SOLUTIONS MANUAL for A SURVEY OF MATHEMATICS WITH APPLICATIONS NINTH EDITION. Allen R. Angel Christi by welldoneassistant

Table of Contents CHAPTER 1 CRITICAL THINKING SKILLS 1.1 Inductive Reasoning 1 1.2 Estimation 3 1.3 Problem Solving 5 Review Exercises 12 Chapter Test 14 Group Projects 16 CHAPTER 2 SETS 2.1 Set Concepts 17 2.2 Subsets 19 2.3 Venn Diagrams and Set Operations 21 2.4 Venn Diagrams With Three Sets and Verification of Equality of Sets 2.5 Applications of Sets 37 2.6 Infinite Sets 40 Review Exercises 41 Chapter Test 44 Group Projects 46 CHAPTER 3 LOGIC 3.1 Statements and Logical Connectives 47 3.2 Truth Tables for Negation, Conjunction, and Disjunction 3.3 Truth Tables for the Conditional and Biconditional 56 3.4 Equivalent Statements 64 3.5 Symbolic Arguments 73 3.6 Euler Diagrams and Syllogistic Arguments 80 3.7 Switching Circuits 82 Review Exercises 85 Chapter Test 92 Group Projects 94

CHAPTER 4 SYSTEMS OF NUMERATION 4.1 Additive, Multiplicative, and Ciphered Systems of Numeration 4.2 Place-Value or Positional-Value Numeration Systems 98 4.3 Other Bases 101 4.4 Computation in Other Bases 108 4.5 Early Computational Methods 111 Review Exercises 115 Chapter Test 119 Group Projects 121

CHAPTER 5 NUMBER THEORY AND THE REAL NUMBER SYSTEM 5.1 Number Theory 123 5.2 The Integers 127 5.3 The Rational Numbers 129 5.4 The Irrational Numbers and the Real Number System 135 5.5 Real Numbers and Their Properties 138 5.6 Rules of Exponents and Scientific Notation 140 5.7 Arithmetic and Geometric Sequences 144 5.8 Fibonacci Sequence 148 Review Exercises 150 Chapter Test 153 Group Projects 153

CHAPTER 6 ALGEBRA, GRAPHS, AND FUNCTIONS 6.1 Order of Operations 155 6.2 Linear Equations in One Variable 157 6.3 Formulas 166 6.4 Applications of Linear Equations in One Variable 176 6.5 Variation 182 6.6 Linear Inequalities 187 6.7 Graphing Linear Equations 193 6.8 Linear Inequalities in Two Variables 203 6.9 Solving Quadratic Equations by Using Factoring and by Using the Quadratic Formula 209 6.10 Functions and Their Graphs 215 Review Exercises 226 Chapter Test 238 Group Projects 241 CHAPTER 7 SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES 7.1 Systems of Linear Equations 243 7.2 Solving Systems of Equations by the Substitution and Addition Methods 250 7.3 Matrices 257 7.4 Solving Systems of Equations by Using Matrices 265 7.5 Systems of Linear Inequalities 268 7.6 Linear Programming 271 Review Exercises 275 Chapter Test 280 Group Projects 283 CHAPTER 8 THE METRIC SYSTEM 8.1 Basic Terms and Conversions Within the Metric System 285 8.2 Length, Area, and Volume 286 8.3 Mass and Temperature 288 8.4 Dimensional Analysis and Conversions to and from the Metric System Review Exercises 295 Chapter Test 297 Group Projects 298 CHAPTER 9 GEOMETRY 9.1 Points, Lines, Planes, and Angles 301 9.2 Polygons 305 9.3 Perimeter and Area 310 9.4 Volume and Surface Area 314 9.5 Transformational Geometry, Symmetry, and Tessellations 9.6 Topology 322 9.7 Non-Euclidean Geometry and Fractal Geometry 323 Review Exercises 325 Chapter Test 329 Group Projects 330 CHAPTER 10 MATHEMATICAL SYSTEMS 10.1 Groups 331 10.2 Finite Mathematical Systems 332 10.3 Modular Arithmetic 336 Review Exercises 340 Chapter Test 343 Group Projects 345

318

290

CHAPTER 11 CONSUMER MATHEMATICS 11.1 Percent 347 11.2 Personal Loans and Simple Interest 350 11.3 Compound Interest 353 11.4 Installment Buying 356 11.5 Buying a House with a Mortgage 363 11.6 Ordinary Annuities, Sinking Funds, and Retirement Investments 367 Review Exercises 368 Chapter Test 372 Group Projects 373 CHAPTER 12 PROBABILITY 12.1 The Nature of Probability 375 12.2 Theoretical Probability 376 12.3 Odds 379 12.4 Expected Value (Expectation) 382 12.5 Tree Diagrams 386 12.6 Or and And Problems 390 12.7 Conditional Probability 394 12.8 The Counting Principle and Permutations 397 12.9 Combinations 399 12.10 Solving Probability Problems by Using Combinations 12.11 Binomial Probability Formula 406 Review Exercises 407 Chapter Test 410 Group Projects 412 CHAPTER 13 STATISTICS 13.1 Sampling Techniques 413 13.2 The Misuses of Statistics 413 13.3 Frequency Distributions and Statistical Graphs 13.4 Measures of Central Tendency 423 13.5 Measures of Dispersion 427 13.6 The Normal Curve 433 13.7 Linear Correlation and Regression 439 Review Exercises 450 Chapter Test 455 Group Projects 457 CHAPTER 14 GRAPH THEORY 14.1 Graphs, Paths, and Circuits 459 14.2 Euler Paths and Euler Circuits 462 14.3 Hamilton Paths and Hamilton Circuits 14.4 Trees 472 Review Exercises 483 Chapter Test 488 Group Projects 489

415

466

CHAPTER 15 VOTING AND APPORTIONMENT 15.1 Voting Systems 491 15.2 Flaws of Voting 496 15.3 Apportionment Methods 500 15.4 Flaws of Apportionment Methods 506 Review Exercises 510 Chapter Test 514

402

CHAPTER ONE CRITICAL THINKING SKILLS Exercise Set 1.1 1. Counting 2. Divisible 3. Hypothesis 4. Counterexample 5. Inductive 6. Deductive 7. Deductive 8. Inductive 9. Inductive reasoning, because a general conclusion was made from observation of specific cases. 10. Inductive reasoning, because a general conclusion was made from observation of specific cases. 11.

5×5 = 25

12.

12×14 = 168

13.

1 5 (= 1 + 4) 10 (= 4 + 6) 10 (= 6 + 4) 5(= 4 + 1) 1

14.

100,000 = 105

15.

16.

17.

18.

19. 21.

23. 25.

10, 12, 14 (Add 2 to previous number.) 3, −3, 3 (Alternate 3 and −3.)

20.

19, 23, 27 (Add 4 to the previous number.)

22.

−3, −5, −7 (Subtract 2 from previous number.) 2500, −12,500, 62,500 (Multiply previous number by –5.)

24.

1 1 1 , , (Increase the denominator value by 1.) 5 6 7 36, 49, 64 (The numbers in the sequence are the squares of the counting numbers.)

26.

21, 28, 36 (15 + 6 = 21, 21 + 7

= 28, 28 + 8 = 36)

CHAPTER 1 Critical Thinking Skills

27.

34, 55, 89 (Each number in the sequence is the sum of the previous two numbers.)

29. There are three letters in the pattern. 39× 3 = 117 , so the 117th entry is the second R in the pattern. Therefore, the 118th entry is Y.

31.

a) 36, 49, 64 b) Square the numbers 6, 7, 8, 9 and 10. c) 8×8 = 64 9 × 9 = 81 72 is not a square number since it falls between the two square numbers 64 and 81.

33. Blue: 1, 5, 7, 10, 12

Purple: 2, 4, 6, 9, 11

28.

243 729 2187 ,− , 256 1024 4096

(Multiply previous number 3 by − .) 4

30.

a) Answers will vary. b) The sum of the digits is 9. c) When a one- or two-digit number is multiplied by 9, repeated summing of the digits in the product yields the number 9. 32. a) 28 and 36 b) To find the 7th triangular number, add 7 to the 6th triangular number. To find the 8th triangular number, add 8 to the 7th triangular number. To find the 9th triangular number, add 9 to the 8th triangular number. To find the 10th triangular number, add 10 to the 9th triangular number. To find the 11th triangular number, add 11 to the 10th triangular number. c) 36 + 9 = 45; 45 + 10 = 55; 55 + 11 = 66; 66 + 12 = 78 72 is not a triangular number since it falls between the consecutive triangular numbers 66 and 78. Yellow: 3, 8

34. a) 19 (Each new row has two additional triangles.) b) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100 35. a) ≈ $200, 000 b) We are using observation of specific cases to make a prediction. 36. a) ≈ $3.7 trillion

b) We are using observation of specific cases to make a prediction. 38.

37.

39. a) You should obtain the original number. b) You should obtain the original number. c) Conjecture: The result is always the original number. 4n + 12 4n 12 = + = n + 3, n + 3 − 3 = n 4 4 4 40. a) You should obtain twice the original number. b) You should obtain twice the original number. c) Conjecture: The result is always twice the original number. d) n, 4n, 4n + 12,

d) n, 4n, 4n + 6,

4n + 6 4n 6 = + = 2n + 3, 2n + 3 − 3 = 2n 2 2 2

SECTION 1.2

41. a) You should obtain the number 5. b) You should obtain the number 5. c) Conjecture: No matter what number is chosen, the result is always the number 5. 2n + 10 2n 10 = + = n + 5, n + 5 − n = 5 d) n, n + 1, n + ( n + 1) = 2n + 1, 2n + 1 + 9 = 2n + 10, 2 2 2 42. a) You should obtain the number 0. b) You should obtain the number 0. c) Conjecture: No matter what number is chosen, the result is always the number 0. n + 10 ⎛⎜ n + 10 ⎞⎟ , 5⎜ = n + 10, n + 10 − 10 = n, n − n = 0 d) n, n + 10, ⎜⎝ 5 ⎠⎟⎟ 5

43. 7 − 5 = 2 is one counterexample. 44. 5 ÷ 2 = 2 12 , which is not a counting number.

5 , which is not an even number. 2 46. 900 is a three-digit number. The product of 900 and 900 is 810,000, which is not a five-digit number. 47. One and two are counting numbers. The difference of 1 and 2 is 1− 2 = −1 , which is not a counting number. 48. The sum of the odd numbers 1 and 5 is 6, which is not divisible by 4. 49. a) The sum of the measures of the interior angles should be 180° . b) Yes, the sum of the measures of the interior angles should be 180° . c) Conjecture: The sum of the measures of the interior angles of a triangle is 180° . 50. a) The sum of the measures of the interior angles should be 360° . b) Yes, the sum of the measures of the interior angles should be 360° . c) Conjecture: The sum of the measures of the interior angles of a quadrilateral is 360° . a b 51. 129, the numbers in positions are found as follows: c a +b+ c

45. Two is a counting number. The sum of 2 and 3 is 5. Five divided by two is

52. 1881, 8008, 8118 (They look the same when looked at in a mirror.) 53. c

Exercise Set 1.2 (Note: Answers in this section will vary depending on how you round your numbers. The answers may differ from the answers in the back of the textbook. However, your answers should be something near the answers given. All answers are approximate.)

1. 2.

Estimation Equal

261 + 127.4 + 273.9 + 16.2 + 81.5 ≈ 260 + 127 + 274 + 16 + 82 = 759

4. 2.57 + 212.6 +176.2 + 83

198, 600×3.072 ≈ 200, 000×3.000 = 600, 000

1854 ×0.0096 ≈ 1900×0.01 = 19

405 400 ≈ = 8000 0.049 0.05

0.63×1523 ≈ 0.6×1500 = 900

≈ 0 + 210 +180 + 80 = 470

CHAPTER 1 Critical Thinking Skills

10. 51, 608× 6981 ≈ 50, 000× 7000 = 350, 000,000

11% of 8221 ≈ 10% of 8000 = 0.10×8000 = 800

11.

18% ×1576 ≈ 20%×1600 = 0.20×1600 = 320

12.

296.3 ÷ 0.0096 ≈ 300 ÷ 0.01 = 30, 000

13.

$10.49 $10 ≈ = $2 5 5

14.

$37.80 $40 ≈ = $2 20 20

15.

12 months ×$120.80 ≈12×$120 = $1440

16.

8% of $11, 250 ≈ 0.08 × $11, 000 = $880

17.

One third of an annual profit of $8795 1 ≈ × $9, 000 = $3000 3

18.

$1.29 + $6.86 + $12.43 + $25.62 + $8.99 ≈ $1+ $7 + $12 + $26 + $9 = $55

19.

95lb +127 lb + 210 lb ≈100 +100 + 200 = 400 lb

21.

20. 22.

15% of $26.32 ≈ 15% of $26 = 0.15×$26 = $3.9

3.25 lb 3.00 lb ≈ = 0.5 lb 6 6 $400 $400 ≈ = 16 $23 $25

24.

23. ($65.99 + $49.99 + $49.95) − $114.99 ≈ ($66 + $50 + $50) − $115 = $166 − $115 = $5

Team A: 189 + 172 + 191 ≈ 190 + 170 + 190 = 550 Team B: 183 + 229 + 167 ≈ 180 + 230 + 170 = 580 580 − 550 = 30 lb

25.

11 × 8 × $1.50 ≈ 10 × 8 × $1.50 = 10 × $12 = $120

26.

6 min, 25 sec×26.2 mi ≈ 6.5 min × 26 mi =169 min 169 min ≈ 3hours 60 min

27. 100 Mexican pesos = 100× 0.083 U.S. dollars

28. $973 + 6 ($61) + 6 ($97) + 6 ($200)

≈ 100× 0.08 U.S. dollars = 8 U.S. dollars $50 − $8 = $42

29. ≈ 60 miles

≈ $970 + 6 ($60) + 6 ($100) + 6 ($200) = $970 + $360 + $600 + $1200 = $3130

30.

≈ 55 miles

SECTION 1.3

32.

a) 100 b) 50 c) 125

33. a) 5 million b) 98 million c) 98 million − 33 million = 65 million d) 19 million + 79 million + 84 million + 65 million + 33 million = 280 million

34.

a) 19% b) 25% c) 20% of 179 lb ≈ 20% of 180 = 0.2 ×180 = 36 lb

35. a) 85% b) 68% − 53% = 15% c) 85% of 70 million acres = 59,500, 000 acres

36.

a) 2 ( 410) + 4 (545)

31. a) 33% of 700 ≈ 30% of 700 = 0.30×700 = 210 b) 9% of 700 ≈10% of 700 = 0.10×700 = 70 c) 24% of 700 ≈ 25% of 700 = 0.25×700 =175

≈ 2 (400) + 4 (550) = 800 + 2200 = 3000 calories b) Running: 4 (920) ≈ 4 (925) = 3700 calories

d) No, since we are not given the area of each state.

Casual bike riding: 4 (300) = 1200 calories , 3700 −1200 = 2500 calories c) 3(545) + 3(545) ≈ 3(550) + 3(550) = 1650 + 1650 = 3300 calories per week , 3300 calories per week (52 weeks) ≈ 3000×50 = 150,000 calories 38.

37.

39. 41. 43. 45. 47.

≈ 120 bananas 150° 10% 9 square units 150 feet

40. 42. 44. 46.

≈ 160 berries 315° 25% 12 square units

48.

4 (60) = 240 in. or

49.-57. Answers will vary. 59. a) Answers will vary. b) Answers will vary.

58.

There are 118 ridges around the edge.

1 in. 20.1 in. = 2.5 yd x yd

Exercise Set 1.3 1 in. 5.75 in. 1. = 12 mi x mi 1x = 12 (5.75)

1x = 2.5 (20.1)

x = 69 mi

x = 50.25 yd

240 ≈ 20 ft 12

CHAPTER 1 Critical Thinking Skills

3 ft x ft = 1.2 ft 19.36 ft 3(19.36) = 1.2 x

x bags 1 bag = 2 4000 ft 35, 000 ft 2 4000 x = 1(35, 000)

58.08 1.2 x = 1.2 1.2 58.08 = 48.4 ft x= 1.2

4000 x 35, 000 = 4000 4000 35, 000 = 8.75 bags x= 4000

5. 6.4% of $7605 = 0.064 × $7605 = $486.72 $7605 + $486.72 = $8091.72 ≈ $8092

30% of $117 = 0.30 × $117 = $35.10 $117 − $35.10 = $81.90

7. a) Ent./Misc.: 19.4% of $1750 = 0.194 × $1750 8. a) 11 − 20 years: 25% of 6.2 million = 0.25 × 6.2 = 1.55 million = $339.50 0 − 3 years: 16% of 48 million = 0.16 × 6.2 Food: 12.7% of $1750 = 0.127 × $1750 = $222.25 = 0.992 million $339.50 − $222.25 = $117.25 1.55 − 0.992 = 0.558 million = 558,000 b) Housing: 33.9% of $1750 = 0.339 × $1750 b) 4 −10 years: 33% of 6.2 million = 0.33 × 6.2 = $593.25 = 2.046 million Transportation: 17% of $1750 = 0.17 × $1750 11− 20 years: 25% of 6.2 million = 0.25 × 6.2 = $297.50 = 1.55 million $593.25 − $297.50 = $295.75 2.046 − 1.55 = 0.496 million = 496,000

54.46% of $200, 000 = 0.5446 × 200, 000 = $108,920 $200, 000 + $108,920 = $308,920

b) 2.9% of $180, 000 = 0.029 ×180,000 = $5220 $180, 000 + $5220 = $185, 220

10. 40 rides × $2 per ride = $80. In order for the cost of rides with the $81 MetroCard to be less than the cost of the rides without the MetroCard, Chandler would have to take 41 rides per month. $81 ≈ $1.98per ride 41rides

c) Flagstaff, AZ: − 1.85% of $200, 000 = −0.0185 × $200, 000 = − $3700 $200, 000 − $3700 = $196,300 Bellingham, WA: − 1.04% of $200,000 = −0.0104×$200, 000 =− $2080 $200, 000 − $2080 = $197,920 $197,920 − $196,300 = $1620 11.

$120 + $80 (15) = $120 + $1200 = $1320 Savings: $1320 − $1250 = $70

12.

2005: $20 × 2 million = $40 million 2006: $20 + 0.25 × $20 = $25 $25 × 2 million = $50 million 50 − 40 = $10 million or $10,000,000

SECTION 1.3

13.

14. 15 year mortgage: $777.83(12)(15) = $140, 009.4

Points needed for 80 average: 80 (5) = 400 points Wallace’s points so far: 79 + 93 + 91 + 68 = 331 points Grade needed on fifth exam: 400 − 331 = 69

30 year mortgage: $521.65(12)(30) = $187, 794.0 Savings: $187, 794.00 − $140, 009.40 = $47, 784.6

15.

a) 10×10×10×10 = 10, 000

16.

b) 1 in 10,000

17.

38,687.0 mi − 38, 451.4 mi = 235.6 mi

18. a) 40 ×$8.50×52 = $17, 680 b) Each week he makes 40×$8.50 = $340. $1275 = 3.75 weeks $340

235.6 mi ≈ 18.698 ≈ 18.7 mpg 12.6 gal

19.

460 = 9.2 min 50 1550 b) = 62 min 25 1400 c) = 40 min 35 1550 2200 3750 d) + = ≈ 47 min 80 80 80

By mail: ($52.80 + $5.60 + $8.56)× 4 = $66.96× 4 = $267.84 Tire store: $324 + 0.08×$324

$885 − $25 (15) = $885 − $375 = $510 20. $510 =17 hours $30

= $324 + $25.92 = $349.92 Savings: $349.92 − $267.84 = $82.08

21.

15,000 ft − 3000 ft = 12,000 ft decrease in elevation. Temperature increases 2.4° F for every 1000 ft decrease in elevation. 2.4° F×12 = 28.8° F −6° F + 28.8° F = 22.8° F The precipitation at the airport will be snow.

22. a) $620 (0.12) = $74.40 b) $1200 (0.22) = $264 c) The store lost $1200 − $1000 = $200 on the purchase. Store's profit: $264 − $200 = $64

CHAPTER 1 Critical Thinking Skills

24. a) 0.1 cm 3 × 60 sec × 60 min × 24 hr × 365 days

23. Steve and Maureen paid more than $9362.50 but less than $26,687.50, so they paid $9362.50 plus 25% of the amount over $68,000.

= 3,153,600 cm 3 b) 30 cm × 20 cm × 20 cm=12,000 cm 3

$13,365 − $9362.50 = $4002.50

0.1 cm3 × 60 sec × 60 min × 24 hr = 8640 cm3

$4002.5 = $16, 01 The amount over $68,000 was 0.25 $68, 000 + $16, 010 = $84, 010

25. a) 1 oz × 60 min × 24 hr × 365 days = 525,600 oz 525,600 = 4106.25 gal 128 b) 4106.25 ×$11.20 = 4.10625× $11.20 = $45.99 1000

27. a)

20,000 20,000 − ≈ 961.538 − 925.926 20.8 21.6 = 35.612 ≈ 35.61 gal

b) 35.61×$3.00 = $106.83 c) 140,000,000 ×35.61 = 4,985, 400, 000 gal

29. Cost after 1 year: $799 + 0.06($799) = $799 + $47.94 = $846.94 Cost after 2 years: $846.94 + 0.06 ($846.94) = $846.94 + $50.82 = $897.76

12, 000 = 1.38 ≈ 1.4 days 8640

26.

a) Short: $33 × 5 = $165 Long: $18 × 5 = $90 $165 − $90 = $75; Jeff saves $75. b) $6 for first hour, plus 6 × $3 for remaining 3 hours, for a total of $24. c) Short: $6 + 8 × $3 = $30 Long: $18 Long term is cheaper by $12.

28. a) Yes, divide the total amount spent by the amount spent per capita.

$45,592.59 ≈ 301.14 million $151.40

$11, 237.53 ≈ 60.78 million $184.90

30. Value after first year: $1000 + 0.10($1000) = $1000 + $100 = $1100 Value after second year: $1100 − 0.10($1100) = $1100 − $110 = $990

31. After paying the $100 deductible, Yungchen must pay 20% of the cost of x-rays. First x-ray: $100 + 0.20 ($540) = $100 + $108 = $208

$990 is less than the intial investment of $1000. 32. $3000 is the difference between one-fourth of the cost and one-fifth of the cost. 1 1 1 − = ; 20 × $3000 = $60,000. 4 5 20

Second x-ray: 0.20 ($920) = $184 Total: $208 + $184 = $392 ⎛1⎞ 3 salt: 3⎜⎜ ⎟⎟⎟ = tsp ⎜⎝ 8 ⎠ 8

33. a) water/milk: 3(1) = 3 cups Cream of wheat: 3(3) = 9 tbsp =

9 cup (because 16 tbsp = 1 cup) 16

SECTION 1.3

2 + 3.75 5.75 7 = = 2.875 cups = 2 cups 2 2 8 0.25 + 0.5 0.75 3 = = 0.375 tsp = tsp salt: 2 2 8 0.5 + 0.75 1.25 5 5 = = 0.625 cups = cup = (16 tbsp) = 10 tbsp cream of wheat: 2 2 8 8 3 15 4 11 3 c) water/milk: 3 −1 = − = = 2 cups 4 4 4 4 4 1 1 4 1 3 3 3 12 3 9 cream of wheat: − = − = cup = 9 tbsp salt: − = − = tsp 2 8 8 8 8 4 16 16 16 16 d) Differences exist in water/milk because the amount for 4 servings is not twice that for 2 servings. 1 Differences also exist in Cream of Wheat because cup is not twice 3 tbsp. 2 1 b) rice: 1(2) = 2 cups 34. a) rice: ( 4) = 2 cups 2 1 4 16 1 1 9 18 2 1 water: 2 (2) = ( 2) = = 4 = 4 cups water: 1 (4) = ( 4) = = 5 cups 3 3 3 3 4 4 4 4 2 1 1 salt: ( 2) = 1 tsp salt: ( 4) = 1 tsp 4 2 b) water/milk:

butter/margarine: 1(4) = 4 tsp

butter/margarine: 2 ( 2) = 4 tsp

1 1 1 3 4 + 1 = + = = 2 cups 2 2 2 2 2 1 1 4 10 14 2 water: 1 + 3 = + = = 4 cups 3 3 3 3 3 3 1 3 4 salt: + = = 1 tsp 4 4 4 butter/margarine: 1 tsp + 1 tbsp = 1 tsp + 3 tsp = 4 tsp d) rice: 3 −1 = 2 cups 1 24 9 15 3 water: 6 − 2 = − = = 3 cups 4 4 4 4 4 1 1 salt: 1 − = 1 tsp 2 2 butter/margarine: 2 tbsp = 2 (3tsp) = 6 tsp

c) rice:

6 tsp − 2 tsp = 4 tsp e)

Differences exist in water because the amount for 4 servings is not twice that for 2 servings.

$425 − $240 (one box of 20 DVDs) = $185 $185 − $180 (one box of 12 DVDs) = $5 One box of 20 DVDs and one box of 12 DVDs are the maximum number of DVDs that can be purchased. b) $240 + $180 = $420

35. a)

36. Mark will win. 38. 1 ft 3 = 12 in.×12 in.×12 in. = 1728 in.3

37. 1 ft 2 would be 12 in. by 12 in. Thus, 1 ft 2 = 12 in.×12 in. = 144 in.2

CHAPTER 1 Critical Thinking Skills

39. Area of original rectangle = lw

20 ft ×20 ft = 400 ft 2

Area of new rectangle = ( 2l )( 2 w) = 4lw Thus, if the length and width of a rectangle are doubled, the area is 4 times as large. 41. Volume of original cube = lwh Volume of new cube = (2l )(2 w)( 2h ) = 8lwh Thus, if the length, width, and height of a cube are doubled, the volume is 8 times as large or increases eight- fold.

43.

5 ft ×5 ft = 25 ft 2

40.

400 ft 2 =16 squares 25ft 2 42. 11 ft is one-sixth of the pole, so the length is 6 × 11 ft = 66 ft.

10 pieces 1000 pieces = $x $10 1000 x = 10 (10) 1000 x 100 = 1000 1000 100 x= = $0.10 = 10¢ 1000

44. Left side: 1(−6) = −6

Right side: 1(2) = 2

2 (−2) = −4

1(3) = 3

− 6 +−4 = −10

1(6) = 6

45. 3

2 + 3 + 6 = 11 Place it at −1 so the left side would total −10 +−1 = −11 46. 10;2002 , 2112 , 2222 , 2332 , 2442 , 2552 ,

47. a) ( 4× 4) + (3×3) + (2 × 2) + (1×1)

2662 , 2772 , 2882 , 2992

= 16 + 9 + 4 + 1 = 30 b) (7×7) + (6×6) + (5×5) + 30 = 49 + 36 + 25 + 30 = 140 49.

48. a) Place the object, 1 g, and 3 g on one side and 9 g on the other side. b) Place the object, 9 g, and 3 g on one side and 27 g and 1 g on the other side. 50. Eight pieces

51.

52.

53. 8 + 6 + 2 + 4 = 20;3 + 7 + 5 + 1 = 16; 10 +14 +12 + 8 = 44

The sum of the four corner entries is 4 times the number in the center of the middle row.

SECTION 1.3

54. 15,12,33 Multiply the number in the center of the middle row by 3. 56. 35 −15 = 20 cubes

45,36,99 Multiply the number in the center of the middle row by 9. 57. 3× 2×1 = 6 ways

58. Each shakes with four people.

59.

55.

Other answers are possible, but 1 and 8 must appear in the center. 61.

60.

(The diagram shows the number of times each part is used.) 62.

With umbrella policy: Mustang reduced premium: $1648 − $90 = $1558 Focus reduced premium: $1530 − 0.12 ($1530) = $1530 − $183.60 = $1346.40 Total for umbrella policy: $1558 + $1346.40 + $450 = $3354.40 Without umbrella policy: $1648 + $1530 = $3178 Net amount for umbrella policy: $3354.40 − $3178 = $176.40 64. 16 + 16 + 4 + 4 + 4 = 44

Other answers are possible. 63.

Mary is the skier.

65.

Areas of the colored regions are: 1×1, 1×1, 2× 2, 3× 3, 5×5, 8×8, 13×13, 21× 21 ; 1 + 1 + 4 + 9 + 25 + 64 + 169 + 441 = 714 square units

66.

Let x be the amount Samantha had to start. 1 1 After first store: x − x − 20 = x − 20 2 2 ⎞⎟ 1 ⎛⎜ 1 1 After second store: ⎜ x − 20⎟⎟ − 20 = x − 30 ⎠ 2 ⎜⎝ 2 4 This is equal to $0, so the original amount was $120.

67.

Thomas would have opened the box labeled grapes and cherries. Because all the boxes are labeled incorrectly, whichever fruit he pulls from the box of grapes and cherries, will be the only fruit in that box. If he pulled a grape, he labeled the box grape. If he pulled a cherry, he labeled the box cherries. That left two boxes whose original labels were incorrect. Because all labels must be changed, there was only one way for Thomas to assign the two remaining labels.

CHAPTER 1 Critical Thinking Skills

Review Exercises 1. 27, 32, 37 (Add 5 to previous number.)

−48, 96, −192 (Multiply previous number by –2.)

10, 4, − 3 (subtract 1, then 2, then 3, ...)

26, 37, 50 (17 + 9 = 26, 26 + 11 = 37, 37 + 13 = 50) 25, 32, 40 (19 + 6 = 25, 25 + 7 = 32, 32 + 8 = 40) 3 3 3 1 , , (Multiply previous number by .) 8 16 32 2

6. 8.

9. c 10. a) The final number is twice the original number. b) The final number is twice the original number. c) Conjecture: The final number is twice the original number. 10n + 5 10n 5 d) n, 10n, 10n + 5, = + = 2n + 1, 2n + 1 − 1 = 2n 5 5 5 11. This process will always result in an answer of 3. n, n + 5,6 (n + 5) = 6n + 30, 6n + 30 −12 = 6n + 18,

6n + 18 6n 18 3n + 9 3n 9 = + = 3n + 9, = + = n + 3, n + 3− n = 3 2 2 2 3 3 3

12. 12 + 22 = 5,5 is an odd number. Other answers are possible. (Note: Answers for Ex. 13 - 25 will vary depending on how you round your numbers. The answers may differ from the answers in the back of the textbook. However, your answers should be something near the answers given. All answers are approximate.) 13. 14. 215.9 + 128.752 + 3.6 + 861 + 792 ≈ 200 + 100 + 0 + 900 + 800 = 2000 210,302 ×1992 ≈ 210,000× 2000 = 420,000,000

15.

19% of 1025 ≈ 20% of 1000 = 0.20×1000 = 200

16.

17.

52 shovels ×$99.97 ≈ 50×100 = $5000

18.

Answers will vary.

7% of $1999 ≈ 7% of 2000 = 0.07 × 2000 = $140 19.

1.1 mi 1 mi 3 mi ≈ = = 3 mph 22 min 20 min 60 min

21. 5 in. =

20.

$2.49 + $0.79 + $1.89 + $0.10 + $2.19 + $6.75

22.

≈ $2 + $1 + $2 + $0 + $2 + $7 = $14.00 2.35 million − 1.95 million = 0.4 million

24.

13 square units

26.

$50 + $40 (12)= $530 Savings: $530 − $500 = $30

⎛1⎞ 20 in. = 20 ⎜⎜ ⎟⎟⎟ in. = 20 (0.1) mi = 2 mi ⎜⎝ 4 ⎠ 4

23.

2.8 million − 1.8 million = 1.0 million

25.

Length = 1.75 in., 1.75(12.5) = 21.875 ≈ 22 ft Height = 0.625 in., 0.625(12.5) = 7.8125 ≈ 8 ft

REVIEW EXERCISES

27.

4 ($2.69) = $10.76 for four six-packs Savings: $10.76 − $9.60 = $1.16

$445 = $89 5 $510 = $85 Cost per person with 6 people: 6 $89 − $85 = $4 savings

29. Cost per person with 5 people:

31.

10% of $1030 = 0.10 × $1030 = $103

30 x 30 × 24,000 = ; x= = 288 lb 2500 24,000 2500 150 b) = 5 bags, and 5 × 2500 = 12,500 ft 2 30

30. a)

32.

Savings: $721 − $60 = $661

1.5 mg x mg = 10 lb 47 lb 10 x = 47 (1.5) 10 x 70.5 = 10 10 x = 7.05 mg

$5000 − 0.30 ($5000) = $5000 − $1500

34.

9 A.M. Eastern is 6 A.M. Pacific, from 6 A.M. Pacific to 1:35 P.M. Pacific is 7 hr 35 min , 7 hr 35 min − 50 min stop = 6 hr 45 min

36.

a) 5280 ft 1hr 1min 5280 ft × × = ≈1.47 ft/sec 1 60 min 60 sec 3600sec

= $3500 take-home 28% of $3500 = 0.28×$3500 = $980 35.

Taylor: $45 for 8 hours Admar: $8 × 6 hours = $48 $48 − $45 = $3 Taylor Rental is cheaper by $3.

28.

$103× 7 = $721

33.

3 P.M. − 4 hr = 11 A.M. July 26, 11:00 A.M.

b) 55 mi 5280 ft 1hr 1min 290, 400 ft × × × = 1hr 1mi 60 min 60sec 3600sec ≈ 80.67 ft/sec 37. Each figure has an additional two dots. To get the hundredth figure, 97 more figures must be drawn, 97 ( 2) = 194 dots added to the third

38.

figure. Thus, 194 + 7 = 201.

39.

40.

59 min 59 sec Since it doubles every second, the jar was half full 1 second earlier than 1 hour.

41. 6 42. Nothing. Each friend paid $9 for a total of $27; $25 to the hotel, $2 to the clerk. $25 for the room + $3 for each friend + $2 for the clerk = $30 43. Let x = the total weight of the four women

CHAPTER 1 Critical Thinking Skills

x 520 + 180 700 = 130, x = 520, = = 140 lb 4 5 5 44. Yes; 3 quarters and 4 dimes, or 1 half dollar, 1 quarter and 4 dimes, or 1 quarter and 9 dimes. 45. 6 cm × 6 cm × 6 cm = 216 cm 3 46. Place six coins in each pan with one coin off to the side. If it balances, the heavier coin is the one on the side. If the pan does not balance, take the six coins on the heavier side and split them into two groups of three. Select the three heavier coins and weigh two coins. If the pan balances, it is the third coin. If the pan does not balance, you can identify the heavier coin.

n (n + 1)

500 (501)

250,500 = 125, 250 2 2 2 48. 16 blue: 4 green → 8 blue, 2 yellow → 5 blue, 2 white → 3 blue 49. 90: 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, … 50. The fifth figure will be an octagon with sides of equal length. Inside the octagon will be a seven sided figure with each side of equal length. The figure will have one antenna. 51. 61: The sixth figure will have 6 rows of 6 tiles and 5 rows of 5 tiles (6× 6 + 5× 5 = 36 + 25 = 61). 52. Some possible answers are given below. There are other possibilities.

47.

53. a) 2 b) There are 3 choices for the first spot. Once that person is standing, there are 2 choices for the second spot and 1 for the third. Thus, 3× 2×1 = 6 . c) 4× 3× 2 ×1 = 24 d) 5× 4×3× 2 ×1 = 120 e) n (n −1)( n − 2)"1, (or n !), where n = the number of people in line

Chapter Test 1. 27, 33, 39 (Add 6 to previous number.)

1 1 1 1 , , (Multiply previous number by .) 16 32 64 2

3. a) The result is the original number plus 1. b) The result is the original number plus 1. c) Conjecture: The result will always be the original number plus 1. 5n + 10 5n 10 = + = n + 2, n + 2 −1 = n + 1 d) n,5n,5n + 10, 5 5 5 (Note: Answers for #4 - #6 will vary depending on how you round your numbers. The answers may differ from the answers in the back of the textbook. However, your answers should be something near the answers given. All answers are approximate.)

CHAPTER TEST

0.21×82, 000 ≈ 0.2×80, 000 = 16, 000

9 square units

a) 3.5 million b) 0.3 million

10.

$15 ≈ 5.79 $2.59 The maximum number of 6 packs is 5.

175, 000 170, 000 ≈ ≈ 1, 700, 000 0.09 0.1 130 lb ≈ 2.0635 a) 63 in. 2.0635 = 0.032754 63 in. 0.032754×703 ≈ 23.03 b) He is in the at risk range.

$85.99 − $59.99 = $26 9. $26 = 65 additional minutes $0.40 11. 1 cut yields 2 equal pieces. Cut each of these 2 equal pieces to get 4 equal pieces. 3 cuts → 3(2.5 min) = 7.5 min

$15.00 − (5×$2.59) = $15.00 − $12.95 = $2.05 $2.05 = 2.5625 $0.80 Thus, two individual cans can be purchased. Number of cans 6 packs Indiv. cans 5 2 32 4 5 29 3 9 27 2 12 24 1 15 21 0 18 18 The maximum number of cans is 32. 12. 2.5 in. by 1.875 in. ≈ 2.5×15.8 by 1.875×15.8 = 39.5 in. by 29.625 in. ≈ 39.5 in. by 29.6 in. (The actual dimensions are 100.5 cm by 76.5 cm.)

13. $12.75 × 40 = $510 $12.75 × 1.5 × 10 = $191.25 $510 + $191.25 = $701.25 $701.25 − $652.25 = $49.00

14.

15.

Mary drove the first 15 miles at 60 mph which took

15 1 = hr, and the second 15 miles at 30 mph which 60 4

15 1 3 = hr for a total time of hr. If she drove the entire 30 miles at 45 mph, the trip would take 30 2 4 30 2 3 = hr (40 min) which is less than hr (45 min). 45 3 4

took

CHAPTER 1 Critical Thinking Skills

16.

6 lb 1 3 1 = 3; 3× tsp = tsp or 1 tsp 2 lb 2 2 2 1

17.

1 1 tsp = tbsp 2 2

Area of lawn including walkway: (10 + 2)×(12 + 2)=12×14 =168 m 2 Area of lawn only: 10×12 = 120 m 2 Area of walkway: 168 −120 = 48 m 2

18. 243 jelly beans; 260 −17 = 243, 234 + 9 = 243, 274 − 31 = 243 19. a) 3 × $3.99 = $11.97 b) 9 ($1.75×0.75) = 11.8125 ≈ $11.81 c) $11.97 − $11.81 = $0.16 Using the coupon is least expensive by $0.16. 20. 24 (The first position can hold any of four letters, the second any of the three remaining letters, and so on. 4 × 3 × 2 × 1 = 24

Group Projects

$325 ≈ $108.33 3 b) Let x = the amount before tax x + 0.07 x = 325

1. a)

1.07 x 325 = 1.07 1.07 x = 303.7383178 ≈ $303.74 $303.74 = 101.246 ≈ $101.25 3 c) Inductive reasoning - arriving at a general conclusion from specific cases d) Combination set: $62.00 − ($62.00×0.10) = $62.00 − $6.20 = $55.80 Individual sets: 2×$36.00 = $72.00,$72.00 −($72.00× 0.20) = $72.00 − $14.40 = $57.60 Therefore, the combination set is cheaper. e) Combination with tax: $55.80 × 1.07 ≈ $59.71 Individual set with tax: $57.60 × 1.07 ≈ $61.63 $61.63 − $59.71 = $1.92 2. a) – d) Answers will vary. 400 mi ÷ 50 mi hr = 8 hrs, 9 A.M. + 8 hrs = 5 P.M. e) f) – h) Answers will vary. 3. Order 1 2 3 4

Name Ernie Zeke Jed Tex

Apparel holster vest chaps Stetson

CHAPTER TWO SETS Exercise Set 2.1 1. Set 2. Ellipsis 3. Description, Roster form, Set-builder notation 4. Finite 5. Infinite 6. Equal 7. Equivalent 8. Cardinal 9. Empty or null 10. { } ,∅ 11. Universal 12. One-to-one 13. Not well defined, “best” is interpreted differently by different people. 14. Not well defined, “most interesting” is interpreted differently by different people. 15. Well defined, the contents can be clearly determined. 16. Well defined, the contents can be clearly determined. 17. Well defined, the contents can be clearly determined. 18. Not well defined, “most interesting” is interpreted differently by different people. 19. Infinite, the number of elements in the set is not a natural number. 20. Finite, the number of elements in the set is a natural number. 21. Infinite, the number of elements in the set is not a natural number. 22. Infinite, the number of elements in the set is not a natural number. 23. Infinite, the number of elements in the set is not a natural number. 24. Finite, the number of elements in the set is a natural number. ⎧San Marino, Scotland, Serbia, Slovakia,⎪ ⎫ 25. { Maine, Maryland, Massachusetts, Michigan, ⎪ ⎪ 26. ⎪⎨ ⎬ ⎪ ⎪ Minnesota, Misssissippi, Missouri, Montana } Slovenia, Spain, Sweden, Switzerland ⎪ ⎪ ⎩ ⎭

27. 29.

{ 11,12,13,14, … ,177 } B = { 2, 4, 6, 8, …}

28.

C = {4}

30.

{ } or ∅

CHAPTER FIVE NUMBER THEORY AND THE REAL NUMBER SYSTEM Exercise Set 5.1 1. Theory 2. Integer 3. Zero 4. Prime 5. Composite 6. Divisor 7. Multiple 8. Mersenne 9. Conjecture 10. Prime 11. Fermat 12. Twin 13. The prime numbers between 1 and 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. 14.

○2 ○3 11 12 13 ○ ○ 21 22 23 ○ 31 32 33 ○ 41 42 43 ○ ○ 51 52 53 ○ 61 62 63 ○ 72 71 73 ○ ○ 81 82 83 ○

102 101 103 ○ ○ 111 112 113 ○ 121

5 ○

6 16

○7 17 ○

37 ○ 47 ○

19 ○ 29 ○

67 ○

59 ○

97 ○

79 ○ 89 ○

104

105

106

107 109 108 110 ○ ○

114

115

116

117

118

112

123

124

125

126

132 131 ○

133

134

135

136

141

143

144

145

146

142

119

80 90 100

120

128 129 130 127 ○ 139 137 138 140 ○ ○ 147 148 150 149 ○

The prime numbers between 1 and 150 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 87, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, and 149.

123

124

CHAPTER 5 Number Theory and the Real Number System

15. True; since 15 ÷ 5 = 3 17. False; 28 is a multiple of 7. 19. False; 56 is divisible by 8 21. True; if a number is divisible by 10, then it is also divisible by 5. 23. False; if a number is divisible by 3, then the sum of the number’s digits is divisible by 3. 25. True; since 2 • 3 = 6. 27. Divisible by 2, 5 and 10. 29. Divisible by 2, 3, 4, 6, 8, and 9. 31. Divisible by 2, 3, 4, 5, 6, 8, and 10. 33. 2 • 3 • 4 • 5 • 6 = 720. (other answers are possible) 35.

2 2 3

36 = 2 • 3 38.

36.

36 18 9 3

2 2

16. False; 24 is a multiple of 3. 18. True; since 18 ÷ 6 = 3. 20. True; since 45 ÷ 15 = 3. 22. False; consider 15. 24. True. 26. True; since 3 • 5 = 15. 28. Divisible by 2, 3, 5, 6, 9, and 10. 30. Divisible by 2, 3, 4, 5, 6, 9, and 10. 32. Divisible by none of the numbers. 34. 3 • 4 • 5 • 9 • 10 = 5400. (other answers are possible) 52 26 13

37.

140 70 35 7 7 140 = 22 • 5 • 7 3 7

52 = 2 • 13

2 2 5

3 3 5

39.

2 2

332 166 83 332 = 22 • 83

40.

399 133 19 399 = 3 • 7 • 19

2 2 2

315 105 35 7 315 = 32 • 5 • 7

41.

3 3 3

513 171 57 19 513 = 33•19

42.

3 13

663 221 17 663 = 3•13•17

43.

1336 668 334 167 3 1336=2 •167

44.

45.

3 23

46.

2 5 11

1313 101 1313 = 13•101

47. The prime factors of 6 and 14 are: 6 = 2 • 3, 14 = 2 •7 a) The common factor is 2, thus, the GCD = 2. b) The factors with the greatest exponent that appear in either are 2, 3, 7. Thus, the LCM = 2 • 3 • 7 = 42.

2001 667 29 2001= 3•23•29

48. The prime factors of 18 and 21 are: 18 = 2 • 32 and 21 = 3 • 7 a) The common factor is 3 thus, the GCD = 3 b) The factors with the greatest exponent that appear in either are 2, 32 and 7; the LCM = 2 • 32 • 7 = 126.

3190 1595 319 29 3190=2•5•11•29

49. The prime factors of 15 and 25 are: 15 = 3 • 5, 25 = 52 a) The common factor is 5; thus the GCD = 5. b) The factors with the greatest exponent that appear in either are: 3 and 52 ; thus, the LCM = 3 • 52 = 75

SECTION 5.1

50. The prime factors of 32 and 224 are: 32 = 25, 224 = 25 • 7 a) The common factor is: 25; thus, the GCD = 25 = 32. b) The factors with the greatest exponent that appear in either are: 25 and 7; thus, the LCM = 25 • 7 = 224

51. The prime factors of 40 and 900 are: 40 = 23 • 5, 900 = 22 • 32 • 52 a) The common factors are: 22, 5; thus, the GCD = 22•5 = 20. b) The factors with the greatest exponent that appear in either are: 23, 32, 52; thus, the LCM = 22•32•52 = 1800

52. The prime factors of 120 and 240 are: 120 = 23 • 3 • 5, 240 = 24 • 3 • 5 a) The common factors are: 23, 3, 5; thus, the GCD = 23•3•5 = 120. b) The factors with the greatest exponent that appear in either are: 24, 3, 5; thus, the LCM = 24•3•5 = 240

53. The prime factors of 96 and 212 are: 96 = 25 • 3,

54. The prime factors of 240 are: 24 ⋅ 3 ⋅ 5 . The prime factors of 285 are: 3⋅ 5 ⋅19 a) The common factors are 3 and 5; thus, the GCD = 3 ⋅ 5 = 15 .

55. The prime factors of 24, 48, and 128 are: 24 = 23 • 3,

212 = 22 • 53 a) The common factors are: 22; thus, the GCD = 22 = 4. b) The factors with the greatest exponent that appear in either are: 25, 3, 53; thus, the LCM = 25•3•53 = 5088

b) The factors with the greatest exponent that appear in either are: 24 , 3, 5, and 19; thus the LCM = 24 ⋅ 3 ⋅ 5 ⋅ 19 = 4560.

48 = 24 • 3, 128 = 27 a) The common factors are: 23; thus, the GCD = 23 = 8. b) The factors with the greatest exponent that appear in any are: 27, 3; thus, LCM = 27•3 = 384

56. The prime factors of 18, 78, and 198 are: 18 = 2 • 32, 78 = 2 • 3 • 13, 198 = 2 • 32 • 11 a) The common factors are: 2, 3; thus, the GCD = 2 i 3 = 6. b) The factors with the greatest exponent that appear in any are: 2, 32, 11, 13; thus, the LCM = 2•32•11•13 = 2574 57. a) The prime factors of 8 and 9 are: 8 = 23 and 9 = 32 Yes, they are relatively prime. b) The prime factors of 21 and 30 are: 21 = 3 • 7 and 30 = 2 • 3 • 5 No, they are not relatively prime. c) The prime factors of 39 and 52 are 39 = 3 • 13 and 52 = 22 • 13 No, they are not relatively prime. d) The prime factors of 177 and 178 are 177 = 3 • 59 and 178 = 2 • 89 Yes, they are relatively prime. 58. No. Any other two consecutive natural numbers will include an even number, and even numbers greater than two are composite. 59. Use the list of primes generated in exercise 13. The next two sets of twin primes are: 17, 19,and 29, 31. 60. Use the formula 2n - 1, where n is a prime number. 22 - 1 = 3, 23 - 1 = 7, 25 - 1 = 31, 27 - 1 = 127, 213 - 1 = 8191. n

61. Fermat number = 22 + 1 , where n is a natural number. 22 + 1 = 5 , 22 + 1 = 24 + 1 = 17 , 3

22 + 1 = 28 + 1 = 257 . These numbers are prime. 62. 4 = 2 + 2, 6 = 3 + 3, 8 = 3 + 5, 10 = 3 + 7, 12 = 5 + 7, 14 = 7 + 7, 16 = 3 + 13, 18 = 5 + 13, 20 = 3 + 17 63. The lcm of 6 and 16 is 48 days. 64. a) The possible committee sizes are: 4, 5, 10, 20, or 25. b) The number of committees possible are:

125

126

CHAPTER 5 Number Theory and the Real Number System

25 committees of 4, 20 committees of 5, 10 committees of 10, 5 committees of 20, or 4 committees of 25. 65. 66. The gcd of 390 and 468 is 78 dolls. 2×60, 3× 40, 4 ×30, 5× 24, 6× 20, 8×15, 10×12,

12×10, 15×8, 20×6, 24×5, 30× 4, 40× 3, 60× 2 67. The gcd of 70 and 175 is 35 cars. 69 The gcd of 150 and 180 is 30 trees. 71. The lcm of 5 and 6 is 30 days.

68. The gcd of 432 and 360 is 72 cards. 70. The lcm of 3500 and 6000 is 42,000 miles.

72. a) 5 = 6 – 1 7 = 6 + 1 11 = 12 – 1 13 = 12 + 1 17 = 18 – 1 19 = 18 + 1 23 = 24 – 1 29 = 30 – 1 b) Conjecture: Every prime number greater than 3 differs by 1 from a multiple of the number 6. c) The conjecture appears to be correct. 73. A number is divisible by 15 if both 3 and 5 divide the number. 74. A number is divisible by 22 if both 2 and 11 divide the number. 75. 40 ÷ 15 = 2 with rem. = 10 15 ÷ 10 = 1 with rem. = 5 10 ÷ 5 = 2 with rem. = 0 Thus, gcd of 15 and 40 is 5. 78. 104 ÷ 78 = 1 with rem. = 26. 78 ÷ 26 = 3 with rem. = 0. Thus, gcd of 104 and 78 is 26.

76. 28 ÷ 12 = 2 with rem. = 4 12 ÷ 4 = 3 with rem. = 0 Thus, gcd of 12 and 28 is 4.

77. 105 ÷ 35 = 3 with rem. = 0. Thus, gcd of 105 and 35 is 35.

79. 180 ÷ 150 = 1 with rem. = 30. 150 ÷ 30 = 5 with rem. = 0. Thus, the gcd of 150 and 180 is 30.

80. 560 ÷ 210 = 2 w/rem. = 140. 210 ÷ 140 = 1 w/rem. = 70. 140 ÷ 70 = 2 w/rem. = 0. Thus, gcd of 210 and 560 is 70.

81. The proper factors of 28 are: 1, 2, 4, 7, and 14. 1 + 2 + 4 + 7 + 14 = 28 Thus, 28 is a perfect number. 83. The proper factors of 56 are:1, 2, 4, 7, 8, 14, and 28. 1 + 2 + 4 + 7 + 8 + 14 + 28 = 64 Thus, 56 is not a perfect #.

82. The proper factors of 32 are: 1, 2, 4, 8, and 16. 1 + 2 + 4 + 8 +16 = 31 Thus, 32 is a not a perfect number. 84. The proper factors of 496 are: 1,2,4,8,16,31,62,124, and 248. 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496 Thus, 496 is a perfect #

85. a) 60 = 22 • 31•51 Adding 1 to each exponent and then multiplying these numbers, we get (2+1)(1+1)(1+1) = 3 • 2 • 2 = 12 divisors of 60 b) They are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. 86. No, 2 and 4 are not unique prime factors since 4 = 2•2. Any number that 4 divides, 2 will also divide, but 8 does not divide all numbers that are divisible by 4. Some examples are: 4, 12, and 20. 87. One of the numbers must be divisible by 3 and at least one must be even, so their product will be divisible by 2 and 3 and thus by 6. 88. The sum of the groups which have the same three digits will always be divisible by three. (i.e. d + d + d = 3d and 3|3d) 89. 54036 = (54,000 + 36); 54,000 ÷ 18 = 3,000 and 36 ÷ 18 = 2 Thus, since 18 | 54000 and 18 | 36, 18 | 54036.

SECTION 5.2

90. 22 – 1 = 3, 23 – 1 = 7, 25 – 1 = 31, 27 – 1 = 127 are prime numbers, but 211– 1 = 2,048 – 1 = 2,047; and since 23 • 89 = 2,047, 2047 is not prime. 91. 8 = 2+3+3, 9 = 3+3+3, 10 = 2+3+5, 11 = 2+2+7, 12 = 2+5+5, 13 = 3+3+7, 14 = 2+5+7, 15 = 3+5+7, 16 = 2+7+7, 17 = 5+5+7, 18 = 2+5+11, 19 = 3+5+11, 20 = 2+7+11. 92. (a) 329 → 32 −18 = 14;7 |14; yes (b) 553 → 55 − 6 = 49; 7 | 49; yes (c) 583 → 58 − 6 = 52; 7 /| 52; no (d) 4823 → 482 − 6 = 476; 476 = 7 i 68; yes Exercise Set 5.2 1. Whole 2. Add 3. a) Positive b) Negative 4. a) Negative b) Positive 5. a) – 3 < 2 b) – 3 < – 2 c) –3 < 0 d) –3 > –4 6. a) – 5 > – 8 b) – 5 < 8 c) – 5 < – 1 d) – 5 < 0 7. –3 + 5 = 2 8. 5 + (−8) = −3

9. −10 + 3 = −7

10. (−5) + (−5) = −10

11. [6 + (−11)] + 0 = −5 + 0 = −5

12. (2 + 5) + ( –4) = 7 + (–4) = 3

13. [(– 3) + (– 4)] + 9 = –7+9=2

14. [8 + (−3)] + (−2) = [5] + (−2) = 3

15. [(−23) + (−9)] + 11 = [−32] + 11 = −21

16. [5 + (−13)] + 18 = [−8] + 18 = 10

17. 3 − 5 = −2

18. −3 − 7 = −10

19. −6 − 2 = −8

20. 9 − (−4) = 13

21. −5 − (−3) = −5 + 3 = −2

22. −4 − 4 = −4 + (−4) = −8

23. 14 − 20 = 14 + (−20) = −6

24. 8 − (−3) = 8 +3 = 11

25. [5 + (−3)] − 4 = 2 − 4 = 2 + (−4) = −2

26. 6 − (8 + 6) = 6 − 14 = 6 + (−14) = −8

27. −7 • 8 = −56

28. 8(−4) = −32

29. (−9)(−9) = 81

30. −6(31) = −78

31. [(−8)(−2)] • 6 = 16 • 6 = 96

32. (4)(−5)(−6) = (−20)(−6) = 120

33. (5 • 6)(−2) = (30)(−2) = −60

34. (−9)(−1)(−2) = (9)(−2) = −18

35. [(−3)(−6)] • [(−5)(8)] = (18)(−40) = −720

37. –16 ÷ (−2) = 8

38. –54 ÷ 6 = −9

36. [(−8)(4)(5)](−2) = [(−32)(5)](−2) = [−160](−2) = 320 39. 15 ÷ (−15) = −1

40. –90 ÷ 9 = −10

56 = −7 −8 186 44. = −31 −6

43.

−210 = −15 14

41.

−75 = −5 15 45. 144 ÷ (– 3) = – 48 42.

127

128

CHAPTER 5 Number Theory and the Real Number System

46. (−900) ÷ (−4) = 225

47. True; every whole number is an integer.

48. False; Negative numbers are not whole numbers.

49. False; the difference of two negative integers may be positive, negative, or zero.

50. True; the sum of two negative Integers is a negative integer.

51. True; the product of two integers with like signs is a positive integer.

52. False; the difference of a positive integer and a neg. integer will always be positive.

53. True; the quotient of two integers with unlike signs is a negative number.

54. False; the quotient of any two integers with like signs is a positive number.

55. False; the sum of a positive integer and a negative integer could be pos., neg., or zero.

56. False; the product of two integers with unlike signs is always a negative integer.

57. (5 + 7) ÷ 4 = 12 ÷ 4 = 3

58. (−15) ÷ [35 ÷ (−7)] = (−15) ÷ [−5] = 3

59. [(−12)(−3)] − 3 = 36 − 3 = 33

60. [7(−4)] − 8 = −28 − 8 = −36

61. (4 − 8)(3) = (−4)(3) = −12

62. [18 ÷ (−2)](−3) = (−9)(−3) = 27

63. [2 + (−17)] ÷ 3 = [−15] ÷ 3 = −5

64. (5 – 9) ÷ (−4) = (−4) ÷ (−4) = 1

65. [(−22)(−3)] ÷ (2 − 13) = = 66 ÷ (−11) = −6 68. –10, −1, 0, 1, 10, 100

66. [15(−4)] ÷ (−6) = (−60) ÷ (−6) = 10 69. −6, −5, −4, −3, −2, −1

71. − 600 + 200 − 400 − 300 = −1100 1100 feet under water 74. 134 − (−80) = 134 + 80=214o F.

72. 10,067 – 23 – 70 + 285 – 122 = 10,137

67. –9, −6, −3, 0, 3, 6 70. −108, −76, −47, 33, 72, 106

73. 14,495 − (−282) = 14,495 + 282 = 14,777 feet

76. 23 − 10 + 8 − 15 +6 − 2 + 17 = 27 people

77. Division by zero is undefined a because = x always leads to a 2 false statement

75. a) + 1 – (– 8) = + 1 + 8 = 9. There is a 9 hr. time diff. b) – 5 – (– 7) = – 5 + 7 = 2. There is a 2 hr. time diff. −a −1 a a = 78. • = −b −1 b b

SECTION 5.3

79.

−1 + 2 − 3 + 4 − 5 +…99 + 100 = 1− 2 + 3 − 4 + 5…+ 99 −100 50 =–1 −50

80. b) continued: and the 4th square number is 16. The sum of 10 and 16 is 26 and 26 – n = 26 – 4 = 22,

80. a) The next 3 pentagonal numbers are 35, 51, and 70. b) The nth pentagonal number is obtained by adding the nth triangular # (see Section 1.1) to the nth square number (see Section 1.1) and subtracting n. For th example, if n = 4, the 4 triangular number is 10

which is the 4 pentagonal #. The next 5 pentagonal numbers 35, 51, 70, 92, and 117. c) Since 70 is the 7th pentagonal th

number and 92 is the 8 pentagonal number, 72 cannot be a pentagonal number.

81. 0 + 1 – 2 + 3 + 4 – 5 + 6 – 7 – 8 + 9 = 1 (other answers are possible) ⎛ 4⎞ 82. (a) 4 ⎜⎜ 4 − ⎟⎟⎟ = 12 ⎜⎝ 4⎠ 4 4 • 4 − = 17 4

4 4 • 4 − = 15 4 ⎛ ⎞ ⎜⎜4 + 4 ⎟⎟ 4 = 20 ⎜⎝ 4 ⎠⎟

4•4•4 = 16 4

(b)

129

44 − 4 = 10 4

Exercise Set 5.3 1. Integers 2. Numerator 3. Denominator 4. Mixed

5. Improper 6. Terminating 7. Repeating 8. Tenths 9. Hundredths 10. Reciprocal 11. Least 12. Equivalent

13. GCD of 3 and 6 is 3. 3 3÷3 1 = = 6 6÷3 2

14. GCD of 15 and 20 is 5. 15 15 ÷ 5 3 = = 20 20 ÷ 5 4

15. GCD of 28 and 63 is 7. 28 28 ÷ 7 4 = = 63 63 ÷ 7 9

16. GCD of 36 and 56 is 4. 36 36 ÷ 4 9 = = 56 56 ÷ 4 14

17. GCD of 95 and 125 is 5. 95 95 ÷ 5 19 = = 125 125 ÷ 5 25

18. GCD of 13 and 221 is 13. 13 13 ÷ 13 1 = = 221 221 ÷ 13 17

19. GCD of 112 and 176 is 16. 112 112 ÷ 16 7 = = 176 176 ÷ 16 11

20. GCD of 120 and 135 is 15. 120 120 ÷ 15 8 = = 135 135 ÷ 15 9

130

CHAPTER 5 Number Theory and the Real Number System

21.

22.

(7)(2) + 3 14 + 3 17 2 = = = 7 7 7 7

(6)(3) + 5 18 + 5 23 3 = = = 6 6 6 6

23. −1

15 16

=−

(7)(5) + 1

24. −7 = − 5

=−

35 + 1 5

(1)(8) + 1

(1)(2) + 1

27. 1 =

30. 1 = 2

33. −

25. −4

=−

8 2 +1 2

−(12i6 + 1)

= −12

1028 1008 + 20 = 21 21 (48)( 21) + 20 20 = 48 = 21 21

6+2

2i 3 + 2

=−

= 2

38.

41.

13 = 2.16 6

44.

115 = 7.6 15

45. 0.25 =

50. .251 =

3 = 0.375 8

878 15

14 + 5 7

=−

131 10, 000

52. .2345 =

251 1000

51. 0.0131 =

54. Let n = 0.7 , 10n = 7.7

8+ 7 8

= −58

8 15

13 100

2 1 = 10 5 2345 469 = 10000 2000

55. Let n = 0.9 , 10n = 9.9

10n = 7.7

10n = 9.9

− n = 0.2

− n = 0.7

− n = 0.9

9n = 2

9n = 7

9n = 9

15 8

23 = 3.285714 7

10n = 2.2

(2)(7) + 5

46. 0.13 =

49. 0.2 =

48. 0.375 =

185

= 0.875

42.

25 1 = 100 4

870 + 8

(58)(15) + 8

176 + 9

=−

(11)(16) + 9

375 375 ÷ 125 3 = = 1000 1000 ÷ 125 8

175 7 = 1000 40

53. Let n = 0.2 , 10n = 2.2

=−

(1)(8) + 7

35. −

= 0.3

1 = 0.16 6

47. 0.175 =

32.

16 + 15

29. 1 =

40.

= 0.6

8 +1

157 156 + 1 =− 12 12 (13)(12) + 1 =− 12 1 = −13 12

37.

36.

43.

34. −

39.

(2)(4) + 1

31.

26. 11

64 + 15

28. 2 =

6 =

(4)(16) + 15

=−

−(72 + 1)

8 +1

−((1)(16) + 15)

5 7

SECTION 5.3

9n 7 = =n 9 9

9n 2 = =n 9 9 56. Let n = 0.2 , 10n = 2.2

57. Let n = 1.36 , 100 n = 136.36

58. Let n = .135, 1000 n = 135.135

100n = 136.36

1000n = 135.135

− n = 0.2

− n = 1.36

− n = .135

9n = 2

99n = 135.0

999n =135.0

99n 23 = =n 99 99

99n 135 15 = = =n 99 99 11

999n 135 5 = = =n 999 999 37

60. Let n = 2.49 , 100 n = 249.9

61. 1 6 1• 6 6 6÷3 2 • = = = = 3 7 3 • 7 21 21÷ 3 7

10 n = 20.5, 100n = 205.5

100n = 249.9

100n = 205.5

−10 n = 24.9

−10 n = 20.5

90n = 225.0

90n = 185.0

90n 245 5 = = =n 90 90 2

90n 185 37 = = =n 90 90 18

62.

9n 9 = = 1= n 9 9

10n = 2.2

59. Let n = 2.05,

1 5 1 8 1i 8 ÷ = i = = 4 8 4 5 4i5

131

63.

−3 −16 48 2 • = = 8 15 120 5

64. ⎛ 3 ⎞⎟ 10 ⎛ 3 ⎞⎟ 21 ⎜⎜− ⎟ ÷ = ⎜⎜− ⎟ • = − 63 ⎜⎝ 5 ⎠⎟ 21 ⎝⎜ 5 ⎠⎟ 10 50

8 8÷ 4 2 = = 20 20 ÷ 4 5

7 8 7 7 49 3 3 3 7 21 ⎛ 3 4 ⎞ 1 12 1 12 3 36 ÷ = • = 66. ÷ = • = = 1 67. ⎜⎜ • ⎟⎟⎟ ÷ = ÷ = • = ⎜⎝ 5 7 ⎠ 3 35 3 35 1 35 8 7 8 8 64 7 7 7 3 21 ⎛4 4⎞ 1 ⎛4 5⎞ 1 5 1 ⎡⎛ 2 ⎞⎛ 5 ⎞⎤ ⎛ 7 ⎞ ⎛ 10 ⎞ ⎛ 16 ⎞ 160 20 5 68. ⎜⎜ ÷ ⎟⎟⎟ • = ⎜⎜ • ⎟⎟⎟ • = • = = 69. ⎢⎜⎜− ⎟⎟⎟⎜⎜ ⎟⎟⎟⎥ ÷ ⎜⎜− ⎟⎟⎟ = ⎜⎜− ⎟⎟⎟ • ⎜⎜− ⎟⎟⎟ = ⎜⎝ 7 5 ⎠ 7 ⎝⎜ 7 4 ⎠ 7 7 7 49 ⎢⎣⎜⎝ 3 ⎠⎝⎜ 8 ⎠⎥⎦ ⎝⎜ 16 ⎠ ⎝⎜ 24 ⎠ ⎝⎜ 7 ⎠ 168 21 65.

⎛ 7 5 ⎞ ⎛ 7 5 ⎞ ⎛ 35 ⎞⎟ ⎛ 35 ⎞⎟ 35 18 18 3 ÷⎜ = • = = 70. ⎜⎜ • ⎟⎟⎟ ÷ ⎜⎜ • ⎟⎟⎟ = ⎜⎜ ⎜⎝15 8 ⎠ ⎝⎜ 9 2 ⎠ ⎝⎜120 ⎠⎟⎟ ⎝⎜⎜ 18 ⎠⎟⎟ 120 35 120 20

71. The lcm of 5 and 3 is 15. 2 1 ⎛⎜ 2 3 ⎞⎟ ⎛⎜ 1 5 ⎞⎟ 6 5 11 + = ⎜ • ⎟⎟ + ⎜ • ⎟⎟ = + = ⎜ ⎜ 5 3 ⎝ 5 3 ⎠ ⎝ 3 5 ⎠ 15 15 15

72. The lcm of 6 and 4 is 12. 5 3 ⎛⎜ 5 2 ⎞⎟ ⎛⎜ 3 3 ⎞⎟ 10 9 1 − = ⎜ • ⎟⎟ − ⎜ • ⎟⎟ = − = ⎜ ⎜ 6 4 ⎝ 6 2 ⎠ ⎝ 4 3 ⎠ 12 12 12

73. The lcm of 11 and 22 is 22. 2 5 ⎛ 2 2⎞ 5 4 5 9 + = ⎜⎜ • ⎟⎟⎟ + = + = ⎜ 11 22 ⎝11 2 ⎠ 22 22 29 22

74. The lcm of 12 and 36 is 36. 5 7 ⎛⎜ 5 3 ⎞⎟ 7 15 7 22 22 ÷ 2 11 + =⎜ • ⎟+ = + = = = 12 36 ⎜⎝12 3 ⎠⎟ 36 36 36 36 36 ÷ 2 18

75. The lcm of 9 and 54 is 54. 5 7 ⎛⎜ 5 6 ⎞⎟ 7 30 7 23 − = ⎜ ⋅ ⎟− = − = 9 54 ⎜⎝ 9 6 ⎠⎟ 54 54 54 54

132

CHAPTER 5 Number Theory and the Real Number System

76. The lcm of 25 and 100 is 100. 17 43 ⎛⎜ 17 4 ⎞⎟ 43 68 43 25 − = ⎜ • ⎟− = − = 25 100 ⎜⎝ 24 4 ⎠⎟ 100 100 100 100 =

77. The lcm of 12, 48, and 72 is 144. 1 1 1 ⎛⎜ 1 12 ⎞⎟ ⎛⎜ 1 3 ⎞⎟ ⎛⎜ 1 2 ⎞⎟ + + =⎜ • ⎟+⎜ • ⎟+⎜ • ⎟ 12 48 72 ⎜⎝12 12 ⎠⎟ ⎝⎜ 48 3 ⎠⎟ ⎝⎜ 72 2 ⎠⎟

1 4

78. The lcm of 5, 15,and 75 is 75.

79. The lcm of 30, 40, and 50 is 600.

3 7 9 ⎛⎜ 3 15 ⎞⎟ ⎛⎜ 7 5 ⎞⎟ 9 + + =⎜ • ⎟+⎜ • ⎟+ 5 15 75 ⎝⎜ 5 15 ⎠⎟ ⎝⎜15 5 ⎠⎟ 75 45 35 9 89 = + + = 75 75 75 75

1 3 7 ⎛ 1 20 ⎞ ⎛ 3 15 ⎞ ⎛ 7 12 ⎞ − − = ⎜⎜ • ⎟⎟⎟ ⎜⎜ • ⎟⎟⎟ ⎜⎜ • ⎟⎟⎟ 30 40 50 ⎝⎜ 30 20 ⎠ ⎝⎜ 40 15 ⎠ ⎝⎜ 50 12 ⎠ 20 45 84 109 = − − =− 600 600 600 600

80. The lcm of 25, 100, and 40 is 200. 4 9 7 ⎛ 4 8 ⎞ ⎛ 9 2 ⎞⎟ ⎛⎜ 7 5 ⎞⎟ − − = ⎜⎜ • ⎟⎟ ⎜⎜ • ⎟ ⎜ • ⎟ 25 100 40 ⎝⎜ 25 8 ⎠⎟ ⎝⎜100 2 ⎠⎟ ⎝⎜ 40 5 ⎠⎟ =

12 3 2 17 + + = 144 144 144 144

32 18 35 21 − − =− 200 200 200 200

81.

5 1 5 • 3 + 8 •1 15 + 8 23 + = = = 8 3 8•3 24 24

82.

4 2 4 • 5 + 2 • 7 20 + 14 34 + = = = 7 5 7•5 35 35 7 5 7 •12 − 3• 5 84 −15 69 23 − = = = = 3 12 3•12 36 36 12

83.

5 7 5 • 4 − 7 • 3 20 − 21 1 - = = =− 6 8 24 24 24

84.

85.

3 5 3•12 + 8 • 5 36 + 40 76 19 + = = = = 8 12 8 •12 96 96 24

⎛ 2 1 ⎞ 3 ⎛ 2 • 4 + 3•1⎞⎟ 3 8 + 3 3 86. ⎜⎜ + ⎟⎟⎟ − = ⎜⎜ − ⎟− = ⎝⎜ 3 4 ⎠ 5 ⎝⎜ 3• 4 ⎠⎟ 5 12 5 =

⎛1 6 ⎞ 1 6 1 ⎛ 6 4 ⎞ ⎛ 1 21⎞ 87. ⎜⎜ • ⎟⎟⎟ + = + = ⎜⎜ • ⎟⎟⎟ + ⎜⎜ • ⎟⎟⎟ = ⎜⎝ 3 7 ⎠ 4 21 4 ⎝⎜ 21 4 ⎠ ⎝⎜ 4 21⎠ =

24 21 45 15 + = = 84 84 84 28

11 3 11• 5 −12 • 3 55 − 36 19 − = = = 12 5 12 • 5 60 60

⎛ 1 2 ⎞ 1 ⎛ 1 3 ⎞ 1 ⎛ 3 7 ⎞ ⎛ 1 12 ⎞ 88. ⎜⎜ ÷ ⎟⎟⎟ − = ⎜⎜ • ⎟⎟⎟ − = ⎜⎜ • ⎟⎟⎟ − ⎜⎜ • ⎟⎟⎟ ⎜⎝ 6 3 ⎠ 7 ⎝⎜ 6 2 ⎠ 7 ⎝⎜12 7 ⎠ ⎝⎜ 7 12 ⎠ =

21 12 9 3 − = = 84 84 84 28

⎛ 3 1⎞ ⎛ 7⎞ ⎛ 3 3 1 2⎞ ⎛ 2 6 7⎞ ⎛ 9 2 ⎞ ⎛12 7 ⎞ 11 5 11 6 66 11 89. ⎜⎜ + ⎟⎟⎟ ÷ ⎜⎜2 − ⎟⎟⎟ = ⎜⎜ • + • ⎟⎟⎟ ÷ ⎜⎜ • − ⎟⎟⎟ = ⎜⎜ + ⎟⎟⎟ ÷ ⎜⎜ − ⎟⎟⎟ = ÷ = • = = ⎜ ⎜ ⎜ ⎝⎜ 4 6 ⎠ ⎝⎜ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 6 4 3 6 2 1 6 6 12 12 ⎠ ⎝⎜ 6 6 ⎠ 12 6 12 5 60 10 ⎛ 1 3 ⎞ ⎛ 3 10 ⎞ 3 30 1 6 1 11 6 7 11 42 53 90. ⎜⎜ • ⎟⎟⎟ + ⎜⎜ • ⎟⎟⎟ = + = + = • + • = + = ⎝⎜ 3 7 ⎠ ⎝⎜ 5 11 ⎠ 21 55 7 11 7 11 11 7 77 77 77 ⎛ 4⎞ ⎛ 2 ⎞ ⎛ 3 9 4 ⎞ ⎛ 4 3 2 ⎞ ⎛ 27 4 ⎞ ⎛12 2 ⎞ 23 14 23 3 69 23 91. ⎜⎜3 ⎟⎟⎟ ÷ ⎜⎜4 + ⎟⎟⎟ = ⎜⎜ • − ⎟⎟⎟ ÷ ⎜⎜ • + ⎟⎟⎟ = ⎜⎜ − ⎟⎟⎟ ÷ ⎜⎜ + ⎟⎟⎟ = ÷ = • = = ⎜⎝ 9 ⎠ ⎝⎜ ⎜ ⎜ ⎜ ⎜ 3⎠ ⎝1 9 9 ⎠ ⎝ 1 3 3⎠ ⎝ 9 9 ⎠ ⎝ 3 3⎠ 9 3 9 14 126 42 ⎛ 2 4 ⎞⎛ 3 ⎞ ⎛ 2 9 ⎞⎛ 3 6 ⎞ 18 18 9 18 162 81 = 92. ⎜⎜ ÷ ⎟⎟⎟⎜⎜ • 6⎟⎟⎟ = ⎜⎜ • ⎟⎟⎟⎜⎜ • ⎟⎟⎟ = • = • = ⎜⎝ 5 9 ⎠⎝⎜ 5 ⎠ ⎝⎜ 5 4 ⎠⎝⎜ 5 1 ⎠ 20 5 10 5 50 25

SECTION 5.3

93.

5 13 → 70 8 8 7 7 −69 → −69 8 8 6 3 1 → 1 inches 8 4 71

94. The LCM of 2, 4, 6 is 12. 1−

1 1 ⎛ 1 6 ⎞ ⎛ 1 3⎞ ⎛ 1 2 ⎞ 6 3 2 11 + + = ⎜ • ⎟⎟⎟ + ⎜ • ⎟⎟⎟ + ⎜ • ⎟⎟⎟ = + + = ; 2 4 6 ⎜⎝ 2 6 ⎠ ⎝⎜ 4 3 ⎠ ⎝⎜ 6 2 ⎠ 12 12 12 12 1

11 1 = musk thistles 12 12

95. ⎛ 5⎞ ⎛ 69 ⎞ 966 966 ÷ 2 483 = = = 120.75" 14 ⎜⎜8 ⎟⎟⎟ = 14 ⎜⎜ ⎟⎟⎟ = ⎝⎜ 8 ⎠ ⎝⎜ 8 ⎠ 8 8÷2 4 96. ⎛ 1 ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜⎜1 ⎟⎟⎜⎜ 1 ⎟⎟ = ⎜⎜ 3 ⎟⎟⎜⎜ 1 ⎟⎟ = 3 cups of snipped parsley ⎟⎜ 4 ⎠⎟ ⎝⎜ 2 ⎠⎝ ⎟⎜ 4 ⎠⎟ 8 ⎜⎝ 2 ⎠⎝

⎛ 1 ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜⎜1 ⎟⎟⎜⎜ 1 ⎟⎟ = ⎜⎜ 3 ⎟⎟⎜⎜ 1 ⎟⎟ = 3 tsp of pepper ⎟ ⎟⎜ 8 ⎠⎟ 16 ⎝⎜ 2 ⎠⎝⎜ 8 ⎠⎟ ⎝⎜ 2 ⎠⎝ ⎛ 1 ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎜⎜1 ⎟⎟⎜⎜ 1 ⎟⎟ = ⎜⎜ 3 ⎟⎟⎜⎜ 1 ⎟⎟ = 3 cups of sliced carrots ⎟⎜ 2 ⎠⎟ ⎝⎜ 2 ⎠⎝ ⎟⎜ 2 ⎠⎟ 4 ⎝⎜ 2 ⎠⎝ 97. ⎛1 1 1⎞ 1 7 1 4 14 4 1−⎜⎜ + + ⎟⎟⎟ 2 + 3 + 4 = 2 + 3 + 4 ⎜⎝ 4 5 2 ⎠ 4 8 4 16 16 16

98. The LCM of 4, 5, 3 is 60. 1 2 1 ⎛⎜ 1 ⎞⎛ 15 ⎞ ⎛ 2 ⎞⎛12 ⎞ ⎛ 1 ⎞⎛ 20 ⎞ + + = ⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟ + ⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟ + ⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟ ⎜ ⎜ 4 5 3 ⎝ 4 ⎠⎝15 ⎠ ⎝⎜ 5 ⎠⎝⎜12 ⎠ ⎝⎜ 3 ⎠⎝⎜ 20 ⎠ =

22 6 = 9 = 10 16 16 5 6 21 6 15 20 −10 = 19 −10 = 9 " 16 16 16 16 16 99. ⎛1 2⎞ ⎛5 4⎞ 9 10 9 1 1−⎜⎜ + ⎟⎟⎟ = 1−⎜⎜ + ⎟⎟⎟ = 1− = − = ⎝⎜ 2 5 ⎠ ⎝⎜10 10 ⎠ 10 10 10 10

15 60

24 60

20 60

59 60

⎛ 1⎞ ⎛ 5 ⎞⎛15 ⎞ 75 3 = 18 cups 100. ⎜⎜1 ⎟⎟⎟(15) = ⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟ = ⎜⎝ 4 ⎠ ⎜⎝ 4 ⎠⎝⎜ 1 ⎠ 4 4

Student tutors represent 0.1 of the budget. 101. 1 1 1 4 2 1 7 4 + 30 + 24 = 4 + 30 + 24 = 58 inches 2 4 8 8 8 8 8

⎛ 7⎞ 199 1 199 7 • = = 12 in. 102. ⎜⎜ 24 ⎟⎟⎟ ÷ 2 = ⎜⎝ 8 ⎠ 8 2 16 16

48 6 3 9 22 103. a) 1 49 60 , 2 60 , 9 60 , 6 60 , 2 60 , 60

48 6 3 9 22 b) 1 49 60 + 2 60 + 9 60 + 6 60 + 2 60 + 60

= (1 + 2 + 9 + 6 + 2 + 0) + = 20 +

49 + 48 + 6 + 3 + 9 + 22 60

137 = 22 17 60 or 22 hours, 17 minutes 60

133

134

CHAPTER 5 Number Theory and the Real Number System

25 1 23 28 104. a) 60 , 7 54 60 , 3 60 , 1 60 , 60 25 23 28 1 b) 60 + 7 54 60 + 3 60 + 1 60 + 60

= (0 + 7 + 3 + 1 + 0) + = 11 +

25 + 54 + 1 + 23 + 28 60

131 = 13 11 60 or 13 hours, 11 minutes 60

⎛ 35 12 ⎞ 3 105. 8 ft = ⎜⎜ ⋅ ⎟⎟⎟ in. = 105 in. ⎜⎝ 4 1 ⎠ 4 ⎡ ⎛ 1 ⎞⎤ ⎡ 840 3 ⎤ 837 1 837 5 5 ⎢105 − (3) ⎜⎜ ⎟⎟⎥ ÷ 4= ⎢ . The length of each piece is 26 − ⎥ ÷ 4= in. • = =26 ⎜⎝ 8 ⎠⎟⎥ ⎢⎣ 8 ⎦⎥ 8 4 32 32 32 ⎣⎢ 8 ⎦

106. width = 8 ft. 3 in. = 96 in. + 3 in. = 99 in.; length = 10 ft. 8 in. = 120 in. + 8 in. = 128 in. a) perimeter = 2L + 2W = 2(128) + 2(99) = 454 in

454" 12"/ft.

= 37

10 5 ft = 37 ft 12 6

3 1 33 8 ft. = 10 2 ft. = 32 ft ft .; length = 10ft. 8in. = 10 12 b) width = 8ft. 3in. = 8 ft. = 8 ft. = 3 4 3 12 4 Area = L× w =

32 3

c) Volume = L ⋅ W ⋅ H =

33 4

1056 12

= 88 sq.ft

32 33 55 58080 2 × × = = 806 cu. ft. 3 4 6 72 3

3 3 3 = 29 in. 8 16 16 1 3 b) 26 + 6 = 33 in. 4 4 1 ⎛ 3 1⎞ 1 2 3 c) 26 + ⎜⎜6 − ⎟⎟⎟⎟ = 26 + 6 = 32 in. ⎜ ⎝ ⎠ 4 4 4 4 4 4

107. a) 20 + 18 ÷ 2 = 20 + 9

1 2

1 4

108. original area = 8 • 9 =

17 37 629 5 1 1 17 41 697 1 = = 78 sq. in.; new area = 8 • 10 = • = = 87 sq.in. • 2 4 8 8 2 4 2 4 8 8

1 5 9 5 4 1 area increase = 87 − 78 = 86 − 78 = 8 = 8 sq. in. 8 8 8 8 8 2 0.21 + 0.22 0.43 4.005 + 4.05 8.055 = = 0.215 = = 4.0275 109. 110. 2 2 2 2 1.3457 + 1.34571 2.69141 −2.176 + (−2.175) − 4.351 = 1.345705 = = − 2.1755 = 111. 112. 2 2 2 2

113.

4.872 + 4.873 9.745 = 4.8725 = 2 2

⎛1 3⎞ 4 1 4 1 115. ⎜⎜ + ⎟⎟⎟ ÷ 2 = • = = ⎜⎝ 4 4 ⎠ 4 2 8 2

−3.7896 + (−3.7895) −7.5791 = = 2 2 −3.78955 ⎛1 2⎞ 3 1 3 116. ⎜⎜ + ⎟⎟⎟ ÷ 2 = • = ⎜⎝ 5 5 ⎠ 5 2 10 114.

SECTION 5.4

⎛ 1 1⎞ 11 1 11 117. ⎜⎜ + ⎟÷ 2 = • = ⎜⎝100 10 ⎠⎟⎟ 100 2 200 119. ⎛ 1 1 ⎞⎟ ⎛ ⎞ ⎜⎜ + ⎟ ÷ 2 = ⎜⎜ 1 + 11 ⎟⎟ • 1 = 12 • 1 = 12 = 2 ⎜⎝ 99 9 ⎠⎟ ⎜⎝ 99 99 ⎠⎟ 2 99 2 198 33 ⎛

3⎞

⎛4

7⎞ 1

⎛7 8⎞ 15 1 15 118. ⎜⎜ + ⎟⎟⎟ ÷ 2 = • = ⎜⎝13 13 ⎠ 13 2 26 ⎛1 2⎞ ⎛ 3 4⎞ 1 7 1 7 120. ⎜⎜ + ⎟⎟⎟ ÷ 2 = ⎜⎜ + ⎟⎟⎟ • = • = ⎜⎝ 2 3 ⎠ ⎜⎝ 6 6 ⎠ 2 6 2 12 11 1

121. a) Water (or milk): ⎜⎜⎜1 + 1 ⎟⎟⎟ ÷ 2 = ⎜⎜⎜ + ⎟⎟⎟ • = • = = 1 cup; ⎝ ⎝4 4⎠ 2 4 2 8 4⎠ 8 ⎛1

⎞

3 1

Oats: ⎜⎜⎜ + 1⎟⎟⎟ ÷ 2 = • = cup ⎝2 ⎠ 2 2 4 1 1 b) Water (or milk): 1 + = 1 cup; 2 2 1 1 1 3 Oats: + ⋅ = cup 2 2 2 4 1 2 1 2 3 122. a) 1 b) 0.9 c) = 0.3, = 0.6, + = = 1, 0.3 + 0.6 = 0.9 3 3 3 3 3 123. a)

1 8

1 2

1 8

1 16

Exercise Set 5.4 1. Irrational 2. Radical 3. Radicand 4. Coefficient 5. Itself 6. Perfect 7. Rationalized 8. a) Exact b) Approximation

10.

irrational

14. Irrational; π is non-terminating, non-repeating. 16. Rational; terminating decimal 5

= 1 1 is an integer.

9 =3

rational

2 rational 3 13. Irrational; non-terminating, non-repeating decimal 15. Rational; quotient of two integers 11.

12. Irrational; non-terminating, non-repeating decimal

18. Rational;

d) 0.9 =1

19.

17. Irrational; non-terminating, non-repeating decimal

0=0

20.

1 =1

21.

36 = 6

22. – 121 = – 11

23. – 169 = – 13

24. 25 = 5 27. 0, Rational, integer

25. – 81 = – 9 28. – 5, Rational, integer

26. – 36 = – 6

3 , Rational 8 33. Rational 36. Irrational

31. Irrational

9 , Rational, integer, natural 32. Rational

34. Rational

35. Rational

30.

39.

45 = 9 5 = 3 5

29.

37.

18 = 9 2 = 3 2

38.

24 = 4 6 = 2 6

40.

50 = 25 2 = 5 2

41.

63= 9 7=3 7

135

136

42.

CHAPTER 5 Number Theory and the Real Number System

75= 25 3=5 3

84 = 4 21 = 2 21

44.

46. 5 3 − 7 3 = (5 − 7) 3 =

47.

43.

45. 2 3 + 5 3 = (2 + 5) 3 = 7 3

3 20 − 2 45 = 3

−2 3

−2

48. 2 18 − 3 50 = 2 −3

( 9 2)

( 25 2 ) = 6 2 −15 2 =

−9 2

51. 5 3+7 12 − 3 75

57.

60.

= −13 3

=12 7

63+13 98 − 5 112 =3 7 +91 2 − 20 7

=(5+14 −15) 3 = 4 3

= −17 7 + 91 2

5 7

50 14

= 29

69.

=2 7+10 7 =(2+10) 7

=5 3+14 3 −15 3

63.

66.

= 4 • 2 3 − 7 • 3 3 = 8 3 − 21 3

=3 7 +13 ⋅ 7 2 − 5 ⋅ 4 7

145

7 5

⋅

50 25 = 14 7

55.

35 5

5 7 7

11 is between 3 and 4 since

6 14 = 6⋅14 = 84

2 7 +5 28=2 7 +5• 2 7

53. 2 8 = 2 ⋅ 8 = 16 = 4

56.

58.

125 = 25= 5 5 1

64.

67.

62.

2 2 7 14 = ⋅ = 7 7 7 7

65.

2 2

60 6

2 15 = 6

15 3

11 is between

⋅

6 =

59.

2 2

2 10 = 2 ⋅ 10 = 20 = 4 5=2 5

= 4 21 = 2 21

61.

( 9 5) = 6 5 − 6 5 = 0

50.

=5 3+7 • 2 3 − 3 • 5 3

20 20 = = 4=2 5 5

( 4 5)

49. 4 12 − 7 27 = 4 4 3 − 7 9 3

52.

54. 3 21 = 3⋅ 21 = 63 = 9 7 = 3 7

90= 9 10=3 10

9 = 3 and 16 = 4 .

closer to 9 than to 16. Using a calculator 11 ≈ 3.3 .

68.

72 = 9= 3 8 2

20 3

2 2 = 2 2

⋅

4 15 2 15 = 3 3

27 =

6 81

2 27 =

⋅

3 3

6 9

11 is between 3 and 3.5 since 11 is

SECTION 5.4

70.

67 is between 8 and 9 since

67 is between

64 = 8 and 81 = 9 .

137

5 is between 8 and 8.5 since 67 is

closer to 64 than to 81. Using a calculator 67 ≈ 8.2 . 71.

107 is between 10 and 11 since

107 is between

100 = 10 and 121 = 11 .

107 is between 10 and 10.5

since 107 is closer to 100 than to 121. Using a calculator 107 ≈ 10.3 . 72.

135 is between 11 and 12 since

135 is between

121 = 11 and

170 is between 13 and 14 since

170 is between

169 = 13 and

200 is between 14 and 15 since

200 is between

196 = 14 .

170 is between 13 and 13.5

170 ≈ 13.04 .

since 170 is closer to 169 than to 196. Using a calculator 74.

135 is between 11.5 and 12

135 ≈ 11.6 .

since 135 is closer to 144 than to 121. Using a calculator 73.

144 = 12 .

196 = 14 and

225 = 15 .

200 is between 14 and 14.5

since 200 is closer to 196 than to 225. Using a calculator 200 ≈ 14.1 . 75. False. The result may be a rational number or an irrational number. 76. False.

p is an irrational number for any prime number p.

77. True

78. True

79. False. The result may be a rational number or an irrational number.

80. False. The result may be a rational number or an irrational number. 81. 3 2+5 2= 8 2 84.

82. π +( − π ) = 0

2 ⋅ 6 = 16= 4

83.

2 ⋅ 3= 6

85. H = 2.9 30 + 20.1 = 36.0 inches

86.

T = 2π

= 2π

35 35 5 7 = 2π = 2π 980 980 5 196 7 2π 7 π 7 = = ≈ 1.2 seconds 14 2•7 7

87. a) s =

= 100 = 10 mph

0.04

b) s =

= 400 = 20 mph

0.04

c) s =

= 1600 = 40 mph

0.04

d) s =

256

= 6400 = 80 mph

0.04

88. a) t =

100 10 = = 2.5 sec 4 4

b) t =

400 20 = = 5 sec 4 4

c) t =

900 30 = = 7.5 sec 4 4

d) t =

1600 40 = = 10 sec 4 4

89. No. 2 ≠ 1.414 since 2 is an irrational number and 1.414 is a rational number.

138

90.

CHAPTER 5 Number Theory and the Real Number System

4.123 is rational, 92.

22 are rational numbers, π is an 7 irrational number.

17 is irrational. Because

91. No. 3.14 and

17 ≠ 4.123 .

9 + 16 ≠ 9 + 16 5

94.

≠

4•9 = 4 9

93.

25 ≠ 3 + 4

36 = 2 • 3

36 36 6 = , 4 = , 2= 2 9 3 9

95. a)

0.04 = 0.2 a terminating decimal and thus it is rational. 7 70 = ; 70 is irrational since 10 10 the only integers with rational square roots are the perfect squares and 70 is not a perfect square. 0.7 =

Thus

70 = 0.7 is irrational. 10

96. No. The sum of two irrational numbers may not be irrational. (i.e. – 3+ 3 = 0)

Exercise Set 5.5 1. Real 2. 3. Closed 4. Commutative 5. Commutative 6. Associative 7. Associative 8. Distributive 11. Not closed. (e.g., 3 − 5 = −2 is not a natural number). 12. Closed. The product of two natural numbers is a natural number. 14. Closed. The sum of two integers is an integer. 15. Closed. The product of two integers is an integer. 16. Closed. The difference of two integers is an integer.

17. Closed 21. Not closed 25. Closed

18. Closed 22. Not closed 26. Closed

19. Not closed 23. Not closed 27. Not closed (for example, 3 ÷ 0 is not a real number) 29. Commutative property of addition. The order 5 + x is changed to x + 5. .

20. Closed 24. Not closed 28. Closed

SECTION 5.5

30. ( x + 5) + 6 = x + (5 + 6) ;

Associative because the grouping of the three terms is changed.

31. (−3) + (−4) = −7 = (−4) + (−3)

32. (−3) • (−4) = 12 = (−4) • (−3)

33. No. 4 − 3 = 1, but 3 − 4 = −1

3 4 34. No. 3 ÷ 4 = , but 4 ÷ 3 = 4 3

35. [(−2) + (−3]) + (−4) = (−5) + (−4) = −9

36. [(−2) • (−3]) • (−4) = (6) • (−4) = −24

(−2) + [(−3]) + (−4)] = (−2) + (−7) = −9

(−2) • [(−3]) • (−4)] = (−2) • (12) = −24 38. No. (10 − 4) − 2 = 6 − 2 = 4, but

37. No.

10 − (4 − 2) = 10 − 2 = 8

(16 ÷ 8) ÷ 2 = 2 ÷ 2 = 1, but 16 ÷ (8 ÷ 2) = 16 ÷ 4 = 4 39. No. (81 ÷ 9) ÷ 3 = 9 ÷ 3 = 3,

40. No. 2 + (3• 4) = 2 + 12 = 14, but ( 2 + 3) • ( 2 + 4)

but 81 ÷ (9 ÷ 3) = 81 ÷ 3 = 27

= 5• 6 = 30 42. 6 + 7 = 7 + 6 Commutative property of addition

41. 3( y + 5) = 3 ⋅ y + 3 ⋅ 5 Distributive property

43. (7 • 8) • 9 = 7 • (8 • 9) Associative property of multiplication 45. ( 24 + 7) + 3 = 24 + (7 + 3)

44. c + d = d + c Commutative property of addition 46. 4 • (11• x ) = (4 •11) • x Associative property of multiplication

Associative property of addition 47.

3•7 = 7• 3 Commutative property of multiplication

48.

3 ⎛⎜ 1 3 ⎟⎞ ⎛⎜ 3 1 ⎞⎟ 3 +⎜ + ⎟ = ⎜ + ⎟+ 8 ⎝⎜ 8 2 ⎟⎠ ⎝⎜ 8 8 ⎠⎟ 2

Associative property of addition 49. −1( x + 4) = (−1) ⋅ x + (−1) ⋅ 4

50. ( r + s ) ⋅ t = ( r ⋅ t ) + ( s ⋅ t ) Distributive property

Distributive property 51. ( r + s ) + t = t + ( r + s ) Commutative property of addition

52. g ⋅ (h + i ) = ( h + i ) ⋅ g Commutative property of multiplication

53. 4 ( z + 1) = 4 z + 4

54. −1(a + b) = −a − b

3 3 3 3 55. − ( x −12) = − x + ⋅ 12 = − x + 9 4 4 4 4

56. − x (− y + z ) = (− x )(− y ) + (− x )( z ) = xy − xz

⎛ x 2 ⎞ 6 x 12 57. 6⎜⎜ + ⎟⎟⎟ = + = 3x + 4 ⎜⎝ 2 3 ⎠ 2 3

⎛ x 1 ⎞ 24 x 24 − = 8x − 3 58. 24 ⎜⎜ − ⎟⎟⎟ = ⎜⎝ 3 8 ⎠ 3 8

⎛1 1 ⎞ 32 x 32 59. 32 ⎜⎜ x − ⎟⎟⎟ = − = 2 x −1 ⎜⎝16 32 ⎠ 16 32

⎛2 4 ⎞ 30 x 60 − = 10 x − 12 60. 15⎜⎜ x − ⎟⎟⎟ = ⎜⎝ 3 5⎠ 3 5

139

140

CHAPTER 5 Number Theory and the Real Number System

( 8 − 2 ) = 16 − 4 = 4 − 2 = 2

61.

63. 5

( 2 + 3) = 5 2 + 5 3

(

)

62. −3 2 − 3 = −6 + 3 3

64.

( 15 − 20 ) = 75 − 100 = 5 3 −10

65. a) Distributive property b) Associative property of addition

66. a) Distributive property b) Associative property of addition

67. a) Distributive property b) Associative property of addition c) Commutative property of addition d) Associative property of addition 69. a) Distributive property b) Commutative property of addition c) Associative property of addition d) Commutative property of addition 71. Yes. You can either feed your cats first or give your cats water first. 73. No. The clothes must be washed first before being dried. 75. No. Pressing the keys will have no effect if there are no batteries in place. 77. Yes. The order does not matter. 79. Yes. The order does not matter 81. Baking pizzelles: mixing eggs into the batter, or mixing sugar into the batter.; Yard work: mowing the lawn, or trimming the bushes 84.

68. a) Distributive property b) Associative property of addition c) Commutative property of addition d) Associative property of addition 70. a) Distributive property b) Commutative property of addition c) Associative property of addition d) Commutative property of addition 72. Yes. Can be done independently; no order needed 74. No. The PC must be turned on first before you can type a term paper. 76. Yes. The order does not matter 78. Yes. The order does not matter. 80. No. The egg cannot be poured before it is cracked. 82. Washing siding/washing windows/washing the car Writing letters to spouse, parents or friends 83. No. 0 ÷ a = 0 but a ÷ 0 is undefined.

85. a) No. (Man eating) tiger is a tiger that eats men, and man (eating tiger) is a man that is eating a tiger. b) No. (Horse riding) monkey is a monkey that rides a horse, and horse (riding monkey) is a horse that rides a monkey. c) Answers will vary.

Exercise Set 5.6 1. Exponent

2. Base

5. x 3

4. x 5 6. 1

8. x 6

3. x

SECTION 5.6

9. 2.013×103 10. 3.4×10−3 11. 0.0000573 12. 1776

13. a) 32 = 3 ⋅ 3 = 9 b) 23 = 2 ⋅ 2 ⋅ 2 = 8

14. a) 25 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 32

15. a) (−5)2 − (−5)(−5) = 25

b) 52 = 5 ⋅ 5 = 25

b) −52 = −(5)(5) = −25

16. a) −32 = −(3)(3) = −9

17. a) −24 = −(2)(2)(2)(2) = −16

b) (−3) 2 = (−3)(−3) = 9

b) (−2) 4 = (−2)(−2)(−2)(−2) = 16 19. a) −43 = −(4)(4)(4) = −64

18. a) (−3)4 = (−3)(−3)(−3)(−3) = 81

b) (−4)3 = (−4)(−4)(−4) = −64

b) −34 = −(3)(3)(3)(3) = −81 20. a) (−2)7 = −( 27 ) = −128 (an odd power of a negative number is negative)

b) −27 = −( 27 ) = −128

⎛ 1 ⎞2 ⎛ 1 ⎞⎛ 1 ⎞ 1 21. a) −⎜⎜ ⎟⎟⎟ = −⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟ = − ⎝⎜ 2 ⎠ ⎝⎜ 2 ⎠⎝⎜ 2 ⎠ 4 ⎛ 1 ⎞2 ⎛ 1 ⎞⎛ 1 ⎞ 1 b) ⎜⎜− ⎟⎟⎟ = ⎜⎜− ⎟⎟⎟⎜⎜− ⎟⎟⎟ = ⎝⎜ 2 ⎠ ⎝⎜ 2 ⎠⎝⎜ 2 ⎠ 4

⎛ 1 ⎞3 ⎛ 1 ⎞⎛ 1 ⎞⎛ 1 ⎞ 1 22. a) −⎜⎜ ⎟⎟⎟ = −⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟ = − ⎜⎝ 2 ⎠ ⎜⎝ 2 ⎠⎝⎜ 2 ⎠⎝⎜ 2 ⎠ 8 ⎛ 1 ⎞3 ⎛ 1 ⎞⎛ 1 ⎞⎛ 1 ⎞ 1 b) ⎜⎜− ⎟⎟⎟ = ⎜⎜− ⎟⎟⎟⎜⎜− ⎟⎟⎟⎜⎜− ⎟⎟⎟ = − ⎜⎝ 2 ⎠ ⎝⎜ 2 ⎠⎝⎜ 2 ⎠⎝⎜ 2 ⎠ 8 24. a) 20121 = 2012 b) 12012 = 1

23. a) 1001 = 100 b) 1100 = 1 25. a) 32 ⋅ 33 = 32 + 3 = 35 = 243

26. a) 23 ⋅ 2 = 23+1 = 24 = 16

b) (−3)2 ⋅ (−3)3 = (−3)2 + 3 = (−3)5 = −243 57 = 57 − 5 = 52 = 25 55 (−5)7 = (−5)7 − 5 = (−5) 2 = 25 b) (−5)5

b) (−2)3 ⋅ (−2) = (−2)3+1 = (−2)4 = 16 45 = 45− 2 = 43 = 64 42 (−4)5 = (−4)5− 2 = (−4)3 = −64 b) (−4)2

27. a)

28. a)

29. a) 60 = 1

30. a) (−6)0 = 1

b) −60 = −1 31. a) (6 x )0 = 1 b) 6 x 0 = 6 1 1 33. a) 3−3 = 3 = 27 3 1 1 −2 b) 7 = 2 = 49 7

b) −(−6)0 = −1 32. a) −6 x 0 = −6 b) (−6 x )0 = 1 1 5 1 1 b) 2−4 = 4 = 16 2

34. a) 5−1 =

141

142

CHAPTER 5 Number Theory and the Real Number System

1 1 35. a) −9−2 = − 2 = − 81 9 1 1 b) (−9)−2 = = 2 81 (−9)

1 1 36. a) −5−2 = − 2 = − 25 5 1 1 b) (−5)−2 = = 2 25 (−5)

37. a) ( 22 ) = 22×3 = 26 = 64

⎛⎛ 1 ⎞2 ⎞⎟3 ⎛ 1 ⎞2×3 ⎛ 1 ⎞6 1 ⎜ 38. a) ⎜⎜⎜⎜ ⎟⎟⎟ ⎟⎟⎟ = ⎜⎜ ⎟⎟⎟ = ⎜⎜ ⎟⎟⎟ = ⎝⎜ 2 ⎠ 64 ⎜⎝⎜⎝ 2 ⎠ ⎠⎟ ⎝⎜ 2 ⎠

b) ( 23 ) = 23×2 = 26 = 64 2

⎛⎛ 1 ⎞2 ⎞⎟3 ⎛ 1 ⎞2×3 ⎛ 1 ⎞6 1 ⎜ b) ⎜⎜⎜⎜− ⎟⎟⎟ ⎟⎟⎟ = ⎜⎜− ⎟⎟⎟ = ⎜⎜− ⎟⎟⎟ = ⎜ ⎜ ⎜ ⎝ 2⎠ 64 ⎜⎝⎝ 2 ⎠ ⎠⎟ ⎝ 2 ⎠

39. a) 43 ⋅ 4−2 = 43− 2 = 41 = 4 b) 2−2 ⋅ 2−2 = 2−2 − 2 = 2−4 =

40. a) (−5)−2 ⋅ (−5)−2 = (−5)−2 − 2 = (−5)−4 =

1 1 = 4 16 2

1 1 = (−5)4 625

b) −5 − 5

(−1)−5 (−1)−5 = (−1)

= (−1)−10 =

1 =1 (−1)10

41. 415000 = 4.15 x 105

42. 923000000 = 9.23 x 108

43. 0.00275 = 2.75 x 10−3

44. 0.000034 = 3.4 x 10−5

45. 0.56 = 5.6 x 10−1

46. 0.00467 = 4.67 x 10−3

47. 19000 = 1.9 x 104

48. 1260000000 = 1.26 x 109

49. 0.000186 = 1.86 x 10−4

50. 0.0003 = 3.0 x 10−4

51. 0.00000423 = 4.23 x 10−6

52. 54000 = 5.4 x 104

53. 4.2×104 = 42, 000

54. 3.9×105 = 390, 000

55. 1.32×10−2 = 0.0132

56. 4.003×107 = 40, 030,000

57. 8.62 x 10−5 = 0.0000862

58. 2.19 x 10−4 = 0.000219

59. 3.12 x 10−1 = 0.312

60. 4.6 x 101 = 46

61. 9.0 x 106 = 9,000,000

62. 7.3 x 104 = 73,000

63. (2 x 103 )(3.1 x 102 ) = 620,000

64.

65. (5.1 x 101 )(3.0 x 10−4 ) = 15.3 x 10−3 = 0.0153 67. 69. 71.

68.

5.2×107 = 1.3×10−2 = 0.013 4 ×109

= 2.1 x 10−3 = 0.0021

70.

25.0 x 103 = 5.0 x 105 = 500,000 5.0 x 10−2

= 2.0 x 101 = 20

72.

16.0 x 103 = 2.0 x 106 = 2,000,000 8.0 x 10−3

7.5×106 = 2.5×102 = 250 3×104 8.4 x 10−6 −3

4.0 x 10

4.0 x 105 2.0 x 10

(4.9 x 104 )(2 x 103 ) = 98,000,000 66. (1.6 x 10−2 )(4.0 x 10−3 ) = 6.4 x 10−5 = 0.000064

73. (1 x 106 )(2.3 x 105 ) = 2.3 x 1011

74. (2 x 106 )(2.5 x 10−4 ) = 5 x 102

75. (3.0 x 10−3 )(1.5 x 10−4 ) = 4.5 x 10−7

76. (2.3 x 105 )(3.0 x 103 ) = 6.9 x 108

77.

5.6×106 = 0.7 ×102 = 7 ×101 8×104

78.

2.8×104 = 0.7×107 = 7×106 4×10−3

79.

4.0×10−5 = 2.0 x 10- 7 2.0×102

80.

1.2×10−3 = 0.2 x 103 = 2.0 x 102 6×10−6

81.

1.5×105 = 0.3 x 109 = 3.0 x 108 5.0×10−4

82.

2.4×104 = 0.3×10-2 = 3.0 ×10-3 8.0×106

SECTION 5.6

83. 3.6×10−3 , 1.7, 9.8×102 , 1.03×104

84. 8.5 x 10–5, 1.3 x 10–1, 8.2 x 103 , 6.2 x 104

85. 8.3 x 10–5; 0.00079; 4.1 x 103 ; 40,000 ; Note: 0.00079 = 7.9 x 10– 4, 40,000 = 4 x 104 86. 1,962,000; 4.79 x 106 ; 3.14 x 107 ; 267,000,000 8.38×1012 ≈ 2.802676×104 = 28, 026.76 or $28, 026.76 2.99×108 b) $42, 261.71− $28, 026.76 = $14, 234.95

87. a)

9.23×1011 ≈ 1.469745×104 = 14, 697.45 or £14,697.45 7 6.28×10 b) £ 14, 697.45× $1.50 ≈ $22, 046.18

88. a)

89.

3.095×108 ≈ 0.45×10−1 ≈ 0.045 6.830×109

90.

1.319×109 ≈ 0.193×100 = 0.193 2.806×109

91.

1.78×107 ≈ 0.2050×104 = 2050 people per square kilometer 8.683×103

92.

1.435×107 ≈ 0.29649×105 = 29, 649 people per square kilometer 4.84×102

93.

1.43×1013 ≈ 0.46429×105 = 46, 429 or $46, 429 8 3.08×10

94.

4.91×1012 ≈ 3.678×103 = 3678 or $3678 1.335×109

95. t =

d 239000 mi = = 11.95 r 20000 mph

11.95 hrs

96. t =

d 4.5×108 = = 1.8×104 r 2.5×104

18,000 hrs.

97 2.5×1013 = (2.5×104 )(t ) 2.5×1013 =t 2.5×104 1.25×109 hours = t 7.6×1065 ≈ 4.343×1050 seconds 1.75×1015 99. (100,000 cu ft/sec) (60 sec/min) (60 min/hr) (24 hr) = 8,640,000,000 ft3 or 8.64 x 109 cu ft

98.

100. (500,000)(40,000,000,000) = (5 x 105)(4 x 1010) = 20 x 1015 = 2 x 1016 2.0 x 1016 drops

143

144

CHAPTER 5 Number Theory and the Real Number System

101. (50)(5,800,000) = (5 x 101)(5.8 x 106) = 29 x 107 = 2.9 x 108 2.9 x 108 cells

102.

4.5×109 = 1.8 x 104 or 18,000 times 2.5×105

103. a) (0.60) (1,200,000,000) = $720,000,000 b) (0.25) (1,200,000,000) = $300,000,000 c) (0.10) (1,200,000,000) = $120,000,000 d) (0.05) (1,200,000,000) = $60,000,000 104. a) (0.40) (3,400,000,000) = $1,360,000,000 b) (0.40) (3,400,000,000) = $1,360,000,000 c) (0.10) (3,400,000,000) = $340,000,000 d) (0.10) (3,400,000,000) = $340,000,000 105. 1,000 times, since 1 meter = 103 millimeters = 1,000 millimeters 106. Since 1 gram = 103 milligrams and 1 gram = 10- 3 kilograms,

10- 3 kilograms = 103 milligrams

10−3 Kilograms 103 milligrams = , Thus, 1 kilogram = 106 milligrams 10-3 10-3 2×1030 = 0.3×106 ≈ 333,333 times 6×1024 108. a) 1,000,000 = 1.0 x 106 ; 1,000,000,000 = 1.0 × 109 ; 1,000,000,000,000 = 1.0 × 1012

107.

1.0×106 = 1.0×103 days or 1,000 days ≈ 2.74 years 1.0×103

1.0×109 = 1.0×106 days or 1,000,000 days ≈ 2,740 years 1.0×103

1.0×1012 = 1.0×109 days or 1,000,000,000 days ≈ 2,740,000 years 1.0×103

1 billion 1.0×109 = = 1.0×103 = 1,000 times greater 1 million 1.0×106 109. a) (1.86 x 105 mi/sec) (60 sec/min) (60 min/hr) (24 hr/day) (365 days/yr) (1 yr) e)

= (1.86 x 105)(6 x 101)(6 x 101)(2.4 x 101)(3.65 x 102) = 586.5696 x 1010 ≈ 5.87 x 1012 miles d 9.3×107 b) t= = = 5.0 x 102 = 500 seconds or 8 min. 20 sec. r 1.86×105 110. a) E(0) = 210 x 20 = 210 x 1 = 1024 bacteria b) E(1/2) = 210 x 21/2 = 210.5 ≈ 1448 bacteria

Exercise Set 5.7 1. Sequence 2. Terms 3. Arithmetic 4. Difference 5. Geometric 6. Ratio 7. a1 = 3, d = 2

9. a1 = 25, d = −5

3, 5, 7, 9, 11 25, 20, 15, 10, 5

8. a1 = 0, d = 3 10. a1 = −11, d = 5

0, 3, 6, 9, 12 −11, −6, −1, 4, 9

SECTION 5.7

11. a1 = 5, d = −2

5, 3, 1, − 1, –3

12. a1 = −3, d = −4

13. a1 = 43 , d = 14

3 5 3 7 4 ,1, 4 , 2 , 4

14. a1 = 52 , d = − 23

− 3, − 7, –11, –15, –19 5 7 1 2 , 1, − 2 , − 2, − 2

15. a4 = 3 + (4 −1)1 = 3 + 3 = 6

16. a7 = 12 + (7 −1)(−2) = 12 + (−12) = 0

17. a11 = −20 + (11−1)(5) = −20 + 50 = 30

18. a12 = 7 + (12 − 1) ( − 3) = 7 + (11) ( − 3) = 7 − 33 = − 26

19. a20 =

4 4 4 95 91 + (19)(−1) = −19 = − = − 5 5 5 5 5

⎛ 5⎞ 21. a22 = −23 + (22 −1) ⎜⎜ ⎟⎟⎟ = −23 + 15 = −8 ⎜⎝ 7 ⎠

⎛ 2 ⎞ 1 26 27 1 =9 20. a14 = + (14 −1)⎜⎜ ⎟⎟⎟ = + = ⎜⎝ 3 ⎠ 3 3 3 3

22. a15 =

⎛ 1 ⎞ 4 14 18 4 + (14)⎜⎜ ⎟⎟⎟ = + = = 6 ⎜⎝ 3 ⎠ 3 3 3 3

23. an = n

(a1 = 1, d = 1)

24. an = 2n −1 ( a1 = 1, d = 2)

25. an = 2n

(a1 = 2, d = 2)

26. a1 = 3, d = 3 an = 3 + (n −1)3 = 3 + 3n − 3 = 3n

27. a1 = 2, d = 4

28. a1 = −7, d = 5

an = 2 + (n −1) ( 4) = 2 + 4n − 4 = 4n − 2

29. a1 = −3, d =

30. a1 = −5, d = 3

3 2

⎛3⎞ 3 3 3 9 an = −3 + (n −1) ⎜⎜ ⎟⎟⎟ = −3 + n − = n − ⎜⎝ 2 ⎠ 2 2 2 2

31. s50 =

33. s50 =

35. s15 =

n (a1 + an )

50(1 + 50)

2 2 = (25)(51) = 1275 50(1 + 99) 2

50(100)

15 (−10 + 60) 2

an = −7 + (n −1)5 = −7 + 5n − 5 = 5n −12

50(51) 2

= (25)(100) = 2500

15 • (50) = 375 2

an = −5 + (n −1)3 = −5 + 3n − 3 = 3n − 8

32. s50 =

34. s34 =

36. s17 =

50(2 + 100) 2

34 (1 + 100) 2

50(102) 2

34 • (101) = 1717 2

17 (100 + (−60)) 2

= (25)(102) = 2550

17 • (40) = 340 2

⎛ 1 24 ⎞ ⎛ ⎞ 24 ⎜⎜ + ⎟⎟⎟ 24 • ⎜⎜ 25 ⎟⎟ ⎜⎝ 5 ⎜⎝ 5 ⎠⎟ 5 ⎠⎟ = = 12 • 5 = 60 37. s24 = 2 2

⎛ 3 17 ⎞ ⎛14 ⎞ 21⎜⎜⎜− + ⎟⎟⎟ 21• ⎜⎜⎜ ⎟⎟⎟ ⎝ 4 4⎠ ⎝ 4 ⎠ 147 38. s21 = = = = 36.75 2 2 4

39. a1 = 1, r = 3

40. a1 = 3, r = 1

1, 3, 9, 27, 81

3, 3, 3, 3, 3

145

146

CHAPTER 5 Number Theory and the Real Number System

41. a1 = 5, r = 2

5, 10, 20, 40, 80

43. a1 = −3, r = −1

–3, 3, –3, 3, –3

1 2 44. a1 = −6, r = −2

1 3

81, − 27, 9, − 3, 1

46. a1 =

45. a1 = 81, r = −

1 2 − 6, 12, − 24, 48, − 96

42. a1 = 8, r =

47. a5 = 7(2)4 = (7)(16) = 112

8, 4, 2, 1,

16 1 ,r= 15 2

16 8 4 2 1 , , , , 15 15 15 15 15

48. a5 = 2(3)4 = (2)(81) = 162

⎛1⎞ ⎛1⎞ 3 49. a3 = 3⎜⎜ ⎟⎟⎟ = 3⎜⎜ ⎟⎟⎟ = ⎝⎜ 2 ⎠ ⎝⎜ 4 ⎠ 4

50. a7 = – 3(–3)6 = – 3(729) = – 2187

51. a7 = (−5) • 36 = (−5) (729) = −3645

⎛1⎞ ⎛1⎞ 7 52. a8 = ⎜⎜ ⎟⎟⎟(−2) = ⎜⎜ ⎟⎟⎟ (−128) = −64 ⎝⎜ 2 ⎠ ⎝⎜ 2 ⎠

53. a10 = (−2)(3) = −39,366

54. a18 = (−5)(−2) = 655,360

an = 5 n

55. 5, 25, 125, 625

n−1

57. −1, 1, −1, 1

an = (−1)(−1)

1 1 59. 2, 1, , 2 4

⎛1⎞ an = ( 2)⎜⎜ ⎟⎟⎟ ⎜⎝ 2 ⎠

1 3

⎛1⎞ an = (9)⎜⎜ ⎟⎟⎟ ⎜⎝ 3 ⎠

61. 9, 3, 1,

63. s6 =

a (1 − r 6 )

65. s6 = 1

1− r = −4095

a1 (1 − r

3(1 − 26 )

1− 2

)

1− r

1− 4

−1

−7 (1 − 3 ) 1− 3

= 189

(−3) (−4095)

−3

−7 (−177,146) −2

= −620,011

(−1)(1 − (−2) ) (−1)(1 + 32,768) = s13 = 3 1 − (−2) (−1)(32,769) =

n−1

⎛2⎞ an = (−4)⎜⎜ ⎟⎟⎟ ⎜⎝ 3 ⎠

64. s7 =

a1 (1 − r 7 )

1− r a1 (1 − r ) 9

66. s9 =

1− r

1(1 − 37 ) 1− 3

−3(1 − 5 )

1(−2186)

1− 5

−2

= 1093

−3(−1,953,124) −4

= −1,464,843 a1 (1 − r ) n

68. s15 =

1− r

(

−1 1 − ( 2 )

1− 2

) = (−1)(1− 32,768) −1

= 1− 32,768 = −32,767 15

69.

n−1

an = (−3)(−2)

8 16 32 62. −4, − , − , − 3 9 27

58. −16, − 8, − 4, − 2 60. −3, 6, −12, 24

3(−63)

(−3) (1 − 46 )

n−1

⎛1⎞ an = a1r n−1 = −16 ⎜⎜ ⎟⎟⎟ ⎜⎝ 2 ⎠

n−1

1− r

67. s11 =

= (−1)

n−1

a1 (1 − r 6 )

an = 5n−1

56. 1, 5, 25, 125

= −10, 923

⎛

10 ⎛1⎞ ⎞ ⎛ 1 ⎞⎟ ⎜ ⎜⎝ ⎝ 2 ⎠ ⎠⎟ (1024)⎝⎜⎜1− 1024 ⎠⎟⎟ = 70. s10 = 1 1 1− 2 2 = 1024 −1 = 1023

(512)⎜⎜⎜1− ⎜⎜⎜ ⎟⎟⎟ ⎟⎟⎟⎟

SECTION 5.7

71. s100 =

(100)(1 + 100) 2

(100)(101)

72. s100 =

= 50 (101) = 5050

73. s100 =

(100)(2 + 200) 2

(100)(1 + 199) 2

147

(100)(200) 2

= 50( 200) = 10000

(100)(202)

74.

s50 =

= 50( 202) = 10100

75. a) a12 = 96 + (11)(–3) = 96 – 33 = 63 in. [12(96 + 63)] (12)(159) = = (6)(159) = 954 in. 2 2 77. a) Using the formula an = a1 + (n – 1)d, we get

(50)(3 + 150)

76. s12 =

(50)(153) 2

= 25(153) = 3825

12(1 + 12) 12 (13) 156 = = = 78 times 2 2 2

78. a11 = 72 + (10)(– 6) = 72 – 60 = 12 in.

a8 = 35,000 + (8 – 1) (1400) = $44,800 8(35000 + 44800) 8(79800) = $319,200 = b) 2 2 79. a10 = (8000)(1.08) = 15,992 students

80. a15 = a1r15 = 1(2)15 = 32,768 layers

81. a15 = 31,000(1.06)14 = $70,088

82. a5 = 30(0.8)4 = 12.288 ft.

83. This is a geometric sequence where a1 =2000 and r = 3. In ten years the stock will triple its value 5 times. a6 = a1r6−1 = 2000(3)5 = $486,000

84. Visitors have scored 85.

8 1(1− 28 ) = 255 runs (1 + 8) = 36 runs; home team has scored 2 1− 2

82[1− (1/ 2)6 ] 82[1− (1/ 64)] 82 63 2 = = y y = 161.4375 1− (1/ 2) 1/ 2 1 64 1

86. The arithmetic sequence 1800, 3600, 5400, 7200,….has a common difference of 180. Thus, an= 180(n – 2) = 180n − 360, n ≥ 3 87. 12, 18, 24, … ,1608 is an arithmetic sequence with a1 = 12 and d = 6. Using the expression for the nth term of an arithmetic sequence an = a1 + (n – 1)d or 1608 = 12 + (n – 1)6 and dividing both sides by 6 gives 268 = 2 + n – 1 or n = 267 88. Since a5 = a1r4 and a2 = a1r, a5/a2 = r3. Thus r3 = 648/24 = 27 or r = 3. Then 24 = a2 = a1r = a1(3) or a1 = 24/3 = 8. 89. The total distance is 30 plus twice the sum of the terms of the geometric sequence having a1 = (30) (0.8) = 24 and r = 0.8. Thus s5 =

24[1− (0.8)5 ] 24[1− 0.32768] 24(0.67232) = 80.6784. = = (1− 0.8) 0.2 0.2

So the total distance is 30 + 2(80.6784) = 191.3568 ft.

148

CHAPTER 5 Number Theory and the Real Number System

90. The sequence of bets during a losing streak is geometric. a) a6 = a1rn−1 = 1(2)6−1 = 1(32) = $32

s5 =

a1 (1− r n ) 1− r

b) a6 = a1rn−1 = 10(2)6−1 = 10(32) = $320

1(1− 25 ) 1− 2

s5 = s10 =

d) a11 = a1rn−1 = 10(2)11−1 = 10(1024) = $10,240 s10 =

−31 = $31 −1

a1 (1− r n ) 1− r

a1 (1− r ) n

c) a11 = a1rn−1 = 1(2)11−1 = 1(1024) = $1,024

1− r a1 (1− r n )

10(1− 25 ) 1− 2

1(1− 2 ) 10

1− 2

10(1− 210 )

= 1− r 1− 2 e) If you lose too many times in a row, then you will run out of money.

10(−31) = $310 −1

1(−1023) = $1,023 −1

10(−1023) = $10, 230 −1

Exercise Set 5.8 1. Fibonacci 2. Golden 3. Ratio 4. Proportion 5. Divine 6. Rectangle

5 +1 = 1.618033989 2 5 −1 b) = .618033989 2 c) Differ by 1

7. a)

8. 3, 5, 8, 13, 21, 34 3 + 5 + 8 = 13 + 21 + 34 = 84 84 ÷ 4 = 21 9. If the first ten are selected;

10.

Fib. No. 1 1 2 3 5 8 13

1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 143 = 13 = 11 11

prime factors ------------prime

Fib. No. 34 55 89

prime prime 23

144 233

prime

prime factors 2 • 17 5 • 11 prime 24 • 32 prime

377 610

13 • 29 2 • 5 • 61

11. If 2, 3, 5, and 8 are selected the result is 52 – 32 = 2y8

→

25 – 9 = 16

→

16 = 16

12. If 5 is selected the result is 2(5) – 8 = 10 – 8 = 2 which is the second number preceding 5.

SECTION 5.8

13. Answers will vary. 16. Answers will vary.

14. Answers will vary.

149

15. Answers will vary. .

17. Not Fibonacci type; it is not true that each term is the sum of the two preceding terms. 18. Fibonacci type; 23 + 37 = 60; 37 + 60 = 97 19. Fibonacci type; 17 + 28 = 45; 28 + 45 = 73 20 Fibonacci type; 3π + 5π = 8π; 5π + 8π =13π 21. Fibonacci type: 1 + 2 = 3, 2 + 3 = 5 Each term is the sum of the two preceding terms. 22. Not Fibonacci; it is not true that each term is the sum of the two preceding terms 23. Fibonacci type; 40 + 65 = 105; 65 + 105 = 170 1 1 1 1 24. Fibonacci type; 1 + 2 = 3 ; 2 + 3 = 5 4 4 4 4 25. a) If 6 and 9 are selected the sequence is 6, 9, 15, 24, 39, 63, 102, … b) 9/6 = 1.5, 15/9 = 1.666, 24/15 = 1.6, 39/24 = 1.625, 63/39 = 1.615, 102/63 = 1.619, … 26. a) If 5, 8, and 13 are selected the result is 82 – (5)(13) = 64 – 65 = – 1. b) If 21, 34, and 55 are selected the result is 342 – (21)(55) = 1156 – 1155 = 1. c) The square of the middle term of three consecutive terms in a Fibonacci sequence differs from the product of the 1st and 2nd term by 1. 27. The sums of the numbers along the diagonals parallel to the one shown is a Fibonacci number. 28. a) Lucas sequence: 1, 3, 4, 7, 11, 18, 29, 47, … b) 8 + 21 = 29; 13 + 34 = 47 c) The first column is a Fibonacci-type sequence. 29. Begin with the numbers 1, 1, then add 1 and 1 to get 2 and continue to add the previous two numbers in the sequence to get the next number in the sequence. 30. Answers will vary. 31. a) Petals on daisies b) Parthenon in Athens 1 32. 89, = .0112359551 , part of Fibonacci sequence 89 33. 1/1 = 1, 2/1 = 2, 3/2 = 1.5, 5/3 = 1.6, 8/5 = 1.6, 13/8 = 1.625, 21/13 = 1.6154, 34/21 = 1.619, 55/34 = 1.6176 89/55 = 1.61818. The consecutive ratios alternate increasing then decreasing about the golden ratio. 34. The ratio of the second to the first and the fourth to the third estimates the golden ratio. 35. −10, x, −10 + x, −10 + 2 x, − 20 + 3 x, − 30 + 5 x, − 50 + 8 x, − 80 + 13 x, − 130 + 21x, − 210 + 34 x a) −10, 4, – 6, – 2, – 8, – 10, – 18, – 28, – 46, –74 b) −10, 5, − 5, 0, − 5, − 5, − 10, −15, − 25, − 40 c) –10, 6, – 4, 2, – 2, 0, – 2, – 2, – 4, – 6 d) –10, 7, − 3, 4, 1, 5, 6, 11, 17, 28 e) –10, 8, –2, 6, 4, 10, 14, 24, 38, 62 f) Yes, because each multiple causes the x term to be greater than the number term. 36.

5 x 5(5 − x ) = x 2 25 − 5 x = x 2 = x 5− x Solve for x using the quadratic formula,

x 2 + 5 x − 25 = 0

150

CHAPTER 5 Number Theory and the Real Number System

⎛ 5 −1 ⎞⎟ −5 ± 25 − 4(1)(−25) −b ± b2 − 4ac −5 ± 125 ⎟⎟ since we want an = = = 5⎜⎜⎜ ⎜⎝ 2 ⎠⎟ 2a 2(1) 2

answer between 0 and 5.

37.

(a+b) a a

Let x =

x + 1 = x2

a b

b 1 = a x

b a 1+ = a b

1 1+ = x x

⎛ 1⎞ x ⎜⎜1+ ⎟⎟⎟ = x ( x ) ⎜⎝ x ⎠

x2 – x – 1 = 0

Solve for x using the quadratic formula,

38. Answers will vary.

{5, 12, 13}

39. a) 3 reflections, 5 paths

−b ± b2 − 4ac 1 ± 1− 4(1)(−1) 1 ± 5 = = 2a 2(1) 2

{16, 30, 34}

{105, 208, 233}

b) 4 reflections, 8 paths

Review Exercises 1. Use the divisibility rules in section 5.1. 60,060 is divisible by 2, 3, 4, 5, 6, and 10.

multiply by x

2 588 2 294 3 147 7 49 7 588=22y3y72

{272, 546, 610}

c) 5 reflections, 13 paths

2. Use the divisibility rules in section 5.1. 400,644 is divisible by 2, 3, 4, 6, and 9 2 2 2 5 3

840 420 210 105 21 7 840 = 23y3y5y7

6. 30 = 2 ⋅ 3 ⋅ 5, 105 = 3 ⋅ 5 ⋅ 7 gcd = 15 lcm = 210 7. 63 = 3 ⋅ 3 ⋅ 5, 108 = 3 ⋅ 4 ⋅ 9 gcd = 9; lcm = 756 8. 45 = 32 ⋅ 5, 250 = 2 ⋅ 53; gcd = 5; lcm = 2 ⋅ 32 ⋅ 53 = 2250 9. 90 = 2 ⋅ 32 ⋅ 5, 300 = 22 ⋅ 3 ⋅ 52; gcd = 2 ⋅ 3 ⋅ 5 = 30; lcm = 22 ⋅ 32 ⋅ 52 = 900 10. 15 = 3 ⋅ 5, 9 = 32; lcm = 32 ⋅ 5 = 45. In 45 days the train stopped in both cities. 11. 6 + (−9) = −3 13. (–2) + (–4) = – 6 15. –5 – 4 = –5 + (–4) = – 9 17. −2 ⋅ 12 = −24

12. –2 + 5 = 3 14. 4 – 8 = 4 + (–8) = –4 16. –3 – (–6) = –3 + 6 = 3 18. (– 2)(−12) = 24

19. −35/−7 = 5 21. [8 ÷ (–4)](–3) = (–2)(–3) = 6 23. 13/4 = 3.25

20. 12/-6 = −2 22. 17/20 = 0.85

25. 7/12 = 0.583 27. 11/16 = 0.6875

24. 6/7 = 0.857142 26. 3/8 = 0.375

2 2 3 11

1452 726 363 121 11 1452 = 22y3y112

REVIEW EXERCISES

28. 1.4 =

29. 0.6666… 10 n = 6.6

14 7 = 10 5

10n = 6.6666…. 9n 6 = 9 9 2 n= 3

− n = 0.6

9n = 6.0

30. 0.5151…

100n = 51.5151…. 99 n 51 100 n = 51.51 = 99 99 − n = 0.51 51 n= 99n = 51 99 5 2 • 7 + 5 19 32. 2 = = 7 7 7

31. 0.083 =

33. −3 1 4 =

11 2 •5 + 1 1 = =2 5 5 5 34. 5 2 • 7 + 5 19 2 = = 7 7 7

83 1000

((−3)(4))−1 4

−13 4

35.

−136 (−27)(5) −1 1 = = −27 5 5 5

36.

11 2 11 2 4 11 8 3 1 − = − • = − = = 12 3 12 3 4 12 12 12 4

37.

1 5 1 2 5 3 2 15 17 + = • + • = + = 6 4 6 2 4 3 12 12 12

38.

7 12 84 84 1 • = = = 16 21 336 4 ⋅ 84 4

39.

5 6 5 7 35 ÷ = y = 9 7 9 6 54

⎛ 4 5 ⎞ 4 28 + 25 5 53 5 53 40. ⎜⎜ + ⎟⎟⎟ ÷ = y = y = ⎜⎝ 5 7 ⎠ 5 35 4 35 4 28

⎛2 1⎞ 4 2 7 1 i = 41. ⎜⎜ i ⎟⎟⎟ ÷ = ⎜⎝ 3 7 ⎠ 7 21 4 6

⎛1⎞ ⎛ 1 ⎞⎛ 71⎞ 71 42. ⎜⎜ ⎟⎟⎟(17 3 4) = ⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟ = = 2 7 32 teaspoons ⎜⎝ 8 ⎠ ⎝⎜ 8 ⎠⎝⎜ 4 ⎠ 32

43.

75 = 25 ⋅ 3 = 25 ⋅ 3 = 5 3

44.

2 − 4 2 = (1− 4) 2 = −3 2

45.

8+ 6 2= 2 2 + 6 2 = 8 2

46.

3 − 7 27 = 3 − 21 3= − 20 3

47.

28 + 63 = 2 7 + 3 7 = 5 7

48.

3 ⋅ 6= 18 =

49.

8⋅ 6 =

50.

300 300 = = 100 = 10 3 3

51.

4 3 4 3 = ⋅ 3 3 3

52.

7 5 = ⋅ 5 5

54.

3(4 + 6)= 4 3+ 18= 4 3+3 2

55. Commutative property of addition

56. Commutative property of multiplication

57. Associative property of addition

58. Distributive property 60. Commutative property of multiplication

59. Associative property of multiplication 61. Distributive property

9⋅2 = 9 ⋅ 2 = 3 2

48 = 16 ⋅ 3 = 16 ⋅ 3 = 4 3

53. 3(2 + 7 ) = 6 + 3 7

35 5

151

152

CHAPTER 5 Number Theory and the Real Number System

62. Natural numbers – not closed for subtraction 2 – 3 = –1 and –1 is not a natural number.

63. Whole numbers – closed for multiplication

64. Not closed; 1 ÷ 2 is not an integer

65. Closed 67. 53 = 5 y 5 y 5 = 125

66. Not closed;

2 • 2 = 2 is not irrational

95 = 95−3 = 92 = 81 69. 93

70. 52 y 51 = 53 = 125

71. 70 = 1

72. (32)2 = 32y2 = 34 = 81

73. 8, 200,000,000 = 8.2 ×109

74. 0.0000158 = 1.58 x 10– 5

75. 2.8×105 = 280,000

76. 1.39 x 10– 4 = 0.000139

77. (3 x 104)(2 x 10– 9) = 6 x 104−9 = 6 x 10 –5

78.

1 1 1 = = 2 5• 5 25 5

68. 5−2 =

1.5×10−3 1.5 10−3 = × = 5 10−4 5×10−4

0.3×101 = 3.0×100 79. (550,000)(2,000,000) = (5.5 x 105)(2 x 106) 8, 400,000 8.4×106 80. = = 1.2×102 = 120 = (5.5)(2) x 105+6 = 11 x 1011 4 70,000 7 ×10 = 1,100,000,000,000

81.

83.

2×10−6 = 0.5×101 =5.0 0.0000004 4 ×10−7 0.000002

20, 000, 000 3, 600

1.49×1011 = .3880208333×103 ≈ 388.02 82. 3.84×108 388 times

84. Arithmetic

2.0×107 3.6×103

17, 21

≈ 0.555556×104 =$5,555.56 8, 16

86. a1 = −6, d = 2

87. a10 = −20 + 9(5) = −20 + 45 = 25

88. 3, 6, 12, 24, 48

85. Geometric

89. - 6, - 12, - 24, - 48

91. s20 =

20 (0.5 + 5.25) 2

(

1 1− (−2)6 93. s6 =

1− (−2)

a4 = - 48

90. s50 =

(20)(5.75) = 57.5 2

92. s4 =

a9 = −6 + (9 −1)2 = 10 a5 = 48

50 (3 + 150) 2 3(1− 24 ) 1− 2

= (25)(153) = 3825

(3)(−15) = 45 −1

) = (1)(1− 64) = (1)(−63) = −21 3

94. Arithmetic: an = 3n

96. Geometric: an = 2(– 1)n – 1

95. Arithmetic: an = 1 + (n – 1)3 = 1 + 3n – 3 = 3n – 2 97. Geometric: an = 5(1/3)n–1

98. No

99. Yes;

−8, −13

CHAPTER TEST Chapter Test

1. 48,395 is divisible by: 5

414 3

207

69 23

414 = 2 y 32 y 23 3. [(−3) + 7] − (−4) = [4] + 4 = 8

5. 4 5 8 =

7. 6.45 =

(8)(4) + 5 8

4. [(– 70)(– 5)] ÷ (8 – 10) = 350 ÷ [8 + (– 10)] =350 ÷ (– 2) = – 175 5 6. = 0.625 8

32 + 5 37 = 8 8

645 129 = 100 20

75+ 48= 25 3+ 16 3= 5 3 + 4 3 = 9 3

10.

11. The integers are closed under multiplication since the product of two integers is always an integer. 45 13. 2 = 45−2 = 43 = 64 4 1 1 15. 3−4 = 4 = 81 3 17. an = – 4n + 2

17 7 17 ⎛ 7 ⎞⎛ 2 ⎞ 17 14 3 1 − = −⎜⎜ ⎟⎟⎟⎜⎜ ⎟⎟⎟ = − = = 24 12 24 ⎜⎝12 ⎠⎝⎜ 2 ⎠ 24 24 24 8 2 2 7 14 14 = ⋅ = = 7 7 7 7 49

12. Associative property of addition 14. 43 y 42 = 45 = 4 ⋅ 4 ⋅ 4 ⋅ 4 ⋅ 4 = 1024

16.

7.2×106 12 11 −6 =0.8×10 = 8.0×10 9.0×10 11 ⎣⎡−2 + (−32)⎦⎤

19. an = 3 y (2)n-1

11(−34)

= −187 2 2 20. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55

18.

Group Projects

1. In this exercise, you may obtain different answers depending upon how you work the problem. 1) a) 2 servings Rice: 2/3 cup, Salt: 1/4 tsp., Butter: 1 tsp. b) 1 serving Rice: 1/3 cup, Salt: 1/8 tsp., Butter: 1/2 tsp. c) 29 servings Rice: 9 2/3 cup, Salt: 4 5/8 tsp., Butter: 14 1/2 tsp. 1 1 2. a) Area of triangle 1 = A1 = bh = (5) 2 5 = 5 5 2 2 1 1 Area of triangle 2 = A1 = bh = (6) 2 5 = 6 5 2 2

(

)

(

)

Area of rectangle = AR = bh = (10) 2 5 = 20 5 Area of trapezoid = AT = 5 5 + 6 5 + 20 5 = 31 5

(

)

1 1 b) Area of trapezoid = AT = h (b1 + b2 ) = 2 5 (10 + 21) = 31 5 2 2 c) Yes, same

153

154

CHAPTER 5 Number Theory and the Real Number System

3. Co-pay for prescriptions = 50% Co-pay for office visits = $10 01/10: $10 + .50 ($44) = $32.00 02/27: $10 + .20 (188) = $47.60 04/19: $10 + .20 (348) + .50 (76) = $117.60 a) Total = $197.20 b) .50 (44) + .80 (188) + .80 (348) + .50 (76) +3(40) = $608.80 c) $500.00 – 197.20 = $302.80 4. a) 1 branch b) 5 branches c) 233 branches

Co-pay for medical tests = 20%

d) Produces the same sequence.

CHAPTER SIX ALGEBRA, GRAPHS, AND FUNCTIONS Exercise Set 6.1 1. Variable 2. Constant 3. Expression 4. 4 5. 3 6. Solution 7. Evaluate 8. Division 9. x = −6, x 2 = (−6) = 36

10. x = 5, x 2 = 52 = 25

11. x = −4, − x 2 = −(−4) = −16

12. x = −2, − x 2 = −(−2) = −4

13. x = −7, − 2 x 3 = −2 (−7) = −2 (−343) = 686

14. x = −4, − x 3 = −(−4) = −(−64) = 64

15. x = 4, x − 7 = 4 − 7 = −3

⎛5⎞ 5 16. x = ,8 x − 3 = 8⎜⎜ ⎟⎟⎟− 3 = 20 − 3 = 17 ⎜ ⎝2⎠ 2

17. x = −2, −3x + 7 = −3(−2) + 7 = 6 + 7 = 13

18. x = −1, 4 x − 3 = 4 (−1) − 3 = −4 − 3 = −7

19. x = 3, x 2 − 5 x + 12 = (3) − 5(3) + 12

20. x = −2, −x 2 + 3x −10 = −(−2) + 3(−2)−10

= −4 − 6 −10 = −20

= 9 −15 + 12 = 6

1 2 2 ⎛1⎞ 1 22. x = , x 2 + x −1 = ⎜⎜ ⎟⎟⎟ + −1 2 3 3 ⎜⎝ 2 ⎠ 2 2

2 ⎛2⎞ 2 1 1 ⎛2⎞ 21. x = , x 2 − 5 x + 2 = ⎜⎜ ⎟⎟⎟ − 5⎜⎜ ⎟⎟⎟ + 2 ⎜ ⎝⎜ 3 ⎠ 3 2 2 ⎝ 3⎠

2 ⎛1⎞ 1 = ⎜⎜ ⎟⎟⎟ + −1 3 ⎜⎝ 4 ⎠ 2

1 ⎛ 4 ⎞ 10 = ⎜⎜ ⎟⎟⎟ − + 2 2 ⎜⎝ 9 ⎠ 3

4 60 36 20 10 − + =− =− 18 18 18 18 9

155

2 1 + −1 12 2 2 6 12 = + − 12 12 12 4 1 =− =− 12 3 =

157

CHAPTER 6 Algebra, Graphs, and Functions

⎛ 1 ⎞3 ⎛ 1 ⎞2 1 23. x = ,8 x 3 − 4 x 2 + 7 = 8⎜⎜ ⎟⎟⎟ − 4 ⎜⎜ ⎟⎟⎟ + 7 ⎜⎝ 2 ⎠ ⎝⎜ 2 ⎠ 2

24. x = 2, y = −3, − x 2 + 5 xy = −(2) + 5( 2)(−3) 2

= −4 − 30 = −34

⎛1⎞ ⎛ 1 ⎞ = 8⎜⎜ ⎟⎟⎟ − 4 ⎜⎜ ⎟⎟⎟ + 7 ⎝⎜ 8 ⎠ ⎝⎜ 4 ⎠ = 1−1 + 7 = 7

25. x = −2, y = 1, 3 x 2 − xy + 2 y 2 = 3(−2) − (−2)(1) + 2 (1) = 12 + 2 + 2 = 16 2

27. x = 2, y = −1, 4 x 2 −10 xy + 3 y 2 = 4 ( 2) −10 ( 2)(−1) + 3(−1) = 16 + 20 + 3 = 39 2

29. 5 x + 2 = 17, x = 3

2 1 26. x = 2, y = 5, 3x 2 + xy − y 2 5 5 2 1 2 2 = 3(2) + (2)(5) − (5) 5 5 = 12 + 4 − 5 = 11 2 2 28. x = 4, y = −2, (2 x − y ) = ⎡⎣ 2 ( 4) − (−2)⎤⎦

= 102 = 100 30. 6 x − 3 = −27, x = −4

5(3) + 2 = 15 + 2 = 17

6 (−4) − 3 = −24 − 3 = −27

17 = 17, x = 3 is a solution.

−27 = −27, x = −4 is a solution. 32. 4 x + 2 y = −2, x = −2, y = 3

31. x − 3 y = 0, x = 6, y = 3 6 − 3(3) = 6 − 9 = −3

4 (−2) + 2 (3) = −8 + 6 = −2

−3 ≠ 0, x = 6, y = 3 is not a solution.

−2 = −2, x = −2, y = 3 is a solution.

33. −x 2 + 3x + 6 = 5, x = 2

34. −2 x 2 + x + 5 = 0, x = 3

−( 2) + 3 ( 2 ) + 6 = − 4 + 6 + 6 = 8

−2 (3) + 3 + 5 = −2 (9) + 3 + 5 = −10

8 ≠ 5, x = 2 is not a solution.

−10 ≠ 20, x = 3 is not a solution.

35. 3x 2 + 2 x = 40, x = −4

36. y = x 2 + 3 x − 6, x = −1, y = −8

3(−4) + 2 (−4) = 3(16) − 8 = 48 − 8 = 40

(−1) + 3(−1)− 6 = 1− 3 − 6 = −8

40 = 40, x = −4 is a solution.

−8 = −8, x = −1, y = −8 is a solution.

37. y = −x 2 + 3x −1, x = 3, y = −1

38. y = x 3 − 3x 2 + 1, x = 2, y = −3

−(3) + 3(3) −1 = −9 + 9 −1 = −1

(2) − 3(2) + 1 = 8 −12 + 1 = −3

−1 = −1, x = 3, y = −1 is a solution.

−3 = −3, x = 2, y = −3 is a solution.

39. x = 120 mi, 0.55 x = 0.55(120 mi ) = $66

40. n = 8,000, 000,000,000 0.000002n = 0.000002 (8,000,000,000, 000) = 16,000,000 sec

41. 2012 is 8 years since 2004. 6.8(8) + 61 = 115.4; 115.4 million taxpayers

42. x = 75, 220 + 2.75 x = 220 + 2.75(75)

43. 2(0.60) + 80(0.60) + 40 = 0.72 + 48 + 40 = 88.72; 88.72 min 2

= 220 + 206.25 = $426.25

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41.

CHAPTER 6 Algebra, Graphs, and Functions

3 7 = x 8 3(8) = 7 x

Cross multiplication

24 = 7 x 24 7 x = 7 7 24 =x 7 42.

Divide both sides of the equation by 7.

x −1 x + 5 = 5 15 15( x −1) = 5 ( x + 5) 15 x −15 = 5 x + 25 15 x − 5 x −15 = 5 x − 5 x + 25 10 x −15 = 25 10 x −15 + 15 = 25 + 15 10 x = 40 10 x 40 = 10 10 x=4

43.

1 1 2 x+ = 2 3 3 ⎛1 ⎛ 2⎞ 1⎞ 6 ⎜⎜ x + ⎟⎟⎟ = 6 ⎜⎜ ⎟⎟⎟ ⎝⎜ 2 ⎝⎜ 3 ⎠ 3⎠

Add 15 to both sides of the equation.

Divide both sides of the equation by 10.

Distributive Property Subtract 2 from both sides of the equation.

Divide both sides of the equation by 3.

1 1 1 y+ = 2 3 4 ⎛1 ⎞ ⎛1⎞ 1 12 ⎜⎜ y + ⎟⎟⎟ = 12 ⎜⎜ ⎟⎟⎟ ⎜⎝ 2 ⎝⎜ 4 ⎠ 3⎠ 6y + 4 = 3 6 y + 4 − 4 = 3− 4 6 y = −1 6 y −1 = 6 6 1 y =− 6

Distributive Property Subtract 5 x from both sides of the equation.

Multiply both sides of the equation by the LCD.

3x + 2 = 4 3x + 2 − 2 = 4 − 2 3x = 2 3x 2 = 3 3 2 x= 3

44.

Cross multiplication

Multiply both sides of the equation by the LCD. Distributive Property Subtract 4 from both sides of the equation.

Divide both sides of the equation by 6.

SECTION 6.2

45.

46.

47.

48.

49.

0.9 x −1.2 = 2.4 0.9 x + 1.2 −1.2 = 2.4 + 1.2 0.9 x = 3.6 0.9 x 3.6 = 0.9 0.9 x=4 5 x + 0.050 = −0.732 5 x + 0.050 − 0.050 = −0.732 − 0.050 5 x = −0.782 5 x −0.782 = 5 5 x = −0.1564 6t − 8 = 4t − 2 6t − 4t − 8 = 4t − 4t − 2 2t − 8 = −2 2t − 8 + 8 = −2 + 8 2t = 6 2t 6 = 2 2 t=3 6t − 7 = 8t + 9 6t − 8t − 7 = 8t − 8t + 9 −2t − 7 = 9 −2t − 7 + 7 = 9 + 7 −2t = 16 −2t 16 = −2 − 2 t = −8

1 x + 2x = 4 3 ⎛x ⎞⎟ ⎛1⎞ 12 ⎜⎜ + 2 x⎟⎟ = 12 ⎜⎜ ⎟⎟⎟ ⎝⎜ 4 ⎠ ⎝⎜ 3⎠ 3 x + 24 x = 4 27 x = 4 27 x 4 = 27 27 4 x= 27

Add 1.2 to both sides of the equation.

Divide both sides of the equation by 0.9.

Subtract 0.050 from both sides of the equation.

Divide both sides of the equation by 5.

Subtract 4t from both sides of the equation. Add 8 to both sides of the equation.

Divide both sides of the equation by 2.

Subtract 8t from both sides of the equation. Add 7 to both sides of the equation.

Divide both sides of the equation by -2.

Mulitply both sides of the equation by the LCD. Distributive Property

Divide both sides of the equation by 27.

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162

50.

CHAPTER 6 Algebra, Graphs, and Functions

r + 2r = 7 3 ⎛r ⎞ 3⎜⎜ + 2r ⎟⎟⎟ = 3(7) ⎜⎝ 3 ⎠

Mulitply both sides of the equation by the LCD.

r + 6r = 21

Distributive Property

7r = 21 7r 21 = 7 7 x=3

51.

Divide both sides of the equation by 7.

x−3 x + 4 = 2 3 3( x − 3) = 2 ( x + 4)

Cross multiplication

3x − 9 = 2 x + 8

Distributive Property

3x − 2 x − 9 = 2 x − 2 x + 8

Subtract 2 x from both sides of the equation.

x−9 = 8 x−9+ 9 = 8+9

Add 9 to both sides of the equation.

x = 17

52.

x −5 x −9 = 4 3 3( x − 5) = 4 ( x − 9)

Cross multiplication

3x −15 = 4 x − 36

Distributive Property

3x − 3x −15 = 4 x − 3x − 36

Subtract 3x from both sides of the equation.

−15 = x − 36 −15 + 36 = x − 36 + 36

Add 36 to both sides of the equation.

21 = x

53.

2 ( x + 3) − 4 = 2 ( x − 4) 2x + 6 − 4 = 2x −8 2x + 2 = 2x −8 2x − 2x + 2 = 2x − 2x −8 2 = −8 No solution

54.

Distributive Property Subtract 2 x from both sides of the equation. False

3( x + 2) + 2 ( x −1) = 5 x − 7 3x + 6 + 2 x − 2 = 5 x − 7 5 x + 4 = 5x − 7 5 x − 5 x + 4 = 5x − 5x − 7 4 = −7 No solution

Distributive Property Subtract 5 x from both sides of the equation. False

SECTION 6.2

55.

4 ( x − 4) + 12 = 4 ( x −1) 4 x −16 + 12 = 4 x − 4 4x − 4 = 4x − 4

Distributive Property

This equation is an identity. Therefore, the solution is all real numbers.

56.

6(t + 2) −14 = 6t − 2 6t + 12 −14 = 6t − 2 6t − 2 = 6t − 2

Distributive Property Combine like terms.

This equation is an identity. Therefore, the solution is all real numbers

57.

1 2 ( x + 3) = ( x + 2) 3 5 ⎛ 1 ⎞⎟ ⎛ 2⎞ 15⎜⎜ ⎟⎟( x + 3) = 15⎜⎜ ⎟⎟⎟( x + 2) ⎝⎜ 3 ⎠ ⎝⎜ 5 ⎠

Multiply both sides of the equation by the LCD.

5 ( x + 3) = 6 ( x + 2) 5 x + 15 = 6 x + 12 5 x − 5 x + 15 = 6 x − 5 x + 12 15 = x + 12 15 −12 = x + 12 −12 3= x

58.

2 1 ( x − 4) = ( x + 1) 3 4 ⎛ 2 ⎞⎟ ⎛1⎞ 12 ⎜⎜ ⎟⎟( x − 4) = 12 ⎜⎜ ⎟⎟⎟( x + 1) ⎜⎝ 3 ⎠ ⎝⎜ 4 ⎠

Distributive Property Subtract 5x from both sides of the equation. Subtract 12 from both sides of the equation.

Multiply both sides of the equation by the LCD.

8( x − 4) = 3( x + 1) 8 x − 32 = 3x + 3 8 x − 3x − 32 = 3x − 3 x + 3 5 x − 32 = 3 5 x − 32 + 32 = 3 + 32 5 x = 35 5 x 35 = 5 5 x=7

Distributive Property Subtract 3 x from both sides of the equation. Add 32 to both sides of the equation.

Divide both sides of the equation by 5.

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164

59.

60.

61.

CHAPTER 6 Algebra, Graphs, and Functions

3x + 2 − 6 x = −x −15 + 8 − 5 x −3 x + 2 = − 6 x − 7 −3 x + 6 x + 2 = − 6 x + 6 x − 7 3 x + 2 = −7 3 x + 2 − 2 = −7 − 2 3 x = −9 3 x −9 = 3 3 x = −3

6 x + 8 − 22 x = 28 + 14 x −10 + 12 x −16 x + 8 = 26 x + 18 −16 x − 26 x + 8 = 26 x − 26 x + 18 −42 x + 8 = 18 −42 x + 8 − 8 = 18 − 8 −42 x = 10 −42 x 10 = −42 −42 10 5 x=− =− 42 21

Subtract 2 from both sides of the equation.

Divide both sides of the equation by 3.

Subtract 26 x from both sides of the equation. Subtract 8 from both sides of the equation. Divide both sides of the equation by − 42.

5(3n + 1) = 2 (5n − 4) + 6n 15n + 5 = 10n − 8 + 6n 15n + 5 = 16n − 8 15n −15n + 5 = 16n −15n − 8 5 = n −8 5 + 8 = n −8 + 8 13 = n

62.

Add 6 x to both sides of the equation.

Distributive Property Subtract 15n from both sides of the equation. Add 8 to both sides of the equation.

4 (t − 3) + 8 = 4 (2t − 6) 4t −12 + 8 = 8t − 24 4t − 4 = 8t − 24 4t − 4t − 4 = 8t − 4t − 24 −4 = 4t − 24 −4 + 24 = 4t − 24 + 24 20 = 4t 20 4t = 4 4 5=t

Distributive Property Subtract 4t from both sides of the equation. Add 24 to both sides of the equation.

Divide both sides of the equation by 4.

SECTION 6.2

1.98 x = 750 27, 000

63.

1.98 200 = 750 x 1.98 x = (200)(750) 1.98 x 150, 000 = 1.98 1.98 x ≈ 75, 757 gal

64.

1.98( 27, 000) = 750 x 1.98(27, 000) 750 x = 750 750 x = $71.28

825 6600 = 1 x 825 x = (6600)(1) 825 x 6600 = 825 825 x = 8 gal

65.

67. a)

6 16 = 9 x 6 x = 9 (16)

1 21 = 1,102, 000 x

66.

x = 1,102, 000 (21) x = 23,142, 000 households

x 40 = 12 480 40(480) = 12 x 40(480) 12 x = 12 12 x = 1600 lb 480 = 40 bags 12

68. a)

6 x = 144 6 x 144 = 6 6 x = 24 oz x 6 = 32 16 16 x = 32 (6)

16 x = 192 16 x 192 = 16 16 x = 12 servings 69. a)

50 1 = 80 x 50 x = 80 50 x 80 = 50 50 x = 1.6 kph x 50 = 80 90 80 x = 50 (90)

70.

40 250 = 0.6 x 40 x = 0.6 (250) 40 x = 150 40 x 150 = 40 40 x = 3.75 mA

80 x = 4500 80 x 4500 = 80 80 x = 56.25 mph

165