Respuestas libro de heussler

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Table of Contents
Chapter 0 Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 1 35 54 89 132 160 177 231 295 333 357 378 423 469 539 614 658 670

Chapter 0
Problems 0.1 1. True; –13 is a negative integer. 2. True, because −2 and 7 are integers and 7 ≠ 0. 3. False,because the natural numbers are 1, 2, 3, and so on.
0 4. False, because 0 = . 1 5 5. True, because 5 = . 1

7. True;

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

⎛ b ⎞ ab 8. True, because a ⎜ ⎟ = . ⎝c⎠ c

9. False; the left side is 5xy, but the right side is
5 x 2 y.

10. True; by the associative and commutative properties, x(4y) = (x ⋅ 4)y = (4 ⋅ x)y = 4xy. 11. distributive 12. commutative 13.associative 14. definition of division 15. commutative and distributive 16. associative 17. definition of subtraction 18. commutative 19. distributive 20. distributive 21. 2x(y − 7) = (2x)y − (2x)7 = 2xy − (7)(2x) = 2xy − (7 · 2)x = 2xy − 14x 22. (a − b) + c = [a + (−b)] + c = a + (−b + c) = a + [c + (−b)] = a + (c − b) 23. (x + y)(2) = 2(x + y) = 2x + 2y 24. 2[27 + (x + y)] = 2[27 + (y + x)] = 2[(27 +y) + x] = 2[(y + 27) + x] 25. x[(2y + 1) + 3] = x[2y + (1 + 3)] = x[2y + 4] = x(2y) + x(4) = (x · 2)y + 4x = (2x)y + 4x = 2xy + 4x 26. (1 + a)(b + c) = 1(b + c) + a(b + c) = 1(b) + 1(c) + a(b) + a(c) = b + c + ab + ac

6. False, since a rational number cannot have 7 is not a number denominator of zero. In fact, 0 at all because we cannot divide by 0. 7. False, because integer. 8. True;
25 = 5,which is a positive

2 is an irrational real number.

9. False; we cannot divide by 0. 10. False, because the natural numbers are 1, 2, 3, and so on, and 3 lies between 1 and 2. 11. True 12. False, since the integer 0 is neither positive nor negative. Problems 0.2 1. False, because 0 does not have a reciprocal. 2. True, because
7 3 21 ⋅ = = 1. 3 7 21

3. False; the negative of 7 is −7because 7 + (−7) = 0. 4. False; 2(3 · 4) = 2(12) = 24, but (2 · 3)(2 · 4) = 6 · 8 = 48. 5. False; –x + y = y + (–x) = y – x. 6. True; (x + 2)(4) = (x)(4) + (2)(4) = 4x + 8.
1

Chapter 0: Review of Algebra 27. x(y − z + w) = x[(y − z) + w] = x(y − z) + x(w) = x[y + (−z)] + xw = x(y) + x(−z) + xw = xy − xz + xw 28. –2 + (–4) = –6 29. –6 + 2 = –4 30. 6 + (–4) = 2 31. 7 – 2 = 5 32. 7 – (–4) = 7 + 4 =11 33. −5 − (−13) = −5 + 13 = 8 34. −a − (−b) = −a + b 35. (–2)(9) = –(2 · 9) = –18 36. 7(–9) = –(7 · 9) = –63 37. (–2)(–12) = 2(12) = 24 38. 19(−1) = (−1)19 = −(1 · 19) = −19 39.
−1 ⎛ 9⎞ = −1⎜ − ⎟ = 9 1 −9 ⎝ 1⎠

ISM: Introductory Mathematical Analysis 51. X(1) = X 52. 3(x – 4) = 3(x) – 3(4) = 3x – 12 53. 4(5 + x) = 4(5) + 4(x) = 20 + 4x 54. –(x – 2) = –x + 2 55. 0(–x) = 0
⎛ 1 ⎞ 8 ⋅1 8 = 56. 8⎜ ⎟ = ⎝ 11 ⎠ 11 11

57.

5 =5 1 14 x 2 ⋅ 7 ⋅ x 2 x = = 21 y 3 ⋅ 7 ⋅ y 3 y 3 3 3 = =− −2 x −(2 x) 2x 2 1 2 ⋅1 2 ⋅ = = 3 x 3 ⋅ x 3x

58.

59.

60.

61.

a a(3b) 3ab (3b) = = c c c

40. –(–6 + x) = –(–6) – x = 6 – x 41. –7(x) = –(7x) = –7x 42. –12(x – y) = (–12)x – (–12)(y) = –12x + 12y (or 12y – 12x) 43. –[–6 + (–y)] = –(–6) – (–y) = 6 + y
−3 3 1⋅ 3 1 =− =− =− 44. −3 ÷ 15 = 15 155⋅3 5 ⎛ 7 ⎞ 62. (5a ) ⎜ ⎟ = 7 ⎝ 5a ⎠

63.

−aby −a ⋅ by by = = −ax −a ⋅ x x 7 1 7 ⋅1 7 ⋅ = = y x y ⋅ x xy 2 5 2 ⋅ 5 10 ⋅ = = x y x ⋅ y xy 1 1 3 2 3+ 2 5 + = + = = 2 3 6 6 6 6 5 3 5 9 5 + 9 14 2 ⋅ 7 7 + = + = = = = 12 4 12 12 12 12 2 ⋅ 6 6 3 7 9 14 9 − 14 −5 5 ⋅1 1 − = − = = =− =− 10 15 30 30 30 30 5⋅6 6

64.

45. −9 ÷ (−27) =

−9 9 9 ⋅1 1 = = = −27 27 9 ⋅ 3 3 −a a = −b b

65.

46. (−a )÷ (−b) =

66.

47. 2(–6 + 2) = 2(–4) = –8 48. 3[–2(3) + 6(2)] = 3[–6 + 12] = 3[6] = 18 49. (–2)(–4)(–1) = 8(–1) = –8 50. (−12)(−12) = (12)(12) = 144
2

67.

68.

ISM: Introductory Mathematical Analysis
4 6 4 + 6 10 + = = =2 5 5 5 5

Section 0.3
a3⋅7 b 4⋅5

69.

7.

(a3 )7 (b 4 )5
5

=

=

a 21 b 20

70.

X
5



Y
5

=

X −Y
5

71.

3 1 1 18 3 2...
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