??
ED
OV
PR
AP
A?_
UVI
ST
, TABLE OF CONTENTS
Solutions Manual: Modern Engineering Mathematics, 5th Edition
Authors: Glyn James, David Burley, Dick Clements, Phil Dyke, Nigel Steele, Jerry Wright
ST
Chapter 1. Numbers, Algebra and Geometry
U
Chapter 2. Functions
Chapter 3. Complex Numbers
VI
Chapter 4. Vector Algebra
Chapter 5. Matrix Algebra
Chapter 6. An Introduction to Discrete Mathematics
A
Chapter 7. Sequences, Series and Limits
?_
Chapter 8. Differentiation and Integration
Chapter 9. Further Calculus
Chapter 10. Introduction to Ordinary Differential Equations
AP
Chapter 11. Introduction to Laplace Transforms
Chapter 12. Introduction to Fourier Series
Chapter 13. Data Handling and Probability Theory
PR
OV
ED
? ?
, CH A P T E R 1
Numbers, Algebra and Geometry
ST
1.2.4 Exercises
U
◼ 1
110110.1012 = 25 + 24 + 22 + 21 + 2−1 + 2−3
= 54.62510
VI
◼ 2
16 321 = 213 + 212 + 211 + 210 + 29 + 28 + 27 + 26 + 20
= 111111110000012
A
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
with the next three digits of the binary expansion to get the 81 term of the octal expansion, and so on.
AP
1012 = 58, 1002 = 48, 0112 = 38, and 12 = 18 so
[10111001011012 = 134558]
◼ 3
PR
30.6 = 24 + 23 + 22 + 21 + 2−1 + 2−4 + 2−5 + 2−8 + 2−9 + 2−12 + 2−13
= 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
OV
The rule works in this case as well: 1002 = 48, 1102 = 68, 0112 = 38 and 0012 = 18.
◼ 4(a)
100011.0112
+ 1011.0012
ED
101110.1002
? ?
1
© 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
ST
0.11110011
1.1110011
11.110011
+ 1111.0011
U
10101.110101012
◼ 5(a)
23 2−4 = 23 24 = 1/2
VI
5(b)
23 2−4 = 23 24 = 23+4 = 27
A
5(c)
(23)−4 = 1/(23)4 = 1/212
?_
5(d)
31/3 35/3 = 3(1/3+5/3) = 32
5(e)
AP
36−1/2 = 1/(36)1/2 = 1/6
5(f)
163/4 = (161/4)3 = 23
6(a)
PR
◼
(21 + ((4 3) 2))
6(b)
(17 − 6(2+3))
OV
6(c)
((4 23) − ((7 6) 2))
6(d)
−5
((2 3) − (6 4) + 3(2 ))
ED
◼ 7(a)
(7 + 5 2 )3 = (7 + 5 2 ) (7 + 5 2 )2
= (7 + 5 2 ) (99 + 70 2 )
= 7 99 + 5 70 2 + (7 70 + 5 99) 2
?
= 1393 + 985
?
2
© Pearson Education Limited 2015
ED
OV
PR
AP
A?_
UVI
ST
, TABLE OF CONTENTS
Solutions Manual: Modern Engineering Mathematics, 5th Edition
Authors: Glyn James, David Burley, Dick Clements, Phil Dyke, Nigel Steele, Jerry Wright
ST
Chapter 1. Numbers, Algebra and Geometry
U
Chapter 2. Functions
Chapter 3. Complex Numbers
VI
Chapter 4. Vector Algebra
Chapter 5. Matrix Algebra
Chapter 6. An Introduction to Discrete Mathematics
A
Chapter 7. Sequences, Series and Limits
?_
Chapter 8. Differentiation and Integration
Chapter 9. Further Calculus
Chapter 10. Introduction to Ordinary Differential Equations
AP
Chapter 11. Introduction to Laplace Transforms
Chapter 12. Introduction to Fourier Series
Chapter 13. Data Handling and Probability Theory
PR
OV
ED
? ?
, CH A P T E R 1
Numbers, Algebra and Geometry
ST
1.2.4 Exercises
U
◼ 1
110110.1012 = 25 + 24 + 22 + 21 + 2−1 + 2−3
= 54.62510
VI
◼ 2
16 321 = 213 + 212 + 211 + 210 + 29 + 28 + 27 + 26 + 20
= 111111110000012
A
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
with the next three digits of the binary expansion to get the 81 term of the octal expansion, and so on.
AP
1012 = 58, 1002 = 48, 0112 = 38, and 12 = 18 so
[10111001011012 = 134558]
◼ 3
PR
30.6 = 24 + 23 + 22 + 21 + 2−1 + 2−4 + 2−5 + 2−8 + 2−9 + 2−12 + 2−13
= 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
OV
The rule works in this case as well: 1002 = 48, 1102 = 68, 0112 = 38 and 0012 = 18.
◼ 4(a)
100011.0112
+ 1011.0012
ED
101110.1002
? ?
1
© 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
ST
0.11110011
1.1110011
11.110011
+ 1111.0011
U
10101.110101012
◼ 5(a)
23 2−4 = 23 24 = 1/2
VI
5(b)
23 2−4 = 23 24 = 23+4 = 27
A
5(c)
(23)−4 = 1/(23)4 = 1/212
?_
5(d)
31/3 35/3 = 3(1/3+5/3) = 32
5(e)
AP
36−1/2 = 1/(36)1/2 = 1/6
5(f)
163/4 = (161/4)3 = 23
6(a)
PR
◼
(21 + ((4 3) 2))
6(b)
(17 − 6(2+3))
OV
6(c)
((4 23) − ((7 6) 2))
6(d)
−5
((2 3) − (6 4) + 3(2 ))
ED
◼ 7(a)
(7 + 5 2 )3 = (7 + 5 2 ) (7 + 5 2 )2
= (7 + 5 2 ) (99 + 70 2 )
= 7 99 + 5 70 2 + (7 70 + 5 99) 2
?
= 1393 + 985
?
2
© Pearson Education Limited 2015
