Issuu

PROLOGUE: Principles of Problem Solving distance 1 1 ; the ascent takes h, the descent takes h, and the rate 15 r 1 1 1 1 1 2  h. Thus we have     0, which is impossible. So the car cannot go total trip should take 30 15 15 r 15 r fast enough to average 30 mi/h for the 2mile trip.

1. Let r be the rate of the descent. We use the formula time 

2. Let us start with a given price P. After a discount of 40%, the price decreases to 06P. After a discount of 20%, the price decreases to 08P, and after another 20% discount, it becomes 08 08P  064P. Since 06P  064P, a 40% discount is better. 3. We continue the pattern. Three parallel cuts produce 10 pieces. Thus, each new cut produces an additional 3 pieces. Since the ﬁrst cut produces 4 pieces, we get the formula f n  4  3 n  1, n  1. Since f 142  4  3 141  427, we see that 142 parallel cuts produce 427 pieces. 4. By placing two amoebas into the vessel, we skip the ﬁrst simple division which took 3 minutes. Thus when we place two amoebas into the vessel, it will take 60  3  57 minutes for the vessel to be full of amoebas. 5. The statement is false. Here is one particular counterexample: Player A First half Second half Entire season

1 1 hit in 99 atbats: average  99 1 hit in 1 atbat: average  11

2 2 hits in 100 atbats: average  100

Player B 0 hit in 1 atbat: average  01

98 hits in 99 atbats: average  98 99

99 99 hits in 100 atbats: average  100

6. Method 1: After the exchanges, the volume of liquid in the pitcher and in the cup is the same as it was to begin with. Thus, any coffee in the pitcher of cream must be replacing an equal amount of cream that has ended up in the coffee cup. Method 2: Alternatively, look at the drawing of the spoonful of coffee and cream

cream

mixture being returned to the pitcher of cream. Suppose it is possible to separate the cream and the coffee, as shown. Then you can see that the coffee going into the

coffee

cream occupies the same volume as the cream that was left in the coffee. Method 3 (an algebraic approach): Suppose the cup of coffee has y spoonfuls of coffee. When one spoonful of cream 1 coffee y cream  and  . is added to the coffee cup, the resulting mixture has the following ratios: mixture y1 mixture y1 1 of a So, when we remove a spoonful of the mixture and put it into the pitcher of cream, we are really removing y1 y spoonful of cream and spoonful of coffee. Thus the amount of cream left in the mixture (cream in the coffee) is y 1 1 y 1  of a spoonful. This is the same as the amount of coffee we added to the cream. y 1 y1 7. Let r be the radius of the earth in feet. Then the circumference (length of the ribbon) is 2r. When we increase the radius by 1 foot, the new radius is r  1, so the new circumference is 2 r  1. Thus you need 2 r  1  2r  2 extra feet of ribbon. 1

FUNDAMENTALS

1.1

REAL NUMBERS

1. (a) The natural numbers are 1 2 3   .

(b) The numbers     3 2 1 0 are integers but not natural numbers. p 5 , 1729 . (c) Any irreducible fraction with q  1 is rational but is not an integer. Examples: 32 ,  12 23 q   p (d) Any number which cannot be expressed as a ratio of two integers is irrational. Examples are 2, 3, , and e. q 2. (a) ab  ba; Commutative Property of Multiplication (b) a  b  c  a  b  c; Associative Property of Addition (c) a b  c  ab  ac; Distributive Property

3. (a) In setbuilder notation: x  3  x  5

(b) In interval notation: 3 5

4. The symbol x stands for the absolute value of the number x. If x is not 0, then the sign of x is always positive. 5. The distance between a and b on the real line is d a b  b  a. So the distance between 5 and 2 is 2  5  7. 6. (a) If a  b, then any interval between a and b (whether or not it contains either endpoint) contains infinitely many ba numbers—including, for example a  n for every positive n. (If an interval extends to infinity in either or both 2 directions, then it obviously contains infinitely many numbers.) (b) No, because 5 6 does not include 5. 7. (a) No: a  b   b  a  b  a in general.

(b) No; by the Distributive Property, 2 a  5  2a  2 5  2a  10  2a  10.

8. (a) Yes, absolute values (such as the distance between two different numbers) are always positive. (b) Yes, b  a  a  b.

9. (a) Natural number: 100 (b) Integers: 0, 100, 8

 9  3, 10  (b) Integers: 2,  100 2  50, 9  3, 10     (c) Rational numbers: 45  92 , 13 , 16666     53 ,  2,  100 2 , 9  3, 10   (d) Irrational numbers: 2, 314

10. (a) Natural numbers: 2,

11. Commutative Property of addition

12. Commutative Property of multiplication

13. Associative Property of addition

14. Distributive Property

15. Distributive Property

16. Distributive Property

17. Commutative Property of multiplication

18. Distributive Property

19. x  3  3  x

20. 7 3x  7  3 x

21. 4 A  B  4A  4B

22. 5x  5y  5 x  y

CHAPTER 1 Fundamentals

23. 2 x  y  2x  2y 25. 5 2x y  5  2 x y  10x y 27.  52 2x  4y   52 2x  52 4y  5x  10y

24. a  b 5  5a  5b   26. 43 6y  43 6 y  8y

28. 3a b  c  2d  3ab  3ac  6ad

15 29 29. (a) 23  57  14 21  21  21

15 1 30. (a) 25  38  16 40  40  40

  31. (a) 23 6  32  23  6  23  32  4  1  3       1 5  4  13  1  13 (b) 3  14 1  45  12 4  4 5 5 4 5 20

2 32. (a) 2  3  2  32  23  12  3  13  93  13  83 2

33. (a) 2  3  6 and 2  72  7, so 3  72

34. (a) 3  23  2 and 3  067  201, so 23  067

5  3  10  9  1 (b) 12 8 24 24 24

15 4 25 (b) 32  58  16  36 24  24  24  24

(b) 6  7 (c) 35  72

3 2  1 2  1 2  1 45 9 (b) 15 23  51 21  51 21  10 10  12  3  3    10 15 10 5 10 5

(b) 23  067

35. (a) False

36. (a) False:

(b) True

(b) False

 3  173205  17325.

37. (a) True

(b) False

38. (a) True

(b) True

39. (a) x  0

(b) t  4

40. (a) y  0

(b) z  3

(d) 5  x  13

(d) 0    17

(e) y    2

41. (a) A  B  1 2 3 4 5 6 7 8

42. (a) B  C  2 4 6 7 8 9 10

(b) A  B  2 4 6

(b) B  C  8

43. (a) A  C  1 2 3 4 5 6 7 8 9 10 (b) A  C  7

44. (a) A  B  C  1 2 3 4 5 6 7 8 9 10 (b) A  B  C   46. (a) A  C  x  1  x  5

45. (a) B  C  x  x  5

(b) A  B  x  2  x  4

(b) B  C  x  1  x  4

48. 2 8]  x  2  x  8

47. 3 0  x  3  x  0 _3

    50. 6  12  x  6  x   12

49. [2 8  x  2  x  8 2

_ _2

SECTION 1.1 Real Numbers

51. [2   x  x  2

52.  1  x  x  1

53. x  1  x   1]

54. 1  x  2  x  [1 2] 1

55. 2  x  1  x  2 1] _2

56. x  5  x  [5  _5

57. x  1  x  1 

58. 5  x  2  x  5 2

59. (a) [3 5]

(b) 3 5]

63. [4 6]  [0 8  [0 6] 0

(b) 13  13 69. (a) 6  4  6  4  2  2

71. (a) 2  6  12  12      (b)   13 15  5  5

73. 2  3  5  5

66.  6]  2 10  2 6]

67. (a) 50  50

1  1  1 (b) 1 1

(b) 2 0]

64. [4 6]  [0 8  [4 8

65.  4  4  _4

60. (a) [0 2

62. 2 0  1   1 0

61. 2 0  1 1  2 1 _2

68. (a) 2  8  6  6 (b) 8  2  8  2  6  6 70. (a) 2  12  2  12  10  10 (b) 1  1  1  1  1  1  1  0  1

       1 1 72. (a)  6 24    4   4        5 (b)  712 127    5   1  1 74. 25  15  4  4

CHAPTER 1 Fundamentals

75. (a) 17  2  15 (b) 21  3  21  3  24  24        3   12 55   67  67 (c)  10  11 8    40  40    40   40

         7 1    49  5    54    18   18 76. (a)  15   21   105 105   105   35  35 (b) 38  57  38  57  19  19.

77. (a) Let x  0777   . So 10x  77777     x  07777     9x  7. Thus, x  79 .

13 (b) Let x  02888   . So 100x  288888     10x  28888     90x  26. Thus, x  26 90  45 . 19 (c) Let x  0575757   . So 100x  575757     x  05757     99x  57. Thus, x  57 99  33 .

78. (a) Let x  52323   . So 100x  5232323     1x  52323     99x  518. Thus, x  518 99 .

62 (b) Let x  13777   . So 100x  1377777     10x  137777     90x  124. Thus, x  124 90  45 .

1057 (c) Let x  213535   . So 1000x  21353535     10x  213535     990x  2114. Thus, x  2114 990  495 .       79.   3, so   3    3. 80. 2  1, so 1  2  2  1.

81. a  b, so a  b   a  b  b  a.

82. a  b  a  b  a  b  b  a  2b

83. (a) a is negative because a is positive.

(b) bc is positive because the product of two negative numbers is positive. (c) a  ba  b is positive because it is the sum of two positive numbers.

(d) ab  ac is negative: each summand is the product of a positive number and a negative number, and the sum of two negative numbers is negative. 84. (a) b is positive because b is negative.

(b) a  bc is positive because it is the sum of two positive numbers.

(c) c  a  c  a is negative because c and a are both negative. (d) ab2 is positive because both a and b2 are positive.

85. Distributive Property 86. (a) When L  60, x  8, and y  6, we have L  2 x  y  60  2 8  6  60  28  88. Because 88  108 the post ofﬁce will accept this package. When L  48, x  24, and y  24, we have L  2 x  y  48  2 24  24  48  96  144, and since 144  108, the post ofﬁce will not accept this package.

(b) If x  y  9, then L  2 9  9  108  L  36  108  L  72. So the length can be as long as 72 in.  6 ft. m2 m1 m m1n2  m2n1 m1 and y  be rational numbers. Then x  y   2  , 87. Let x  n1 n2 n1 n2 n1 n2 m m n  m 2 n1 m m m m m , and x  y  1  2  1 2 . This shows that the sum, difference, and product xy 1  2  1 2 n1 n2 n1 n2 n1 n2 n1n2 of two rational numbers are again rational numbers. However the product of two irrational numbers is not necessarily   irrational; for example, 2  2  2, which is rational. Also, the sum of two irrational numbers is not necessarily irrational;     for example, 2   2  0 which is rational.        88. 12  2 is irrational. If it were rational, then by Exercise 6(a), the sum 12  2   12  2 would be rational, but this is not the case.  Similarly, 12  2 is irrational. (a) Following the hint, suppose that r  t  q, a rational number. Then by Exercise 6(a), the sum of the two rational numbers r  t and r is rational. But r  t  r  t, which we know to be irrational. This is a contradiction, and hence our original premise—that r  t is rational—was false.

SECTION 1.1 Real Numbers

a for some nonzero integers a and b. Let us assume that rt  q, a rational b a c bc c , implying number. Then by deﬁnition, q  for some integers c and d. But then r t  q  t  , whence t  d b d ad that t is rational. Once again we have arrived at a contradiction, and we conclude that the product of a rational number and an irrational number is irrational.

(b) r is a nonzero rational number, so r 

89. x

100

1000

1 x

1 2

1 10

1 100

1 1000

As x gets large, the fraction 1x gets small. Mathematically, we say that 1x goes to zero. x 1 x

05

01

001

0001

1 05  2

1 01  10

1 001  100

1 0001  1000

As x gets small, the fraction 1x gets large. Mathematically, we say that 1x goes to inﬁnity. 90. We can construct the number

 2 on the number line by

Ï2

transferring the length of the hypotenuse of a right triangle with legs of length 1 and 1.  Similarly, to locate 5, we construct a right triangle with legs of length 1 and 2. By the Pythagorean Theorem, the length   of the hypotenuse is 12  22  5. Then transfer the

length of the hypotenuse to the number line.

1 1 Ï5

Ï2

2 Ï5

The square root of any rational number can be located on a number line in this fashion. The circle in the second ﬁgure in the text has circumference , so if we roll it along a number line one full rotation, we have found  on the number line. Similarly, any rational multiple of  can be found this way. a  b  a  b abab   a. 2 2 On the other hand, if b  a, then max a b  b and a  b   a  b  b  a. In this case, a  b  a  b abba   b. 2 2 If a  b, then a  b  0 and the result is trivial. a  b  b  a a  b  a  b   a. (b) If a  b, then min a b  a and a  b  b  a. In this case 2 2 a  b  a  b Similarly, if b  a, then  b; and if a  b, the result is trivial. 2

91. (a) Suppose that a  b, so max a b  a and a  b  a  b. Then

92. Answers will vary. 93. (a) Subtraction is not commutative. For example, 5  1  1  5. (b) Division is not commutative. For example, 5  1  1  5.

(c) Putting on your socks and putting on your shoes are not commutative. If you put on your socks ﬁrst, then your shoes, the result is not the same as if you proceed the other way around. (d) Putting on your hat and putting on your coat are commutative. They can be done in either order, with the same result. (e) Washing laundry and drying it are not commutative.

CHAPTER 1 Fundamentals

94. (a) If x  2 and y  3, then x  y  2  3  5  5 and x  y  2  3  5. If x  2 and y  3, then x  y  5  5 and x  y  5. If x  2 and y  3, then x  y  2  3  1 and x  y  5. In each case, x  y  x  y and the Triangle Inequality is satisﬁed. (b) Case 0: If either x or y is 0, the result is equality, trivially.   xy Case 1: If x and y have the same sign, then x  y    x  y

 if x and y are positive   x  y. if x and y are negative 

Case 2: If x and y have opposite signs, then suppose without loss of generality that x  0 and y  0. Then x  y  x  y  x  y.

1.2

EXPONENTS AND RADICALS

1. (a) Using exponential notation we can write the product 5  5  5  5  5  5 as 56 .

(b) In the expression 34 , the number 3 is called the base and the number 4 is called the exponent.

2. (a) When we multiply two powers with the same base, we add the exponents. So 34  35  39 .

35 (b) When we divide two powers with the same base, we subtract the exponents. So 2  33 . 3 3. To move a number raised to a power from numerator to denominator or from denominator to numerator change the sign of

1 1 a 3 1 6a 2 the exponent. So a 2  2 , 2  b2 , 2  3 2 , and 3  6a 2 b3 . a b b a b b  3 13 4. (a) Using exponential notation we can write 5 as 5 .  (b) Using radicals we can write 512 as 5. 2  12  2   (c) No. 52  52  5212  5 and 5  512  5122  5. 3  12   6412  8 5. 412  23  8; 43

6. Because the denominator is of the form 7. 513  523  51  5

    a, we multiply numerator and denominator by a: 1  1  3  33 .

  8. (a) Yes, 54  625, while 54   54  625.  3 (c) No, 2x 4  23 x 43  8x 12 .   9. (a) 26   26  64

(b) 26  64  2 27 33 (c) 15  33   25 52  0 1 5 11. (a)  21  3 2 23 1 1  3  8 30 2  2  2 2 3 9 (c)   3 2 4

(b)

 3 (b) No, x 2  x 23  x 6 .

(d) No; if a is negative, then

 4a 2  2a.

10. (a) 53  125   (b) 53   53  125  2 (c) 52  25  4

  12. (a) 23  20   23  1  8 1 1 (b) 23  20   3  1   8 2   53 125 3 3 (c)   5 27 33

SECTION 1.2 Exponents and Radicals

13. (a) 53  5  531  625

14. (a) 38  35  385  313

(b) 54  52  542  25  3 (c) 22  223  64

     15. (a) 3 3 16  3 3 8  2  3 3 8 3 2  6 3 2     18 18 2 2   (b)   81 9 3 81     27 93 3 3 3 3 (c)     4 2 2 4      17. (a) 3 15  45  9  5  3 5   48 48  (b)    16  4 3 3      3 (c) 3 24 3 18  3 24  18  63  2  6 3 2 19. (a) t 5 t 2  t 52  t 7  2  2 (b) 4z 3  42 z 3  16z 32  16z 6 (c) x 3 x 5  x 35  x 2

1 21. (a) x 5  x 3  x 53  x 2  2 x (b) 2 4 5  245  1  (c)

1 

x 16  x 1610  x 6 x 10

a 9 a 2  a 921  a 6 a 3  3  3  (b) a 2 a 4  a 24  a 6  a 63  a 18

23. (a)

 x 3  5x 6  x 3  5x 9 5x 6   3 2 8 2    3 2 3 3 2 2y  3  2x y y 3  6x 3 y 5 25. (a) 3x y (c)

2    z3  (b) 52 z 2



107  1074  1000 104  4 (c) 35  354  320      16. (a) 2 3 81  2 3 27  3  2 3 27 3 3  6 3 3    3 2 18 92  (b)   5 5 25    12 43 2 3 (c)    49 7 49      18. (a) 10 32  320  64  5  8 5   54 54 (b)   3 6 6      3 (c) 3 15 3 75  3 15  75  53  9  5 3 9 (b)

52 z2

2

20. (a) a 4 a 6  a 46  a 10 3  3  8 (b) 2b3  23 b3  8b33   9 b 2 (c) 2y 10 y 11  2y 1011  2y 1   y 1 22. (a) y 2  y 5  y 25  y 3  3 y 1 (b) z 5 z 3 z 4  z 534  z 2  2 z y7 y0 1 (c) 10  y 7010  y 3  3 y y z2 z4 24. (a) 3 1  z 2431  z 4 z z 4  4   4 (b) 2a 3 a 2  24 a 32  16 a 5  16a 20 3    2z 3  33  z 23  2z 3  54z 9 (c) 3z 2

 2   52 2 z 3 254 z3    2 z z2

   4n 2 8m 2 n 4 12 n 2  8  12 m 2 n 42  2 m 3       27a 14 a 2 b1  33 a 43 b23 a 2 b1  33 a 122 b61  (b) 3a 4 b2 b7

26. (a)

x 2 y 1 x7  x 25 y 1  5 y x 3  3 33 9 a a a  (b)  3  6 2 3 2 2b 8b 2 b

27. (a)

CHAPTER 1 Fundamentals

y 2 z 3 y 1  2 3  3 y 1 y z yz 2  2  2  x 3 y 2 y8 33 y 22 6 y 4 (b)  x  x  x 3 y 2 x 12 3  5  a 19 b a 3 b2 a2 29. (a)  a 2533 b523 c33  9 3 b c c  2 10 u 1  2 1233  2223   (b)  3  u u 11 u 3  2 28. (a)

 2 22 432 522 232 x 10 2x 3 y 2 x 4 z2 x 30. (a)  y z  4 5 3 4 4y z yz  3 rs 2 332 s 2322  r 9 s 2 (b)  2  r r 3 s 2 

31. (a)

8 8a 3 b4 35 b45  4a  4a 2a 5 b5 b9

3 125 y  513 x 23 y 3  6 3 2 5x x y 3  a9 2a 1 b (b)  23 a 1323 b333  12 2 3 a b 8b   4 4 (b) 16x 8  24 x 8  2x 2 13   (b) 3 x 3 y 6  x 3 y 6  x y2 (b)



5x y 2  5x 11 y 23  5x 2 y x 1 y 3  4 33. (a) x 4  x  15  5 34. (a) x 10  x 10  x2     3 3 35. (a) 3 8x 9 y 3  23 x 3 y 3  2x 3 y        4 4 (b) 4 8x 6 y 2 4 2x 2 y 2  4 8x 6 y 2  2x 2 y 2  4 16x 8 y 4  24 x 2 y 4  2x 2 y       19   3  3 2 3 2 2 (b) y z  4x 2 y3 z 512x 9  512x 9  2x 36. (a) 16x 4 y 6 z 2  42 x 2            37. (a) 32  18  16  2  9  2  16 2  9 2  4 2  3 2  7 2            (b) 75  48  25  3  16  3  25 3  16 3  5 3  4 3  9 3        38. (a) 125  45  25  5  9  5  5 5  3 3  8 5          (b) 3 54  3 16  3 27  2  3 8  2  3 27 3 2  3 8 3 2  3 2         39. (a) 9a 3  a  9 a 2  a  a  3a a  a  3a  1 a          x (b) 16x  x 5  16 x  x 4 x  x 2  4      3 3 3 40. (a) x 4  3 8x  x 3  x  23 x  x  2 3 x               (b) 4 18rt 3  5 32r 3 t 5  4 32  2 r t 2  t  5 42  2 r 2  r t 4  t  4 3 t 2rt  5 4 r t 2 2rt    20rt 2  12t 2rt           41. (a) 36x 2  36x 4  62 x 2 1  x 2  6x 1  x 2 (b) 81x 2  81y 2  92 x 2  y 2  9 x 2  y 2       42. (a) 27a 3  63a 2  32 a 2 3a  7  3a 3a  7 (b) 25t 3  100t 2  52 t 2 t  4  5t t  4   44. 5 6  615 43. 10  1012 32. (a)

SECTION 1.2 Exponents and Radicals

 15  5 45. 735  73  73

1 1 1 46. 652  52   12   6 65 65

1 1 47.   12  512 5 5 1 49. y 15  y 32   y3  51. (a) 1614  4 16  2

 (b) 813  3 8  2

 52. (a) 2713  3 27  3

 (b) 813  3 8  2

53. (a) 3225 

  2 5 32  22  4

1 1 1 48. 534  34     4 3 4 5 125 5 1 1 50.   23  x 23 3 2 x x

  12 3 1 9 4   (b)  9 4 2 49

  2 3 125  52  25  32   3  3 125 25 5 25     (b) 64 8 512 64

1 1 (c) 912    3 9  13 1 1 1  (c)    3 8 2 8 3  34   4 8 16 16 (c)    4 81 27 81

54. (a) 12523 

1 1 1 (c) 2743    4  4  81 3 3 27

55. (a) 523  513  52313  51  5

 335  33525  315  5 3 25 3 3   3  (c) 3 4  413  4133  41  4 (b)

56. (a) 327  3127  327127  32  9

1 723  72353  71  53 7 7   10  10 1 1 5 15 1510 (c) 6  6 6  62  2  36 6 (b)

57. (a) x 34 x 54  x 3454  x 2 (b) y 23 y 43  y 2343  y 2

43 23  432313  53 13 6  (b) 3x 12 y 13  36 x 126 y 136  36  729x 3 y 2

59. (a)

23  61. (a) 8a 6 b32  823 a 623 b3223  4a 4 b

  58. (a) 4b12 8b14  412  8b1214  16b34 2    5a 12  32  5a 34212  45a 2 (b) 3a 34 23  1 60. (a) 8y 3  823 y 323  2 4y 13  1 4 6 413 613 (b) u  u   43 2 u 

32  8b9  432 a 432 b632  8a 6 b9  6 (b) 4a 4 b6 a 35  3 x 62. (a) x 5 y 13  x 535 y 1335  15 y 13  12  16u 6  23 (b) u 3  2  u 313  213 1612 u 612  2312  4u 23 u 3  13  4u 4 

CHAPTER 1 Fundamentals

12 32 x 142 x 112 9x 12 x 12 9 x 1    4 y4 y2 y 412 y2 y2 y 4  2  44323 34 134 22 132 43 423 42 213 313  (b)    1 12 14 4 5 122 z z z z z 81z 81z 81z 5  3 3    y 23 y 2 x 1 x 1 x 1 y 23 64. (a)    x2     3 1 x 1 y 2 x 3 y 2 x 1 y 2  2  2  13  13 2  13  42 y 3 z 23 x 3 y6 x 3 y 6 4y 3 z 23 16y 6 z 43 x 1 y 2 8y 8 (b)        2 13 x x 12 8z 4 2z 43 x2 x 12 813 z 4       12 12 65. (a) x 3  x 3  x 312  x 32 66. (a) x 5  x 5  x 512  x 52  15  14   5 4 (b) x 6  x 6  x 615  x 65 (b) x 6  x 6  x 614  x 32     4 67. (a) 6 y 5 3 y 2  y 56  y 23  y 5623  y 32 68. (a) b3 b  b3412  b54           3 2 (b) 5 3 x 2 4 x  5  2x 1314  10x 712 (b) 2 a a  2a 1223  2a 76     6 70. (a) 5 x 3 y 2 10 x 4 y 16  x 35410 y 251610  x y 2 69. (a) 4st 3 s 3 t 2  412 s 1236 t 3226  2st 116   4 7 3 x 8x 2 7434  x (b)   x (b)   813 x 2312  2x 16 4 3 x x   13 12     72. (a) s s 3  s 132 71. (a) 3 y y  y 112  y 3213  y 12  s 54     2 4 3 3y 18u 5  9u 2 3u 3 27y 3 54x y (b)   (b)   5 3 3 3 2 x  2x y x 2u        12 1 6 6 3 1 12 12 3 74. (a)       4 3 73. (a)       6 3 6 6 6 3 3 3         2 15 3 3 2 6 12 43 5 (b) (b)     2 2 2 2 5 5 5 5   4 34 13 9 2 9 8 9 8 5 835 8       (c)  (c)  4 3 2 2 5 214 234 523 513 2 5       5x 5x s s 3t 3st 1 1   76. (a) 75. (a)       5x 3t 3t 3t 3t 5x 5x 5x       3 2 3 x x 5 5x a a b2 b a (b)     (b)    6 2 3 5 5 5 5 b b 3 b2 b    5 5 2 25 2 25 1 x 1 x c 1 1 c (c) 5 3  35  25  (c) 35  35  25  x x x x c c c c    1 1 1 1 77. (a) 4 4  4  4 64 4 4  43      5 5 1 2 2    (b)   40 8 16 4 40  4 x 32 x 324 x6 78. (a)  124  2  x 6 y 2 12 y y y 12  5 u 5u (b) 25u 2  4  2512 u 212  412  5 u  2  2 . Since u  0, this is equivalent to  2 .   63. (a)



3x 14 y

2 

SECTION 1.2 Exponents and Radicals

  y 4 y 2  y 12 y 24  y 12   4  8112 8412 z 8412  9  z. Since   0 and z  0, this is equivalent to 9z. (b) 81 8 z 8

79. (a)

80. (a)



(b)



x 32 y 12

4 

4  3 3 z 6

x 2 y3

12









x 32

4

y 12

4 

x 2 x 6  x 2 x4   y y3 y 2  y 3

412 2 32    z z 12 z 212

81. (a) 69,300,000  693  107

82. (a) 129,540,000  12954  108

(b) 7,200,000,000,000  72  1012

(b) 7,259,000,000  7259  109

(d) 00001213  1213  104

(d) 00007029  7029  104

83. (a) 319  105  319,000

84. (a) 71  1014  710,000,000,000,000

(b) 2721  108  272,100,000

(b) 6  1012  6,000,000,000,000

(d) 9999  109  0000 000 009 999

(d) 6257  1010  0000 000 000 625 7

85. (a) 5,900,000,000,000 mi 59  1012 mi

(b) 0000 000 000 000 4 cm  4  1013 cm

86. (a) 93,000,000 mi  93  107 mi

(b) 0000 000 000 000 000 000 000 053 g  53  1023 g

(c) 5,970,000,000,000,000,000,000,000 kg  597  1024 kg    87. 72  109 1806  1012  72  1806  109  1012  130  1021  13  1020    88. 1062  1024 861  1019  1062  861  1024  1019  914  1043

1295643 1295643  109     109176  01429  1019  1429  1019 89.  3610  2511 3610  1017 2511  106     731  10 16341  1028 731 16341  1028 731  16341   101289  63  1038 90.  9 00000000019 19 19  10    5 2 1582  10 162  10 162  1582 00000162 001582     105283  0074  1012   91. 594621  58 594621000 00058 594621  108 58  103  74  1014

 9 3542  106 8774796 35429  1054  105448  319  104  10102  319  10106 92.   12  12  1048 27510376710 4 505 505  10 93. (a) b5 is negative since a negative number raised to an odd power is negative.

(b) b10 is positive since a negative number raised to an even power is positive. (c) ab2 c3 we have positive negative2 negative3  positive positive negative which is negative.

CHAPTER 1 Fundamentals

(d) Since b  a is negative, b  a3  negative3 which is negative. (e) Since b  a is negative, b  a4  negative4 which is positive. (f)

a 3 c3 negative positive3 negative3 positive negative which is negative.    6 6 positive positive positive b c negative6 negative6

94. (a) Since 12  13 , 212  213 .  12  13  12  13 (b) 12  212 and 12  213 . Since  12   13 , we have 12  12 .  112  112  343112 ; 413  4412  44  256112 . So 714  413 . (c) We ﬁnd a common root: 714  7312  73  16  16    2516 ; 3  312  336  33  2716 . So (d) We ﬁnd a common root: 3 5  513  526  52   3 5  3. 95. Since one light year is 59  1012 miles, Centauri is about 43  59  1012  254  1013 , or 25,400,000,000,000 miles away. 93  107 mi t st  s  500 s  8 13 min. s 186 000    103 liters   3 14 2  133  1021 liters 97. Volume  Average depth Area  37  10 m 36  10 m m3

96. 93  107 mi  186 000

98. Each person’s share is equal to

2670  1013 National debt   80555  104  $80,555. Population 33145  108

99. First, we estimate the total mass of the stars in the observable universe:     Number of stars Total star mass  Number of galaxies Mass of typical star  15  2  177 1012  1011  1030 Galaxy  531  1053 kg

Thus, the number of hydrogen atoms in the observable universe is 531  1053 Total star mass   318  1080 Mass of a single hydrogen atom 167  1027

1 mile  0215 mi. Thus the distance you can see is given 100. First convert 1135 feet to miles. This gives 1135 ft  1135  5280 feet    by D  2r h  h 2  2 3960 0215  02152  17028  413 miles.   101. (a) Using f  04 and substituting d  65, we obtain s  30 f d  30  04  65  28 mi/h.    (b) Using f  05 and substituting s  50, we ﬁnd d. This gives s  30 f d  50  30  05 d  50  15d  2500  15d  d  500 3  167 feet.

102. Since 1 day  86,400 s, 36525 days  31,557,600 s. Substituting, we obtain d   23 315576  107  15  1011 m  15  108 km.



667  1011  199  1030 4 2

103. Since 106  103  103 it would take 1000 days  274 years to spend the million dollars.

Since 109  103  106 it would take 106  1,000,000 days  273972 years to spend the billion dollars.  5 18 185  25  32 104. (a) 5  9 9

13



SECTION 1.3 Algebraic Expressions

(b) 206  056  20  056  106  1,000,000

105. (a)

100

21n

211  2

212  1414

215  1149

2110  1072

21100  1007

So when n gets large, 21n decreases to 1. (b) n  1n 1 2

1  11 1 2

So when n gets large, 12 m factors

1 2

 05

 1n

 12

 0707

increases to 1.

 15 1 2

 0871

10  110 1  0933 2

100  1100 1  0993 2

   am a  a    a 106. (a) n  . Because m  n, we can cancel n factors of a from numerator and denominator and are left with a a  a       a n factors

m  n factors of a in the numerator. Thus, n factors

(b)

107. (a)

 a n

1.3

n factors

      a an a a a  a    a        n b b b b b   b     b

 a n b

am  a mn . an

n factors



1 1 bn  n n  n n a b a ab

m a n 1a n 1 m  b (b) m    b b 1bm an an

ALGEBRAIC EXPRESSIONS

  1. The greatest common factor in the expression 18x 3  30x is 6x, and the expression factors as 6x 3x 2  5 .

2. (a) The polynomial 2x 3  3x 2  10x has three terms: 2x 3 , 3x 2 , and 10x.   (b) The factor x is common to each term, so 2x 3  3x 2  10x  x 2x 2  3x  10 .

3. To factor the trinomial x 2  8x  12 we look for two integers whose product is 12 and whose sum is 8. These integers are 6 and 2, so the trinomial factors as x  6 x  2. 4. The Special Product Formula for the “square of a sum” is A  B2  A2  2AB  B 2 . So 2x  32  2x2  2 2x 3  32  4x 2  12x  9.

5. The Special Product Formula for the “product of the sum and difference of terms” is A  B A  B  A2  B 2 . So 6  x 6  x  62  x 2  36  x 2 .

6. The Special Factoring Formula for the “difference of squares” is A2  B 2  A  B A  B. So 49x 2  9  7x  3 7x  3.

7. The Special Factoring Formula for a “perfect square” is A2  2AB  B 2  A  B2 . So x 2  10x  25  x  52 .

8. (a) No; x  52  x 2  2 5x  25  x 2  25. (b) Yes; x  a2  x 2  2xa  a 2 .

(d) No, x  a x  a  x 2  a 2 , by a Special Product Formula.

CHAPTER 1 Fundamentals

9. Type: binomial. Terms: 5x 3 and 6. Degree: 3. 10. Type: trinomial. Terms: 2x 2 , 5x, and 3. Degree: 2. 11. Type: monomial. Term: 8. Degree: 0.

12. Type: monomial. Terms: 12 x 7 . Degree: 7.

13. Type: fourterm polynomial. Terms: x, x 2 , x 3 , and x 4 . Degree: 4.   14. Type: binomial. Terms: 2x and  3. Degree: 1. 15. 12x  7  5x  12  12x  7  5x  12  7x  5

16. 5  3x  2x  8  x  3     17. 2x 2  3x  1  3x 2  5x  4  2x 2  3x  1  3x 2  5x  4  x 2  2x  3     18. 3x 2  x  1  2x 2  3x  5  3x 2  x  1  2x 2  3x  5  x 2  4x  6     19. 5x 3  4x 2  3x  x 2  7x  2  5x 3  4x 2  3x  x 2  7x  2  5x 3  3x 2  10x  2

20. 3 x  1  4 x  2  3x  3  4x  8  7x  5

21. 8 2x  5  7 x  9  16x  40  7x  63  9x  103     22. 4 x 2  3x  5  3 x 2  2x  1  4x 2  12x  20  3x 2  6x  3  x 2  6x  17     23. x 3 x 2  3x  2x x 4  3x 2  x 5  3x 4  2x 5  6x 3  x 5  3x 4  6x 3     24. 4x 1  x 3  3x 3 x 3  x  4x  4x 4  3x 6  3x 4  3x 6  7x 4  4x   25. 4x x  2  2 x 2  4x  2x 2 x  1  4x 2  8x  2x 2  8x  2x 3  2x 2  2x 3     26. 6x 3 x 2  1  2x 2  3x 2  2 2x  4  6x 5  6x 3  4x  6x 3  4x  8  6x 5  12x 3  8 27. 3t  2 7t  4  21t 2  12t  14t  8  21t 2  26t  8 28. 4s  1 2s  5  8s 2  18s  5

29. 3x  5 2x  1  6x 2  10x  3x  5  6x 2  7x  5 30. 7y  3 2y  1  14y 2  13y  3 31. x  3y 2x  y  2x 2  5x y  3y 2

32. 4x  5y 3x  y  12x 2  19x y  5y 2

33. 4x  32  16x 2  24x  9

34. 2  7y2  49y 2  28y  4

35. y  3x2  y 2  6x y  9x 2

36. 5x  y2  25x 2  10x y  y 2

37. 2x  3y2  4x 2  12x y  9y 2

38. r  2s2  r 2  4rs  4s 2

39.   7   7  2  49

40. 5  y 5  y  25  y 2

41. 3x  4 3x  4  3x2  42  9x 2  16 42. 2y  5 2y  5  4y 2  25           x 2 x 2  x 4 44. y 2 y  2  y 2 43.   45. y  23  y 3  3y 2 2  3y 22  23  y 3  6y 2  12y  8

46. x  33  x 3  9x 2  27x  27   47. x  2 x 2  2x  3  x 3  2x 2  3x  2x 2  4x  6  x 3  4x 2  7x  6   48. x  1 2x 2  x  1  2x 3  x 2  1   49. 2x  5 x 2  x  1  2x 3  2x 2  2x  5x 2  5x  5  2x 3  7x 2  7x  5   50. 1  2x x 2  3x  1  2x 3  5x 2  x  1