Solution Manual for Basic Technical Mathematics, 11th Edition

INSTRUCTOR’S SOLUTIONS MANUAL MATTHEW G. HUDELSON B ASIC T ECHNICAL M ATHEMATICS AND B ASIC T ECHNICAL M ATHEMATICS WITH C ALCULUS ELEVENTH EDITION Allyn J. Washington Dutchess Community College Richard S. Evans Corning Community College The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson from electronic files supplied by the author. Copyright © 2018, 2014, 2009 Pearson Education, Inc. Publishing as Pearson, 330 Hudson Street, NY NY 10013 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-13-443589-3 ISBN-10: 0-13-443589-3 Instructor’s Solutions Manual for Basic Technical Mathematics and Basic Technical Mathematics with Calculus, 11th Edition Chapter 1 Basic Algebraic Operations ………………………………………………………………………….1 Chapter 2 Geometry………………………………………………………………………………………………..104 Chapter 3 Functions and Graphs ………………………………………………………………………………171 Chapter 4 The Trigonometric Functions ……………………………………………………………………259 Chapter 5 Systems of Linear Equations; Determinants………………………………………………..346 Chapter 6 Factoring and Fractions…………………………………………………………………………….490 Chapter 7 Quadratic Equations…………………………………………………………………………………581 Chapter 8 Trigonometric Functions of Any Angle………………………………………………………666 Chapter 9 Vectors and Oblique Triangles ………………………………………………………………….723 Chapter 10 Graphs of the Trigonometric Functions………………………………………………………828 Chapter 11 Exponents and Radicals ……………………………………………………………………………919 Chapter 12 Complex Numbers …………………………………………………………………………………1001 Chapter 13 Exponential and Logarithmic Functions……………………………………………………1090 Chapter 14 Additional Types of Equations and Systems of Equations…………………………..1183 Chapter 15 Equations of Higher Degree…………………………………………………………………….1290 Chapter 16 Matrices; Systems of Linear Equations …………………………………………………….1356 Chapter 17 Inequalities……………………………………………………………………………………………1477 Chapter 18 Variation ………………………………………………………………………………………………1598 Chapter 19 Sequences and the Binomial Theorem………………………………………………………1634 Chapter 20 Additional Topics in Trigonometry ………………………………………………………….1696 Chapter 21 Plane Analytic Geometry………………………………………………………………………..1812 Chapter 22 Introduction to Statistics …………………………………………………………………………2052 Chapter 23 The Derivative ………………………………………………………………………………………2126 Chapter 24 Applications of the Derivative …………………………………………………………………2310 Chapter 25 Integration …………………………………………………………………………………………….2487 Chapter 26 Applications of Integration ……………………………………………………………………..2572 Chapter 27 Differentiation of Transcendental Functions ……………………………………………..2701 Chapter 28 Methods of Integration……………………………………………………………………………2839 Chapter 29 Partial Derivatives and Double Integrals …………………………………………………..2991 Chapter 30 Expansion of Functions in Series……………………………………………………………..3058 Chapter 31 Differential Equations…………………………………………………………………………….3181 Chapter 1 Basic Algebraic Operations 1.1 Numbers −7 12 and . 1 1 1. The numbers –7 and 12 are integers. They are also rational numbers since they can be written as 2. The absolute value of –6 is 6, and the absolute value of –7 is 7. We write these as −6 = 6 and −7 = 7 . 3. −6 < −4 ; –6 is to the left of –4. –7 –6 –5 –4 –3 –2 –1 0 4. 5. 6. The reciprocal of 1 1 2 2 3 = 1× = . is 3/ 2 3 3 2 3 3 is an integer, rational   , and real. 1 −4 is imaginary. 7 is irrational (because 3 7 is an irrational number) and real.  −6  −6 is an integer, rational   , and real.  1  7. − π 6 is irrational (because π is an irrational number) and real. 1 is rational and real. 8 8. − −6 is imaginary. −233 −2.33 = is rational and real. 100 9. 3 =3 −3 = 3 − 10. π 2 = π 2 −0.857 = 0.857 2 = 2 − 19 19 = 4 4 Copyright © 2018 Pearson Education, Inc. 1 2 11. Chapter 1 Basic Algebraic Operations 6 5 ; 7 is to the right of 5. 3 4 5 6 7 8 9 10 11 13. π < 3.1416; π (3.1415926 …) is to the left of 3.1416. (π ) (3.1416) 3.141592 14. 3.1416 −4 < 0 ; –4 is to the left of 0. –6 –5 –4 –3 –2 –1 0 1 2 15. −4 −1.42; ( − 2 = − (1.414….) = −1.414 …), − 2 is to the right of –1.42. –1.44 17. − 2 3 2 3 > − ; − = −0.666… is to the right of − = −0.75 . 3 4 3 4 –0.8 18. –1.43 –1.42 –1.41 –1.40 –0.7 –0.6 –0.5 –0.4 −0.6 0, then a = a . If b > a and a > 0 , then b = b . If b > a then b − a > 0 , then b − a = b − a . Therefore, b − a = b − a = b − a . The two sides of the expression are equivalent, one side is not less than the other. 29. List these numbers from smallest to largest: −1, 9, π = 3.14, –3.1 − −3 -1 5 –4 –3 –2 –1 0 1 π −8 2 3 4 5 6 7 List these numbers from smallest to largest: –6 –4 − 10 1 5 0.25 5, π , −8 , 9 . 1 = 0.20, − 10 = −3.16…, − −6 = −6, − 4, 0.25, −π = 3.14… . 5 −π –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 So, from smallest to largest, they are − −6 , − 4, − 10, 31. 9 8 9 So, from smallest to largest, they are −3.1, − −3 , − 1, 30. 5 = 2.236, −8 = 8, − −3 = −3, −3.1 . 6 7 1 , 0.25, −π . 5 If a and b are positive integers and b > a , then (a) (b) (c) b − a is a positive integer. a − b is a negative integer. b−a , the numerator and denominator are both positive, but the numerator is less than the denominator, so the b+a answer is a positive rational number than is less than 1. 32. If a and b are positive integers, then (a) a + b is a positive integer (b) a / b is a positive rational number (c) a × b is a positive integer 33. (a) Is the absolute value of a positive or a negative integer always an integer? x = x , so the absolute value of a positive integer is an integer. -x = x , so the absolute value of a negative integer is an integer. (b) Is the reciprocal of a positive or negative integer always a rational number? 1 If x is a positive or negative integer, then the reciprocal of x is . Since both 1 and x are integers, the reciprocal x is a rational number. Copyright © 2018 Pearson Education, Inc. Section 1.1 Numbers 34. (a) Is the absolute value of a positive or negative rational number rational? x = x , so if x is a positive or negative rational number, the absolute value of it is also a rational number. (b) Is the reciprocal of a positive or negative rational number a rational number? 5 A rational number is a number that can be expressed as a fraction where both the numerator and denominator are integer a integers and the denominator is not zero. So a rational number has a reciprocal of integer b 1 integer b = , which is also a rational number if integer a is not zero. integer a integer a integer b 35. (a) If x > 0 , then x is a positive number located to the right of zero on the number line. x –4 –3 –2 –1 (b) 0 1 2 3 4 If x < −4 , then x is a negative number located to the left of –4 on the number line. x –6 –5 –4 –3 –2 –1 0 1 2 36. (a) If x < 1 , then −1 < x 2 , then x 2 . x x –4 –3 –2 –1 0 37. If x > 1, then 1 2 3 4 1 1 is a positive number less than 1. Or 0 < < 1 . x x 1 x –4 –3 –2 –1 0 38. 3 4 1 2 3 4 If x 100V 43. N= a bits 1000 bytes × × n kilobytes bytes 1 kilobyte N = 1000 an bits 44. x L y x = length of base in m y = the shortened length in centimetres. 100 x = length of base in cm y + L = 100 x, all dimensions in cm L = 100 x − y 45. Yes, −20 °C > −30 °C because −30 °C is found to the left of −20 °C on the number line. −30℃ 46. −20℃ −10℃ 0℃ 10℃ For I 12 Ω . Copyright © 2018 Pearson Education, Inc. Section 1.2 Fundamental Operations of Algebra 1.2 Fundamental Operations of Algebra 1. 16 − 2 × ( −2 ) = 16 − ( −4 ) = 16 + 4 = 20 2. −18 + 5 − ( −2 )( 3) = 3 + 5 − ( −6 ) = 8 + 6 = 14 −6 3. −12 5 − 1 −12 4 + = + = −2 + ( −2 ) = −4 8 − 2 2(−1) 6 −2 4. 7 × 6 42 = = is undefined , not indeterminate. 0×0 0 5. 5 + ( −8 ) = 5 − 8 = −3 6. −4 + ( −7 ) = −4 − 7 = −11 7. −3 + 9 = 6 or alternatively −3 + 9 = + ( 9 − 3) = + ( 6 ) = 6 8. 18 − 21 = −3 or alternatively 18 − 21 = −(21 − 18) = −(3) = −3 9. −19 − ( −16 ) = −19 + 16 = −3 10. −8 − ( −10 ) = −8 + 10 = 2 11. 7 ( −4 ) = −(7 × 4) = −28 12. −9 ( 3) = −27 13. −7 ( −5 ) = + (7 × 5) = 35 14. −9 = −3 3 15. −6(20 − 10) −6(10) −60 = = = 20 −3 −3 −3 16. −28 −28 −28 = = = −4 −7(5 − 6) −7(−1) 7 17. −2 ( 4 )( −5 ) = −8 ( −5 ) = 40 18. −3 ( −4 )( −6 ) =12 ( −6 ) = −72 Copyright © 2018 Pearson Education, Inc. 7 8 Chapter 1 Basic Algebraic Operations 19. 2 ( 2 − 7 ) ÷ 10 = 2 ( −5 ) ÷ 10 = −10 ÷ 10 = −1 20. −64 −64 −64 −64 = = = =8 −2 4 − 8 −2 −4 −2(4) −8 21. 16 ÷ 2(−4) = 8(−4) = −32 22. −20 ÷ 5(−4) = −4(−4) = 16 23. −9 − 2 − 10 = −9 − −8 = −9 − 8 = −17 24. ( 7 − 7 ) ÷ ( 5 − 7 ) = 0 ÷ ( −2 ) = 0 25. 17 − 7 10 = is undefined 7−7 0 26. (7 − 7)(2) 0(2) 0 = = is indeterminate (7 − 7)(−1) 0(−1) 0 27. 8 − 3 ( −4 ) = 8 + 12 = 20 28. −20 + 8 ÷ 4 = −20 + 2 = −18 29. −2 ( −6 ) + 30. | −2 | 2 = = −1 −2 −2 31. 10 ( −8)( −3) ÷ (10 − 50) = 10( −8)( −3) ÷ ( −40) 8 = 12 + −4 = 12 + 4 = 16 −2 = −80( −3) ÷ ( −40) = 240 ÷ ( −40) = −6 32. 7 − −5 −1(−2) = 7−5 2 = =1 2 2 33. 24 24 − 4 ( −9 ) = + (4 × 9) = −12 + 36 = 24 3 + (−5) −2 34. −18 4− | −6 | −18 4 − 6 −2 − = − = −6 − = −6 − 2 = −8 −1 −1 −1 3 3 Copyright © 2018 Pearson Education, Inc. Section 1.2 Fundamental Operations of Algebra 35. −7 − −14 2 ( 2 − 3) − 3 6 − 8 = −7 − 14 − 3 −2 2 ( −1) 14 − 3(2) −2 = −7 − ( −7 ) − 6 = −7 − = −7 + 7 − 6 = −6 36. 37. −7 ( −3) + −6 − | −9 |= +(7 × 3) + 2 − 9 −3 = 21 + 2 − 9 = 14 3 | −9 − 2( −3) | 3 | −9 + 6 | = 1 − 10 −9 3 | −3 | = −9 9 = −9 = −1 38. 20 ( −12 ) − 40(−15) 39. 6 ( 7 ) = ( 7 ) 6 demonstrates the commutative law of multiplication. 40. 6 + 8 = 8 + 6 demonstrates the commutative law of addition. 41. 6 ( 3 + 1) = 6 ( 3) + 6 (1) demonstrates the distributive law. 42. 4 ( 5 × π ) = (4 × 5)π demonstrates the associative law of multiplication. 43. 3 + ( 5 + 9 ) = ( 3 + 5 ) + 9 demonstrates the associative law of addition. 44. 8 ( 3 − 2 ) = 8 ( 3) − 8 ( 2 ) demonstrates the distributive law. 45. ( 5 × 3) × 9 = 5 × (3 × 9) demonstrates the associative law of multiplication. 46. ( 3 × 6 ) × 7 = 7 × (3 × 6) demonstrates the commutative law of multiplication. 47. −a + ( −b ) = −a − b , which is expression (d). 48. b − ( −a ) = b + a = a + b , which is expression (a). 49. −b − ( − a ) = −b + a = a − b , which is expression (b). 98 − −98 = −240 + 600 360 = = is undefined 98 − 98 0 Copyright © 2018 Pearson Education, Inc. 9 10 Chapter 1 Basic Algebraic Operations 50. −a − ( −b ) = −a + b = b − a , which is expression (c). 51. Since | 5 − (−2) |=| 5 + 2 |=| 7 |= 7 and | −5 − (−2) |=| −5 + 2 |=| −3 |= 3 , | 5 − (−2) |>| −5 − (−2) | . 52. Since | −3− | −7 ||=| −3 − 7 |=| −10 |= 10 and || −3 | −7 |=| 3 − 7 |=| −4 |= 4 , | −3 − | −7 || > || −3 | − 7 | . 53. (a) The sign of a product of an even number of negative numbers is positive. Example : −3 ( −6 ) = 18 (b) The sign of a product of an odd number of negative numbers is negative. Example: − 5 ( −4 )( −2 ) = −40 54. Subtraction is not commutative because x − y ≠ y − x . Example: 7 − 5 = 2 does not equal 5 − 7 = −2 55. Yes, from the definition in Section 1.1, the absolute value of a positive number is the number itself, and the absolute value of a negative number is the corresponding positive number. So for values of x where x > 0 (positive) or x = 0 (neutral) then x = x . Example : 4 = 4 . The claim that absolute values of negative numbers x = − x is also true. Example: if x is − 6, then −6 = − ( −6 ) = 6. 56. The incorrect answer was achieved by subtracting before multiplying or dividing which violates the order of operations. 24 − 6 ÷ 2 × 3 ≠ 18 ÷ 2 × 3 = 9 × 3 = 27 The correct value is: 24 − 6 ÷ 2 × 3 = 24 − 3 × 3 = 24 − 9 = 15 57. (a) (b) 58. (a) 1 , providing that the x 1  1 and − xy = − (12 )  −  = 1 . number x in the denominator is not zero. So if x = 12 , then y = − 12  12  x− y = 1 is true for all values of x and y, providing that x ≠ y to prevent division by zero. x− y − xy = 1 is true for values of x and y that are negative reciprocals of each other or y = − x + y = x + y is true for values where both x and y have the same sign or either are zero: x + y = x + y , when x ≥ 0 and y ≥ 0 or when x ≤ 0 and y ≤ 0 Example: 6 + 3 = 6 + 3 = 9 and 6 + 3 = 6+3 = 9 Also, −6 + (−3) = −9 = 9 −6 + −3 = 6 + 3 = 9 x + y = x + y is not true however, when x and y have opposite signs x + y ≠ x + y , when x > 0 and y < 0 ; or x 0 . Copyright © 2018 Pearson Education, Inc. Section 1.2 Fundamental Operations of Algebra Example: −21 + 6 = −15 = 15, −21 + 6 = 21 + 6 = 27 ≠ 15 4 + (−5) = −1 = 1, 4 + −5 = 4 + 5 = 9 ≠ 1 (b) In order for x − y = x + y it is necessary that they have opposite signs or either to be zero. Symbolically, x − y = x + y when x ≥ 0 and y ≤ 0 ; or when x ≤ 0 and y ≥ 0 . Example: 6 − (−3) = 6 + 3 = 9 and 6 + −3 = 6 + 3 = 9 Example: −11 − 7 = −18 = 18 −11 + −7 = 11 + 7 = 18 x − y = x + y is not true, however, when x and y have the same signs. x − y ≠ x + y , when x > 0 and y > 0; or x < 0 and y < 0 . Example: 21 − 6 = 15 = 15, 21 + 6 = 27 ≠ 15 59. The total change in the price of the stock is −0.68 + 0.42 + 0.06 + (−0.11) + 0.02 = −0.29 . 60. The difference in altitude is −86 − (−1396) = 1396 − 86 = 1310 m 61. The change in the meter energy reading E would be: Echange = Eused − Egenerated Echange = 2.1 kW ⋅ h − 1.5 kW ( 3.0 h ) Echange = 2.1 kW ⋅ h − 4.5 kW ⋅ h Echange = − 2.4 kW ⋅ h 62. Assuming that this batting average is for the current season only which is just starting, the number of hits is zero and number of hits 0 the total number of at-bats is also zero giving us a batting average = = which is indeterminate, not at − bats 0 0.000. 63. The average temperature for the week is: −7 + (−3) + 2 + 3 + 1 + (−4) + (−6) °C Tavg = 7 −7 − 3 + 2 + 3 + 1 − 4 − 6 °C Tavg = 7 −14 °C = −2.0 °C Tavg = 7 Copyright © 2018 Pearson Education, Inc. 11 12 Chapter 1 Basic Algebraic Operations 64. The vertical distance from the flare gun is d = ( 70 )( 5 ) + ( −16 )( 25 ) d = 350 + ( −400 ) d = 350 − 400 d = −50 m The flare is 50 m below the flare gun. 65. The sum of the voltages is Vsum = 6V + ( −2V ) + 8V + ( −5V ) + 3V Vsum = 6V − 2V + 8V − 5V + 3V Vsum = 10V 66. (a) (b) (c) The change in the current for the first interval is the second reading – the first reading Change1 = −2 lb/in 2 − 7 lb/in 2 = −9 lb/in 2 . The change in the current for the middle intervals is the third reading – the second reading Change2 = −9 lb/in 2 − ( −2 lb/in 2 ) = −9 lb/in 2 + 2 lb/in 2 = −7 lb/in 2 . The change in the current for the last interval is the last reading – the third reading Change3 = −6 lb/in 2 − ( −9 lb/in 2 ) = −6 lb/in 2 + 9 lb/in 2 = 3 lb/in 2 . 67. The oil drilled by the first well is 100 m + 200 m = 300 m which equals the depth drilled by the second well 200 m + 100 m = 300 m . 100 m + 200 m = 200 m + 100 m demonstrates the commutative law of addition. 68. The first tank leaks 12 69. The total time spent browsing these websites is the total time spent browsing the first site on each day + the total time spent browsing the second site on each day minutes minutes t = 7 days × 25 + 7 days × 15 day day t = 175 min + 105 min t = 280 min OR minutes t = 7 days × (25 + 15) day minutes t = 7 days × 40 day t = 280 min which illustrates the distributive law. L L ( 7 h ) = 84 L .The second tank leaks 7 (12h ) = 84L. h h 12 × 7 = 7 × 12 demonstrates the commutative law of multiplication. Copyright © 2018 Pearson Education, Inc. Section 1.3 Calculators and Approximate Numbers 70. 13 Distance = rate × time km km 3h d = 600 + 50 h h km km d = 600 (3h ) + 50 (3h ) h h d = 1800 km + 150 km = 1950 km OR d = 600 km km 3h + 50 h h km 3h h . d = 1950 km This illustrates the distributive law. d = 650 1.3 Calculators and Approximate Numbers 1. 0.390 has three significant digits since the zero is after the decimal. The zero is not necessary as a placeholder and should not be written unless it is significant. 2. 35.303 rounded off to four significant digits is 35.30. 3. In finding the product of the approximate numbers, 2.5 × 30.5 = 76.25 , but since 2.5 has 2 significant digits, the answer is 76. 4. 38.3 − 21.9(−3.58) = 116.702 using exact numbers; if we estimate the result, 40 − 20(−4) = 120 . 5. 8 cylinders is exact because they can be counted. 55 km/h is approximate since it is measured. 6. 0.002 mm thick is a measurement and is therefore an approximation. $7.50 is an exact price. 7. 24 hr and 1440 min (60 min/h × 24 h =1140 min) are both exact numbers. 8. 50 keys is exact because you can count them; 50 h of use is approximate since it is a measurement of time. 9. Both 1 cm and 9 g are measured quantities and so they are approximate. 10. The numbers 90 and 75 are exact counts of windows while 15 years is a measurement of time, hence it is approximate. 11. 107 has 3 significant digits; 3004 has 4 significant digits; 1040 has 3 significant digits (the final zero is a placeholder.) 12. 3600 has 2 significant digits; 730 has 2 significant digits; 2055 has 4 significant digits. 13. 6.80 has 3 significant digits since the zero indicates precision; 6.08 has 3 significant digits; 0.068 has 2 significant digits (the zeros are placeholders.) 14. 0.8730 has 4 significant digits; 0.0075 has 2 significant digits; 0.0305 has 3 significant digits. 15. 3000 has 1 significant digit; 3000.1 has 5 significant digits; 3000.10 has 6 significant digits. Copyright © 2018 Pearson Education, Inc. 14 Chapter 1 Basic Algebraic Operations 16. 1.00 has 3 significant digits since the zeros indicate precision; 0.01 has 1 significant digit since leading zeros are not significant; 0.0100 has 3 significant digits, counting the trailing zeros. 17. 5000 has 1 significant digit; 5000.0 has 5 significant digits; 5000 has 4 significant digits since the bar over the final zero indicates that it is significant. 18. 200 has 1 significant digit; 200 has 3 significant digits; 200.00 has 5 significant digits. 19. (a) (b) 0.010 has more decimal places (3) and is more precise. 30.8 has more significant digits (3) and is more accurate. 20. (a) (b) Both 0.041 and 7.673 have the same precision as they have the same number of decimal places (3). 7.673 is more accurate because it has more significant digits (4) than 0.041, which has 2 significant digits. 21. (a) (b) Both 0.1 and 78.0 have the same precision as they have the same number of decimal places. 78.0 is more accurate because it has more significant digits (3) than 0.1, which has 1 significant digit. 22. (a) (b) (a) (b) 0.004 is more precise because it has more decimal places (3). 7040 is more accurate because it has more significant digits (3) than 0.004, which has only 1 significant digit. 0.004 is more precise because it has more decimal places (3). Both have the same accuracy as they both have 1 significant digit. 23. 24. The precision and accuracy of −8.914 and 8.914 are the same. (a) (b) Both 50.060 and 8.914 have the same precision as they have the same number of decimal places (3). 50.060 is more accurate because it has more significant digits (5) than 8.914, which has 4 significant digits. 25. (a) (b) 4.936 rounded to 3 significant digits is 4.94. 4.936 rounded to 2 significant digits is 4.9. 26. (a) (b) 80.53 rounded to 3 significant digits is 80.5. 80.53 rounded to 2 significant digits is 81. 27. (a) (b) -50.893 rounded to 3 significant digits is -50.9. -50.893 rounded to 2 significant digits is -51. 28. (a) (b) 7.004 rounded to 3 significant digits is 7.00. 7.004 rounded to 2 significant digits is 7.0. 29. (a) (b) 5968 rounded to 3 significant digits is 5970. 5968 rounded to 2 significant digits is 6000 . 30. (a) (b) 30.96 rounded to 3 significant digits is 31.0. 30.96 rounded to 2 significant digits is 31. 31. (a) (b) 0.9449 rounded to 3 significant digits is 0.945. 0.9449 rounded to 2 significant digits is 0.94. 32. (a) (b) 0.9999 rounded to 3 significant digits is 1.00. 0.9999 rounded to 2 significant digits is 1.0. 33. (a) (b) Estimate: 13 + 1 − 2 = 12 Calculator: 12.78 + 1.0495 − 1.633 = 12.1965, which is 12.20 to 0.01 precision Copyright © 2018 Pearson Education, Inc. Section 1.3 Calculators and Approximate Numbers 34. (a) (b) Estimate: 4 × 17 = 68 Calculator: 3.64(17.06) = 62.0984, which is 62.1 to 3 significant digits 35. (a) (b) Estimate 0.7 × 4 − 9 = −6 Calculator: 0.6572 × 3.94 − 8.651 = −6.061632, which is −6.06 to 3 significant digits 36. (a) (b) Estimate 40 − 26 ÷ 4 = 40 − 6.5 = 34 Calculator: 41.5 − 26.4 ÷ 3.7 = 34.3648649, which is 34 to 2 significant digits 37. (a) (b) Estimate 9 + (1)(4) = 9 + 4 = 13 Calculator: 8.75 + (1.2)(3.84) = 13.358, which is 13 to 2 significant digits 38. (a) Estimate 30 − (b) 39. (a) (b) 40. (a) (b) 42. 20 = 30 − 10 = 20 2 20.955 = 18.475, which is 18 to 2 significant digits Calculator: 28 − 2.2 9(15) 135 = = 6, to 1 significant digit 9 + 15 24 8.75(15.32) = 5.569173, which is 5.57 to 3 significant digits Calculator: 8.75 + 15.32 Estimate 9(4) 36 = = 5, to 1 significant digit 2+5 7 8.97(4.003) = 5.296, which is 5.3 to 2 significant digits Calculator: 2.0 + 4.78 (a) Estimate (b) 41. 15 (a) (b) 2(300) = 3.0, to 2 significant digits 400 2.056(309.6) = 2.9093279, which is 2.91 to 3 significant digits Calculator: 4.52 − 395.2 Estimate 4.5 − 15 = 12, to 2 significant digits 2+2 14.9 Calculator: 8.195 + = 12.1160526, which is 12 to 2 significant digits 1.7 + 2.1 Estimate 8 + 43. 0.9788 + 14.9 = 15.8788 since the least precise number in the question has 4 decimal places. 44. 17.311 − 22.98 = −5.669 since the least precise number in the question has 3 decimal places. 45. −3.142(65) = −204.23 , which is -204.2 because the least accurate number has 4 significant digits. 46. 8.62 ÷ 1728 = 0.004988 , which is 0.00499 because the least accurate number has 3 significant digits. 47. With a frequency listed as 2.75 MHz, the least possible frequency is 2.745 MHz, and the greatest possible frequency is 2.755 MHz. Any measurements between those limits would round to 2.75 MHz. 48. For an engine displacement stated at 2400 cm3, the least possible displacement is 2350 cm3, and the greatest possible displacement is 2450 cm3. Any measurements between those limits would round to 2400 cm3. Copyright © 2018 Pearson Education, Inc. 16 Chapter 1 Basic Algebraic Operations 49. The speed of sound is 3.25 mi ÷ 15 s = 0.21666… mi/s = 1144.0… ft/s . However, the least accurate measurement was time since it has only 2 significant digits. The correct answer is 1100 ft/s. 50. 4.4 s − 2.72 s = 1.68 s , but the answer must be given according to precision of the least precise measurement in the question, so the correct answer is 1.7 s. 51. (a) (b) 2.2 + 3.8 × 4.5 = 2.2 + (3.8 × 4.5) = 19.3 (2.2 + 3.8) × 4.5 = 6.0 × 4.5 = 27 52. (a) (b) 6.03 ÷ 2.25 + 1.77 = (6.03 ÷ 2.25) + 1.77 = 4.45 6.03 ÷ (2.25 + 1.77) = 6.03 ÷ 4.02 = 1.5 53. (a) (b) (c) (d) (e) 2+0 = 2 2−0 = 2 0 − 2 = −2 2×0 = 0 2 ÷ 0 = error; from Section 1.2, an equation that has 0 in the denominator is undefined when the numerator is not also 0. 54. (a) (b) (c) 2 ÷ 0.0001 = 20 000 ; 2 ÷ 0 = error 0.0001 ÷ 0.0001 = 1 ; 0 ÷ 0 = error Any number divided by zero is undefined. Zero divided by zero is indeterminate. 55. π = 3.14159265… (a) π < 3.1416 (b) 22 ÷ 7 = 3.1428 π < (22 ÷ 7) 56. (a) (b) π = 3.14159265… 57. (a) (b) (c) 1 ÷ 3 = 0.333… It is a rational number since it is a repeating decimal. 5 ÷ 11 = 0.454545… It is a rational number since it is a repeating decimal. 2 ÷ 5 = 0.400… It is a rational number since it is a repeating decimal (0 is the repeating part). 58. 124 ÷ 990 = 0.12525…. the calculator may show the answer as 0.1252525253 because it has rounded up for the next 5 that doesn’t fit on the screen. 59. 32.4 MJ + 26.704 MJ + 36.23 MJ = 95.334 MJ . The answer must be to the same precision as the least precise measurement. The answer is 95.3 MJ. 60. We would compute 8(68.6) + 5(15.3) = 625.3 and round to three significant digits for a total weight of 625 lb. The values 8 and 5 are exact. 61. We would compute 12(129) + 16(298.8) = 6328.8 and round to three significant digits for a total weight of 6330 g. The values 12 and 16 are exact. 62. V = (15.2 Ω + 5.64 Ω + 101.23 Ω) × 3.55 A V = 122.07 Ω × 3.55 A 8 ÷ 33 = 0.2424… = 0.24 V = 433.3485 V V = 433 V to 3 significant digits Copyright © 2018 Pearson Education, Inc. Section 1.4 Exponents and Unit Conversions 63. 100(40.63 + 52.96) = 59.1386 % = 59.14 % to 4 signficiant digits 105.30 + 52.96 64. T= 50.45(9.80) = 91.779 N = 92 N to 2 significant digits 1 + 100.9 ÷ 23 65. (a) (b) Estimate 8 × 5 − 10 = 30, to 1 significant digit. Calculator: 7.84 × 4.932 − 11.317 = 27.34988 which is 27.3 to 3 significant digits. 66. (a) (b) Estimate 20 − 50 ÷ 10 = 15 to 2 significant digit. Calculator: 21.6 − 53.14 ÷ 9.64 = 16.0875519 which is 16.1 to 3 significant digits. 1.4 Exponents and Unit Conversions 2 1. (− x 3 ) 2 = (−1) x 3  = (−1) 2 ( x 3 ) 2 = (1) x 6 = x 6 2. 2 x 0 = 2(1) = 2 3. x 3 x 4 = x 3+ 4 = x 7 4. y 2 y 7 = y 2+7 = y9 5. 2b 4 b 2 = 2b 4 + 2 = 2b 6 6. 3k 5 k = 3k 5 +1 = 3k 6 7. m5 = m5− 3 = m 2 m3 8. 2 x6 = −2 x 6 −1 = −2 x 5 −x 9. −n5 n5− 9 n −4 1 = − = − =− 4 9 7 7 7n 7n 10. 3s 3 = 3s1− 4 = 3s −3 = 3 4 s s 11. (P ) = P 12. (x ) = x 13. ( aT ) = a T 14. ( 3r ) = (3) r 2 4 8 3 2 30 2 3 2(4) 8(3) = P8 = x 24 30 3 2(30) 2(3) = a 30T 60 = 27r 6 Copyright © 2018 Pearson Education, Inc. 17 18 Chapter 1 Basic Algebraic Operations 15. (2)3 8 2   = 3 = 3 b b b 16. F 20 F =   t 20  t  17.  x2  x 2(4) x8 = =   (−2) 4 16  −2  18. (3)3 27  3 = =  3 n3(3) n9 n  19. ( 8a ) = 1 20. −v 0 = −1 21. −3x 0 = −3(1) = −3 22. −(−2)0 = −1(1) = −1 23. 6−1 = 24. − w−5 = − 25. 1 = R2 R −2 26. 1 = −t 48 −t −48 27. (−t 2 )7 =  (−1)(t 2 )  = (−1)7 t 2(7) = (−1)t14 = −t 14 28. (− y 3 )5 = (−1)( y 3 )  = (−1)5 y 3(5) = (−1) y15 = − y15 29. − 30. 2 2i 40 i −70 = 2i 40 + ( −70) = 2i −30 = 30 i 31. 2v 4 2v 4 2v 4 1 = = = 4 4 4 (2v ) (2) (v ) 16v 4 8 3 20 4 3 0 1 1 = 61 6 1 w5 7 5 L−3 = − L−3− ( −5) = − L2 L−5 Copyright © 2018 Pearson Education, Inc. Section 1.4 Exponents and Unit Conversions 32. x 2 x3 x 2+3 x5 1 = = = ( x 2 )3 x 2(3) x 6 x 33. (n 2 ) 4 n 2(4) n8 = = =1 (n 4 ) 2 n 4(2) n8 34. (3t ) −1 (3) −1 t −1 t 1 = = = 3 ( 3) t 9 3t −1 3t −1 35. (π 0 x 2 a −1 ) −1 = π 0( −1) x 2( −1) a −1( −1) = π 0 x −2 a1 = 36. (3m −2 n 4 ) −2 = (3) −2 m −2( −2) n 4( −2) = 3−2 m 4 n −8 = 37. (−8 g −1 s 3 ) 2 = (−8) 2 g −1(2) s 3(2) = 38. ax −2 (−a 2 x)3 = ax −2 (−1)3 (a 2(3) ) x 3 = − 39.  4 x −1  (4) −3 x −1( −3) x3 =  −1  = −1( −3) a 64a 3  a  40.  2b 2   5   y  41. 15n 2T 5 5n 2 − ( −1) 5n3 = = T T 3n −1T 6 42. (nRT −2 )32 n32 R 32 − ( −2)T −2(32) n32 R 34T −64 n32 R 34 n32 R 34 = = = = R −2T 32 T 32 T 32 T 32 − ( −64) T 96 43. 7 ( −4 ) − (−5) 2 = −28 − 25 = −53 44. 6 − −2 − (−2)(8) = 6 − 32 − (−16) = 6 − 32 + 16 = −10 45. −(−26.5) 2 − ( −9.85)3 = −(702.25) − (−955.671625) = 253.421625 which gets rounded to 253 because 702.25 and –955.671625 are both accurate to only 3 significant digits due to the original numbers having only 3 significant digits. 46. −0.7112 − ( − −0.809 ) 6 = ( −1)(0.711) 2 − ( −0.809) 6 = ( −1)(0.505521) − (0.2803439122) = −0.7858649122 a x2 m4 9n8 64 s 6 g2 a(a 6 ) x3 = −a1+ 6 x 3− 2 = −a 7 x x2 −3 −2 = (2) −2 b 2( −2) b −4 y10 = −10 = 4 5( −2) y 4y 4b 5 which gets rounded to 3 significant digits: –0.786. Copyright © 2018 Pearson Education, Inc. 19 20 47. 48. 49. 50. Chapter 1 Basic Algebraic Operations 3.07(− −1.86 ) −5.7102 −5.7102 = = −0.420956185 (−1.86) + 1.596 11.96883216 + 1.596 13.56483216 which gets rounded to 3 significant digits: –0.421. 4 = 15.66 2 − (−4.017) 4 245.2356 − 260.379822692 −15.144222692 = = = 3.941837074 −3.84192 −3.84192 1.044(−3.68) which gets rounded to 3 significant digits: 3.94. 254 254 = 2.38(3684.49) − 3 1.17 1.601613 = 8769.0862 − 158.5901213339 = 8610.4960786661 which gets rounded to 3 significant digits: 8610. 2.38(−60.7) 2 − 0.889 1.89 − 1.092 0.889 = 19.32 + 1.89 − 1.1881 0.889 = 19.32 + 0.7019 = 19.32 + 0.889880728 4.2(4.6) + = 20.209880728 which gets rounded to 2 significant digits: 20 . −1 51. 1−1 1  1   −1  = −1( −1) = , which is the reciprocal of x. x x x   52. 1  0 0.2 −   0.2 − 5   5 =  0  = 00 ≠ 1 , since a 0 = 1 requires that a ≠ 0 . =       −2  10   1   0.01   100  0 53. 0 −1 If a 3 = 5 , then a12 = a 3(4) a12 = ( a 3 ) a12 = ( 5 ) 4 4 a12 = 625 54. For any negative value of a , a will be negative, and a 2 will be positive, making all values of Therefore, it is never the case for negative values of a , a −2 x when x > 1 . Any number greater than 1 will have a square root that is smaller than itself. For example, 2 > 2 = 1.41 x = x when x = 1 or x = 0 because the only numbers that are their own squares are 0 and 1 (i.e., 0 2 = 0 and 12 = 1 ). x < x when 0 < x < 1 . Any number between 0 and 1 will have a square root larger than itself. For example, 0.25 < 0.25 = 0.5 (a) (b) 3 (a) (b) 7 f = 3 7 2140 = 12.8865874254 ,which is rounded to 12.9 −0.214 = −0.59814240297 ,which is rounded to –0.598 0.382 = 0.87155493458 ,which is rounded to 0.872 −382 = −2.33811675837 ,which is rounded to –2.34 1 2π LC = 1 2(3.1416) 0.250(40.52 × 10−6 ) 1 = 6.2832 10.0625 × 10−6 1 = 6.2832(0.003172144385) 1 = 0.0199312175998 = 50.172549 which is rounded to 50.2 Hz 60. standard deviation = = 80.5 kg variance 2 = 8.972179222 kg which is rounded to 8.97 kg 1.7 Addition and Subtraction of Algebraic Expressions 1. 3x + 2 y − 5 y = 3x − 3 y 2. 3c − (2b − c) = 3c − 2b + c = −2b + 4c Copyright © 2018 Pearson Education, Inc. 32 Chapter 1 Basic Algebraic Operations 3. 3ax − [( ax − 5s ) − 2ax ] = 3ax − [ax − 5s − 2ax ] = 3ax − [ − ax − 5s ] = 3ax + ax + 5s = 4ax + 5s 4. 3a 2 b − {a − [2a 2 b − (a + 2b)]} = 3a 2 b − {a − [2a 2 b − a − 2b]} = 3a 2 b − {a − 2a 2 b + a + 2b} = 3a 2 b − {2a − 2a 2 b + 2b} = 3a 2 b − 2a + 2a 2 b − 2b = 5a 2 b − 2a − 2b 5. 5x + 7 x − 4 x = 8x 6. 6t − 3t − 4t = −t 7. 2 y − y + 4x = y + 4x 8. −4C + L − 6C = −10C + L 9. 3t − 4s − 3t − s = 0t − 5s = −5s 10. −8a − b + 12a + b = 4a + 0b = 4a 11. 2 F − 2T − 2 + 3F − T = 5F − 3T − 2 12. x − 2 y − 3x − y + z = −2 x − 3 y + z 13. a 2 b − a 2 b 2 − 2a 2 b = − a 2 b − a 2 b 2 14. − xy 2 − 3 x 2 y 2 + 2 xy 2 = xy 2 − 3x 2 y 2 15. 2 p + ( p − 6 − 2 p) = 2 p − 6 − p = p − 6 16. 5 + (3 − 4n + p ) = 5 + 3 − 4n + p = −4n + p + 8 17. v − (7 − 9 x + 2v) = v − 7 + 9 x − 2v = −v + 9 x − 7 18. 1 1 1 3 1 −2a − (b − a ) = −2a − b + a = − a − b 2 2 2 2 2 19. 2 − 3 − (4 − 5a ) = −1 − 4 + 5a = 5a − 5 20. A + (h − 2 A ) − 3 A = A + h − 2 A − 3 A = −4 A + h 21. (a − 3) + (5 − 6a ) = a − 3 + 5 − 6a = −5a + 2 22. (4 x − y ) − (−2 x − 4 y ) = 4 x − y + 2 x + 4 y = 6 x + 3 y Copyright © 2018 Pearson Education, Inc. Section 1.7 Addition and Subtraction of Algebraic Expressions 23. −(t − 2u ) + (3u − t ) = −t + 2u + 3u − t = −2t + 5u 24. −2(6 x − 3 y ) − (5 y − 4 x) = −12 x + 6 y − 5 y + 4 x = −8 x + y 25. 3(2r + s ) − (−5s − r ) = 6r + 3s + 5s + r = 7r + 8s 26. 3(a − b) − 2(a − 2b) = 3a − 3b − 2a + 4b = a + b 27. −7(6 − 3 j ) − 2( j + 4) = −42 + 21 j − 2 j − 8 = 19 j − 50 28. −(5t + a 2 ) − 2(3a 2 − 2st ) = −5t − a 2 − 6a 2 + 4st = −7a 2 + 4st − 5t 29. −[(4 − 6n ) − ( n − 3)] = −[4 − 6n − n + 3] = −[ −7n + 7] = 7n − 7 30. −[( A − B ) − ( B − A)] = −[ A − B − B + A] = −[2 A − 2 B ] = −2 A + 2 B 31. 2[4 − (t 2 − 5)] = 2[4 − t 2 + 5] = 2[ − t 2 + 9] = −2t 2 + 18 32. 2 2 8 −3[ −3 − ( − a − 4)] = −3[ −3 + a + ] 3 3 3 2 1 = −3[ a − ] 3 3 = −2 a + 1 33. −2[ − x − 2a − ( a − x )] = −2[− x − 2a − a + x ] = −2[ −3a ] = 6a 34. −2[−3( x − 2 y ) + 4 y ] = −2[ −3x + 6 y + 4 y ] = −2[ −3x + 10 y ] = 6 x − 20 y 35. aZ − [3 − ( aZ + 4)] = aZ − [3 − aZ − 4] = aZ − [ − aZ − 1] = aZ + aZ + 1 = 2aZ + 1 36. 9v − [6 − ( − v − 4) + 4v ] = 9v − [6 + v + 4 + 4v ] = 9v − [5v + 10] = 9v − 5v − 10 = 4v − 10 Copyright © 2018 Pearson Education, Inc. 33 34 Chapter 1 Basic Algebraic Operations 37. 5z − {8 − [4 − (2 z + 1)]} = 5z − {8 − [4 − 2 z − 1]} = 5z − {8 − 4 + 2 z + 1} = 5z − {5 + 2 z} = 5z − 5 − 2 z = 3z − 5 38. 7 y − { y − [2 y − ( x − y )]} = 7 y − { y − [2 y − x + y ]} = 7 y − { y − [3 y − x ]} = 7 y − { y − 3 y + x} = 7 y − {−2 y + x} = 7y + 2y − x = −x + 9 y 39. 5 p − ( q − 2 p ) − [3q − ( p − q )] = 5 p − q + 2 p − [3q − p + q ] = 5 p − q + 2 p − [4q − p ] = 7 p − q − 4q + p = 8 p − 5q 40. − (4 − LC ) − [(5 LC − 7) − (6 LC + 2)] = −4 + LC − [5 LC − 7 − 6 LC − 2] = −4 + LC − [ − LC − 9] = −4 + LC + LC + 9 = 2 LC + 5 41. −2{−(4 − x 2 ) − [3 + (4 − x 2 )]} = −2{−4 + x 2 − [3 + 4 − x 2 ]} = −2{−4 + x 2 − 3 − 4 + x 2 } = −2{2 x 2 − 11} = −4 x 2 + 22 42. −{−[ −( x − 2a ) − b] − ( a − x )} = −{−[ − x + 2a − b] − a + x} = −{x − 2a + b − a + x} = −{−3a + b + 2 x} = 3a − b − 2 x 43. 5V 2 − (6 − (2V 2 + 3)) = 5V 2 − (6 − 2V 2 − 3) = 5V 2 − ( −2V 2 + 3) = 5V 2 + 2V 2 − 3 = 7V 2 − 3 44. −2 F + 2((2 F − 1) − 5) = −2 F + 2(2 F − 1 − 5) = −2 F + 2(2 F − 6) = −2 F + 4 F − 12 = 2 F − 12 Copyright © 2018 Pearson Education, Inc. Section 1.7 Addition and Subtraction of Algebraic Expressions 45. − (3t − (7 + 2t − (5t − 6))) = −(3t − (7 + 2t − 5t + 6)) = −(3t − ( −3t + 13)) = −(3t + 3t − 13) = −(6t − 13) = −6t + 13 46. a 2 − 2( x − 5 − (7 − 2( a 2 − 2 x ) − 3x )) = a 2 − 2( x − 5 − (7 − 2a 2 + 4 x − 3x )) = a 2 − 2( x − 5 − (7 − 2a 2 + x )) = a 2 − 2( x − 5 − 7 + 2a 2 − x ) = a 2 − 2(2a 2 − 12) = a 2 − 4a 2 + 24 = −3a 2 + 24 47. −4[4 R − 2.5( Z − 2 R ) − 1.5 ( 2 R − Z )] = −4[4 R − 2.5Z + 5R − 3R + 1.5Z ] = −4[6 R − Z ] = −24 R + 4 Z 48. −3{2.1e − 1.3[− f − 2(e − 5 f )]} = −3{2.1e − 1.3[ − f − 2e + 10 f ]} = −3{2.1e − 1.3[ −2e + 9 f ]} = −3{2.1e + 2.6e − 11.7 f } = −3{4.7e − 11.7 f } = −14.1e + 35.1 f 49. 3D − ( D − d ) = 3D − D + d = 2 D + d 50. i1 − (2 − 3i2 ) + i2 = i1 − 2 + 3i2 + i2 = i1 + 4i2 − 2 51. 4 2 B+ α +2 B− α 3 3 − 4 2 B+ α − B− α 3 3 4 4 4 2 = B + α + 2B − α − B + α − B + α 3 3 3 3 = [ 3B ] − 6 α 3 = 3B − 2α 52. Distance = 30 km/h × (t − 1)h + 40 km/h × (t + 2) h = 30(t − 1) km + 40(t + 2) km = (30t − 30 + 40t + 80) km = (70t + 50) km 53. Memory = x (4 terabytes) + (x + 25)(8 terabytes) = (4 x + 8 x + 200) terabytes = (12 x + 200) terabytes Copyright © 2018 Pearson Education, Inc. 35 36 Chapter 1 Basic Algebraic Operations 54. Difference = 2[(2n + 1)($30) − (n − 2)($20)] = $ 2[60n + 30 − 20n + 40] = $ 2[40n + 70] = $ (80n + 140) 55. (a) (2 x 2 − y + 2a ) + (3 y − x 2 − b) = 2 x 2 − y + 2a + 3 y − x 2 − b = x 2 + 2 y + 2a − b (b) (2 x 2 − y + 2a ) − (3 y − x 2 − b) = 2 x 2 − y + 2a − 3 y + x 2 + b = 3 x 2 − 4 y + 2a + b 56. (3a 2 + b − c 3 ) + (2c 3 − 2b − a 2 ) − (4c 3 − 4b + 3) = 3a 2 + b − c 3 + 2c 3 − 2b − a 2 − 4c 3 + 4b − 3 = 2a 2 + 3b − 3c 3 − 3 57. The final y should be added and the final 3 should be subtracted. The correct final answer is − 2 x − 2 y + 2. 58. The final occurrence of 2c should be added rather than subtracted, resulting in the final answer of 7a − 6b − 2c. 59. a − b = −( − a + b) = −(b − a ) = −1 × (b − a ) = −1 × (b − a ) = 1× b − a = b−a 60. ( a − b) − c = a − b − c However, a − (b − c ) = a − b + c Since they are not equivalent, subtraction is not associative. For example, (10 − 5) − 2 = 5 − 2 = 3 is not the same as 10 − (5 − 2) = 10 − 3 = 7 . 1.8 Multiplication of Algebraic Expressions 1. 2 s 3 ( − st 4 )3 (4 s 2 t ) = 2 s 3 ( −1)3 s 3 t 12 (4 s 2 t ) = −2 s 6 t12 (4 s 2 t ) = −8s 8 t13 2. −2ax (3ax 2 − 4 yz ) = ( −2ax )(3ax 2 ) − ( −2ax )(4 yz ) = ( −6a 2 x 3 ) − ( −8axyz ) = −6a 2 x 3 + 8axyz 3. ( x − 2)( x − 3) = x ( x ) + x ( −3) + ( −2)( x ) + ( −2)( −3) = x 2 − 3x − 2 x + 6 = x 2 − 5x + 6 Copyright © 2018 Pearson Education, Inc.