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Solution Manual for Basic Technical Mathematics, 11th Edition

Solution Manual for Basic Technical Mathematics, 11th Edition simplifies even the toughest textbook questions with step-by-step solutions and easy explanations.

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    S OLUTIONS M ANUAL M ATTHEW G. H UDELSON B ASIC T ECHNICAL M ATHEMATICS AND B ASIC T ECHNICAL M ATHEMATICS WITH C ALCULUS E LEVENTH E DITION Allyn J. Washington Dutchess Community College Richard S. Evans Corning Community College

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    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

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    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

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    1 Chapter 1 Basic Algebraic Operations 1.1 Numbers 1. The numbers –7 and 12 are integers. They are also rational numbers since they can be written as 7 1 − and 12 1 . 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 1 4. The reciprocal of 3 2 is 1 2 2 1 3 / 2 3 3 = × = . 5. 3 is an integer, rational 3 1       , and real. 4 − is imaginary. 6. 7 3 is irrational (because 7 is an irrational number) and real. 6 − is an integer, rational 1 6       − , and real. 7. 6 π − is irrational (because π is an irrational number) and real. 1 8 is rational and real. 8. 6 − − is imaginary. 233 2.33 100 − − = is rational and real. 9. 3 3 = 3 3 − = 2 2 π π − = 10. 0.857 0.857 − = 2 2 = 19 19 4 4 − =

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    2 Chapter 1 Basic Algebraic Operations 11. 6 8 < ; 6 is to the left of 8. 3 4 5 6 7 8 9 10 11 12. 7 5 > ; 7 is to the right of 5. 3 4 5 6 7 8 9 10 11 13. 3.1416; π < (3.1415926 ) π … is to the left of 3.1416. ( π ) (3.1416) 3.141592 3.1416 14. 4 0 − < ; –4 is to the left of 0. –6 –5 –4 –3 –2 –1 0 1 2 15. 4 3 − < − − ; –4 is to the left of 3 − − , ( ) ( ) 3 3 3 − − = − = − . –6 –5 –4 –3 –2 –1 0 1 2 16. 2 1.42; − > − ( ( ) 2 1.414 . 1.414 ), 2 − = − … = − … − is to the right of –1.42. –1.44 –1.43 –1.42 –1.41 –1.40 17. 2 3 2 ; 0.666 3 3 4 − > − − = − … is to the right of 5 3 4 0.7 − = − . –0.8 –0.7 –0.6 –0.5 –0.4 18. 0.6 0.2 − < ; –0.6 is to the left of 0.2. –0.6 –0.5 –0.4 –0.3 –0.2 –0.1 0 0.1 0.2

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    Section 1.1 Numbers 3 19. The reciprocal of 3 is 1 3 . The reciprocal of 4 1 3 is 4 3 4 3 − − = − . The reciprocal of 1 is y b b y b y = . 20. The reciprocal of 1 1 3 is 3 3 1/ 3 1 − − = − = − . The reciprocal of 1 1 4 0.25 is 4 4 1/ 4 1 = = = . The reciprocal of 2 x is 1 2 x . 21. Find 2.5, 12 3 2.4; 0.75; 3 1.732... 5 4 − = − − = − = 12 5 − 3 4 − 3 2.5 –4 –3 –2 –1 0 1 2 3 4 22. Find 7 2 1.414 123 2.333...; 0.707; 2 2 3.14 6.28; 6.47 3 2 2 19 π … − = − − = − = − = × … = = . 3 7 − 2 2 − 2 π 123 19 –3 –2 –1 0 1 2 3 4 5 6 7 23. An absolute value is not always positive, 0 0 = which is not positive. 24. Since 217 2.17 100 − = − , it is rational. 25. The reciprocal of the reciprocal of any positive or negative number is the number itself. The reciprocal of n is 1 n ; the reciprocal of 1 1 is 1 1/ 1 n n n n = ⋅ = . 26. Any repeating decimal is rational, so 2.72 is rational. It turns out that 30 2.72 11 = . 27. It is true that any nonterminating, nonrepeating decimal is an irrational number.

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    4 Chapter 1 Basic Algebraic Operations 28. No, b a b a − = − , as shown below. If 0, a > then a a = . If b a > and 0 a > , then b b = . If b a > then 0 b a − > , 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, 3.14, 5 2.236, 8 8, 3 3, 3.1 π − = = − = − − = − − . –3.1 3 − − -1 5 π 8 − 9 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 So, from smallest to largest, they are 3.1, 3 , 1, 5, , 8 , 9 π − − − − − . 30. List these numbers from smallest to largest: 1 0.20, 10 3.16..., 6 6, 4, 0.25, 3.14... 5 π = − = − − − = − − − = . –6 –4 10 − 1 5 0.25 π − –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 So, from smallest to largest, they are 1 6 , 4, 10, , 0.25, 5 π − − − − − . 31. If a and b are positive integers and b a > , then (a) b a − is a positive integer. (b) a b − is a negative integer. (c) b a b a − + , the numerator and denominator are both positive, but the numerator is less than the denominator, so the 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? If x is a positive or negative integer, then the reciprocal of x is 1 x . Since both 1 and x are integers, the reciprocal is a rational number.

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    Section 1.1 Numbers 5 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? A rational number is a number that can be expressed as a fraction where both the numerator and denominator are integers and the denominator is not zero. So a rational number integer integer a b has a reciprocal of 1 integer integer integer integer b a a b = , which is also a rational number if integer a is not zero. 35. (a) If 0 x > , then x is a positive number located to the right of zero on the number line. x –4 –3 –2 –1 0 1 2 3 4 (b) If 4 x < − , 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 1 x < , then 1 1 x − < < . x –4 –3 –2 –1 0 1 2 3 4 (b) 2 x > , then 2 or 2 x x < − > . x x –4 –3 –2 –1 0 1 2 3 4 37. If 1 1, then x x > is a positive number less than 1. Or 1 0 1 x < < . 1 x –4 –3 –2 –1 0 1 2 3 4 38. If 0 x < , then x is a positive number greater than zero. x –4 –3 –2 –1 0 1 2 3 4 39. 1 a bj a b + = + − is a real number when 1 − is eliminated, which is when b = 0. So a bj + is a real number for all real values of a and b = 0.

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    6 Chapter 1 Basic Algebraic Operations 40. The variables are w and t . The constants are c , 0.1, and 1. 41. 1 2 1 1 1 T C C C = + . Find T C , where 1 0.0040 F C = and 2 0.0010 F C = . 1 1 1 0.0040 0.0010 T C = + 1 1(0.0040) 1(0.0010) 0.0040 0.0010 T C + = × 0.0040 0.0010 0.0000040 0.0040 0.0010 0.0050 T C × = = + 0.00080 F T C = 42. 100 100 V V = 200 200 V V − = 200 100 V V − > 43. bits 1000 bytes kilobytes bytes 1 kilobyte a N n = × × 1000 bits N an = 44. x y L length of base in m the shortened length in centimetres. 100 length of base in cm 100 , all dimensions in cm 100 x y x y L x L x y = = = + = = − 45. Yes, 20 C 30 C − ° > − ° because 30 C − ° is found to the left of 20 C − ° on the number line. 30 − ℃ 20 − ℃ 10 0 − ℃ ℃ 10 ℃ 46. For 4 A, 12 I R < > Ω .

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    Section 1.2 Fundamental Operations of Algebra 7 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 0 0 0 × = = × , not indeterminate. 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 1 6 2 72 − − − − = = −

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    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. ( ) 8 2 6 12 4 12 4 16 2 − − + = + − = + = − 30. | 2 | 2 1 2 2 − = = − − − 31. ( ) ( ) 10 8 3 (10 50) 10( 8)( 3) ( 40) 80( 3) ( 40) 240 ( 40) 6 − − ÷ − = − − ÷ − = − − ÷ − = ÷ − = − 32. 7 5 7 5 2 1 1( 2) 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 3 1 3 1 1 − − − − − − − = − = − − = − − = − − − −

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    Section 1.2 Fundamental Operations of Algebra 9 35. ( ) ( ) ( ) 14 14 7 3 6 8 7 3 2 2 2 3 2 1 14 7 3(2) 2 7 7 6 7 7 6 6 − − − − − = − − − − − − = − − − − = − − − − = − + − = − 36. ( ) 6 7 3 | 9 | (7 3) 2 9 3 21 2 9 14 − − − + − − = + × + − − = + − = 37. 3 | 9 2( 3) | 3 | 9 6 | 1 10 9 3 | 3 | 9 9 9 1 − − − − + = − − − = − = − = − 38. ( ) 20 12 40( 15) 240 600 360 is undefined 98 98 98 98 0 − − − − + = = = − − − 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 (4 5) π 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).

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    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 0 x > (positive) or 0 x = (neutral) then x x = . Example : 4 4 = . The claim that absolute values of negative numbers x x = − is also true. Example: ( ) if is 6, then 6 6 6. x − − = − − = 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) 1 xy − = is true for values of x and y that are negative reciprocals of each other or 1 y x = − , providing that the number x in the denominator is not zero. So if 12 x = , then 1 12 y = − and ( ) 1 12 1 12 xy   − = − − =     . (b) 1 x y x y − = − is true for all values of x and y , providing that x y ≠ to prevent division by zero. 58. (a) 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 0 and 0 x y ≥ ≥ or when 0 and 0 x y ≤ ≤ 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 0 ; or 0 and 0 y x y > < < > .

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    Section 1.2 Fundamental Operations of Algebra 11 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 0 and 0 x y ≥ ≤ ; or when 0 and 0 x y ≤ ≥ . 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 0; or 0 and 0 y x y > > < < . 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: ( ) 2.1 kW h 1.5 kW 3.0 h 2.1 kW h 4.5 kW h 2.4 kW h change used generated change change change E E E E E E = − = ⋅ − = ⋅ − ⋅ = − ⋅ 62. Assuming that this batting average is for the current season only which is just starting, the number of hits is zero and the total number of at-bats is also zero giving us a number of hits 0 batting average at bats 0 = = − which is indeterminate, not 0.000. 63. The average temperature for the week is: 7 ( 3) 2 3 1 ( 4) ( 6) C 7 7 3 2 3 1 4 6 C 7 14 C 2.0 C 7 avg avg avg T T T − + − + + + + − + − = ° − − + + + − − = ° − = ° = − °

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    12 Chapter 1 Basic Algebraic Operations 64. The vertical distance from the flare gun is ( )( ) ( )( ) ( ) 70 5 16 25 350 400 350 400 50 m d d d d = + − = + − = − = − The flare is 50 m below the flare gun. 65. The sum of the voltages is ( ) ( ) 6V 2V 8V 5V 3V 6V 2V 8V 5V 3V 10V sum sum sum V V V = + − + + − + = − + − + = 66. (a) The change in the current for the first interval is the second reading – the first reading 2 2 2 1 2 lb/in 7 lb/in 9 lb/in Change = − − = − . (b) The change in the current for the middle intervals is the third reading – the second reading ( ) 2 2 2 2 2 2 9 lb/in 2 lb/in 9 lb/in 2 lb/in 7 lb/in Change = − − − = − + = − . (c) The change in the current for the last interval is the last reading – the third reading ( ) 2 2 2 2 2 3 6 lb/in 9 lb/in 6 lb/in 9 lb/in 3 lb/in Change = − − − = − + = . 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 ( ) L 12 7 h 84 L h = .The second tank leaks ( ) L 7 12h h 84L. = 12 7 7 12 × = × demonstrates the commutative law of multiplication. 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 7 days 25 7 days 15 day day 175 min 105 min 280 min OR minutes 7 days (25 15) day minutes 7 days 40 day 280 min t t t t t t = × + × = + = = × + = × = which illustrates the distributive law.

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