Solutions Manual for Modern Engineering Mathematics, 5th Edition (James, Burley, Clements, Dyke, Steele & Wright, 2018) | All Chapters 1–13 Covered

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, TABLE OF CONTENTS
Solutions Manual: Modern Engineering Mathematics, 5th Edition
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Authors: Glyn James, David Burley, Dick Clements, Phil Dyke, Nigel Steele, Jerry Wright



Chapter 1. Numbers, Algebra and Geometry
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Chapter 2. Functions
Chapter 3. Complex Numbers
Chapter 4. Vector Algebra
Chapter 5. Matrix Algebra
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Chapter 6. An Introduction to Discrete Mathematics
Chapter 7. Sequences, Series and Limits
Chapter 8. Differentiation and Integration
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Chapter 9. Further Calculus
Chapter 10. Introduction to Ordinary Differential Equations
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Chapter 11. Introduction to Laplace Transforms
Chapter 12. Introduction to Fourier Series
Chapter 13. Data Handling and Probability Theory
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, CH A P T E R 1
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Numbers, Algebra and Geometry

1.2.4 Exercises
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◼ 1
110110.1012 = 25 + 24 + 22 + 21 + 2−1 + 2−3
= 54.62510

◼ 2
16 321 = 213 + 212 + 211 + 210 + 29 + 28 + 27 + 26 + 20
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= 111111110000012
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16 321 = 3  84 + 7  83 + 7  82 + 80
= 377018

To convert from binary to octal: take the first three entries immediately to the right and include the
20 term in the binary expansion; this may be considered as a three-digit binary number; convert this
number into octal; the resulting octal number is the 80 term of the octal expansion. Now do the same
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with the next three digits of the binary expansion to get the 81 term of the octal expansion, and so on.

1012 = 58, 1002 = 48, 0112 = 38, and 12 = 18 so
[10111001011012 = 134558]

◼ 3
30.6 = 24 + 23 + 22 + 21 + 2−1 + 2−4 + 2−5 + 2−8 + 2−9 + 2−12 + 2−13
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= 11110.1001100110011 . . . 2

30.6 = 3  8 + 6  80 + 4  8−1 + 6  8−2 + 3  8−3 + 8−4 + 4  8−5
+ 6  8−6 + 3  8−7 + 8−8 + . . .
= 36.46314631 . . . 8

The rule works in this case as well: 1002 = 48, 1102 = 68, 0112 = 38 and 0012 = 18.
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◼ 4(a)
100011.0112
+ 1011.0012
101110.1002
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© Pearson Education Limited 2015

, James, Burley, Clements, Dyke, Searl and Wright, Modern Engineering Mathematics, 5th Edition,
Solutions Manual on the Web

4(b)
111.100112
 10.1112
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0.11110011
1.1110011
11.110011
+ 1111.0011
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10101.110101012

◼ 5(a)
23  2−4 = 23  24 = 1/2

5(b)
23  2−4 = 23  24 = 23+4 = 27
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5(c)
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(23)−4 = 1/(23)4 = 1/212

5(d)
31/3  35/3 = 3(1/3+5/3) = 32

5(e)
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36−1/2 = 1/(36)1/2 = 1/6

5(f)
163/4 = (161/4)3 = 23

◼ 6(a)
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(21 + ((4  3)  2))

6(b)
(17 − 6(2+3))

6(c)
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((4  23) − ((7  6)  2))

6(d)
−5

((2  3) − (6  4) + 3(2 ))

◼ 7(a)

(7 + 5 2 )3 = (7 + 5 2 ) (7 + 5 2 )2
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= (7 + 5 2 ) (99 + 70 2 )
= 7  99 + 5  70  2 + (7  70 + 5  99) 2
= 1393 + 985
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© Pearson Education Limited 2015