Solutions Manual for Basic Technical Mathematics with Calculus, SI Version, 11th edition By Allyn J. by welldoneassistant

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

The numbers –7 and 12 are integers. They are also rational numbers since they can be written as

The absolute value of –6 is 6, and the absolute value of –7 is 7. We write these as −6 = 6 and −7 = 7 .

−6 < −4 ; –6 is to the left of –4.

–7 –6 –5 –4 –3 –2 –1 0 4. 5.

The reciprocal of

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 

−

π 6

is irrational (because π is an irrational number) and real.

1 is rational and real. 8

− −6 is imaginary. −233 −2.33 = is rational and real. 100

3 =3

−3 = 3 −

10.

π 2

−0.857 = 0.857 2 = 2 −

19 19 = 4 4

2 11.

Chapter 1 Basic Algebraic Operations 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.

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 < − −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 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 < 0.2 ; –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

Section 1.1 Numbers

19.

1 . 3

The reciprocal of 3 is The reciprocal of − The reciprocal of

4 3

is −

=−

3 . 4

y 1 b is = . b y b y

1 1 3 is − = − = −3 . 3 1/ 3 1 1 1 4 The reciprocal of 0.25 = is = = 4. 4 1/ 4 1 1 The reciprocal of 2x is . 2x

20.

The reciprocal of −

21.

Find 2.5, −

12 3 = −2.4; − = −0.75; 3 = 1.732... 5 4 12 3 − 3 2.5 − 5 4

–4 –3 –2 –1 0 22.

1 4

Find −

2 3 4

7 2 1.414 … 123 = −2.333...; − =− = −0.707; 2π = 2 × 3.14… = 6.28; = 6.47 . 3 2 2 19 −

7 3

–3 –2 –1

−

2 2

2π

3 4

123 19

5 6 7

23.

An absolute value is not always positive, 0 = 0 which is not positive.

24.

Since −2.17 = −

25.

The reciprocal of the reciprocal of any positive or negative number is the number itself. 1 1 1 n The reciprocal of n is ; the reciprocal of is = 1⋅ = n . n n 1/ n 1

26.

Any repeating decimal is rational, so 2.72 is rational. It turns out that 2.72 =

27.

It is true that any nonterminating, nonrepeating decimal is an irrational number.

217 , it is rational. 100

30 . 11

Chapter 1 Basic Algebraic Operations

28.

No, b − a = b − a , as shown below. If a > 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

–4 –3 –2 –1 0

−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.

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.

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?

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)

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 < 1 . x –4 –3 –2 –1

(b)

x > 2 , then x < −2 or x > 2 .

–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

2 3 4

If x < 0 , then x is a positive number greater than zero. x

–4 –3 –2 –1 0 39.

2 3 4

a + bj = a + b −1 is a real number when real values of a and b = 0.

−1 is eliminated, which is when b = 0. So a + bj is a real number for all

Chapter 1 Basic Algebraic Operations

40.

The variables are w and t. The constants are c, 0.1, and 1.

41.

1 1 1 = + . Find CT , where C1 = 0.0040 F and C2 = 0.0010 F . CT C1 C2 1 1 1 = + CT 0.0040 0.0010 1 1(0.0040) + 1(0.0010) = CT 0.0040 × 0.0010 CT =

0.0040 × 0.0010 0.0000040 = 0.0040 + 0.0010 0.0050

CT = 0.00080 F

42.

100V = 100V −200V = 200V

−200V > 100V

43.

a bits 1000 bytes × × n kilobytes bytes 1 kilobyte

N = 1000 an bits 44.

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 < 4 A, R > 12 Ω .

Section 1.2 Fundamental Operations of Algebra

1.2 Fundamental Operations of Algebra 1.

16 − 2 × ( −2 ) = 16 − ( −4 ) = 16 + 4 = 20

−18 + 5 − ( −2 )( 3) = 3 + 5 − ( −6 ) = 8 + 6 = 14 −6

−12 5 − 1 −12 4 + = + = −2 + ( −2 ) = −4 8 − 2 2(−1) 6 −2

7 × 6 42 = = is undefined , not indeterminate. 0×0 0

5 + ( −8 ) = 5 − 8 = −3

−4 + ( −7 ) = −4 − 7 = −11

−3 + 9 = 6 or alternatively −3 + 9 = + ( 9 − 3) = + ( 6 ) = 6

18 − 21 = −3 or alternatively 18 − 21 = −(21 − 18) = −(3) = −3

−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

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

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

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 and y > 0 .

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

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.

Section 1.3 Calculators and Approximate Numbers 70.

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.

35.303 rounded off to four significant digits is 35.30.

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.

38.3 − 21.9(−3.58) = 116.702 using exact numbers; if we estimate the result, 40 − 20(−4) = 120 .

8 cylinders is exact because they can be counted. 55 km/h is approximate since it is measured.

0.002 mm thick is a measurement and is therefore an approximation. $7.50 is an exact price.

24 hr and 1440 min (60 min/h × 24 h =1140 min) are both exact numbers.

50 keys is exact because you can count them; 50 h of use is approximate since it is a measurement of time.

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.

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

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.

(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.

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

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.

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

(− x 3 ) 2 = (−1) x 3  = (−1) 2 ( x 3 ) 2 = (1) x 6 = x 6

2 x 0 = 2(1) = 2

x 3 x 4 = x 3+ 4 = x 7

y 2 y 7 = y 2+7 = y9

2b 4 b 2 = 2b 4 + 2 = 2b 6

3k 5 k = 3k 5 +1 = 3k 6

m5 = m5− 3 = m 2 m3

2 x6 = −2 x 6 −1 = −2 x 5 −x

−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

2(30)

2(3)

= a 30T 60

= 27r 6

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 = −t14

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

1 1 = 61 6

1 w5

L−3 = − L−3− ( −5) = − L2 L−5

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) ) x3 = −

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 R32 − ( −2)T −2(32) n32 R 34T −64 n32 R 34 n32 R34 = = = 32 − ( −64) = 32 32 −2 32 R T T T T 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 x3− 2 = −a 7 x x2

−3

−2

(2) −2 b 2( −2) b −4 y10 = −10 = 4 5( −2) y 4y 4b

which gets rounded to 3 significant digits: –0.786.

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 

53.

−1

If a 3 = 5 , then a12 = a 3(4) a12 = ( a 3 ) a12 = ( 5 )

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 < a −1 .

55.

( x a ⋅ x − a )5 = ( x a − a )5 = ( x 0 )5 = x 0(5) = x 0 = 1 , provided that x ≠ 0 .

56.

(− y a − b ⋅ y a + b ) 2 = ((−1) y a −b + ( a + b ) ) 2 = (−1) 2 ( y 2 a ) 2 = y 2 a (2) = y 4 a .

1 1 greater than . a2 a

Section 1.4 Exponents and Unit Conversions

57.

kT hc

(GkThc ) 2 c =

k 3T 3 ⋅ (G 2 k 2 T 2 h 2 c 2 ) c h3c3

k 3T 3 ⋅ (G 2 k 2 T 2 h 2 c 3 ) h3c3 ( G 2 k 2 + 3 T 2 + 3 c 3− 3 ) = h1 2 5 5 G kT = h =

GmM GM = 3 mr1+ 2 r

58.

GmM (mr ) −1 (r −2 ) =

59.

π   =  = π   = 2 3π r 2 8 3π r 2 24 6

60.

61.

 r3   4    

r  4     

gM (2π fM ) −2 gM = 2π fC 2π fC (2π fM ) 2 gM = 2π fC (4π 2 f 2 M 2 ) gM = 3 3 8π f CM 2 g = 3 3 8π f CM 0.042 2500 1 + 4

= $2500 (1.0105)

= $2500(1.28490602753) = $3212.26700688 = $3212.27

62.

6.85(1000 − 20(6.85) 2 + (6.85) 3 ) 6.85(1000 − 20(46.9225) + 321.419125) = 1850 1850 6.85(1321.419125 − 938.45) = 1850 6.85(382.969125) = 1850 2623.33850625 = 1850 = 1.418020814 = 1.42 cm

63.

If f ( n ) = 1 +

1 n

then f (1) = 21 = 2.000, f (10) = 1.110 = 2.594,

f (100) = 1.01100 = 2.705, and f (1000) = 1.0011000 = 2.717.

Chapter 1 Basic Algebraic Operations

64.

We have 1 TB = 210 GB = 210 (210 MB) = 210 (210 (220 bytes)) = 210 +10 + 20 bytes = 240 bytes

65.

ft    28.2  ( 9.81 s ) = 276.642 ft which is rounded to 277 ft. s 

66.

 mi   40.5  ( 3.7 gal ) = 149.85 mi which is rounded to 150 mi. gal  

67.

m   1 ft  60 s  ft ft  which is rounded to 85,600 .  7.25 2     = 85, 629.92 2 s   0.3048 m  1 min  min min 2 

68.

kg   1000 g   1 m  g  .   238 3    = 0.238 3 m 1 kg 100 cm cm    

69.

 1L  15.7 qt = 15.7 qt ×   = 14.8533586 L which is rounded to 14.9 L.  1.057 qt 

70.

 1 hp  7.50 W = 7.50 W ×   = 0.01005362 hp which is rounded to 0.0101 hp.  746.0 W 

71.

 1 in  2 2 245 cm 2 = 245 cm 2 ×   = 37.975076 in which is rounded to 38.0 in .  2.54 cm 

72.

 1 km  2 2 85.7 mi 2 = 85.7 mi 2 ×   = 221.941401 km which is rounded to 222 km .  0.6214 mi 

73.

65.2

m m  60 s   1 ft  ft ft which is rounded to 12800 . = 65.2 ×  ×  = 12834.6457 s s  1 min   0.3048 m  min min

74.

25.0

mi mi  1 km   1 gal  km km which is rounded to 10.6 . = 25.0 × ×  = 10.6292562 gal gal  0.6214 mi   3.785 L  L L

75.

 2.54 cm  15.6 in = 15.6 in ×   = 39.624 cm which is rounded to 39.6 cm.  1 in 

76.

 1 km  12,500 mi = 12,500 mi ×   = 20,115.8674 km which is rounded to 20,100 km.  0.6214 mi 

77.

575, 000

78.

79.

1130

gal gal  1 day   3.785 L  L L which is rounded to 90,700 . = 575, 000 ×  = 90, 682.2917 × day day  24 hr   1 gal  hr hr

gal gal  1 min   3.785 L  L L which is rounded to 5.4 . = 85 ×  = 5.3620833 × min min  60 s   1 gal  s s ft ft  60 s   60 min   0.3084 m   1 km  km km which is rounded to 1250 . = 1130 ×  × × ×  = 1254.5712 s s  1 min   1 hr   1 ft   1000 m  hr hr

Section 1.5 Scientific Notation

80.

7200

81.

14.7

82.

62.4

km km  1 hr   1 min   1000 m  m = 7200 × × ×  = 2000 . hr hr  60 min   60 s   1 km  s 2

lb lb  4.448 N   1 in   100 cm  N = 14.7 2 ×  ×  ×  = 101,347.883 2 which is rounded to 101,000 Pa. in 2 in  1 lb   2.54 cm   1 m  m

lb lb 1 kg 1 ft = 62.4 3 × × 3 2.205 lb 0.3048 m ft ft kg (The actual value is 1000 3 .) m

= 999.381

kg kg which is rounded to 999 3 . 3 m m

1.5 Scientific Notation 1.

8.06 × 103 = 8060

750 000 000 000-1 = (7.5 × 1011 )-1 = 7.5-1 × 10-11 = 0.1333... × 10-11 = 1.33 × 10-12 rounded to 3 significant digits.

4.5 × 104 = 45,000

6.8 × 107 = 68,000,000

2.01× 10−3 = 0.00201

9.61× 10−5 = 0.0000961

3.23 × 100 = 3.23 × 1 = 3.23

8 × 100 = 8 × 1 = 8

1.86 × 10 = 18.6

10.

1× 10−1 = 0.1

11.

4000 = 4 × 103

12.

56 000 = 5.6 × 104

13.

0.0087 = 8.7 × 10 −3

14.

0.00074 = 7.4 × 10 −4

15.

609, 000, 000 = 6.09 × 108

Chapter 1 Basic Algebraic Operations

16.

10 = 1× 101

17.

0.0528 = 5.28 × 10 −2

18.

0.0000908 = 9.08 × 10 −5

19.

28,000(2,000,000,000) = 2.8 ×10 4 (2 ×109 ) = 5.6 × 1013

20.

50,000(0.006) = 5 × 104 (6 × 10−3 ) = 300 = 3 × 102

21.

88,000 8.8 × 104 = = 2.2 × 108 0.0004 4 × 10−4

22.

0.00003 3 × 10−5 = = 5 × 10−12 6,000,000 6 × 106

23.

35, 600, 000 = 35.6 × 106

24.

0.0000056 = 5.6 × 10−6

25.

0.0973 = 97.3 × 10−3

26.

925, 000, 000, 000 = 925 × 109

27.

0.000000475 = 475 × 10 −9

28.

370, 000 = 370 × 103

29.

2 × 10−35 + 3 × 10−34 = 0.2 × 10 −34 + 3 × 10 −34 = 3.2 × 10 −34

30.

5.3 × 1012 − 3.7 ×1010 = 530 ×1010 − 3.7 × 1010 = 526.3 ×1010 = 5.263 × 1012

31.

(1.2 × 1029 )3 = 1.23 × 1029(3) = 1.728 × 1087

32.

(2 × 10−16 ) −5 = 2−5 × 10−16( −5) = 0.031 25 × 1080 = 3.125 × 1078

33.

1320(649,000)(85.3) = 7.3074804 × 1010

which gets rounded to 7.31× 1010 . 34.

0.0000569(3,190, 000) = 181.511

which gets rounded to 1.82 × 102 . 35.

0.0732(6710) 491.172 = = 1.5867803 × 107 0.00134(0.0231) 0.000030954

which gets rounded to 1.59 × 107 .

Section 1.5 Scientific Notation

36.

0.00452 0.00452 = = 1.915635741× 10−11 2430(97,100) 235,953,000

which gets rounded to 1.92 × 10−11 . 37.

(3.642 × 10−8 )(2.736 × 105 ) = 9.964512 × 10−3

which gets rounded to 9.965 × 10−3 . 38.

(7.309 × 10−1 ) 2 0.53421481 = = 3.56656723394 × 1018 5.9843(2.5036 × 10 −20 ) 1.49787029 × 10−19

which gets rounded to 3.567 × 1018 . 39.

40.

(3.69 × 10−7 )(4.61× 1021 ) 1.70109 × 1015 = = 3.37517857142 × 1016 0.0504 0.0504 which gets rounded to 3.38 × 1016 . (9.907 × 107 )(1.08 × 1012 ) 2 (9.907 × 107 )(1.1664 × 1024 ) 1.01555248 × 1032 = = = 1.56143882045 × 1033 (3.603 × 10 −5 )(2054) 0.07400562 0.07400562

which gets rounded to 1.56 × 1033 . 41.

500,000,000 tweets = 5 ×108 tweets

42.

17,200,000,000 bytes = 1.72 ×1010 bytes

43.

0.000003 W = 3 × 10 −6 W

44.

0.0075 mm = 7.5 × 10−3 mm

45.

1,200,000,000 Hz = 1.2 ×109 Hz

46.

1.84 × 1012 = 1,840, 000, 000, 000

47.

12, 000, 000, 000 m 2 = 1.2 × 1010 m 2

48.

3.086 × 1016 m = 30,860,000,000,000,000 m

49.

1.6 × 10 −12 W = 0.0000000000016 W

50.

2.4 × 10 −43 = 0.00000000000000000000000000000000000000000024

51.

(a) (b) (c)

2300 = 2.3 × 103 = 2.3 kW 0.23 = 230 × 10−3 = 230 mW 2,300, 000 = 2.3 × 106 = 2.3 MW

(d)

0.00023 = 230 × 10−6 = 230 μ W

(a) (b) (c)

8090000 = 8.09 × 106 = 8.09 MΩ 809 000 = 809 × 103 = 809 kΩ 0.0809 = 80.9 × 10−3 = 80.9 mΩ

52.

Chapter 1 Basic Algebraic Operations

53.

(a)

googol = 1 × 10100 = 10100

(b)

googolplex = 10googol = 1010

54.

100

10100 = 10100− 79 = 1021 1079 A googol is 10 21 times larger than the number of electrons in the universe. googol = 10100 , so to find the ratio

sun's diameter 1.4 × 109 m = = 1.27272 × 107 m which is rounded to 1.3 × 107 m. 110 110

55.

earth's diameter =

56.

230 = 1, 073, 741,824 = 1.073741824 × 109 ≈ 1× 109

57.

7.5 × 10−15 s × 5.6 × 106 additions = 4.2 × 10−8 s addition

58.

0.000000039 % = 0.00000000039 0.00000000039 × 0.085 mg = 3.315 ×10−11 mg = 3.3 × 10−11 mg

59.

0.078 s × 2.998 × 108

60.

(a)

1 day ×

(b)

100 year ×

m = 2.3384400 ×107 m which rounds to 2.3 ×107 m s

24 h 60 min 60 s = 86400 s = 8.64 × 104 s × × day h min 365.25 day 24 h 60 min 60 s = 3 155 760 000 s = 3.155 760 0 ×109 s × × × year day h min

61.

1.66×10-27 kg 1.6×101 amu × 1.25 × 108 oxygen atoms = 3.32 × 10−18 kg × amu oxygen atoms

62.

W = kT 4 W = 5.7 × 10-8 W K 4 ×(3.03×102 K) 4 W = 5.7 × 10-8 W K 4 × 8.428892481×109 K 4 W = 4.80446871417×102 W W = 4.8×102 W

63.

64.

k 2.196×10-8 Ω ⋅ m 2 2.196×10-8 Ω ⋅ m 2 = = = 3.432 966 268 57 Ω = 3.433 Ω 2 2 -9 2 d ( 7.998×10-5 m ) 6.396 800 4×10 m

1.496 × 108 km AU × = 2.99799599198×105 km s = 2.998 × 105 km s AU 4.99 × 102 s This is the same speed mentioned in Question 56 as the speed of radio waves.

Section 1.6 Roots and Radicals

1.6 Roots and Radicals 1.

− 3 64 = − 3 (4)3 = −4

(15)(5) Neither 15 nor 5 is a perfect square, so this expression is not as useful. However, if we further factor the 15

to (3)(5)(5) = 3(5) 2 = 5 3 , the result can still be obtained. 16 × 9 = 144 = 122 = 12

3. 4.

− −64 is still imaginary because an even root (in this case n = 2) of a negative number is imaginary, regardless of the numerical factor placed in front of the root.

49 = 7 2 = 7

225 = (25)(9) = 25 × 9 = 5 × 3 = 15

− 121 = − 112 = −11

− 36 = − 62 = −6

− 64 = − 82 = −8

10.

0.25 =

1 1 1 = = = 0.5 4 4 2

11.

0.09 =

9 9 3 = = = 0.3 100 10 100

12.

− 900 = − (9)(100) = − 9 × 100 = −3 × 10 = −30

13.

125 = 3 53 = 5

14.

16 = 4 24 = 2

15.

81 = 4 34 = 3

16.

− 5 −32 = − 5 (−2)5 = − ( −2 ) = 2

17.

( 5) = 5 × 5 = 5

18.

( 31) = 31 × 31 × 31 = 31

Chapter 1 Basic Algebraic Operations

19.

( − −47 ) = ( −1) ( −47 ) = (−1)(−47) = 47

20.

( −23 ) = −23

21.

( − 53 ) = ( −1) ( 53 ) = (1)(53) = 53

22.

75 = (25)(3) = 25 × 3 = 5 3

23.

18 = (9)(2) = 9 × 2 = 3 2

24.

− 32 = − (16)(2) = − 16 × 2 = −4 2

25.

1200 = (100)(4)(3) = 100 × 4 × 3 = 10 × 2 × 3 = 20 3

26.

50 = (25)(2) = 25 × 2 = 5 × 2 = 5 2

27.

2 84 = 2 (4)(21) = 2 × 4 × 21 = 2 × 2 × 21 = 4 21

28.

(36)(3) 108 36 × 3 6 × 3 = = = =3 3 2 2 2 2

29.

80 80 = = 20 = 4 × 5 = 4 × 5 = 2 × 5 = 2 5 3−7 4

30.

81× 102 = 81 × 10 2 = 9 × 10 = 90

31.

−82 = 3 −64 = 3 (−4)3 = −4

32.

92 = 4 81 = 4 34 = 3

33.

34.

7 2 81 (−3)

25 5 243 −3 144

=−

(49)(9) (49) (9) =7 = (9)(7) (9) (7) 32 5 35 3 12

=−

(32) (3) (3) (12)

=−

8 3

35.

36 + 64 = 100 = 102 = 10

36.

25 + 144 = 169 = 132 = 13

37.

32 + 92 = 9 + 81 = 90 = (9)(10) = 9 × 10 = 3 10

Section 1.6 Roots and Radicals

38.

82 − 42 = 64 − 16 = 48 = (16)(3) = 16 × 3 = 4 3

39.

85.4 = 9.24121204171 ,which is rounded to 9.24

40.

3762 = 61.3351449007 ,which is rounded to 61.34

41.

0.8152 = 0.9028842672 ,which is rounded to 0.9029

42.

0.0627 = 0.25039968051 ,which is rounded to 0.250

43.

44.

45.

(a) (b)

1296 + 2304 = 3600 = 60 , which is expressed as 60.00

(a) (b)

10.6276 + 2.1609 = 12.7885 = 3.57610122899 ,which is rounded to 3.57610

(a)

0.0429 2 − 0.01832 = 0.00184041 − 0.00033489

1296 + 2304 = 36 + 48 = 84 , which is expressed as 84.00

10.6276 + 2.1609 = 3.26 + 1.47 = 4.73 ,which is expressed as 4.7300

= 0.00150552 = 0.03880103091 = 0.0388

46.

(b)

0.0429 2 − 0.01832 = 0.0429 − 0.0183 = 0.0246

(a)

3.6252 + 0.614 2 = 13.140625 + 0.376996 = 13.517621 = 3.67663174658 = 3.677

(b)

3.625 + 0.614 = 3.625 + 0.614 = 4.239

47.

24 s = (24)(150) = 3600 = 60 mi/h

48.

Z 2 − X 2 = (5.362 Ω) 2 − (2.875 Ω) 2 = 28.751044 Ω 2 − 8.265625 Ω 2 = 20.485419 Ω 2 = 4.52608208056 Ω = 4.526 Ω

49.

Chapter 1 Basic Algebraic Operations

B 2.18 × 109 Pa = d 1.03 × 103 kg/m 3 = 2116504.85436

N/m 2 kg/m 3

( kg ⋅ m/s ) /m 2

= 2116504.85436

kg/m 3

= 2116504.85436 m 2 /s2 = 1454.82124481 m/s = 1450 m/s

50.

40m  (40)(75)  3000  54.7722557505  55 m/s

51.

w2  h 2  (52.3 in)2  (29.3 in)2

 2735.29 in 2  858.49 in 2  3593.78 in 2  59.948144258 in  59.9 in

52.

  V 24000   100 1  100 1   C 38000   



 100 1  0.63157895



 100 1  0.79471941589   100(0.2052805841)  20.52805841 %  21 %

53.

gd  (9.8)(3500)

 34300  185.20259  190 m/s

54.

1.27  104 h  h 2  1.27  104 (9500)  (9500) 2 = 1.2065  108  9.025  107 = 2.109  108 =14522.3965 which is rounded to 15000 km

Section 1.7 Addition and Subtraction of Algebraic Expressions

55.

a 2 = a is not necessarily true for negative values of a because a2 will be a positive number, regardless whether a is negative or positive.The principal root calculated is assumed to be positive, but there are always two solutions to a

square root, a 2 = ± a since (+ a ) 2 = a 2 and (−a ) 2 = a 2 (see the introduction to this chapter section),so it is sometimes true and sometimes false for negative values of a, depending on which root solution is desired. If only principal roots are considered, then it will not be true for negative values of a. For example, (−4) 2 = 16 = 4 ≠ −4 . 56.

(a) (b) (c)

57.

58.

59.

x > 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)

f =

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 )

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

= 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

3c − (2b − c) = 3c − 2b + c = −2b + 4c

Chapter 1 Basic Algebraic Operations

3ax − [( ax − 5s ) − 2ax ] = 3ax − [ax − 5s − 2ax ] = 3ax − [ − ax − 5s ] = 3ax + ax + 5s = 4ax + 5s

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

5x + 7 x − 4 x = 8x

6t − 3t − 4t = −t

2 y − y + 4x = y + 4x

−4C + L − 6C = −10C + L

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

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 + 4 st = −7 a 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

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

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 − 3 x ))

= 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

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 t12 (4 s 2 t ) = −2 s 6 t12 (4 s 2 t ) = −8s 8 t13

−2ax (3ax 2 − 4 yz ) = ( −2ax )(3ax 2 ) − ( −2ax )(4 yz ) = ( −6a 2 x 3 ) − ( −8axyz ) = −6a 2 x 3 + 8axyz

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

Section 1.8 Multiplication of Algebraic Expressions

(2a − b) 2 = (2a − b)(2a − b) = (2a )(2a ) + (2a )( − b) + (2a )( − b) + ( −b)( −b) = 4a 2 − 2ab − 2ab + b2 = 4a 2 − 4ab + b2

(a 2 )(ax) = a 3 x

(2 xy )( x 2 y 3 ) = 2 x 3 y 4

−a 2 c 2 (a 2 cx3 ) = − a 4 c3 x3

( −2cs 2 )( −4cs ) 2 = ( −2cs 2 )( −4cs )( −4cs ) = ( −2cs 2 )(16c 2 s 2 ) = −32c 3 s 4

(2ax 2 ) 2 ( −2ax ) = (2ax 2 )(2ax 2 )( −2ax ) = (4a 2 x 4 )( −2ax ) = −8a 3 x 5

10.

(6 pq 3 )(3 pq 2 ) 2 = (6 pq 3 )(3 pq 2 )(3 pq 2 ) = (6 pq 3 )(9 p 2 q 4 ) = 54 p 3 q 7

11.

i 2 ( Ri + 2ri ) = (i 2 )( Ri ) + (i 2 )(2ri ) = i 3 R + 2i 3 r

12.

2 x( − p − q) = (2 x )( − p ) − (2 x )( q) = −2 px − 2qx

13.

−3s( s 2 − 5t ) = ( −3s )( s 2 ) + ( −3s )( −5t )

14.

= −3s 3 + 15st −3b(2b2 − b) = ( −3b)(2b2 ) + ( −3b)( −b) = −6b3 + 3b2

15.

5m( m 2 n + 3mn ) = (5m)( m 2 n ) + (5m )(3mn ) = 5m 3 n + 15m 2 n

16.

a 2 bc(2ac − 3b2 c ) = ( a 2 bc)(2ac) + ( a 2 bc )( −3b2 c ) = 2a 3 bc 2 − 3a 2 b3 c 2

17.

3M ( − M − N + 2) = (3M )( − M ) + (3M )( − N ) + (3M )(2) = −3M 2 − 3MN + 6 M

18.

−4c 2 ( −9 gc − 2c + g 2 ) = ( −4c 2 )( −9cg ) + ( −4c 2 )( −2c ) + ( −4c 2 )( g 2 ) = 36c 3 g + 8c 3 − 4c 2 g 2

Chapter 1 Basic Algebraic Operations

19.

xy (tx 2 )( x + y 3 ) = tx 3 y ( x + y 3 ) = (tx 3 y )( x ) + (tx 3 y )( y 3 ) = tx 4 y + tx 3 y 4

20.

−2( −3st 3 )(3s − 4t ) = 6st 3 (3s − 4t ) = (6st 3 )(3s ) + (6st 3 )( −4t ) = 18s 2 t 3 − 24 st 4

21.

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

22.

( a + 7)( a + 1) = ( a )( a ) + ( a )(1) + (7)( a ) + (7)(1) = a 2 + a + 7a + 7 = a 2 + 8a + 7

23.

( x + 5)(2 x − 1) = ( x )(2 x ) + ( x )( −1) + (5)(2 x ) + (5)( −1) = 2 x 2 − x + 10 x − 5 = 2x2 + 9x − 5

24.

(4t1 + t2 )(2t1 − 3t2 ) = (4t1 )(2t1 ) + (4t1 )( −3t2 ) + (t2 )(2t1 ) + (t2 )( −3t2 ) = 8t12 − 12t1t2 + 2t1t2 − 3t2 2 = 8t12 − 10t1t2 − 3t2 2

25.

( y + 8)( y − 8) = ( y )( y ) + ( y )( −8) + (8)( y ) + (8)( −8) = y 2 − 8 y + 8 y − 64 = y 2 − 64

26.

( z − 4)( z + 4) = ( z )( z ) + ( z )(4) + ( −4)( z ) + ( −4)(4) = z 2 + 4 z − 4 z − 16 = z 2 − 16

27.

(2a − b)( −2b + 3a ) = (2a )( −2b) + (2a )(3a ) + ( − b)( −2b) + ( −b)(3a ) = −4ab + 6a 2 + 2b2 − 3ab = 6a 2 − 7ab + 2b2

28.

( −3 + 4 w2 )(3w2 − 1) = ( −3)(3w2 ) + ( −3)( −1) + (4 w2 )(3w2 ) + (4 w)( −1) = −9 w2 + 3 + 12 w4 − 4 w2 = 12 w4 − 13w2 + 3

29.

(2 s + 7t )(3s − 5t ) = (2 s )(3s ) + (2 s )( −5t ) + (7t )(3s ) + (7t )( −5t ) = 6s 2 − 10st + 21st − 35t 2 = 6s 2 + 11st − 35t 2

Section 1.8 Multiplication of Algebraic Expressions 30.

(5 p − 2q)( p + 8q) = (5 p )( p ) + (5 p )(8q) + ( −2 q)( p ) + ( −2q)(8q) = 5 p 2 + 40 pq − 2 pq − 16q 2 = 5 p 2 + 38 pq − 16q 2

31.

( x 2 − 1)(2 x + 5) = ( x 2 )(2 x ) + ( x 2 )(5) + ( −1)(2 x ) + ( −1)(5) = 2 x 3 + 5x 2 − 2 x − 5

32.

(3 y 2 + 2)(2 y − 9) = (3 y 2 )(2 y ) + (3 y 2 )( −9) + (2)(2 y ) + ( −9)(2) = 6 y 3 − 27 y 2 + 4 y − 18

33.

( x − 2 y − 4)( x − 2 y + 4) = ( x )( x ) + ( x )( −2 y ) + ( x )(4) + ( −2 y )( x ) + ( −2 y )( −2 y ) + ( −2 y )(4) + ( −4)( x ) + ( −4)( −2 y ) + ( −4)(4) = x 2 − 2 xy + 4 x − 2 xy + 4 y 2 − 8 y − 4 x + 8 y − 16 = x 2 + 4 y 2 − 4 xy − 16

34.

(2a + 3b + 1)(2a + 3b − 1) = (2a )(2a ) + (2a )(3b) + (2a )( −1) + (3b)(2a ) + (3b)(3b) + (3b)( −1) + (1)(2a ) + (1)(3b) + (1)( −1) = 4a 2 + 6ab − 2a + 6ab + 9b2 − 3b + 2a + 3b − 1 = 4a 2 + 9b2 + 12ab − 1

35.

2( a + 1)( a − 9) = 2[( a )( a ) + ( a )( −9) + (1)( a ) + ( −9)(1)] = 2[a 2 − 9a + a − 9] = 2[a 2 − 8a − 9] = 2a 2 − 16a − 18

36.

−5( y − 3)( y + 6) = −5[( y )( y ) + ( y )(6) + ( −3)( y ) + ( −3)(6)] = −5[ y 2 + 6 y − 3 y − 18] = −5[ y 2 + 3 y − 18] = −5 y 2 − 15 y + 90

37.

−3(3 − 2T )(3T + 2) = −3[(3)(3T ) + (3)(2) + ( −2T )(3T ) + ( −2T )(2)] = −3[ −6T 2 + 9T − 4T + 6] = −3[ −6T 2 + 5T + 6] = 18T 2 − 15T − 18

38.

2n( − n + 5)(6n + 5) = 2n[( − n )(6n ) + ( − n )(5) + (5)(6n ) + (5)(5)] = 2n[ −6n 2 − 5n + 30n + 25] = 2n[ −6n 2 + 25n + 25] = −12n 3 + 50n 2 + 50n

39.

2 L( L + 1)(4 − L) = 2 L[( L)(4) + ( L)( − L) + (1)(4) + (1)( − L)] = 2 L[ − L2 + 4 L − L + 4] = 2 L[ − L2 + 3L + 4] = −2 L3 + 6 L2 + 8L

Chapter 1 Basic Algebraic Operations

40.

ax( x + 4)(7 − x 2 ) = ax[( x )(7) + ( x )( − x 2 ) + (4)(7) + (4)( − x 2 )] = ax[ − x 3 − 4 x 2 + 7 x + 28] = − ax 4 − 4ax 3 + 7ax 2 + 28ax

41.

(3x − 7) 2 = (3x − 7)(3x − 7) = (3x )(3x ) + (3x )( −7) + ( −7)(3x ) + ( −7)( −7) = 9 x 2 − 21x − 21x + 49 = 9 x 2 − 42 x + 49

42.

( x − 3 y ) 2 = ( x − 3 y )( x − 3 y ) = ( x )( x ) + ( x )( −3 y ) + ( −3 y )( x ) + ( −3 y )( −3 y ) = x 2 − 3xy − 3xy + 9 y 2 = x 2 − 6 xy + 9 y 2

43.

( x1 + 3 x2 ) 2 = ( x1 + 3x2 )( x1 + 3x2 ) = ( x1 )( x1 ) + ( x1 )(3x2 ) + (3x2 )( x1 ) + (3x2 )(3x2 ) = x12 + 3x1 x2 + 3x1 x2 + 9 x2 2 = x12 + 6 x1 x2 + 9 x2 2

44.

( −7m − 1) 2 = ( −7m − 1)( −7m − 1) = ( −7m )( −7m) + ( −7m)( −1) + ( −1)( −7m) + ( −1)( −1) = 49m 2 + 7m + 7m + 1 = 49m 2 + 14m + 1

45.

( xyz − 2) 2 = ( xyz − 2)( xyz − 2) = ( xyz )( xyz ) + ( xyz )( −2) + ( −2)( xyz ) + ( −2)( −2) = x 2 y 2 z 2 − 2 xyz − 2 xyz + 4 = x 2 y 2 z 2 − 4 xyz + 4

46.

( −6 x 2 + b)2 = ( −6 x 2 + b)( −6 x 2 + b) = ( −6 x 2 )( −6 x 2 ) + ( −6 x 2 )(b) + (b)( −6 x 2 ) + (b)(b) = 36 x 4 − 6bx 2 − 6bx 2 + b2 = 36 x 4 − 12bx 2 + b2

47.

2( x + 8) 2 = 2[( x + 8)( x + 8)] = 2[( x )( x ) + ( x )(8) + (8)( x ) + (8)(8)] = 2[ x 2 + 8 x + 8 x + 64] = 2[ x 2 + 16 x + 64] = 2 x 2 + 32 x + 128

Section 1.8 Multiplication of Algebraic Expressions

48.

3(3R − 4) 2 = 3[(3R − 4)(3R − 4)] = 3[(3R )(3R ) + (3R )( −4) + ( −4)(3R ) + ( −4)( −4)] = 3[9 R 2 − 12 R − 12 R + 16] = 3[9 R 2 − 24 R + 16] = 27 R 2 − 72 R + 48

49.

(2 + x )(3 − x )( x − 1) = [(6 − 2 x + 3x − x 2 )]( x − 1) = ( x − 1)[ − x 2 + x + 6] = ( x )( − x 2 ) + ( x )( x ) + (6)( x ) + ( −1)( − x 2 ) + ( −1)( x ) + ( −1)(6) = − x3 + x2 + 6 x + x2 − x − 6 = − x3 + 2 x 2 + 5x − 6

50.

( − c 2 + 3x )3 = ( − c 2 + 3x )( − c 2 + 3x )( − c 2 + 3x ) = [(3x )(3x ) − 3c 2 x − 3c 2 x + c 4 )]( − c 2 + 3x ) = ( − c 2 + 3x )[9 x 2 − 6c 2 x + c 4 ] = ( − c 2 )(9 x 2 ) + ( − c 2 )( −6c 2 x ) + ( − c 2 )( + c 4 ) + (3x )(9 x 2 ) + (3 x )( −6c 2 x ) + (3 x )( + c 4 ) = −9c 2 x 2 + 6c 4 x − c 6 + 27 x 3 − 18c 2 x 2 + 3c 4 x = − c 6 + 9c 4 x − 27c 2 x 2 + 27 x 3

51.

3T (T + 2)(2T − 1) = 3T [(T )(2T ) + (T )( −1) + (2)(2T ) + (2)( −1)] = 3T [2T 2 − T + 4T − 2] = 3T [2T 2 − T + 4T − 2] = 3T [2T 2 + 3T − 2] = 6T 3 + 9T 2 − 6T

52.

[( x − 2) 2 ( x + 2)]2 = [( x − 2)( x − 2)( x + 2)][( x − 2)( x − 2)( x + 2)] = [( x − 2)[( x )( x ) + ( −2)( x ) + (2)( x ) + ( −2)(2)]][( x − 2)[( x )( x ) + ( −2)( x ) + (2)( x ) + ( −2)(2)]] = [( x − 2)[ x 2 − 2 x + 2 x − 4]][( x − 2)[ x 2 − 2 x + 2 x − 4]] = [( x − 2)[ x 2 − 4]][( x − 2)[ x 2 − 4]] = [( x )( x 2 ) + ( −4)( x ) + ( −2)( x 2 ) + ( −2)( −4)][( x )( x 2 ) + ( −4)( x ) + ( −2)( x 2 ) + ( −2)( −4)] = [ x 3 − 2 x 2 − 4 x + 8][ x 3 − 2 x 2 − 4 x + 8] = ( x 3 )( x 3 ) + ( x 3 )( −2 x 2 ) + ( x 3 )( −4 x ) + ( x 3 )(8) + ( −2 x 2 )( x 3 ) + ( −2 x 2 )( −2 x 2 ) + ( −2 x 2 )( −4 x ) + ( −2 x 2 )(8) +( −4 x )( x 3 ) + ( −4 x )( −2 x 2 ) + ( −4 x )( −4 x ) + ( −4 x )(8) + (8)( x 3 ) + (8)( −2 x 2 ) + (8)( −4 x ) + (8)(8) = x 6 − 2 x 5 − 4 x 4 + 8 x 3 − 2 x 5 + 4 x 4 + 8 x 3 − 16 x 2 − 4 x 4 + 8 x 3 + 16 x 2 − 32 x + 8 x 3 − 16 x 2 − 32 x + 64 = x 6 − 4 x 5 − 4 x 4 + 32 x 3 − 16 x 2 − 64 x + 64

53.

(a)

( x + y ) 2 = (3 + 4) 2 = 72 = 49 x 2 + y 2 = 32 + 42 = 9 + 16 = 25 ( x + y )2 ≠ x 2 + y 2 49 ≠ 25

Chapter 1 Basic Algebraic Operations

(b)

( x − y ) 2 = (3 − 4) 2 = ( −1) 2 = 1 x 2 − y 2 = 32 − 42 = 9 − 16 = −7 ( x − y )2 ≠ x 2 − y 2 1 ≠ −7

54.

One can write ( x + 3)5 = ( x + 3)( x + 3)( x + 3)( x + 3)( x + 3) and then perform the multiplications using the rightmost pair of terms at each step.

55.

( x + y )3 = ( x + y )( x + y )( x + y ) = ( x + y )[( x )( x ) + ( x )( y ) + ( y )( x ) + ( y )( y )] = ( x + y )[ x 2 + xy + xy + y 2 ] = ( x + y )[ x 2 + 2 xy + y 2 ] = ( x )( x 2 ) + ( x )(2 xy ) + ( x )( y 2 ) + ( y )( x 2 ) + ( y )(2 xy ) + ( y )( y 2 ) = x3 + 2 x2 y + y2 x + x2 y + 2 y 2 x + y3 = x 3 + 3x 2 y + 3 y 2 x + y 3 This differs from x 3 + y 3 by the presence of 3 x 2 y + 3 y 2 x in the center.

56.

( x + y )( x 2 − xy + y 2 ) = ( x )( x 2 ) + ( x )( − xy ) + ( x )( y 2 ) + ( y )( x 2 ) + ( y )( − xy ) + ( y )( y 2 ) = x3 − x2 y + y 2 x + x2 y − y 2 x + y3 = x3 + y3

57.

P (1 + 0.01r ) 2 = P (1 + 0.01r )(1 + 0.01r ) = P[(1)(1) + (1)(0.01r ) + (0.01r )(1) + (0.01r )(0.01r )] = P[1 + 0.01r + 0.01r + 0.0001r 2 ] = 0.0001r 2 P + 0.02rP + P

58.

1000(1 + 0.0025r ) 2 = 1000(1 + 0.0025r )(1 + 0.0025r ) = 1000[(1)(1) + (1)(0.0025r ) + (0.0025r )(1) + (0.0025r )(0.0025r )] = 1000[1 + 0.0025r + 0.0025r + 0.0000625r 2 ] = 1000 + 2.5r + 0.0625r 2

59.

The room will be 5 + w + 5 = w + 10 feet wide and 5 + 2 w + 5 = 2 w + 10 feet long. Its area is ( w + 10)(2 w + 10) = ( w)(2 w) + ( w)(10) + (10)(2 w) + (10)(10) = 2 w2 + 10w + 20w + 100 = 2 w2 + 30w + 100

60.

R = xp = x(30 − 0.01x ) = 30 x − 0.01x 2

Section 1.9 Division of Algebraic Expressions

61.

(2 R − X ) 2 − ( R 2 + X 2 ) = (2 R − X )(2 R − X ) − ( R 2 + X 2 ) = [(2 R )(2 R ) + (2 R )( − X ) + (2 R )( − X ) + ( − X )( − X )] − ( R 2 + X 2 ) = 4 R 2 − 2 RX − 2 RX + X 2 − R 2 − X 2 = 3R 2 − 4 RX

62.

(2T 3 + 3)(T 2 − T − 3) = (2T 3 )(T 2 ) + (2T 3 )( −T ) + (2T 3 )( −3) + (3)(T 2 ) + (3)( −T ) + (3)( −3) = 2T 5 − 2T 4 − 6T 3 + 3T 2 − 3T − 9

63.

Number of switches for n elements = n 2 Number of switches for n + 100 elements = ( n + 100) 2 = ( n + 100)( n + 100) = ( n )( n ) + ( n )(100) + (100)( n ) + (100)(100) = n 2 + 100n + 100n + 10000 = n 2 + 200n + 10000

64.

(T 2 − 100)(T − 10)(T + 10) = (T 2 − 100) T 2 + 10T − 10T − 100 = (T 2 − 100) T 2 − 100 = T 4 − 100T 2 − 100T 2 + 10 000 = T 4 − 200T 2 + 10 000

65.

( R1 + R2 ) 2 − 2 R2 ( R1 + R2 ) = [( R1 + R2 )( R1 + R2 ) ] − 2 R2 ( R1 + R2 )

= [( R1 )( R1 ) + ( R1 )( R2 ) + ( R2 )( R1 ) + ( R2 )( R2 ) ] − 2 R1 R2 − 2 R2 2 = R12 + R1 R2 + R1 R2 + R2 2 − 2 R1 R2 − 2 R2 2 = R12 − R2 2

66.

27 x 2 − 24( x − 6) 2 − ( x − 12)3 = 27 x 2 − 24( x − 6)( x − 6) − ( x − 12)( x − 12)( x − 12) = 27 x 2 − 24 x 2 − 6 x − 6 x + 36 − ( x − 12) x 2 − 12 x − 12 x + 144 = 27 x 2 − 24 x 2 − 12 x + 36 − ( x − 12) x 2 − 24 x + 144 = 27 x 2 − 24 x 2 + 288 x − 864 − ( x )( x 2 ) + ( x )( −24 x ) + ( x )(144) + ( −12)( x 2 ) + ( −12)( −24 x ) + ( −12)(144) = 3x 2 + 288 x − 864 − x 3 + 24 x 2 − 144 x + 12 x 2 − 288 x + 1728 = − x 3 + 39 x 2 − 144 x + 864

1.9 Division of Algebraic Expressions 1.

−6a 2 xy 2  −6  a 2 − 2 x1−1 3 =  = 3 −2a 2 xy 5  −2  y 5− 2 y

Chapter 1 Basic Algebraic Operations

4 x3 y − 8x3 y 2 + 2 x2 y 4 x3 y 8x3 y 2 2 x2 y = − + 2 xy 2 2 xy 2 2 xy 2 2 xy 2

x 2−1 2 x 3−1 − 4 x 3−1 y 2 − 2 + 2−1 2 −1 y y

x 2x2 − 4x2 + y y

3x − 2 2 x − 1 6x2 − 7x + 2 6 x 2 − 3x − 4x + 2 −4 x + 2 0

2x − 1 4 x2 − 1 8x3 − 4 x2 + 0x + 3 8x3

- 2x 2

− 4x + 2x + 3 − 4x2

1 2x + 2

8x3 − 4 x2 + 3 2x + 2 = 2x − 1 + 2 2 4x − 1 4x − 1

8 x3 y 2 = −4 x3−1 y 2 −1 = −4 x 2 y −2 xy

−18b 7 c3 = −18b7 −1c3− 2 = −18b 6 c bc 2

−16r 3t 5 4t 5−1 4t 4 = 5−3 = 2 −4 r 5 t r r

51mn5 3n5 − 2 3n3 = 2 −1 = 2 2 m 17 m n m

(15 x 2 y )(2 xz ) 30 x3 yz = = 3x3−1 y1−1 z = 3 x 2 z 10 xy 10 xy

10.

(5sT )(8s 2T 3 ) 40 s 3T 4 = = 4s 3− 3T 4 − 2 = 4T 2 10 s 3T 2 10 s 3T 2

11.

(4a 3 )(2 x) 2 4a 3 (4 x 2 ) = = 1a 3− 2 x 2 − 2 = a (4ax) 2 16a 2 x 2

Section 1.9 Division of Algebraic Expressions

12.

12a 2 b 12a 2 b 4a 2 − 2 4 = 2 4 = 4 −1 = 3 2 2 (3ab ) 9a b 3b 3b

13.

3a 2 x + 6 xy 3a 2 x 6 xy 3a 2 x1−1 6 x1−1 y = + = + = a2 + 2 y 3x 3x 3x 3 3

14.

2m 2 n − 6mn 2m 2 n 6mn = − = − m 2 −1n + 3m1−1n = − mn + 3n −2m −2m −2m

15.

3rst − 6r 2 st 2 3rst 6r 2 st 2 = − = r1−1 s1−1t − 2r 2 −1 s1−1t 2 = −2rt 2 + t 3rs 3rs 3rs

16.

−5a 2 n − 10an 2 −5a 2 n 10an 2 = − = −a 2 −1n1−1 − 2a1−1n 2 −1 = −a − 2n 5an 5an 5an

17.

4 pq 3 + 8 p 2 q 2 − 16 pq5 4 pq 3 8 p 2 q 2 16 pq5 = + − 4 pq 2 4 pq 2 4 pq 2 4 pq 2 = p1−1 q 3− 2 + 2 p 2 −1 q 2 − 2 − 4 p1−1 q5− 2 = −4 q 3 +2 p + q

18.

a 2 x1 x22 + ax13 − ax1 a 2 x1 x22 ax13 ax1 = + − ax1 ax1 ax1 ax1 = a 2 −1 x11−1 x22 + a1−1 x13−1 − a1−1 x11−1 = ax22 + x12 − 1

19.

2π fL − π fR 2 2π fL π fR 2 = − π fR π fR π fR 2 f 1−1 L − f 1−1 R 2 −1 R 2L = −R R =

20.

9( aB ) 4 − 6aB 4 9( aB ) 4 6aB 4 = − −3aB 3 −3aB 3 −3aB 3 9a 4 B 4 6aB 4 =− + 3aB 3 3aB 3 4 −1 4 − 3 = −3a B + 2a1−1 B 4 − 3 = −3a 3 B + 2 B

21.

−7a 2 b + 14ab2 − 21a 3 7a 2 b 14ab2 21a 3 =− + − 2 2 2 2 2 2 14a b 14a b 14a b 14a 2 b2 a 2− 2 b2− 2 3 = − 2 −1 + 2−1 − a 3− 2 b −2 2 2b a 1 1 3a =− + − 2b a 2b 2

22.

Chapter 1 Basic Algebraic Operations

2 x n + 2 + 4ax n 2 x n + 2 4ax n = + 2xn 2 xn 2xn n−n+2 =x + 2ax n − n = x 2 + 2a

23.

6 y 2 n − 4ay n +1 6 y 2 n 4ay n +1 = − 2 yn 2 yn 2 yn = 3 y 2 n − n − 2ay n − n +1 = 3 y n − 2ay

24.

3a ( F + T )b2 − ( F + T ) 3a ( F + T )b2 (F + T ) = − a( F + T ) a( F + T ) a( F + T ) =

3a1−1 ( F + T ) b2 (F + T )

= 3b2 −

25.

−

(F + T )

a (F + T )

1 a

x+5 x + 4 x 2 + 9 x + 20 x2 + 4 x 5x + 20 5 x + 20 0 2

x + 9 x + 20 = x+5 x+4

26.

x+9 x − 2 x 2 + 7 x − 18 x2 − 2x 9x − 18 9x − 18 0 2

x + 7 x − 18 = x+9 x−2

Section 1.9 Division of Algebraic Expressions

27.

2x + 1 x + 3 2x2 + 7x + 3 2x2 + 6x x+3 x+3 0 2

2x + 7x + 3 = 2x + 1 x+3

28.

3t − 4 t − 1 3t 2 − 7t + 4 3t 2 − 3t − 4t + 4 −4 t + 4 0 3t 2 − 7t + 4 = 3t − 4 t −1

29.

x −1 x − 2 x 2 − 3x + 2 x2 − 2 x −x+2 −x + 2 0 x 2 − 3x + 2 = x −1 x−2

30.

2x − 7 x + 1 2 x 2 − 5x − 7 2x2 + 2x − 7x − 7 −7 x − 7 0 2 x 2 − 5x − 7 = 2x − 7 x +1

Chapter 1 Basic Algebraic Operations

31.

4 x2 − x − 1 2 x − 3 8 x − 14 x 2 + x + 0 3

8 x 3 − 12 x 2 − 2 x2 + x −2 x 2 + 3 x − 2x + 0 −2 x + 3 −3 8 x 3 − 14 x 2 + x 3 = 4 x2 − x − 1 − 2x − 3 2x − 3

32.

3y + 2 2y +1 6y + 7y + 6 2

6 y2 + 3y 4y + 6 4y + 2 4 2

6y + 7y + 6 4 = 3y + 2 + 2y +1 2y +1

33.

Z −2 4 Z + 3 4 Z 2 − 5Z − 7 4 Z 2 + 3Z − 8Z − 7 − 8Z − 6 −1 4 Z 2 − 5Z − 7 1 = Z −2− 4Z + 3 4Z + 3

34.

2x + 1 3x − 4 6 x − 5 x − 9 2

6 x2 − 8x 3x − 9 3x − 4 −5 6 x 2 − 5x − 9 5 = 2x + 1 − 3x − 4 3x − 4

Section 1.9 Division of Algebraic Expressions

35.

x2 + x − 6 x + 2 x + 3x − 4 x − 12 3

x3 + 2 x2 x2 − 4x x2 + 2 x − 6 x − 12 −6 x − 12 0 x 3 + 3x 2 − 4 x − 12 = x2 + x − 6 x+2

36.

x2 + 7 x + 9 3x − 2 3x + 19 x + 13x − 20 3

3x 3 − 2 x 2 21x 2 + 13x 21x 2 − 14 x 27x − 20 27 x − 18 −2 3x 3 + 19 x 2 + 13x − 20 2 = x2 + 7 x + 9 − 3x − 2 3x − 2

37.

2a 2 + 8 a − 2 2a + 0a + 4a 2 + 0a − 16 2

2a 4

− 4a 2 8a 2

− 16

0 2a 4 + 4a 2 − 16 = 2a 2 + 8 2 a −2

38.

2T + 1 3T 2 − T + 2 6T 3 + T 2 + 0T + 2 6T 3 − 2T 2 +4T 3T 2 − 4T + 2 3T 2 − T + 2 − 3T 6T 3 + T 2 + 2 3T = 2T + 1 − 2 2 3T − T + 2 3T − T + 2

Chapter 1 Basic Algebraic Operations

39.

y2 − 3y + 9 y + 3 y 3 + 0 y 2 + 0 y + 27 y3 + 3y2 − 3y2 + 0 y −3 y 2 − 9 y 9y + 27 9y + 27 0 y 3 + 27 = y2 − 3y + 9 y+3

40.

D2 + D + 1 D − 1 D + 0D 2 + 0D − 1 3

D 3 − 1D 2 D 2 + 0D D 2 − 1D D −1 D −1 0

D3 − 1 = D2 + D + 1 D −1

41.

x− y x − y x − 2 xy + y 2 2

x 2 − xy − xy + y 2 − xy + y 2 0 x 2 − 2 xy + y 2 = x− y x− y

42.

3r + 4 R r − 3R 3r − 5rR + 2 R 2 2

3r 2 − 9rR 4rR + 2 R 2 4rR − 12 R 2 14 R 2 2

3r − 5rR + 2 R 14 R 2 = 3r + 4 R + r − 3R r − 3R

Section 1.9 Division of Algebraic Expressions

43.

t−2 t 2 + 2t + 4 t 3 + 0t 2 + 0t − 8 t 3 + 2t 2 + 4t − 2t 2 − 4t − 8 −2t 2 − 4t − 8 0 t3 − 8 =t−2 t 2 + 2t + 4

44.

a 2 + 2ab + 2b2 a − 2ab + 2b a + 0a b + 0a b2 + 0ab3 + b4 2

a 4 − 2a 3 b + 2a 2 b 2 2a 3 b − 2a 2 b2 + 0ab3 2a 3 b − 4a 2 b2 + 4ab3 2a 2 b2 − 4ab3 + b4 2a 2 b2 − 4ab3 + 4b4 − 3b4 3b4 a +b 2 2 2 2 a ab b = + + − a 2 − 2ab + 2b2 a 2 − 2ab + 2b2 4

45.

We know that 2 x + 1 multiplied by x + c will give us 2 x 2 − 9 x − 5 , so 2 x 2 − 9 x − 5 divided by 2 x + 1 will give us x + c : x−5 2x + 1 2 x2 − 9 x − 5 2 x2 + x − 10 x − 5 −10 x − 5 0 x+c = x −5 c = −5

46.

2x − 3 3x + 4 6 x − x + k 2

6x2 + 8x − 9x + k −9 x − 12 k − ( −12) = 0 k + 12 = 0

k = −12

Chapter 1 Basic Algebraic Operations

47.

x3 − x2 + x − 1 x + 1 x + 0x + 0x2 + 0x + 1 4

x4 + x3 − x3 + 0x2 − x3 − x2 x2 + 0x x2 + x − x +1 −x −1 2 2 x4 + 1 = x3 − x2 + x − 1 + ≠ x3 x +1 x +1

48.

x 2 − xy + y 2 x + y x3 + 0x2 y + 0 y 2 x + 0x + y3 x3 + x2 y − x2 y + 0 y 2 x − x2 y − y2 x y2 x + y3 y2 x + y3 0 x3 + y 3 = x 2 − xy + y ≠ x 2 + y 2 x+ y

49.

V1 1 +

T2 − T1 T T = V1 1 + 2 − 1 T1 T1 T1 = V1 1 + = V1 =

50.

T2 −1 T1

T2 T1

V1T2 T1

3x + 2 2 x + 5 6 x + 19 x + 10 2

6 x 2 + 15 x 4 x + 10 4 x + 10 0 The width is 3x + 2

Section 1.9 Division of Algebraic Expressions

51.

52.

8 A5 + 4 A3 μ 2 E 2 − Aμ 4 E 4 8 A5 4 A3 μ 2 E 2 Aμ 4 E 4 = 4 + − 8 A4 8A 8 A4 8 A4 2 2 4 4 μ E μ E = A5− 4 + 4− 3 − 4−1 2A 8A 2 2 4 4 μ E μ E = A+ − 2A 8 A3

6 R1 + 6 R2 + R1 R2 6 R1 6 R2 RR = + + 1 2 6 R1 R2 6 R1 R2 6 R1 R2 6 R1 R2 = =

53.

54.

R1 R1 R2

R2 R1 R2

R1 R2 6 R1 R2

1 1 1 + + R1 R2 6

GMm[( R + r ) − ( R − r )] GMm[ R + r − R + r ] = 2rR 2rR GMm[2r ] = 2rR GMm [2r ] = 2r R GMm = R 3T 2 − 2T − 4 T − 2 3T − 8T 2 + 0T + 8 3

3T 3 − 6T 2 − 2T 2 + 0T −2T 2 + 4T − 4T + 8 −4T + 8 0

Chapter 1 Basic Algebraic Operations

−1

s2 − 2s − 2 s4 + 4

55.

s4 + 4 s2 − 2s − 2

s2 + 2s + 6 s − 2 s − 2 s + 0s + 0s 2 + 0s + 4 2

s4 − 2s3 − 2s 2 2 s 3 + 2 s 2 + 0s 2s3 − 4s2 − 4s 6s 2 + 4 s + 4 6s 2 − 12 s − 12 16s + 16 16s + 16 s4 + 4 = s2 + 2s + 6 + 2 s − 2s − 2 s − 2s − 2 2

t 2 − 3t + 5 2t + 100 2t + 94t − 290t + 500 3

56.

2t 3 + 100t 2 − 6t 2 − 290t −6t 2 − 300t 10t + 500 10t + 500 0

1.10

Solving Equations

x − 3 = −12 x − 3 + 3 = −12 + 3

(a)

x = −9

(b)

x + 3 = −12 x + 3 − 3 = −12 − 3 x = −15 x = −12 3

x = 3( −12) 3 x = −36

Section 1.10 Solving Equations (d)

3x = −12 3x −12 = 3 3 x = −4

7 − 2t = 9 7 − 7 − 2t = 9 − 7 −2t = 2 −2t 2 = −2 −2 t = −1 Check: 7 − 2t = 9 7 − 2( −1) = 9 7 − ( −2) = 9 9=9

x − 7 = 3x − (8 − 6 x ) x − 7 = 3x − 8 + 6 x x − 7 = 9x − 8 −8 x = −1 x=

1 8

L 7 = 3.80 4 7 L 3.80 = 3.80 3.80 4 L = 6.65 m x−2=7 x = 7+2 x=9

x − 4 = −1 x = −1 + 4 x=3

x+5= 4 x = 4−5 x = −1

s + 6 = −3 s = −3 − 6 s = −9

10.

Chapter 1 Basic Algebraic Operations

t = −5 2 t = 2( −5) t = −10 x =2 −4 x = −4(2) x = −8

11.

12.

y −8 =4 3 y − 8 = 4(3) y = 12 + 8 y = 20 7−r =3 6 7 − r = 6(3) − r = 18 − 7 − r = 11

r = −11

13.

4 E = −20 −20 4 E = −5

14.

2 x = 12 12 2 x=6

15.

5t + 9 = −1 5t = −1 − 9 −10 t= 5 t = −2

16.

5D − 2 = 13 5D = 13 + 2 15 D= 5 D=3

Section 1.10 Solving Equations 17.

5 − 2 y = −3 −2 y = −3 − 5 −8 −2 y=4

18.

−5t + 8 = 18 −5t = 18 − 8 10 t= −5 t = −2

19.

3x + 7 = x x − 3x = 7 −2 x = 7 x=−

20.

7 2

6 + 4 L = 5 − 3L 4 L + 3L = 5 − 6 7 L = −1 L=−

21.

1 7

2(3q + 4) = 5q 6q + 8 = 5q 6q − 5q = −8 q = −8

22.

3(4 − n ) = − n − n = 12 − 3n − n + 3n = 12 2n = 12 12 2 n=6 n=

23.

− ( r − 4) = 6 + 2r − r + 4 = 6 + 2r − r − 2r = 2 −3r = 2 r=−

2 3

Chapter 1 Basic Algebraic Operations

24.

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

25.

3 1 = 6 2

8( y − 5) = −2 y 8 y − 40 = −2 y 8 y + 2 y = 40 10 y = 40 40 y= 10 y=4

26.

4(7 − F ) = −7 28 − 4 F = −7 −4 F = −7 − 28 −35 35 F= = −4 4

27.

0.1x − 0.5( x − 2) = 2 x − 5( x − 2) = 2(10) x − 5 x + 10 = 20 −4 x = 20 − 10 10 5 x= =− −4 2

28.

1.5 x − 0.3( x − 4) = 6 15 x − 3( x − 4) = 6(10) 15 x − 3x + 12 = 60 12 x = 60 − 12 48 x= 12 x=4

29.

−4 − 3(1 − 2 p ) = −7 + 2 p −4 − 3 + 6 p = −7 + 2 p −7 + 6 p − 2 p = −7 4 p = −7 + 7 0 p= 4 p=0

Section 1.10 Solving Equations 30.

31.

3 − 6(2 − 3t ) = t − 5 3 − 12 + 18t = t − 5 −9 + 18t − t = −5 17t = 4 4 t= 17 4 x − 2( x − 4) =8 3 4 x − 2 x + 8 = 3(8) 2 x = 24 − 8 16 2 x=8 x=

32.

−5(7 − 3x ) + 2 4 4(2 x ) = −35 + 15 x + 2 2x =

8 x − 15 x = −33

33.

−7 x = −33 −33 33 x= = 7 −7 x −9 = 2 x = 2 + 9 = 11 x = 11 or x = −11

34.

2− x = 4 − x = 4−2 2 −1 x = −2 x =

There is no real solution for x. 35.

2x − 3 = 5 2 x − 3 = 5 or 2 x − 3 = −5 2x − 3 = 5

2 x − 3 = −5

2x = 5 + 3

2 x = −5 + 3

2x = 8 8 x= 2 x = 4 or

2 x = −2 −2 x= 2 x = −1

Chapter 1 Basic Algebraic Operations

36.

7− x =1 7 − x = 1 or 7 − x = −1 −x = 1− 7 − x = −6 x=6

37.

− x = −1 − 7 or

− x = −8 x=8

5.8 − 0.3( x − 6.0) = 0.5 x 0.5 x = 5.8 − 0.3 x + 1.8 0.5 x + 0.3x = 7.6 0.8 x = 7.6 7.6 0.8 x = 9.5

38.

1.9t = 0.5(4.0 − t ) − 0.8 1.9t = 2.0 − 0.5t − 0.8 1.9t + 0.5t = 1.2 2.4t = 1.2 1.2 2.4 t = 0.50 t=

39.

−0.24(C − 0.50) = 0.63 −0.24C + 0.12 = 0.63 −0.24C = 0.63 − 0.12 −0.24C = 0.51 0.51 −0.24 C = −2.125 C=

C = −2.1

40.

27.5(5.17 − 1.44 x ) = 73.4 142.175 − 39.6 x = 73.4 −39.6 x = 73.4 − 142.175 −39.6 x = −68.775 −68.775 −39.6 x = 1.736742424 x = 1.74 x=

41.

17 x = 2.0 6.0 17 6.0 x = 5.6666666... x = 5.7 x = 2.0

Section 1.10 Solving Equations

42.

3.0 R = 7.0 42 R = 42

3.0 7.0

R = 18

43.

44.

165 13V = 223 15 15 165 15 13V = 13 223 13 15 2475 V = 2899 V = 0.85374267 V = 0.85 276 x 1360 = 17.0 46.4 1360 276 x = 17 46.4 498.2758621 x= 276 x = 1.805347326 x = 1.81

45.

(a)

2x + 3 = 3 + 2x 2x + 3 = 2x + 3 Is an identity, since it is true for all values of x.

(b)

2x − 3 = 3 − 2x 4x = 6 6 3 = 4 2 Is conditional as x has one answer only. x=

46.

There are no values of a that result in a conditional equation. If a = 0 , then the identity 2 x = 2 x results. If a ≠ 0 , then a contradiction results.

47.

x − 7 = 3x − (6 x − 8) 0 = 3x − 6 x + 8 − x + 7 0 = −4 x + 15 x = 3.75

Chapter 1 Basic Algebraic Operations

48.

0.0595 − 0.525i − 8.85(i + 0.0316) = 0 0.595 − 0.525i − 8.85i − 0.27966 = 0 −9.375i + 0.31534 = 0 i = 0.033636266 i = 0.0336

49.

0.03x + 0.06(2000 − x ) = 96 0.03x + 120 − 0.06 x = 96 −0.03x = 96 − 120 −0.03x = −24 −24 −0.03 x = $800 x=

50.

15(5.5 + v ) = 24(5.5 − v ) 82.5 + 15v = 132 − 24v 15v + 24v = 132 − 82.5 39v = 49.5 49.5 39 v = 1.269230769 km/h v=

v = 1.3 km/h

51.

(T − 76) 40 40(1.1) = T − 76 1.1 =

44 = T − 76 T = 44 + 76 T = 120 °C

52.

1.12V − 0.67(10.5 − V ) = 0 1.12V − 7.035 + 0.67V = 0 1.79V − 7.035 = 0 1.79V = 7.035 7.035 1.79 V = 3.930167598 V V =

V = 3.9 V

Section 1.11 Formulas and Literal Equations 53.

0.14n + 0.06(2000 − n ) = 0.09(2000) 0.14n + 120 − 0.06n = 180 0.14n − 0.06n = 180 − 120 0.08n = 60 60 n= 0.08 n = 750 L 210(3x ) = 55.3x + 38.5(8.25 − 3x )

54.

630 x = 55.3x + 317.625 − 115.5 x 630 x − 55.3 x + 115.5 x = 317.625 690.2 x = 317.625 317.625 690.2 x = 0.4601927 m x = 0.460 m x=

55.

56.

x 30 kW ⋅ h = 350 mi 107 mi 30 kW ⋅ h x = 350 mi × 107 mi x = 98 kW ⋅ h 20 min x = 250 cal 400 cal x = 400 cal

20 min 250 cal

x = 32 min

1.11

Formulas and Literal Equations

v = v0 + at v − v0 = at a=

v − v0 t

L( wL + 2 P ) 8 8W = L( wL + 2 P ) W =

8W = wL2 + 2 LP

wL2 = 8W − 2 LP 8W − 2 LP w= L2

Chapter 1 Basic Algebraic Operations

V = V0 [1 + b(T − T0 )] V = V0 [1 + bT − bT0 ] V = V0 + bTV0 − bT0V0 bT0V0 = V0 + bTV0 − V T0 =

V0 + bTV0 − V bV0

V = V0 + V0 βT V − V0 = V0 βT

β=

V − V0 V0 T

E = IR E R= I

pV = nRT pV T= nR

rL = g 2 − g1 g1 + rL = g 2 g1 = g 2 − rL

W = Sd T − Q Q + W = Sd T Q = Sd T − W

10.

11.

nTWL 12 12 B = nTWL 12 B n= TWL B=

P = 2π Tf P T= 2π f p = pa + dgh p − pa = dgh h=

p − pa dg

Section 1.11 Formulas and Literal Equations 12.

2Q = 2 I + A + S 2 I = 2Q − A − S I=

13.

mv 2 r rFc = mv 2 Fc =

14.

2Q − A − S 2

mv 2 Fc

π D2

Pπ D 2 = 4 F F=

15.

16.

A + 0.05d 5T A ST − 0.05d = 5T A = 5T ( ST − 0.05d ) ST =

eL 2m eL = −2mu u=−

2mu e

L=−

17.

Pπ D 2 4

ct 2 = 0.3t − ac ac + ct 2 = 0.3t ac = 0.3t − ct 2 a=

18.

− ct 2 + 0.3t c

2 p + dv 2 = 2d (C − W ) 2 p + dv 2 = 2Cd − 2dW 2Cd = dv 2 + 2 p + 2dW C=

19.

dv 2 + 2dW + 2 p 2d

c+d v c + d = Tv T=

d = Tv − c

20.

Chapter 1 Basic Algebraic Operations

μ0 I 2π R μ I BR = 0 2π μ I R= 0 2π B B=

K1 m1 + m2 = K2 m1

21.

K 2 ( m1 + m2 ) = K1 m1 K 2 m1 + K 2 m2 = K1 m1 K 2 m2 = K1 m1 − K 2 m1 m2 =

22.

K1 m1 − K 2 m1 K2

F d−F f (d − F ) = F fd − fF = F f =

fd = F + fF F + fF d= f

23.

2mg M + 2m a ( M + 2m ) = 2 gm aM + 2am = 2 gm aM = 2 gm − 2am a=

M =

24.

25.

2 gm − 2am a

V (m + M ) m mv = mV + MV MV = mv − mV mv − mV M = V v=

C02 = C12 (1 + 2V ) C02 = C12 + 2C12V

2C12V = C02 − C12 V =

C02 − C12 2C12

Section 1.11 Formulas and Literal Equations

26.

A1 = A( M + 1) A1 = AM + A AM = A1 − A A1 − A A

M =

27.

N = r( A − s) N = Ar − rs rs + N = Ar rs = Ar − N s=

28.

Ar − N r

T = 3(T2 − T 1 ) T = 3T2 − 3T 1

3T1 + T = 3T2 3T1 = 3T2 − T T1 =

29.

3T2 − T 3

h 100 100T2 = 100T1 − h T2 = T1 −

h + 100T2 = 100T1 h = 100T1 − 100T2

30.

p2 = p1 + rp1 (1 − p1 ) p2 − p1 = rp1 (1 − p1 ) r=

31.

p2 − p1 p1 (1 − p1 )

Q1 = P (Q2 − Q1 ) Q1 = PQ2 − PQ1 PQ2 = Q1 + PQ1 Q2 =

32.

Q1 + PQ1 P

p − pa = dg ( y2 − y1 ) p − pa dg p − pa − y1 = − y2 dg p − pa y1 = y2 − dg

y2 − y1 =

33.

Chapter 1 Basic Algebraic Operations

N = N1T − N 2 (1 − T ) N1T = N + N 2 (1 − T ) N1 =

34.

N + N 2 − N 2T T

ta = tc + (1 − h )tm ta = tc + tm − htm ta + htm = tc + tm htm = tc + tm − ta t +t −t h= c m a tm

35.

L = π ( r1 + r2 ) + 2 x1 + 2 x2 L = π r1 + π r2 + 2 x1 + 2 x2 π r1 = L − π r2 − 2 x1 − 2 x2 L − π r2 − 2 x1 − 2 x2 r1 = π

36.

VR2 + VR1 (1 + μ ) R1 R2

IR1 R2 = VR2 + VR1 + VR1 μ VR1 μ = IR1 R2 − VR2 + VR1

μ=

37.

IR1 R2 − VR2 + VR1 VR1

V1 (V2 − V1 ) gJ

gJP = V1V2 − V12 V1V2 = V12 + gJP V2 =

38.

V12 + gJP V1

W = T ( S1 − S2 ) − Q W + Q = TS1 − TS2 TS 2 = TS1 − W − Q S2 =

39.

TS1 − W − Q T C=

2eAk1k 2 d ( k1 + k2 )

Cd ( k1 + k2 ) = 2eAk1k2 e=

Cd ( k1 + k2 ) 2 Ak1k 2

Section 1.11 Formulas and Literal Equations

40.

3LPx 2 − Px 3 6 EI 6dEI = 3LPx 2 − Px 3 d=

3LPx 2 = 6dEI + Px 3 L=

41.

6dEI + Px 3 3Px 2

n N Cn V =C− N Cn V+ =C N Cn = C −V N Cn = CN − NV V = C 1−

42.

CN − NV C

p AI = P B + AI p( B + AI ) = AIP pB + AIp = AIP pB = AIP − AIp B=

43.

AIP − AIp p

p (C − n ) + n = A pC − pn + n = A

( − p + 1)n = A − pC (1 − p )n = A − pC A − pC n= 1− p 13.0 L − 0.25(15.0 L) n= 1 − 0.25 13.0 L − 3.75 L T1 = 0.75 9.25 L T1 = 0.75 T1 = 12.333333 L T1 = 12 L

Chapter 1 Basic Algebraic Operations

44.

Pt = Pc (1 + 0.500m 2 )

2 Pt Pc = 1 + 0.500m 2 680 W Pc = 1 + 0.500(0.925) 2 680 W Pc = 1 + 0.500(0.855625) 680 W Pc = 1 + 0.4278125 680 W Pc = 1.4278125 Pc = 476.253009 W Pc = 476 W 9 C + 32 5 9 90.2 = C + 32 5 F=

45.

5 (90.2 − 32 ) = C 9 5 C = × 58.2 9 C = 32.3 C

46.

1 L( B + b) 2 2V = BL + bL bL = 2V − BL 2V − BL b= L 2(38.6 ft 3 ) − (2.63 ft 2 )(16.1 ft) b= 16.1 ft 3 77.2 ft − 42.343 ft 3 b= 16.1 ft 34.857 ft 3 b= 16.1 ft b = 2.16503106 ft 2 V =

b = 2.16 ft 2

Section 1.11 Formulas and Literal Equations

47.

V1 =

VR1 R1 + R2

V1 ( R1 + R2 ) = VR1 R1 + R2 = R2 =

VR1 V1

VR1 − R1 V1

(12.0 V)(3.56 Ω) − (3.56 Ω) 6.30 V R2 = 6.780952381 Ω − 3.56 Ω R2 =

R2 = 3.220952381 Ω R2 = 3.22 Ω

η=

48.

1 q + p(1 − q)

η [ q + p(1 − q) ] = 1 ηq + η p(1 − q) = 1 η p(1 − q) = 1 − ηq 1 − ηq p= η(1 − q)

1 − (0.66)(0.83) 0.66(1 − 0.83) 1 − 0.5478 p= 0.66(0.17) 0.4522 p= 0.1122 p = 4.03030303 p = 4 processors p=

d = v2 t2 + v1t1

49.

d = v2 (4 h) + v1 (t + 2 h) tv1 + v1 (2 h) = d − v2 (4 h) tv1 = d − v2 (4 h) − v1 (2 h) t=

50.

d − v2 (4 h) − v1 (2 h) v1

C = x + 15 y 15 y = C − x y=

C−x 15

Chapter 1 Basic Algebraic Operations

1.12 1.

Applied Word Problems

Let x = the number of 25 W lights. Let 31−x = the number of 40 W lights. 25 x + 40(31 − x) = 1000 25 x + 1240 − 40 x = 1000 −15 x = 1000 − 1240 −15 x = −240 x = 16 There are 16 of the 25 W lights and(31– 16)=15 of the 40 W lights. Check: 25 ⋅16 + 40(31 − 16) = 1000 400 + 40(15) = 1000 400 + 600 = 1000 1000 = 1000

Letx = the number of slides with 5 mg. Let x−3 = the number of slides with 6 mg. (5 mg)x = (6 mg)( x − 3) (5 mg)x = (6 mg)x − 18 mg − x = −18 x = 18 slides There are 18 slides with 5 mg and (18−3)=15 slides with 6 mg. Check: 5 mg(18) = 6 mg(15) 90 mg = 90 mg

Let t = the time for the shuttle to reach the satellite. (29 500 km/h)t = 6000 km + (27 100 km/h)t (2400 km/h)t = 6000 km 6000 km t= 2400 km/h t = 2.500 h It will take the shuttle 2.500 h to reach the satellite. Check: (29500 km/h)(2.500 h) = 6000 km + (27100 km/h)(2.500 h) 73750 km = 6000 km + 67750 km 73750 km = 73750 km

Let x = the number of litres of 50% methanol blend that must be added. 0.0600(7250 L) + 0.500( x) = 0.100(7250 L + x) 435 L + 0.500( x) = 725 L + 0.100 x 0.400( x) = 290 L 290 L 0.400 x = 725 L 725 L of the 50% methanol blend must be added. x=

Section 1.12 Applied Word Problems

Check: 0.0600(7250 L) + 0.500(725 L) = 0.100(7250 L + 725 L) 435 L + 362.5 L = 0.1(7975 L) 797.5 L = 797.5 L 5.

Let x = the cost of the car 6 years ago. Let x + $5000 = the cost of the car model today. x + ( x + $5000) = $49 000 2 x = $44 000 $44 000 2 x = $22 000 The cost of the car 6 years ago was $22 000, and the cost of the today’s model is ($22 000 + 5000) = $27 000. Check: $22 000 + ($22 000 + $5000) = $49 000 $22 000 + $27 000 = $49 000 $49 000 = $49 000 x=

Let x = the flow from the first stream in m3/s. Let x – 1700 ft3/s = the flow from the second streamin m3/s. 1.98 × 107 ft 3 x + ( x − 1700 ft 3 /s) = 3600 s 2 x − 1700 ft 3 /s = 5500 ft 3 /s 2 x = 7200 ft 3 /s 7200 ft 3 /s 2 x = 3600 ft 3 /s x=

The first stream flows 3600 ft 3 /s and the second stream flows 3600 ft3/s – 1700 ft3/s = 1900 ft3/s. Check: 1.98 × 107 ft 3 3600 ft 3 /s + (3600 ft 3 /s − 1700 ft 3 /s) = 3600 s 7200 ft 3 /s − 1700 ft 3 /s = 5500 ft 3 /s 5500 ft 3 /s = 5500 ft 3 /s

Let x = the number of cars recycled the first year. Let x+500000 = the number of cars recycled the second year. x + ( x + 500000 cars) = 6900000 cars 2 x + 500000 cars = 6900000 cars 2 x = 6400000 cars 6400000 cars x= 2 x = 3200000 cars The first year,3.2 × 106 cars were recycled, and the second year (3200000 + 500000)= 3.7×106 cars were recycled. Check: 3200000 cars + (3200000 cars + 500000 cars) = 6900000 cars 3200000 cars + 3700000 cars = 6900000 cars 6900000 cars = 6900000 cars

Chapter 1 Basic Algebraic Operations

Let x = the number of hits to the website on the first day. Let 1/2 x = the number of hits on the second day. x + 1/ 2 x = 495000 hits 3 / 2 x = 495000 hits 495000 hits x= 3/ 2 x = 330000 hits The first day there were 330000 hits, the second day there were 1/2(33000 hits= 165000 hits. Check: 330000 hits + 1/ 2(330000 hits) = 495000 hits 330000 hits + 165000 hits = 495000 hits 495000 hits = 495000 hits

Let x = the number acres of land leased for $200 per acre. Let 140 – x = the number of acres of land leased for $300 per acre. $200 / acre x + $300 / acre(140 acre − x ) = $37 000 −$100 / acre (x) = −$5 000 −$5000 x= −$100 / acre x = 50 acres There are 50 acres leased at $200 per acre and (140 acres – 50 acres) = 90 hectares leased for $300 per hectare. Check: $200 / acre (50 acres) + $300 / acre(140 acres − 50 acres) = $37 000 $10 000 + $27 000 = $37 000 $37 000 = $37 000

10.

Let x = the first dose in mg. Let x + 660 mg = the second dose in mg. x + x + 660 mg = 2000 mg 2 x = 1340 mg 1340 mg x= 2 x = 670 mg The first dose is 670 mg, and the second dose is (670 mg + 660 mg) = 1130 mg. Check: 670 mg + 670 mg + 660 mg = 2000 mg 670 mg + 1330 mg = 2000 mg 2000 mg = 2000 mg

11.

Let x = the amount of pollutant after modification in ppm/h. (5 h)x = (3 h)150 ppm/h 450 ppm x= 5h x = 90 ppm/h The amount of pollutant after modification is 90 ppm/h. The device reduced emissions by (150 ppm/h – 90 ppm/h) = 60 ppm/h. Check: (5 h)90 ppm/h = (3 h)150 ppm/h 450 ppm = 450 ppm

Section 1.12 Applied Word Problems 12.

Let x – 13 = the number of teeth that the first meshed spur has. Let x = the number of teeth that the second meshed spur has. Let x + 15 = the number of teeth that the third meshed spur has. x − 13 teeth + x + x + 15 teeth = 107 teeth 3x + 2 = 107 teeth 3x = 105 teeth 105 teeth x= 3 x = 35 teeth The first spur has (35 – 13)= 22 teeth, the second spur has 35 teeth, and the third spur has (35 + 15) = 50 teeth. Check: 35 teeth − 13 teeth + 35 teeth + 35 teeth + 15 teeth = 107 teeth 107 teeth = 107 teeth

13.

Let x = amount paid per month for first six months. Let x + 10 = amount paid per month for final four months. (6 mo)x + (4 mo)( x + $10 / mo) = $890 (10 mo)x + $40 = $890 (10 mo)x = $850 $850 x= 10 mo x = $85 / mo The bill was $85/mo for the first six months and $95/mo for the next four months. Check: (6 mo)$85/mo + (4 mo)($85 / mo + $10 / mo) = $890 $510 + (4 mo)($95 / mo) = $890 $510 + $380 = $890 $890 = $890

14.

Let x = amount paid per month for first year. Let x + 30 = amount paid per month for next two years. Let (x + 30)+20= x + 50= amount paid per month for final two years. (12 mo) x + (24 mo)( x + $30 / mo) + (24 mo)( x + $50 / mo) = $7320 (12 mo) x + (24 mo) x + $720 + (24 mo) x + $1200 = $7320 (60 mo) x + $1920 = $7320 (60 mo) x = $5400 $5400 x= 60 mo x = $90 / mo For the first year, the bill was $90/mo, during years 2 and 3, the bill was $120/mo, and during years 4 and 5, the bill was $140/mo. Check: (12 mo)($90 / mo) + (24 mo)($90 / mo + $30 / mo) + (24 mo)($90 / mo + $50 / mo) = $7320 $1080 + (24 mo)($120 / mo) + (24 mo)($140 / mo) = $7320 $1080 + $2880 + $3360 = $7320 $7320 = $7320

Chapter 1 Basic Algebraic Operations

15.

Let x = the first current in μ A . Let 2x = the second current in μ A . Let x + 9.2 μ A = the third current in μ A x + 2 x + x + 9.2 μ A = 0 μ A 4 x = −9.2 μ A −9.2 μ A x= 4 x = −2.3 μ A The first current is −2.3 μ A ,the second current is 2(−2.3 μ A ) = −4.6 μ A , and the third current is (−2.3 μ A +9.2 μ A ) = 6.9 μ A . Check: −2.3 μ A + 2(−2.3 μ A) + (-2.3) μ A + 9.2 μ A = 0 μ A −2.3 μ A − 4.6 μ A − 2.3μ A + 9.2 μ A = 0 μ A 0 μA = 0 μA

16.

Let x = the number of trucks in the first fleet. Let x + 5 = the number of trucks in the second fleet. (8 h)x + (6 h)( x + 5) = 198 h (8 h)x + (6 h)x + 30 h = 198 h (14 h)x = 168 h 168 h x= 14 h x = 12 trucks There are 12 trucks in the first fleet and (12 trucks + 5 trucks) = 17 trucks in the second fleet. Check: (8 h)(12) + (6 h)(12 + 5) = 198 h 96 h + (6 h)(17) = 198 h 96 h + 102 h = 198 h 198 h = 198 h

17.

Let x = the length of the first pipeline in km. Let x + 2.6 km = the length of the 3 other pipelines. x + 3( x + 2.6 km) = 35.4 km x + 3x + 7.8 km = 35.4 km 4 x = 27.6 km 27.6 km x= 4 x = 6.9 km The first pipeline is 6.9 km long, and the other three pipelines are each (6.9 km + 2.6 km) = 9.5 km long. Check: 6.9 km + 3(6.9 km + 2.6 km) = 35.4 km 6.9 km + 3(9.5 km) = 35.4 km 6.9 km + 28.5 km = 35.4 km 35.4 km = 35.4 km

Section 1.12 Applied Word Problems 18.

Let x = the power of the first generator in MW. Let 750 MW−x = the power of the second generator in MW. 0.65 x + 0.75(750 MW − x) = 530 MW 0.65 x + 562.5 MW − 0.75 x = 530 MW −0.1x = −32.5 MW −32.5 MW x= −0.1 x = 325 MW The first generator produces 325 MW of power, and the second generator produces (750 MW – 325 MW) = 425 MW of power. Check: 0.65(325 MW) + 0.75(750 MW − (325 MW)) = 530 MW 211.25 MW + 0.75(425 MW) = 530 MW 211.25 MW + 318.75 MW = 530 MW 530 MW = 530 MW

19.

Let x = the number of deluxe systems. Let 2x = the number economy systems. Let x+75= the number of econo-plus systems. $140 x + $40(2 x) + $80( x + 75) = $42000 $140 x + $80 x + $80 x + $6000 = $42000 $300 x = $36000 x = 120 systems There are 120 deluxe systems, 240 economy systems, and 195 econo-plus systems sold. Check: $140(120) + $40(240) + $80(195) = $42000 $16800 + $9600 + $15600 = $42000 $42000 = $42000

20.

The amount of lottery winnings after taxes is $20 000 ×(1−0.25) = $15 000. Let x = the amount of money invested at a 40% gain. Let $15 000− x = the amount of money invested at a 10% loss. 0.40 x − 0.10($15 000 − x) = $2000 0.40 x − $1500 + 0.10 x = $2000 0.50 x = $3500 $3500 x= 0.50 x = $7000 The 40% gain investment had $7000 invested, and the 10% loss investment had ($15 000− $7000) = $8000 invested. Check: 0.40($7000) − 0.10($15 000 − $7000) = $2000 $2800 − $1500 + 0.10($7000) = $2000 $2800 − $1500 + $700 = $2000 $2000 = $2000

Chapter 1 Basic Algebraic Operations

21.

Let x = the amount of time in seconds between when the start of the trains pass each other to when the end of the trains pass each other. The total distance the ends must travel in this time is 960 feet. We first convert mi/hr into ft/sec. 5280 ft 22 ft 22 1 mi/hr = = = ft/s 3600 s 15 s 15 Therefore, train A travels at 60(22/15)=88 ft/s and train B travels at 40(22/15)=176/3 ft/s. (88 ft/s)x + (176 / 3 ft/s) x = 960 ft (440 / 3 ft/s)x = 960 ft 960 ft x= 440 / 3 ft/s 72 s x= 11 The trains completely pass each other in about 6.55 seconds. Check: 72 72 (88 ft/s) s + (176 / 3 ft/s) s = 960 ft 11 11 576 ft + 384 ft = 960 ft 960 ft = 960 ft

22.

Let x = the mortgage payment and x/0.23=the monthly income. x / 0.23 − x = $3850  1  − 1 = $3850 x  0.23  0.23   1 − x  = $3850 0.23 0.23    0.77  x  = $3850  0.23 

x = $3850 ⋅

0.23 0.77

x = $1150 The mortgage payment is $1150 and the monthly income is $5000. Check: $1150 / 0.23 − $1150 = $3850 $5000 − $1150 = $3850 $3850 = $3850

23.

Let x = the amount of time the skier spends on the ski lift in minutes. Let 24 minutes −x = the amount of time the skier spends skiing down the hill in minutes. (50 m/min)x = (150 m/min)(24 min − x) (50 m/min)x = 3600 m − (150 m/min)x (200 m/min)x = 3600 m 3600 m x= 200 m/min x = 18 min The length of the slope is 18 minutes × 50 m/minute = 900m.

Section 1.12 Applied Word Problems

Check: (50 m/min)18 min = (150 m/min)(24 min − 18 min) 900 m = 3600 m − (150 m/min)(18 min) 900 m = 3600 m − 2700 m 900 m = 900 m 24.

Let x = the speed of sound. Let x – 120 mi/h = speed travelled for 1 h. Let x + 410 mi/h = the speed travelled for 3 h. 1 h( x − 120 mi/h) + 3 h( x + 410 mi/h) = 3990 mi (1 h)x − (1 h)(120 mi/h) + (3 h)x + (3 h)(410 mi/h)= 3990 mi (1 h)x − 120 mi + (3 h)x + 1230 mi = 3990 mi (4 h)x = 2880 mi 2880 mi x= 4h x = 720 mi/h The speed of sound is 720 mi/h. Check: 1 h(720 mi/h − 120 mi/h) + 3 h(720 mi/h + 410 mi/h) = 3990 mi (1 h)(600 mi/h) + (3 h)(1130 mi/h)= 3990 mi 600 mi + 3390 mi = 3990 mi 3990 mi = 3990 mi

25.

Let x = the speed the train leaving England in km/h. Let x + 8 km/h = speed of the train leaving France in km/h. The distance travelled by each train is speed ×time.  17 min   17 min  x  + ( x + 8 km/h)   = 50 km  60 min/ h   60 min/ h  (0.28333 h) x + ( x + 8 km/h)(0.28333 h)= 50 km (0.28333 h) x + (0.28333 h) x + 2.26667 km = 50 km (0.56666 h) x = 47.73333 km 47.73333 km x= 0.56666 h x = 84.23529421 km/h x = 84.2 km/h The train leaving England was travelling at 84.2 km/h, and the train leaving France was travelling at (84.2 km/h + 8 km/h) = 92.2 km/h. Check:  17 min   17 min  84.23529421 km/h   + (84.23529421 km/h + 8 km/h)   = 50 km  60 min/ h   60 min/ h   17 min  23.86666 km + (92.23529421 km/h)   = 50 km  60 min/ h  23.86666 km + 26.13333 km = 50 km 50 km = 50 km

Chapter 1 Basic Algebraic Operations

26.

Let x = time left until the appointment. Let x– 10.0 min = time taken to get to the appointment travelling at 60.0 mi/h. Let x– 5.0 min = time taken to get to the appointment travelling at 45.0 mi/h. The distance travelled by the executive in each scenario is the same. Distance = speed × time 10.0 min  5.0 min    60.0 mi/h  x −  = 45 mi/h  x −  60 min/h 60 min/h      10.0 min   5.0 min  (60.0 mi/h)x − 60 mi/h   = (45 mi/h)x − 45 mi/h    60 min/h   60 min/h  (60.0 mi/h)x − 10 mi = (45.0 mi/h)x − 3.75 mi (15.0 mi/h)x = 6.25 mi 6.25 mi x= 15.0 mi/h x = 0.416666667 h x = 0.416666667 h x 60 min/h x = 25 min There is 25 minutes left until the executive’s appointment. Check: 10.0 min  5.0 min    60.0 mi/h  0.41667 h −  = 45 mi/h  0.41667 h −  60 min/h  60 min/h    60.0 mi/h(0.25 h) = 45 mi/h(0.33333 h) 15 mi = 15 mi

27.

Let x – 30.0 s = time since the first car started moving in the race in seconds. Let x= time since the second car started the race in seconds. The distance travelled by each car will be the same at the point where the first car overtakes the second car. Distance = speed × time. 260.0 ft/s( x − 30.0 s) = 240.0 ft/s( x) (260.0 ft/s)x − (260.0 ft/s)(30.0 s) = (240.0 ft/s)x (260.0 ft/s)x − 7800 ft = (240.0 ft/s)x (20.0 ft/s)x = 7800 ft 7800 ft x= 20.0 ft/s x = 390 s The first car will overtake the second car after 390 s. The first car travels 260 ft/s × (390 s – 30 s) = 93600 ft by this point. 8 laps around the track is 2.5 mi/lap. 8 laps × 5280 ft/mi = 105,600 ft, so the first car will already be in the lead at the end of the 8th lap. Check: 260.0 ft/s(390.0 s − 30.0 s) = 240.0 ft/s(390 s) 260.0 ft/s(360.0 s) = 240.0 ft/s(390 s) 93, 600 ft = 93, 600 ft

Section 1.12 Applied Word Problems 28.

Let x = the number of the first chips that is defective 0.50%. Let 6100 −x = the number of the second chips that is defective 0.80%. 0.0050( x) + 0.0080(6100 chips − x) = 38 chips (0.0050) x + 48.8 chips − (0.0080) x = 38 chips −(0.0030) x = −10.8 chips −10.8 chips x= −0.0030 x = 3600 chips There are 3600 chips that are 0.50% defective and (6100 chips – 3600 chips) = 2500 chips that are defective 0.80%. Check: 0.0050(3600 chips) + 0.0080(6100 chips − 3600 chips) = 38 chips 18 chips + 0.0080(2500 chips) = 38 chips 18 chips + 20 chips = 38 chips 38 chips = 38 chips x mi

29.

$2.90/gal

(228 – x) mi

228 mi

$2.70/gal

Assuming that the customer is located between the two gasoline distributors: Let x = the distance in km to the first gasoline distributor that costs $2.90/gal. Let 228 mi− x= the distance in km to the second gasoline distributor that costs $2.70/gal. $2.90 + $0.002( x) = $2.70 + $0.002(228 − x) $2.90 + $0.002( x) = $2.70 + $0.456 − $0.002( x) $0.004( x) = $0.256 $0.256 $0.004 x = 64 mi The customer is 64 mi away from the first gas distributor ($2.90/gal) and (228 mi – 64 mi)= 164 mi away from the second gas distributor ($2.70). Check: $2.90 + $0.002(64) = $2.70 + $0.002(228 − 64) $2.90 + $0.128 = $2.70 + $0.002(164) $3.028 = $2.70 + $0.328 $3.028 = $3.028 x=

30.

?L (100 % Gas)

75% Gas 8.0 L gas can (needs to be full of 93.75% gas/oil mixture)

A 15:1 gas/oil mixture is 15/16 gasoline = 93.75%. Let x = the amount of 100% gasoline added in L. Let 8.0 L – x = the amount of 75% gasoline mixture in L.

Chapter 1 Basic Algebraic Operations

1.00( x) + 0.75(8.0 L − x) = 0.9375(8.0 L) 1.00( x) + 6.0 L − 0.75( x) = 7.5 L 0.25( x) = 1.5 L 1.5 L 0.25 x = 6.0 L 6.0 L of 100% gasoline must be added to the 75% gas/oil mixture to make 8 L of 15:1 gasoline/ oil. Check: 1.00(6.0 L) + 0.75(8.0 L − 6.0 L) = 0.9375(8.0 L) x=

6 L + 0.75(2.0 L) = 7.5 L 6 L + 1.5 L = 7.5 L 7.5 L = 7.5 L

31.

(x) L of 25% antifreeze 100% Antifreeze

12.0 L radiator (needs to be filled with 50% mixture)

Let x = the amount in L of 25% antifreeze left in radiator Let 12.0 L – x = the amount of 100% antifreeze added in L. 0.25( x) + 1.00(12.0 L − x) = 0.5(12.0 L) 0.25( x) + 12.0 L − 1.00( x) = 6.0 L −0.75( x) = −6.0 L −6.0 L x= −0.75 x = 8.0 L There needs to be 8L of 25% antifreeze left in radiator, so (12.0 L – 8.0 L) = 4.0 L must be drained. Check: 0.25(8.0 L) + 1.00(12.0 L − 8.0 L) = 0.5(12.0 L) 2 .0L + 1.00(4.0 L) = 6.0 L 2.0 L + 4.0 L = 6.0 L 6.0 L = 6.0 L 32.

(x) lb Sand 250 lb Cement (22% Sand Mixture)

Let x = the amount of sand added. Let 250 lb + x = the amount in lb of the final 25% sand mixture.

Section 1.12 Applied Word Problems

1.00( x) + 0.22(250 lb) = 0.25(250 lb + x) 1.00( x) + 55 lb = 62.5 lb + 0.25( x) 0.75( x) = 7.5 lb 7.5 lb 0.75 x = 10 lb x=

Check: 1.00(10 lb) + 0.22(250 lb) = 0.25(250 lb + 10 lb) 10 lb + 55 lb = 62.5 lb + 2.5 lb 65 lb = 65 lb (x) km/h m

33.

70 km/h 5.0 m 20.0 m Let x = the speed the car needs to travel in km/h to pass the semi in 10 s. Speed = distance/time. 10 s is 10s/3600 s/h = 0.002777777 h. distance needed to pass truck + distance travelled by truck in 10s x= 10s 0.025 km + 70 km/h ( 0.0027777 h ) x= 0.0027777 h 0.025 km + 0.19444 km x= 0.0027777 h 2.19444 km x= 0.0027777 h x = 79 km/h The car needs to travel at a speed of 79 km/h to pass the semitrailer in 10s. Check: 0.025 km + 70 km/h ( 0.0027777 h ) 79 km/h = 0.0027777 h 0.025 km + 0.19444 km 79 km/h = 0.0027777 h 79 km/h = 79 km/h 34.

5km/s

8 km/s Seismic Station

(?) km Let x = the time the first wave takes to travel to the seismic station in s. Let x + 120 s = the time the first wave takes to travel to the seismic station in s. Distance =speed × time. The distances travelled by both waves to the seismic station are the same. 2.0 min is (2.0 min × 60 s/min) = 120 s.

Chapter 1 Basic Algebraic Operations

8.0 km/s(x) = 5.0 km/s( x + 120 s) 8.0 km/s(x) = 5.0 km/s( x ) + (5 km/s)(120 s) 3.0 km/s(x) = 600 km 600 km 3.0 km/s x = 200 s The distance to the seismic station is (200 s × 8.0 km/s) = 1600 km. Check: 8.0 km/s(200 s) = 5.0 km/s(200 s + 120 s) 1600 km = 5.0 km/s(320 s) x=

1600 km = 1600 km

Review Exercises 1.

False, because 0 = 0 which is not a positive value.

True. The order of operations dictates performing the division first, then the subtraction.

False. The reported answer should have only two significant digits.

False. Had the problem been given as (2a )3 = 8a 3 , then it would be true.

True.

False. The left-hand side, − −4 , is not a real number, in fact.

False. The left-hand side simplifies to 4 x − (2 x + 3) = 4 x − 2 x − 3 = 2 x − 3 .

True.

False. The left-hand side simplifies to

10.

True.

11.

False. Solving for c yields

6x + 2 6x 2 = + = 3x + 1 . 2 2 2

a − bc = d −bc = d − a d −a c= −b d −a c=− b

12.

False. It is likely that one should set up a phrase such as ‘let x be the number of gears of the first type…’

13.

(−2) + (−5) − 3 = −2 − 5 − 3 = −10