-38.

a. g(h(4)) = g 12 !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 1

2 "

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 = ± 2 x + 4 has two outputs forevery 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 = x 3

flipped over y -axis y = ! x 3

shifted down 3 units y = ! x 3 ! 3

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

b. exponential function y = 2 x

shifted down 3 units y = 2 x ! 3

shifted right 2 units y = 2 x ! 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(8 2 + 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 + 2 x + 1 + 2 x + 2 = x2 + 4 x + 3

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




1-42.h( j ( x)) = 3(ax + b) ! 2 = 3ax + 3b ! 2 j (h( x)) = a (3 x ! 2) + b = 3ax ! 2a + b3b ! 2 = ! 2 a + b2 a + 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





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 !=

15c

0.788 = 15

c

c = 150.788

= 19.04 cm

sin 38 !=

b19.04

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 rectanglesc. 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) = 9 2 ! 1 = 81 ! 1 = 80

c. y = 2 x + 5 x = 2 y + 5

x ! 5 = 2 y x! 5

2 = y

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




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

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




1-53.The graph does not give a full line. The line starts at ( ! 2 , 3) and then follows the line y = 2 x + 7 . Since h( x) is defined for only values of x ! " 2 , the composite function is onlydefined for x ! " 2 .

1-54.a. Opposite = ! 54

Reciprocal = 5 ! 4 = 154

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 = ! 7

13( )! 5

= ! 713( )

! 5

Reciprocal = 713( )5

1-55. x

m

x n

= 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 =

36

= 12 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 ! x 1 h. y ! y1

x ! x1

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

23

! 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




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

27 5 = 1

(3 3 )5 = 1

315 = 3! 15

b. 18( )

x

= 1

2 3( ) x

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

c. 16 x ! 132( )

2 " x ( )= 2 4( ) x ! 1

2 5( ) 2 " x ( )

= 2 4 x ! 2 " 5( ) 2 " x ( )

= 2 4 x !2 " 10 + 5 x = 2 4 x " 10 + 5 x = 2 9 x " 10

1-70.

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

= 12

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

2 x + 4 = x ! 1

x + 4 = ! 1

x = ! 5

1-71.

a. (2 3 )( x + 3) = 2 5

23 x + 9

= 25

! 3 x + 9 = 5

3 x = " 4

x = " 43

b. (3 3 )2 x = 132( )

( x ! 1)

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

36 x = 3 ! 2 x + 2

" 6 x = ! 2 x + 2

8 x = 2

x = 14

c. 15 3( )

(2 x ! 3)= 1

52

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

" 5 ! 6 x + 9 = 5 ! 2

! 6 x + 9 = ! 2

! 6 x = ! 11

x = 116

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

x ! 20 = 2 x + 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. x 2 ! 7 x + 12 = 0

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

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

x = 3, 4

d. x 3 + x 2 ! 6 x = 0

x ( x 2 + x ! 6) = 0 x ( x + 3)( x ! 2) = 0

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




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

= x2 + 2 x + 1 + x

= x2 + 3 x + 2

c. ( x + 1)2 + ( x + 1) = 2

x 2 + 2 x + 1 + x + 1 = 2

x 2 + 3 x = 0

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

1-74.sin ! P =

816

= 12

sin " 1 sin ! P( ) = sin " 1 12( )

! P = 30 !

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

sin 60!

=

r16

32

#16 = r

r = 8 3 $ 13.86 cm

Lesson 1.2.2

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

b. ! b + D2 a

= R and ! b ! D2 a

= 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)/2 ! Y, and Y/2 ! Y.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 up 3 units: ! 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 , 6

2( )= 5, 3( )

Slope of perpendicular line = ! 12

Equation of line y ! 3 = ! 12 x ! 5( )

1-82.a. p = 6, q = 2 b. not in p q form c. p = x + 3 y , q = 2 ! r d. not in p q form

1-83.a. 6 2 !(6 2 )" 3 !1 = 6 2 !6 " 6 = 6 " 4

b.(5 2 )2 !5 " 3

(5 3 )" 2 =

5 4 !5 " 3

5 " 6 =

51

5 " 6 = 51" (" 6) = 5 7

c. 3 !19 97

8 !19 94 =

3

8!19 97 " 94

=

3

8!19 3

d.3 !19 -97

8 !19 " 94 =

3 !19 94

8 !19 97 =

3

8!19 94 " 97

=

3

8!19 " 3

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

! (2 + 3)2 " 4 + 925 " 13

b. Example: p = 2, q = 3

! 22 + 32 " 2 + 3

13 " 5

Solution continues on next page. "




Lesson 1.3.1

1-90.a. 1

2 K ah= b. sin C =

hb

! h = b sin C

c.K =

1

2 ah ! K =

1

2 ab sin C

1-91.Area =

12 (6)(4) sin 76 !

= 12 !0.970 = 11.644 cm 2

1-92.K =

12

bh

sin A = hc

! h = c sin A

K =12

bc sin A

1-93.SA of one side =

12 (10)(10) sin 40 !

= 50 !0.643 = 32.139 ft 2

Total surface area of sides = 4(32.139) = 128.558 ft 2

1-94.a. sin A =

hb

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

! h = a sin B

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

b sin A = a sin B

sin Aa

= sin B

b

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

b sin C = c sin B

sin Bb

= sin C

c

e. sin Aa

= sin B

b =

sin C c

1-95.

1-96.

a. ! NAT = 180 ! " 100 ! " 38 != 42 ! b. sin 42 !

200 =

sin100 !

y

sin P p

=

sin Qq

= sin R

r




sin 42 !

200 =

sin 38 !

x

0.0033 x = 0.6157

x = 184.018 ft

0.0033 x = 0.985

x = 294.354 ft

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. Notethat (c) is the only diagram in which you’re given exactly one side.


1-98.a. b. ! G = 180 ! " 64 ! " 38 !

= 78 !

8

sin 78 !

= OG

sin 64 !

8 #0.8988 = OG #0.9781

7.19040.9781

= OG

7.351 in = OG

8

sin 78 !

= DG

sin 38 !

8 !0.6157 = DG !0.9781

4.92560.9781

= 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 + 4 x + 4 + 2 x + 4

0 = x2 + 6 x + 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.




y n = x

y = xn




1-102.

a. 64 1/3( )2

= 4 2= 16 b. 125 1/3( )

4= 5 4

= 625

c. 81 3/ 4= 81 1/ 4( )

3= 3 3

= 27 d. 278( )

1/ 3! "

# $

%2

= 3

2( )%2

= 2

3( )2

= 49

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

x = 2 y ! 13

x3 = 2 y ! 1

x3 + 12

= y12 ( 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 = 2 x3/2

x = 2 y3/2

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

102 "

= r # r = 5"

= 1.592 feet

1-105.a. n 2

+ n 2= d 2

2 n 2= d 2

2 n 2= d

n 2 = d

b. 12 !90 !

= 45 !

c. (2 n )2 = n 2 + k 2

4 n 2 = n 2 + k 2

3n 2 = k 2

3n 2 = k

n 3 = k

d. y = 60 ! (equilateral triangle)




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.a 2 ! d 2

= c 2 ! e 2

a 2 ! d 2+ e 2

= c 2

e. c 2 = a 2 ! d 2 + (b ! d )2

c 2 = a 2 ! d 2 + b 2 ! 2bd + d 2

c 2 = a 2 + b 2 ! 2bd

f. cos C = d a

! d = a cos C

g. c 2 = a 2 + b 2 ! 2b (a cos C )

1-107.a. c 2 = 110 2 + 126 2 ! 2(110)(126) cos 74 !

c 2 = 12100 + 15876 ! 7640.668

c 2 = 20335.332

c = 20335.332 = 142.6 ft

Yes, 150 feet of fencing is enough.

b. sin 74 !

142.6 =

sin B

126

0.0067 = sin B

126

0.849 = sin B

sin! 1

0.849 =

sin! 1

(sin B

)58.1 !

= " B

sin 74 !

142.6 =

sin C

110

0.0067 = sin C

110

0.7415 = sin C

sin! 1

0.7415 =

sin! 1

(sin C )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

c 2 = a 2 + b 2 – 2 ab 0( )c 2 = a 2 + b 2




1-110.c 2 = 10 2 + 14 2 ! 2(10)(14) cos 60 °

c 2 = 296 ! 280 " 12

c 2 = 156

c = 156 = 12.490

1-111. BE 2 = 3.5 2 + 2.8 2 ! 2(2.8)(3.5) cos 43 !

BE 2 = 20.09 ! 14.3345

BE 2 = 5.7555 BE = 2.399 km

1-112.

c2

= 102

+ 202

! 2(10)(20) cos 30!

c 2 = 500 ! 346.4102

c 2 = 153.5898

c = 12.393 cm

1-113. x 2 = 28 2 + 42 2 ! 2(28)(42) cos X !

x 2 = 784 + 1764 ! 2352 cos X !

x 2 = 2548 ! 2352 cos X !

x 2 + 2352 cos X !

= 2548

If X = 0 !

x 2 + 2352 = 2548

x 2 = 196

x = 196 = 14

If X = 180 !

x 2 ! 2352 = 2548

x 2= 4900

x = 4900 = 70

14 < x < 70 inches


1-114.a. 6 2 = 5 2 + 8 2 ! 2 " 5 " 8 cos A

36 = 89 ! 80 cos A

5380 = cos A

0.6625 = cos A

48.51 !

= # A

6

sin 48.51 !

= 5sin B

6 sin B = 5 ! 0.7491

sin B = 3.74546

= 0.6242

" B = 38.62 !

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

b. A = 12 8 !6 !(sin 38.62) =

8!6!0.62422

= 29.959

2 = 14.98 square meters




1-115. a. x

2 + 1 + 2 x ! 8 = x 2 + 2 x ! 7

b. (2 x - 8) 12

x + 1( )= x 2 + 2 x ! 4 x ! 8 = x 2 ! 2 x ! 8

c. 2( 12

x + 1) ! 8 = x + 2 ! 8 = x ! 6

d. ( x 2 + 1)2 + 1 = x 4 + 2 x 2 + 1 + 1 = x 4 + 2 x 2 + 2

1-116.

a. 12 !360 !

= 180 ! b. 12 !2 " = " meters c. AB

!

= 16 !2 " =

"

3 meters

1-117.

a. 5 2 ! 5 2( )" 3 !5 3 = 5 2 + "6( )+ 3

= 5 " 1

b.3 2( )

2 !3 " 3

33

( )" 2

=

3 4 !3 " 3

3 " 6 =

3

3 " 6 = 31 " " 6( ) = 3 7

c.5 !14 98

8 !14 95 =

5

8!14 98 " 95

=

5

8!14 3

d.5 !14 -98

8 !14 " 95 =

5

8!14 " 98 " " 95( ) =

5

8!14 " 3

1-118.

a. 8 1/ 3( )2

= 2 2= 4 b. 100 1/2( )

3= 10 3

= 1000

c. 125 1/3( )2

= 5 2= 25

1-119.3 x ! 7 y = 42! 7 y = ! 3 x + 42

y = 37 x ! 6

perpendicular slope = ! 73

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

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

3 y = 4 x ! 36

y = 4

3 x ! 12

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

y = 66 x +

16

y = x + 16

x ! " 16

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




1-122.3 ! x " 0

! x " ! 3

x # 3

Lesson 1.4.1

1-123.d. 2 ! radius lengths = circumference

1-124.

Length of AB !

= 1 unit .

1-125.

C = 2

! "1 = 2

!

1-126.a. 360°b. 2 ! radiansc. 2 ! = 6.2832 , nearest whole number = 6

1-127.

1-128.a. Degrees in half a circle: 180 ! b. ! 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.




b. 8: x 2 ! 8 x + 16 = ( x ! 4)( x ! 4) = ( x ! 4)2




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 = " 4 x " 12

y = " 4 x " 14

1-138.a. tan A = 4

7

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. 5 x ( x ! 3) + 4( x ! 3) = ( x ! 3)(5 x + 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




1-142.a. In the center of each quadrant. b. y-axisc. 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 thesame distance as a point on the outside of the CD.

c. 200(2 ! " 5.25) # 6597 cm d. Distance around innermost track = 2 ! "2

65972 ! "2

= 524.97

525 rotations

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

63360 in902 rev

= 70.2439 inches in one revolution

C = 70.2439 = " d

d = 22.36 inch diameter

b. 902 !" !26 = 73676.63 inches

73676.6312

= 6140 feet61405280

= 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 1.4.21-147.

a. 120 !

1 ! "

180 !

=120 "

180 =

2 "

3 b. ! 225 !

1 " #

180 !

=! 225 #

180 = ! 5 #

4

c. 80 !

1 ! "

180 !

=80 "

180 =

4 "

9




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

9.8480.5736

= 17.17 = x

c. x 2 = 6.5 2 + 7.1 2 ! 2 "6.5 "7.1cos119 !

x 2 = 42.25 + 50.41 ! 92.3( ! 0.4848)

x 2 = 92.66 + 44.75 = 137.41

x = 137.41 = 11.72

d. 60 = 12 !15 ! x !sin 28 !

607.5

= 0.4695 x

80.4695

= 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( )= 2 3 ! 2 = 6

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

Composing 1and ! f f in either order returns the original number.

1-150.ax 2 + bx + c = d ( x2 ! 2ex + e2 ) + f

ax 2 + bx + c = dx2 ! 2dex + (de 2 + f )

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

1-151.a. Parent Graph ! y = x 3

Transformation

Flip over y -axis ! y = " x 3

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)




1-152. f (g( x)) = ( x ! 3)2 ! 2( x ! 3) + 5

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

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

x2 ! 8 x + 20 = 8

x2 ! 8 x + 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 !=

PQ12

1

2 !12 = 6 =

PQ

c. cos 30 !=

QR12

32

!12 = 6 3 = QR

d. sin P =6 312

=3

2

1-154. a. b.

c.

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

12

k ( x ) + 1




Closure Problems

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

c. g ( x+ 2) f ( x+ 2)

= 3( x+ 2) ! 1

2( x+ 2) 2 ! ( x+ 2)= 3 x+ 6 ! 1

2( x2 + 4 x+ 4) ! x! 2= 3 x+ 5

2 x2 + 8 x+ 8 ! x! 2= 3 x+ 5

2 x2 + 7 x+ 6

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

2 ( x ! 2)

CL 1-157.a. Third angle = 180 ! ! 50 ! ! 45 !

= 85 ! b. Law of Sinessin 85 !

7 =

sin 50 !

a

0.1423 = 0.766

a

a = 0.7660.1423

= 5.383

sin 85 !

7 =

sin 45 !

b

0.1423 = 0.707

b

b = 0.7070.1423

= 4.969

c. ASA

CL 1-158.a. c 2 = 42 + 7 2 ! 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. 25 x 4

= 5 !5 ! x 2 ! x

2= 5 x

2 b. ( xy 2 )3

( x 2 y 3 )1/2 =

x 3 y 6

xy 3/2 = x 3! 1 y 6 ! 3/2 = x 2 y 9/ 2

c. x 3 + ( x 2 )1/2 = x 3 + x

CL 1-160.




a. 3 !2 " = 6 " b. 262

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



CL 1-161.

a. sin 45 != 4

hyp.

hyp = 41/ 2

= 4 2

Isosceles triangle

leg = 4

b. sin 45 !=

hyp.6

hyp = 12

!6 = 62

=6 2

2 = 3 2

Isosceles triangle

leg = 3 2

c. sin 30 != 2

hyp. cos 30 !

=leg4

hyp = 21 2

= 4 leg =3

2 !4 = 2 3

CL 1-162. f ( x) = 2 x ! 3 + 1

Parent graph: y x=

Stretched: 2 y x=

Shifted right three units: 2 3 y x= !

Shifted up one unit: 2 3 1 y 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

=1253.5

= 35.7 mph

Cpm Precalculus Chapter 01 Solutions

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Transcript of Cpm Precalculus Chapter 01 Solutions