SOLUTIONS MANUAL - F IFTH EDITION PRECALCULUS: A RIGHT TRIANGLE AP...

SOLUTIONS MANUAL

JUDITH A.

PENNA ALGEBRA & TRIGONOMETRY F

IFTH EDITION

PRECALCULUS: A RIGHT

TRIANGLE APPROACH

FIFTH EDITION

Judith A.Beecher Judith A. Pen na Marvin

• Bittinger 1 / 4

4 2 2 4 42 2 4x y

( 1, 4)

( 3, 5)

(0, 2)

(4, 0)

(2, 2)

4 2 2 4 42 2 4x y

( 5, 0)

( 4, 2)

(1, 4)

(4, 0)

(2, 4)

4 2 2 4

• 2 4x

y

( 5, 1)

(2, 3)

(0, 1)(5, 1)

(2, 1)

4 2 2 4 42 2 4x y

( 5, 2)

( 5, 0)

( 1, 5)

(4, 0)

(4, 3)

Chapter 1 Graphs, Functions, and Models Exercise Set 1.1 1.Point A is located 5 units to the left of they-axis and

• units up from thex-axis, so its coordinates are (−5,4).

Point B is located 2 units to the right of they-axis and

• units down from thex-axis, so its coordinates are (2,−2).

Point Cis located 0 units to the right or left of they-axis and 5 units down from thex-axis, so its coordinates are

(0,−5).

Point D is located 3 units to the right of they-axis and

• units up from thex-axis, so its coordinates are (3,5).

Point E is located 5 units to the left of they-axis and

• units down from thex-axis, so its coordinates are

(−5,−4).

Point F is located 3 units to the right of they-axis and

• units up or down from thex-axis, so its coordinates are

(3,0).

2.G: (2,1); H: (0,0); I: (4,−3); J: (−4,0); K: (−2,3);

L: (0,5)

3.To graph (4,0) we move from the origin 4 units to the right of they-axis. Since the second coordinate is 0, we do not move up or down from thex-axis.To graph (−3,−5) we move from the origin 3 units to the left of they-axis. Then we move 5 units down from the x-axis.To graph (−1,4) we move from the origin 1 unit to the left of they-axis. Then we move 4 units up from thex-axis.To graph (0,2) we do not move to the right or the left of they-axis since the first coordinate is 0. From the origin we move 2 units up.To graph (2,−2) we move from the origin 2 units to the right of they-axis. Then we move 2 units down from the x-axis.

4.

5.To graph (−5,1) we move from the origin 5 units to the left of they-axis. Then we move 1 unit up from thex-axis.To graph (5,1) we move from the origin 5 units to the right of they-axis. Then we move 1 unit up from thex-axis.To graph (2,3) we move from the origin 2 units to the right of they-axis. Then we move 3 units up from thex-axis.To graph (2,−1) we move from the origin 2 units to the right of they-axis. Then we move 1 unit down from the x-axis.To graph (0,1) we do not move to the right or the left of they-axis since the first coordinate is 0. From the origin we move 1 unit up.

6.

7.The first coordinate represents the year and the second co- ordinate represents the number of Sprint Cup Series races in which Tony Stewart finished in the top five. The or- dered pairs are (2008, 10), (2009, 15), (2010, 9), (2011, 9), (2012, 12), and (2013, 5).

8.The first coordinate represents the year and the second coordinate represents the percent of Marines who are women. The ordered pairs are (1960, 1%), (1970, 0.9%), (1980, 3.6%), (1990, 4.9%), (2000, 6.1%), and (2011, 6.8%).CopyrightcP2016 Pearson Education, Inc. 2 / 4

2Chapter 1: Graphs, Functions, and Models

9.To determine whether (−1,−9) is a solution, substitute −1 forxand−9 fory.y=7x−2

−9?7(−1)−2

−7−2

−9

−9TRUE

The equation−9=−9 is true, so (−1,−9) is a solution.To determine whether (0,2) is a solution, substitute 0 for xand 2 fory.y=7x−2

2?7·0−2

0−2 2

−2 FALSE

The equation 2 =−2 is false, so (0,2) is not a solution.

10.For P 1 2 ,8 o

:

y=−4x+10

8?−4·

1 2 +10

−2+10

8

8TRUE P 1 2 ,8 o is a solution.

For (−1,6):y=−4x+10

6?−4(−1)+10

4+10 6

14 FALSE

(−1,6) is not a solution.

11.To determine whether i 2 3 , 3 4 n is a solution, substitute 2 3 forxand 3 4 fory.6x−4y=1 6· 2 3

−4·

3 4 ?1

4−3

1

1TRUE The equation 1 = 1 is true, so i 2 3 , 3 4 n is a solution.To determine whether i 1, 3 2 n is a solution, substitute 1 for xand 3 2 fory.6x−4y=1

6·1−4·

3 2 ?1

6−6

1 FALSE

The equation 0 = 1 is false, so i 1, 3 2 n is not a solution.

12.For (1.5,2.6): x

2 +y 2 =9 (1.5) 2

+(2.6)

2 ?9

2.25+6.76

9.01

9 FALSE

(1.5,2.6) is not a solution.

For (−3,0):x

2 +y 2 =9

(−3)

2 +0 2 ?9 9+0

9

9TRUE (−3,0) is a solution.

13.To determine whether i − 1 2 ,− 4 5 n is a solution, substitute − 1 2 foraand− 4 5 forb.2a+5b=3 2 i − 1 2 n +5 i − 4 5 n ?3

−1−4

−5

3 FALSE

The equation−5 = 3 is false, so i − 1 2 ,− 4 5 n is not a solu- tion.To determine whether i 0, 3 5 n is a solution, substitute 0 for aand 3 5 forb.2a+5b=3

2·0+5·

3 5 ?3

0+3

3

3TRUE The equation 3 = 3 is true, so i 0, 3 5 n is a solution.

14.For i 0, 3 2 n

: 3m+4n=6

3·0+4·

3 2 ?6

0+6

6

6TRUE i 0, 3 2 n is a solution.For i 2 3 ,1 n

: 3m+4n=6

3· 2 3

+4·1?6

2+4

6

6TRUE The equation 6 = 6 is true, so i 2 3 ,1 n is a solution.CopyrightcP2016 Pearson Education, Inc. 3 / 4

y x

• 224

4 2 2 4 5x 3yPoP 15

( 3, 0)

(0, 5)

y x

• 224

4 2 2 4 2x 4yPoP8

(4, 0)

(0, 2)

(0, 4)

y x

• 224

4 2 2 4 2x i yPoP4

(2, 0)

y x

• 224

2 2 4 6

(0, 6)

3x i yPoP6

(2, 0)

Exercise Set 1.13 15.To determine whether (−0.75,2.75) is a solution, substi- tute−0.75 forxand 2.75 fory.x 2 −y 2 =3

(−0.75)

2

−(2.75)

2 ?3

0.5625−7.5625

−7

3 FALSE

The equation−7 = 3 is false, so (−0.75,2.75) is not a solution.To determine whether (2,−1) is a solution, substitute 2 forxand−1 fory.x 2 −y 2 =3 2 2

−(−1)

2 ?3 4−1

3

3TRUE The equation 3 = 3 is true, so (2,−1) is a solution.

16.For (2,−4): 5x+2y

2 =70

5·2+2(−4)

2 ?70

10+2·16

10+32

42

70 FALSE

(2,−4) is not a solution.

For (4,−5): 5x+2y

2 =70

5·4+2(−5)

2 ?70

20+2·25

20+50

70

70 TRUE

(4,−5) is a solution.

17.Graph 5x−3y=−15.To find thex-intercept we replaceywith 0 and solve for x.5x−3·0=−15 5x=−15 x=−3 Thex-intercept is (−3,0).To find they-intercept we replacexwith 0 and solve for y.5·0−3y=−15 −3y=−15 y=5 They-intercept is (0,5).We plot the intercepts and draw the line that contains them. We could find a third point as a check that the intercepts were found correctly.

18.

19.Graph 2x+y=4.To find thex-intercept we replaceywith 0 and solve for x.2x+0=4 2x=4 x=2 Thex-intercept is (2,0).To find they-intercept we replacexwith 0 and solve for y.2·0+y=4 y=4 They-intercept is (0,4).We plot the intercepts and draw the line that contains them. We could find a third point as a check that the intercepts were found correctly.

20.CopyrightcP2016 Pearson Education, Inc.

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