Answers to All Exercises - Issaquah Connectconnect.issaquah.wednet.edu/.../ALG-1-Book-Answers.pdf6 ANSWERS TO ALL EXERCISES LESSON 0.5 1a. 8.0 cm 1b. 4.3 cm 1c. 7.2 cm 2a. 2.8 cm segment

ANSWERS TO ALL EXERCISES 1

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Answers to All Exercises7b.

7c. �19�; �

18

71�; �

27

12

79� 7d. �

89�; �

68

41�; �

57

12

29�

8a. 8 � 4 � 2 8b. 8 � �14� � 2

8c. Essentially, they are the same.

8d. 8 � �34� � 6

9a. �196�

9b. sample answers:

12 � �196� � 6�

34�; (12 � 16) � 9 � 6.75

10a. �19� 10b. �8

11�

10c. 11�59�; ��

49� � 24� � ��8

31� � 24� or 24 � ��

49� � �8

31��

11. Answers should be equivalent to these:

11a.

�14� � �

14� � 32 � �

31

26� � 2

11b.

�34� � �

14� � �

14� � 32 � �6

964� � �

32� � 1�

12�

11c.

1�2� � �

12� � �

14� � 32 � �

31

26� � 2

12. �14� � �1

16� � �3

12� � �

13

12�

13. 1 � �13

12� � �

23

12�

Stage 3

LESSON 0.1

1a. �18�; �1

16� � �1

16� or 2 � �1

16�

1b. �634�; �6

14� � �6

14� � �6

14� or 3 � �6

14�

1c. �35�; �2

15� � �2

15� � �2

15� � �2

15� � �2

15� � �2

15� � �2

15� � �2

15�

� �215� � �2

15� � �2

15� � �2

15� � �2

15� � �2

15� � �2

15� or 15 � �2

15�

1d. �6725�; �6

125� � �6

125� � �6

125� � �6

125� � �6

125� � �6

125�

� �6125� or 7 � �6

125�

2a. �14� � �1

16� � �1

56� 2b. �1

26� � �6

34� � �

16

14�

2c. 9 � �811� � �

19� 2d. �

19� � �8

21� � �

18

11�

3a.

3b. 32 3c. 48 3d. 56

4a. �59� 4b. �1

129�

4c. �365� 4d. �

194� or 1�

59�

5. sample answers:5a. 5b.

5c. 5d.

6. Answers should include creation by a repetitiveprocess, and smaller parts look like the whole.7a. Sample description: Divide each side of thesquare into thirds, and connect those points with linesparallel to the sides. A square is formed in the middle.Erase everything except the center square. To get thenext stage, do the same thing in all eight squaresformed around the middle square.

Stage 4

CHAPTER 0 • CHAPTER CHAPTER 0 • CHAPTER 0

LESSON 0.2

1a. 54

1b. 75

1c. 37

1d. 23

2a. 3 � 3 � 3 � 3; 3 � 3 � 3 � 3; 3(3)(3)(3)2b. 5 � 5 � 5 � 5 � 5 � 5; 5 � 5 � 5 � 5 � 5 � 5;5(5)(5)(5)(5)(5)

2c. �12� � �

12� � �

12�; �

12� � �

12� � �

12�; �

12��

12��

12��

3a. 33

3b. 25

3c. 54

3d. 73

4a. 8

4b. �11

45�

4c. �58�

4d. �373� or 4�

57�

4e. �12�

4f. �66

34�

5a. 4 or 22; 8 or 23

5b. 25

6a. 25 or 52

6b. 125 or 53

6c. 55

7a. 647b. 5127c. 82; 83

7d. 262,144 or 86

7e. The exponent is always one less than the stagenumber.7f. yes; because 80 � 18a. 108b. 508c. Stage 3: 2 � 52; Stage 4: 2 � 53; Stage 5: 2 � 54

8d. The 2 is the number of new branches at Stage 1.The 5 is the number of smaller segments created inStage 1. At each subsequent stage, new branches areadded by multiplying by 5 again.9b.

9c. The area of one shaded triangle is �14� the area of one

of the previous shaded triangles. The total area of theshaded triangles in each figure is �

34� the shaded area in

the previous figure.10. 2 � 2 � 2 � 2 � 2 � 2 � 2 � 2 � 28 � 256 or $25611. One possibility: The restaurant had �

34� of a pie left.

Five people wanted pie. After cutting the pie into fifths,how much of the pie did each person get? �2

30�

12a. �26

14�

12b. �62

49�

2 ANSWERS TO ALL EXERCISES

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One Total area of theStage triangle shaded triangles

1 �14� �

34�

29 � �1

16�

or �34� � �

34� � �1

96�

3 �614� �

34� � �1

96� � �

26

74�

1�16


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LESSON 0.3

1a. �12

85

�; 15.63

1b. �295�; 2.78

1c. �24

801

1�; 29.64

1d. �76249

�; 11.39

2. �53

2

2� � �53� � �

190� or 2.78 � 1.67 � 1.11

3. ��53��

4� 7.72; ��

53��

5� 12.86; Stage 5

4a. �21

09�

4b. 25

4c. �77

25� or �22

45�

4d. �31

58

� or 63�35�

5b. Stage 5: ��54��5

� 3.05; Stage 11: ��54��11

� 11.64.

See below.

6a. See below.6b. Stage 5

6c. Stage 11: ��64��

11

� 86.50

Total length

Stagenumber Multiplication form Exponent form Decimal form

0 1 10 1

1 5 � �14� � �

54� ��

54��

1

1.25

2 5 � 5 � �14� � �

14� � �

21

56�

3

52 � ��14��2

� ��54��2

53 � ��14��3

� ��54��3

1.56

1.955�5�5� �14� � �

14� � �

14� � �

16245

�

5b. (Lesson 0.3)

Total length


0 1 10 1

1 6 � �14� � �

64� � �

32� 61 � ��

14��

1

� ��64��

1

� ��32��

1

1.5

2

3

6a. (Lesson 0.3)

6 � 6 � �14� � �

14� � �

31

66� � �4

9� 62 � ��

14��

2

� ��64��

2

� ��32��

2

2.25

6 � 6 � 6 � �14� � �

14� � �

14�

� �26146

� � �287�

63 � ��14��

3

� ��64��

3

� ��32��

3

3.38

7a. See below.7b. Stage 57c. No; Stage 6 has a length of slightly more than 161,and Stage 7 has a length of over 376.8a. See below.8b. Estimates will vary; ��

83��4 or about 50.

8c. After 23 stages, many calculators resort toscientific notation or get an overf low error.9. 2.8


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7 � 7 � �13� � �

13� � �

499� 72 � ��

13��2

� ��73��

25.44

7 � 7 � 7 � 7 � �13� � �

13� � �

13�

� �13� � �

248

01

1�

74 � ��13��4

� ��73��4

29.64

7 � 7 � 7 � �13� � �

13� � �

13� � �

32473

� 73 � ��13��3

� ��73��3

12.70

Total length


0 1 10 1

1 7 � �13� � �

73� 71 � ��

13��

1

� ��73��

1

2.33

2

3

4

7a. (Lesson 0.3)

8 � 8 � �13� � �

13� � �

694� 82 � ��

13��2

� ��83��2

7.11

8 � 8 � 8 � �13� � �

13� � �

13� � �

52172

� 83 � ��13��3

� ��83��3

18.96

Total length


0 1 10 1

1 8 � �13� � �

83� 81 � ��

13��

1

� ��83��

1

2.67

2

3

8a. (Lesson 0.3)

10. �43�, or 1�

13�

11a. 411b. 1611c. 6411d. 41, 42, 43

11e. The exponent is 1 less than the stage number;for Stage 1, 40 � 1.

Starting value 2 �1 10

First recursion

Second recursion

Third recursion

�


LESSON 0.4

1a. 3 1b. �3 1c. �7 1d. �32a. �10 2b. �24 2c. 12 2d. �42e. �6 2f. 103a. �8 3b. �25 3c. �17 3d. �11 4a. i; 2 4b. iv; �14 4c. ii; 14 4d. iii; �14 4e. v; �2 5a. Add the two absolute values. The result will be anegative number.5b. Subtract the number with the smaller absolutevalue from the number with the larger absolute value.The sign of the answer is the sign of the number withthe larger absolute value.5c. Subtracting a negative number is the same asadding a positive number, so you just add the twoabsolute values. The result will be a positive number.5d. Subtracting a negative number is the same asadding a positive number, so the problem actuallyinvolves adding a negative number and a positivenumber.5e. Multiply the two absolute values. When youmultiply two numbers with different signs, the result isa negative number.5f. Multiply the two absolute values. When youmultiply two numbers with the same sign, the result isa positive number.5g. Divide the two absolute values. When you dividetwo numbers with different signs, the result is anegative number.5h. Divide the two absolute values. When you dividetwo numbers with the same sign, the result is a positivenumber.6a. In the first recursion, he should get �0.2 � 2 ��0.4, not �0.4. His arithmetic when evaluating 0.4 �4was correct. In the second recursion, he used thewrong value (�3.6 instead of �4.4) because of hisprevious error. His arithmetic was also incorrect,because �0.2 � �3.6 � �0.72, not �0.72. Hisarithmetic when evaluating �0.72 � 4 was correct.6b. �4.4, �3.12, �3.376, �3.32486c. �3.8, �3.24, �3.3526d. Yes; the calculations seem to be approaching avalue close to �3.3.7a.

7b. yes; about �2.2227c. Entering �2.222222222222 as a starting valuereturns the same value as an answer.8a.

8b. No; the values get farther and farther apart.8c. The result is �

13�. The value �

13� is a fixed point for this

expression.9a. i. 129a. ii. �169a. iii. �89b. When the coefficient of the box is 0.5, theattractor value is twice the constant. In general the attractor value is

9c. One possibility is 0.5 � � 3.

10a. i. 7.5 or �125�

10a. ii. �1010a. iii. 6.2510b. When the coefficient is 0.2, the attractor value is1.25 times the constant. In general the attractor value

is

10c. One possibility is 0.2 � � 1.8.11. All involve repeating a process. Each time, theresult becomes the starting value or figure for the nextrepetition.12a. 2112b. 0.212c. 1312d. 3

13. �22.5, or �22�12�

14. 20

15. �176�

constant term��1 � coefficient of the box

constant term��1 � coefficient of the box

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�1.80 �2.1 �1

�2.18 �2.21 �2.1

�2.218 �2.221 �2.21

Starting value 2 �1 10

First recursion

Second recursion

Third recursion

�

�3 �3 �19

�17 �5 �39

�13 11 �77


LESSON 0.5

1a. 8.0 cm 1b. 4.3 cm 1c. 7.2 cm2a. 2.8 cm segment 2b. 5.7 cm segment2c. about 5.1 cm segment3a–c. Answers should look proportional to this:

3d. points D and B; no

4a. 10 4b. 17 4c. ��15

1� 4d. 11

4e. 16 4f. 19 4g. �6�12�

5d. The resulting figure should slightly resemble aright-angle Sierpinski triangle.6a–d. The resulting figure should slightly resemblethe Sierpinski carpet from Lesson 0.1, Exercise 7.6e. Answers will vary. You could ignore any rolls of 5or 6 and move toward corner 1, 2, 3, or 4 on those rolls.7a. This game fills the entire square.7b. This game creates a small Sierpinski triangle ateach corner of the triangle.7c. This game creates four small Sierpinski carpets,one at each corner of the square.

A C DE B

7d. This game creates a pattern like the Sierpinskitriangle.

8. Possible answers: Use a die and ignore rolls of 6, orchoose one of five playing cards to indicate the move.The answer should describe a process by which allcorners are equally likely to be chosen.9. Point should divide segment into an 8 cm and a4 cm segment.10i. 2 10ii. 4 10iii. �610b. The attractor is two-thirds of the constant.10c. One possibility is �0.5 � �15.11. See below.

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Total length


3 6 � 6 � 6 � ��14��

14��

14�� 63 � ��

14��

3

� ��64��

3

� �287� 3.38

4 6 � 6 � 6 � 6 � ��14��

14��

14��

14�� 64 � ��

14��

4

� ��64��

4

� �81

16� 5.06

Total length


0 1 10 1

1 6 � �14� 61 � ��

14��

1

� ��64��

1

� ��32��

1

2 6 � 6 � �14� � �

14�

1.5

2.2562 � ��14��2

� ��64��2

� �49

�

11. (Lesson 0.5)


CHAPTER 0 REVIEW

1a. iii1b. v1c. ii1d. iv1e. i2a. 722b. 2902c. �10 2d. 3122e. 2.16–

2f. �34

3a. �13� ��

13� ��

13�

3b. �23� ��

23� ��

23� ��

23�

3c. 1.2 �1.23d. 16 � 16 � 16 � 16 � 163e. 2 � 2 � 2 � 2 � 2 � 2 � 2

4a. �116� � �1

16� � �1

16� � �1

36�

4b. �19� � �

19� � �8

11� � �8

11� � �

28

01�

4c. �14� � �1

16� � �6

14� � �6

14� � �

13

12�

5a.

A branch is added at the midpoint of each of thenewest segments, with half the length, at a 45°clockwise rotation.

Stage 3

5b.

A “bottomless” equilateral triangle is built on the“right” half of segments.

5c.

Each new segment is crossed at its midpoint by acentered perpendicular segment of equal length.

5d.

Each unshaded square is divided horizontally andvertically to create four congruent squares; thebottom-right square is shaded.6a. See below.

6b. ��75��20

� 836.68

7. The attractor is 5.

Stage 3Stage 3

Stage 3

Stage 3

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Total length

Stage Multiplication Exponent Decimalnumber form form form

0

1

2

1 1 1

7 � ��15�� 71 � ��

15��

1

� ��75��

1

� �75� 1.4

7 � 7 � ��15��

15�� 72� ��

15��

2

� ��75��

2

� �24

59� 1.96

6a. (Chapter 0 Review)


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Answers to All Exercises6a. iii6b. ii6c. iv6d. i7. 3 in classical8. a bar graph; because the information falls intocategories, is not numerical data, and cannot be scaledon a number line9a. possible answer: Jonesville’s Varsity BasketballTeam9b. possible answer: Jonesville’s Kindergarten Classand Their Teacher9c. possible answer: Jonesville’s Algebra Class9d. possible answers: Group A: “Everyone in theschool is very tall”; Group B: “Almost everyone in theschool is very short”; Group C: “Jonesville School is ahigh school only.”9e. Answers will vary.10a. 104

10b. 23 � 56

10c. �38

6

3�

11a. 911b. 711c. �4.211d. �111e. 3.812a. 18;

12b. 9�12�;

13. �116� � �1

16� � �

14� � �

38�

LESSON 1.1

1. max: 93 bpm; min: 64 bpm; range:93 bpm � 64 bpm � 29 bpm2.

3a. Jupiter3b. Mercury and Venus have no satellites.3c. 303d. 15 times4a.

4b. 304c. 241 min

4d. �23401

� � 8 min

5a. 80 bpm

5b. 29 bpm5c. She counted her pulse for 1 full minute.5d. Any whole number could occur, not just multiplesof 4.5e. a full minute, sometimes longer; to ensureaccuracy

0 4 8 12 16Time (min)

Travel Time to School

O OtherPhC NH Ca

Elements in the Human Body

Per

cen

tage

20

40

60

Elements


Doubles of 225 450 900 1800 3600 7200Doubles of 1 2 4 8 16 32

Doubles of 6 6 12 24 48 96

Doubles of 1 1 2 4 8 16

Half of 6 3

Half of 1 �12�


LESSON 1.2

1a. mean and median: 6; mode: 51b. mean: 5.1; median: 5; modes: 3, 81c. mean: 10.25; median: 9; no mode1d. mean: 17.5; median and mode: 20

2a. mean: �61

55� � 4.3; median: 4; no mode

2b. mean: �51

35� � 3.5; median: 2; modes: 1, 2

3a. 203b. 83c. 14a. mean: 262.2 ft; median: 215 ft4b. mean: 251.2 ft; median: 206 ft5. $18.24; if you multiply the mean by the number ofitems, you get the sum of the items.6a. mean: 97.485 m; median: 60 m6b. Median; the mean is affected by the extremevalue 525.7. The first three averaged 53 s each; together theytook 159 s. The total time for the whole team must be50(5) or 250 s. The two remaining members must havea total of 250 � 159 or 91 s. So the last two membersmust average 91 � 2 or 45.5 s each.8. The mean, 83.8, is lower than all but one of hisscores; the median, 88, is more representative.

9. Answers will vary. If the speaker is talking aboutthe entire state, this cannot occur. All scores cannotbe greater than the middle score. In fact, exactly halfof the scores will be above the median and half willbe below.10a. Multiply the mean by 10; together they weighapproximately 15,274 lb.10b. Five of the fish caught weigh 1449 lb or less, andfive weigh 1449 lb or more.10c. 2664 lb11a. sample answer: {8, 10, 12, 32, 33} years11b. sample answer: {2, 3, 3, 5, 5, 5} people11c. sample answer: {7, 14, 20, 21, 24, 27, 27} points12a. See below.12b. mean: 32.65; median: 30; mode: 2812c. The median is probably best; the mean isdistorted by one extremely high value.13a. A dot plot may be most appropriate for thenumeric data.However,if each value was translated intoyears (divide by 12),you could make a bar graph orpictograph with ages as categories.13b. See below.14a. 12 cm14b. 8 cm14c. 2.4 cm

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Answers may vary.

2015 25 30 35 40 45 50 55 60 65 70 75Age (yr)

Highest-Paid Athletes

180175170165160155150145140Age (months)

Student Ages

12a. (Lesson 1.2)

13b. (Lesson 1.2)


LESSON 1.3

1a. 5, 10, 23, 37, 50 1b. 10, 22, 31.5, 37, 501c. 14, 22.5, 26, 41, 47 1d. 5, 10, 19, 34.5, 472a. i.

2a. ii.

2b. i. 0, 1, 1.5, 3, 7; 2b. ii. 64, 75, 80, 86, 933. 1, 4, 6, 7, 9

4. a, d5a. Quartiles are the boundaries dividing a data setinto four groups, or quarters, with the same numberof values.5b. the range5c. the interquartile range, or IQR5d. Outliers are at or near the minimum andmaximum values, which are the endpoints of thewhiskers.6a. The prediction should be less than the mean.6b. 23 points6c. 23, 23, 25, 28, 296d. Because there are only four values, a five-numbersummary may be inappropriate—the valuesthemselves illustrate the spread of the data.7a. 173, 292, 360, 915, 13837b. The mean for the 1997–98 Bulls is about 675points, and the mean for the 2003–04 Bulls is about564 points; the medians are 416 points and 360 points,respectively. The means are both much higher than themedians because both teams have a few players whoscore very high. However, the 1997–98 mean is muchhigher than the 2003–04 mean because MichaelJordan scored so many total points.

7c. The median probably best represents the total-points-scored data for the 1997–98 Bulls. Students canjustify choosing either the mean or the median for the2003–04 Bulls. As a team owner, you might think themean better reflects your team’s talents.

7d. The box for the 2003–04 Bulls is longer, indicatingthat more of the team members are grouped towardthe center. The minimum number of points scored bya player is about the same, but the maximum is muchhigher for the 1997–98 Bulls.

1 3 5 87 92 64

76 80 84 88 9272686460Pulse rate (bpm)

80 2 4 7531 6

Number of pets

7e. Without Jordan, the range of the data is muchsmaller, and the box is a little shorter. If Jordan’s pointsscored are eliminated, the 2003–04 Bulls have thehigher-scoring players. There is more variation in thenumber of points scored by individual 2003–04 Bullsthan for individual 1997–98 Bulls without Jordan.

8a. For men, the mean salary is $639.56, and the five-number summary is 342, 495, 629, 718.5, 1001; forwomen, the mean salary is approximately $466.69, andthe five-number summary is 288, 353, 445, 563.5, 708.

8b. Women received less pay than men for the sametype of work.8c. The box plots highlight the discrepancy in payand spread of the data; however, dollar amountcomparisons within a single profession are possibleonly in the table.8d. Answers will vary. In 2000, gender equity did notexist for pay.9a. 35 ft9b. More information is needed. The length isbetween 17.5 and 25 ft.9c. 65 mi/h9d. The ten longest snakes vary in length from about8 ft to 35 ft.About half of the snakes have lengths fromabout 11 ft to 25 ft.Running speeds of the ten fastestmammals vary from 42 mi/h to 65 mi/h.About half ofthe speeds are between 43 mi/h and 50 mi/h.The cheetahappears to run much faster than most other mammals.9e. No; the units of these data sets are different.9f. about 47 mi/h10a. 829.5 10b. �541.5 10c. 1670.510d. An outlier would have to score fewer than�541.5 points or more than 1670.5 points. MichaelJordan is an outlier.11. One possibility is a family with ages 4, 10, 14, 39,and 43. The 14 is fixed, and the total of all ages must be110 years.12a. 76 million 12b. 15 million

12c. 10�12� pawprints

200 400 600 800500300 900 1000700Weekly wages ($)

Median Weekly Earnings, 2000

Women

Men

1997–98 Bulls (without Jordan)

Points scored0 500 1000 1500 2000 2500

2003–04 Bulls

1997–98 Bulls

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LESSON 1.4

1a. matinee: 29; evening: 301b. matinee1c. none1d. The number is less than or equal to 4.2a. possible graphs:

2b. One observation is that student scores tended tobe in the middle of the range.2c. 1, 6, 10, 12, 18

2d. 7, 123a. 763b. Approximately �

14� of the countries had a life

expectancy between approximately 69 yr and 74 yr.3c. 243d. There are no bins to the right of 85 in thehistogram. Also, the maximum point in the box plot islocated at approximately 83 yr.4.

907050 80604030

Life Expectancy

Nu

mb

er o

f co

un

trie

s

15

20

25

30

35

5

10

0

Age (yr)

0 8 164 12 20Number correct

Number of Problems Correctin Math Competition

208 1240 16


Nu

mb

er o

f st

ud

ents 8

76543210

Number correct

2 20166 100 14 188 124


Nu

mb

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f st

ud

ents

56

43210

Number correct

5a. minimum: 6.0 cm; maximum: 8.5 cm; range:2.5 cm5b.

6a. 240,000 cars

6b. Two models sold between 80,000 and119,999 cars, inclusive.

6c. [0, 480000, 40000, 0, 9, 1]

6d.

[0, 480000, 40000, 0, 9, 1]

6e. An approximate five-number summary is 115000,131000, 157000, 241000, 434000. The actual five-number summary is 115428, 130650, 157278.5,240712, 434145.

6f.

7. Answers will vary.7a. The bin heights should be about the same, withabout 16 or 17 in each of six bins.7b. This graph will have a few short bins on the endsand several tall ones around the actual height.7c. This plot might be fairly flat except for severalshort bins on the right end.7d. This graph will have fairly flat bins on the leftgetting taller as you go to the right, with the last twobins (8–9 and 9–10) both 25 units tall.8a. {1, 7, 7, 7, 7, 7, 7, 9}; the Q1 value, the median, andthe Q3 value are all 7.8b. {5, 5, 5, 5, 5, 5, 5, 5}; each value could be anynumber from 5 up to, but not including, 6.8c. {1, 2, 3, 4, 5, 6, 7, 8}; these values could have anynumber of digits after the decimal point.

2002 Sales

8 16 24 32 40

Cars sold (times 10,000)

Nu

mb

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f m

odel

s

0

1

2

4

3

5

6

7

8

0 48

0 5 50 0 50

5

678

0 means 6.0 cm6

Key

Ring Finger Length

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8d. {2, 2, 2, 2, 2, 4, 4, 4} or {2, 2, 2, 4, 4, 4, 4, 4}; theminimum and Q1 are 2, Q3 and the maximum are 4,and the median is either 2 or 4; so the mediancoincides with Q1 or Q3, respectively.9a. Perhaps the top two get As, and the next seven(down to 15 points) get Bs. Those with 9 to 15 pointswould get Cs, and the bottom three would get Ds.Students who score in the center (mean, median, andmode) get a C grade.9b. The outlier gets an A.9c. Answers will vary.10a. $1.5010b. chips priced $1.75 and $1.7910c. 1410d. $1.9910e. $1.0911a. Hospital A’s histogram is mounded toward theleft. Hospital B’s histogram is mounded toward theright. Hospital C’s histogram has all bins of equalheight. Hospital D’s histogram is mounded inthe middle.11b. Hospital A had more short waiting times andfewer long waiting times. Patients at Hospital B hadfewer short waiting times and more long waitingtimes. An equal number of patients were in each timerange at Hospital C. At Hospital D, the largest number

of patients had waiting times of 20–25 min; as thewaiting time gets farther from that central value, thereare fewer and fewer patients.11c. Answers will vary. No patient at Hospital B hadto wait longer than 30 min, while approximately 40%of patients at Hospital A had to wait longer than30 min. However, few patients at Hospital B waited lessthan 10 min, while almost 20% of patients at HospitalA waited less than 10 min.11d. Answers will vary. Students may preferHospital B because all patients were seen in less than30 min. At Hospital A, many patients were seen in lessthan 20 min, but some patients had to wait a verylong time.12a. Ida weighed the apples from the market, whichare more uniform in weight, and Mac weighed thebackyard apples, whose weights vary more widely.12b. Mac’s apples had a greater variety of weights;they were less uniform than Ida’s apples.12c. Both histograms are shaped like mounds, higherin the middle and decreasing in height toward thesides; more apples have weights near the mean, andfewer apples have weights farther from the mean.13. See below.14. One possibility is {6

�4�

, 70, 74, 80, 8�

2�

, 8�

2�

, 8�

2�

, 8�

2�

, 9�

5�

};the underlined values are fixed.

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140 145 150 155 160 165 170 175Age (months)

180

Student Ages

13. (Lesson 1.4)


LESSON 1.6

1.

2a. positive x-axis2b. Quadrant II2c. Quadrant I2d. Quadrant III2e. the origin2f. negative y-axis4a. about 2 m4b. about 5 s4c. about 2.7 m4d. between 0 and 1 s, and between 4.5 and 5.5 s5a. A(�7, �4), B(0, �2), C(3, 6), D(0, 3), E(4, �4),F(�2, �6), G(4, 0), H(7, 1), I(�5, 2), J(�5, 4)5b.

[�9, 9, 1, �6, 6, 1]5c. points B, D, and G5d. I: C and H; II: I and J; III: A and F; IV: E

6. Location scenarios should include informationsuch as (positive, positive) in Quadrant I, (negative,positive) in Quadrant II, (negative, negative) inQuadrant III, (positive, negative) in Quadrant IV,(0, any number) on y-axis, (any number, 0) on x-axis,and (0, 0) as the origin.7a. approximate answers (the second coordinates arein millions): (1984, 280), (1985, 320), (1986, 340),(1987, 415), (1988, 450), (1989, 445), (1990, 440), (1991,360), (1992, 370), (1993, 340), (1994, 345), (1995, 275),(1996, 225), (1997, 175), (1998, 160), (1999, 125), (2000,75), (2001, 45), (2002, 30), (2003, 15)

x

y

9–9

6

–6

C

DH

I

F

G

BA

J

E

7b.

[1984, 2003, 1, 0, 500, 50]7c. Answers will vary. Shipments increased until 1988and then decreased. The introduction of compact discsmay have influenced the decrease.8a.

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Average Miles per Gallon for All U.S. Automobiles

Years Year elapsed mpg

1960 0 14.3

1970 13.5

1980 15.9

1990 20.2

1995 21.1

1996 21.2

1997 21.5

1998 21.6

1999 21.4

2000 21.9

2001 22.1

(U.S. Department of Transportation,www.dot.gov) [Data set: AMPG]

10

20

30

35

36

37

38

39

40

41

8b.

8c.

4010 30200Years elapsed (since 1960)

Mil

es p

er g

allo

n 20

Average Miles per Gallonfor All U.S. Automobiles

(since 1960)

10

0


Mil

es p

er g

allo

n 20


(since 1960)

10

0


8d. 19.5 mpg8e.

The horizontal line separates those data points that areabove the mean of the averages from those that arebelow it.8f. Answers will vary.9a. (�1.5, 2.6), (�3, 0), (�1.5, �2.6), (1.5, �2.6)9b. Answers will vary.10a. 8:0610b. 40 mi/h10c. 12 min10d. Answers will vary. That he never fully stoppedduring the trip to school is unusual.


Mil

es p

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allo

n 20


(since 1960)

10

0

11a. One possibility is {5�

, 1�

2�

, 14, 1�

5�

, 20, 3�

0�

, 4�

7�

}; theunderlined values are fixed.11b. One possibility is {5

�, 10, 1

�2�

, 13, 14, 16, 20, 3�

0�

, 40,4�

7�

}; the underlined values are fixed.11c. One possibility is {5

�, 10, 12, 12, 13, 14, 16, 20, 28,

32, 40, 4�

7�

}; the underlined values are fixed.12a. 476, 496, 506.5, 536, 60512b.

12c. The greatest spread is between the third quartileand the maximum, that is, the right whisker. The leastspread occurs between the first quartile and themedian.12d. 40 points12e. Scotland, England, Sweden, Lithuania, UnitedStates, Australia; Belgium, Japan, Taiwan, Hong Kong,Korea, Singapore

400 600 700500Mean scores

Result of TIMSS

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LESSON 1.7

1a. The lower Ymin value allows trace numbers toappear on the calculator screen without interferingwith plotted points.

[0, 70, 10, �10, 60, 10]

1b. y � x1c. No; in no case is the estimated number ofdinosaurs more than the actual number.1d. Yes; the estimated count of five different specieswas less than the actual count.2a. Lucia’s estimates are all below the line—points B, C, E, F, and J.2b. Malcolm’s estimates are all above the line—points A, D, G, H, and I.3. overestimates: B, C, D, E, G; underestimates:A, F, H, I4a. possible approximations:

4b. Answers will vary. A minimum window is [0, 5, 1, 0, 5, 1].4c. y � x4d. The distance is increasing at about 1 m/s.The person may be walking away from the sensor.5a. The more the rubber band is stretched, the fartherit flies.5b. Answers will vary between 400 and 600 cm.5c. Answers will vary between 7 and 12 cm.6a. (12, 16)6b. (18, 13)6c.

Point C is an overestimate of $2, point D anoverestimate of $5, and point E an overestimate of $2.

6d. See graph for 6c.

28201240 8 16 24Actual price ($)

Est

imat

ed p

rice

($)

Estimated Prices vs. Actual Prices

28

24

20

16

12

8

4

0

C

A G

D

E

BF

6e. on a horizontal line through the estimate $16 onthe vertical axis6f. on a vertical line through the actual $16 on thehorizontal axis6g. They are points on the line shown that bisectsQuadrant I; these points indicate when estimate equalsactual price.7a. It provides the differences between the estimatednumber of each species and the actual count. This helpsidentify over- and underestimates.7b.

[0, 60, 10, �40, 10, 10]7c. 5; underestimates7d. (8, �29); 8 is the estimated number ofvelociraptors; the number of velociraptors wasunderestimated by 29.8a. (�4, 1)8b. (�2, �5)8c. (1, 5)8d. Plotted points lie on the line that bisects Quadrant II and Quadrant IV; the equation y � �x fits these points.9a. The verbal score is equal to the mathematicsscore.9b. These states also have high verbal scores. Theverbal scores are not as high as the math scores.9c. “More states had students with highermathematics scores than verbal scores”10a. about 1 m/s, because the distances in meters areabout equal to the times in seconds10b. between 0.5 and 1.0, between 2.5 and 3.0,between 3.5 and 4.0, between 4.0 and 4.510c. between 1.0 and 1.5, between 1.5 and 2.0,between 2.0 and 2.510d. between 0 and 0.5, between 3.0 and 3.511a. Answers will vary. The mean of the values is125.0 cm, and the median is 125.3 cm.11b. Answers will vary. The string is longer than thestick, so you would need to measure twice or fold thestring. The string will have some stretch, so it would behard to keep it straight without stretching it. The endsof the string are never “clean” cut, so you would haveto choose “the end.”11c. It means that this measurement is accuratewithin 0.2 cm.11d. Answers will vary. The range of measures is123.3 to 126.5. This could be written 124.9 ± 1.6 cm.

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Time 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

Distance 0.5 0.8 1.4 2.1 2.7 3.0 3.6 3.9 4.25


12. Possible answers are shown.12a. {1, 3, 3, 3, 4, 5, 6}12b. {1, 4, 4, 4, 4, 4, 7}12c. {1, 6, 6, 7, 7, 8, 9}12d. {1, 2, 3, 4, 5, 6, 6}13a. Rocky made several mistakes. The data valuesneed to be organized in increasing order. The keyshould show actual values from the data.

13b.

13c. 18.4 � 13.5, or 4.9 cm

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2 2 43 3 3 61

2

2 6 7 95 7 7 8 8

4 5 82

5 8131415161718

5 means 13.5 cm13

Key

4


LESSON 1.8

1. Randall Cunningham threw 19 touchdown passesin 1992.2. Steve Young made 322 pass completions in 1998.3. 3 � 44. Answers will vary.

5. � �This matrix gives the totals from the two years.6. Yes; this result should always be true if the matriceshave the same dimensions.

7. � �This matrix gives the difference between the 1998totals and the 1992 totals.

8. ([A] � [B]) • ��12��

9a. � �9b. � �9c. � �9d. � �

10. � �11. A possibility would be to find the labor costs ofbuilding an item using 16 hours billed at $5.25 perhour and 30 hours billed at $8.75 per hour. Theproduct is

[5.25 8.75]� � � [346.50]

12a. Quantity: � �; profit: � �;

the number of columns in the quantity matrix must bethe same as the number of rows in the profit matrix.

0.901.252.15

371634

253852

7432

120

1630

�2.4�2.2

1.5

�4.83.5

3.25

2.252.5

�2.54.75

4�3

63

�310.5

15�12

�2.5�4

4�6

�32

4.55

�59.5

8�6

�9�1

5

�111511

�1152654

�15841

115

788 489 35 19809 492 53 21919 590 61 19

12b. Atlanta: 74($0.90) � 25($1.25) � 37($2.15) �$177.40; Decatur: 32($0.90) � 38($1.25) � 16($2.15) � $110.70; Athens: 120($0.90) � 52($1.25) � 34($2.15)� $246.10

12c. � �; 3 � 1

12d. The answer matrix gives the profit at eachlocation.

12e. The error message—ERR: DIM MISMATCH—means that the dimensions of the matrices in theproduct don’t match.13a. The minimum in the data set must be 0, and themaximum must be 7. Also, the data value 2 must occurmore frequently than any other. A sample solution is{0, 1, 2, 2, 2, 2, 3, 4, 6, 7}.13b. The minimum must be 22.2, and the maximummust be 30.4. No values in the list should occur moreoften than any other values. A sample solution is {22.2,24.5, 25.1, 26.2, 28.3, 28.7, 29.4, 30.4}.14a. Shapiro: 40; Chin: 4114b. Shapiro: 1.35; Chin: 1.2514c. 1314d. In Mrs. Shapiro’s class, one student had $1.35and one student had $1.30.14e. Based on the stem plot, you might expectMrs. Shapiro’s class to have more money. There aremore students with more than a dollar in her class andmore with less than 50¢ in Mr. Chin’s class.14f. Shapiro: $24.96; Chin: $20.34

Profit

Atlanta $177.40

Decatur $110.70

Athens $246.10

177.40110.70246.10

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CHAPTER 1 REVIEW

1a. Mean: 41.5; divide the sum of the numbers by 14.Median: 40; list the numbers in ascending order andfind the mean of the two middle numbers. Mode: 36;find the most frequently occurring number.1b. 27, 36, 40, 46, 58

2. possible answer: {9, 11, 14, 16, 19, 21, 22}3a.

3b. greatest jump: from a master’s degree to adoctorate; smallest difference: from not finishinghigh school to a high school diploma

4a.

4b. 157 points (Diana Taurasi)4c. Choices will vary; mean: 83.9; median: 73.5;modes: 66, 74.

50 70 90 110 130 150 170Points

2003 NCAA Women's Tournament Top Scorers

Mean Annual Wages, 1998

Did not finish high school

80503010 70Amount ($ thousand)

0 20 40 60

High school diploma only

Two-year degree (AA/AS)

Bachelor’s degree (BA/BS)

Master’s degree (MA/MS)

Doctorate degree

25 35 45 55 65Hours of use

Battery Life

5a. Mean: approximately 154; median: 121; there isno mode.5b. Bin widths may vary.

5c.

5d. Possible answer: Most of the students questionedhad read fewer than 200 pages, with a fairly evendistribution between 0 and 200.6a.

1860

2000

1980

1960

1940

1920

1900

1880

Actual year

Est

imat

ed y

ear

Invention Dates2000

1980

1960

1940

1920

1900

0 500400300200100Number of pages

Pages Read in Current Book

5002000 300100 400


Nu

mb

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f st

ud

ents

6

5

4

3

2

1

0

Number of pages

5002000 300100 400


Nu

mb

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f st

ud

ents 10

8

6

4

2

0

Number of pages

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6b. (1952, 1945), (1985, 1980)6c. y � x, where x represents actual year and yrepresents estimated year

7a. � �, � �, [43 81 37]

7b. [A] � [B] � � �7c. [C] � ([A] � [B]) � [708.5 938.5];matinee: $708.50, evening: $938.508a. between points A and B8b. Kayo was not moving; perhaps she was resting.8c. Possible answer: Kayo started out jogging fast buthad to rest for a few minutes. Then she jogged muchslower until she had to rest again. She finally got theenergy to jog all the way home at a steady pace without stopping.9a. 2,900,000

9b. See below.

9c.

01

69

19 22 32 52 959014950

12345678

90 means 2.90 million2

Key

The Ten Most PopulatedU.S. Cities, 2000

8.755.004.25

5.504.004.00

0.750.250.25

0.500.500.50

8.004.754.00

5.003.503.50

9d.

9e. The bar graph helps show how each city compareswith the others, because they remain identified byname. The stem plot shows distribution but also showsactual values. The box plot shows distribution and aclustering between 1 and 1.4 million but does not showindividual city names or populations.10a. 416.875 min10b. 425 min10c. 480 min

0 1 2 3 4 5 6 7 8 9Population (millions)

The Ten Most Populated U.S. Cities, 2000

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City

Pop

ula

tion

(m

illi

ons)

0

10

8

6

4

2

The Ten Most Populated U.S. Cities, 2000

Chi

cago

Dal

las

Det

roit

Hou

ston

Los

An

gele

s

New

Yor

k

Phi

lade

lphi

a

Ph

oen

ix

San

An

ton

io

San

Die

go

9b. (Chapter 1 Review)


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Answers to All Exercises6a. �1

4.5

gg� � �5

x5

kkgg�; x � 146.6� kg

6b. �01

.

.82

47

cc

mm� � �1.

x65

mm�; x � 1.09 m

6c. �00.0

.412

k7m

m� � �1x.6

k5mm�; x � 52 km

7a. �52

� � �21

50�, �1

20� � �2

55�, �

255� � �

120�

7b. �a9

� � �21

72�, �1

a2� � �2

97�, �

297� � �

1a2�

7c. �kj� � �

ml�, �

jl� � �m

k�, �

mk� � �j

l�

8a. 85%

8b. �7.t38� � �1

8050�, t � $6.27

9. �18� � �

30P00� ; P � 24,000

10. 2�23� cups water and �

23� cup oatmeal; 6�

23� cups

water and 1�23� cups oatmeal

11a. 3 carbon, 6 hydrogen, 1 oxygen11b. You will need 3(470), or 1410 atoms of carbonand 6(470), or 2820 atoms of hydrogen.11c. 500 molecules; use all the hydrogen atoms,1500 atoms of carbon, and 500 atoms of oxygen.12.

13. Mean: $104.9 million; median: $77.5 million;modes: $210 million, $80 million. The high incomes ofMel Gibson and Oprah Winfrey make the mean muchhigher than the median. The modes are meaninglessfor a data set this small.14a. �1214b. �414c. �8

14d. ��23�

LESSON 2.1

1. b, a, d, c1a. 0.8751b. 0.651c. 2.61d. 2.08

2a. �194�

2b. �12

10�

2c. �43�

2d. �330�, or �1

10�

3a. �24

10

hmi�

3b. , or

3c. , or

3d.

4a. 304b. 324c. 165a. T � 185b. R � 285c. S � 73.55d. x � 2.15e. M � 65f. n � 215g. c � 31.25h. W � 9

35,500 dollars��1 person

7 women-owned firms��20 firms

350 women-owned firms��1000 firms

1 part capsaicin��100,000 parts water

10 parts capsaicin��1,000,000 parts water


0 4 8 12 16 20


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LESSON 2.2

1a. 32% of what number is 24?1b. 48% of 450 is what number?1c. What percent of 117 is 98? or 98 is what percentof 117?

2a. �8d0� � �

11

20

50�

2b. �4k6� � �

01.0205

�

2c. �47720� � �10

r0�

3. 269

4a. �25

f0

� � �755�; f � 3750 fish

4b. �5255000� � �

1s5�; s � 330 fish

5a. Marie should win over half the games.

5b. � �1M

2�;

M � 7.15 or 7 games

5c. �41

79� � �

3G0�; G � 74 games

6a. Slightly fewer than 600 pieces; 16 oz is almost10 times 1.69 oz.

6b. �

16 � �16.609� � 568; about 568 candies in a 1 lb bag

6c. �1,00

506

,8000� � 1761 lb

7. There are approximately six errors that they bothmissed.

N pieces�16 oz

60 candy pieces��1.69 oz

28 games won by Marie��28 � 19 total games

8. Bass � �32640

� � 235 � 3525;

trout � �21253

� � 147 � 2185.4;

perch � �21068

� � 151 � 1963;

total � �75951

� � 533 � 7665.5;

these values are estimates, so the total should be closeto the sum of the species totals but not exactly equalto it.

9a. �53�

9b. �58�

9c. �38�

10a. 54 or 55

10b. 8

10c. younger than 42, 44, 45, 66, 67, and older than 69

10d.

11. Matt; by the order of operations, you multiplybefore you subtract.

Age

Nu

mb

er o

f p

resi

den

ts

42 50 5846 6654 62 700

16

14

12

10

8

6

4

2

Presidents’ Agesat Inauguration


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LESSON 2.3

1a. x � 49.41b. x � 401c. x � 2161d. x � 583.3�

2. 22.62 ft/s

3a. �5

10

sm� � �1

100

k0m

m� � �16m0

isn� � �

601mh

in� � 180 km/h

3b. 0.025 day � �124

dahy� � �

601mh

in� � �1

6m0

isn� � 2160 s

3c. 1200 oz � �116

lob

z� � �2100

to0

nlb� � 0.0375 ton

4a. 227 g4b. 1.76 oz4c. 4544 g4d. 3.52 oz5a. 159 cm5b. 244 cm5c. 4.72 in.5d. 1.28 in.

6a. �303

dlbays� � 0.1 lb per day

6b. �455

dlbays� � 0.1� lb per day

6c. Crystal’s cat7a. 1 m � 1.1 yd or 1 yd � 0.9 m7b. 90 m7c. 222 yd7d. 111 yd7e. 13.5 m8a.

8b. For each additional yard there are 3 more feet.

8c. �yf� � �

31�

8d. i. 450 ft8d. ii. 128 yd

9. 1500 m � �1100

k0m

m� � �1.16

mkm

i� � 0.9375 mi; a 1 mi race

is longer.10a. fifteen 12 oz cans to make 960 oz

10b. �16

24� or 0.1875 oz

10c. � �16

24�

10d. �16Loz� � �

16

24� ; L � 85 oz

11. 120 mL ��10010LmL��

1.01

6L

qt� � �

41cu

qpt

s� �

0.5088 cup or about �12� cup

12. If the profits are divided in proportion to thenumber of students in the clubs, the Math Club wouldget $288, leaving $192 for the Chess Club.13. (Lesson 2.3)

Laughing kookaburra: 46 cmGreen kingfisher: 22 cmBelted kingfisher: 33 cmPygmy kingfisher: 10 cmRinged kingfisher: 41 cm

10 15 20 25 30 35 40 45

Kingfisher length (cm)

number of ounces of concentrate��number of ounces of lemonade

Yards 1 2 3 4 5

Feet

Measurement in Yards and Feet

3 6 9 1512


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7a.

7b. y � 2.2x7c. 2.95 kg7d. 7920 lb7e. 100 lb � 45.45� kg; 100 kg � 220 lb8a. Thu calculated the ratio �

19530

� � 1.61; 1.61 is therate of kilometers per mile; x represents miles, andy represents kilometers.8b. Sabrina calculated the ratio �1

9530� � 0.62; her

constant is the rate of miles per kilometer;x represents kilometers, and y represents miles.8c. Thu’s equation may be more convenient becauseyou can directly multiply the number of miles by 1.61.8d. Sabrina’s equation may be more convenient whenconverting kilometers to miles.9a. A sample answer is to use U.S. coins anddenominations in dollars. {100, 50, 20, 10, 5, 1, 0.50,0.25, 0.10, 0.05, 0.01}

9b. Multiply the list by the exchange rate. Forexample, to convert to Japanese yen, multiply the listby 104.160. {10416, 5208, 2083.2, 1041.6, 520.8, 104.16,52.08, 26.04, 10.416, 5.208, 1.0416}

9c. Divide list L2 by the exchange rate to obtain the original values.

9d. Using dimensional analysis:

� � 14.633 pesos per euro;

multiply the number of euros by this exchange rate.

1 dollar��0.772 euro

11.297 pesos��1 dollar

LESSON 2.4

1a. 401b. 752a. 882b. 2813.

4a. Divide by 3.5 to undo the multiplication; x � 4.4b. Divide by 8 to undo the multiplication;x � 3.4875.4c. Multiply by 7 to undo the division; x � 2.625.

4d. Change the proportion to �1x2� � �0

1.8�, and then

multiply by 12 to undo the division; x � 15.5a. $3.135b. 4 yd5c. $1.256a.

6b.

[0, 80, 10, 0, 18, 1]

Market A: y � 0.179x; Market B: y � 0.212x6c. Market A: $0.18; Market B: $0.21. These are theconstants of variation in the equations, rounded to thenearest hundredth.6d. The steeper graph with the higher rate representsthe more expensive market (Market B). The cheapermarket is associated with the less steep graph.

Distance (mi) Distance (km)

4.5

7.8

650.0

1500.0

2.8

937.5

12.5

1040.0

Market A

Ears 7 14 21 28 35 42

Cost

Market B

Ears 13 26 39 52 65 78

Cost

1.25 2.50 3.75 5.00 6.25 7.50

2.75 5.50 8.25 11.00 13.75 16.50

Bernard Lavery’s Vegetables

Weight WeightVegetable (kg) (lb)

Cabbage 56

Summer squash 108

Zucchini 64

Kohlrabi 28

Celery 21

Radish 28

Cucumber 9

Brussels sprout 18

Carrot 5

(The Top 10 of Everything 1998, p. 98)

123

49

29

13

8

62

46

20

11


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10a. 2 mi10b. 4 mi10c. 3 h10d.

10e. 2 mi/h; this represents the constant walkingspeed.10f. d � 2t, where d is distance traveled in miles andt is travel time in hours.11a. D � 5t, where D is the distance traveled in inchesand t is the time elapsed in minutes.11b. The constant of variation is 5; this represents theconstant rate of 5 inches every minute.11c. 300 in.11d. 163.2 min, or 2.72 h11e.

t

D

Time (min)

Dis

tan

ce (

in.)

0

200

400

600

800

50 100 150

(163.2, 816)

x

y

4

8

12

16

20

2 6 8 104

Hours

Mil

es

0

12a. 81.25 mi/h12b. 40 km/h12c. 65 mi/h is 104 km/h. A speed limit sign mightpost 100 km/h.13a. 11 children13b. approximately 17%13c. $6 per child14a. $2.49 per box, 42¢ per bar, $2.99 per box,25¢ per ounce14b. yes14c. 1.495 oz per bar14d. about 25¢14e. Possible answers: If comparing price per ounce,neither is a better value because both bars are 25¢ perounce. If comparing price per bar, they should buy theChewy Granola Bars because they are cheaper per bar(but the bars are smaller). If Marie and Tracy preferfewer, larger bars, they should buy Crunchy GranolaBars. If they prefer more, smaller bars, they should buyChewy Granola Bars.


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LESSON 2.5

1a. y � �1x5� 1b. y � �

3x5� 1c. y � �x

3�

2a. (4, 3) 2b. (6, 2) 2c. (1, 12) 2d. (0.5, 24)3. Possible choices include (4, 5), (2, 10), (5, 4),(10, 2), and (2.5, 8).4. This is not an inverse variation. The product of thequantities (time spent watching TV, time spent doinghomework) is not constant. Instead, the sum isconstant. This is a relationship of the form x � y � k,or y � k � x, not an inverse variation xy � k, or y � �x

k�.

5a. 3 h 5b. 2 h 5c. 60 mi/h6a. inverse variation; y � �

2x4� or xy � 24

6b. direct variation; y � 12x6c. neither; neither x nor y can be zero in an inversevariation, and when x � 0, y � 0 in a direct variation.

6d. inverse variation; y � �19

x.5� or xy � 19.5

7a. 62.3� N, 93.5 N, and 187 N7b. As you move closer to the hinge, it takes moreforce to open the door. Moving from 15 cm to 10 cmneeds an increase of about 31.2 N. Moving 5 cm closerrequires an increase of 93.5 N. As you move closer, theforce needed increases more rapidly. When you getvery close to the hinge, the force needed becomesextremely large.7c. On the graph, the curve goes up very steeply near the y-axis.8a. 104 lb8b. 130�4 � 104�D; Emily needs to sit approximately5 ft from the center.The seesaw is only 4 ft long from thecenter to the seat,so she can’t balance the two boys as longas they stay on the seat.However,if the boys move andEmily sits on the seat,it can be done. 130 � D � 104 � 4;the boys must sit 3.2 ft from the center.9a. If the balance point is at the center, then theweight of an unknown object will be exactly the sameas the weight that balances it on the other side. If thebalance point is off-center, you must know the twodistances and do some calculation.9b. 15 � M � 20 � 7; M � 9.3 kg10a. The table should include points such as (100, 100), (200, 50), (250, 40), (400, 25).

10b. y � �10,

x000�

[�50, 600, 50, �10, 140, 10]

10c. The graph should stop at x � 500 because thereare only that many students.

11a. Answers will vary. On this graph, x representsfrequency and y represents tube length.

[400, 1000, 100, 30, 90, 10]

11b. Possible answer: y � �37,2

x27.1�; the constant

37,227.1 is the mean of the products of the frequenciesand tube lengths.

11c. y � �37

8,8202.70.1

�; y � 42.3 cm

12a. 2 atm 12b. 4 atm 12c. 0.1 L12d. Possible answer: You would have to increase thevolume of the container. If you kept the same volume,you would have to suck some of the air out of thecontainer.

12e.

13. s � 0.85p; the sale price is $11.86.

14a. 1.5 lb14b. 2.6 lb of calcium and 1.56 lb of phosphorus15a. 12 apartments15b. 72 gal16a. one sulfur atom, two hydrogen atoms, and fouroxygen atoms16b. 200 hydrogen atoms would combine with 100sulfur atoms and 400 oxygen atoms.16c. Use all 400 atoms of oxygen, 200 atoms ofhydrogen, and 100 atoms of sulfur to make 100 molecules of sulfuric acid.

x

y

Pressure (atm)

Vol

um

e (L

)

842 6

4

2

8

6

0

x

y

Number of students

Mon

ey to

rai

se

500

20

0

40

60

80

100

Daxun’s Lacy’s Claudia’s Al’s Description sequence sequence sequence sequence

Pick the starting number. 14 �5 �8.6 x

19 0

76 0

64 �12

16 �3

2 2

Add 5.

Multiply by 4.

Subtract 12.

Divide by 4.

Subtracttheoriginalnumber.

�3.6

�14.4

�26.4

�6.6

2

x � 5

4(x � 5)

4(x � 5) � 12

�4(x �

45) � 12�

�4(x �

45) � 12�� x

4a–c. (Lesson 2.7)


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LESSON 2.7

1.

2. Seija; Peter incorrectly added before multiplying

�45 times �62� is 135�.

3a. First multiply 16 by 4.5, then add 9.3b. First divide 18 by 3, then add 15.3c. First square 6, then add �5, then multiply by 4and subtract the result, 124, from 3.4a–c. See below.5a. See bottom of page 27.5b. At Stages 6 and 7; the original number has beensubtracted.

5d. �2(n �

23) � 4� � n � 4 or �3 ��2(n �

23) � 4�� n�

1 2

3

9

10

11

12

6

54

7 8

1557 855

67 9 4 9

1 4 3 / 4 2

0 6 2 173 67

401

40�529 31581

013� 1 4

9

33

8

8Jack’s Nina’s

Description sequence sequence

Pick the starting number. 5 3

10 6

30 18

36 24

12 8

7 5

2 2

6a.

Multiply by 2.

Multiply by 3.

Add 6.

Divide by 3.

Subtract your original number.

Subtract your original number again.

Jack’s Nina’sDescription sequence sequence

�10 10

�8 12

�24 36

�15 45

�30 30

�60 60

�10 10

6b.

Add 2.

Pick the starting number.

Multiply by 3.

Add 9.

Subtract 15.

Multiply by 2.

Divide by 6 (you should have your original number).

7a. Possible answers: (3 � 2)(5) � 7 � 18. First add 3 and 2. Then multiply this sum by 5 and subtract 7.Or, 3(2) � 5 � 7 � 18. Multiply 3 by 2 and add theproduct, 6, to 5 plus 7.


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7b. 8 � 5(6 � 7) � 13. First subtract 7 from 6. Thenmultiply this difference by 5. Then subtract this answerfrom 8.8a. Pick a number. Subtract 5. Multiply by 4. Add 8.Divide by 2. Subtract the original number. Add 6.8b. Solutions will vary. The trick always produces theoriginal number.9. other solutions:1 � 2 � 3 � 4 � 5 � 6 � 7 �8 � 9 � 100; 12 � 34 � 5 � 6 � 7 � 8 � 9 � 100;�1 � 2 � 3 � 4 � 5 � 6 � 7 � 8 � 9 � 10010a. Sample answer: Pick a number. Subtract 3.Multiply by 2. Add 10. Divide by 2. Subtract theoriginal number. Subtract 6.10b. Sample answer: Start with �4 and make upoperations, then work backward.

10c. �2(x �

23) � 10� � x � 6

11a. about 28.5 mpg11b. about 26.3 gal11c. $61.8111d. Portia’s gas mileage is more than 19% or 6.5 mpglower than the higher estimate. It is about 5% or1.5 mpg lower than the lower estimate.12a.

12b.

12c. Inverse variation; the product of the length andwidth is a constant.

12d. l � w � 24 or w � �2l4�; yes, the situation requires

the dimensions to be whole numbers of inches, so youcan’t have a box that is 16 in. by 1.5 in., even though itsatisfies the equation.

13a.

13b. 15 mi/h13c. y � 15x13d. downhill during the intervals of 1 to 2.25 hand 3.5 to 4 h (the bike’s velocity is above average),and uphill during the interval of 2.25 to 3.5 h (the bike’s speed is slower than average)

x

y

40

20

21 40 3

60

Time (h)D

ista

nce

(m

i)

Distance Traveled

x

y

10

20

30

300 2010Length

Wid

th

Possible Boxes

Stage Picture Description

1

2

3

4

5

6 Subtract the original number.

7 +1+1+1

–1

–1n

–1–1nn

–1–1–1–1–1–1nn

–1–1–1n

n

5a. (Lesson 2.7)

Pick a number.

Subtract 3.

Multiply your result by 2.

Add 4.

Divide by 2.

Add 4 or multiply by �3.

Length (in.) Width (in.)

1 24

2 12

3 8

4 6

6 4

8 3

12 2

24 1


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LESSON 2.8

1a. �121b. 321c. �241d. 351e. 131f. 31g. �191h. �61i. �42a. Subtract 32.2b. Divide by 9.2c. Multiply by 9.2d. Add 32.3a. 153b. 183c. 173d. 54a. 154b. 54c. 65. Multiply by 60, multiply by 60 again, and thendivide by 5280.6. Work backward by adding 2 to 33 and thendividing by 5; 7.7a. 37b. Start with 3 and see if you get the answer 3.7c. 157d. The final result is always the original number nomatter what number you choose.8a. 8.88b. 158c. Undo the operations shown in reverse order.

8d. �2(x � 1

50) � 12�

8e. x � �4; yes; sequence: �4, 6, 12, 0, 09a. 25; add 7, multiply by 5, divide by 3.

9b. x � �17.85(x � 7)_______

3� �18Equation:

Operationson x

Undooperations

x � �17.8

� (7)

� (5)

� (3)

� (7)

� (5)

� (3)

�10.8

�54

�18

Results

10a. �2.610b. 11.210c. �210d. 7511a. �2.411b. x � 23.6

12a. 28 in.12b. 80 min12c. t � �

D0�.4

6� � 5

13.

2.5(x � 4.2)__________5

� 4.3 � 5.4Equation:

Operationson x

� (4.2)

� (2.5)

� (5)

� (4.3)

� (4.2)

� (2.5)

� (5)

� (4.3)

19.4

48.5

9.7

5.4

Undooperations Results

x � 23.6

Equation: �3(2 � 4x)

4� � 7 � 14

Description Undo Result

Pick x. 52

13

27

81

21

� �(4)

� (4)

� (3)

� (7) 14

/ (�4)

� 6.5

26

28

84

� (2)� (2) � (2)

� (�4)

� (2)

� (3)

/ (4)

� (7)

/ (3) � (3)

14a. about 202 ft/min14b. about 103 cm/s15a. 18.7 gal (0.036 gal/mi)15b. 189 mi (21 mi/gal)

16a. 1�11

12� cups

16b. $13.05


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CHAPTER 2 REVIEW

1a. n � 8.751b. w � 84.6

1c. k � 5�16�, or 5.16�

2. possible answers:

�75

bhh

� � �30

xbh

h�; �3

70

bbhh� � �x

5 hh�; �7

5bhh� � �30

xbh

h�;

�370

bbhh

� � �x5 h

h�

3a. Possible points include (2, 1), (3, 1.5), (4, 2),(5, 2.5), (6, 3), (7, 3.5), (8, 4).3b. All points appear to lie on a line.4a. 75 ft4b. 0.52 ft/mo5. 1365 shih rice; 169 shih millet6a. If x represents the weight in kilograms and yrepresents weight in pounds, one equation is y � 2.2xwhere 2.2 is the data set’s mean ratio of pounds tokilograms.6b. about 13.6 kg6c. 55 lb7a. about 7.5 cm7b. approximately 17 days7c. H � 1.5 � D, where H represents height incentimeters and D represents time in days8a. Because the product of the x- and y-values isapproximately constant, it is an inverse relationship.8b. One possibility: y � �

45x.5�; the constant 45.5 is the

mean of the products.

8c. y � �4352.5�, y � 1.4

9a. directly; d � 50t9b. directly; d � 1v, or d � v

9c. inversely; 100 � vt, or t � �10

v0

�

10a. 2.1875 L10b. 2.3� atm

10c. y � �1.

x75�

10d.

11a. Start with a number. Double it. Subtract 1.Multiply by 3. Add 1.

11b. x; 2x; 2x � 1; 3(2x � 1); 3(2x � 1) � 1

11c. 4.5, 9, 8, 24, 25

11d. The starting value is 4.

12. Start with 1. Add 4, to get 5. Multiply by �3, to get�15. Add 12, to get �3. Divide by 6, to get �0.5. Add 5,to get 4.5.

13.

x

y

Pressure (atm)

Vol

ume

(L)

4 521 3

4

2(0.8, 2.2)

(2.3, 0.75)

8

6

0


� (4)

� (�3)

� (12)

/ (6)

� (5)

� (4)

� (6)

/ (�3)

� (12)

� (5)

2

6

� 6

� 1

4

� 18

Equation: 12 � 3(x � 4)

6�� 5 � 4

Pick x.


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Answers to All Exercises6e. The 48th term is 95; students might press 48 times or compute 2(48) � 1.7a. The table for six figures of the L-shaped puzzlepieces is

7b. To find the number of toothpicks, press 8 and then Ans � 6 . To find the perimeter,press 8 and then Ans � 4 . For the area, press 3 and then Ans � 3 .

7c. Figure 10 has 62 toothpicks, a perimeter of 44, andan area of 30.

7d. Figure 25, made from 152 toothpicks, has aperimeter of 104 and an area of 75.8a. 4 m8b. Press 101 and then Ans � 4 . The19th term represents the height of the 7th floor. Theheight is 29 m.8c. 26 terms8d. 19 m

9a. One routine is press �16 and thenAns � 12 . Another is press 2 and then Ans � �2 .9b. Two possible sequences are {�16, �4, 8, 20, 32, 44,56, . . .} and {2, �4, 8, �16, 32, �64, 128, . . .}.9c. More numbers are needed to uniquely determinea recursive routine.10a. 17 � 7, or 11910b. 1410c. Possible answer: There are 14 multiples between100 and 200. There are also 14 multiples of 7 between200 and 300, but there are 15 between 300 and 400.10d. Possible answer: The 4th multiple of 7 is 4 � 7, or28; the 5th multiple of 7 is 5 � 7, or 35; and so on.Recursively, you start with 7 and then continue adding 7.11a. Press 6.8 and then Ans � 1.5 ,

. . . .11b. Press 7.2 , and then Ans � 1.5 ,

. . . .ENTER

ENTERENTER

ENTER

ENTERENTER

ENTER

ENTERENTER

ENTER

ENTERENTER

ENTERENTER

ENTERENTER

ENTER

ENTER

ENTER

LESSON 3.1

1a. 15 1b. �161c. �52a.

2b. 5 , Ans + 3 , , . . .2c. 322d. Figure 153. �14.2, �10.5, �6.8, �3.1, 0.6, 4.3 4a. Start with 3, then apply the rule Ans � 6; 10thterm � 57.4b. Start with 1.7, then apply the rule Ans � 0.5;10th term � �2.8.4c. Start with �3, then apply the rule Ans � �2;10th term � 1536.4d. Start with 384, then apply the rule Ans/2 or Ans � 0.5; 10th term � 0.75.5a. The recursive routine is 0 and then Ans� 12.35 . The starting value is 0, the height ofground level (the first floor). Add the average floorheight for the next 85 floors: 12.35 ft.

5b. The recursive routine is 1050 and then Ans � 10.875 . The starting value is the heightof the 86th floor. Add 10.875, the average floor heightof floors 86 through 101.5c. When you are 531 ft high, you are 43 floors up andthus on the 44th floor.5d. 1093.5 ft; 94th floor6a. Possible explanation: The smallest square has anarea of 1. The next larger white square has an area of 4,which is 3 more than the smallest square. The nextlarger gray square has an area of 9, which is 5 morethan the 4-unit white square.6b. The recursive routine is 1 , Ans � 2

, , and so on.6c. 17, the value of the 9th term in the sequence6d. 39

ENTERENTER

ENTER

ENTER

ENTER

ENTER

ENTER

ENTERENTERENTER


Figure number Perimeter

1 5

2 8

3 11

4 14

5 17

Figure Toothpicks Perimeter Area

1 8 8 32 14 12 63 20 16 94 26 20 125 32 24 156 38 28 18


11c. The starting terms differ; the rule itself is thesame.11d. See below.11e. Each baby always increases by 1.5 lb, and thedifference between the babies’ weights is always 0.4 lb;the starting values are different.12a. Press 1 , Ans � 3 , . . . ; the9th term is 6561.12b. Press 5 , Ans � (�1) , . . . ;the 123rd term is 5.12c. Press �16.2 , Ans � 1.4 , . . . ;the 13th term is the first positive term.12d. Press �1 , Ans � (�2) , . . . ;the 8th term, 128, is the first to be greater than 100.13a. The top box plot is Portland, the middle is SanFrancisco, and the bottom is Seattle.

[0, 7, 0.5, 0, 12, 1]

ENTERENTERENTER

ENTERENTERENTER

ENTERENTERENTER

ENTERENTERENTER

San Francisco has the least precipitation and isthe only city in which there is a month with noprecipitation. One indicator that the weather is muchdrier in San Francisco is that the month with noprecipitation is not an outlier.13b. You lose information about what time of year iswettest; a bar graph or scatter plot would show trendsover the months of the year more clearly.14. x � �2.6

Equation: 8 � 3(x � 5) � �14.8


� (5)

� (3)

� (8)

� (5)

� (3)

� (8)

� 2.6

� 7.6

� 14.8

� 22.8

Pick x.

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Age (mo) 0 1 2 3 4 5 6

Weight of Baby A (lb) 6.8 8.3 9.8 11.3 12.8 14.3 15.8

Weight of Baby B (lb) 7.2 8.7 10.2 11.7 13.2 14.7 16.2

11d. (Lesson 3.1)


LESSON 3.2

1a. negative; �15171b. positive; 4721c. positive; 12.3�

1d. positive; 3261e. negative; �3.3�

1f. negative; �14642a. {0.5, 1, 1.5, 2, 2.5, 3}; 0.5, Ans � 0.52b. {4, 3, 2, 1, 0}; 4, Ans � 1 2c. {�1, �0.75, �0.5, �0.25, 0, 0.25}; �1,Ans � 0.252d. {�1.5, 0, 1.5, 3}; �1.5, Ans � 1.5 3.

4a.

[�5, 10, 1, 0, 40, 10]

4b.

[�5, 10, 1, 0, �30, 10]

4c. the y-axis; 04d. In 4a, the y-coordinates increase by 7. In 4b, the y-coordinates decrease by 6.5a. {0, 0} , {Ans(1) � 1, Ans(2) � 2.54}5b.

ENTER

6a. {0, 272} , {Ans(1) � 1, Ans(2) � 68}, , , ,

6b and 6e.

6c. The starting value is the point (0, 272) on thegraph.6d. On the graph, you move right 1 unit and down68 units to get from one point to the next. In therecursive routine, you add 1 to the first number andsubtract 68 from the second number.6e. This is a linear graph relating a distance to anytime between 0 and 5 h. The line represents thedistances at all possible times; points representdistances only at certain times.6f. The car is within 100 mi of San Antonio after2.53 h have elapsed. Explanations will vary.Graphically, it is the time after which the line crossesthe horizontal line y � 100.6g. The car takes 4 h to reach San Antonio.Answers will vary. The answer is the fourth entry inthe table. Graphically, it is where the line crosses the x-axis.7a. Possible answer: {1, 1.38} , {Ans(1) � 1,Ans(2) � 0.36} , , . . . . The recursiveroutine keeps track of time and cost for each minute.Apply the routine until you get {7, 3.54}. A 7 min callcosts $3.54.7b. Possible answer: The graph should consist ofpoints that lie on a line. It should include the point(1, 1.38). Each subsequent point should be 1 unit to theright and $0.36 higher than the point before it.8a.

[�10, 35, 5, �60, 20, 10]8b. The points for each submarine appear to lie on aline; the USS Dallas surfaces at a faster rate.8c. Yes; each line means that any time in this rangecorresponds to depth below the surface.8d. The submarine’s nose rises slightly above thewater when surfacing.

ENTERENTER

ENTER

x

y

200

–100 31 420

Time (h)

Dis

tan

ce fr

omSa

n A

nto

nio

(m

i)

ENTERENTERENTERENTERENTER

ENTER

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x y

0 4

1 3

2 2

3 1

4 0

x y

0 �1.5

1 0

2 1.5

3 3

2b. 2d.

Inches Centimeters

0 0

1 2.54

2 5.08

14 35.56

17 43.18


9a. See below.9b. Number of tiles: The starting value is 1; the rule is add 1.Triangle: The starting value is 3; the rule is add 1.Rhombus: The starting value is 4; the rule is add 2.Pentagon: The starting value is 5; the rule is add 3.Hexagon: The starting value is 6; the rule is add 4.To generate the sequences for all tiles simultaneously,enter {1, 3, 4, 5, 6} and {Ans(1) � 1, Ans(2) � 1,Ans(3) � 2, Ans(4) � 3, Ans(5) � 4}.9c. triangle: 52; rhombus: 102; pentagon: 152;hexagon: 2029d.

9e. The points of each graph appear to lie on a line,and each graph starts at 1; the graphs increase insteepness from the triangle tile to the hexagon tile.9f. No; there must be a whole number of tiles and awhole number of edges.10a. Answers will vary. The graph starts at (0, 5280).The points (0, 5280), (1, 4680), (2, 4080), and (3, 3480)will appear to lie on a line. From (3, 3480) to(8, �1520), the points will appear to lie on a steeperline. The bicyclist ends up 1520 ft past you.

x

y

50

90 71 3 542 8 106

40

30

10

20

Number of tiles

Nu

mb

er o

f ed

ges

10b.

10c. Sample answer: What place on the graph showswhen the bicyclist passes you? The answer is on the x-axis between 6 and 7 min.11a. 13.93�

11b. x � 3.4

12a. �9(C �

540)

� � 40

12b. Add 40, multiply by 5, divide by 9, thensubtract 40.

12c. �5(F �

940)

� � 40

13a. 0.118, about �18� L water; 170.25, about 170 g flour

13b. 218.3�, or about 220°C

42 86

6000

4000

–2000

2000

0

Time (min)

Bicyclist

Dis

tan

ce (

ft)

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Number of tiles Triangle Rhombus Pentagon Hexagon

1 3 4 5 6

2

3

Tile Edges on the Perimeter

4

5

6

8

8

11

10

14

64 10 14 18. . .1210 22 32 42

Undo Operations operations Results

x � 3.4� (2.8) � (2.8) 0.6

· (3.2) � (3.2) 1.92� (5.4) � (5.4) 7.32

� (1.2) · (1.2) 6.1� (2.3) � (2.3) 3.8

9a. (Lesson 3.2)


14a.

14b. Quad I: H; Quad II: B, I; Quad III: D, J; Quad IV:A, E, F; x-axis: C; y-axis: G

x

y

10–10

10

–10

F

JG

D

C

A

E

B

I

H

14c. Sample answer: If the coordinates are both 0, thenthe point is on the origin. If the x-coordinate is 0,then the point is on the y-axis. If the y-coordinateis 0, then the point is on the x-axis.

If the first coordinate is positive, then the pointwill be in Quadrant I or IV. To tell which quadrant,look at the y-coordinate. If the y-coordinate is positive,the point is in Quadrant I. If the y-coordinate isnegative, the point is in Quadrant IV.

If the first coordinate is negative, then the pointwill be in Quadrant II or III. To tell which quadrant,look at the y-coordinate. If the y-coordinate is positive,the point is in Quadrant II. If the y-coordinate isnegative, the point is in Quadrant III.

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LESSON 3.3

1. {0, 4.0} and {Ans(1) � 1, Ans (2) � 0.4}2.

3. Start at the 0.8 m mark and walk away from thesensor at a constant rate of 0.2 m/s.

4a. The walker starts 2.5 m away from the motionsensor and walks toward it very slowly at a rate of 1 min 6 s.4b. The walker starts 1 m away from the motionsensor and walks away from it at a rate of 2.5 m in 6 s.5a. The walker starts 6 m away from the motionsensor and walks toward it at a rate of 0.2 m/s for 6 s.5b. The walker starts 1 m away from the motionsensor and walks away from it at a rate of 0.6 m/s for 6 s.6. The first graph, which shows a line, because thewalk is a continuous process; the walker is somewhereat every possible time in the 6 s.7. Convert 1 mi/h to ft/s:

�11

mh

i� � �60

1mh

in� � �16m0

isn

� � �52

18m0

ift

� � 1.46� ft/s

8a. 4 s8b. Away; the distance is increasing.8c. approximately 0.5 m

8d. �2.9 �

40.5

� � 0.6 m/s

8e. �05.6.5

mm

/s� � 9.16� s, or approximately 9 s

8f. The graph is a straight line.

9.

Time (s)

Dis

tan

ce (

m)

0

2

4

6

2 4 6 8

x

y

Time (s)

Dis

tan

ce (

m)

0

2

4

2 4 6

0x

y

Time (s)

Dis

tan

ce (

m)

2

4

2 4 6

10a. The rate is negative, so the line slopes down tothe right.10b. The rate is neither negative nor positive, it iszero, so the line is horizontal.10c. The line is not very steep.11a. ii11b. iv11c. iii11d. i12. Start walking at the 0 mark when the sensor startsand walk 1 ft every second. Start walking at the 0 markwhen the sensor starts and walk 1 m every second.1 m/s is a faster rate, because more distance is coveredper second.13a. Not possible; the walker would have to be at morethan one distance from the sensor at the 3 s mark.13b. Possible; the walker simply stands still about2.5 m from the sensor.13c. Not possible; the walker can’t be in two places atany given time.

14a. x � �251�, or 4.2

14b. x � �292�, or 2.4�

14c. x � �ced�

15a. � 31.4 mi/day

15b. �3

11.

d4

amy

i� ��

(1.5 � 3165) days� � 17,191.5 mi

15c. �311.

d4

amy

i� � �60,00t0 mi� ;

t � 1,911 days, or more than 5 yr16a. � 13 mi/gal or 0.077 gal/mi16b. 65 mi16c. 7.7 gal

24,901.55 mi��(2 � 365 � 2 � 30.4 � 2) days

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LESSON 3.4

1a. ii1b. iv1c. iii1d. i2a. t � 0.18 h2b. t � 0.47 h2c. 24 represents the initial number of miles thedriver is from his or her destination.2d. 45 means the driver is driving at a speed of45 mi/h.

2e. t � �485�, or 0.17�

3a. d � 38.3 ft3b. d � 25.42 ft3c. The walker started 4.7 ft away from the motionsensor.3d. The walker was walking at a rate of 2.8 ft/s.4a. x � 7.2674b. x � 11.25a. 35 � 0.8(25) � 55 mi5b. 50 min; students might use a graph or the undomethod.6a. Louis has burned 400 calories before beginningto run. His calorie-burning rate is 20.7 calories perminute, and he wants to burn 700 total calories.6b. 400 , Ans � 20.7 6c. Y1 � 400 � 20.7x6d. 700 , Ans � 0 6e. Y2 � 700 � 0x or Y2 � 7006f. The y-intercept of Y1, which is 400, is the numberof calories burned after 0 min of running (beforeLouis begins to run).

[0, 30, 5, 0, 800, 100]6g. The approximate coordinates of the point wherethe lines meet are (14.5, 700). This means that after14.5 min of running, Louis will have burned off hisdesired total of 700 calories.7a. One possible scenario: Jo has an initial start-upcost of $300 for equipment and expenses. She makes$15 for every lawn she mows, N.

ENTERENTER

ENTERENTER

7b. Sample questions: How many lawns will Jo have to mow to break even? [Solve the equation�300 � 15N � 0; Jo must mow 20 lawns.] How muchprofit will Jo earn if she mows 40 lawns? [Substitute 40for N; $300.]

7c. N � �(P �

15300)�

7d. It tells you the number of lawns you have to mowto make a certain amount of profit.8a. s � 5 � 9.8t or s � 9.8t � 58b. 34.4 m/s8c. 8 s8d. It doesn’t account for air resistance and terminalspeed.9a. y � 45 � 0.12x, where x represents dollaramounts customers spend and y represents Manny’sdaily income in dollars

9b.

[0, 840, 120, 0, 180, 30]9c. 45 � 0.12 · 312 � $82.449d. between $500 and $62510a. y � 114 � 6.9x10b. y � 207 � 7.3x10c. y � 160.5 � 11.3x10d. Monday: 321 calories; Wednesday: 426 calories;Friday: 499.5 calories11. Partial answer: Write the percent as one ratioof a proportion. Put the part over the whole in theothe ratio.

11a. �n8

� � �11050�, n � 53.3

11b. �11050� � �18

n.95�, n � 2.8

11c. �1

p00� � �

36246

� , p � 509.4

11d. �11000

� � �4n0� , n � 400

12a.

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Miles traveled Gallons �gma

illleo

sn�

363 16.2 22.4

342 15.1 22.6

285 12.9 22.1

Carl’s Purchases

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12b. 22.4 mi/gal12c. 383 mi12d. approximately 189 gal13. Sample explanation: I matched the rate of changeto each graph. I assumed the starting value was the y-intercept.13a. ii13b. iv13c. iii13d. i14a. 14 m/s14b.

14c. {0,0} , {Ans(1) + 1, Ans(2) + 14} 14d. The points lie on a line.14e. 50,400 m, or 50.4 km15a. The expression equals �4.

15b. y � 14

ENTERENTER

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Time (s) Distance (m)

1 14

2 28

3 42

4 56

5 70

6 84

7 98

8 112

9 126

10 140

Ans � 8 �3

Ans � 4 �12

Ans/3 �4


LESSON 3.5

1a.

1b.

2a. w � 15.8°F2b. w � 15°F2c. The wind chill temperature changes by 1.4° foreach 1° change in actual temperature.2d. If the actual temperature is 0°F, the wind chilltemperature is �29°F.3a. The rate is negative, so the line goes from theupper left to the lower right.3b. The rate is neither negative nor positive but zero.The line is a horizontal line.3c. The rate is positive, so the line goes from the lowerleft to the upper right.3d. The rate for the speedier walker will be greaterthan the rate for the person walking more slowly, sothe graph for the speedier walker will be steeper thanthe graph for the slower walker.4. A sample:

5a. i. 3.5; ii. 8; iii. �1.4

5b. i. �6; ii. 1; iii. 23; the y-intercept5c. i. y � �6 � 3.5x; ii. y � 1 � 8x;iii. y � 23 � 1.4x6a. The input variable x is the temperature in °F, andthe output variable y is the wind chill in °F.6b. The rate of change is 1.4°. For every 10° increasein temperature, there is a 14° increase in wind chill.6c. y � �28 � 1.4x6d. Both graphs show linear relationships withidentical rates of change and y-intercepts. The graphsare different in that the points are discrete and theequation continuous.

[�10, 40, 5, �40, 30, 10]7a. distance from sensor � 3.5 � 0.25 � time7b. 14 s after she begins walking8. Because height times width gives area, 7.3 and xrepresent the height and width, respectively. Thenumber 200 represents the area of the rectangle insquare units. The solution is about 27.4 units. Therectangle is not drawn to scale. The length should beabout 3.8 times the width.9a. 990 square units9b. possible answers: 33x � 990; x � �

93930

�

9c. 30 units10. A sample:

11a. �15 �15Ans � 52 37Ans/1.6 23.125�52 � 1.6(23.125) � �15 Check.

11b. 52 52Ans � 7 45Ans/�3 �157 � 3(�15) � 52 Check.

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L1 L2

x y

0 �5.2

�8 74.8

24 �245.2

�35 344.8

�5.2 46.8

Input Outputx y

20 100

�30 �25

16 90

15 87.5

�12.5 18.75


12a.

ABC: {1,5}, {Ans(1) � 1, Ans(2) � 2};Cozy: {1,3}, {Ans(1) � 1, Ans(2) � 3};Corner: {1,15}, {Ans(1) � 1, Ans(2) � 0}12b.

35

862 4

25

15

20

30

10

5

10

Downtown Parking

Hours parked

Cos

t

ABC ParkingCozy CarCorner Lot

12c. For less than 3 h, Cozy Car is the least expensiveoption because on the graph its points are lower thanthe points of the others. For exactly 3 h, ABC and CozyCar cost the same. For 3 to 5 h, ABC Parking has thebest price, because its graph has a lower cost in thattime frame. For exactly 6 h, ABC and The Corner Lotcost the same. For more than 6 h, The Corner Lot is theleast expensive option.12d. No; because you have to pay for a whole hour forany fraction of the hour, the price of parking does notincrease continuously.13a. 70.4 lengths13b. 2.2 ft/s13c. about 44 lengths13d. 40 lengths for a kilometer; about 64 lengths fora mile14a. y � 6 � 1.25x14b.

[0, 60, 10, 0, 100, 10]14c. 44 movies

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Hours ABC Cozy Corner

1 5 3 15

2 7 6 15

3 9 9 15

4 11 12 15

5 13 15 15

6 15 18 15

7 17 21 15

8 19 24 15

9 21 27 15

10 23 30 15


LESSON 3.6

1a. 2x � 61b. x � 2 � 51c. 2x � 1 � 31d. 2 � 2x � 32. See below.3a. 0.1x � 12 � 2.2 Original equation.

0.1x � 12 � 12 � 2.2 � 12 Subtract 12 from bothsides.

0.1x � �9.8 Remove the 0 andsubtract.

x � �98 Divide both sides by 0.1.

3b. �12 �

33.12x�� 100 Original equation.

12 � 3.12x � �300 Multiply both sidesby 3.

12 � 12 � 3.12x � �300 � 12 Subtract 12 fromboth sides.

3.12x � �312 Remove the 0.

x � �100 Divide both sidesby 3.12

4. See bottom of next page.

5a. ��15�

5b. �175c. 2.35d. x

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Picture Action taken Equation

Original equation.

Add 2 to both sides.

Divide both sides by 2. �22x� � �

62�

Reduce.

�1 �1

�1 �1

�1

�1

�xx

�1 �1�1

�1

�1 �1�

xx

Remove 0 from left side.

2x � 2 � 4

2x � 2 � 2 � 4 � 2

2x � 6

x � 3

+1 +1

+1 +1

+1

+1

–1 –1

+1 +1

=xx

+1 +1+1x=

+1 +1

+1 +1

+1

+1

xx=

2. (Lesson 3.6)


6a. �112�

6b. 66c. 506d. �2

7a. x � �112�

7b. x � 368a. Add 10 to both sides, divide both sides by 3.8b. (5, 5)

[�10, 10, 1, �5, 20, 1]8c. (5, 15)

8d. (5, 5)

8e. Even though the lines are different in each graph,in all three graphs the x-coordinate of the intersectionis the same: x � 5. This illustrates that transformingthe equation by doing the same thing to both sidesdoes not change the solution.

9c.

[�10, 10, 1, �5, 20, 1]9d.

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9a. 4 � 1.2x � 12.44 � 4 � 1.2x � 12.4 � 4

1.2x � 8.4

�11..22x

� � �81

.

.42�

x � 7

9b. Start with 12.4. 12.4Ans � 4 8.4Ans/1.2 7

Original equation.

Subtract 4 from both sides.Remove the 0 and subtract.Divide both sides by 1.2.Reduce.

Description Undo Result Equation

Pick x. 53

45

27

x � 53

Subtract 8.

Multiply by 3. 3(x � 8) � 135� (3)

� (5)

34 �3(x

5� 8)� � 7 � 34

Equation: ��3(x � 8)

� 7 � 345

Undo

Pick x.

Description Result Equation

Equation: 7 ��2 � x

4 � 5 � 16� �

Add 2.

Divide by 4.

Multiply by 7.

Subtract 5.

� (2)

� (4)

� (7)

� (5)

10

12

3

21

16

2 � x � 12

�2 �

4x

� � 3

7��2 �4

x�� 21

7��2 �4

x�� 5 � 16

x � 10

Divide by 5.

Add 7.

� (8)

� (7)

135

x � 8 � 45

�3(x

5� 8)� � 27

4a. (Lesson 3.6) 4b. (Lesson 3.6)


10a. 3 � 2x � 17

3 � 3 � 2x � 17 � 32x � 14

�22x� � �

124�

x � 710b. 0.5x � 2.2 � 101.0

0.5x � 2.2 � 2.2 � 101.0 � 2.2

0.5x � 98.8

�00..55x

� � �908..58

�

x � 197.610c. x � 307.2 � 2.1

x � 307.2 � 307.2 � 2.1 � 307.2x � �305.1

10d. 2(2x � 2) � 7

�2(2x

2� 2)� � �

72�

2x � 2 � 3.52x � 2 � 2 � 3.5 � 2

2x � 1.5

�22x� � �

12.5�

x � 0.75

10e. �4 �

60.2

.01x� � 6.2 � 0

�4 �

60.2

.01x� � 6.2 � 6.2 � 0 � 6.2

�4 �

60.2

.01x� � 6.2

�4 �

60.2

.01x�� 6.2� 6.2 � 6.2

4 � 0.01x � 38.444 � 4 � 0.01x � 38.44 � 4

0.01x � 34.44

�00..0011x

� � �304..0414

�

x � 3444

11a. r � �2Cπ�

11b. h � �2bA�

11c. l � �P2� � w

11d. s � �P4�

11e. t � �dr�

11f. h � �a2�A

b�

12a. See below.

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Picture Action taken Equation

Original equation. 2 � 4x � x � 8

3x � 6

Divide both sides by 3.

�1�

�1

xx x

x x �1 �1�1

�1 �1

�1 �1

�1

�1�

�1

x x

x �1 �1�1

�1 �1

�1 �1

�1

Subtract 1x from both sides.

Subtract 2 from both sides.

2 � 3x � 8

x � 2

+1 +1

+1 +1

+1

+1

=xx x

+1+1=

x

12a. (Lesson 3.6)


12b. 5x � 4 � 2x � 55x � 2x � 4 � 2x � 2x � 5

3x � 4 � 53x � 4 � 4 � 5 � 4

3x � 9x � 3

Check: 5(3) � 4 �? 2(3) � 5

15 � 4 �? 6 � 5

11 � 11

13. �2$.9205� � �

x3� , x � $120

15a. See below.15b.

[0, 72, 6, 0, 30, 5]

The line with the square markers is the bagel store, andthe line with the crosses is the grocery store.

15c. y represents cost; x represents number of bagels.bagel store: y � �

61.439

�x (or y � 0.50x)grocery store: y � �

2.650� x (or y � 0.42x)

15d. Bagel store: about 50¢ per bagel; grocery store:about 42¢ per bagel; these are the coefficients of x orconstants of variation in the equations.15e. the grocery store, because its line is lower15f. Bernie’s routine calculates each price by doublingthe last. It works the first time, because if you buytwice as many bagels, you pay twice as much. But usingBernie’s routine, if you buy 36 bagels at the bagel store,you pay $25.96 instead of $19.47, which amounts topaying four times as much as a single dozen instead ofthree times the price of a dozen. The routine shouldbe: 6.49 , Ans � 6.49, , , . . . .ENTERENTERENTER

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Bagels 13 26 39 52 65 78

Cost 6.49 12.98 19.47 25.96 32.45 38.94

Bagels 6 12 18 24 30 36 42 48 54 60

Cost 2.50 5.00 7.50 10.00 12.50 15.00 17.50 20.00 22.50 25.00

Bagel Store

Grocery Store

15a. (Lesson 3.6)


CHAPTER 3 REVIEW

1a. x � �71b. x � �23.42a. 1; 3; add 1; y � 3 � x2b. 0.01; 0; add 0.01; y � 0.01x2c. 2; 5; add 2; y � 5 � 2x

2d. ��12�; 3; subtract �

12�; y � 3 � �

12�x

3a. iii3b. i3c. ii4a. y � �68.994b. y � 4289.834c. y � 0.140324d. y � 238,7235a. y � x5b. y � �3 � x5c. y � �4.3 � 2.3x5d. y � 16a. 0 represents no bookcases sold; �850 representsfixed overhead, such as start-up costs; Ans(1)represents the previously calculated number ofbookcases sold; Ans(1) � 1 represents the currentnumber of bookcases sold, one more than theprevious; Ans(2) represents the profit for the previousnumber of bookcases; Ans(2) � 70 represents theprofit for the current number of bookcases—thecompany makes $70 more profit for each additionalbookcase sold.6b.

6c. Sample answer: The graph crosses the x-axis atapproximately 12.1 and is positive after that; thecompany needs to make at least 13 bookcases to makea profit.6d. �850, the profit if the company makes zerobookcases, is the y-intercept; 70, the amount ofadditional profit for each additional bookcase, is therate of change; y goes up by $70 each time x goes up byone bookcase.6e. No; partial bookcases cannot be sold.

x

y

4 6 8 10 12 142

200

–200

–400

–600

–800

Pro

fit (

$)

Number of bookcases

7a. 37b.

7c. 4 , Ans � 3 , , . . .7d. 216 m8a. Let v represent the value in dollars and y representthe number of years; v � 5400 � 525y.8b. The rate of change is �525; in each additionalyear, the value of the computer system decreasesby $525.8c. The y-intercept is 5400; the original value of thecomputer system is $5,400.8d. The x-intercept is approximately 10.3; this meansthat the computer system no longer has value afterapproximately 10.3 yr.9a. 50 � 7.7t

t � �75.07� � 6.5 s

9b. 50 � 5 � 6.5t

t � �50

6�.5

5� � 6.9 s

9c. Andrei wins; when Andrei finishes, his youngerbrother is 50 � [5 � 6.5(6.5)] � 2.8 m from thefinish line.10a. x � 4.510b. x � �4.13�

10c. x � 0.6�

10d. x � 12.810e. x � 6.3�

11a. L2 � �5.7 � 2.3 · L1

11b. L2 � �5 � 8 · L1

11c. L2 � 12 � 0.5 · L1

12a. y � 1 � �12�x ; the output value is half the input

value plus 1.

ENTERENTERENTER

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Number 1 2 3 4 . . . 30 . . . 50of sections

Number 4 7 10 13 . . . 91 . . . 151of logs

x y

0 1

1 1.5

2 2

3 2.5

4 3


12b. y � �x; the output value is the additive inverse(or opposite) of the input value, or the sum of theinput value and the output value is 0.

13. No, they won’t fit; 210 cm is 6.89 ft.14a.

14b. 11.6 letters15a. �5415b. 515c. 815d. �1816a. The starting value is 12; Ans � 55.Possible assumptions: Tom’s home is 12 mi closerto Detroit than to Traverse City. He travels at aconstant speed. We are measuring highway distance.

40 8 12 16Number of letters

Nu

mb

er o

f st

ud

ents

1

0

2

3

4

5

6

Name Length

x

y

1–1 2–2 3–3

1

–1

–2

–3

2

3

y � �x

16b.Hours 0 1 2 3 4 5Distance (mi) 12 67 122 177 232 287

16c. Tom traveled 55 mi each additional hour. Therate of change is 55 mi/h.17a. approximately 1061 thousand (or 1,061,000)visitors17b. 404, 482, 738, 1131, 3379

17c.

[0, 3500, 500, 0, 2, 1]17d. Yosemite; the number of visitors exceeds 1131 bymore than 1.5(1131 � 482).18a. 9 amperes18b. 6 ohms19a. Solution methods will vary; x � 3.5.19b. 2(3.5 � 6) � 2(�2.5) � �5

20a. �50

60

� � 83.3 h

20b. �0.57

05

0�6� � 111.1 h

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Answers to All Exercises7a. Use the slope to move backward from (40, 16.55):(40 � 10, 16.55 � 0.29 � 10) � (30, 13.65), or $13.65for 30 h; (30 � 10, 13.75 � 0.29 � 10) � (20, 10.75), or$10.75 for 20 h.7b. Continuing the process in 7a leads to (0, 4.95), or$4.95 for 0 h. This is the flat monthly rate for Hector’sInternet service.7c. y � 4.95 � 0.29x, where x is time in hours and y istotal fee in dollars7d. Substitute 280 for x and solve for y:y � 4.95 � 0.29(280) � 86.15, or $86.15 for 280 h.

8. y � e � �ac� x

9a. The change in y and the change in x are the samefor any slope triangle on the line.9b. A steeper line would have a greater change iny than its change in x. Numerically, the slope wouldbe greater than 1.9c. A less steep line would have a greater change inx than its change in y. Numerically, the slope would bebetween 0 and 1.9d. The line would go down from left to right,because the slope is negative; the line would be verysteep, because 15 is significantly greater than 1.10a. 30 m/min10b. m/min; the hot-air balloon rises at a rate of30 m/min.10c. y � 14 � 30x10d. 254 m10e. between 0 and 16.2 min11a. i. Line 2 is a better choice. A majority of pointsare closer to line 2 than to line 1.11a. ii. Line 4 is a better choice. Line 3 passes throughor is close to a good number of points, but too manypoints are above this line and too few are below it. Eventhough line 4 does not intercept any points, it is thebetter choice because about the same number ofpoints are above the line as below it.11b. i. y

x

LESSON 4.1

1a. 2

1b. �23�

1c. ��43�

2a. �32�, or 1.5; one possible point is (6, 10).

2b. ��32�, or �1.5; one possible point is (4, 2).

2c. 0; any point with a y-coordinate of 4 will be onthe line.2d. �

185�, or 1.875; one possible point is (17, 27).

3. Possible answers:3a. (1, 7), (�1, 1)3b. (3, 3), (1, 13)3c. (12, 3), (4, 9)3d. (6, 7.2), (4, 6.8)4. Answers will vary.5a. i. The x-values don’t change, so the slope isundefined.

5a. ii. The y-values decrease as the x-values increase,so the slope is negative.5a. iii. The y-values don’t change,so the slope is zero.5a. iv. The y-values increase as the x-values increase,so the slope is positive.5b. i. Using the points (4, 0) and (4, 3), the slope is�43 �

�04� � �

30�. You can’t divide by 0, so the slope is

undefined.5b. ii. Using the points (1, 3) and (4, �3), the slope is ��

43��

13

� � �2.

5b. iii. Using the points (�4, �5) and (�3, �5), theslope is�

��

53

��

((��

54

))

� � 0.5b. iv. Using the points (0, �2) and (4, 1), the slope is �1 �

4 �(�

02)

� � �34�.

5c. i. x � 4; ii. y � 5 � 2x; iii. y � �5;iv. y � �2 + �

34�x

6a. The lines are parallel, so they have the same slope;the y-intercepts are different.6b. line b6c. y � 1 � �

25�x

6d. The slope, �25�, is the same in each equation; the

y-intercepts, �3 and 1, are different.



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11b. ii.

11c. Answers will vary. The line should reflect thedirection of the data, and about the same number ofpoints should be above the line as below it.12a. 0.5(18.3)(7.4) � 67.7 cm2

12b. 0.5(18.2)(7.3) � 66.4 cm2

y

x

12c. 0.5(18.4)(7.5) � 69.0 cm2

12d. 67.7 � 1.3 cm2

13. {3, 3, 6, 16, 22}; no14a. L 2 � 2.5(L 1 � 14); {27.5, 32.5, 40, 55, 60}

14b. L3 � �(L2

2�.5

35)�, or L3 � �2

L.2

5� � 1415a. 85%15b. 150%15c. 6.5%15d. 107%16. See below.

2x � 10 � 7x � 3 1. Original equation.

2x � 2x � 10 � 7x � 2x � 3 2. Subtract 2x from both sides.

�10 � 5x � 3 3. Combine like terms.

�10 � 3 � 5x � 3 � 3 4. Subtract 3 from both sides.

�13 � 5x 5. Combine like terms.

��

513� � �

55x� 6. Divide both sides by 5.

x � �2.6 7. Reduce.

2(�2.6) � 10 �? 7(�2.6) � 3

�5.2 � 10 �?

�18.2 � 3

�15.2 � �15.2 Solution checks.

16. (Lesson 4.1)


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LESSON 4.2

1a. No; although this line goes through four points,too many points are below the line.1b. No; although the slope of the line shows thegeneral direction of the data, too many points arebelow the line.1c. Yes; about the same number of points are abovethe line as below the line, and the slope of the lineshows the general direction of the data.1d. No; although the same number of points areabove the line as below the line, the slope of the linedoesn’t show the direction of the data.2. vertical; x � 2

3a. y � �2 � �23�x

3b. y � 2 � �23�x

3c. x � �33d. y � 34a. There is a linear pattern.

[0, 36, 6, 200, 1200, 100]4b. Answers will vary. Using the points (8, 376) and(19, 684), the slope is 28.4c. The slope represents the number of quartersPenny collects per month.4d. y � 28x; the line needs to move up (the y-intercept needs to increase).

4e. A possible equation is y � 152 � 28x.

4f. The y-intercept represents the number of quartersPenny’s grandmother gave her.4g. Possible answer: 1160 quarters. The predictionmay not be reliable because it extrapolates 10 monthsbeyond the data.

5a. The number of representatives depends on thepopulation.5b. Let x represent population in millions, and lety represent the number of representatives.

[0, 10, 1, 0, 16, 5]5c. Answers will vary. Two possible points are (2.8, 4)and (6.1, 9). The slope between these points isapproximately 1.5. The equation y � 1.5x appears tofit the data with a y-intercept of 0. The slope representsthe number of representatives per 1 million people.The y-intercept means that a state with no populationwould have no representatives.

[0, 10, 1, 0, 16, 5]5d. The equation y � 1.5x gives y � 1.5(33.9) � 50.85, or 51 representatives.(For 2001–2010, California actually has 53 representatives.)5e. The equation y � 1.5x gives 8 � 1.5x;x � �1

8.5� � 5.3�; 5.3 million. (The estimated population

of Minnesota in the 2000 census was 4.9 million.)5f. The relationship should be a direct variationbecause it should go through the point (0, 0). A statewith no population would have no representatives.6a. No; each state has two senators regardless of itspopulation.6b. y � 2, where x represents population in millionsand y represents the number of senators

6c. The graph is a horizontal line because there’s nochange in y, the number of senators.


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10c. All lines are parallel to the x-axis,or horizontal.10d. All lines are parallel to the y-axis, or vertical.11a. neither11b. inverse variation; y � �

1x00�

11c. direct variation; y � �2.5x11d. direct variation; y � �1

13�x

12a. 8 � 12m � 17�12m � 9

m � �0.7512b. 2r � 7 � �24

2r � �31r � �15.5

12c. �6 � 3w � 42�3w � 48

w � �1613a. mean: 24.86�; median: 2113b. mean: 44.5; median: 4013c. mean: approximately 140.1; median: 14513d. mean: 85.75; median: 86.5

7a. �xy2

2

�

�

yx

1

1� � �

44.4.5

��

32.4

�; the slope is 0.4 m/s.

7b. The y-intercept is 2.6 m; students can find this byworking backward with the slope or by estimatingfrom a graph.7c. y � 2.6 � 0.4x8a. The slope is negative because the distancedecreases as the time increases.8b. The y-intercept represents the start distance forthe walk; the x-intercept represents the time elapsedwhen the walker reaches the detector.8c. Answers will vary. Quadrant II could indicatewalking before you started timing. Quadrant IV couldindicate that the walker walks past you; the distancesbehind you are considered negative.9a. Answers will vary. y � �8 � 4x is one possibility.9b. Answers will vary. y � �2x is one possibility.9c. y � 6 � x9d. y � 1010a. All lines have a slope of 3; they are all parallel.10b. All lines cross the y-axis at 5; they radiate aroundthe point (0, 5).


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LESSON 4.3

1a. 4; (5, 3)1b. 2; (�3.1, 1.9)1c. �3.47; (7, �2)1d. �1.38; (2.5, 5)2a. y � 5 � 3(x � 2)2b. y � �4 � 5(x � 1)3a. 23b. y � �1 � 2(x � 2)3c. y � 13 � 2(x � 5)3d. The graphs coincide, and the tables are identical.

4a. �65.5� � 1.3; the data are exactly linear, so any two

points will give this slope.4b. Answers will vary. Using the point (5, �15), theequation is y � �15 � 1.3(x � 5).4c. Answers will vary. Using the point (20, 4.5), theequation is y � 4.5 � 1.3(x � 20).4d. The two equations should give the same graphand table.4e. �21.5°F; this is the graph’s y-intercept.6a. y � 0.5 � 1(x � 0) or y � 2.5 � 1(x � 2)6b. y � 2.5 � 0.75(x � 2) or y � 1 � 0.75(x � 4)6c. y � 1 � 1(x � 4) or y � 3 � 1(x � 6)7a. AD: y � 2 � 0.2(x � 1) or y � 3 � 0.2(x � 4)BC: y � �2 � 0.2(x � 3) or y � �1 � 0.2(x � 2)AB: y � 2 � 2(x � 1) or y � �2 � 2(x � 3)DC: y � 3 � 2(x � 4) or y � �1 � 2(x � 2)7b. The slopes are the same; the coordinates of thepoints are different.7c. ABCD appears to be a parallelogram because eachpair of opposite sides is parallel; the equal slopes in 7bmean that AD�� and BC� are parallel. AB� and DC�� areparallel because they both have slope 2.8a. The data appear linear.

[0, 6, 1, 0, 1.5, 0.5]8b. $0.23/oz; this is the cost per additional ounce afterthe first.8c. Answers will vary. Using the point (1, 0.37), theequation is y � 0.37 � 0.23(x � 1).8d. $2.448e. The rates are given for weights not exceeding thegiven weights, so a letter weighing 3.5 oz would cost thesame as a 4 oz letter, or $1.06; a letter weighing 9.1 ozwould cost the same as a 10 oz letter, or $2.44.

8f. Answers will vary. A continuous line includespoints whose x-values are not whole numbers andwhose y-values are not possible rates.9a. The data are approximately linear.

[0, 45, 5, 0, 650, 50]

9b. Answers will vary. Using the points (28, 450) and(13, 310), the equation is y � 310 � 9.3(x � 13).

[0, 45, 5, 0, 650, 50]9c. y � 310 � 9.3(41 � 13) � 570.4;approximately 570 calories9d. The actual data point lies above the graph ofy � 310 � 9.3(x � 13); if a point lies above the line, thesandwich has more calories than the model predicts.9e. Answers will vary. Using y � 310 � 9.3(x � 13) asa model, three points are above the line, two points areon the line, and three points are below the line.9f. Answers will vary. The line y � 310 � 9.3(x � 13)appears to be a good fit.

9g. Answers will vary. Using y � 310 � 9.3(x � 13),approximately 189 calories; this makes sense, becausenot all calories in food come from fat.10a. y � 205 � 1.8(x � 1990) or y � 214 � 1.8(x � 1995)10b and 10c.

[1955, 2010, 5, 85, 250, 10]The point (2000, 223) is somewhat close to the line, butthe predicted value is too low.


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10d.

[1955, 2010, 5, 85, 250, 10]

10e. The data are generally linear, but the line doesn’tfit them very well; a line with a steeper slope would bea better fit.10f. Answers will vary. y � 200 � 3.7(x � 1990) givesa reasonable fit.

[1955, 2010, 5, 85, 250, 10]

10g. Answers will vary. Using y � 200 + 3.7(x � 1990), 274 million tons.11a. 4.125 L11b. 180 K

12a. (�1, 5) and (3, 1); (5, )

12b. (2, �5) and (2, �2); ( , 3)

12c. (�10, 22) and (�2, 2); ( , �3)

13. 4x � 3 � 2x � 7 Original equation.

4x � 2x � 3 � 2x � 2x � 7 Subtract 2x from bothsides.

2x � 3 � 7 Combine like terms.

2x � 3 � 3 � 7 � 3 Subtract 3 from bothsides.

2x � 4 Combine like terms.

�22x� � �

42� Divide both sides by 2.

x � 2 Reduce.

�1�1

0 ��52�

2 undefined


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LESSON 4.4

1a. not equivalent; �3x � 91b. equivalent1c. equivalent1d. not equivalent; �2(x � 4) or 2(�x � 4)2a. y � �1 � 3x2b. y � �15 � 2x2c. y � 12 � 3x3a. x � 4; division property3b. �x � 92; addition property x � �92; multiplication property3c. x � �7; subtraction property3d. x � 112; multiplication property4a. 3x � 64b. �4x � 204c. �2x � 165a. (�5, 25)5b. x � 0

6a. x � �3y

� � 86b. y � 3 � 10(x � 4)

6c. y � , or

y � �18�x � 4

7a. 3(x � 4)7b. �5(x � 4)7c. 4(8 � x)7d. �7(x � 4)8a. y � 5(2 � x)8b. y � 5(x � 2)8c. The y1-value is missing, which means it is zero;y � 0 � 5(x � 2).8d. (�2, 0); this is the x-intercept.9a. Equations i and ii are equivalent.9b. Equations i and iii are equivalent.9c. Equations ii and iii are equivalent.9d. Equations i and iii are equivalent.10a. x � 2; the point (2, 0) is the x-intercept.10b. y � 3; the point (0, 3) is the y-intercept.

�x �

412

� � 5��2

10c.

10d. The slope is ��32�; y � 3 � �

32�x.

10e.

[�10, 10, 1, �10, 10, 1]The two lines are the same; hence the equations areequivalent.10f. 3x � 2y � 6 Original equation.

2y � 6 � 3x Subtract 3x from both sides.

y � 3 � �32� x Divide both sides by 2.

11a. y � �4.411b. y � �4.4 � 4.2(x � 2)11c. x � �0.511d. y � 6.1 � 4.2(x � 0.5)11e. Answers will vary. You could rewrite each point-slope equation in slope-intercept form.11f. (4, �12) is not on the line; (�3, 16.6) is on theline. Possible answer: Substitute the x- and y-values into the equation and check whether you get atrue statement when you evaluate the equation. Orsubstitute the given x-value into the equation, evaluate,and see if it is equivalent to the given y-value.12a. y � 15.20 � 0.85(x � 20)12b. $19.4512c. The equation is used to model the bill only whenDorine is logged on for more than 15 h. Substituting15 for x gives the flat rate of $10.95 for all amounts oftime less than or equal to 15 h.12d. 30 h

x

y

2

6

8

2 4 6 8–2–2


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13a. The possible answers are y � 568 � 4.6(x � 5);y � 591 � 4.6(x � 10); y � 614 � 4.6(x � 15);y � 637 � 4.6(x � 20).

13b. y � 545 � 4.6x13c. The slope represents the number of caloriesburned per minute; the y-intercept represents thenumber of calories Avery burned from the time shewent to sleep Friday night until she started hiking.

13d. Yes; it is equivalent to the slope-interceptequation y � 545 � 4.6x.

13e. The point (60, 821) tells you that if Avery hikes for 60 min, she will have burned a cumulative total of821 calories since she went to sleep Friday night.14a. possible answer: (0, 15); $0.45/min14b. y � 15 � 0.45x

x

y

10

20

30

302010

Time (min)

Cos

t ($)

14c. The line will be parallel to the original line, but5 units higher.14d. The line will be parallel to the original line, but15 units lower (passing through the origin).14e. The line will be steeper but will have the same y-intercept.

15a. ��cchh

aa

nn

gg

ee

iinn

xy

�� 13.5� � �0.5

15b. �cchh

aa

nn

gg

ee

iinn

xy

�� 36� � �0.5

15c. Possible answers: The slope triangle side lengthsfor 15b are twice as long, but the slopes are equal.15d. Possible answer: You would get a larger triangle,but the ratio of the side lengths would equal �0.5,giving a slope of �0.5.

16. z � �3.8

0�.2

5.4� � 6.2; z � 39.8


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LESSON 4.5

1a. y � 1 � 2(x � 1) or y � 5 � 2(x � 3)

1b. y � 3 � �23�(x � 1) or y � 5 � �

23�(x � 4)

1c. y � 6 � �43�(x � 1) or y � 2 � �

43�(x � 4)

2. y � �1 � 2x

y � �73� � �

23� x or y � 2.3� � 0.6�x

y � �232� � �

43� x or y � 7.3� � 1.3�x

3a. 33b. �43c. 6The x-intercept of y � b�x � x1� is at x � x1.4a. Answers will vary. Using the points (1982, 341)and (1996, 363) gives the equation y � 341 �1.6(x � 1982), where x is the year and y is theconcentration of CO2 in parts per million.4b. All graphs should look approximately like this

[1975, 2005, 5, 325, 380, 10]

4c. The equation y � 341 � 1.6(x � 1982) gives402 ppm.4d. Using the equation in 4c, the x-intercept is about1769. It represents the year when the concentrationof CO2 would have been zero. This is not reasonable,because plants depend on CO2 and there would havebeen some concentration of CO2 for as long asthere have been plants. The model is limited; itcannot be extended much before or after the timeperiod of the data.4e. about 1.6 ppm/yr5a.

[10, 45, 5, 40, 120, 20]

5b. Using the points (20, 67) and (31.2, 88.6), theslope is approximately 1.9 and a possible equation is y � 67 � 1.9(x � 20).

5c. one possible answer:

5d. y � 32 � 1.8(x � 0) or y � 212 � 1.8(x � 100)5e. The sample equation in 5b gives y � 29 � 1.9x;the equations in 5d both give y � 32 � 1.8x.5f. The difference could be a result of measurementerror or faulty procedures.6a.

[0, 20, 5, 0, 20, 5]

6b. Using (13, 11) and (8, 14), the equation is y � 11 � 0.6(x � 13) or y � 14 � 0.6(x � 8).

6c. one possible answer:

6d. Using y � 11 � 0.6(x � 13), the concentration ofdissolved oxygen is 17.6 ppm.

6e. Using y � 11 � 0.6(x � 13), the temperature isabout 11.3°C.7a. Using the slope �0.6, y � 14 � 0.6(x � 11).7b. y � 13 � 0.6(x � 7)7c. y � 18.8 � 0.6x; y � 20.6 � 0.6x; y � 17.2 � 0.6x7d. The equation has prediction accuracy within1.8 ppm.8a. y � 30 � 1.4(x � 67)8b.

[60, 85, 5, 0, 70, 10]8c. Equations will vary. The graph with a larger y1-value is parallel but higher, and the graph with asmaller y1-value is parallel but lower.


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8d. Equations will vary. The graphs pass through thepoint (67, 30), but the one with a larger value of b issteeper and the one with a smaller value of b is lesssteep.

8e. possible equation: y � 26 � 2(x � 67)

[60, 85, 5, 0, 70, 10]

9a. �39b. y � 106 � 3(x � 10)

9c. After 45 full days, there will be only one biscuitleft, so the box will be empty at some time on the46th day.9d. When the box was new, before Anchor had anybiscuits, there were 136 biscuits.10. See below.11a. 4.4� h, or about 4 h 27 min 11b. 90 min

Description Undo Equation

Pick y.

�1 �1

�(�3)

�2x

/ (�3)

�2x

�3(y �1) �12 �2x

2x �3(y �1) �12

y � �12

��

32x

� � 1, or y ��5 ��23x�

y �1 ��12

��

32x

�, or y �1 ��4 ��23x�

10. (Lesson 4.5)


LESSON 4.6

1a. 166, 405, 623, 1052, 14831b. 204, 514, 756, 1194, 19911c.

[0,1650, 100, 0, 2500, 250]

1d. The slope will be positive because as the flyingdistance increases so does the driving distance.1e.

Q-points: (405, 514), (1052, 1194)

1f.

The slope is approximately 1.05; y � 1194 �1.05(x � 1052) or y � 514 � 1.05(x � 405).1g. approximately 1054 mi1h. approximately 535 or 536 mi2a. 12 g saturated fat2b. 38 g total fat3a. (5, 4), (10, 9)3b. (3, 8), (8, 2)4a. Let x represent years, and let y represent winningtime in minutes. The five-number summary for xis 1952, 1962, 1976, 1990, 2000. The five-numbersummary for y is 27.12, 27.5, 27.78, 28.65, 29.46. TheQ-points are (1962, 28.65) and (1990, 27.5). The slopeof the line through these two points is about �0.0411,so the possible equations are y � 28.65 � 0.0411(x �1962) and y � 27.5 � 0.0411(x � 1990).

4b.

[1945, 2005, 10, 26, 30, 0.5]

4c. The slope, �0.0411, means that the winning time decreases by an average of 0.0411 min (2.47 s)each year.4d. The prediction is 26.92 min, which is 0.16 min(9.6 s) less than the actual winning time.4e. Answers will vary. However, there is a physicallimit to how fast a runner can run. Eventually, thetimes will have to level off.5. Answers will vary. One example is {(2, 22), (4, 30),(6, 28), (8, 35), (10, 42), (12, 47), (14, 53)}.6. Reasons will vary. The data pattern has a negativeslope, and the Q-points that lie on the line are (11, 1.3)and (6, 2.2). The point (6, 1.3) is not one of the Q-points used to draw the line, so equation iv is thecorrect equation.7a. y � 1.3 � 0.625(x � 4) or y � 6.3 � 0.625(x � 12)7b. The elevator is rising at a rate of 0.625 s per floor.7c. 36.3 s after 2:00, or at approximately 2:00:367d. almost at the 74th floor8a. y � 1.3 � 0.625(x � 92) ory � 6.3 � 0.625(x � 84)8b. The elevator is moving down at 0.625 s per floor.8c. after 52.55 s, or at approximately 2:00:538d. between the 39th and 40th floors9a. Estimates will vary.9b. At 28.8 s, or at about 2:00:29, the elevators willpass at the 48th floor.10a. Start with 370, then use the rule Ans � 54.

10b.

[0, 10, 1, 0, 400, 50]

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Time (h) Distance from Mt. Rushmore (mi)

0 370

1 316

2 262

3 208

4 154

5 100

6 46


10c.

The line represents the distance remaining at any timeduring the trip. With the line, you can see how far awayfrom Mt. Rushmore you are at any time, instead of justat the top of each hour.10d. �54; the real-world meaning of the slope is thatyour distance from Mt. Rushmore decreases by 54 mieach hour.10e. Answers will vary. The car will reach the WallDrug Store in the first half hour of the fifth hour of thetrip.You can see this on the graph if you look at the linewhere it has a y-value of about 80.

10f. The car will reach Mt. Rushmore after almost 7 h of travel. You can see this on the graph or in thetable, because after 7 h, the car would have gone 8 mitoo far.11. The size and cost are almost directlyproportional; the 4 oz bottle costs $0.22/oz, the 7.5 ozbottle costs $0.22/oz, and the 18 oz bottle costs$0.2217/oz. If you change the price of the 18 oz bottleto $3.96, then it also will cost exactly $0.22/oz.12. Answers will vary.

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LESSON 4.7

1a. (6, 6)1b. (5, 9)1c. y � 9 � 3(x � 5)1d. y � 24 � 3x1e. (8, 0)2a. x � 102b. x � �7.52c. x � 2.52d. x � 41.5

3a. y � �18 �

52x

�, or y � 3.6 � 0.4x

3b. y � ��12

��2

5x�, or y � 6 � 2.5x

4a. Let x represent years, and let y represent distancein meters. The Q-points are (1964, 61.00) and (1992, 68.82). The slope of the line through thesepoints is about 0.28, so the equation is y � 61.00 � 0.28(x � 1964) or y � 68.82 �0.28(x � 1992). The slope, 0.28, means that thewinning distance increases an average of 0.28 m,or 28 cm, each year. The y-intercept, �489 m, ismeaningless in this situation because it would indicatethat a negative distance was the winning distance inyear 0. The model cannot predict that far out from thedata range.4b. 46.44 m using y � 61.00 � 0.28(x � 1964); thepredicted distance is 1.23 m more than the actualdistance.

4c. 2032 using 80 � 61.00 � 0.28(x � 1964)5a. Let x represent distance from Los Angeles inmiles, and let y represent elapsed time from Seattlein minutes; y � 1439 � 1.51(x � 411.5) or y � 273 � 1.51(x � 1181.5); the slope means thedistance from Los Angeles decreases by 1 mi each 1.51 min.5b. approximately 1758, or 29 h 18 min, by the firstequation or approximately 1755 min, or 29 h 15 min,by the second equation5c. approximately 967 mi by the first equation or965 mi by the second equation

6a. 4t � 6�t � �12�� 7

6b. t � 1; Ellen jogged for 1 hour and Eric joggedfor �

12� hour.

6c. propeller airplane: 130 km/h; jet airplane: 650 km/h 7. 15.4321 grains per gram8. 4x � 2 � x � 7 Original equation.

4x � x � 2 � x � x � 7 Subtract x from bothsides.

3x � 2 � 7 Combine like terms.

3x � 2 � 2 � 7 � 2 Subtract 2 from bothsides.

3x � 5 Combine like terms.

�33x� � �

53� Divide both sides by 3.

x � �53�, or 1.6� Reduce.

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CHAPTER 4 REVIEW

1. x2 � 4 2a. slope: �3; y-intercept: �42b. slope: 2; y-intercept: 72c. slope: 3.8; y-intercept: �2.43. Line a has slope �1, y-intercept 1, and equation y � 1 � x. Line b has slope 2, y-intercept �2, andequation y � �2 � 2x.

4a. y � 13.6x � 25,7094b. y � �37 � 5.2(x � 10)5a. (�4.5, �3.5)5b. y � 2x � 5.55c. y � 2(x � 2.75); the x-intercept is �2.75.5d. The x-coordinate is 5.5; y � 16.5 � 2(x � 5.5).5e. Answers will vary. Possible methods are graphing,using a calculator table, and putting all equations inintercept form.6a.

6b.

6c.

6d.

7a. y � 12,600 � 1,350x7b. �1,350; the car’s value decreases by $1,350 each year.7c. 12,600; Karl paid $12,600 for the car.

7d. 9�13�; in 9�

13� years the car will have no monetary

value.8a. 43 � 30 � 0.375(x � 36)8b. x � 71 s

220 � 12.5(x � 6) � 470�12.5(x � 6) � 470 � 220 � 250

x � 6 � ��

21520.5� � �20

x � �20 � 6 � �14

11 � 3(x � 8) � 413(x � 8) � 41 � 11 � 30

x � 8 � �330� � 10

x � 10 � 8 � 18

38 � 0.35x � 27�0.35x � 27 � 38 � �11

x � ��

01.315� � 31.4

4 � 2.8 � 512.8x � 51 � 4 � 47

x � �24.78� � 16.8

[0, 80, 10, 0, 50, 10]

8c. x � �43

0.�375

30� � 36 � 70.6�

9a. 1956, 1966, 1980, 1994, 2004; 1.76, 1.875, 1.97,2.025, 2.069b. The Q-points are (1966, 1.875) and (1994, 2.025).9c. y � 1.875 � 0.00536(x � 1966) or y � 2.025 � 0.00536(x � 1994)9d. Answers will vary. There are more points abovethe line than below the line.

[1950, 2005, 10, 1.6, 2.2, 0.1]

9e. Using y � 1.875 � 0.00536(x � 1966), theprediction is 2.12 m.10a. y � 2.25 � 0.13(x � 1976.5) or y � 4.025 � 0.13(x � 1990.5)

10b. The slope means the minimum hourly wageincreased approximately $0.13 per year.

10c. Using the equation y � 2.25 � 0.13(x � 1976.5),the prediction is $6.61; if the other equation is used,the prediction is $6.56.

10d. Using either equation from 10a, the predictionis 1967.11. Answers will vary. Possible answers:

11a. In an equation written as y � a � bx, b is theslope and a is the y-intercept.

11b. If the points are �x1, y1� and �x2, y2�, then the slope

of the line is given by the equation �xy2

2

��

yx

1

1� � b. The

equation of the line is y � y1 � b�x � x1�.

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Answers to All Exercises6b. P represents profit, N represents hits.Widget.kom’s start-up costs are $5,000, and itsadvertisers pay $1.60 per hit. Because Widget.komspent less in start-up costs, its website might be lessattractive to advertisers, hence the lower rate.6c. When N � 7778, P � 7445 in both equations.6d. Graphing windows will vary. The one shown is [0, 12000, 1000, �15000, 15000, 6000].

6e. Use the table to find (7778, 7445). Tracing on thisgraph is not precise.6f. This intersection point indicates that for 7778 hits totheir websites, the two companies make a profit of about$7,445.7a. P � �5000 � 2.5N7b.

[0, 7000, 500, �13000, 2000, 2000]7c. Sally will always profit more than Gizmo.kom forthe same number of website hits. Because their linesnever intersect, there is no solution to the system ofequations, and their profits will never be equal.7d. 2000 hits; after 2000 hits, Sally will have earnedback her start-up costs.8a. y � 25 � 30x, where y is tuition for x credits atUniversity College; y � 15 � 32x, where y is tuition forx credits at State College8b. (5, 175); check: 175 � 25 � 30(5),175 � 15 � 32(5)8c. Answers will vary. The table is more accurate thantracing on the calculator graph.8d. When a student takes 5 credit hours, the tuition ateither college is $175.8e. It is cheaper to attend University if taking morethan 5 credits; for fewer than 5 credits, it is cheaper toattend State. For 5 credits, they cost the same.

LESSON 5.1

1a. yes, because 47 � 3(�15.6) � 0.2 and 8 � 0.5(�15.6) � 0.21b. No, because 23 � 12 � (�4); the point satisfiesonly one of the equations.1c. No, because 12.3 � 4.5 � 5(2); furthermore, thelines are parallel, so the system has no solution.2a. table iv2b. table iii2c. table i2d. table ii3a. (8, 7)

3b. (1.5, 0.5)

In this case, the calculator gives exact solutions thatsatisfy each system.4a. (3.4, 15.5)

4b. (7.3, �5.6)

5a. y � 3 � 2x; (1, 1): 4(1) � 2(1) � 65b. y � �4 � 0.4x; (1, �3.6): 2(1) � 5(�3.6) � 20The point satisfies both forms of the linear equation.6a. Let P represent profit in dollars and N representthe number of hits; P � �12,000 � 2.5N.



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9a. d � 9 � t, where d is the drill team member’sdistance from the end zone; d � 3 � 0.5t, where d isthe tuba player’s distance from the end zone9b. (4, 5); after 4 s, the tuba player bumps into thedrill team member at the 5 yd mark.10a. The equations give winning times of 28.239 minand 28.2398 min; the difference is 0.0008.10b. The equations give winning times of 26.7594min and 26.7602 min; the difference is 0.0008.10c.

[1945, 2005, 10, 26, 30, 0.5]10d. �The graph in 10c appears to show one line; however,the y-values are 0.0008 unit apart. While the two linesare not identical, they are well within the accuracy ofthe model, so you could say they are the same model.11a. Because lines with different slopes alwaysintersect, the y-intercept a can equal any number, andb can be any number except �5.11b. a � 2 and b � �5; same slope, different y-intercept, lines do not intersect11c. a � 2 and b � �5; same slope and y-intercept,lines overlap

y � 109.2882 � 0.0411xy � 109.289 � 0.0411x

12a. Spirit of the Tri-Cities: 5.172 min;Miss B: 4.597 min12b. Spirit of the Tri-Cities: 22.241 gal;Miss B: 19.766 gal12c. Spirit of the Tri-Cities: 24.167 mi;Miss B: 27.194 mi12d. Spirit of the Tri-Cities: 0.562 mpg;Miss B: 0.632 mpg13a. x � 8513b. x � �8.213c. x � 313d. x � 3.513e. x � 1.514. 2x � 9 � 6x � 1 Original equation.

2x � 2x � 9 � 6x � 2x � 1 Subtract 2x from both sides.

9 � 4x � 1 Combine like terms.

9 � 1 � 4x � 1 � 1 Subtract 1 from both sides.

8 � 4x Combine like terms.

�84� � �

44x� Divide both sides by 4.

x � 2 Reduce.

15a. � �15b. � �16a. y � 5x � 216b. y � 0.8 � 1.4x16c. y � 1.5 � 3x

�18

137

�118

1�6


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LESSON 5.2

1. See below.2a. no, because the point satisfies only the first equation 2b. yes, because 19.25 � 32 � 3(4.25) and 19.25 � 15 � 4.252c. No, because the point satisfies only the secondequation; furthermore, the lines have the same slope,so they are parallel and there is no solution.3a. 2x � 3x � 4 � 14

5x � �10x � �2

3b. �2y � y � �3 � 7 �y � �10

y � 103c. 5d � 2d � 9

3d � 9 d � 3

3d. t � 4t � �12 �3t � �12

t � 44. (5, 175); check: 175 � 25 � 30(5) and 175 � 15 � 32(5)5a. 5x � 2(4 � 3x) � 5x � 8 � 6x � �x � 85b. 7x � 2(4 � 3x) � 7x � 8 � 6x � 13x � 86a. (1, 1); check: 1 � 4 � 3(1) and 1 � 2(1) � 1

6b. ��74�, ��

14�� or (1.75, �0.25); check: 2(1.75) �

2(�0.25) � 4 and 1.75 � 3(�0.25) � 17a. See below.7b. The approximate solution, N � 7778 and P � 7444, is more meaningful because there cannotbe a fractional number of website hits.

8a. The total admission price for two adults and threestudents is $13.50.8b. x � 4.5 and y � 1.58c. An adult ticket costs $4.50, and a student ticketcosts $1.50.9a. A � C � 2009b. 8A � 4C � 13049c. A � 126 and C � 74, so the theater sold 126 adulttickets and 74 child tickets.10a. The first walker starts at the 0.5 m mark andwalks away at 0.75 m/s. The second walker starts at the2.5 m mark and walks away at 0.75 m/s.10b. no solution10c. The walkers will never meet.

11a. �1.1t � 35 � 0.8t; �116 �

23�, 128 �

13��

The pickup passes the sports car roughly 128 mi fromFlint after approximately 117 min.

11b. �220 � 1.2t � 1.1t, ��22

203

0�, �

242

23

0�� (95.7, 105.2)

The minivan meets the pickup truck about 105 mifrom Flint after approximately 96 min.

11c. �35 � 0.8t � 220 � 1.2t ; (92.5, 109) The minivan meets the sports car 109 mi from Flintafter 92.5 min.

d � 220 � 1.2td � 35 � 0.8t

d � 220 � 1.2td � 1.1t

d � 35 � 0.8td � 1.1t

Answers will vary. A sample solution:

�12,000 � 2.5N � �5,000 � 1.6N�12,000 � 0.9N � �5,000

0.9N � 7,000

N � �70,

9000� � 7,777�

79�

P � �12,000 � 2.5 ��70,9000�� 7,444�

49�

Set equations equal to each other.Subtract 1.6N from both sides.Add 12,000 to both sides.

Divide both sides by 0.9.

d � 12 � 2.5t 1. Original equation.

1.5t � 12 � 2.5t 2. Substitute 1.5t for d.

1.5t � 2.5t � 12 � 2.5t � 2.5t 3. Add 2.5t to both sides.4t � 12 4. Combine like terms.

�44t� � �

142� 5. Divide both sides by 4.

t � 3 6. Reduce.

7a. (Lesson 5.2)

1. (Lesson 5.2)


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11d. 220 � 1.2t � 2(35 � 0.8t), t � 53.6 min;minivan is about 156 mi, sports car is about 78 mi.12a. women: y � 71.16 � 0.1715(x � 1976) or y � 67.73 � 0.1715(x � 1996); men: y � 63.44 �0.142 (x � 1976) or y � 60.60 � 0.142(x � 1996)12b. x � 2238, y � 26.2312c. Answers will vary. The window shown is [1950, 2300, 100, 0, 80, 10].

12d. The solution means that in the year 2238 (a littlemore than 230 years from now), both men and womenwill swim this race in 26.23 s. This is not likely. Themodel may be a good fit for the data, but extrapolatingthat far into the future produces unlikely predictions.13. 5 lb of sour cherry worms and 15 lb of sour limebugs

14. 4�23�L of bottled fruit juice and 5�

13� L of natural

orange soda15a. y � �4.515b. x � 316a. 12.1 ft/s16b. 50 s16c. y � 100 � 12.1x, where x represents the time inseconds and y represents her height above groundlevel. To find out how long her ride to the observationdeck is, solve the equation 520 � 100 � 12.1x.

17a. 2�16�

17b. �23�

17c. �16�

17d. �697

0�, or 1�36

70�

18a. i18b. iii18c. ii


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LESSON 5.3

1a. y � �10 �

25x

�, or y � 5 � �52x�

[�9.4, 9.4, 1, �6.2, 6.2, 1]

1b. y � �30 �

615x�, or y � 5 � �

52x�

The graph is the same as the graph for 1a. Bothequations are equivalent to y � 5 � �

52�x.

2a. (6, 10)2b. (�4, �15)2c. (12 , 25)2d. (0, �5)3a. You can simply add the equations as they areto eliminate the x-terms: �5y � 5, y � �1;6x � �15, x � �2.5. The solution is (�2.5, �1).3b. You can multiply the first equation by 2to eliminate the y-terms: 17x � 51, x � 3;8y � �16, y � �2. The solution is (3, �2).4a. substitution4b. The y-value is missing from her solution.4c. y � �1; (4, �1)5a. Multiply the first equation by �5 and the secondequation by 3, or multiply the first equation by 5 andthe second equation by �3.5b. Multiply the first equation by �8 and the secondequation by 7, or multiply the first equation by 8 andthe second equation by �7.6. The solution is (2, �2). You can (1) solve for y and graph, then look for the point wherethe lines intersect; (2) solve for y, create tables, andzoom in to where the y-values are equal; (3) solve oneequation for y (or x) and substitute into the other; or (4) multiply the equations and add them to eliminate x or y.7a. (4, 2)7b. (3, �1)7c. (�3, �1)

8.

8a. y � �3 � 0.5x8b. y � 2 � 0.75x8c. y � 7 � 2x8d. The solution of the system is also a solution of thesum of the equations.9a. y � 163 � x and y � �33 � x9b. 2y � 130, y � 65

9c. 2x � 196, x � 98

9d. The four lines intersect at the same point, (98, 65);the solution to the system must satisfy all theequations—the original equations in the system andany new equations created by combining pairs ofequations.10. Answers will vary. Substitute (5, 2) for x and y in4x � ay � b to get 20 � 2a � b. One possibility is 4x � 3y � 14.11a. Multiply the first equation by �3;�6x � 15y � �36.11b. 0 � 011c. There are infinitely many solutions.11d. One equation is a multiple of the other.12a. y � 3.5 � 1.5x and y � �4 � 2x

x

y

5421

2

12b 12c

12d

–2

x

y

–10

20

10

–6


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12b. y � 11 � 5x; this line passes through the pointwhere the two original equations intersect.12c. x � �

175�; this line passes through the point where

the two original equations intersect.12d. y � �

27�; this line passes through the point where

the two original equations intersect.

12e. ��175�, �

27��; this is the intersection point of all the

lines in 12a–d.12f. Answers will vary. If two equations intersect in apoint, any combination of multiples of the twoequations intersects in the same point. That’s why theelimination method works.

13a. �13b. They bought six wallet-size pictures and fourportrait-size.14a. Let c represent gallons burned in the city and hrepresent gallons burned on the highway.

�14b. (6.875, 4.125); 6.875 gal in the city, 4.125 gal onthe highway

14c. �17

gaml

i� � 6.875 gal � 117 city mi, �25

gaml

i� �

4.125 gal � 103 hwy mi

14d. check: �and 117 � 103 � 220

15a. �58�

6.875 � 4.125 � 1117(6.875) � 25(4.125) � 220

c � h � 1117c � 25h � 220

w � p � 103.25w � 10.50p � 61.50

15b. �34�

15c. ��490�

15d. �23�

15e. Sample answer: Find a common denominator,select a new numerator between the other two, andreduce.16a.

16b. T � 95.2 � 0.004E; the slope is the rate ofchange in temperature for each increase of 1 ft inelevation, and the y-intercept (in this case, T-intercept)is the temperature that day at sea level in thesame area.16c. At the summit the temperature was 13.9°F.17a. y � �3 � 2(x � 5)17b. y � 7 � 2.5(x � 3)18a. Walker A: y � 0.5 � x; Walker B: y � 10.5 whenx � 1 and y � 10.5 � 0.5(x � 1), or y � 11 � 0.5x,when x � 118b. They meet 7.5 ft from the sensor, when 7 s havepassed.18c. Walker B is farther from the sensor than Walker Afor all times up to, but not including, 7 s.

Elevation (ft) Temperature (°F)

Start 4,300 78

Rest station 7,800 64

Highest point 11,900 47.6

Marsha’s Climb


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LESSON 5.4

1a. �1b.

1c. �2a. � �2b. � �2c. � �3a. (8.5, 2.8)

3b. ��12�, �

11

36��

3c. (0, 0)4. Divide row 1 by 4.2:

� �.

Multiply row 2 by �1:

� �.

Solution: (3, �5.25).

5a. �5b. � �6. See below.7a. Adults Children Total (kg)

Monday 40 15 10.8Tuesday 35 22 12.29

721

11

32

3x � y � 72x � y � 21

3�5.25

01

10

35.25

0�1

10

36

11

11

33

�1�2.1

70.1

39

42

1�1

2x � 3y � 12y � 0

�12�x � �

12�

�x � 2y � 0

2x � 1.5y � 12.75�3x � 4y � 9

7b. Let x represent the average weight of chips anadult eats and y represent the average weight of chips achild eats. The system is

� .

7c. � �7d. Add �35 times row 1 to 40 times row 2 and

put the result in row 2: � �.

Divide row 2 by 355: � �.

Add �15 times row 2 to row 1:

� �.

Divide row 1 by 40: � �.

7e. Each adult ate an average of about 0.15 kg (150 g)of chips, and each child ate an average of 0.32 kg(320 g) of chips.8.

9a. Let x represent the number of small trucks andy represent the number of large trucks. The system

is � .

9b. � �532284

124

57

5x � 12y � 5327x � 4y � 284

0.150.32

01

10

60.32

01

400

10.80.32

151

400

10.8113.6

15355

400

10.812.29

1522

4035

40x � 15y � 10.835x � 22y � 12.29

Description Matrix System equations

The matrix for � � �Add 8 times row 1 to �3 times row 2and put the result in row 2.

� �

� �4.77.4

01

10

14.17.4

01

30

3x � 2y � 28.98x � 5y � 74.6

28.974.6

25

38

3x � 2y � 28.98x � 5y � 74.6

Add �2 times row 2 to row 1 and putthe result in row 1.

Divide row 1 by 3. The solution is (4.7, 7.4).

3x � 2y � 28.9y � 7.4

3x � 14.1y � 7.4

x � 4.7y � 7.4

6. (Lesson 5.4)

28.97.4

21

30� �


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9c. Solution steps will vary; � � →

� � → � �.

9d. Zoe should order 20 small trucks and 36 largetrucks.10a. x represents the number of grams of Flour Xused in each loaf, and y represents the number ofgrams of Flour Y used in each loaf. The first equationsums the amount of each type of flour to get the totalamount of flour in the loaf, and the second equationsums the amount of calcium contributed by each typeof flour to get the total amount of calcium.

10b. � �10c. � �10d. Will should mix 225 g of Flour X with 75 g ofFlour Y.

11a.

11b. � �The rows represent each equation. The columnsrepresent the coefficients of each variable and theconstants.11c. The sequence of row operations will vary; thesolution matrix is

� �.108101

77

001

010

100

2867

24

10

�1

1�1

1

110

m � t � w � 286m � t � 7

t � w � 24

22575

01

10

30030

10.04

10.12

2036

01

10

�320�2304

0�64

�160

532284

124

57

11d. They cycled 108 km on Monday, 101 km onTuesday, and 77 km on Wednesday.

12a. � � � � �� 12b. If you are planning to be in the park for 3 days,then the 3-day ticket is a much better deal. If youbought three 1-day tickets, the cost would be

� �.

12c. If you are going to be in the park for 2 days, thecost of two 1-day tickets would be

� �. This is less than the cost of the 3-day ticket,

so if you are going for only 2 days, you should buy two1-day tickets.13a. 4, Ans � 0.513b. �3, Ans � 213c. 1/2, Ans � 113d. 0, Ans � 114a. Slope: 0.75; the slope is the cost per drink onceyou’ve bought the mug.14b. y � 49.75 � 0.75(x � 33)14c. y � 25 � 0.75x; the y-intercept is the cost ofbuying the mug.

15. � � → →

�7y � �14, y � 2; x � 7

0�7

�14

�3�1

�23

3�6

9

31

23

1�2

3

564832

625242

847248

937821

373119

412924

282416

312621

655535

725545


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LESSON 5.5

1a. Multiply by 4; 12 � 28.1b. Multiply by �3; �15 � �36.1c. Add �10; �14 � x � 10.1d. Subtract 8; b � 5 � 7.1e. Divide by 3; 8d � 10 �

23�.

1f. Divide by �3; �8x � �10 �23�.

2a. Answers will vary, but the values must be � 8.2b. Values must be � �7.2c. Values must be � 7.92.2d. Values must be � �

11230

� or 9 �133� (� 9.2308).

3a. x � �13b. x � 03c. x � �23d. �2 � x � 13e. 0 � x � 24a. 3 � x4b. y � �24c. z � 124d. n � 7

5a. y � �5.2 �

43x

� � 1.3 � 0.75x

5b. y � �23x� � 5, or �2x �

315

�

6a. x � 4.34375, or �13329

�

6b. b � 1.3�

6c. x � �66d. x � �5.67a. x � 3

7b. x � �2

7c. x � �3

7d. x � 0

8. 50 � 7.5w � 120; w � 9.3�; Ezra has been saving forat least 10 wk.9a. Add 3 to both sides; 4 � 5.9b. Divide both sides by 2 (or multiply by 0.5); 3 � 1.9c. Multiply both sides by �3; 3 � �3.9d. Multiply both sides by 2; 0 � 6.10a. �9 � 9 is true.10b. 21 � 51 is false.

–5 50

–5 50

–5 50

–5 50

10c. 7 � 7 is false.10d. 24 � 18 is true.11a. The variable x drops out of the inequality, leaving�3 � 3, which is never true. So the original inequalityis not true for any number x. The graph would be anempty number line, with no points filled in.11b. The variable x drops out of the inequality,leaving �6.6 � �15, which is always true. So theoriginal inequality is true for any number x. The graphwould be a line with arrows on both ends.

12. 2.834 � 0.002 � x � 2.834 � 0.002;2.832 � x � 2.836 m13a. d � 30 (d for dollars spent on CDs)13b. h � 48 (h for height of riders)13c. p � 3 (p for people in carpool)13d. a � 17 (a for age of person admitted)14a. When is the sports car 131 or more miles awayfrom Flint?14b. x � 120 14c. When is the minivan closer than the sports carto Flint?14d. x � 92.5 15. 12 � x � 32

16a. Multiply 12 by 3.2 to get 38.4. Subtract 38.4 from72 to get 33.6.16b. Square 5 to get 25. Subtract 25 from 3 to get �22.Multiply �22 by 1.5 to get �33.Add �33 to 2 to get �31.16c. Divide 21 by 7 to get 3 and divide 6 by 2 to get 3.Subtract 3 from 3 to get 0.17a. 0.37 Ans � 0.23 , , . . .ENTERENTER

ENTER

–2 20

6 9 12 15 18 21 24 27 30 33 36

Weight Rate(oz) ($)

1 0.372 0.60

3 0.834 1.065 1.29

6 1.527 1.758 1.989 2.21

10 2.4411 2.67


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17b.

17c. A line would mean that the cost would passthrough each amount between the differentincrements. For example, if a package weighed 0.5 oz,you would pay $0.185. However, the cost increasesdiscretely. To show this, draw segments for eachintegral ounce. Note the open and closed circles.

x

y

1210860 42

32.5

21.5

10.5

Weight (oz)

Postage Costs

Cos

t ($)

x

y

1210860 42

32.5

21.5

10.5

Weight (oz)

Postage Costs

Cos

t ($)

17d. $2.6718a. �2x � 1618b. 3 � 4y18c. �z � 5


LESSON 5.6

1a. iii 1b. ii 1c. i 1d. iv2a. y � �12x � 10 2b. y � 40x � 603a.

3b.

3c.

3d.

4a–c.

5a–c.

6a.

6b.

6c.

7a. y � 1 � 2x 7b. y � �2 � �23�x

7c. y � 1 � 0.5x 7d. y � �2 � �13�x

7e. y � 2 7f. x � 2

–3 6 93x

y

–6 –3

36

–9

–3–6

6 93x

y

–6 –3

36

–9

–3–6

6 93x

y

–6 –3

36

–9

5

5

–5

x

y

–5

(–4, –3) T

(–2, 2) T (3, 2) F

(–1, –1) T

y

6

–6

x–6

(2, –3) T

(–2, –1) T

(1, 2) F

(4, 0) F

–2 20

–3 30

–10 10–5 50

–10 10–5 50

8a. y � �3 � 1.5x

8b. (1, 3); above8c. If the coefficient of y is negative, then shade theside opposite what the inequality symbol indicates.9a.

9b.

9c.

9d.

10a. F � 2S � 8410b. F � 2S � 8410c.

400 302010S

F

10

60708090

20304050

Two-point shots

Free

thro

ws

–3–6

6 9x

y

–6 –3

36

–9

–3–6

9x

y

36

–9

–3–6

6 93x

y

–6

36

–9 –3

–3–6

6 93x

y

–6 –3

36

–9

5

5

–5

x

y

–5

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12b. minimum wage12c. Q-points: (1956, 1.00), (1981, 3.35);y � �182.864 � 0.094x12e. The minimum wage increases 9¢ every year onaverage, but the actual dollar value was highest in 1968and has decreased almost every year since then.13a. about 27 mi/h13b. Because d � r � t and the distance was the samefor both Ellie and her grandmother, you can set theseproducts equal to each other. If you let r represent Ellie’sgrandmother’s speed, then 2.5(65) � 6r.

14a. y � �73� x � �

232�

14b. y � ��54� x � 3

10d. possible answer: (0, 50), (10, 30), (25, 0)11.

12a, d.Minimum Wage

Year

Dol

lars

1

0

2

3

4

5

6

7

8

9

1930 1940 1950 1960 1970 1980 1990 2000 2005

Minimum wage2003 dollars


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LESSON 5.7

1a. iii 1b. i 1c. ii2a. Yes; (1, 2) satisfies both inequalities.2b. No; only one inequality is satisfied.

2c. No; for both inequalities, �43� � �

43� is not true.

2d. No; neither inequality is satisfied.3a. y � �x � 2; y � x �23b.

4a. 4b.

5.

6a. y � �1250 � 0.40x, y � �1250 � 1.00x, x � 06b.

7a.

7b.

All the points in the dark-shaded triangular regionsatisfy the two inequalities. The point (50, 10) representsthe situation in which 50 children escort 10 adults.

20

0

40

60

80

80604020C

A

Number of children

Nu

mb

er o

f ad

ult

s

A � CA � C � 75A � 0C � 0

–3000

0

3000

x

y

Rev

enu

e

5000

Number of boxes

y � 2 � xy � 2x � 3

–3 93x

y

–6 –3

3

–9–3–6

6 93x

y

–6 –3

36

–9

4

4

–4

x

y

–4

7c. Answers will vary. It is possible to have all childrenand no adults at the restaurant. One possibleadditional constraint is that there must be at least oneadult per five children, or A � �

15�C. The solution for this

set of constraints is the triangular region bounded byA � C, A � C � 75, and A � 0.2C.8a. r � 220 � a, where a represents age in years andr represents the heart rate in beats per minute

8b. � or �8c.

8d. a � 14 and a � 40

8e.

9. x � 3 and y � �2 � �12� x

10. AB: y � �23� x � �

53� ; BC: y � ��

35� x � �

559� ;

AC: y � �111� x � �

31

11�

11. The region is a pentagon.

12. Region 1: Region 2:

13a. $713.15 13b. $957.8014a. 4 14b. 10

14c. , which simplifies to x

15a. x � 6, y � 21 15b. x � �2, y � �116. 19% acid

10��3x �5

12� � 1.4� � 10

��6

y � 3y � x � 2

y � �13� x

y � 3y � x � 2

y � �13� x � �

83�

x

y

6

80

r

200

160

a

120

80

40

0 6040 8020Age (yr)

Pu

lse

rate

(b

pm

)

r

200

160

a

120

80

40

0 6040 8020Age (yr)

Pu

lse

rate

(b

pm

)

r � 198 � 0.90ar � 121 � 0.55a

r � 0.90(220 � a)r � 0.55(220 � a)


CHAPTER 5 REVIEW

1. line a: y � 1 � x; line b: y � 3 � �52�x;

intersection: ��47�, �

171��

2. The lines meet at the point (4, 1); the equations3(4) � 2(1) � 10 and (4) � 2(1) � 6 are both true.3.

The point of intersection is (3.75, 4.625).4. See below.5a. . . . the slopes are the same but the intercepts aredifferent (the lines are parallel).5b. . . . the slopes are the same and the intercepts arethe same (the lines coincide).5c. . . . the slopes are different (the lines intersect in asingle point).

x

y

7

100

6a. x � �16b. x � 26c. �2 � x � 17. x � �1

8. �9a. 10 m2/min; 7 m2/min9b. No; he will cut 156 m2, and the lawn measures396 m2.9c. 10h � 7l � 396

9d. �310� L/min; �2

300� L/min

9e. �3h0� � �2

30l0� � 1.2

9f. l � 14.4 min, h � 29.52 min; if Harold cuts for29.52 min at the higher speed and 14.4 min at thelower speed, he will finish Mr. Fleming’s lawn and useone full tank of gas.

10. � ��3�8

01

10

y � x � 4y � �1.25x � 8.5y � 1

–2 0

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Set the right sides of the two equations equal to each other.Apply the distributive property.Subtract.Add �4.2x and 5.5 to both sides.Divide both sides by 0.1.Substitute �15 for x to find y.Multiply and add.

The solution is x � �15 and y � �70.

16 � 4.3(x � 5) � �7 � 4.2x16 � 4.3x � 21.5 � �7 � 4.2x

�5.5 � 4.3x � �7 � 4.2x0.1x � �1.5

x � �15y � �7 � 4.2(�15)y � �70

4. (Chapter 5 Review)


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Answers to All Exercises6c. 17 wk (the 18th term of the sequence)7a. Start with 7.1, then apply the rule Ans � (1 � 0.117).7b. See below.7c.

7d. Answers will vary. The graph implies a smooth,ever-increasing amount of Medicare spending, whichis probably not realistic.8a. Start with 115, then apply the rule

Ans � (1 � 0.03).8b. 12 min9a. 1.7 m9b. Start with 2, then apply the rule Ans � 0.85.

9c. approximately 0.75 m9d. approximately 1.97 ft9e. 19 times9f. Answers will vary. The mean is 1.7 m, or 85% of2 m, and the median and mode are both 1.68 m, or84% of 2 m. However, only two of the balls tested metor exceeded 85% of the drop height.10a. increasing; 75; 2%10b. decreasing; 1000; 18%10c. increasing; P; 100r%10d. decreasing; 75; 2%10e. decreasing; 80; 24%10f. decreasing; A; 100r%

x

y

Elapsed time (yr)

Spen

ding

($ b

illi

on)

2010 30

100

200

300

400

0

LESSON 6.1

1a. starting value: 16; multiplier: 1.25;7th term: 61.035

1b. starting value: 27; multiplier: �23�, or 0.6�;

7th term: 2.370��, or �62

47�

2. Start with 100, then apply the rule Ans � �1.6;first six terms are 100, �160, 256, �409.6, 655.36,�1048.576.

3a. �11

00

80�; 1 � 0.08

3b. �18090�; 1 � 0.11

3c. �11

10

20

50�, or �

111020.5

�; 1 � 0.125

3d. �190,,307050�, or �

9130.705

�; 1 � 0.0625

3e. �10

10

0�0

x�; 1 � �1

x00�

3f. �10

10

0�0

y�; 1 � �10

y0�

4a. 75(1 � 0.02), or 75(1.02)4b. 1000(1 � 0.18), or 1000(0.82)4c. P(1 � r)4d. 75 � 75 � 0.024e. 80 � 80 � 0.244f. A � A � r5. Start with 32, then apply the rule Ans � 0.75; Stage 2has a shaded area of 18 square units; Stage 5 has ashaded area of 7.59375 square units.6a. Start with 20,000, then apply the rule Ans � (1 � 0.04).6b. 5th term: 16,986.93; $16,982.93 is the selling price of the car after four price reductions.


Medicare Spending

Year 1970 1975 1980 1985 1990 1995 2000 2005

Elapsed time (yr) 0 5 10 15 20 25 30 35x

Spending ($ billion) 7.1 12.3 21.5 37.3 64.9 112.9 196.3 341.3y

7b. (Lesson 6.1)


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11a. See below.11b. See below.11c. The graph of the first plan is linear. The graphof the second is not; its slope increases betweenconsecutive points.

[0, 15, 3, 0, 4750, 500]11d. Possible answer: Grace’s charity will receivemore total money with option 2. This option will alsogive the charity more income later in the year when thebudget may be tighter.12a. $7.7612b. $7.4912c. Her wage has dropped by $0.01/h because theincrease was calculated as 3.5% of $7.50, but thedecrease was based on $7.76.

13. y � �1.2(x � 600); y-intercept: (0, 720)

14. y � 4.7 � 1.87(x � 2.2),or y � �3.9 � 1.87(x � 6.8)15a. i15b. iii15c. ii15d. iv16a. Let x represent minutes of use and y representcost; y � 50.16b. y � 50 � 0.35(x � 500)16c. y � 45 for 600 min or less of use;y � 45 � 0.55(x � 600) for more than 600 min of use.16d. First plan: $67.50; second plan: $45.00 (she paysonly the flat rate of $45.00). She should sign up for thesecond plan.16e. First plan: $172.50; second plan: $182.50. Heshould sign up for the first plan.16f. The plans cost the same for 800 min of use. A newsubscriber who will use more than 800 min shouldchoose the first plan. If she will use 800 min or less,then the second plan is better.

11a. (Lesson 6.1)

Jan Feb Mar Apr May June July Aug Sep Oct Nov DecOption 1 $50 $25 $25 $25 $25 $25 $25 $25 $25 $25 $25 $25Option 2 $1 $2 $4 $8 $16 $32 $64 $128 $256 $512 $1,024 $2,048

11b. (Lesson 6.1)

Jan Feb Mar Apr May June July Aug Sep Oct Nov DecOption 1 $50 $75 $100 $125 $150 $175 $200 $225 $250 $275 $300 $325Option 2 $1 $3 $7 $15 $31 $63 $127 $255 $511 $1,023 $2,047 $4,095


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LESSON 6.2

1a. 78 1b. 34 � 55

1c. (1 � 0.12)4

2a. 450(1 � 0.2) � 540 bacteria2b. 450(1 � 0.2)7 � 1612 bacteria3a. ii 3b. iii3c. iv 3d. i4a. iv 4b. ii4c. i 4d. iii5a. y � 1.2 � 2x

5b. y � 500 � 0.2x

5c. y � 125 � 0.4x

6. The initial deposit is $500. The account earns 4%interest per year; thus the constant multiplier is (1 � 0.04). The variable x represents the number ofyears since the initial deposit. The variable y representsthe balance after x years.7. Students will self-check their work as they run theprogram.8. 100(1 � 0.0175)(1 � 0.0175)(1 � 0.0175)(1 � 0.0175) � 100(1 � 0.0175)4; about $107.199a. $9,2009b. Start with 11,500, then apply the rule Ans � (1 � 0.2).

13b.

[0, 10, 1, 0, 12000, 2000]The intersection point represents the time and thevalue of both cars when their value will be the same.By tracing the graph shown, students should see thatboth cars will be worth approximately $5,800 after alittle less than 3 years 1 month.14. Possible answer: The first equation could model aprincipal of $400 to which $20 is added each timeperiod; the second equation could model a startingbank balance of $400, with 5% interest added to thebalance each time period. Both models have the samestarting value, 400. In both models, y � 420 when x � 1. For x greater than 1, y increases much morequickly in the second model.

[0, 50, 10, 400, 1500, 100]15a. Answers will vary.15b. Answers will vary.15c. Answers will vary. The solution should use

y � 100(1 � 0.035)x.16a.

16b.

16c. y � 4x16d. 188 cm16e. A perimeter of 74 cm is not possible because itwould require 18.5 steps.

x

y

Number of steps

Per

imet

er (

cm)

521 43

18

16

14

12

10

8

6

4

2

0

Time elapsed (yr) 0 1 2 3 4

Value ($) 11,500 9,200 7,360 5,888 4,710.40

9c.

Number ofsteps x 1 2 3 4

Perimeter(cm) y 4 8 12 16

9d. y � 11,500(1 � 0.2)x

9e.

[0, 10, 1, 0, 12000, 2000]

10a. 6 cm10b. 18 cm10c. 2(3)3

10d. 2(3)7

11. Students will self-check their work as they run theprogram.12a. The number of layers doubles with each fold.12b. Estimates will vary.12c. Methods will vary. Eight folds give 256 layers(512 pages), and nine folds give 512 layers (1024 pages).13a. y � 5000(1 � 0.05)x


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11. Possible answers:11a. x3 � x5 � x8

11b. �x3�5� x15

11c. (3x)5 � 35x5 � 243x5

11d. Exponents are added when you multiply twoexponential expressions with the same base.Exponents are multiplied when an exponentialexpression is raised to a power. An exponent isdistributed when a product is raised to a power.12a. 500(1 � 0.015)6; $546.7212b. $46.7212c. 500(1 � 0.015)12; $597.8112d. $51.0912e. Answers will vary. The increase is greaterbetween 6 and 12 months because the interest eachmonth is a percentage of a greater current balance.13a. x4 � x5

13b. �2x4 � 2x6

13c. 17x7 � 8.5x8

14a. 15x4

14b. x6

14c. 4x6

14d. 3.5(x � 0.15)6

14e. 8x9

14f. 9(x � 0.05)6

15a. 4.5x � 4715b. 4.5x � 47 � 0; x � 10.4�; he must shovel11 sidewalks to pay for his equipment.15c. 4.5x � 47 � 100; x � 32.6�; he must shovel33 sidewalks to pay for his expenses and buy alawn mower.16a. (5.3625, 0.70625)16b. approximately (3.095, 0.762)

LESSON 6.3

1a. 5x4

1b. 15x10

1c. 8x10

1d. �2x4 � 2x6

2a. (3 � 3 � 3 � 3 � 3)(3 � 3 � 3 � 3 � 3 � 3 � 3 � 3) � 313

2b. (7 � 7 � 7)(7 � 7 � 7 � 7) � 77

2c. (x � x � x � x � x � x)(x � x) � x8

2d. ( y � y � y � y � y � y � y � y)( y � y � y � y � y) � y13

2e. (x � x � y � y � y � y)(x � y � y � y) � (x � x � x) �( y � y � y � y � y � y � y) � x3y7

3a. 340

3b. 712

3c. x12

3d. y40

4a. r2t2

4b. x6y3

4c. 1024x5

4d. 8x12y6z15

5. Student 2 was correct; according to the orderof operations, squaring should be done beforemultiplication.6. a, d, and g; b and f; c and h; e has no match.7. a, d, and g: 27,521.40084; b and f: 11,711.2344;c and h: 1,060.32; e: 129,350.58398a. 6x6

8b. 20x6y8

8c. 6x5y4

8d. 4x7

9. Enclose the �5 in parentheses.

10a. 2810b. 6610c. 310d. 1


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LESSON 6.4

1a. 3.4 � 1010

1b. �2.1 � 106

1c. 1.006 � 104

2a. 74,0002b. �2,134,0002c. 40103a. 12x6

3b. 7y16

3c. 2b6 � b5

3d. 10x4 � 6x2

4a. 12x5

4b. 81y12

4c. 50x11

4d. 27m6n9

5. 3.5 � 107 � 3.5 � 10 � 10 � 10 � 10 � 10 � 10 � 10 � 35,000,000;

3.57 � 3.5 � 3.5 � 3.5 � 3.5 � 3.5 � 3.5 � 3.5� 6433.9296875

6. 1.674 � 1025

7a. 3.3411 � 1025

7b. approximately 3.6 � 1047

8. Answers will vary based on the model of calculatorused.8a. 2.5 � 102; 2.5E28b. 7.42 � 1012; 7.42E128c. �1.8 � 101; �1.8E19a. yes, because they are both equal to 51,800,000,0009b. Al’s answer9c. possible answer: 518 � 108

9d. Rewrite the digits before the 10 in scientificnotation, then use the multiplication property ofexponents to add the exponents on the 10’s. In thiscase, 4.325 � 102 � 103 � 4.325 � 105.10i. 6 � 1013

10ii. 1.3 � 109

10b. Regroup, multiply the numbers, and multiply the powers of 10 by adding the exponents.10c. �4 � 105��6 � 107� � 4 � 6 � 105 � 107 �24 � 1012 � 2.4 � 101 � 1012 � 2.4 � 1013

11a. 2,000,000,000; 2 � 109

11b. 7.3 � 1011 calls per year12a. 1.5 � 106 cells per hour12b. 1.314 � 1010 cells per year13a. 9.46 � 1012 km; 1.0 � 105 light-years13b. 9.46 � 1017 km

13c. �((91..4267�

�

1100

1

4

7

))

�� 7.45 � 1013

14a. 3.8 is the population (in millions) in 1900; 0.017is the annual growth rate; t is the elapsed time in yearssince 1900; P is the population (in millions) t yearsafter 1900.14b. Answers will vary depending on the currentyear; 0 � t � (current year � 1900).

14c.

[0, 100, 5, 0, 20, 5]14d. approximately 8.8 million

14e. Answers will vary depending on the currentyear; P � 3.8(1 � 0.017)current year�1900.

15. y

6

–6

x–7 7


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LESSON 6.5

1. x3y2a. 78

2b. x6

2c. 4x3

2d. 0.5x3y2

3. Possible answer: �33

6

2� means there are six factors of 3 in the numerator and two factors of 3 in thedenominator. So there are two factors of 1, or �

33�,

in the entire expression, leaving four factors of 3 in the numerator, or 34.4a. A represents the starting value.4b. 10,000 � A(1 � 0.1)20

4c. 10,000 � A(1 � 0.1)20

�(1

1�0,0

00.1

0)20�� A

1486.43 � AThe furniture was worth about $1,486 twenty years ago.5a. 648x11

5b. 25x2

5c. �4x2

5d. �108x5y14

6. It takes 500 s, or approximately 8 min.7a. about 132 people per square mile7b. about 867 people per square mile7c. The population of Japan was about 6.6 timesdenser than that of Mexico.

8a. 0.25%8b. $5,100.888c. $5,062.818d. $5,126.429. Four days earlier;Method 1: Use a recursive routine:864 , Ans/3 ,

, , .Method 2: Use an equation: y � 864��

13��

x; look at the

table to find x when y is less than 20.10. approximately 108,68811a. approximately 61 yr11b. approximately 1.3 yr11c. approximately 25.4 yr12. approximately 8.3 � 10�1 T-shirt per person13. approximately 2.272 � 105 flowers14. 14 yr15a. (answers recorded to tenths) Mercury: 4.1 cm;Venus: 10 cm; Earth: 10.5 cm; Mars: 5.6 cm;Jupiter: 117.3 cm; Saturn: 94.7 cm; Uranus: 69.3 cm;Neptune: 41.3 cm; Pluto: 2 cm; Sun: 1152 cm15b. Possible answer: Halley should make her modelsmuch smaller because Jupiter is 1 m in diameter andthe Sun is greater than 11 m in diameter; it would bebetter to leave the Sun out of her models altogether. Ifshe makes Pluto with a diameter of 0.2 cm, Jupiter willbe only about 12 cm in diameter.

ENTERENTERENTER

ENTERENTER


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LESSON 6.6

1a. �21

3�

1b. �51

2�

1c. �11.03

45

�

2a. �

2b. �

2c. �

2d. �

3a. �53b. �43c. �44a. 45,647(1 � 0.028)0

4b. the population 12 yr ago4c. 45,647(1 � 0.028)�8 � 36,599

4d. �(1 �45

0,6.0

4278)12�;�(1 �

450,6.04

27

8)8�

5. Possible answer: Negative exponents mean touse a reciprocal base with the exponent positive;6�3 � �6

13� � �2

116�, �63 � �216.

6a. 12x10

6b. 2x2

6c. �83x�

6d. �82x7

9�

7a. 3500(1 � 0.04)�4; about $2,9927b. 250(1 � 0.04)�3; about $2227c. 25(1 � 0.04)�5; about $217d. 187,000(1 � 0.04)�30; about $57,6568a. approximately 1.2 � 10�3 square mile per person8b. approximately 3.35 � 104 square feet per person(or 3.22 � 104 square feet per person if the answerfrom 8a is used without rounding)9a. true; �23�2

� 23 � 23 � 2 � 2 � 2 � 2 � 2 � 2 � 26

9b. false; �30�4� �1�4

� 1

9c. false; �10�2�4� ��

1102��

4

�

��101� 10��10

1� 10��10

1� 10��10

1� 10��

1108� � 10�8

9d. true; �5�3��4� ��

51

3��4

� �

�

� �5�

112� � 5121

��5112�

1��

��5 �15 � 5��5 �

15 � 5��5 �

15 � 5��5 �

15 � 5��

1�

��51

3��4

10a. original: 1 � 100; Abigail: 1 � 10�1;Barbara: 1 � 10�2; Cruz: 1 � 10�3

10b. 1 � 10�4 mi10c. Convert to inches; Damien’s string is too short(6.3 in.).

11. 36(1 � 0.5)4�6, or 36(1 � 0.5)�2

12a. $1,050.63; $1,103.8112b. $1,050; $1,102.5012c. Possible answer: In the savings account, interestis added at 6 mo, so the interest earns interest. The 1 yr interest is (1 � 0.025)2, or 1.050625; that is morethan 5%.13. exponential: y � �3(1 � 0.4)x, y � 2 � 3x;polynomial: y � 4x3, y � 2x5 � 3x2 � 4x � 2,y � 2x � 7, y � �6 � 2x � 3x2, y � 3;neither: y � xx � x2

14a. The voltage is decreasing because the base,0.957, is less than 1.14b. 9.4 is the voltage at time 0. Each second, 4.3% ofthe previous voltage is lost.14c.

[0, 60, 10, 0, 10, 1]14d. When x � 15.77 s; possible answer: Theexponential graph is below the graph of y � 4.7 for x � 15.77 s. (The exponential graph intersects thegraph of y � 4.7 at x � 15.7706, but students will notsolve to this precision.)15a. i. 4 � 105

15a. ii. 3.1 � 1010

15a. iii. 1.21 � 105

15a. iv. 2 � 10�4

15b. Possible answer: Divide the coefficients of thepowers of 10, and then divide the powers of 10(subtract the exponents).15c. 0.6 � 105, or 6 � 104 in scientific notation


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LESSON 6.7

1a. 1 � 0.15; rate of increase: 15%1b. 1 � 0.08; rate of increase: 8%1c. 1 � 0.24; rate of decrease: 24%1d. 1 � 0.002; rate of decrease: 0.2%1e. 1 � 1.5; rate of increase: 150%2a. decreasing2b. 12%2c. y � 8.922d. The y-values approach zero.3. B � 250(1 � 0.0425)t

4a. 108x18

4b. 4x5y3

4c. 724d. 1

4e. �xy4

3�

4f. �8

1x3�

4g. �x2

3�

4h. �5

x4

4y6�

5a. The ratios are 0.957, 0.956, 0.965, 0.964, 0.963,0.961, 0.959, 0.958, 0.971, and 0.955.5b. approximately 0.965c. 1 � 0.045d. y � 47(1 � 0.04)x

5e.

[�1, 11, 1, �10, 50, 5]

Adjustments to A or r are not necessary—the fit isgood as is.

5f. 55 min

[�10, 70, 10, �10, 50, 5]

6a. possible answer: y � 431(1 � 0.26)x, where 0.26 isderived from the mean ratio of about 0.746b. With each layer of plastic, the amount of light isreduced 26%.6c. With 9 layers, the reading would be below 30.7a. 50%7b. y � 32(1 � 0.5)x

7c. 243 mosquitoes; 9,342 mosquitoes;2,727,126 mosquitoes7d. Answers will vary. Possibilities include lack ofresources, overcrowding, and extermination.8a. y � 2(1 � 0.5)x

8b. y � 2(1 � 0.5)0.5 � 2.45; approximately 2.45 L8c. approximately 115 L8d. after about 30.4 min, or 30 min 24 s9a. Possible answer: Let x represent years since 2000 and y represent median price in dollars. An equation is y � 135,500(1 � 0.06)x, where 0.06 is derived from themean ratio of about 1.06.

9b–d. Answers will vary depending on year. See thetable for possible answers.

10. Note 75 above middle C (a D#) would be thehighest audible note; note �44 (an E 44 notes belowmiddle C) would be the lowest audible note.11a. y gets closer and closer to zero.11b. No, because y can never equal zero. The numberis just smaller than the calculator is able to represent.11c. y approaches infinity.12. 1000 nanoseconds per microsecond13a. Answers are rounded to the nearest $10: $2,970,$3,710, $8,540, $9,810, $11,840, $23,330, $27,770.13b. Answers are rounded to the nearest $10: $3,430,$4,290, $9,880, $11,340, $13,690, $26,980, $32,110.

14. 5(3 � 108)2 � 5(9 � 1016) � 45 � 1016, or 4.5 � 1017 joules

9b. Median 9c. Down 9d. SavingsYear price ($) payment ($) ($/mo)

2011 257,219 25,722 429

2012 272,653 27,265 454

2013 289,012 28,901 482

2014 306,352 30,635 511

2015 324,734 32,473 541

2016 344,218 34,422 574

2017 364,871 36,487 608


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CHAPTER 6 REVIEW

1a. 34

1b. 33

1c. 32

1d. 3�1

1e. 3�2

1f. 30

2a. x2

2b. �x2

�

2c. 1.23x5

2d. �31

x�

2e. 32f. x7

2g. 34x

2h. x2

3a. Possible answer: A $300 microwave depreciates ata rate of 15% per year.3b. the years (x) for which the depreciating value ofthe microwave is at least $753c. Answers will vary given the context of 3a. x � 8 or0 � x � 8 (some integers may be excluded by the real-life situation).4. Answers will vary. Possible answer: �

33

x

x� � 3x�x � 30.The result of any number divided by itself is 1.5a.

5b.

6a. �2,400,0006b. 0.0003256c. 3.714 � 1010

6d. 8.011 � 10�8

7. approximately 1.17 � 100 yr8. after 24 yr, or in 20289a. False; 3 to the power of 3 is not 9; 27x6.

9b. False; you can’t use the multiplication property ofexponents if the bases are different; 9 � 8, or 72.9c. False; the exponent �2 applies only to x; �x

22�.

9d. False; the power property of exponents saysto multiply exponents; �

xy9

6�.

10a. Possible answer: y � 80(1 � 0.17)x, where x isthe time elapsed in minutes and y is the maximumdistance in centimeters; (1 � 0.17) is derived from themean ratio of approximately 0.83.10b. approximately 15.0 cm10c. 15 min

x y

0 200.80

1 280.80

2 392.80

3 548.80

4 768.32

5 1075.648

6

x y

�2

�1

0 850.50000

1 722.50000

2 614.12500

3 522.00625

4

y � 200(1 � 0.4)x

1505.9072

1176.4706

1000.0000

443.7053

y � 850(1 � 0.15)x


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Answers to All Exercises8d. Answers will vary. The graphs show that the 26letters of the alphabet are coded into 13 letters. Sylvanawill have difficulty decoding because each coded letterhas two possible inputs.9. ALGEBRA10a. Each input codes to a single output, but eachoutput does not decode to a single input. There are twodecoding choices for B.10b. Coding would be easier.10c. Every output letter on the vertical axis shouldmatch only one input letter on the horizontal axis.Possible answer: Change the code so that D encodesas A.11a. Domain: {0, 1, �1, 2, �2}; range: {0, 1, 2}; therelationship is a function.11b. Domain: {1, 4, 9}; range: {1, �1, 2, �2, 3, �3};the relationship is not a function.12a. Double the position of the letter and add 1. If theresult is greater than 26, subtract 26. This number isthe position of the coded letter.12b. GEIK12c. You cannot decode SPY because the letter P isnot in the range.13. Yes, it could represent a function even thoughdifferent inputs have the same output; domain:{�2, 0, 1, 3}; range: {�2, 3}.14. No; the input �2 has two different outputs, 3 and0, and the input 3 has two different outputs, 1 and �2.15a. Subtract the input letter’s position from 27 to getthe output letter’s position.15b. Yes; each input matches no more than oneoutput.15c. Sample: LISA codes as ORHZ.16. Subtract 9 from the letter’s position, or add 17, sothat your result is between 1 and 26, inclusive. Theoriginal message is CRYPTOGRAPHY.17a. 28b6

17b. 11a3

17c. 1.7517d. 1372d9

17e. not possible17f. 4x5

18. �4.118

c6a8l

J� � �4x

7c0al

J�, so x � �4.4178068� � 112 cal

LESSON 7.1

1a. SBOHF 1b. EPNBJO1c. UBCMF 1d. HSBQI2a. INPUT 2b. OUTPUT2c. RELATIONSHIP 2d. RULE3a. SECRET CODES3b. The coding scheme is a letter shift of �1.4a. letters on the horizontal axis with shaded squaresabove them—A, B, C, E, G, H, I, K, L, M, N, O, Q,U, V, Y, Z

4b. all 26 letters of the alphabet4c. No; the letters B, E, G, I, K, M, and Q each havemore than one output.5a. {1:00, 2:00, 3:00, 4:00, 5:00, 6:00, 7:00, 8:00,9:00, 10:00, 11:00, 12:00} or {1:00 A.M., 1:00 P.M., . . .,12:00 A.M., 12:00 P.M.}5b. range: {0100, 0200, 0300, 0400, 0500, 0600, 0700,0800, 0900, 1000, 1100, 1200, 1300, 1400, 1500, 1600,1700, 1800, 1900, 2000, 2100, 2200, 2300, 2400}5c. It is not a function because each standard timedesignation has two military time designations. Ifstudents distinguish A.M. from P.M. times, then it isa function.6a. G 6b. M 6c. A 6d. A7a. L1 � {6, 21, 14, 3, 20, 9, 15, 14, 19}

L2 � {15, 30, 23, 12, 29, 18, 24, 23, 28}7b. You have to subtract 26 from the numbers greaterthan 26. The new list L2 is {15, 4, 23, 12, 3, 18, 24, 23, 2}.

7c. ODWLCRXWB

7d.

[�10, 37, 0, �5, 31, 0]

7e. Avoid any multiple of 26, such as 0, �26, �52, andso on.8a. L1 � {1, 2, . . . , 26}8b. L2 � {2, 4, . . . , 26, 2, 4, 6, 8, . . . , 26}8c.


LESSON 7.2

1a.

1b.

2.

3.

4. Answers will vary. In the table, every input valueproduces exactly one output value. Both graphs inExercises 2 and 3 pass the vertical line test. Both rulesare functions.5. Sample answer: Start at the 2 m mark and stand stillfor 2 s. Walk toward the 4 m mark at 2 m/s for 1 s. Standstill for another second. Walk toward the 8 m mark at4 m/s for 1 s. Then stand still for 3 s.Yes, the graphrepresents a function.6a. yes6b. No; many input values have two different output values.6c. No; there is a vertical segment. All the points onthe vertical segment have the same input value butdifferent output values.6d. yes

x

y

8–4 1612

4

–4

–8

x

y

840–4

12

8

6e. No; each input value has infinitely many outputvalues.7a. No; Los Angeles, for example, has more than oneZIP Code (90001, 90002, . . . ).7b. Yes; each person has only one birth date.7c. No; the same last name will correspond to manydifferent first names.7d. Yes; each state has only one capital.8a. i, ii, iii, iv, and vi8b. v and vii8c. Sample conclusion: It is not possible to walk agraph that does not represent a function.9a. Not a function; the x-value 3 has two different y-values, 10 and 8.9b. A function; each x-value corresponds to only oney-value.9c. A function; each x-value corresponds to only oney-value.10. Graphs must pass the vertical line test, have thecorrect domain and range, and pass through thepoints (�2, 3) and (3, �2).11. Graphs will not pass a vertical line test, but theyshould include points (�2, 3) and (3, �2) and have thecorrect domain and range. Graphs of inequalities arepossible.12a.

The graph is a line. This is a function; each x-value is paired with only one y-value.

x

y

84–4

–4

Input Outputx y

�4

�1

1.5

6.4

9

Domain Rangex y

�4

�1

2.4

�7.6

�10

1

3.4

5.4

9.32

11.4

4.4

2

�0.72

11

14

x 2 �4 0

y 1 �2 0

x �2 3 0 �3 �1

y 9

12b.

This is a function; each x-value is paired with only oney-value.

x

y

40–4

4

8

12

16

20

8 �1 5

�1 �3 ��53�

�2

9 19 1 19 3


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12c.

The equation does not represent a function becausethere are two different y-values for some x-values.12d. Tables will vary. Sample: (0, 0), (2, 3), (�2, �3),(4, 6). The graph is a line and represents a functionbecause each x-value is paired with only one y-value.

13a. domain: {�5, �4, �3, �2, �1, 0, 1, 2, 3, 4, 5};range: {0, 1, 2, 3, 4, 5}13b. domain: 0 � x � 360; range: �1 � y � 113c. Answers may vary; possible domain: all numbersx; possible range: all numbers y.

x

y

4–4

4

–4

x

y

4–4–8

4

–4

14a. Letters such as A, B, and C are not functionsbecause a vertical line could intersect them at morethan one point.14b. Letters such as V and W are functions becauseno vertical line could intersect them at more thanone point.15a. When x � 10, y � �15. Answers will vary. Zoomin on the table by changing the start values and thetable increments (�Tbl).15b. Answers will vary. The lines intersect at thesolution point.

[0, 40, 5, �40, 10, 10]16a. Multiply by 3, add 8, divide by 4, add 7; x � 24.16b. Invert the proportion to get �x4

�.5

3� � �

32�, multiply

by 4.5, add 3; x � 9.75.

17a. ��21

81�, �1

219�� (2.55, 2.64)

17b. ��51

63�, ��

21

13�� (�4.31, �1.62)

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x �7 �2 2

y 1 �2 �3

1 �2 �7

�3 �2 0


LESSON 7.3

1. Sample answers:1a.

The graph shows an inverserelationship. It is not possible to take0 hr to decorate, no matter how manystudents help.

1b.This graph shows an inverserelationship. There is a price so highthat no one will buy T-shirts.

1c.This graph shows a linear relationship.Some product will be sold without anyadvertising. Eventually the amount ofproduct sold will level off, but thatisn’t shown on the graph.

2a.

2b.

2c.

Advertisingbudget

Pro

du

ct s

old

T-shirtprice

Nu

mb

er s

old

Number ofstudents

Tim

e

2d.

3a. 0 � x � 43b. 4 � x � 63c. 4 � x � 103d. 6 � x � 123e. 0 � x � 124a.

4b.

4c.

4d.

5a. The reading on the scale depends on the weight ofthe dog, so the dog’s weight is the independent variableand the reading on the scale is the dependent variable.5b. The amount of time you spend in the planedepends on the distance you fly, so the distancebetween the cities is the independent variable and theamount of time in the plane is the dependent variable.

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5c. The wax sticks to the candle wick each time youdip it, so the number of dips is the independentvariable and the diameter of the candle is thedependent variable.6a. Yes; the graph passes the vertical line test andthe student cannot be at two different places at thesame time.6b. The domain is 0 � x � 8.6c. The range is 2 � y � 8.6d. Before any time has elapsed, the distance is 2 m.6e. y � 4; y � 8; x � 2.57.

Student graphs should consist of three line segments,each less steep than the one before it.8a. about 11:00 A.M.8b. between 10:10 and 10:40 A.M.8c. between 9:00 and 9:45 A.M. and then again after11:00 A.M.9. The temperature dropped slowly from midnightuntil 6 A.M. and was fairly constant through themorning.At about 10 A.M., the temperature began adramatic rise for 2 hr. Then the temperature continuedto rise more slowly for another 7 hr until 7 P.M. Thetemperature dropped only slightly from 7 P.M. untilmidnight.10a. Erica won in about 13.5 s.10b. Eileen10c. They were tied at approximately 3 s, at 5.5 s, from10 to 10.5 s, and just before the end of the race.10d. from approximately 0 to 3 s, from 5.5 to 10 s, andfrom 10.5 to about 13.2 s.11. Graph B12a. Answers will vary.A sample graph is shown. Itshould be made up of at least three horizontal segmentsat heights 0, 2, and �2.

12b. A sample graph is shown.As the school bus leavesa stop, its speed increases. It travels for a while at aconstant speed. Then it slows down briefly and speeds upagain to a faster speed than before it slowed down. Ittravels at a constant speed for a while, then slows down

x

y

8

2

–2

0642V

eloc

ity

Time

Time

Wat

er d

epth

and stops as it lets some students off. It speeds up againand levels off at a constant speed.

12c. Graphs and descriptions will vary. One possiblegraph is shown.

12d. A sample graph is shown. The height of the grass isquite short after it has been mowed, then grows taller untilit is mowed again.

12e. A sample graph is shown. The number ofstudents and the number of buses can only be integers,so the graph is made up of dots rather than lines. Aftera bus is full, students get into the next bus.

13a. moving away, speeding up13b. moving toward, speeding up13c. moving away, constant speed13d. moving toward, slowing down13e. moving toward, constant speed13f. moving away, slowing down14a. y � 8 � 2x14b. The equation is y � 8(1 � 0.25)x.15a. �2.515b. �415c. �

53�

16a. y � �25.9 � 2.5x16b. The slope is �2.5, and the y-intercept is �25.9.

Students

Bus

es

1.0

2.0

3.0

5 10 15 20 25 30 35 40 45 50

DaysG

rass

hei

ght

Time

Hei

ght

Time

Spee

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LESSON 7.4

1a. 3(3) � 2 � 11; Y1(3) � 111b. 3x � 2 � 2, x � 0; Y1(0) � 21c. (5)2 � 1 � 24; Y2(5) � 241d. (�3)2 � 1 � 8; Y2(�3) � 82a. �2(6) � 5 � �17 2b. �2(0) � 5 � �52c. 3.75(2.5)2 � 23.4375 2d. 3.75(2.5)�2 � 0.63a. f(4) � 0 3b. f(6) � 43c. f(2) � 2, f(5) � 23d. f (0.5) � 1, f (3) � 1, f (4.5) � 13e. three 3f. x � 53g. domain: �1 � x � 7; range: 0 � y � 64a. The dependent variable, y, is temperature indegrees Fahrenheit; the independent variable, x, is timein hours.4b. domain: 0 � x � 24; range: 5 � y � 354c. f(10) 4d. f(x) � 105a. f(x) � 7x � 5 5b. f(x) � 5 � 7(x � 1)6a. the level of the lake on the 60th day of the year6b. At a certain time, the level was 3 in. below lastyear’s mean.6c. On certain days the level was the same as it was onday 150.7a. amount of medication in milligrams7b. time in hours7c. 0 � x � 10; all real numbers x7d. 53 � y � 500, y � 07e. 500 7f. about 4 h8a. f (x) � 650(1 � 0.085)x

8b. domain: x � 0, range: 0 � y � 650

[0, 9.4, 1, 0, 700, 100]8c. The point of intersection is (7.8, 325).

8d. After 7.8 h, there are 325 bacteria present. Thehalf-life is about 7.8 h.9a. f(72) � 22.2°C 9b. f(�10) � �23.3°C

9c. f(x) � 20; x � 68°F 9d. f(x) � �5; x � 23°F10a. 6 10b. 21 10c. 14 10d. 410e. 1 10f. 25 10g. 19 10h. 110i. 8 10j. 5 10k. 1 10l. 411. Answers will vary.11a. Domain: all real numbers and lists and squarematrices of real numbers, range: nonnegative realnumbers, lists of nonnegative real numbers, and squarematrices of real numbers; the same input always givesthe same output—function.11b. Domain: all nonnegative numbers or listscontaining nonnegative numbers, range: nonnegativenumbers or lists of nonnegative numbers; the sameinput always gives the same output—function.11c. Domain: any size lists of real numbers, range: allreal numbers; the same input always gives the sameoutput—function.11d. Two answers are possible: (1) This commanddoes not take an input, but it gives one output, such as0.471359732. It will give a different output each time.It is not a function. (2) This command takes one input,the seed. It gives one output that depends on the seed.It will always give the same output for the same seed.It is a function.12a. f (x): Independent variable x is time in seconds;dependent variable y is height in meters. g(x):Independent variable x is time in seconds; dependentvariable y is velocity in meters per second.12b. for f (x): domain 0 � x � 3.2, range 0 � y � 50;for g(x): domain 0 � x � 3.2, range �31 � y � 012c. Answers will vary. For the graph of f(x), the ballis dropped from an initial height of 50 m. It hits theground after about 3.2 s. At the moment the ball isdropped, its velocity is 0 m/s. For the graph of g(x), thevelocity starts at 0 m/s and changes at a constant rate,becoming more and more negative.12d. In the 1st second, the ball falls about 5 m, from50 m at x � 0 to about 45 m at x � 1.

12e. In the 2nd second, the ball falls about 15 m, fromabout 45 m at x � 1 to about 30 m at x � 2.

12f. From the graph of f(x), the ball hits the groundafter about 3.2 s. From the graph of g(x), at x � 3.2 sthe velocity is about �31 m/s.

13a. �a1

9� 13b. b10 13c. a12b6 13d. �c81d12�

14a. �1 14b. �151� 14c. 0

15a. x � �4 15b. x � �0.75 15c. x � 4

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LESSON 7.5

1a. 71b. 0.51c. 51d. 91e. �51f. �51g. 121h. 32a. x � 10 or x � �102b. x � 4 or x � �43a. 12 � 123b. 40 � 403c. 15 � 93d. 9 � 133e. 4 � 43f. 16 � �1

16�

4a. 104b. �12.54c. 84d. 25. |2.8| � 2.8, |�1.5| � 1.5

[�4.7, 4.7, 1, �3.1, 3.1, 1]

6.

y � x, the absolute-value function7. The solutions are (2.85, 2.85) and (�2.85, 2.85).

[�4.7, 4.7, 1, �3.2, 3.2, 1]

x

y

5–5

6

–2

8a. x � �12 or x � 128b. x � �6 or x � 68c. x � 2 or x � �28d. x � 09. The walker starts 5 m away from the motion sensorand walks toward the motion sensor at a rate of 1 m/sfor 4 s and then walks away from the motion sensor atthe same rate.10a. Y2 � 710b. Y2 � any number less than 711a. g (5) � 5� 6 � 1111b. h(1) � 18(1.5) � 2711c. g (�2) � �2� 6 � 8; h(�2) � 18(1.5)�2 � 811d. f(3) � 7 � 4 3 � 1912a. Answers will vary. Possible answer: The graphsof y � 17 and y � |x � 4| � 3 intersect twice, at x � �10 and x � 18.12b. �10 � x � 18; when �10 � x � 18, the graph ofy � |x � 4| � 3 is at or below the graph of y � 17.12c. x � �10 or x � 18; when x � �10 or x � 18, thegraph of y � |x � 4| � 3 is above the graph of y � 17.13. Find the mean of the absolute values of thedeviations for each lake. Spider Lake: 0.675 lb;Doll Lake: 0.375 lb. The fish weights varied more inSpider Lake.14a. x � 6 or x � �8

14b. x � 1 or x � ��13�

14c. x � 2.114d. no solution15a. (�4.5, 1)15b. (3, �3)

16a. �1�23� � x, or x � �1�

23�

16b. x � �1

16c. 15 � x, or x � 15

150–15

20–2

50–3

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LESSON 7.6

1a.

2a. x � �6 2b. x � �62c. x � �3.8 2d. x � �3.83a. x � 7.5 or x � �7.5 3b. no real solution3c. x � 5 or x � �54a. x � 6 or x � �2

[�9.4, 9.4, 1, �6.2, 6.2, 1]4b. x � 6 or x � �2

[�9.4, 9.4, 1, �2.2, 18.6, 1]4c. x � 4 or x � �10

[�12.4, 6.4, 1, �2.2, 10.2, 1]4d. x � 4 or x � �10

[�12.4, 6.4, 1, �11, 60, 5]

5. �1 � x � 1. Good graphing window answers willvary. One possibility is a TI-83 friendly window like[�2.35, 2.35, 1, 0, 3.1, 1].6a. y � 0 6b. y � 0 6c. y � 07. y � x8. Answers will vary. One explanation is that the slopebetween x � 0 and x � 1 is 1, but the slope from x � 1to x � 2 is 3. The rate of change varies.9a. 1 � 3 � 5 � 7 � 9 � 25, or 52

9b. 1 � 3 � 5 � 7 � 9 � � � � � 29 � 225, or 152

9c. The sum of the first n positive odd integers is n2.9d. Each larger square is created by adding a borderof small squares on two sides. The number of smallsquares added on each time is the next odd number.The resulting figure is the next square number.10a. y � 8 � 2x 10b. y � x2 10c. y � x11a. sixteen 1-by-1 squares, nine 2-by-2 squares, four3-by-3 squares, and one 4-by-4 square11b. A 3-by-3 grid has fourteen squares: nine 1-by-1squares, four 2-by-2 squares, and one 3-by-3 square. A2-by-2 grid has five squares: four 1-by-1 squares andone 2-by-2 square. A 1-by-1 grid has one 1-by-1 square.11c. Answers will vary. One possible response: Thereare n2 1-by-1 squares, (n � 1)2 2-by-2 squares, (n � 2)2

3-by-3 squares, and so on. There would be 55 squaresin a 5-by-5 grid.12. Answers will vary. It’s impossible for x2 to benegative. If x is a negative number, x2 is a negativetimes a negative, which is positive. If x is a positivenumber, x2 is a positive times a positive, which is alsopositive. If x � 0, then x2 � 0. So x2 must be positive orzero, no matter what x is.13a. y � 400(0.75)x

13b.

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Side Perimeter Area(cm) (cm) (cm2)

1

2

12.8

16

14

60.8

441

100.8

2209

4

8

16

56

84

188

1

4

9

196

225

635.04

3

4

15

21

25.2

47

14a. 48x9 14b. 30x8 14c. 24x9

14d. 62.5x14 14e. �2.5x3 14f. �48x11

14g. �6xy

6

6� 14h. 375x3y6

15.

[�9.4, 9.4, 1, �6.2, 6.2, 1]Answers will vary. The graphs tell us that there isexactly one solution and that the x-value that makesthe equation true is x � 2.

x 0 4 3 1 � �3.19

y 400 0126.5625 168.75 300 1000.0000


CHAPTER 7 REVIEW

1a. �2 � x � 4 1b. 1 � f (x) � 31c. 1 1d. �1 and 32a. A function; each x-value corresponds to only oney-value.2b. Not a function; the input x � 3 has two differentoutput values, 5 and 7.2c. A function; each x-value corresponds to only oney-value.

3. The graph is a horizontal line segment at 0.5 m/s.

4a. DESCARTES 4b. HYPATIA4c. EUCLID4d. This code shifts 20 spaces forward, or 6 spacesback, in the alphabet.

5a. Stories will vary. At the 20 s mark, each girl ismoving at the same velocity. Bea’s velocity increasessteadily in a linear fashion. Caitlin’s velocity increasesvery slowly at first and then becomes faster and faster.Abby’s velocity increases very quickly at first and thenincreases at a slower rate.5b. No; because Abby starts out moving faster thanboth Bea and Caitlin, even when she slows down totheir speed she stays ahead.6a. y � 4.25x � 1.006b.

6c. It shifts the graph up 0.50 unit on the y-axis.6d. y � 4.25x � 1.50

y

x54321

25

20

15

10

5

0

A B C D E F G H I J K L M N O P Q R S T U V W X Y ZABCDEFGHIJKLMNOPQRSTUVWXYZ

Original input

Cod

ed o

utp

ut

t

v

6

1

Elapsed time (s)

Vel

ocit

y (m

/s)

7a. Answers will vary. The graph will pass the verticalline test.7b. Answers will vary. The graph will fail the verticalline test.8. The domain of the 26 letters is coded to a range ofthe 13 even-number-positioned letters—{B, D, F, . . . , Z}.The code is a function because every original letter iscoded to a unique single letter. The rule for decoding isnot a function because there are two choices for everyletter in the coded message. For example, the letter Bcould be decoded to either A or N.9a. f (�3) � �3 � 39b. f (2) � 2 � 29c. 10 and �1010a.

[�4.7, 4.7, 1, �3.1, 3.1, 1]

10b. The graph of y � �x� looks like half of the graphof y � x2 lying on its side.10c. The graph of y � �x� has only one branchbecause it gives only positive solutions.10d. This equation does not represent a function,because a given input can have two different outputs.For example, if x � 4, then y � 2 or y � �2.

11a. Let t be the number of T-shirts, and let s be thenumber of sweatshirts.t � s � 126t � 10s � 8811b. 8 T-shirts and 4 sweatshirts12a.

71 2 6 6483 3

4 8

3 4 6 62

2

123456789

101112

7 means 1.7 cm1

Key

1

Praying Mantis Length

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12b. 10.4 cm12c. Mean: approximately 5.4 cm; median: 5.3 cm;mode: 5.3 cm. Choice and explanations will vary.13a. Start with 21, then apply the rule Ans � 4;10th term � �15.13b. Start with �5, then apply the rule Ans (�3);10th term � 98,415.13c. Start with 2, then apply the rule Ans � 7;10th term � 65.14a. x � 1.875, y � 8.2514b. infinitely many solutions15a. y � 1.6x, where x is a measurement in miles andy is a measurement in kilometers15b. 400 km15c. 3.2 km15d. approximately 168 m16. right triangle

17a. �4 17b. 13 17c. �12 17d. 218a. slope: �

52�; y-intercept: ��

121�

18b. slope: undefined; y-intercept: none

18c. slope: ��12�; y-intercept: ��

52�

19a. $19,777 19b. $22,104

19c. y � 23,039(1 � 0.0225)�8 (1 � 0.035)�2;approximately $18,00020. Answers will vary depending on the method used.The following possible answers used the Q-pointmethod and a decimal approximation of the slope.20a. y � 96 � 5.7(x � 5)

[0, 25, 5, 0, 125, 25]

y

–4

–6

–2

x

2

8

10

6

2–4 4 6 8 10

20b. y � 124.5 � 5.7x20c. approximately 16 days20d. The y-intercept would become 200; y � 200 � 5.7x.20e. 86 g21a. independent: number of licks; dependent: mass;decreasing, discrete

21b. independent: number of scoops; dependent:cost; increasing, discrete

21c. independent: amount of stretch; dependent:flying distance; increasing, continuous

21d. independent: number of coins flipped;dependent: number of heads; increasing, discrete

Number of coins flipped

Num

ber

of h

eads

Amount of stretch

Flyi

ng d

ista

nce

Number of scoops

Cos

t

Number of licks

Mas

s

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Answers to All Exercises7b.

7c.

8a. a translation right 12 units and up 7 units8b. (x � 12, y � 7)

8c. �x � �12

20�, y � �2

70��, or (x � 0.6, y � 0.35)

9a. (4.5, 1.5), (4.5, 2.5), (5.5, 1.5), (5.5, 2.5)9b. (0.75, 2.25), (0.75, 3.25), (1.75, 2.25), (1.75, 3.25)9c. (�3, 3), (�3, 4), (�2, 3), (�2, 4)10. See below.11a. 17 11b. x � �411c. 8 � 3x 11d. �1 � 6x12a. x � 4 12b. x � 3.3�

12c. x � �413a. y � � 2 � x 13b. y � 1 � 0.5(x � 1)13c. y � x 13d. y � 1.5x

x

y

5

5

x

y

5

5

LESSON 8.1

1a. (�2, 3), (4, 1), (2, �5)1b. (�5, 1), (3, 4), (6, �3), (�3, �5)2a. a translation left 5 units2b. a translation right 1 unit and up 2 units3a. a translation up 4 units3b. The x-coordinates are unchanged.3c. The y-coordinates are increased by 4.4a.

4b. (x � 2, y)5a. a translation right 10 units and down 8 units5b. L3 � L1 � 10, L4 � L2 � 85c. The signs would change: L3 � L1 � 10, L4 �L2 � 8.6a. Possible answer: Enter the x-coordinates into listL1 and the y-coordinates into list L2: L1 � {2, 5, 1, 2}and L2 � {�1, 0, 2, �1}. Then make a connectedgraph.6b. i. a translation up 4 units6b. ii. a translation left 5 units6b. iii. a translation right 3 units and down 2 units7a.

x

y

5

5

x

y

5–5

5


x 0 1 2 3 4 5 6

g (x)

h(x)

3 2 1 0 1 2 3

9 4 1 0 1 4 9

10. (Lesson 8.1)

LESSON 8.2

1a. 19 1b. 51c. 8 1d. 2x � 6 � 12a. L3 � L1 � 8; L4 � L2 � 42b. a translation right 8 units and down 4 units3a. (1, �3) 3b. (�5, �3)3c. (6, 4) 3d. (�1, 0)4a. a translation of y �x right 1.5 units and down2.5 units

4b. a translation of y � x2 left 3 units

4c. a translation of y �xup 3.5 units

4d. a translation of y � 3x left 1 unit and up 2 units

5a. y � x2 � 25b. y � 4x�5

5c. y � x � 4 � 16a. a translation of y � x2 right 1 unit and down3 units; y � (x � 1)2 � 36b. a translation of y �x left 5 units and down3 units; y �x � 5� 36c. a translation of y �x right 6 units and up4 units; y �x � 6� 46d. a translation of y � x2 left 1 unit; y � (x � 1)2

7a. The input variable is time; the output variable isdistance.

7b. Time is in seconds and distance is in meters.7c. domain: 0 � t � 5, where t is time in seconds;range: 1 � d � 4, where d is distance in meters7d. Possible answer: Beth starts 3 m from her teacherand walks toward the teacher at 1 m/s for 2 s. When sheturns in the test, she is 1 m from the teacher. Beth thenturns and walks away from the teacher at 1 m/s for 3 s.7e. d � t � 2 � 18a. a translation down 4 units

8b. a translation right 4 units

9a. a translation right 3 units9b. a translation left 2 units9c. a translation down 2 units9d. a translation up 3 units10a. y � a � b x�10

10b. approximately 0.955 volt/s; the value of bis 0.955.10c. a � 6.579; y � 6.579(0.955)x�10

10d. y � 6.579(0.955)0�10 � 10.426 volts10e. approximately 51 s after the capacitor isdisconnected11a. Let x represent time in minutes, and let yrepresent temperature in degrees Celsius. The scatterplot suggests an exponential function.

[�1, 10, 1, �10, 100, 10]

11b. Ratios to the nearest thousandth:0.765, 0.788, 0.829, 0.882, 0.900, 0.926; the ratios arenot approximately constant and do not support anexponential function.


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11c.

A translation down 21 units; the long-run value willnow be 0°C.

11d. Ratios to the nearest thousandth: 0.660, 0.645,0.650, 0.692, 0.667, 0.667; the ratios are approximatelyconstant; the mean is approximately 0.66.11e. y � 47(1 � 0.34)x

11f. a translation up 21 units11g. y � 47(1 � 0.34)x � 21

12a. The year is the input variable, x, and populationin billions is the output variable, y.12b. The graph should be an increasing exponentialfunction passing through the point (2004, 6.4).

12c. (2004, 6.4)

12d. y � 6.4(1 � 0.0114)x�2004

[1995, 2015, 1, 5, 8, 1]12e. 1995: 5.8 billion; 2015: 7.2 billion13a. (4, 8)13b. y � b(x � 4) � 813c. (H, V )13d. y � b(x � H) � V14a. 90.6%14b. Answers will vary.14c. 91.6%15a. x � 0.75, y � 6.515b. x � �3.5, y � �24.515c. x � 3, y � 1

Temperature Time (min) (°C)

0 471 312 203 134 95 66 4


LESSON 8.3

1a. �11b. 37.51c. �101d. 0.5(�x � 3)2 � 31e. �0.5(x � 3)2 � 32a. a translation of the graph of y � x2 down 2 units;y � x2 � 22b. a translation of the graph of y � x right 3units; y � x � 32c. a translation of the graph of y � x2 left 2 units anddown 1 unit; y � (x � 2)2 � 12d. a translation of the graph of y � x right 1 unitand up 1 unit; y � x � 1 � 13a. a reflection across the x-axis3b. a translation right 6 units or a reflection acrossthe y-axis3c. a translation left 2 units3d. a translation left 2 units and a reflection acrossthe x-axis4a. a reflection across the y-axis

4b. a reflection across the x-axis

5a. a reflection across the x-axis

[�9.4, 9.4, 1, �6.2, 6.2, 1]5b. a translation left 3 units and a reflection acrossthe x-axis

5c. a reflection across the x-axis followed by atranslation up 3 units

5d. a reflection across the y-axis and a translation up3 units (or just a translation up 3 units)

6a. Possible answer: Enter the x-coordinates into list L1

and the y-coordinates into list L2: L1 � {2, 7, 7, 2} and L2 � {2, 2, 4, 2}. Then make a connected graph.6b. i. Define L3 � �L1 and L4 � L2.6b. ii. Define L3 � �L1 and L4 � �L2.6b. iii. Define L3 � L1 and L4 � �L2.6b. iv. Define L3 � L1 � 2 and L4 � �L2.7a. a reflection across the y-axis7b. a reflection across the x-axis7c. a translation left 8 units and a reflection across thex-axis7d. a translation right 2 units and down 4 units7e. a reflection across the x- and y-axes7f. a reflection across the line y � x8a. y � x � 4 � 18b. y � �x � 3 � 3.58c. domain: 0 � x � 6; range: 0.5 � y � 3.59. (x � 1, �y)10a. possible answer: y � �(x � 3)2 � 610b. possible answer: y � �f (�(x � 2)) � 411a. i. y � �x2 � 411a. ii. y � �x � 711a. iii. y � 2�(x�6)

11a. iv. y � 2[�(x � 8)] � 4;y � [4 � 2(�x)] � 16;y � �(4 � 2x) � 8; ory � �[4 � 2(x � 4)]

11b. i. y � �2

11b. ii. y � 3.5

11b. iii. x � 3

11b. iv. x � �4 or y � �4

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11c. If the line of reflection is x � a or y � b, theamount of translation is twice a or twice b.

11d. y � �f (x) � 2b

11e. y � f (�x � 2a)12a. There is a 12% decrease per minute;(1 � 0.12 � 0.88).12b. 0 g; all the reactant will be used.12c. 100 g; not all the reactant will be used.12d. Both graphs are decreasing exponential graphs.The graph of y � 500(0.88)x shows a starting amountof 500 g reactant; the graph of y � 500(0.88)x � 100

shows a starting amount of 600 g reactant. The secondgraph is a translation of the first graph up 100 units.

13. 47 T � �116

cuTp

� � �14

qcuu

aprst

�

� �46

74� quarts � 0.734 quart

14a. possible answers using Q-points:y �36 � 3.6(x � 4) or y � 72 � 3.6(x � 14)14b. approximately 22 min

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LESSON 8.4

1. y � x � 52a.

2b. y � 2x � 53a. y � �1.2x � 5 � 63b. Ted began at the motion sensor and walked awayat 1.2 m/s for 5 s. At 6 m away, he turned and walkedback at the same speed.4. Answers will vary.5. Answers will vary.6. (2, 0)7a. a vertical stretch of y � x2 by a factor of 2

[�9.4, 9.4, 1, �6.2, 6.2, 1]

x

y

0 2 6 104 8

10

8

6

4

2

Time (s)

Dis

tan

ce fr

om s

enso

r (m

)

7b. a vertical shrink of y �x by a factor of 0.25,then a translation right 2 units and up 1 unit

7c. a reflection of y � x2 across the x-axis, then atranslation left 4 units and down 1 unit

7d. a vertical stretch of y �xby a factor of 2 anda reflection across the x-axis, then a translation right3 units and up 4 units

8. The absolute-value graph is stretched vertically bya factor of 3. Its vertex remains at (0, 0).9. See below.

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Change to the equation New equation in Transformation of they � x y � form graph of y � x

Replace x with x � 3 y � x � 3 Translation right 3 units

Translation down 2 units

y � �xReplace y with y � 2

Replace y with �0y.5� Vertical shrink by a factor of 0.5

Translation left 4 units

y � 1.5xTranslation right 1 unit

Replace y with �3y

�

Replace y with y � 2

Replace y with �y

Replace x with x � 4

Replace y with �1y.5�

Replace x with x � 1

y � x � 2

y � x � 2

y � 0.5x

y � x � 4

y � x � 1

y � 3x

Reflection across the x-axis

Translation up 2 units

Vertical stretch by a factor of 1.5

Vertical stretch by a factor of 3

9. (Lesson 8.4)


10a. a reflection across the x-axis and a translationleft 3 units10b. possible answer: a vertical shrink by a factor of0.5, then a translation right 2 units and up 1 unit11a. a horizontal stretch by a factor of 3

[0, 28.2, 2, 0, 18.6, 2]

11b. a horizontal and vertical stretch by a factor of 3

11c. a horizontal and vertical shrink by a factor of 0.5

11d. These transformations increase or decrease theoverall size of the object without changing its shape.12a. a vertical shrink by a factor of 0.5 and areflection across the x-axis

12b. a vertical stretch by a factor of 2 and atranslation right 4 units

12c. a vertical stretch by a factor of 3 and a reflectionacross the x-axis, then a translation left 2 units and up4 units

13a. possible answer: f (x) � �25x � 3.2� 80

[�1, 7, 1, �10, 100, 10]13b. Using the equation in 13a: f (2.5) � 62.5;the depth of the snow after 2.5 months (mid-December) would be about 62.5 cm.13c. Using the equation in 13a: x � 1.88 or x � 4.52; the depth of the snow would be 47 cm after1.88 months (end of November) or after 4.52 months(mid-February).13d. after 3.2 months (early January); about 80 cm14. (x � 1, 0.8y)15a. Yes; when you substitute 1 for x, you get y � a � 12 � a.15b. Yes; when you substitute 1 for x, you get y � a � 1 � a.15c. No; unless f (1) � 1, a will not be the same as they-value. For example, if f (1) � 3, then y � a� f (1) �a�3,but a � �3

y�.

16a. �21

9� 16b. 510

16c. 212 � 36 16d.

17a. x represents actual temperature, and y representswind chill temperature.17b. 10; when the wind chill temperature is �15°Fwith a wind speed of 40 mi/h, the actual temperature isapproximately 10°F.18a. x � 3 18b. x � 16

18c. x � �170�, or about 1.4

18d. x � 11

1�38 � x12

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LESSON 8.6

1a. a reflection of the graph of y � x2 across the x-axis, then a translation up 2 units (or translationthen reflection); y � �x2 � 21b. a vertical shrink of the graph of y �xby a factor of �

13� and a translation right 2 units;

y � �13�x � 2

1c. possible answer: a vertical shrink of the graph ofy � x2 by a factor of 0.5, then a translation right 1 unitand down 1 unit; y � 0.5(x � 1)2 � 11d. possible answer: a vertical stretch of the graph ofy � x by a factor of 2 and a reflection across the x-axis, then a translation left 2 units and up 3 units; y � �2x � 2� 3

2. y � �x2

�

3. y � ��x5

�

4. The red curve is described by f (x) � �x4

�,and the blue curve is described by g(x) � �x

8�. Explanations will vary. Possible

explanations: The red curve is closer to theorigin, and f (x) � �x

4� indicates less vertical

stretch; or f (1) � 4, which is a point on the red curve,whereas g (1) � 8.5a. a vertical stretch by a factor of 4; domain: x � 0

[�9.4, 9.4, 1, �6.2, 6.2, 1]5b. a translation right 5 units and down 2 units;domain: x � 5

5c. a vertical shrink by a factor of 0.5, then atranslation up 3 units; domain: x � 0

5d. a vertical stretch by a factor of 3, a reflectionacross the x-axis, and a translation left 3 units; domain:x � �3

6. If the expressions were equivalent, the two graphswould be the same. The graph shows that they are not.You can only cancel factors, and 3x � 7 and x � 7 don’thave any common factors.

7a. y � �x �1

3�

7b. y � �x �1

2�

7c. y � �x �1

1� � 1

7d. y � �x �1

1� � 2

8a. y � � f (x), or y � ��x1

�

8b. y � f (�x), or y � ��

1x�; see sketch above.

8c. Either reflection produces the same graph because��

1x� � ��x

1�.

9. Let x represent the amount of water to add and yrepresent the concentration of salt.The amount of salt is0.05(0.5) � 0.025.

y � �00.5

.0�25

x� ; 0.01 � �00.5

.0�25

x�; x � 2; 2 L

x

y

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10a. $70010b.

10c. y � �535

�00

x�, where x is the number of additionalbusinesses that have signed up and y is the cost perbusiness10d. 150 � �5

35�

00x�; x � 18.3�; 19 additional businesses

11a. �x5

�, where x � 011b. x2, where x � 011c. �

x4�x2

5�, where x � 5

11d. �1 �

5x5x

�, where x � 011e. 1 � 3x3, where x � 012a. x12b. �12

7x�, where x � 0

12c. �15 �

6x10x�, where x � 0

12d. �(x � 57)x(x � 2)�, where x � 5 and x � �2

13a. �125�, where x � 0

13b. �x

1�

26

�, where x � 6

13c. �4yx2�, where x � 0 and y � 0

13d. 2, where x � 0 and x � �414a. x � �214b. x � �214c. x � 415. vertex: (3, 1); point to the left: (2, 3); point to theright: (4, 3)

16a. � �; these are the total numbers of baked

goods sold, in dozens, for the past two years.

16b. � �; the first row is the total cost for

fall and spring this year, and the second row is the totalincome from sales for fall and spring this year.17a. i17b. ii17c. iv17d. Answers will vary as students create a scenariofor graph iii.

20.50 24.1090.25 102.75

39 3417 2411 13

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Additionalbusinesses 0 5 10 15 20Cost perbusiness ($) 700.00 350.00 233.33 175.00 140.00


LESSON 8.7

1a. (�2, 2), (1, 2), (�2, 6)

1b. � � 1c. � �2a. possible answer: � �2b. � � � � � � � �2c. � � � � � �

� �3a. [6 15] 3b. � �3c. [64] 3d. � �4a. � � 4b. � �5a. a rectangle5b. Possible answer: For the x-coordinate, multiplyrow 1 of the transformation matrix by column 2 of the

quadrilateral matrix: [1 0] � � � � 2; this goes in

row 1, column 2 of the image matrix. For the y-coordinate, multiply row 2 of the transformationmatrix by column 2 of the quadrilateral matrix:

[0 2] � � � � �2; this goes in row 2, column

2 of the image matrix.

5c. � � 5d. a parallelogram

x

y

5

–5

5–5

�22

1�4

2�2

�14

2�1

2�1

0�1

10

01

�10

3256

814

615

�6�3

�61

�8�1

�4�4

�4�4

�4�4

�21

�25

�43

71

75

53

90

90

90

�21

�25

�43

�21

�25

�43

�23

1�1

�2�1

0�3

0�3

0�3

x

y

5

–5

5–5

6a. [S] � � �6b. i. a vertical stretch by a factor of 3; ii. a verticalstretch by a factor of 3 and a reflection across the x-axis; iii. a horizontal stretch by a factor of 4 and avertical stretch by a factor of 2; iv. a translation right4 units and up 2 units

7a. possible answer: [Q] � � �7b. � � � [Q] � � �7c. � � � [Q] � � �7d. � �; � � � [Q]

� � �;

a reflection across both the x- and y-axes; alone,

� � results in a reflection across the x-axis, and

� � results in a reflection across the y-axis.

7e. � �; a quarter-turn

counterclockwise

8a. y � x2

8b. [P ] � � �8c. � �; y � (x � 3)2 � 2

8d. � �; y � (�x)2�24

�11

00

11

24

56

43

32

23

16

24

11

00

�11

�24

x

y

5

5–5

�17

�56

�43

�22

x

y

5

–5

5–5

01

�10

0�1

10

�7�1

�6�5

�3�4

�2�2

0�1

�10

0�1

�10

3.50.5

32.5

1.52

11

00.5

0.50

70.5

62.5

32

21

00.5

10

71

65

34

22

01

11

10

00

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9a. Answers will vary; sample answer: � �.

9b. Answers will vary; sample answer:

� � � � � � �.

9c. a rotation through 60° about the origin, or the point (0, 0)

9d. � �; sample answer:

� � � �� ;a rotation through 120° about the origin

9e. Possible answer: Multiply � � by

itself three times (3 � 60° � 180°). Then multiply theresult by matrix [R].

�0.866�0.5

0.50.866

x

y

5

–5

5–5

�23.464

�3.2321.598

00

40

32

00

�0.866�0.5

�0.5000.866

�0.866�0.5

�0.5000.866

x

y

5

–5

5–5

23.464

�0.2323.598

00

40

32

00

�0.866�0.5

0.50.866

x

y

5

–5

5–5

40

32

00 9f. Possible answer: Multiply � � by

itself six times (6 � 60° � 360°). Then multiply the

result by matrix [R] or just multiply [R] by � �.

10a.

10b. y � 5 � 0.5x � 611a. in millions: 127, 141, 171.5, 293, 129911b.

11c. yes, China and India12. x � 0 or x � 6

Population (millions)

Most Populated Countries, 2004

0 1500500 1000

x

y

654321

8641 3 5 7 92Time (s)

Dis

tan

ce fr

om m

otio

nse

nso

r (m

)

01

10

�0.866�0.5

0.50.866

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Pick x.

37

( )2 ��

Equation: 4(x � 3)2 � 1 � 37

�(3)

�(4)

�(1)

�(3)

/(4)

�(1)

0, 6

� 3

9

36


CHAPTER 8 REVIEW

1a. a translation left 2 units and up 1 unit1b. (x � 2, y � 1)2a. i. a vertical shrink by a factor of 0.5 and atranslation left 6 units2a. ii. possible answer: a reflection across the x-axis,then a translation up 2 units2a. iii. possible answer: a horizontal stretch by afactor of 2 and a reflection across the y-axis, then atranslation right 5 units and down 3 units2b. i. L3 � L1 � 6, L4 � 0.5 � L2

2b. ii. possible answer: L3 � L1, L4 � �L2 � 22b. iii. possible answer: L3 � �2 � L1 � 5, L4 � L2 � 33. Answers will vary. For these possible answers, listL3 and list L4 are used for the x- and y-coordinates,respectively, of each image.3a. L3 � L1, L4 � �L2

3b. L3 � �L1, L4 � L2

3c. L3 � L1 + 3, L4 � �L2

4a. a vertical stretch of the graph of y �xby afactor of 2, then a translation up 1 unit

[�9.4, 9.4, 1, �6.2, 6.2, 1]4b. a reflection of the graph of y �xacross the x-axis, then a translation left 2 units and up 2 units

4c. possible answer: a vertical shrink of the graph ofy � x2 by a factor of 0.5, then a reflection across the y-axis (which doesn’t show, because the graph issymmetrical) then a translation down 1 unit

4d. possible answer: a reflection of the graph ofy � x2 across the x-axis, then a translation right 2 units and up 1 unit

5. g(x) � f (x � 1) � 26a. y � �x� 3 6b. y � (x � 4)2 � 26c. y � 0.5x2 � 5 6d. y � �2x � 3� 17a. The graph should have the same x-intercept asf (x). The y-intercept should be the opposite of that forf (x).

7b. Answers will vary. Possible answer for a friendlywindow with a factor of 1: If Y1 � �x � 2, then Y2 � �Y1 reflects the graph across the x-axis (becausethe calculator interprets �Y1 as �(�x � 2), or x � 2);this supports the answer to 7a.8a. a translation right 3 units; asymptotes: x � 3,y � 08b. a vertical stretch by a factor of 3 and then atranslation left 2 units; asymptotes: x � �2, y � 08c. a translation right 5 units and down 2 units;asymptotes: x � 5, y � �29a. 5.625 lumens9b. approximately 2.12 m10a. a translation of the graph of y � �x

1� right 3 units

and down 2 units; y � �x �1

3� � 2

10b. a translation of the graph of y � 2x right 4 unitsand down 2 units; y � 2x�4 � 210c. possible answer: a reflection of the graph ofy � 2x across the x-axis and across the y-axis, followedby a translation up 3 units; the reflection across the y-axis can be done at any time, but the translation upmust be done after the reflection across the x-axis;y � �2�x � 3

x�f(x)

y

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10d. possible answer: a vertical stretch of the graph ofy � �x

1� by a factor of 4 and a reflection across the x-axis,

followed by a translation up 1 unit and left 2 units;y � ��x �

42� � 1

11a. �14�, where x � �

32�

11b. 28x2, where x � 312a. possible answer:

[A] � � ��1�1

1�1

11

�11

12b. i. Nothing; the image is identical to the originalsquare.12b. ii. a reflection across the x-axis and across the y-axis, or a rotation through 180°12b. iii. a vertical stretch by a factor of 312b. iv. a translation right 1 unit and up 1 unit

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Answers to All Exercises3d. 2(x � 1)2 � 14

(x � 1)2 � 7 x � 1 � ��7�

x � �1 ��7�4. sample answers:4a. 4b.

4c. 4d.

5a. h(0) � �4.9(0)2 � 147 � 147; it is the startingheight of the ball, when t � 0 s.5b. Solve the equation 20 � �4.9t2 � 147 symbol-

ically to get ��, or approximately �5.09.

The graph shows two solution points: (5.09, 20) and(�5.09, 20).

[�10, 10, 1, �50, 150, 25]5c. No; only the solution (5.09, 20) makes sense,because the time t must be positive.5d. t � 5.09 s5e. The ball hits the ground when t � 5.48 s becausethe positive x-intercept is near the point (5.48, 0).6a. (0, 0), (4.7, 108), (9.4, 0)6b. sample answer: [�1, 10, 1, �10, 120, 10]6c. (4.7, 108)6d. h(t) � �4.9(t � 4.7)2 � 108Start with h(t) � �t2 and a vertical stretch of 108,ahorizontal stretch of 4.7,a horizontal translation of�4.7,and a vertical translation of �108.The equation is

h(t) � �108��41.7�(t � 4.7)�

2� 108 � �4.9(t � 4.7)2 � 108.

20 � 147�

�4.9

x

y

x

y

x

y

x

y

LESSON 9.1

1. Enter one side of the equation into Y1 and the otherinto Y2 on the calculator. Then find the x-coordinate ofeach intersection point by tracing on the graph orzooming in on the table.1a. x � �6 or x � 3

[�10, 10, 1, �10, 15, 1]1b. no solution

[�10, 10, 1, �10, 10, 1]1c. x � 3

[�10, 10, 1, �10, 10, 1]1d. x � �2.14 or x � 0.47

[�10, 10, 1, �10, 10, 1]2a. real, rational2b. real, rational, integer2c. real, irrational2d. real, rational, integer, whole number,natural number3a. x � ��18�3b. x2 � 49, x � �73c. x � 2 � �5, x � 7 or x � �3



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6e. approximately 94 m; at 3 s, the flare is about 94 mabove the ground

6f. Answers are t � 4.7 � ��46.19��. At about 1.17 s and

8.23 s, the flare will be 47 m above the ground.7a. After the ball has gone 2 m horizontally, it will beapproximately 3.75 m above the ground.7b. The ball will be 2 m above the ground once it hasgone about 0.31 m horizontally and again once it hasgone about 6.49 m horizontally. Exact solutions are

x � 3.4 ��.

7c. When x � 0, the height is 1.5412 m.7d. When p(x) � 0, the horizontal distance is 3.4 � ��, or approximately 7.67 m.

8a. The graphs intersect at (3, 4), so the solution isx � 3.

[0, 6, 1, �5, 5, 1]8b. The table shows the solution to be at x � 3.

8c. 4 � �2(x � 3)2 � 40 � �2(x � 3)2

0 � (x � 3)2

0 � x � 3

3 � x9a. The x-intercepts indicate when the projectile is atground level.

420�23

220�23

9b. 2.63 s and 7.58 s9c. Find the average of 2.63 and 7.58, which is 5.105;h(5.105) � 30.051.9d. After 5.105 s, the projectile is 30.051 m above theground, its maximum height.9e. h(3.2) � 12.324 m; this is the height at 3.2 s.9f. Answers will vary. The horizontal line y � 12.5intersects the parabola twice—when x � 3.2 s andwhen x � 7.0 s.

[0, 10, 1, �5, 35, 5]10a. i. a vertical stretch of 16 units, a reflection acrossthe x-axis, a horizontal translation right 3 units, and avertical translation of 20 units10a. ii. (3, 20)10a. iii. �16 or �

12�(�32) ft/s2 is the effect of gravity, the

maximum height occurs at 3 s, and the maximumheight is 20 ft.

10b. i. a vertical stretch of 4.9 units, a reflectionacross the x-axis, a horizontal translation right4.2 units, and a vertical translation of 12 units

10b. ii. (4.2, 12)10b. iii. �4.9 or �

12�(�9.8) m/s2 is the effect of gravity,

the maximum height occurs at 4.2 s, and themaximum height is 12 m.

11. �3x � 4 � 16 The given inequality.

�3x � 12 Subtract 4 from both sides.

x � �4 Divide both sides by �3 andreverse the inequality symbol.

12. The slope is �16��

60� � �

�6

5�. An intercept form of the

inequality is y � 6 � �56�x.


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LESSON 9.2

1. The average of 3 and �2 is �3 �

2�2�, or 0.5. So the

axis of symmetry is x � 0.5, and the vertex has an x-coordinate of 0.5.2. y � 0.4(0.5)2 � 0.4(0.5) � 2.4, or y � �2.5.3. 0 � (x � 1.5)2 � 7.25

7.25 � (x � 1.5)2

��7.25� � x � 1.5�1.5 ��7.25� � x

x � 1.193 or x � �4.1934a. x � �2.732 and x � 0.7324b. x � �1.869 and x � 0.5355a. (x � 3) � ��7�

x � �3 ��7�x � �5.646 orx � �0.354

5b. (x � 2)2 � 21x � 2 ��21�x � �2.583 orx � 6.583

6. Answers will vary. The graph of y � (x � 3)2

intersects the graph of y � 7 at (�5.646, 7) and(�0.354, 7).

[�7, 1, 1, �1, 10, 1]7a. The ball is on the ground when h � 0. Thishappens at t � 0 s and t � 3 s.7b. The ball is at its highest point at t � 1.5 s, halfwaythrough its flight.7c. When t � 1.5, h � 36, so the ball goes 36 ft high.7d. domain: 0 � t � 3; range: 0 � h � 368a. between 3.67 s when the height is still positive and3.68 s when the height is negative

8b. Starting the table at 3.67 and setting �Tbl equal to0.001 gives the answer 3.676 s.8c. The maximum height occurs between 1.81 and1.82 s when the height of the ball is at least 55.562 ft.9a. Answers will vary. The ball is thrown from aninitial height of 8 m. It reaches its maximum height,about 25 m, in 2 s and hits the ground at about 4.3 s.9b. The ball starts at a velocity of 20 m/s and slowsdown at a constant rate. At 2 s it is not moving. Then itstarts falling and is moving down at 22 m/s when ithits the ground.9c. When the velocity is negative, the ball is falling.9d. This is when the ball is at its maximum heightand, at that instant not moving. Its velocity is zero.9e. These are the two times when the height of theball is about 13 m.9f. domain: 0 � t � 4.3; range: 0 � h(t) � 26;�22 � v(t) � 2010a. Bo’s ball is at its highest point 0.5(3.4) s, or 1.7 s,after it is hit. Gale’s is at its highest 0.5(4.7) s, or 2.35 s,after it is hit.10b. You can’t tell whose ball goes farther. Eventhough Gale’s is in the air longer, it might just gohigher and not as far as Bo’s.11a.

[0, 9.4, 1, �10, 20, 5]11b. x � 4.511c. (4.5, 19)11d. y � �3(x � 4.5)2 � 1912. y � �16(x � 2)2 � 67

13a. y � 2 � �13�x

13b. y � 3.5 � �14�x


LESSON 9.3

1a. yes; three terms (trinomial)1b. yes; two terms (binomial)1c. No; the first term has a negative exponent.1d. No; the first term is equivalent to 3x�2, which hasa negative exponent.1e. yes; one term (monomial)1f. yes; three terms (trinomial)1g. yes; two terms (binomial)1h. Not a polynomial as written, but it is equivalent to3x � 6, a binomial.2a. x2 � 10x � 252b. x2 � 14x � 492c. 3(x2 � 4x � 4) or 3x2 � 12x � 123a.

(x � 2)2 � x2 � 4x � 43b.

(x � 12)2 � x2 � 24x � 1443c.

(x � 7)2 � x2 � 14x � 494a. y � x2 � 10x � 294b. y � 2x2 � 28x � 904c. y � �3x2 � 24x � 474d. y � 0.5x2 � 3x5a.

(x � 2)(x � 4) � x2 � 6x � 85b.

(x � 3)(x � 5) � x2 � 8x � 15

3x

x x2 3x

5 5x 15

2x

x x2 2x

4 4x 8

�7x

x x2 �7x

�7 �7x 49

12x

x x2 12x

12 12x 144

2x

x x2 2x

2 2x 4 An

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5c.

(x � 2)(x � 5) � x2 � 3x � 105d.

x(x � 3) � x2 � 3x5e.

(x � 2)(2x � 5) � 2x2 � 9x � 105f.

(3x � 1)(2x � 3) � 6x2 � 7x � 36a. (2, 3)6b. y � (x � 2)2 � 36c. x2 � 4x � 4 � 3 � x2 � 4x � 77. No; by squaring the values inside the parentheses,Heather is accounting for only two of the fourrectangles in a rectangle diagram. She needs to add theterms in the two rectangles that sum to the middleterm.8a. 70 m8b. 50 � 0.0056x2 � 0.14x; x � 83 km/h9a. The vertex is at (0.4, 2.5), so the ball reaches amaximum height of 2.5 m.9b. h(t) � �4.9t 2 � 3.92t � 1.7169c. The pitcher released the ball at a height of1.716 m.9d. domain: 0 � t � 1.1; range: 0 � h � 2.510a. yes10b. No; the right side should be 2x2 � 18.8x � 46.98.10c. No; the right side should be �3.5x2 � 11.2x � 11.10d. yes

�13x

2x 6x2 �2x

3 9x �3

2x

2x 2x2 4x

5 5x 10

x

x x2

�3 �3x

2x

x x2 2x

�5 �5x �10


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11a. meaningful domain: 0 � x � 6.5;meaningful range: 0 � y � 897.81

[0, 9.4, 1, �100, 1000, 100]11b. Possible answer: Average the two x-intercepts,then substitute this value into the equation to findthe y-coordinate of the vertex. The vertex is at(3.25, 897.8125).11c. The price $3.25 gives the maximum income,$897.81.11d. The prices $0.00 and $6.50 produce no income.11e. If the warehouse charges $5 per item, incomewill be $637.50.12a.

x 2 � y 2 � 2xy � 6x � 6y � 9

yx

x x2 xy

y xy y2

3

3x

3y

3 3x 3y 9

12b.

4x 2 � y2 � 4xy � 20x � 10y � 2513. y � x2 � 8x � 16; answers will vary.14. No; it doesn’t pass a vertical line test.15. The message is POLYNOMIALS.15a. 16; P 15b. 15; O15c. 12; L 15d. 25; Y15e. 14; N 15f. 15; O15g. 13; M 15h. 9; I15i. 1; A 15j. 12; L15k. 19; S16a. possible answer: [0, 400, 50, �20, 100, 10]16b. 0 � x � 34016c. 0 � y � 70.3

�y2x

2x 4x2 �2xy

�y �2xy y2

5

10x

�5y

5 10x �5y 25


LESSON 9.4

1a. x � 4 � 0 or x � 3.5 � 0, so x � �4 or x � �3.51b. x � 2 � 0 or x � 6 � 0, so x � 2 or x � 61c. x � 3 � 0 or x � 7 � 0 or x � 8 � 0, so x � �3 or x � 7 or x � �81d. x � 0 or x � 9 � 0 or x � 3 � 0, so x � 0 or x � 9 or x � �32a. y � (x � 3)(x � 1)

[�9.4, 9.4, 1, �6.2, 6.2, 1]2b. y � (x � 8)(x � 3)

[�9.4, 9.4, 1, �35, 5, 5]2c. y � (x � 3)(x � 9)

[�14.1, 14.1, 1, �9.3, 9.3, 1]2d. y � (x � 10)(x � 3)

[�20, 20, 2, �50, 10, 2]3a. x � 7 and x � �2

[�10, 10, 1, �25, 10, 5]

3b. x � �1 and x � �8

[�10, 10, 1, �25, 10, 5]3c. x � 11 and x � �7

[�15, 15, 1, �250, 250, 50]3d. x � �5 and x � 9

[�15, 15, 1, �25, 25, 5]4a. y � (x � 2.5)(x � 1)4b. y � (x � 4)(x � 4), or y � (x � 4)2

4c. y � (x � 2)(x � 2)4d. y � �x � r1��x � r2�5a. two intercepts, x � �1 and x � 35b. (1, �4)5c. y � (x � 1)2 � 4; there is a translation right 1 unitand down 4 units.6a. yes6b. No; change to (x � 6)(x � 5).6c. yes

6d. No; change to 4(x � 1)2.6e. yes6f. No; change to (x � 6)(x � 6).7a. (x � 6)(x � 1) 7b. (x � 5)(x � 2)

7c. (x � 7)(x � 6) 7d. (x � 6)(x � 3)�6x

x x2 �6x

3 3x �18

7x

x x2 7x

�6 �6x �42

5x

x x2 5x

2 2x 10

6x

x x2 6x

1 x 6

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7e. (x � 4)(x � 6) 7f. (x � 12)(x � 4)

8a. See below.8b. The sum of the roots needs to be �2, and theproduct needs to be �8; 2 and �4 satisfy theserequirements.9a. y � (x � 3)(x � 7)9b. (5, �4)9c. y � �(x � 3)(x � 7); x-intercepts: x � 3 and x � 7;vertex: (5, 4)9d. y � 2(x � 3)(x � 7); x-intercepts: x � 3 and x � 7;vertex: (5, �8)9e. Possible answer: There are infinitely manyquadratic equations with roots 3 and 7. Each is createdby substituting a unique value of a into y � a(x � 3)(x � 7). Hence, all are vertical stretches orshrinks and/or reflections across the x-axis. The x-coordinate of each vertex is always 5, but they-coordinate depends on the value of a.10. y � 0.25(x � 3)(x � 9)11a. length: 140 ft; area: 4200 ft2

11b. l � 200 � 2w11c. A � w l � w(200 � 2w); 30(200 � 2 30) �30(140) � 420011d. w � 0 ft and w � 100 ft11e. width: 50 ft; maximum area: 5000 ft2

12a. two x-intercepts: x � 3 and x � �3

[�18.8, 18.8, 2, �12.4, 12.4, 2]12b. y � (x � 3)(x � 3)

12x

x x2 12x

�4 �4x �48

�4x

x x2 �4x

�6 �6x 24

12c. The x-intercepts are the positive and negativesquare roots of the number subtracted from x2.12d. i. y � (x � 7)(x � 7)12d. ii. y � (4 � x)(4 � x)12d. iii. y � (x � �47�)(x � �47�)12d. iv. y � (x � �28�)(x � �28�)12e. a2 � b2 � (a � b)(a � b)12f. There are no x-intercepts. The graph is above thex-axis.

[�9.4, 9.4, 1, �6.2, 6.2, 1]12g. Because there are no x-intercepts, there are noroots, and hence there is no factored form of theequation.13. Answers will vary. Possible answer: Derek iscorrect because substituting 4 or �4 for x gives 0 � 16 � 16, or 32, and not zero; the roots shouldsatisfy the equation.

14a. �xx

��

23�, where x �2 and x �3

14b. �xx

��

14�, where x 4 and x �2

14c. �x �

x2

�, where x 0 and x 5

14d. �xx

��

12�, where x �3 and x �2

14e. �xx

��

23�, where x �3

15a. x2 � 19x � 4215b. 3x2 � 13x � 415c. 4x2 � 2x � 1216a. possible answer: s � �14.5(w � 8.6)2 � 82716b. week 16 or 1716c. approximately 8283 or 8284 games16d. approximately 1619 games

Factored form Roots Sum of roots Product of roots General form

y � (x � 3)(x � 4) �3 and 4 �3 � 4 � 1 (�3)(4) � �12 y � x2 � 1x � 12

5 and �2

�5 6

y � (x � 5)(x � 5) 0 �25

y � (x � 5)(x � 2)

y � (x � 2)(x � 3) �2 and �3

5 and �5

5 � 2 � 3 5(�2) � �10 y � x2 � 3x � 10

y � x2 � 5x � 6

y � x2 � 25

8. (Lesson 9.4)


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LESSON 9.6

1a. x � �3 ��2�1b. x � 5 ��2�

1c. x � �8 ��73��

1d. x � �6 ��75��

2a. x � 5 or x � �32b. x � �3 or x � 7

2c. x � ��43� or x � �1

2d. x � 0 or x � �6 or x � �9

3a. ��128��

2

; x2 � 18x � 81 � (x � 9)2

3b. ��

210��

2

; x2 � 10x � 25 � (x � 5)2

3c. ��32��

2

; x2 � 3x � �49

� � �x � �32��

2

3d. ��

21

��2

; x2 � x � �14� � �x � �

12��

2

3e. ��12� �

23��2

; x2 � �23� x � �

19� � �x � �

13��2

3f. ��21.4��2

; x2 � 1.4x � 0.49 � (x � 0.7)2

4a. x2 � 4x � 8 � 0x2 � 4x � 8

x2 � 4x � 4 � 12

(x � 2)2 � 12

x � 2 � ��12�x � 2 ��12�

4b. x2 � 2x � 1 � �5x2 � 2x � �4

x2 � 2x � 1 � �3

(x � 1)2 � �3x � 1 � ��3�

x � �1 ��3�There are no real roots.4c. x � �5 ��34�4d. x � �1 ��8�5a. y � a(x � 2)2 � 31.55b. Solve the equation 0 � a(5 � 2)2 � 31.5; a � 3.5.5c. y � 3.5(x � 2)2 � 31.55d. y � �2(x � 2)2 � 326a. Let w represent the width in meters. Let l representthe length in meters. Then l � w � 4. The areaequation is w(w � 4) � 12.

6b. w2 � 4w � 12w2 � 4w � 4 � 12 � 4

(w � 2)2 � 16w � 2 � �4

w � �2 �4w � �6 or w � 2

6c. The width cannot be negative, so it must be 2 m.The length is 4 m more than the width, so the lengthis 6 m.

7a. y � x2 � 6x � ��62��

2

� 32 � 10

y � x2 � 6x � 9 � 9 � 10y � (x � 3)2 � 1

7b. vertex: (�3, 1);

[�9.4, 9.4, 1, �6.2, 6.2, 1]7c. x2 � 6x � 10 � 0

x2 � 6x � �10

x2 � 6x � 9 � �10 � 9

(x � 3)2 � �1

x � 3 � ��1�x � �3 ��1�

Sample answer: There are no real roots because thegraph doesn’t cross the x-axis.8a. 2.2 s; 26.9 yd (80.7 ft)

8b. t � 2.2 ��26.9� ��

316��, t � �0.046 or t � 4.446;

4.446 s8c. The general form is ��3

16� t2 � 23.46�t � 1.086�,

so the football is about 1 yd high.8d. The vertex is the maximum height of the ball. The y-intercept is the height of the ball when the punter kicksit. The positive x-intercept is the hang time. The other x-intercept has no real-world meaning.

[0, 5, 1, 0, 40, 10]9a. p � 2500 � 5x, where p represents the price indollars of a single ticket and x represents the numberof tickets sold.


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9b. C � xp � x(2500 � 5x)9c. Let C represent the total price of the grouppackage. C � �5(x � 250)2 � 312,5009d. C(20) � 20(2,500 � 100) � 48,000; $48,0009e. 2,500x � 5x2 � 200,000

500x � x2 � 40,000x2 � 500x � �40,000

x2 � 500x � 2502 � 2502 � 40,000(x � 250)2 � 22,500

x � 250 � �150x � 100 or x � 400

9f. To earn a profit, the cruise company should onlyallow between 100 and 400 people.10a. P(10) � 0.9; this means that when there are10 bears in the park, the population grows at a rate of0.9 bear per year.10b. P(b) � 0 when b � 0 or b � 100; when there areno bears, the population does not grow, and when thereare 100 bears, the population does not grow but remainsat that level.10c. The vertex lies halfway between the roots, 0 and100, so the population is growing fastest when thereare 50 bears.

10d. Because the population does not grow whenthere are 100 bears, this value must be the maximumpopulation.10e. It means that if 120 bears were brought in, thepopulation would shrink, due to overpopulation.11a. 2x3 � 5x2 � 4x � 111b. 6x3 � 11x2 � 18x � 2012a. 2x2 � 4x � 212b. 3x2 � 4x � 912c. �2x2 � 2x12d. �3x2 � 113a. x � 0 or x � 413b. x � 1 or x � �313c. x � 4 or x � �1

13d. x � 3 or x � �52�

13e. x � 1 or x � �85�

13f. x � 1 or x � ��83�


LESSON 9.7

1a. 25 � 24 � 11b. 9 � (�12) � 9 � 12 � 211c. 36 � 24 � 121d. 81 � 0 � 812a. a � 2, b � 3, c � �72b. x2 � 6x � 11 � 0; a � 1, b � 6, c � 112c. a � �3, b � �4, c � 122d. �4.9x2 � 47x � 18 � 0; a � �4.9, b � 47, c � 182e. �16x2 � 28x � 47 � 0; a � �16, b � 28, c � �472f. 5x2 � 6x � 7 � 0; a � 5, b � �6, c � �7

3a. x � ; there are no real solutions.

3b. Use the quadratic equation or complete the square;x � �

12� or x � 3.

3c. Complete the square.x2 � 6x � 8 � 0

x2 � 6x � 8x2 � 6x � 9 � 17

(x � 3)2 � 17x � 3 � ��17�

x � 3 ��17�3d. Use the quadratic formula;

x � � .

4a. The graph does not cross the x-axis.

[�10, 10, 1, �10, 10, 1]

4b. To use the quadratic formula, a � 1, b � 3, and

c � 5; x � .

There are no real square roots of negative numbers, sothere are no real roots.4c. If the discriminant is negative, there are no realroots. If it is positive or zero, there are real roots.5a. �4.9t2 � 6.2t � 1.9 � 0; t � �0.255 s or t � 1.52 s; the ball hits the ground 1.52 s after Brandiheads it.5b. �4.9t2 � 6.2t � 1.9 � 3; t � 0.21 s or t � 1.05 s;the ball is 3 m above the ground after 0.21 s (on the wayup) and after 1.05 s (on the way down).

�3 � ��11��2

�1 � �19��3�2 � �76��6

3 � ��23��4

5c. �4.9t2 � 6.2t � 1.9 � 4; t � ; this

equation has no real solution, so the ball is never 4 m high.6a. Sample equation: 0 � x2 �14x � 49. Thex-intercept is 7.

6b. Sample equation: 0 � 2x2 � 3x � 2. There are nox-intercepts.

6c. Sample equation: 0 � 2x2 � 3x � 2. The x-intercepts are �0.5 and 2.

7. When b2 � 4ac is negative, there are no real roots.When it’s zero, there is one root. When it’s positive,there are two roots.7a. i; 12 � 4(1)(1) � �3; no x-intercept7b. iii; 22 � 4(1)(1) � 0; one x-intercept7c. ii; 32 � 4(1)(1) � 5; two x-intercepts8. The average of the roots is ��2a

b�; it is the

x-coordinate of the vertex of the parabola.9. You need to know the t-value when h � 0.

Solve the equation 0 � �4.9t2 � 17t � 2.2 with a � �4.9, b � 17, c � 2.2 to get t � �0.125 or t � 3.59. The positive solution, 3.59 s, makes sense inthis situation.

x

y

5

5

x

y

5

5

10

x

y

1050

5

10

�6.2 ��2.72��

�9.8

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10a. See below.10b. The perimeter does not change.10c. The area increases and then decreases.10d. A � (4 � x)(7 � x)10e. The largest area occurs at the vertex of theparabola, which is at x � 1.5. At this point therectangle has dimensions 5.5 m by 5.5 m, which meansthe rectangle is a square.11. (0.5, 4.25) and (�4, 2)12a. x � 2; x 312b. x � 6; x �112c. x � 1; x �0.5

12d. �xx

��

32�; x 5 and x �2

12e. �xx

��

44�; x �6 and x 4

13a.

[�15, 15, 1, �10, 10, 1]13b.

[�15, 15, 1, �10, 10, 1]

Increase (x) Width Length Area Perimeter(m) (m) (m) (m2) (m)

0 4 7

0.5

1.0

1.5

2.0

4.5

5

5.5

6

6.5

6

5.5

5

28

29.25

30

30.25

30

22

22

22

22

22

10a. (Lesson 9.7)


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LESSON 9.8

1a. perfect square; 472 � 2,2091b. perfect cube; 183 � 5,8321c. neither1d. perfect square; 1012 � 10,2012a. (2x)3 � 5,832; 2x � 18; x � 9 cm2b. (3.5x)3 � 21,952; 3.5x � 28; x � 8 cm2c. 2(2.3x)3 � 3,309; 2.3x � 11.83; x � 5.143a. 4x(x � 3) 3b. 2x(3x � 2)3c. 7x�2x3 � x � 3� 3d. 3x2�4x3 � 2x � 1�4a. quadratic4b. linear4c. exponential4d. cubic5a. y � 0.5(x � 4)(x � 2)(x � 1)5b. y � �(x � 2)(x � 1)2

6a. Answers will vary. Three possibilities are 02 � 03 � 0, 12 � 13 � 1, 82 � 43 � 64.6b. Answers will vary. (a3)2 � (a2)3 for any integer a.For example, (43)2 � (42)3 � 4096, which is both aperfect cube and a perfect square. Or enter Y1 � x3,Y2 � x2, and Y3 � x6 into your calculator and look at the table. Y3 will be Y1 squared and Y2 cubed.7a. If the width is w, the length is w � 6 and theheight is w � 2, so the volume is given by the equation V � w(w � 6)(w � 2).7b. Three solutions to the equation 47 � w(w � 6)(w � 2) are shown on the graph. However, only onesolution is a positive value of w. A table gives theanswer: w � 3.4. Widths greater than about 3.4 cmgive volumes greater than 47 cm3.

8a. true 8b. false8c. true 8d. true9a. x3 � 6x2 � 11x � 6

w

V

7�7 0

100

w � 3.4w � �3.3

V � 47

w � �4.2

9b. x3 � 3x2 � 4x � 1210a. 50 cm10b. 70 cm10c. 10 50 70 � 35,000 cm3

10d. The width is 45 cm, and the length is 65 cm; thevolume is 15 45 65, or 43,875 cm3.

10e. w � �120

2� 2x� � 60 � x

10f. l � �160

2� 2x� � 80 � x

10g. V � x(60 � x)(80 � x)10h. x � 0 cm, x � 60 cm, x � 80 cm; these x-valuesmake boxes with no volume.10i. height 22 cm, width 38 cm, and length 58 cm,or height 23.265 cm, width 36.735 cm, and length56.735 cm11a. 3x3 � 8x2 � x � 2011b. 2x � 411c. 2x � 311d. 2x2 � 4x � 312a. 11x3 � 2x2 � 2x � 1212b. 5x3 � 2x2 � 12x � 1212c. �6x2 � 13x � 2012d. �16x4 � 34x3 � 28x2 � 131x � 99

13a. �xx

��

24�; x �2, x 4, and x �4

13b. �x �

x4

�; x 0, x 2, x �2, and x 4

13c. �(2x

x��

33)2�; x �3

13d. �x��

31�; x 1 and x �1

14a. Answers will vary.14b. Japan is the country with the largest increase($16.48), and Mexico is the country with the leastincrease (�$0.13).14c.

[0, 25, 1, 0, 10, 1]The top box is 1980,the middle is 1990,and the bottom is2000.Possible answer: There is a much larger range in thedata for 2000 than for 1980.It appears that compensationchanged less between 1990 and 2000 than between 1980and 1990.Overall,the lowest compensation has notchanged much,but the top end has moved considerably.The median has moved upward as well.


CHAPTER 9 REVIEW

1a. false; (x � 3)(x � 8)1b. false; 2x2 � 4x � 51c. false; x2 � 6x � 91d. true2. Sample response: There is a reflection across the x-axis � y � �x2 and a vertical stretch by a factor of2 � y � �2x2. Finally, there is a translation left 5 units and up 4 units � y � �2(x � 5)2 � 4.3a. y � �(x � 2)2 � 3; vertex form3b. y � 0.5(x � 2)(x � 3); factored form4a. y � �3(x � 1.5)2 � 18.754b. y � �1.6(x � 5)2 � 305a. 2w � 9 � 0 or w � 3 � 0; w � �4.5 or w � 35b. 2x � 5 � 0 or x � 7 � 0; x � �2.5 or x � 76a. y � (x � 1)2 � 46b. sample answers:

y � (x � 1.5)�x � �13��, y � (2x � 3)(3x � 1)

7a. x2 � 6x � 9 � 13x2 � 6x � 22

x2 � 6x � 9 � 22 � 9(x � 3)2 � 31

x � 3 � ��31�x � �3 ��31�

7b. 3x2 � 24x � 27 � 03x2 � 24x � �27

x2 � 8x � �9x2 � 8x � 16 � �9 � 16

(x � 4)2 � 7x � 4 � ��7�

x � 4 � �7�

8a. x � ; no real number solutions

8b. x �

9a. f (60) � 8.1; when there are 60 fish in the tank,the population is growing at a rate of about 8 fishper week.9b. f (x) � 0 for x � 0 and x � 150; when there are nofish, the population does not grow; when there are 150fish, the number of fish hatched is equal to the numberof fish that die, so the total population does not change.9c. There are 75 fish when the population is growingfastest.9d. The population no longer grows once there are150 fish, so that is the maximum number of fish thetank has to support.

�7 � �157��6

13 � ��191��10

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9e.

[�10, 200, 10, �1, 10, 1]10. The roots are at 0 s and 1.6 s, so start with theequation y � x(x � 1.6). Then reflect the graph acrossthe x-axis. When x � 0.5, y � 0.55. You need the valueof y to be 8.8, so apply a vertical stretch with a factorof �0

8..585�, or 16. The final equation is y � �16x(x � 1.6).

11a. No x-intercepts means taking the square root ofa negative number. So (�6)2 � 4(1)(c) � 0; �4c � �36;c � 9. Or translate the graph of y � x2 � 6x verticallyto see that for c � 9, the parabola does not cross the x-axis.11b. One x-intercept implies a double root,so x2 � 6x � c must be a perfect-square trinomial.Make a rectangle diagram to find ��

�2

6��2

� 9, so x2 � 6x � 9 is a perfect-square trinomial, and c � 9. The graph touches the x-axis once. You can alsosolve b2 � 4ac � 36 � 4c � 0 to get c � 9.11c. For c � 9, b2 � 4ac � 0, so there are two realroots, and therefore two x-intercepts. The parabola y � x2 � 6x � c crosses the x-axis twice for values of cless than 9.12a. x � �5 � �31� and x � �5 � �31�

12b. x � 1 and x � �53�

13a. x � �2, x � �1, x � 1, and x � 3;y � 2(x � 3)(x � 2)(x � 1)(x � 1)13b. x � �2 (double root) and x � 3;y � �3(x � 2)2(x � 3)14a. (x � 3)(x � 4) 14b. (x � 7)2

14c. (x � 7)(x � 4) 14d. (x � 9)(x � 9)�9x

x x2 �9x

9 9x �81

�4x

x x2 �4x

7 7x �28

�7x

x x2 �7x

�7 �7x 49

3x

x x2 3x

4 4x 12


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Answers to All ExercisesChloe’s bag of candy had the same dominant color asthe graph from the manufacturer, and her leastcommon color was one of the least manufactured. Butthe distributions are not very close.7a. 9th: 27%; 10th: 26%; 11th: 25%; 12th: 22%7b. 9th: 189; 10th: 172; 11th: 170; 12th: 147; 0.3%;2 students7c.

The ninth grade increased to 28% and the tenth gradedecreased to 25%.8a. iii8b. i8c. iv8d. ii

9. �1664�� 79%

10. y � 2x2 � 3x � 511a. 180 pulses per second11b. 18.6� m/s

12th grade9th grade

10th grade 11th grade

22%

25%25%

28%

12th grade9th grade

27%

26%25%

22%

10th grade 11th grade

LESSON 10.1

1. type AB � 3,750; type B � 9,000;type A � 30,000; type O � 32,2502a. 40%2b. 43%2c. 95%3. c4. No; the total height of all the bars must be 100%.5.

6a.

6b.

Y BlO R G Br

Manufacturer’s candyChloe’s candy

Comparing Chloe’s Candy with the Manufacturer’s

Per

cen

t

30

25

20

15

10

5

0

Candy colors

Candy colors

Per

cen

t

Y GRBl BrO0

30

25

20

15

10

5

Chloe’s Candy Distribution

Orange(36°)

Yellow(72°)

Blue(36°)

Red(72°)

Green(36°)

Brown(108°)

CHAPTER 10 • CHAPTER CHAPTER 10 • CHAPTER10


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12a. See below. Douglas County had the largestpercent of growth.12b. Los Angeles County grew by 118,156, a changeof 1.2%.

12c. Los Angeles County grew by over 4.9 times morepeople than did Douglas County, but it increased byonly 1.2% rather than 13.6%. The population growthof Los Angeles County was more manageable from thepoint of view of the stress on the infrastructure, suchas roads, schools, and utilities.

2000 2001 Change from 2000 PercentCounty population population to 2001 growth

Douglas County, CO 175,766 199,753

Loudoun County,VA 169,599 190,903

Forsyth County, GA 98,407 110,296

Rockwall County, TX 43,080 47,983

Williamson County, TX 249,967 278,067

Fastest-Growing Counties between 2000 and 2001

(U.S. Census Bureau, www.census.gov)

12a. (Lesson 10.1)

23,987 13.6

12.6

12.1

11.4

11.2

21,304

11,889

4,903

28,100


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LESSON 10.2

1a. heads or tails1b. 1, 2, 3, 4, 5, or 61c. 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, or 121d. A, B, C, D, or E2a. See below. 2b. 0.55 2c. 0.652d. P(C) � 0.25. All probabilities would be the sameas for the original pack because the ratios wouldn’tchange.

3a. �12070� � 0.27

3b. P(landing in shaded area) � 0.25 because theshaded area is �

14� of the circle.

4. �18�

5. Answers will vary. The probability of 5f is 0 or verynearly 0; the probability of 5g is 1.6. Total area: 77 � 63 � 4851; �48

a51� � �

13

35

60�; a � 1885; the

area is about 1900 square units.7a. Finding and counting a litter is a trial; an outcomemay be having one cub (or two or three or four).7b. No; if outcomes were equally likely, then thenumber of litters of each size would have been almostthe same, with about nine litters of each size.

7c. �23

25� � 0.63

8a.

8b. �250�, or �

14�

8c. 0.75 � 4500, or 3375

9a. �18�

9b. �90

3�

60180� � �

23

76

00�, or �

34�

9c. �45

3�60

45� � �3

9600�, or �

14�

10a. See below.10b. Answers will vary. A routine that works for somecalculators is randInt(1,100,50).10c. Answers will vary. Sorting the list in ascendingorder is suggested.10d. Answers will vary.11. Outcomes: Cup lands on its bottom, cup lands onits top, cup lands on its side; answers will vary, but theoutcomes are probably not equally likely; flip or toss acup many times, record the results, and calculate theexperimental probability of each outcome.12. Answers could be any parallelogram withhorizontal sides; one answer might be (�4, 1), (�1, 3),(4, 3), (1, 1).13. 3

10 20 30 40 50 8060 70 90

Student responses

Outcomes

Vanilla Orange Strawberry Cinnamon Winter Total

Number 4 2 6 5 3 20

Theoretical probability 0.20 0.10 0.30 0.25 0.15

2a. (Lesson 10.2)

Outcomes

Red Orange Yellow Green Light blue Dark blue Total

Interval 1 to 12

Number of candies 50

Probability

10a. (Lesson 10.2)

13 to 30 31 to 45 46 to 63 64 to 75 76 to 100


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LESSON 10.3

1. Theoretical probability: �17840� � 0.411; experimental

probability: �15

50�, or 0.30. Possible answers: You can

expect a wide variation in survey results. Perhaps yourmethod of selecting students was not random. Forexample, your results could be biased because youtalked only to students who were participating inafter-school activities or only to students in aparticular class. Perhaps the question was worded insuch a way that students were biased in their responseor reluctant to answer it honestly.

2a. �3150000� � 0.0286

2b. You have to assume that the population is 3500, itremains stable (no fish die and no new fish hatch), andthe fish are well mixed.

2c. �1300�, or 0.03

3a. �21550�

3b. �22

35

50�

3c. �21550�, or 0.06

4. �13226�x � 25; x � 98

5a. H, H, T, H, H, T5b. Find the cumulative sum of list L1.

5c. After many steps, you may still be close to 0.6a. Answers will vary.6b. Answers will vary. Experiments using a tack witha 9 mm diameter head and a 6 mm point gave anexperimental probability of “point up” of about 0.55.6c. In one actual experiment, there was no changebetween a hard surface and a soft one.7. One possible routine is randInt(1,6). Assignstudents in order to groups 1 to 6, skipping a numberonce that group is full.

8a. x � 1.75 or x � 3.258b. x � 2 � �6�8c. 1.75 � x � 3.25, or x � 1.75 and x � 3.25. When1.75 � x � 3.25, the graph of the absolute-valuefunction y � 22x � 5� 4 is at or below the graph ofthe line y � 7.

[�9.4, 9.4, 1, �2.2, 10.2, 1]8d. x � 2 � �6� � 4.45 or x � 2 � �6� � �0.45. Whenx � 2 � �6� or x � 2 � �6�, the graph of the quadraticfunction y � �0.5(x � 2)2 � 7 is below the graph ofthe line y � 4.

[�9.4, 9.4, 1, �2.2, 10.2, 1]9. 22910a. 64

10b. �316�

10c. �196�

10d. 33

10e. ��15��

3

10f. ��29��

2

, or �23

2

4�

11. Answers will vary. You could cover the rectanglewith a grid, count the squares in the shaded area, andcompare that number to the total number of squaresin the rectangle. Or you could cover the area withbeans and compare the number of beans inside theshaded area to the total number of beans. Multiply theratio by 150,000 cm2.number of beans in shaded area

��total number of beans


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LESSON 10.4

1. combination, because the order doesn’t matter andno dish can be chosen more than once2a. combination2b. permutation2c. combination2d. neither, because repeats are allowed3a. 5 � 4 � 3 � 60

3b. �5

3

PP

3

3� � �

5

3��

4

2��

3

1� � 10

3c. 5 � 4 � 3 � 2 � 120

3d. �5

4��

4

3��

3

2��

2

1�� 5

4a and b.

4c. neither, because repetition of the objects isallowed

4d. �18�, or 0.125

4e. �14�, or 0.25

5a. Possible answer: I own eight pairs of shoes andam deciding which two pairs to take. I will wear onepair and pack the other. There are 56 ways to do this.5b. Possible answer: I have 12 pairs of socks and amdeciding which 4 pairs to take. The order in which I packthem doesn’t matter. There are 495 ways to do this.5c. Possible answer: I have six pairs of pants and amdeciding which two pairs to take. I can pack them ineither order. There are 15 ways to do this.6a. 7206b. 120

6c. �16� � 0.167

6d. �7120� � 0.001

D

T

D

D

D

D

D

T

T

T

T

T

D

T

6e. �77

21

09

� � 0.9997. The answer to 6e can be found by subtracting theanswer to 6d from 1. (That is, the answers to 6d and esum to 1.) The probabilities of complementaryoutcomes always sum to 1.8a. 1208b. 248c. permutations9a. 8! � 8 � 7 � 6 � 5 � 4 � 3 � 2 � 1 � 40,3209b. 10! � 3,628,8009c. nPn � n!10a. 3,307,80010b. 551,30011. 6C2 � 6C4 � 15; possible answer: Every time youselect a group of four students from six students, youare also selecting a group of two students who areleft out.12a. 20C6 � 38,76012b. 18C4 � 3060

12c. �338,,076600�, or 0.08

13a. �11080�, or 0.18

13b. �16�, or 0.17

13c. �15010�, or 0.51

13d. �16� � �

16� � �

16� � �

12�, or 0.5

13e. Yes, the results are close; the die is probably fair.14a. �3x3 � 3x2 � 5x � 1014b. x5 � 7x4 � 4x2 � 28x14c. 10x � 39y � 4xy15a. 20 units2

15b. 31.5 units2

15c. 94 units2

16a. x � ��23�, y � 0

16b. x � �1.3�, y � �0.583�

16c. x � 0, y � �1

17. �166�, or �

38�


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LESSON 10.5

1a. �18�

1b. This is the probability that Cheryl makes thesecond shot if she misses the first shot.2a.

2b. 1; the sum of the probabilities of all the outcomesshould be 1, because the probability that one of theoutcomes happens is 100%, or 1.3.

4.

5a. independent5b. dependent5c. independent6.

a b

0.75

0.8

e

f

c

d

0.5625

g

R

B

R

R

B

B

4_92_9

2_91_9

1_3

2_3

1_3

2_3

1_3

2_3

R

B

R

R

B

B

1_31_3

1_3

2_3

1_3

1_2

1_2

1

00

0.7

0.8

b

d

e

f

g

c

0.3

a

7.

8. Dependent; the first student selected affects theprobabilities for the second choice.

9a. �14�

9b. �14�

9c. �116�

10a. 5040

10b. �17�

10c. �50140�

10d. �55

00

43

09

�

10e. �17�

11a. 15,890,700

11b. �15,8910,700� � 0.00000006

11c. P(W) � �1581,907� � 0.000006;

P(L) = �1155

88

,,99

00

67� � 0.999994

11d. See below.11e. P(L for 52 weeks) � ��115

588

,,99

00

67��

52

� 0.999673

11f. $5,200

P (F1) � 2_3

P (F2 | F1) � 13__20 P(F1 and F2) �

13__30

P(M1 and F2) � 7__

30

P(M1 and M2) � 1__

10

P(F1 and M2) � 7__

30

P (F2 | M1) � 7__

10

P (M2 | F1) � 7__

20

P (M2 | M1) � 3__

10

P (M1) � 1_3

0.3

0.2

0.7

0.560.14

0.210.09

0.250.2

0.75

0.25

0.20.05

0.1875

P(L for four weeks) � ��1155

88

,,99

00

67��

4

� 0.999975

158,906______158,907

158,906______158,907

158,906______158,907

158,906______158,907

1______158,907

1______158,907

1______158,907 W

L

W

L

W

L

W

L 1______158,907

11d. (Lesson 10.5)


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12a. Possible answers: Pupil diameter decreases as aperson gets older. Pupils are larger in the dark. Pupildiameter decreases faster in darkness than in daylight.As a person ages, daylight and darkness pupildiameters get closer.

12b.

[0, 80, 10, 0, 10, 1]

12c. lines through Q-points: for daylight,d � 4.3 � 0.04(a � 30); for darkness,d � 7.0 � 0.097(a � 30)12d. (77.37, 2.41); at about age 77, a person’s pupildiameter stays the same in daylight and darkness,about 2.41 mm.

13. Answers will vary. About 90% of the population isright-handed. The theoretical chance of a single headis four in sixteen, or �

14�.

14. She can give A’s to seven students.

13 othergrades

7 A’s

9 othergrades 11 A’s


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LESSON 10.6

1. See below.2a.

2b. The expected number of red marbles drawn is �43�,

or about 1.3.

3. The expected value for concert income is $142,500.

4a. 29 4b. �229�, or 0.07 4c. 151.6 cm

5a. $5.505b. Yes, it is a good deal, because over the long runyou’ll come out $0.50 ahead per spin.6a.

1_3

1st draw 2nd draw1_2P (R2 | R1) �

2_3P (R1) �

P (B1) �

1_2P (B2 | R1) �

2_2P (R2 | B1) �

0_2

P (B2 | B1) �

1_3

P (R1 and B2) �

1_3

P (R1 and R2) �

1_3

P (B1 and R2) �

0P (B1 and B2) �

6b. 77a. P�G1 and G2� � 0.6

7b. 1.4757c. 7.375

8a. $9 8b. �16� 8c. $15.17

8d. The store will receive a mean payment of $15.17for each DVD player sold.9a. $4,750 9b. $6,00010a.

10b. 311a. 48

11b. �14

08� � 0.208

11c. �14

08� � 0.208

11d. �12

66� � 0.615

12a. 2512b. (0.7)25 � 0.00013

1__16

1_8

1_8

1_4

1_4

. . .

. . .

Meal

1_2P (A) �

1_2P (A) �

1_2P (A) �

1_2P (A) �

1_2P (A) �

1_2P (A) �

1_2P (B) �

1_2P (B) �

1_2P (B) �

1_2P (B) �

1_2P (B) �

1_2P (B) �

0.75

0.8G

G

M

M

G

M

0.5

0.2

0.5

0.25

1st throw 2nd throw

The expected value is $4.83.

Outcome 0 1 2

Probability 0 �23� �

13� Sum

Product 0 �23� �

23� �

43�

Outcome $200,000 �$30,000

Probability 0.75 0.25 Sum

Product 150,000 �7,500 142,500

1 2 3 4 5 6

1 2 3

2 3

3

4

5

6

Firs

t D

ie

Second Die

4 5 6 7

4 5 6 7 8

4 5 6 7 8 9

5 6 7 8 9 10

6 7 8 9 10 11

7 8 9 10 11 12

Outcome $2 $5 $10

Probability �13� �

12� �

16� Sum

Product 0.6� 2.5 1.6� 4.83�

1. (Lesson 10.6)


CHAPTER 10 REVIEW

1a. �499

9�

1b. �39

39�, or �

13�

1c. �199

9�

1d. �999�, or �1

11�

2. 105 students have blue eyes, about 52 or 53 havegray eyes, 70 have green eyes, and about 122 or 123have brown eyes.3. Degrees for each sector, rounded to the nearestdegree: China 75°, India 60°, United States 16°,Indonesia 13°, Brazil 10°, Other 185°; degrees add upto less than 360° because of rounding.

4a. $254b. 6%, or 0.064c. One person is $500 ahead, 5 people are $100ahead, 10 people are even, and 84 people are $25behind. This is a net loss of $1,100, or $11 per person.

5a. 12.5 cm2; �1420.5�, or 0.3125

5b. 32.5 cm2; �3425.5�, or 0.72�

India

China

Other

16.7%20.7%

51.5%

United States 4.5%

Indonesia 3.7%

Brazil 2.8%

6. 158,184,0007a. 4 � 3 � 2 � 1, or 247b. permutation, because the order is important andno person can have more than one role7c. 188a. 4495

8b. �4443955� � 0.097

9a.

9b. 0.001156

10a. �23

66�, or 0.72�

10b. �13

06�, or 0.27�

10c. �23

66�(3) � �

13

06�(0) � 2.16�

10d. 21.6�

10e. Nozomi, by 5 points

Twins

Twins

Not twins

Not twinsTwins

Not twins

0.068

0.017

0.017

0.932

0.983

0.983

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Answers to All Exercises2c. parallel

[�9.4, 9.4, 1, �6.2, 6.2, 1]

2d. perpendicular

[�9.4, 9.4, 1, �6.2, 6.2, 1]3. 1.2

4a. ��1.2

1� � ��

56�� 0.83�

4b. �1

5. y � ��13�(x � 8) � 2

6. See below.

7. right trapezoid; slopes: �23�, ��

15�, �

23�, ��

32�

8. rectangle; slopes: �12�, �2, �

12�, �2

LESSON 11.1

1a. 0.8 1b. �21c. �1.25 1d. 2

1e. �32� 1f. ��

32�

1g. �32� 1h. �

23�

2a. perpendicular

[�9.4, 9.4, 1, �6.2, 6.2, 1]

2b. neither

[�9.4, 9.4, 1, �6.2, 6.2, 1]

CHAPTER 11 • CHAPTER CHAPTER 11 • CHAPTER11

parallelogram

right trapezoid rectangle trapezoid

A

B E

C D

6. (Lesson 11.1)

square


9. trapezoid; slopes: 0, ��43�, undefined, ��

43�

10. parallelogram; slopes: 1, �3, 1, �3

11. quadrilateral; slopes: 4, ��14�, �3, �

47�

12. trapezoid; slopes: 2, ��52�, ��

14�, ��

52�

13. rectangle; slopes: �32�, ��

23�, �

32�, ��

23�

14. quadrilateral; slopes: 1, �1, �53�, ��

53�

15. possible answer:

x

y

4–4

8

(–2, 5) (2, 5)

Slope � 2Slope � �2

(0, 1)

Slope � �1_2Slope � 1_

2(0, 6)

16a. deductive; hypothesis: Tyrannosaurus rex hadsharp teeth; conclusion: Tyrannosaurus rex was acarnivore.16b. inductive; hypothesis: two consecutive numbersare added; conclusion: the result is an odd number.16c. deductive; hypothesis: 2(x � 3) � 10;conclusion: x � 8.17a. 2x3 � 3x2 � 2x17b. 0.01x2 � 4.4118a. 718b. 1.518c. �6.518d. 24.5

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LESSON 11.2

1a. (0.5, 1.5)1b. (6, �4.5)

2. ��a �2

c�, �

b �2

d��

3a. (16, 9)3b. (12.5, 5)3c. (0.5, 3)4a. possible answer: y � 9 � 6(x � 16)4b. possible answer: y � 3 � �

15

65�(x � 0.5)

5. Yes. Possible answer: The slope of AB� is 5 and theslope of BC� is ��

15�, so angle B is a right angle.

6. One line is horizontal and the other is vertical. Ahorizontal line has slope 0 and a vertical line has anundefined slope, so the product is also undefined.7a. (3, 3.5)7b. y � 3.5 � �

25�(x � 3)

8a. midpoint of AB�: (10, 3); midpoint of BC�: (15, 8);midpoint of CD��: (9, 10); midpoint of DA��: (4, 5)8b. Parallelogram; the opposite sides are parallelbecause the slopes are 1, ��

13�, 1, and ��

13�.

8c. No; the slopes of the diagonals are �131� and �7.

9a. D(7.5, �1), E(�1, �1), F(2.5, 6)

9b. slope of median AE� : �172�; slope of median BF� : ��

238�;

slope of median CD��: ��12

47�

9c. possible answers:

median AE� : y � 6 � �172�(x � 11);

median BF� : y � 6 � �238��x � �

52��;

median CD��: y � 6 � �12

47�(x � 6)

x

y

4

–4

–8

–4 10

B

AFC

E D

9d. �3, �43��

9e. �3, �43��

9f. Possible answer: The medians of a triangle meet ata single point.9g. inductive10a. (12, 4)10b. (3, 5)10c. (�5, 3.5)

11a. �AB

�

11b. ��AB�

12a. (�1, 3)12b. possible answer: x � y � 2 and y � 2x � 5 orany line of the form y � 3 � m(x � 1), where m is anynumber, or x � �112c. Possible answer: y � (x � 1)2 � 3; any parabolaof the form y � a(x � 1)2 � 3 will have its vertex atthis point.13a. possible answers:

13b. Answers will vary. Any method that uses onlyone line segment will form congruent polygons. Someother methods may also produce congruent polygons.13c. Yes; if you imagine a vertical segment through theupper vertex of the triangle, you can see that thetriangles on either side of the vertical line are congruent.

Same area

Same area

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LESSON 11.3

1a. ��47�1b. 4 � �28�1c. �2 � �14�1d. 1 � �7�2a. �6.8562b. �1.292, 9.2922c. �5.742, 1.7422d. �1.646, 3.6463a. 4 units2

3b. 12 units2

3c. 2 units2

3d. 6 units2

3e. 20 units2

3f. 18 units2

4. �18� units5. polygon 3a: �8� units and �2� unitspolygon 3b: �8� units and �18� unitspolygon 3e: �50� units, �50� units, and �40� units6. Each triangle has area 18 units2.7a. 36 units2, 18 units2, 18 units2

7b. length of AB�: 6 units; length of BC�: �18� units;length of AC�: �18� units8. (6, 11) and (12, 7), or (�2, �1) and (4, �5)9a. AC� is the hypotenuse; AB� and BC� are the legs.9b.

9c. 20 units2, 16 units2, 4 units2

9d. length of AB�: 2 units; length of BC�: 4 units; lengthof AC�: �20� units9e. The areas of the two smaller squares sum to thearea of the larger square.10a. City A: 58,571; City B: 52,72010b. in 27 years10c. 63,18611. y � 2(x � 7)(x � 1)

–1 7x

y

7

–3

B C

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LESSON 11.4

1. 576 cm2

2. a � 24 cm, b � 45 cm, c � 51 cm3. b � �300� cm4a. the two legs4b. c � �128� cm5a. 13.5 ft5b. 22.5 ft5c. 2160 ft2

6. Al is right. Answers will vary. As an example,�9 � 4� does not equal �9� � �4�.7a. approximately 21.6 cm by 27.6 cm7b. approximately 35.0 cm7c. Answers will vary but should be close to 35 cm.7d. Answers will vary. The two results should beapproximately the same.8. Answers will vary. If x � 1, then Miya is claimingthat 12 � 42 � 52, but 17 does not equal 25. You have to isolate x before you take the square root:x2 � 16 � 25 ⇒ x2 � 9 ⇒ x � �3.9. approximately 7010 ft10a. right triangles: i, ii, iv; not a right triangle: iii10b. Yes, the theorem works in reverse.11a. L1 � {1, 2, 3, . . . , 26}11b. approximately 26.98 in.11c. L2 � ��272 � L12�11d. approximately 53.85 in.2

11e. L3 � L1 * L2, or L3 � L1 * ��272 � L12�

11f. A model that works is y � x�272 ��x2�, where x isthe width and y is the area.

[0, 27, 5, �50, 450, 50]11g. Trace the graph or use a table to find that a 19-by-19 in. square gives maximum area.12a. Possible answer: The ratios �

150�, �

21

42�, and �

21

63� all equal

2, so the sides are proportional and the triangles aresimilar.12b. Yes. Possible answer: Similar figures havecongruent angles. Another possible answer: Thenumbers satisfy the Pythagorean Theorem.13a.

13b. �12

.

.71� � �8

x.5�; x � 6.88 or 6.9 m high

13c. Measure the length of the tree’s shadow and writea proportion using a person’s height and the length ofhis or her shadow.14. (x � 3)(x � 2) � 210, x2 � 5x � 6 � 210,x2 � 5x � 204 � 0, x � �17 or x � 12; the originaldeck is 12 ft on a side.

2.1 m

8.5 m

1.7 m

x

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LESSON 11.5

1a. 3�3�1b. 5�2�1c. 2 � �6�1d. 4�5� � 5�2�1e. 6�6�1f. 2�42�1g. �5�1h. 202a. a � �91�2b. b � �10�2c. c � 42d. d � �13�3a. y � x2 � 3

[�9.4, 9.4, 1, �6.2, 6.2, 1]

3b. y � x2 � 2x�5� � 5

[�9.4, 9.4, 1, �6.2, 6.2, 1]

4a. x � ��3� � �1.7324b. x � ��5� � �2.236

5a. (0, �3)5b. ��5�, 0� � (�2.236, 0)6a. y � x2 � 112

[�120, 120, 10, �120, 120, 10]

6b. y � 2x2 � 2x�6� � 72

[�100, 100, 10, �100, 100, 10]

6c. y � x2 � 6x � 7

[�9.4, 9.4, 1, �6.2, 6.2, 1]

7a. �4�7� � �10.5837b. 2�6� � 4.899 and �3�6� � �7.3487c. �3 � �2� � �4.414 and �3 � �2� � �1.586

8a. vertex: (0, �112)

8b. � , �75� or � (�1.225, �75)

8c. (�3, �2)

��6��2

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9a. �112� 9b. �550�9c. �972� 9d. �4500�10a. 6�2� 10b. 3�3�10c. 30�2� 10d. 7�3�

11a. ��5 �

3�3�� 11b. 3 ��2� 11c. 1, �2

12a. The red line creates two right triangles, eachwith leg length �

12�(8) cm, or 4 cm, and hypotenuse

length 8 cm. By the Pythagorean Theorem, the lengthof the other leg is 4�3� cm.12b. 4�2� cm 12c. approximately 566 ft13. See below. 14. See below.15a. 5 cm 15b. 13 cm16a. 3 16b. 5 16c. 4 16d. m16e. x25y4

17. a � 2�2� cm, b � 2�3� cm, c � �8 �3� �� 12� cm

18a. 93 mm2

18b. 92.16 mm2; this is about 1 mm2 too low.

18c. 744 mm3 and 737.28 mm3; the first volumecalculation is accurate, while the second is almost7 mm3 too low.18d. 744,000 mm3 and 737,280 mm3; the first volumecalculation is accurate, while the second is 6,720 mm3

too low.18e. 743,436.8 mm3

18f. When you estimate a square root value and thenuse that value in calculations,if you estimate with moreprecision,the errors in the calculated values are smaller.19a. y � (x � 2)2 � 3 19b. y � �x2 � 419c. y � 3(x � 1)2

20a. no x-intercept 20b. two x-intercepts20c. one x-intercept

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0 � 0.5x2 � 6x � 8 The original equation.

0 �? 0.5�6 � �20��2� 6�6 � �20�� 8 Substitute 6 � �20� for x.

0 �? 0.5�6 � �20��2� 36 � 6�20� � 8 Distribute the �6 over 6 � �20�.

0 �? 0.5�36 � 6�20� � 6�20� � 20� � 36 � 6�20� � 8 Square the expression.

0 �? 18 � 3�20� � 3�20� � 10 � 36 � 6�20� � 8 Distribute the 0.5 over the expression in parentheses.

0 �? 18 � 10 � 36 � 8 Combine the radical expressions.

0 � 0 Add and subtract.

14. (Lesson 11.5)

0 � 0.5x2 � 6x � 8 Original equation.

0 �?

0.5�6 � �20��2 � 6�6 � �20�� 8 Substitute 6 � �20� for x.

0 �?

0.5�6 � �20��2 �36 � 6�20� � 8 Distribute the �6 over 6 � �20�.

0 �?

0.5�36 � 6�20� � 6�20� � 20� � 36 � 6�20� � 8 Use a rectangle diagram to square the expression6 � �20�.

0 �?

18 � 3�20� � 3�20� � 10 � 36 � 6�20� � 8 Distribute the 0.5 over the expression inparentheses.

0 �?

18 � 10 � 36 � 8 Combine the radical expressions.

0 � 0 Add and subtract.

366

6

20

20

6 20

20 6 20

13. (Lesson 11.5)


LESSON 11.6

1. no, because 92 � 162 � 252

2a. possible answer:

2b. horizontal length: 5 units; vertical length: 8 units

2c. �89� units3a. 53b. �65� � 8.063c. �(2 � s)�2 � t2�, or �(s � 2)�2 � t2�4. possible answer: (6, 3) and (1, 7)

5a. slope of AB� and DC��: �45�; slope of AD�� and BC� : ��

23�

5b. parallelogram5c. length of AB� and DC��: �41� units; length of AD��and BC� : �13� units6a. �26� units, or approximately 0.5 mi6b. 5 units, or 0.5 mi7a. possible answer: y � �3 � �

43�(x � 2)

7b. d � �(x � 2�)2 � (y�� 3)2�

7c. d � �(x � 2�)2 � ��3 � �43�(�x � 2)� � 32�

� �(x � 2�)2 � ��43�(x �� 2)�

2�7d. 6 � �(x � 2�)2 � ��

43�(x �� 2)�

2�;

x � �1.6 or x � 5.6

x

y

(�2, �3) (3, �3)

(3, 5)

7e. (�1.6, 1.8)8. 10�38� cm9a. �20 � x�� x

20 �x � x2

0 � x2 � x �200 � (x � 5)(x � 4)x � �5 or x � 4

9b. Possible answer: The intersection of the graphs ofY1 � ��(20 � x) and Y2 � x occurs once, at x � 4.

[�9.4, 9.4, 1, �6.2, 6.2, 1]9c. x � 4 is a solution; x � �5 is not a realisticsolution because a square root cannot be negative.10a. a � 6310b. b � 710c. c � 3

11a. �11

32� � �

a8�, a � 8.6�; 8 ft 8 in. long

11b. �152� � �

148� b�, b � 10�

23�; 10 ft 8 in. tall

12a. 10�2�12b. 6�17�12c. 3�5�12d. 9�3�

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LESSON 11.7

1a. x � 271b. x � 351c. x � 1601d. x � �42. 180 mi

3a. sinD � �275�

3b. cos E � �275�

3c. tan D � �274�

4a. Yes. Possible answer: Angle A is common to bothtriangles and angles B and D are both right angles, soangles C and E must also be congruent. Because allthree angles are congruent, the triangles are similar.

4b. �84� � 2

4c. 6 cm and 10 cm4d. 24 cm2 and 6 cm2

4e. �264� � 4 � (2)2

5a. x � 125b. x � 145c. x � 356a. �60� m, or approximately 7.75 m

6b. �28� m, or approximately 5.29 m7a. cosine7b. sine7c. tangent

7d. cos 65° � 0.4226; sin 65° � 0.9063;tan 65° � 2.14457e. approximately 5.6 m8a. tan 28° � �x

y�, or y � x � tan 28°

8b. The graph is a direct variation.

[0, 200, 25, 0, 125, 25]8c. y � 53.28d. x � 1519a.

9b. See below.10a and b.

2 2

30°

60°60°

30°

1 1

3

1

45°

45°1

2

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Trigonometric Functions for a 45° Angle

Sine Cosine Tangent

Exact value of ratio

Decimal approximation of exact value

Value by trigonometric function keys

0.7071

0.7071

1��2�

0.7071

0.7071

1��2�

�11�

1.0000

1.0000

9b. (Lesson 11.7)


10c. See below.11a. rectangle

–8 8x

y

7

–5

11b. (1.2, 1.4), (2.4, 3.8), (0.8, 4.6), (�0.4, 2.2)

11c. y � 11 � 8x, y � �177� � �

47�x

11d. (1, 3)12. a � 26

b � 13�2� � 18.38c � 13�3� � 13 � 9.52

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10c. (Lesson 11.7)


Sine Cosine Tangent





Sine Cosine Tangent




�12

�

0.5000

0.5000

0.8660

0.8660

�3��

2

0.5774

0.5774

1��3�

0.8660

0.8660

�3��

2�12

�

0.5000

0.5000

1.7321

1.7321

�3��

1


LESSON 11.8

1a. d1b. A1c. A

1d. �dc�

1e. A1f. A2a.

2b. Possible answer:

A � sin�1��35�� 37°,

B � sin�1��45�� 53°

2c. The angles should measure approximately 37°and 53°.3. x � 44.6 m4. b � 14.6

5. See below.

b

6.8

25°

3

45

A

CB

6a. 17 cm6b. 60 cm2

6c. approximately 28°6d. approximately 62°6e. 180°7. approximately 81.1 m8. Answers will vary. An average rise-to-run ratiofor stairs is �1

72�. The answers provided are based on this

ratio.8a. approximately 30°8b. rise: 7; run: 12

8c. �172�

8d. tan�1��172�� 30.26°

8e. 0.583�, or �172�

9a. approximately 2.86°9b. about 148 ft10a. 68.7 cm2

10b. 59.7 cm2

11a. approximately 19 ft11b. approximately 51 ft11c. approximately 70 ft

12. 13. y � 1.5x2 � 3x � 4.5

y � 2x � �3y ��

83� � �

13�x

y � �2 � x

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e

dc

b

a

H

F

20 cm

2 cm

17 cm

12 cm

18 cm

5 cm

28°

64°9.4 cm

17.7 cm

4.6 cm

4.1 cm

44.9°

15.5°

�145� � 12.0 cm

�349� � 18.7 cm

5. (Lesson 11.8)


CHAPTER 11 REVIEW

1a. 8�5�1b. 4�17�1c. 123�3�1d. �15�1e. 801f. 17001g. 3�10�1h. 80�5�1i. �2� � �3�1j. 3�2�1k. �6�1l. 4�3�2. The area is 5 square units. Here are two possiblestrategies:i. Draw a square around the tilted square using thegrid lines. Subtract the area of the outer triangles fromthe area of the larger square: 9 � 4(1) � 5.ii. Find the length of the side between (1, 0) and (3, 1): �(3 � 1�)2 � (1� � 0)2� � �5�. Square the side length to find the area: ��5��2 � 5.3. The slopes of the sides are �

12�, �2, �

12�, and �2. The

slopes of each pair of adjacent sides are oppositereciprocals, so the sides are perpendicular. Answerswill vary. Possible hypothesis: The given figure is asquare. Possible conclusion: Its sides areperpendicular.4. Possible answer: Draw a 7-by-7 square on graphpaper and remove triangles with areas of 5 squareunits (legs 2 units and 5 units) from each corner.The area of the remaining square is 49 � 4 � 5, or29 square units.5. Possible answer: Sides of length 5 ft, 12 ft, and 13 ftsatisfy the Pythagorean Theorem and form a righttriangle. Side lengths of 10 ft, 24 ft, and 26 ft, whichsum to 60 ft, also form a right triangle. Stretch 10 ft ofthe rope along one wall and 24 ft of the rope along theadjacent wall; the remaining 26 ft of rope shouldexactly fit along the hypotenuse if the foundationcorners are right angles.6a. A(�4, 2), B(0, 5), C(6, �3), D(2, �6)6b. slope of AB�: �

34�; slope of BC� : ��

43�; slope of CD�� : �

34�;

slope of AD�� : ��43�

6c. It is a rectangle; the product of the slopes ofadjacent sides is �1, so each pair of adjacent sides isperpendicular.

6d.

6e. Each side length is �31.25� units, orapproximately 5.59 units.6f. The slopes are �0.5, �5.5, �0.5, and �5.5.6g. It is a rhombus; the sides are all the same lengthand opposite sides have equal slope, so they areparallel.7. a � 3.38 m

b � 7.25 m8a. sample answer:

8b. approximately 23°8c. sin�1��1

53�� 23°, cos�1��

11

23�� 23°, and

tan�1��152�� 23°

8d. Possible answer: Subtract 23° from 90°;approximately 67°.9a. �116� cm, or approximately 10.77 cm9b. �141� cm, or approximately 11.87 cm10. Possible answer: If a triangle has base 8 cm andheight 4 cm, its area is 16 cm2. If the triangleis enlarged by a factor of 3, its base will be 24 cm, itsheight will be 12 cm, and its area will be 144 cm2,which equals 32 � 16 cm3. For any triangle, the area isgiven by A � �

12�bh. If the sides are enlarged by a factor

of k, the area is enlarged by k2: A � �12�(kb)(kh), or

A � �12�bh � k2.

11a. y � 61 � 1.08(x � 40) or y � 34 � 1.08(x � 15)11b. approximately 51°F11c. approximately 38°F

12a. , where a is the price per

pound for dried apricots and p is the price per poundfor dried papaya

12b. apricots: $2.79; papaya: $3.58

3a � 1.5p � 13.742a � 3p � 16.32

x

y

5

12

13

x

y

4

6

–4

–6

42–4

2A

B

C

D

6–6

(�2, 3.5)

(3, 1)

(�1, �2)

(4, �4.5)

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13a. P(0) � �210�, or 0.05

13b. P(less than zero) � �230�, or 0.15

13c. �18�, or 0.125

13d. �129�, or about 0.105

14a. Inverse variation. Possible explanation: Theproduct of x and y is constant; xy � 2 or y � �x

2�.

14b. Neither. Possible explanation: The product is notconstant, so it is not an inverse variation. The y-valuefor x � 0 is not 0, so it is not a direct variation.14c. Direct variation. Possible explanation: The ratioof y to x is constant; y � 0.25x.14d. Neither. Possible explanation: The graph is not acurve, so the relationship is not an inverse variation.The line does not pass through the origin, so it is not adirect variation.14e. Inverse variation. Possible explanation: Theproduct of the x- and y-coordinates for any point onthe curve is 8; xy � 8 or y � �x

8�.

14f. Direct variation. Possible explanation: The graphis a straight line through the origin; y � 1.5x.15a. Possible answer: For 0 � x � 3, f is nonlinearand increasing at a slower and slower rate. For 3 � x � 5, f is linear and decreasing. For 5 � x � 7, fis linear and increasing. For 7 � x � 9, f is linear andconstant (neither increasing nor decreasing). For 9 � x � 12, f is nonlinear and decreasing at a slowerand slower rate.15b. 0 � y � 5 15c. 315d. 1, 5, 12 15e. 7 � x � 916a. 27x6y3 16b. 5p4q2

16c. �yx

2� 16d. �mn4

2� � �

m1

4�

17a. mean: 108.4; median: 105; mode: 10517b. five-number summary: 82, 99, 105, 112, 179

17c. Bin widths may vary.

17d. Sample answers: (1) About 75% of the newreleases have running times of 112 min or less.(2) None of the new releases have running timesbetween 140 and 169 min. (3) Most of the runningtimes are between 90 and 119 min.

10080 120 140 160 180

Running time (min)

Freq

uen

cy

1

0

2345678

Running time (min)18016014012010080

18a. y � (x � 3)(x � 1)18b. y � (x � 1)2 � 418c. y � x2 � 2x � 319a. approximately $1,19719b. approximately $4,10220a. 2620b. 220c. 3620d. 12820e. �2421a. $62.3921b. $65.5122a. possible answer: a reflection across the y-axisand a vertical shrink by a factor of 0.522b. (�x, 0.5y)23a. x � �5 or x � 223b. x � �423c. x � �3 or x � 1023d. x � ��5�24a. y � (x � 2)2 � 424b. y � �0.5x � 325a. $3525b. $22525c. {0, 225} ; {Ans(1) � 1, Ans(2) � 35}

, , . . .25d. y � 225 � 35x25e. $64525f. 826a.

26b. Isosceles triangle; two sides have equal length.26c. D(2, 1)26d. Right triangles. Possible explanation: BD� has anundefined slope, so it is vertical; AC� has slope 0, so itis horizontal.26e. A drawing should confirm 26a–d.

x

y

–5 5 10

5

10

A CD

B

ENTERENTER

ENTER

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Segment Length Slope

AB� 10 �34�

BC� 10 ��34�

AC� 16 0

Documents

Answers to All Exercises - Issaquah Connectconnect.issaquah.wednet.edu/.../ALG-1-Book-Answers.pdf6 ANSWERS TO ALL EXERCISES LESSON 0.5 1a. 8.0 cm 1b. 4.3 cm 1c. 7.2 cm 2a. 2.8 cm segment