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---

Chapter 9

Basic algebra
9
...

For example, if the length of a football pitch is L and its
width is b, then the formula for the area A is given by
A= L ×b
This is an algebraic equation
...

The total resistance, RT , of resistors R1 , R2 and R3
connected in series is given by
RT = R1 + R2 + R3
This is an algebraic equation
...
3 k, R2 = 2
...
5 k, then
RT = 6
...
4 + 8
...
2 k
The temperature in Fahrenheit, F, is given by
9
F = C + 32
5
where C is the temperature in Celsius
...

9
If C = 100◦C, then F = × 100 + 32
5
= 180 + 32 = 212◦F
...


9
...

If, say, a, b, c and d represent any four numbers then in
algebra:
DOI: 10
...
00009-0

(a) a + a + a + a = 4a
...

(b) 5b means 5 × b
...

(c) 2a + 3b + a − 2b = 2a + a + 3b − 2b = 3a + b
Only similar terms can be combined in algebra
...

In addition, with terms separated by + and − signs,
the order in which they are written does not matter
...
(Note that the first term, i
...

2a, means +2a
...
(Note
that − × − = +)
(e) (a)(c)(d) means a × c × d
Brackets are often used instead of multiplication
signs
...

(f ) ab = ba
If a = 2 and b = 3 then 2 × 3 is exactly the same
as 3 × 2, i
...
6
...
For
32 = 3 × 3 = 9
...

Here are some worked examples to help get a feel for
basic operations in this introduction to algebra
...
2
...


Find the sum of 4x, 3x, −2x and −x

4x + 3x + −2x + −x = 4x + 3x − 2x − x
(Note that + ×− = −)
= 4x
Problem 2
...


Simplify 3 + x + 5x − 2 − 4x
...


Add x − 2y + 3 to 3x + 4y − 1
...


Subtract a − 2b from 4a + 3b
...


From a + b − 2c take 3a + 2b − 4c
...
From x 2 + x y − y 2 take x y − 2x 2
...
2
...


Simplify bc × abc

5x+3y + z + −3x + −4y + 6z
= 5x + 3y + z − 3x − 4y + 6z

bc × abc = a × b × b × c × c

= 5x − 3x + 3y − 4y + z + 6z

= a × b 2 × c2

= 2x − y + 7z

= ab2 c 2

Note that the order can be changed when terms are separated by + and − signs
...


Problem 6
...


Simplify 4x 2 − x

− 2y + 5x + 3y
Problem 7
...


ab × b2c × a = a × a × b × b × b × c
= a 2 × b3 × c

Simplify 3x y − 7x + 4x y + 2x

3x y − 7x + 4x y + 2x = 3x y + 4x y + 2x − 7x
= 7xy − 5x

Simplify ab × b2c × a

= a2 b3 c
Problem 8
...


Find the sum of 4a, −2a, 3a and −8a
...


Find the sum of 2a, 5b, −3c, −a, −3b and
7c
...


Simplify 2x − 3x 2 − 7y + x + 4y − 2y 2
...


Simplify 5ab − 4a + ab + a
...


Simplify 2x − 3y + 5z − x − 2y + 3z + 5x
...


5 pq 2r 3 = 5 × p × q 2 × r 3
 2  3
1
2
= 5×2×
× 2
5
2

Basic algebra
 2  3
2
5
= 5×2×
×
5
2

1 5
since 2 =
2 2

5 2 2 2 5 5 5
× × × × × ×
1 1 5 5 2 2 2
1 1 1 1 1 5 5
= × × × × × ×
1 1 1 1 1 1 1
= 5×5

Problem 13
...
Multiply 2a + 3b by a + b
Each term in the first expression is multiplied by a, then
each term in the first expression is multiplied by b and
the two results are added
...

2a + 3b
a+b
Multiplying by a gives
Multiplying by b gives

2a 2 + 3ab
2ab + 3b2

Adding gives

2a 2 + 5ab + 3b2

Thus, (2a + 3b)(a + b) = 2a2 + 5ab + 3b2
Problem 11
...
The usual layout is shown below with the
dividend and divisor both arranged in descending
powers of the symbols
...


...
e
...

(iii) The divisor is then multiplied by 2x, i
...

2x(x − 1) = 2x 2 − 2x, which is placed under the
dividend as shown
...

(iv) The process is then repeated, i
...
the first term
of the divisor, x, is divided into 3x, giving +3,
which is placed above the dividend as shown
...
The remainder, on subtraction, is zero,
which completes the process
...
Simplify 2x ÷ 8x y
2x
8x y
2x
2×x
=
8x y 8 × x × y
1×1
=
4×1× y
1
=
4y

Problem 14
...


2x ÷ 8x y means

(A check can be made on this answer by
multiplying (2x + 3) by (x − 1), which equals
2x 2 + x − 3
...
Simplify

x 3 + y3
x+y

64 Basic Engineering Mathematics
(i) (iv) (vii)

Alternatively, the answer may be expressed as

2
2
x − xy + y
3
x + y x + 0 + 0 + y3
x3 + x2 y

−x 2 y

4b 3
4a 3 − 6a 2 b + 5b3
= 2a 2 − 2ab − b2 +
2a − b
2a − b

+ y3

Now try the following Practice Exercise

−x 2 y − x y 2

Practice Exercise 36 Basic operations in
algebra (answers on page 343)

x y2 + y3
x y2 + y3

...


1
...


2
...


(i)

x into x 3 goes x 2
...


3
...


(ii)

x 2 (x + y) = x 3 + x 2 y

4
...


5
...

(iv)

x into −x y goes −x y
...

2

(v) −x y(x + y) = −x 2 y − x y 2

6
...

(vii)

x into x y 2 goes y 2
...


(viii)

y 2 (x + y) = x y 2 + y 3

(ix) Subtract
...

Thus,
x+y
The zeros shown in the dividend are not normally shown,
but are included to clarify the subtraction process and
to keep similar terms in their respective columns
...


Divide 4a 3 − 6a 2 b + 5b3
2a 2 −

by 2a − b

8
...


9
...


25x 2 y z 3
5x yz

10
...

11
...

12
...

13
...

14
...


2ab − b2


2a − b 4a 3 − 6a 2 b
4a 3 − 2a2 b

+ 5b3

−4a 2 b

+ 5b3

15
...

16
...

17
...


−4a 2 b + 2ab2
−2ab2 + 5b3
−2ab2 + b3
4b3
4a 3 − 6a 2 b + 5b3
= 2a 2 − 2ab − b2 , remain2a − b
der 4b3
...
3

Laws of indices

The laws of indices with numbers were covered in
Chapter 7; the laws of indices in algebraic terms are
as follows:
(1) am ×an = am + n

For example, a 3 × a 4 = a 3+4 = a 7

Basic algebra
(2)

am
= am−n
an

(3) (am )n = amn
√
n

m

(4) a n =
(5) a−n =

am

1
an

(6) a0 = 1

c5
= c5−2 = c3
c2
 3
For example, d 2 = d 2×3 = d 6

For example,

4

For example, x 3 =

√
3

For example, 3−2 =

x4

1
1
=
32
9

a 3 b 2 c4
Problem 20
...

Problem 17
...
Simplify ( p3 )2 (q 2 )4

a 2 × a 1 × b 3 × b 2 × c1 × c5
Using law (3) of indices gives

Using law (1) of indices gives
a 2+1 × b3+2 × c1+5 = a 3 × b5 × c6
a 2 b3c × ab2 c 5 = a 3 b5 c 6

i
...


1

3

1

1

Problem 18
...
Simplify

Problem 22
...
Using law (3) of indices
gives
(mn 2 )3
m 1×3 n 2×3
m3n6
=
=
(m 1/2 n 1/4)4
m (1/2)×4n (1/4)×4 m 2 n 1
m3 n6
= m 3−2 n 6−1 = mn5
m2 n1
Problem 23
...
Simplify

66 Basic Engineering Mathematics
Using law (3) of indices gives

p1/2 q 2r 2/3
and evaluate
p1/4 q 1/2r 1/6
when p = 16, q = 9 and r = 4, taking positive roots
only

Problem 25
...

1
...


a × a2 × a5

3
...


b4 × b7

5
...


c5 × c3 ÷ c4
(x 2 )(x)
x6
 2 −3
y
 −7 −2
c
 4
1
b3
1
 3
s3

9
...

13
...


17
...

12
...

16
...


x 3 y2 z
x 5 y z3

19
...

2
a 5 bc3

and
20
...


Simplify

when

Here are some further worked examples on the laws of
indices

x 2 y3 + x y2
xy
a +b
can be split
c

Algebraic expressions of the form
a b
into +
...

Problem 27
...

Dividing each term by x y gives

x2 y
evaluate

1

= (2)(33 )(2) = 108

Practice Exercise 37 Laws of indices
(answers on page 343)

8
...


When p = 16, q = 9 and r = 4,

x y2 − x y

Problem 28
...
Dividing
each term by a 1/2b gives
a 2b
a3/2
a 1/2b
=
=
2
1/2
3
2
1/2
3
ab − a b
a b
ab
a1/2 b−b2
−
a 1/2b
a 1/2b
a2b

√ √
√ √
3
Problem 29
...
Hence,
7

7

a2 b6 c

11
2

=


 
6
a7 b7 c11

1
When a = , b = 64 and c = 1,
4

 7 √
√

 

1
6
6
7
7
11
a b c =
647
111
4
 7
1
(2)7 (1) = 1
=
2
√ 
(x 2 y 1/2 )( x 3 y 2 )
Problem 30
...

Now try the following Practice Exercise
Practice Exercise 38 Laws of indices
(answers on page 343)



1
...

In problems 2 to 5, simplify the given expressions
...


a 2b + a 3 b
a 2b2

3
...

5
...


(abc)2
(a 2 b−1c−3 )3
p3 q 2
pq 2 − p2 q
√  √
√  
3
( x y 3 z 2 )( x y 3 z 3 )

7
...


8
...

The n’th term is: a + (n − 1)d
n
Sum of n terms, Sn = [2a + (n − 1)d]
2

Geometric progression:
If a = first term and r = common ratio, then the geometric progression is: a, ar, ar2 ,
...

4
...

10
...

16
...


Exercise 5 (page 11)

19 kg
2
...
479 mm
−66
5
...
−225
−2136
8
...
£10 7701
−4
11
...
5914
189 g
14
...
$15 333
89
...

3
...

5
...

9
...

9
...

17
...

(a) 8613 kg (b) 584 kg
(a) 351 mm (b) 924 mm
(a) 10 304 (b) −4433
6
...

(a) 8067 (b) 3347
10
...

4
...

8
...


(a) 48 m (b) 89 m
(a) 1648 (b) 1060
18 kg

1
...
14
7
...
88
8
...
1016/B978-1-85617-697-2
...


3 4 1 3 5
, , , ,
7 9 2 5 8
9
10
...
1
15
16
18
...


6
...

21
...
2
3
...
11
8
...

13
...

18
...

11
...

19
...

8
...

16
...


4
...
2

5
...
5

5
12
1
9
...

23
4
...
15

3
28
8
10
...


15
...
400 litres
22
...

15

1
...
59
6
...
−1

2
5

Exercise 6 (page 13)

11
...
7

(a) £1827 (b) £4158

Exercise 3 (page 6)
1
...

5
...

9
...
2

11
...
4
20
17
12
...
−

3
...
2

1
6

3
4
1
9
...
4

13
20
1
10
...


Answers to practice exercises
Exercise 14 (page 25)

Chapter 3

1
...
571
5
...
96
8
...
0871

Exercise 8 (page 17)
1
...
23
8
...


13
20
21
6
...
(a) 1
50
5
...
0
...


4
...
6

7
16

(e) 16

17
80

1
...

7
...

13
...
182
2
...
122
3
...
82
0
...
0
...
2
...
273
8
...
256
9
...
30366
6
1
...
3
...
37
...
2 × 10
14
...
767 ×10
15
...
32 ×106

12
...
6875 13
...
21875 14
...
1875

Exercise 16 (page 27)
1
...
4667

Exercise 9 (page 18)
1
...
18
5
...
297

2
...
785
3
...
38
6
...
000528

2
...
3
6
...
3

4
...
27

3
...
54
7
...
52 mm

4
...
83

13
14

3
...
458

6
...
7083

7
...
2
...
3

1
3

10
...
0776

1
...
9205
5
...
4424
9
...
6992

2
...
7314
6
...
0321
10
...
8452

3
...
9042
7
...
4232

4
...
2719
8
...
1502

Exercise 18 (page 28)

4
...
47
...
385
...
582
...
9 6
...
82
7
...
1
8
...
6
0
...
0
...
1
...
53
...
84 14
...
69
15
...
81 (b) 24
...
00639 (b) 0
...
(a) 8
...
6˙
2400

1
...
995
5
...
6977
9
...
520

Exercise 12 (page 23)
3
...
62
7
...
330

4
...
832
8
...
45

Exercise 13 (page 24)
1
...
25
2
...
0361 3
...
923 4
...
296 × 10−3
5
...
4430 6
...
197 7
...
96 8
...
0549
9
...
26 10
...
832 × 10−6

2
...
782
6
...
92
10
...
3770

3
...
72
7
...
0

4
...
42
8
...
90

Exercise 19 (page 29)
1
...

7
...


Chapter 4

2
...
1
...
12
...


Exercise 17 (page 27)

Exercise 11 (page 20)

1
...
797
5
...
42
9
...
59

1
21
9
...
567
5
...

5
...

13
...

18
...
40
3
...
13459
4
...
9
6
...
4481 7
...
36 × 10−6
9
...
625 × 10−9
10
...
70

Exercise 15 (page 25)

3
125

15
...
28125

1
...
3
5
...
3

341

A = 66
...
144 J
14 230 kg/ m3

2
...

8
...


C = 52
...
407 A
628
...
1 m/s

3
...

9
...


R = 37
...
02 mm
224
...
526 

Exercise 20 (page 30)
1
...

7
...

12
...
27
2
...
1 W
3
...
61 V
F = 854
...
I = 3
...
T = 14
...
96 J
8
...
77 A 9
...
25 m
A = 7
...
V = 7
...
53 h (b) 1 h 40 min, 33 m
...
h
...
02 h (d) 13
...
£556 2
...
264 kg 4
...
14
...
(a) 0
...
(a) 440 K (b) 5
...
0
...
173
...
5
...
37
...
128
...
0
...
0
...
0
...
38
...
(a) 21
...
2% (c) 169%
13
5
9
11
...

13
...

20
16
16
1
15
...
25, D = 25%, E = 0
...
60, H = 60%, I = 0
...
(a) 2 mA (b) 25 V 2
...
685
...
83 lb10 oz
5
...
1 litres (b) 16
...
29
...
584
...
$1012

Exercise 28 (page 46)
Exercise 22 (page 36)
1
...

5
...

10
...


21
...
9
...
4 t (b) 8
...
67%
14 minutes 57 seconds
37
...
39
...
7%
15
...
60 m

(c) 20
...

(c) 5
...

8
...

12
...

16
...
5%

2
...
8 g
£611
38
...
3
...
20 days
3
...
18 (b) 6
...
3375 4
...
(a) 300 × 103 (b) 0
...

5
...

13
...

15
...


2
...
18%
3
...
£175 000
£260
6
...
£9116
...
£50
...
60 10
...
70 11
...
7
...
6 kg, B 0
...
5 kg
54%, 31%, 15%, 0
...
5 mm, 11
...
600 kW

Chapter 6

1
...
±5

2
...
±8

3
...
100

5
...
64

4
...
1

Exercise 30 (page 50)
1
5
...
8
3
...

9
1
or 0
...
5 11
...
100 8
...

100
12
...
36
14
...
1 16
...
5 or
18
...

or 0
...
1
3
243
2
1
...
39

Exercise 24 (page 41)
1
...
3
...
47 : 3
1
4
...
5 hours or 5 hours 15 minutes
4
6
...
12 cm
8
...


1
3 × 52

5
...
1 : 15 2
...
25% 4
...
6 kg
5
...
3 kg 6
...
−
18
9
...


1
3
7 × 37

6
...
−1
14
...
±
2

3
...
−3
15
...


1
210 × 52

8
...

45
9

2
3

344 Basic Engineering Mathematics
Chapter 10

Chapter 11

Exercise 39 (page 69)
1
...

5
...

9
...

13
...

17
...

21
...

25
...


x 2 + 5x

+6
+9
4x 2 + 22x + 30
a 2 + 2ab + b2
a 2 − 2ac + c2
4x 2 − 24x + 36
64x 2 + 64x + 16
3ab − 6a 2
2a 2 − 3ab − 5b2
7x − y − 4z
x 2 − 4x y + 4y 2
0
4ab − 8a 2
2 + 5b2
4x 2 + 12x

Exercise 42 (page 75)

2
...

6
...

10
...

14
...

18
...

22
...

26
...


2x 2 + 9x

+4
− 12
2 pqr + p2 q 2 + r 2
x 2 + 12x + 36
25x 2 + 30x + 9
4x 2 − 9
r 2 s 2 + 2rst + t 2
2x 2 − 2x y
13 p − 7q
4a 2 − 25b2
9a 2 − 6ab + b2
4−a
3x y + 9x 2 y − 15x 2
11q − 2 p
2 j2 +2 j

Exercise 40 (page 71)
2(x + 2)
p(b + 2c)
4d(d − 3 f 5)
2q(q + 4n)
bc(a + b2 )
3x y(x y 3 − 5y + 6)
7ab(3ab − 4)


2x y x − 2y 2 + 4x 2 y 3
3x
17
...
(a + b)(y + 1)
22
...

3
...

7
...

11
...

15
...
0 19
...
( p + q)(x + y)
23
...

4
...

8
...

12
...

16
...
1

2
...
1

7
...
2
16
...
−4

3
...

2
13
...
−3

12
...
6

9
...
−2

2
...
4 + 3a
6
...
10y 2 − 3y +
9
...


1
− x − x2
5

1
7

1
4

8
...
5

2
...
−4

6
...
−4

8
...
−10

12
...
9

17
...
±12

22
...
−15t
12
...
2
...
2

5
...
2

10
...
3

14
...
−6

18
...
4

20
...
±3

24
...

4
...

6
...
8 m/s2
3
...
472
(a) 1
...
30 m/s2

Exercise 45 (page 80)
1
...
45◦ C
7
...
0
...
50
8
...
30
6
...
3
...
d = c − e − a − b

1
− 4x
3

10
...
2x + 8x 2
3
...

− 4x
2

5
...
R =
I
c
7
...
v =

y
7
v −u
4
...
y = (t − x)
3
y−c
8
...
x =

Answers to practice exercises
I
PR
E
11
...
C = (F − 32)
9
9
...
L =

XL
2π f

12
...
x = a(y − 3)
14
...
64 mm

1
2π CX C

*

1
√

Z2 −

14
...
1 × 10−6

aμ
ρCZ 4 n

2

Chapter 13
Exercise 47 (page 87)

Exercise 49 (page 92)

S −a
a
1
...
x =

yd
d
(y + λ) or d +
λ
λ

3
...
D =

AB 2
5E y

5
...
R2 =

R R1
R1 − R

E −e
E − e − Ir
or R =
−r
I
I

y

ay
8
...
x = 
2
4ac
(y 2 − b2 )

7
...
R =
πθ
√
Z 2 − R2
, 0
...
L =
2π f
10
...
u =




xy
1
...
r =
(1 − x − y)
c
5
...
b =
2( p2 + q 2 )
9
...
L =

8S 2
3d

Q
, 55
mc
+ d, 2
...
L =
μ−m


x−y
6
...
R = 4

uf
, 30
u− f


2dgh
, 0
...
v =
0
...
v =

x = 4, y = 2
x = 2, y = 1
...
5, n = 0
...

4
...

8
...

12
...

16
...
a = N 2 y − x

Exercise 48 (page 89)

1
...

5
...

9
...

13
...


1
...

5
...


p = −1, q = −2
a = 2, b = 3
x = 3, y = 4
x = 10, y = 15

2
...

6
...


x = 4, y = 6
s = 4, t = −1
u = 12, v = 2
a = 0
...
40

Exercise 51 (page 96)
1
1
1
...
p = , q =
4
5
5
...
x = 5, y = 1
4

1
1
2
...
x = 10, y = 5
1
6
...
1

Exercise 52 (page 99)
1
...

5
...

8
...
2, b = 4
u = 12, a = 4, v = 26
m = −0
...
00426, R0 = 22
...
I1 = 6
...
62
4
...
a = 12, b = 0
...
F1 = 1
...
5

Exercise 53 (page 100)
1
...
x = 5, y = −1, z = −2

2
...
x = 4, y = 0, z = 3

346 Basic Engineering Mathematics
5
...

9
...

11
...
x = 1, y = 6, z = 7
x = 5, y = 4, z = 2 8
...
5, y = 2
...
5
i1 = −5, i2 = −4, i3 = 2
F1 = 2, F2 = −3 F3 = 4

Exercise 57 (page 109)
1
...

6
...

10
...
191 s 2
...
345 A or 0
...
619 m or 19
...
066 m
1
...
165 m
12 ohms, 28 ohms

3
...

7
...


7
...
0133
86
...
4 or −4
4
...
5 or 1
...

10
...

16
...
−2 or −

2
3

2
...
0 or −
3
8
...
−3 or −7
14
...
−3
20
...
5

1
...
2 or −2

2
1
2
1
2
...

12
...

18
...
−1 or 1
...

or −
2
5
2
or −3
28
...
1 or −
3
7
4
1
29
...

2
3
27
...

4
21
...
4 or −7

31
...
x 2 + 5x + 4 = 0
35
...
2 or −6

or −
or

or −

Chapter 15
Exercise 59 (page 112)

1
3

1
3
1
1
6
...
2
8
...
1 10
...
2
12
...
100 000 14
...

32
1
16
...
01 17
...
e3
16
1
...
x 2 + 3x − 10 = 0
34
...
x 2 − 1
...
68 = 0

2
...
3

4
...


Exercise 60 (page 115)

Exercise 55 (page 106)
1
...
732 or −0
...
1
...
135
5
...
443 or 0
...
x = 0, y = 4 and x = 3, y = 1

2
...
137 or 0
...
1
...
310
6
...
851 or 0
...
log 6
5
...
log 15
6
...
log 2
7
...
log 3
8
...
log 10 10
...
log 2
12
...
log 16 or log24 or 4 log2
14
...

3
...

7
...

11
...


0
...
137
2
...
719
3
...
108
0
...
351
1
...
081
4 or 2
...
562 or 0
...

4
...

8
...

12
...
296 or −0
...
443 or −1
...
434 or 0
...
086 or −0
...
176 or −1
...
141 or −3
...
0
...
1
...
b = 2 20
...
x = 2
...
t = 8
21
...
x = 5

Exercise 61 (page 116)
1
...
690 2
...
170 3
...
2696 4
...
058 5
...
251
6
...
959 7
...
542 8
...
3272 9
...
2

Answers to practice exercises
Chapter 16

Chapter 17

Exercise 62 (page 118)
1
...

3
...

5
...
1653
(a)
5
...
55848
(a) 48
...
739

(b)
(b)
(b)
(b)
6
...
4584
0
...
40444
4
...
7 m

(c)
(c)
(c)
(c)

22030
40
...
05124
−0
...
2
...
(a) 7
...
7408
8 3
3
...
2x 1/2 + 2x 5/2 + x 9/2 + x 13/2
3
1 17/2
1
+ x
+ x 21/2
12
60

1
...
1 V
(c) Horizontal axis: 1 cm = 10 N, vertical axis:
1 cm = 0
...
(a) −1 (b) −8 (c) −1
...
14
...
(a) −1
...
4
5
...
3
...
05
3
...
1
...
30
4
...

2
...

7
...

14
...


(a) 0
...
91374 (c) 8
...
2293 (b) −0
...
13087
−0
...
−0
...
2
...
816
...
8274 8
...
02
9
...
522 10
...
485
1
...
3
13
...
9 15
...
901 16
...
095
a
t = eb+a ln D = eb ea ln D = eb eln D i
...
t = eb D a
 
U2
18
...
W = PV ln
U1

Exercise 68 (page 140)
1
...
75, 0
...
75, 2
...
75;
1
Gradient =
2
2
...
(a) 6, −3 (b) −2, 4 (c) 3, 0 (d) 0, 7
3
...
(a) 2, − (b) − , −1 (c) , 2 (d) 10, −4
2
3
3
18
3
3
5
6
...
(a) and (c), (b) and (e)
8
...
(1
...
(1, 2)

11
...
4 (d) l = 2
...
P = 0
...
5

13
...
(a) 40◦ C (b) 128 
2
...
5 V

Exercise 66 (page 127)
1
...
5◦C

3
...
25 (b) 12 (c) F = 0
...
99
...
(a) 29
...
31 × 10−6 s
4
...
993 m (b) 2
...
(a) 50◦ C (b) 55
...
30
...
(a) 3
...
46 s
8
...
45 mol/cm3
10
...
(a) 7
...
966 s

(d) 89
...
−0
...
73
5
...
5 m/s (b) 6
...
7t + 15
...
m = 26
...
63
7
...
31 t (b) 22
...
09 W + 2
...
(a) 96 × 109 Pa (b) 0
...
8 × 106 Pa

348 Basic Engineering Mathematics
1
1
(b) 6 (c) E = L + 6 (d) 12 N (e) 65 N
5
5
10
...
85, b = 12, 254
...
5 kPa, 280 K
9
...
(−2
...
2), (0
...
8); x = −2
...
6
10
...
2 or 2
...
6
...
9
...

4
...

9
...

13
...
36
...
48 m
3
...
5 m 4
...
1 m
5
...
0 m
6
...
50 m 7
...
8 m
8
...
43 m, 10
...
60 m

9 cm
2
...
9
...
81 cm 5
...
21 m
6
...
18 cm
24
...
82 + 152 = 172
(a) 27
...
20
...
35 m, 10 cm
12
...
7 nautical miles
2
...
24 mm

Chapter 22
Exercise 87 (page 198)

Exercise 83 (page 185)
40
40
9
9
1
...
(a) 42
...
22◦ (b) 188
...
47◦
2
...
08◦ and 330
...
86◦ and 236
...
(a) 44
...
21◦ (b) 113
...
12◦

350 Basic Engineering Mathematics
4
...
α = 218◦41 and 321◦19
6
...
5
2
...
30
4
...
1, 120
6
...
3, 90
8
...
, 960◦ 10
...
4, 180◦ 12
...
40 Hz
14
...
1 ms
15
...
leading 17
...
p = 13
...
35◦, R = 78
...
7 cm2
2
...
127 m, Q = 30
...
17◦ ,
area = 6
...
X = 83
...
62◦, Z = 44
...
8 cm2
4
...
77◦, Y = 53
...
73◦ ,
area = 355 mm2

Exercise 92 (page 210)
Exercise 89 (page 203)
1
...
04 s or 40 ms (c) 25 Hz
(d) 0
...
62◦) leading 40 sin 50πt

1
...

5
...


193 km 2
...
6 m (b) 94
...
66◦, 44
...
4 m (b) 17
...
163
...
9 m, EB = 4
...
6
...
37 m
32
...
31◦

2
...
37 Hz (c) 0
...
54 rad (or 30
...
(a) 300 V (b) 100 Hz (c) 0
...
412 rad (or 23
...
(a) v = 120 sin100πt volts
(b) v = 120 sin (100πt + 0
...
i = 20 sin 80πt −
6
i = 20 sin(80πt − 0
...
3
...
488) m
7
...
75◦ lagging
(b) −2
...
363 A (d) 6
...
423 ms

Chapter 23
Exercise 90 (page 207)
1
...
1 mm, c = 28
...
A = 52◦2 , c = 7
...
152 cm,
area = 25
...
D = 19◦48 , E = 134◦12 , e = 36
...
E = 49◦ 0 , F = 26◦ 38 , f = 15
...
6 mm2
5
...
420 cm,
area = 6
...
811 cm, area = 0
...
K = 47◦ 8 , J = 97◦ 52 , j = 62
...
2 mm2 or K = 132◦52 , J = 12◦8 ,
j = 13
...
0 mm2

Exercise 93 (page 212)
1
...
42◦, 59
...
20◦ 2
...
23 m (b) 38
...
40
...
05◦
4
...
8 cm 5
...
2 m
6
...
3 mm, y = 142 mm 7
...
13
...

2
...

4
...

6
...

8
...
83, 59
...
83, 1
...
61, 20
...
61, 0
...
47, 116
...
47, 2
...
55, 145
...
55, 2
...
62, 203
...
62, 3
...
33, 236
...
33, 4
...
83, 329
...
83, 5
...
68, 307
...
68, 5
...
294, 4
...
(1
...
960)
(−5
...
500)
4
...
884, 2
...
353, −5
...
(−2
...
207)
(0
...
299)
8
...
252, −4
...
04, 12
...
04, 12
...
51, −32
...
51, −32
...
47
...

3
...

7
...


Answers to practice exercises
Exercise 103 (page 234)

Chapter 25
Exercise 96 (page 221)
1
...
t = 146◦

351

2
...
(i) rhombus (a) 14 cm2 (b) 16 cm (ii) parallelogram
(a) 180 mm2 (b) 80 mm (iii) rectangle (a) 3600 mm2
(b) 300 mm (iv) trapezium (a) 190 cm2 (b) 62
...
35
...
(a) 80 m (b) 170 m 4
...
2 cm2
5
...
1200 mm
7
...
560 m2
2
9
...
4 cm
10
...
43
...
32

1
...

7
...

11
...

16
...

20
...
2376 mm2
3
...
1709 mm
6

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