MATHEMATICS
N2
Mathematics Lecturer Guide
Hands-On!
Jolandi Daniels & Maria Kropman
© Future Managers 2015
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ISBN 978-1-77581-335-4
First edition 2015
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Contents
Module 1
Exponents and logarithms
Answers to activities . ..................................................................................... 1
Summative assessment answers . ................................................................ 20
Module 2
Factorisation, HCF, LCM and algebraic fractions
Answers to activities . ................................................................................... 24
Summative assessment answers . ................................................................ 36
Module 3
Equations, word problems and manipulation of technical formulae
Answers to activities . ................................................................................... 38
Summative assessment answers . ................................................................ 51
Module 4
Algebraic graphs
Answers to activities . ................................................................................... 54
Summative assessment answers . ................................................................ 90
Module 5
easuring of angles, angular and peripheral velocity
M
and sectors of circles
Answers to activities . ................................................................................... 96
Summative assessment answers . ............................................................. 106
Module 6
Trigonometry
Answers to activities . ................................................................................ 110
Summative assessment answers . ............................................................. 128
Module 7
Mensuration
Answers to activities . ................................................................................ 132
Summative assessment answers . ............................................................. 138
A note to the lecturer
Dear Lecturer
The textbook is extremely student-friendly and provides a source of information
for classwork, homework and revision.
The logical compilation of each module of the textbook
Each module is compiled as follows:
• Introduction
• Pre-knowledge
• Examples with explanations
• Assessment activities
• Summary of the module
• Summative assessment(s)
• Answers to assessment activities.
This Lecturer Guide provides complete solutions to the activities and assessments in
the textbook.
We sincerely hope that you will enjoy working through these books and that your
students will pass Mathematics N2 with flying colours!
The authors
MODULE
1 Exponents and logarithms
Answers to activities
Activity 1.1
1.
x2 × x5 = x7
2.
(a3)(a–2) = a
3.
4x9 × 2x–6 = 8x3
4.
6x6 ÷ 2x3 = 3x3
5.
a9
a3
6.
a18
a−5
= a6
7.
(a4)2 = a8
9.
(a2b3)4 = a8b12
11.
1 xy 2 2
3
2
=
= a18 – (–5)
= a18 + 5
= a23
8. x4 × (x2)3 × x
= x4 × x6 × x
= x11
10. (2x2y)3 = 23x6y3
= 8x6y3
t7
x6
y4
3 2
12. 2t −3 = 12 t7 – (–3)
= 12 t10
2
8x y
13. 16xy = x2y
14. xn + 1.xn – 1 = x n + 1 + n – 1
15. (–2xy3z)2 = 4x2y6z2
17.
(x 4)3
(x −2)3
× (x2)2 =
x12
x −6
× x4
=x2n
16. (a .b )6 = a3b2
1
2
1
3
18. (3x–2)3 = 33x–6
= x12 + 4 – (–6)
= x22
19.
x 2y3 × x 4y4
x 2y2
=
x6y7
x 2y2
20.
(x 3)2 × y 4
x 2y2
a
23.
3x 2 y 2
1 2xy 4 2
= a9
=
32x 4 y 2
2 2x 2 y 8
=
9x 2
4y 6
x6y4
x 2y2
= x4y2
= x4y5
6
21. a5 × a2 = a5 × a4
=
3 6
2 2
12x y
22. 16xy 4 = 3x4y
24. [(22)3]2 = [26]2
= 212
= 4 096
2
Module 1 • Exponents and logarithms
Activity 1.2
1
a2
1.
a–2 =
3.
(x3)–2 = x–6
=
5.
–4
1 x3 2
1
x6
4
= 1 x3 2
=
34
x4
=
81
x4
2.
1
2−3
4.
1 13 2
6.
1 6y −2 2
23
=
=8
–3
33
=
= 27
3x −1 –1
−2
3x
=
2x
y2
2x + y.2x – y = 2x + y + x – y
= 22x or 4x
7.
3(2x)0 = 3
8.
9.
4a0 – 2 = 4 – 2
=2
3
10. 3 64 = 43
3
= 43
=4
1
16x 2 = (16x2) 2
11.
6 3
12. 3 27x y
1
= (42x2) 2
= 4x
81a15
a5
13.
81a10
=
= 92a10
= 9a5
15.
x −3 y 0
32
= 91x 3
1
17. (3x 2 y)−3
= (3x2y)3
= 27x6y3
=
3 33x 6 y 3
= 3x2y
14. –(xy)0 × 2x2y0
= –(1) × 2x2
= –2x2
16.
2 –2
31 14 2 4
1 –2
= 3 16
4
= [16]2
= 256
(3x 2)2
18. (2x 3)−3
=
9x 4
2−3x −9
= 9x4.8.x9
= 72x13
4 33
19. 3 (x y )
=
3
x12 y 9
= x4y3
1
= 1 6y −1 2
20. x3.2x2.3x.x0.3x0
= 18x6
N2 Mathematics Lecturer Guide|Hands-On
21.
−1
1 32ba−2 2–3
22.
−2
= 1 32ba−1 23
=
=
23.
27b−6
8a−3
27a3
8b6
x+y
1
= (x + y) 2
=1
27. (6x)0 + 2–1
=
1 12
= (3a2)4
= 81a8
24. –(2a)0
= –1
26. –a(x – y)0
25. (–3a)0
=1+
1
(3a2)−4
1
2
= –a
1
28. x–1 + y–1 + z −1
=
or 1,5
1
x
+
1
y
+z
1
3a+b
30. (−3xy 2)−2
29. 3a−b
= 3a + b – (a – b)
= 3a + b – a + b
= 32b or 9b
= (–3xy2)2
= 9x2y4
Activity 1.3
1.
a)
c)
e)
g)
(2x3)2 × (x4)–2
= 4x6 × x–8
= 4x–2
= 42
x
4x 2 y −5
8xy 3
= 2xy 8
(–2x–3)4 × 2(x3)3
= 24x–12 × 2x9
= 32x–3
= 323
b)
d)
f)
x
(a2 – b2)2
= (a2 – b2)2
or (a2 – b2)(a2 – b2)
or a4 – 2a2b2 + b4
h)
3(2ab)2 × 2(a2b2)–2
= 3(4a2b2) × 2(a–4b–4)
= 12a2b2 × 2a–4b–4
= 24a–2b–2
= 24
2 2
ab
x −3 y 2
2xy −4
6
= y4
2x
1
3x −2
= 13 x2
or
x2
3
27 p3q−6
3p−5q−7
= 9p8q
3
4
Module 1 • Exponents and logarithms
x4y2
xy 6
x3
x4y–3 × y 4
2x 7 y −5
y4
2x 7
y9
x4y–3 ×
× 2x0y–2
=
× 2y–2
=
=
i)
k)
(2x–3y2)4 ×
(x 2 y 4)−2
(2xy)3
= 16x–12y8 ×
=
=
m)
2.
a)
c)
e)
2x −16 y 0
x 3y 3
2
x19 y 3
x −4 y −8
8x 3 y 3
(3ab)–1 × 3(ab)–1
=
1
3ab
=
1
a2b 2
1
×
3
ab
2
x 8 y16 2
= (x4y8)2
= x8y16
x a − b 21 x a + b 2
=x
l)
.x
a+b
2
a−b + a+b
2
2
=x
a−b+a+b
=x 2
2a
=x2
= xa
16a−5b−3
64a3b−4
b
= 4a8
81x −2 y −2
33x 2 y −3
34x −2 y −2
33x 2 y −3
=
= 3x–4y
=
n)
3y
x4
–2
−2
1 43xy−1xy 2
–2
= 1 4.33 2
=1
y
12 –2
y3
2
3
2
y
= 1 12
2
=
b)
y6
144
–2
−4
1 xy 2 2
2 2
= 1 y−4 2
x
y4
x −8
=
= x8y4
d)
= a6x – 4 – 5x + 2
= ax – 2
a−b
2
a6x − 4
a5x − 2
1
j)
(x −3 y 2)−3
(x 3 y −2)−2
=
x 9 y −6
x −6 y 4
=
x15
y10
f)
82 × 3 64
512 × (8a)0
÷
1
3 125
=
82 × 4
83 × 1
=
4
8
=
20
8
= 2,5 or 2 12 or
×5
×5
5
2
N2 Mathematics Lecturer Guide|Hands-On
5
243x10 y15
=
5 35x 10 y15
= 3x2y3
g)
i)
4
x −4n + 3n −3 y −8n
2x 3n y 3n − n
=
=
=
j)
a2x + y × a3(x − y)
a3x − 2y
=
a2x + y + 3x − 3y
a3x − 2y
=
a5x − 2y
a3x − 2y
= a5x – 2y – 3x + 2y
= a2x
(xy)15 × x2y × 3xy0
= (xy)3 × x2y × 3xy0
= x3y3 × x2y × 3x
= 3x6y4
5
27x −4 y −2
3−2x 2 y −3
33x −4 y −2
3−2x 2 y −3
=
= 35x–6y
=
243y
x6
l)
9a4
−27a9
– 15
3a
2ab
–2
1 (ab)−3 2
)−3 2
= 1 (ab
2ab 2
−3 −3 2
b
= 1 a 2ab
2
−4 −4 2
= 1 a 2b 2
=
=
n)
x −n − 3 y −8n
2x 3n y 2n
1
2.x 3n.x n + 3 y 2n.y 8n
1
2x 4n + 3 y10n
(–3a2)2 ÷ (–3a3)3
x 2y2
x 2y3
1
y
o)
=
=
=
x −4n y −8n × x 3n − 3
2x 3n y 3n × y −n
x8y8
x 2y3
=
=
=
m)
=
(x 2 y 4)−2n × x 3(n − 1)
2(xy)3n × y −n
(x 2 y 2)4
4 6 −2 −3
x y .x y
4
k)
h)
a−8b−8
4
1
4a8b8
(x 2 y 2)8
x y × x −2 y −4
4
−3 6
4
x16 y16
x −5 y 2
=
=
= x9y2
x4y4
x −5 y 2
5
6
Module 1 • Exponents and logarithms
3.
6x × 3x
= (6 × 3)x
= 18x
a)
2–2 × 3–2
= (2 × 3)–2
= 6–2
1
= 62
1
= 36
c)
e)
52x
32x
g)
x3 =
4
2x
–1
9–1 × 1 13 2
d)
= (32)–1 × (3–1)–1
= 3–2 × 31
= 3–1
= 13
x
= 1 53 2 or 1 25
9 2
3
64x ÷ 16x
64 x
= 1 16
2
= 4x
b)
f)
a2x – y =
h)
a5
1
= 2
=
x4
−2
Activity 1.4
1.
81 =
92
=9
100 2
= (102)
= 10
5.
16 4 5
= 16 4 5
= (24) 4
= 25
= 32
1
1
2
1
= (x2 + y2) 2 or
64
x2 + y2
−2
4.
27 3 2
−
= (33) 3
= 3–2
= 19
6.
x 3y 3
= xy
8.
81
1
x2 + y2
3
= 3 43
=4
1
3.
7.
2.
3
−3
4
= (34)
= 3–3
=
1
27
−3
4
a5
1
5 2
a
a2x
ay
N2 Mathematics Lecturer Guide|Hands-On
9.
3
8a6b9
=
3
10.
23a6b9
= 2a2b3
1
1 161 2 4
1
= 1 214 2 4
1
= (2–4) 4
= 2–1
1
2
=
11.
1
−3
2
1 259 2
2 −3
= 1 32 2 2
5
=
3−3
5−3
=
53
33
=
125
27
12.
x3 . 4 x
1
x6
1
=
1
x 3. x 4
1
x6
1+1 −1
4 6
= x3
or 4,63 or
=x
4 17
27
4+3–2
12
5
= x 12
1
13. [(642)] 3
2
= [64] 3
2
= [43] 3
= 42
= 16
14.
−2
8
1 125
23
3 −2
= 1 23 2 3
5
52
22
25
4
=
=
2
1
x4
2+3
2
3
x4
x3
2+3−3
2 4
= x3
=x
8 + 18 − 9
12
17
12
x 4y6
or 6 4 or 6,25
3
=x
x12 y18
= x2y3
3
2
16. x ×3 x
=
=
2−2
5−2
=
3
15.
17.
−3
4
1 16
81 2
4 −3
= 1 24 2 4
3
=
=
=
2−3
3−3
33
23
27
8
or 383 or 3,375
7
8
Module 1 • Exponents and logarithms
18.
3
=
=
=
20.
x 36 y 30
x7y4
3
2
8
= 35 . 3
6 5
x y
x7y4
y
x
125x 6
−3
5
= (34) 5 × 3
x18 y15
x7y4
1 64y 9 2
−3
5
−3
5
5
= 35
=3
−2
3
3 6
= 1 53x 9 2
4 y
=
2
19. 81 5 × 3
21.
1
3
0
27x 3 2
=1
−2
3
5−2x −4
4−2 y −6
42 y 6
5 2x 4
16y 6
= 25x 4
=
22.
1
27a−3 2
252n − 1
−2
3
1
2
= [(33a–3) ]
3
−2
3
23. 53n − 1.5n − 3. 1
5
2
−3 −3
= [3 2 .a 2 ]
24.
=
3–1a
=
a
3
=
(52)2n − 1
5
.5 .5
=
54n − 2
53n − 1 + n − 3 − 1
=
3n − 1 n − 3 −1
54n − 2
54n − 5
= 54n – 2 – 4n + 5
= 53
= 125
a6b8
a b
4 16 20
=
=
a3b 4
a4b5
1
ab
Activity 1.5
1.
3.
3x – 2 = 10
3x = 12
x=4
2.
x3 = 27 or x3 = 33
1
1
(x3) 3 = (33) 3 ∴ x = 3
∴x=3
4.
–4 – 2(x – 3) = 16
–4 –2x + 6 = 16
–2x = 16 – 6 + 4
–2x = 14
x = –7
1
x4 = 2
1
(x 4 )4 = (2)4
x = 16
N2 Mathematics Lecturer Guide|Hands-On
5.
x5 =
x5 =
1
1
32
1
25
6.
−1 − 3
1
x = 2–1
= 12
3
8.
4x 2 = 256
3
4
2
2
4
−4 −3
(x 3 ) 4 = (3–4)
x = 33
x = 27
2
x = (43) 3
x = 42
x = 16
10. x3 = 64
3
8x 4 = 27
3
4
x =
3
4
4
27
8 4
33 3
23
34
24
81 or
16
(x ) 3 = 1
x=
x=
1
81
1
34
x − 3 = 3–4
(x 2 ) 3 = (64) 3
9.
4
x −3 =
x−3 =
x 2 = 64
3
3
(x 3 ) 1 = (2) −1
x = 2–3
x = 18
(x5) 5 = (2–5) 5
7.
−1
x 3 =2
x3 = 43
x=4
2
1
5 16
Activity 1.6
1.
3x = 9
x=3
3.
4.2x = 256
2x = 64
2x = 26
x=6
5.
32x – 3 = 81
32x – 3 = 34
2x – 3 = 4
2x = 7
x = 72 or 3 12
2.
64x = 4
43x = 4
3x = 1
x = 13
4.
103x =
103x =
6.
1
100
1
102
103x = 10–2
3x = –2
x = – 23
6x – 1 = 1
6x – 1 = 60
x–1=0
x=1
−3
4
9
10
Module 1 • Exponents and logarithms
7.
102x – 1 = 0,001
102x – 1 = 10–3
2x – 1 = –3
2x = –2
x = –1
8.
3.104x = 0,0003
104x = 0,0001
104x = 10 1000
104x = 10–4
4x = –4
x = –1
271 – 2x – 33 = 0
271 – 2x = 33
(33)1 – 2x = 33
33 – 6x = 33
3 – 6x = 3
–6x = 0
x=0
11. 16x + 5 = 82x – 2
(24)x + 5 = (23)2x – 2
24x + 20 = 26x – 6
4x + 20 = 6x – 6
26 = 2x
x = 13
1
10. 256
= 4x – 3
13. 25(1 – 3x) – 54 = 0
14. 21 12 2
9.
52(1 – 3x)
54
=
52–6x = 54
2 – 6x = 4
6x = –2
x = – 13
15.
3 – 2x
1
31 27
2
=
1
81
3.(3–3)3 – 2x = 3–4
3.3–9 + 6x = 3–4
3–8 + 6x = 3–4
–8 + 6x = –4
6x = 4
x=
2
3
1
44
= 4x – 3
4–4 = 4x – 3
–4 = x – 3
–1 = x
∴ x = –1
12.
3x – 4
1 12 2
× 642x + 3 =
1
2−3
(2–1) 3x–4 × (26)2x + 3 = 23
2–3x + 4 × 212x + 18 = 23
29x + 22 = 23
9x + 22 = 3
9x = –19
x = – 19
or –2,111
9
or –2 19
3x – 4
=
1
128
2.2–3x + 4 = 2–7
2–3x + 5 = 2–7
–3x + 5 = –7
–3x = –12
x=4
x–5
16. 493x + 1 × 1 17 2
=
1
7−3
(72)3x + 1 × 7–x + 5 = 73
76x + 2 – x + 5 = 73
5x + 7 = 3
5x = –4
x = – 54
N2 Mathematics Lecturer Guide|Hands-On
17. 33x – 1 = 92x + 4
18.
33x – 1 = (32)2x + 4
33x – 1 = 34x + 8
3x – 1 = 4x + 8
–x = 9
x = –9
1 = 4 1 6 – 2x
19. 64
14 2
4–3= 4.4–6 + 2x
4–3= 4–5 + 2x
–3 = –5 + 2x
2 = 2x
x=1
21.
1
8x − 2
(22)5 – 4x = 23x1− 6
45 – 4x =
210 – 8x = 2–3x + 6
10 – 8x = –3x + 6
–5x = –4
x = 54
4.8x – 1 =
2
16 x
22.23x – 3 = 2.2–4x
23x – 1 = 21 – 4x
3x – 1 = 1 – 4x
7x = 2
x = 72
20.
3x
= 81
1 3x
= 34
1 19 2
1 32 2
(3–2)3x = 34
3–6x = 34
–6x = 4
x = – 23
22. 252x – 1 = 1253x + 2
(52)2x – 1 = (53)3x + 2
54x – 2 = 59x + 6
4x – 2 = 9x + 6
5x = –8
x = – 58 or –1 53
Activity 1.7
1.
Exponential form
25 = 32
Logarithmic form
log2 32 = 5
5.
a = by
102 = 100
10–3 = 0,001
61 = 6
logb a = y
logb 100 = 2
log10 0,001 = –3
log6 6 = 1
6.
8–2 =
1 = –2
log8 1 64
2
7.
ax = b
2.
3.
4.
1
64
x = loga b
8.
6–2 =
9.
71 = 7
106 = x3
10.
2
11.
1 14 2
12.
mk
13.
a4
14.
1
36
=x
=x
= a4
60 = 1
1 = –2
log61 36
2
log7 7 = 1
log x3 = 6
2 = log 1 x
4
logm x = k
loga a4 = 4
log6 1 = 0
11
12
Module 1 • Exponents and logarithms
Activity 1.8
1.
3.
5.
x = log2 64
2x = 64
2x = 26
∴x=6
log2 x = 6
25 = x
∴ x = 32
log 0,0001 = x
10x = 0,0001
10x =
7.
9.
11.
1 = logx 4
x1 = 4
∴x=4
= log 81 x
1
81 4 = x
1
(34) 4 = x
31 = x
∴x=3
4.
log3 27 = x
3x = 27
3x = 33
∴x=3
x = log5 5
∴x=1
6.
x = log100 102
100x = 102
102x = 102
∴ 2x = 2
x=1
8.
logx
1
10 000
10x = 10–4
∴ x = –4
logx 49 = 2
x2 = 49
x2 = 72
∴x=7
1
4
2.
1
9
x–2
= –2
=
1
9
x–2 = 3–2
∴x=3
10.
x = log5
5x =
1
125
1
125
5x = 5–3
∴ x = –3
12. log (x – 1) = 4
104 = x – 1
10 000 = x – 1
∴ x = 10 001
N2 Mathematics Lecturer Guide|Hands-On
13
Activity 1.9
1.
a)
log 5 + log 2
= log (5 × 2)
= log 10
=1
b)
c)
log2 64 + log2 4
= log2 (64 × 4)
= log2 256
= log2 28
=8
1
1 000
log 13
10
d)
f)
log
+ log4 64 + logee2
=
+ log4 43 + loge e2
= log 103 + 3log4 4 + 2loge e
= –3log 10 + 3log4 4 + 2loge e
= –3 + 3 + 2
=2
e)
g)
i)
k)
log 80 + log 5 – log 4
= log 1 804× 5 2
= log 100
= log 102
= 2log 10
=2
log3 3 – log3 27
1
= log3 3 2 – log3 33
= 12 log3 3 – 3 log3 3
= –2 12
log2 82 – log7 49 + log5 1
= 2log2 23 – log7 72 + log5 50
= (2)(3)log2 2 – 2log7 7 + 0log5 5
=6–2+0
=4
1
h)
log3 81 – 5 log10 100
– loge e
1
= log3 34 – 5 log10 10–2 – loge e 2 = 4 log3 3 + 10log10 10 – 12 loge e = 4 + 10 – 12
= 13 12
(log3 81)(log5 25)
= (log3 34)(log5 52)
= (4log3 3)(2log5 5)
= (4)(2)
=8
log2 (log3 81)
= log2 (log3 34)
= log2 (4log3 3)
= log2 4
= log2 22
=2
log3 (27 × 81)
= log3 (33 × 34)
= log3 37
= 7 log3 3
=7
log 100 + log9 81 – 3 log5 125
= log 102 + log9 92 – 3 log5 53
= 2 log 10 + 2 log9 9 – 9 log5 5
=2+2–9
= –5
j)
2 log 6 + 2 log 2 – 2 log 3
= log 62 + log 22 – log 32
= log
62 × 22
32
= log
32.22.22
32
= log 22.22
= log 16
l)
14
Module 1 • Exponents and logarithms
m)
o)
p)
log3 13 – log4 256 + log2 128
= log3 3–1 – log4 44 + log2 27
= –1 – 4 + 7
=2
log 0,001 – log7
1
49
+ log9
2.
log 250 + log x = 4
log 250x = 4
250x = 104
000
x = 10250
c)
e)
=
=
=
=
log 3 27
9
log 3 (81 × 1)
log 3 3
log 3 81
1
log 3 34
1
4 log 3 3
=
1
4
+ log2
1
128
log3 27 + log4 16 – log5 25
1
1
= log3 (33) 2 + log4 42 – log5 (52) 2
3
= log3 3 2 + 2 log4 4 – log5 5
= 32 + 2 – 1
= 2 12
x = 40
b)
= 103
x = 103 × 120
= 120 000
log8 2 + log8 x = 2
log8 2x = 2
2x = 82
x = 64
2
x = 32
d)
log x + log 20 – log 3 = 2
log 203x = 2
f)
20x = 100 × 3 20x = 300
x = 300
20
x
120
20x
3
log x – log 120 = 3
x
log 120
=3
log 3 27 − log 3 9
log 3 81 + log 3 1
= log 10–3 – log7 7–2 + log9 9–2 + log2 2–7
= –3 – (–2) + (–2) + (–7)
= –3 + 2 – 2 – 7
= –10
a)
1
81
n)
= 102
x = 15
x = log2 64
8
x = log2 8
= log2 23
=3
log2 16 – log2 x = 4
log2 16
=4
x
16
x
16
x
= 24
= 16
16 = 16x
∴x=1
N2 Mathematics Lecturer Guide|Hands-On
Activity 1.10
1.
log5 5
=
2.
log 5
log 5
=
=1
3.
=
log 5
log 25
=
log 5
log 52
log 5
2 log 5
1
2
=
5.
2 log4 32
=
=
=
=
log 16
log 8
log 24
log 23
4 log 2
= 3 log 2
= 43 or 1 13
log64 18
=
log25 5
=
log8 16
4.
=
log 18
log 64
=
log 8−1
log 82
=
=
6.
−1 log 8
2 log 8
– 12
log3 2 187
log 2 187
log 3 2 log 32
log 4
=
2 log 25
log 22
=
log 3
log 3
=
7 log 3
log 3
10 log 2
2 log 2
10
2
7
=7
=5
7.
9.
log 6 36
log 6 216
8.
log 1 64
4
=
=
log 6 62
log 6 63
=
2 log 6 6
3 log 6 6
=
=
2
3
=
3log9 27
log 64
log 1
4
log 43
log 4−1
3 log 4
−1 log 4
= –3
10. log100 0,1
=
3 log 27
log 9
=
log 0,1
log 100
=
3 log 33
log 32
=
log 10−1
log 102
=
9 log 3
2 log 3
=
−1 log 10
2 log 10
=
9
2
or 4 12
or
= – 12
log3 37
=7
15
16
Module 1 • Exponents and logarithms
11. log9 81
or
log9 92
=
log 3 81
log 3 9 = 2 log9 9
=
log 3 34
log 3 32
=2
=
4 log 3 3
2 log 3 3
12. log49 17
=
log e 17
log e 49
=
log e 7−1
log e 7 2
=
=2
=
13. log512 8
−1 log e 7
2 log e 7
– 12
14. log8 128
=
log 8 8
log 8 512
=
log 2 128
log 2 8
=
log 8 8
log 8 83
=
7 log 2 2
3 log 2 2
=
log 8 8
3 log 8 8
=
7
3
=
1
3
or = 2 13
Activity 1.11
1.
a)
c)
0,507
log2 e
b)
log2 3
log 3
log 2
=
= 1,585
d)
loge 3
=
log e
log 2
=
= 1,443
= 1,099
e)
2.
a)
c)
log10 1 324
= 3,122
x = antilog 4 – antilog 3
= 104 – 103
= 9 000
log3 x = 8
x = 38
x = 6 561
f)
loge x = 1,345
x = e1,345
= 3,838
log4 9
log 9
log 4
=
= 1,585
b)
d)
e)
log 3
log e
f)
log x = 0,451
x = 100,451
∴ x = 2,825
4 = log2 2x
2x = 24
x = 16
2
x=8
log 4,56 = x
∴ x = 0,659
N2 Mathematics Lecturer Guide|Hands-On
Activity 1.12
1.
a)
b)
c)
d)
e)
f)
2.
a)
b)
x = 2,59 × 3,2 × 1,522
log x = log 2,59 + log 3,2 + log 1,522
= 1,101
x = 101,101
= 12,614
x = 6,41 ÷ 2,42
log x = log 6,41 – log 2,42
= 0,423
x = 2,649
2
) × 2,5
x = (2,314,1
log x = 2 log 2,31 + log 2,5 – log 4,1
= 0,512
∴ x = 3,254
x = 5 495
log x = 51 log 495
= 0,539
x = 3,459
x=
3
2,55
4,13
log x = 13 log 2,55 –
= –0,0698
x = 0,852
x=
log 4,13
2,34 × 3 4,15
1
(0,35)2
log x = log 2,34 +
= 0,803
x = 6,356
x=
1
3
3
64,5
1,14
1 log
3
1
3
log 4,15 –
1
2
log 0,35
× (3,23)2
log x =
64,5 – log 1,14 + 2 log 3,23
= 1,565
x = 36,702
x=
4,13 × 0,123
13,4 × 1,82
log x = log 4,13 +
x = –1,477
= 0,033
1
2
log 0,123 – log 13,4 – 2 log 1,8
17
18
Module 1 • Exponents and logarithms
3, 83 × (2, 41)2
x=
c)
1
1
x = [(3,83) 2 × 2,412] 2
x = 3,83 4 × 2,41
log x = 14 log 3,83 + log 2,41
= 0,528
∴ x = 3,371
1
x=
4
6,12
3,45 × 1,34
log x =
1
4
log 3,45 +
d)
log 6,12 –
1
4
log 1,34
= 0,299
x = 1,992
x=
e)
1
4
(5,32)3 × (8,42)3
0,352 × 1,5
log x = 32 log 5,32 + 3 log 8,42 – 2 log 0,35 – log 1,5
= 4,601
x =39 863,587
Activity 1.13
1.
a)
log 3,045
= –0,484
b)
loge 4,1
= 1,411
c)
2 loge 3,5 = 2,506
d)
ln 0,142 = –1,952
e)
loge 100 = 4,605
f)
log 100 = 2
g)
log2 3,142
h)
loge e = 1
log 3,142
log 2
=
= 1,652
i)
ln e = 1
j)
log e = 0,434
k)
loge 4 2,156
= 0,192
l)
1
2
2.
a)
c)
x = antilog 6 – antilog 3
= 106 – 103
= 999 000
loge x = 3,192
x = e3,192
∴ x = 24,337
b)
d)
ln 1 432
= 1,817
x = antilog 5 – antilog 3
= e5 – e3
= 128,328
loge 2x = 4
2x = e4
∴ x = 27,299
N2 Mathematics Lecturer Guide|Hands-On
ln x = 0,515
x = e0,515
∴ x = 1,674
e)
Activity 1.14
x=
1.
2.
0,842
1
x = 0,842 2
loge x = 12 loge 0,842
= –0,086
x = 0,918
3.
(3,64)4 × 3 1,34
ln x = 4 ln 3,64 +
= 5,265
x = 193,541
x=
5.
4.
1
3
ln 1,34
15,4 × (0,4)5
0,312
x=
ln x =
1
2
x=
6.
x=
4
ln x =
1
4
0,325 × (3,85)4
2,3 × 0,007
ln 0,325 + 4 ln 3,85 – ln 2,3 – ln 0,007
2
(63,2)3 × (4,5)5
13,8
ln x = 23 ln 63,2
+
5
2
ln 4,5 – ln 13,8
= 3,8997
x = 49,389
9.
3
(4,32)2 × 1,934
32,1 × 1,52
ln x = 23 ln 4,32 + 12 ln 1,934 – ln 32,1 – 2ln 1,5
= –2,974
x = 0,051
10.
x=
3 15,4 × 3,15
2,06
1
1
3 ln 15,4 – 3
ln x =
= 1,053
x = 2,866
ln 2,06 +
0,143 × 3,245
1
3
ln 3,15
384 × (12,3)3
108
ln 384 + 3 ln 12,3 – ln 108
= 4,334
x = 76,274
= 8,959
x = 7 779,626
8.
3
ln x = 13 ln 0,143 + 13 ln 3,245
= –0,256
x = 0,775
ln x = ln 15,4 + 5 ln 0,4 – ln 0,312
= –0,682
x = 0,505
7.
x = (3,4)3 ÷ 6,148
ln x = 3 ln 3,4 – ln 6,148
= 1,855
x = 6,393
19
20
Module 1 • Exponents and logarithms
Module 1: Summative assessment answers
Question 1
4
−1
2
1.1 1.1.1
64 3 × 4
= (43) 3 × 4
= 44 × 4
= 44− 2
= 42
7
= (22) 2
= 27
= 128
1.1.2
4
−1
2
1
7
(3)
9x 2 y −2
27xy −3
=
32x 2 y −2
33xy −3
=
xy
3
1.1.3
(3)
3 (x 3 y 3)4
4 6 −3 −4
x .y .x .y
3
x12 y12
xy 2
=
=
= x3y2
1.1.4
1.1.5
−1
2
x4y4
xy 2
(3)
1
[(81)2] 2
1
= [94] 2
= 92
= 81
(2)
−2
8
1 125
23
3
= 1 23 2
=
2−2
5−2
=
52
22
=
25
4
−2
3
5
or 6 14 or 6,25
(3)
N2 Mathematics Lecturer Guide|Hands-On
1.1.6
(x 4)n − 1 × (x 2 y)−2n
xy −n × (xy)3n
=
x 4n − 4 × x −4n y −2n
xy −n × x 3n y 3n
=
x 4n − 4 − 4n ⋅ y −2n
x1 + 3n ⋅ y 2n
=
x −4 y −2n
⋅ y 2n
x
= x–4 – 1 – 3n. y–2n – 2n
= x–5 – 3n.y–4n
=
1 + 3n
1
x 5 + 3n ⋅ y 4n
1.2 1.2.1
103x – 4 = 0,0001
103x – 4 =
1
1 000
103x – 4 =
1
103
1.2.2
1.2.3
21
(4)
103x – 4 = 10–3
∴ 3x – 4 = –3
3x = 1
x = 13 3x – 1
1
51 25
2
=
(3)
1
125
5(5–2)3x – 1 = 5–3
5.5–6x + 2 = 5–3
51–6x + 2 = 5–3
53–6x = 5–3
∴ 3 – 6x = –3
–6x = –6
x = 1
1
4−2
(3)
2x – 1
= 643x – 2 × 1 14 2
42 = (43)3x – 2 × (4–1)2x – 1
42 = 49x – 6 × 4–2x + 1
42 = 49x – 6 – 2x + 1
42 = 47x – 5
∴ 2 = 7x – 5
7x = 7
x = 1
(4)
(28)
22
Module 1 • Exponents and logarithms
Question 2
2.1 2.1.1
1 = logx 4
∴ x = 4
(1)
2.1.2
log 104 = x
∴ x = 4 log 10
= 4
(1)
2.1.3
log (x + 5) = 3
103 = x + 5
x = 1 000 – 5
x = 995
(2)
2.2 2.2.1
log x = 0,541
x = 100,541
= 3,475
(2)
2.2.2
loge x = 0,145
x = e0,145
= 1,156
(2)
2.3 2.3.1
log8 16 – log8 4
= log8 16
4
= log8 4
=
=
=
=
2.3.2
2 log3 (log3 27)
= 2 log3 (log3 33)
= 2 log3 (3log3 3)
= 2 log3 3
= 2
log 4
log 8
log 22
log 23
2 log 2
3 log 2
2
3
(4)
(3)
N2 Mathematics Lecturer Guide|Hands-On
log 100 + log7
2.3.3
1
49
= log 102 + log7
23
– log3 81
1
72
– log3 34
= 2 log 10 – 2 log7 7 – 4 log3 3
=2–2–4
= –4
(3)
2.4 log4 1 024
=
log 1 024
log 4
=
log 45
log 4
= 5
2.5
log x 27 + log x 9 − log x 3
log x 2 187
=
log x 273.9
log x 2 187
=
log x 81
log x 2 187
=
log x 34
log x 37
=
=
2.6
(3)
4 log x 3
7 log x 3
4
7 = RHS
(4)
3 16,3 × 3,2462
x=
log x =
1
3
1,08 × 210
log 16,3 +2 log 3,246 – log 1,08 – log 210
= –0,939
x = 0,118
2.7
x=
3
(6)
3,74 × (15,1)3
1
1
x = [3,74 2 × (15,1)3] 3
1
x = 3,74 6 × 15,1
ln x =
1
6
ln 3,74 + ln 15,1
= 2,935
x = 18,813
(6)
[37]
Total: [65]
MODULE
2
Factorisation, HCF, LCM and
algebraic fractions
Activity 2.1
1.
2mpx – 3mpy
= mp(2x – 3y)
2.
3x – 9y
= 3(x – 3y)
3.
a3 – a2 + a
= a(a2 – a + 1)
4.
x2yz – xy2z + xyz2
= xyz(x – y + z)
5.
81a2 + 9ab2c
= 9a(9a + b2c)
6.
15abc2 – 25ab3c2
= 5abc2(3 – 5b2)
7.
–18x2 – 36x
= –18x(x + 2) or 18x(–x – 2)
8.
25x3 + 125x
= 25x(x2 + 5)
9.
4πr2 + 2πrh
= 2πr(2r + h)
10. 3x2(x – y) + 4x(x – y) – (x – y)
11. 6a(b – 1) + (1 – b)
= 6a(b – 1) – (b – 1)
= (b – 1)(6a – 1)
13. (3x – y)(3x – y) – 3x + y
= (3x – y)(3x – y) – (3x – y)
= (3x – y)(3x – y – 1)
15. (6x2 – y)2 – 6x2 + y
= (6x2 – y)2 – (6x2 – y)
= (6x2 – y)(6x2 – y – 1)
17. 7m3p – 14m2p2 + 49mp2
= 7mp(m2 – 2mp + 7p)
= (x – y)(3x2 + 4x – 1)
12. 4xy2(a – b) – 12xy(a – b)
= 4xy(a – b)(y – 3)
14. 12 xy + 14 x2 – 14 xy2
= 14 x(2y + x – y2)
2
2
16. πd4 – πD
4
= 14 π(d2 – D2) or
π
4
(d2 – D2)
= 14 π(d – D)(d + D)
2 2
18. 6x3y4 – 6x5y
= 6x2y21xy2 –
1
5
2
19. 3(a – b) x – y (b – a)
20. 2x(a – b) + y(b – a) + 4z(a – b)
21. 6a x + 9b x – 3c x
= 3 x (2a + 3b – c)
22. 8a4b3 – 4ab2 + 64a2b4
= 3x(a – b) + y(a – b)
= (a – b)(3x + y)
= 2x(a – b) – y(a – b) + 4z(a – b)
= (a – b) (2x – y + 4z)
= 4ab2(2a3b – 1 + 16ab2)
N2 Mathematics Lecturer Guide|Hands-On
Activity 2.2
1.
x3 + x2 + x + 1
= (x3 + x2) + (x + 1)
= x2(x + 1) + (x + 1)
= (x + 1)(x2 + 1)
2.
xy + 3a + 3x + ay
= (xy + ay) + (3a + 3x)
= y(x + a) + 3(a + x)
= (x + a) (y + 3) or (a + x)(y + 3)
3.
1 – x – x2 + x3
= (1 – x) + (–x2 + x3)
= (1 – x) + x2(–1 + x)
= (1 – x) – x2(1 – x)
= (1 – x)(1 – x2)
= (1 – x)(1 – x)(1 + x)
4.
2a + 8b + 2ab + 8
= (2a + 2ab) + (8b + 8)
= 2a(1 + b) + 8(b + 1)
= (1 + b)(2a + 8)
= 2(1 + b)(a + 4)
or
2[a + 4b + ab + 4]
= 2[(a + ab) + (4b + 4)]
= 2[a(1 + b) + 4(b + 1)]
= 2[(1 + b)(a + 4)]
= 2(1 + b)(a + 4)
5.
35ex + 20dx + 21ey + 12dy
= (35ex + 20dx) + (21ey + 12dy)
= 5x(7e + 4d) + 3y(7e + 4d)
= (7e + 4d)(5x + 3y)
6.
3ax – 3ay + 2bx – 2by
= (3ax – 3ay) + (2bx – 2by)
= 3a(x – y) + 2b(x – y)
= (x – y)(3a + 2b)
7.
5ay – 3ax + 3bx – 5by
= (5ay – 5by) + (–3ax + 3bx)
= 5y(a – b) + 3x(–a + b)
= 5y(a – b) – 3x(a – b)
= (a – b)(5y – 3x)
8.
20x2y3 – 8xz2 – 6z2 + 15xy3
= (20x2y3 – 8xz2) + (–6z2 + 15xy3)
= 4x(5xy3 – 2z2) + 3(–2z2 + 5xy3)
= 4x(5xy3 – 2z2) + 3(5xy3 – 2z2)
= (5xy3 – 2z2)(4x2 +3)
9.
x3y2 – a2x2y + 2axy – 2a3
= (x3y2 – a2x2y) + (2axy – 2a3)
= x2y(xy – a2) + 2a(xy – a2)
= (xy – a2)(x2y + 2a)
10. 2pq – 3pr + 6sr – 4sq
= (2pq – 3pr) + (6sr – 4sq)
= p(2q – 3r) + 2s(3r – 2q)
= p(2q – 3r) – 2s(2q – 3r)
= (2q – 3r)(p – 2s)
11. –8a – 64a2 – 1 – 8a
or
12. 3a + 3b – 5ac – 5bc
13. 3xy – xyz + 24x – 8xz
= (–8a – 64a2) + (–1 – 8a)
= –8a(1 + 8a) – (1 + 8a)
= (1 + 8a)(–8a – 1)
= (3a + 3b) + (–5ac – 5bc)
= 3(a + b) + 5c (–a – b)
= 3(a + b) – 5c (a + b)
= (a + b)(3 – 5c)
= (–8a – 64a2) + (–1 – 8a)
= 8a(–1 – 8a) + (–1 – 8a)
= (–1 – 8a)(8a + 1)
= (3xy – xyz) + (24x – 8xz)
= xy(3 – z) + 8x(3 – z)
= (3 – z)(xy + 8x)
= x(3 – z)(y + 8)
25
26
Module 2 • Factorisation, HCF, LCM and algebraic fractions
Activity 2.3
1.
3.
5.
7.
9.
11.
13.
15.
17.
x2 + 6x + 8
= (x + 4)(x + 2)
x2 + 5x – 6
= (x + 6)(x – 1)
a2 – 7a + 12
= (a – 4)(a – 3)
x2 – 17x + 30
= (x – 15)(x – 2)
8 – 9a + a2
= a2 – 9a + 8
= (a – 8)(a – 1)
b2 – 2b – 35
= (b – 7)(b + 5)
x2 – 3x – 18
= (x – 6)(x + 3)
x2 + 2xy + y2
= (x + y)(x + y)
x2 + 6 + 5x
= x2 + 5x + 6
= (x + 3)(x + 2)
x2 – 5x + 6
= (x – 3)(x – 2)
4. x2 – x – 2
= (x – 2)(x + 1)
6. x2 + 10x + 25
= (x + 5)(x + 5)
8. m2 + 3m – 10
= (m + 5)(m – 2)
10. p2 – 6p – 27
= (p – 9)(p + 3)
2.
12. x2 + 8x – 9
= (x + 9)(x – 1)
14. m2 + 5m – 14
= (m + 7)(m – 2)
16. x2 – 2xy – 15y2
= (x – 5y)(x + 3y)
18. 21 – 4y – y2
= (7 + y)(3 – y)
Activity 2.4
1.
y2 – 4y – 5
= (y – 5)(y + 1)
3.
6a2 – 17a + 12
= (2a – 3)(3a – 4)
5.
3b3 + 3b – 36
= 3(b2 – b – 12)
= 3(b – 4)(b + 3)
6x2 – 5x – 6
= (3x + 2)(2x – 3)
7.
9.
5a2 + 9a – 2
= (5a – 1)(a + 2)
3x2 + 36x + 96
= 3(x2 + 12x + 32)
= 3(x + 8)(x + 4)
4. 3x2 – 6x + 3
= 3(x2 – 2x + 1)
= 3(x – 1)(x – 1)
6. 5a2 – 45a + 90
= 5(a2 – 9a + 18)
= 5(a – 6)(a – 3)
8. 4b2 – 12b – 40
= 4(b2 – 3b – 10)
= 4(b – 5)(b + 2)
10. 6x2 – 7x – 20
= (3x + 4)(2x – 5)
2.
N2 Mathematics Lecturer Guide|Hands-On
11. 12a2 – 35a + 18
= (3a – 2)(4a – 9)
13. 2 – 3x + x2
= x2 – 3x + 2
= (x – 2)(x – 1)
15. 4x4 – 34x2 – 18
= 2(2x4 – 17x2 – 9)
= 2(2x2 + 1)(x2 – 9)
= 2(2x2 + 1)(x – 3)(x + 3)
12. 3x2 – 18x – 21
= 3(x2 – 6x – 7)
= 3(x – 7)(x + 1)
14. 36 – 21x + 3x2
= 3x2 – 21x + 36
= 3(x2 – 7x + 12)
= 3(x – 4)(x – 3)
16. 16y2 + 56y + 49
= (4y + 7)(4y + 7)
Activity 2.5
1.
x2 – y2 = (x – y)(x + y)
2.
x2 + y2
3.
x2 – 25 = (x – 5)(x + 5)
4.
a2 – 100 = (a – 10)(a + 10)
4x2 – 9 = (2x – 3)(2x + 3)
7. 6x2 – 24 = 6(x2 – 4)
= 6(x – 2)(x + 2)
2
9. 16x – 25y2 = (4x – 5y)(4x + 5y)
11. 64x2 – 9y2 = (8x – 3y)(8x + 3y)
6.
9a2 – 49 = (3a – 7)(3a + 7)
1 – 16b2 = (1 – 4b)(1 + 4b)
2
13. D4 – d 2 = ( D2 – d)( D2 + d)
2
y2
14. x9 – 481
= 1 x3 – 29y 21 x3 + 29y 2
4
a2 = b2 – a b2 + a
15. b25 – 49
1 5 7 21 5 7 2
17. x4 – y8 = (x2 – y4)(x2 + y4)
16. a2b2 – c2 = (ab – c)(ab + c)
5.
= (x – y2)(x + y2)(x2 + y4)
19. 3xy2 – 27x3 = 3x(y2 – 9x2)
= 3x(y – 3x)(y + 3x)
21.
9x 2
16y 2
– 4a6 = 1 43xy – 2a321 43xy + 2a32
Activity 2.6
1. a)
x2 – 8x + 15 = (x – 5)(x – 3)
x2 – 25 = (x – 5)(x + 5)
HCF = (x – 5)
8.
10. (x – y)2 = (x – y)(x – y)
1 = 3x – 1 3x + 1
12. 9x2 – 16
1
4 21
42
18. 4x2 – 64y2 = 4(x2 – 16y2)
20.
2a8
–
= 4(x – 4y)(x + 4y)
2(a8 – 4b8)
= 2(a4 – 2b4)(a4 + 2b4)
8b8 =
22. 81y4 – 1 = (9y2 – 1)(9y2 + 1)
= (3y – 1)(3y + 1)(9y2 + 1)
27
28
Module 2 • Factorisation, HCF, LCM and algebraic fractions
b)
x(x + y)
x2 – y2 = (x – y) (x + y)
x2 + xy = x(x + y)
HCF = (x + y)
c)
x2 – 16 = (x – 4)(x + 4)
2x2 – 2x – 24 = 2(x2 – x – 12) = 2(x – 4)(x + 3)
x2 – 7x + 12 = (x – 3)(x – 4)
HCF = (x – 4)
d)
x2 + 3x – 28 = (x + 7)(x – 4)
2x + 14 = 2(x + 7)
x2 – 49 = (x – 7)(x + 7)
HCF = (x + 7)
2. a)
4x2 – 16 = 4(x2 – 4) = 2.2(x – 2)(x + 2)
3x + 6 = 3(x + 2)
x2 + x – 2 = (x + 2)(x – 1)
LCM = 2.2.(x – 2)(x + 2) 3(x – 1)
= 12(x – 2)(x + 2)(x – 1)
b)
4x – 2y = 2(2x – y)
4x2 – y2 = (2x – y)(2x + y)
LCM = 2(2x – y)(2x + y)
c)
x2 – 2x – 15 = (x – 5)(x + 3)
x2 – 5x + 6 = (x – 3)(x – 2)
LCM = (x – 5)(x + 3)(x – 3)(x – 2)
d)
4a2 – 64 = 4(a2 – 16) = 2.2(a – 4)(a + 4)
2a – 8b + 2ab – 8 = (2a + 2ab) + (–8b – 8)
= 2a(1 + b) – 8(b + 1)
= (1 + b)(2a – 8)
= 2(1 + b)(a – 4)
2
b – 4b – 5 = (b – 5)(b + 1)
LCM = 2.2(a – 4)(a + 4)(1 + b)(b – 5)
= 4(a – 4)(a + 4)(1 + b)(b – 5)
3. a)
(x – y)(a – 1)2 = (x – y)(a – 1)(a – 1)
(x – y)(x + y)(a – 1) = (x – y)(x + y)(a – 1)
(a2 – 1)(m + n) = (a – 1)(a + 1)(m + n)
LCM = (x – y)(a – 1) (a – 1)(x + y) (a + 1)(m + n)
HCF = (a – 1)
N2 Mathematics Lecturer Guide|Hands-On
b)
c)
4)
a2 – 4 = (a – 2)(a + 2)
8a2 – 14a – 4 = 2(4a2 – 7a – 2) = 2(4a + 1)(a – 2)
12a2 – 21a – 6 = 3(4a2 – 7a – 2) = 3(4a + 1)(a – 2)
LCM = (a – 2)(a + 2) . 2(4a + 1) . 3
= 6(a – 2)(a + 2)(4a + 1)
HCF = (a – 2)
(x + y)2 = (x + y)(x + y)
2x2 – 2y2 = 2(x2 – y2) = 2(x – y)(x + y)
x2 + 2xy + y2 = (x + y)(x + y)
LCM = (x + y)(x + y) . 2(x – y)
= 2(x + y)2(x – y)
HCF = (x + y)
(2x – 3y)2 – 2x + 3y = (2x – 3y)(2x – 3y) – (2x – 3y)
= (2x – 3y)(2x – 3y – 1)
4x2 – 9y2 = (2x – 3y)(2x + 3y)
8x + 12y = 4(2x + 3y) = 2.2(2x + 3y)
a) LCM = (2x – 3y)(2x – 3y – 1)(2x + 3y) . 2.2
= 4(2x – 3y)(2x – 3y – 1)(2x + 3y)
b) HCF = 1
Activity 2.7
1.
6x 2
3xy
3.
−12pqr 3
18p 4q 2r
=–
8x 3 + 2x
2x
=
5.
7.
9.
=
2x
y
a2 − b2
a2 − ab
4x 2 − 64
x 2 − 16
2.
2r 2
3p3q
4.
2a2b 4
= 4ab2
8ab6
10x − 2 = 2(5x − 1)
2
2
= 5x – 1
2x(4x + 1)
2x
6.
=
(a − b)(a + b)
a(a − b)
8.
=
a+b
a
= 4x + 1
3y 2 + 21y − 36
3
(x − 1)2
x2 − 1
=
=
=
4(x 2 − 16)
(x − 4)(x + 4)
=
4(x − 4)(x + 4)
(x − 4)(x + 4)
=
3( y 2 + 7 y − 12)
3
= y2 + 7y – 12
(x − 1)(x − 1)
(x − 1)(x + 1)
x −1
x +1
2
a + 3)
10. a −a −a −4 12 = (a −(a4)(
− 4)
=a+3
=4
11.
rp 2 − rq 2
p − 2pq + q 2
2
=
r( p − q)( p + q)
( p − q)( p − q)
=
r( p + q)
( p − q)
12.
25x − 20
5x 2 + 6x − 8
=
=
5(5x − 4)
(5x − 4)(x + 2)
5
x+2
29
30
Module 2 • Factorisation, HCF, LCM and algebraic fractions
13.
x 2 + 2xy + y 2
x2 − y2
=
x 2 + 2xy + y 2
(x 2 − y 2)
(x + y)(x + y)
(x − y)(x + y)
=
x+y
x−y
=
2
2(x 2 − 1)
15. 2x − 22 =
2
(x − 1)
(x − 1)
=
2(x − 1)(x + 1)
(x − 1)(x − 1)
=
2(x + 1)
x −1
2
x−y
x−y
14. y − x = −(x − y)
= –1
2
2
16. xx ++ 16
= xx ++ 16
4
4
=
3(x − 3)(x + 3)
6(x − 3)
=
2(9x 2 − 1)
(3x − 1)(3x − 1)
2(3x − 1)(3x + 1)
(3x − 1)(3x − 1)
=
x+3
2
=
2(3x + 1)
(3x − 1)
2
27 = 3(x − 9)
17. 36xx −−18
6(x − 3)
18.
18x 2 − 2
9x 2 − 6x + 1
=
Activity 2.8
1.
3.
21x 7
5xy 3
×
=
7.3x 7
5xy 3
=
9x 8 y 2
x6y3
=
9x 2
y
6a2 − 2a
6a
15xy 2
7x 5
×
×
2.
5.3xy 2
7x 5
12a2
18a − 6
4.
12a2
6(3a − 1)
a2 − ab
3a − 3b
=
a(a − b)
3(a − b)
=
a
3
9x 2 − 3x
6x
=
9x 2 − 3x
6x
24a3
36a
=
3x(3x − 1)
6x
2a2
3
=
12x 2
24x
x
2
=
2a(3a − 1)
6a
=
=
×
=
5.
÷
(a2 + b 2)2
a4 − b4
=
(a2 + b 2)(a2 + b 2)
(a2 − b 2)(a2 + b 2)
=
a2 + b2
a2 − b2
6.
2x 2 − 8x
x 2 − 6x + 8
=
12x − 4
4x
×
4x
12x − 4
4x
4(3x − 1)
×
÷
x 2 + 2x
x2 − 4
2x ( x − 4)
(x − 4)(x − 2)
=2
×
(x − 2)(x + 2)
x(x + 2)
N2 Mathematics Lecturer Guide|Hands-On
7.
9.
11.
a2 + a − 6
a2 − 9
÷
a2 − 3a + 2
a2 − a
=
(a − 3)(a + 2)
(a − 3)(a + 3)
=
a(a + 2)
(a + 3)(a − 2)
a2 + 2a
a2 + a − 6
×
a(a + 2)
(a + 3)(a − 2)
=
a(a + 1)
(a + 3)(a − 2)
rp 2 − rq 2
p − 2pq + q 2
a(a − 1)
(a − 2)(a − 1)
×
×
p−q
rp − rq
×
(x + y)2
x 2 + 2x + 1
=
31
( − x 2)
2(x + y)
=
31
( − x)(1 + x )
2(x + y )
=
31
( − x)(x + y)
2(x + 1)
×
(x + y)(x + y)
(x + 1)(x + 1)
×
(x + y )(x + y)
(x + 1)(x + 1)
2
2
x + xy
3x + 3y
10. x 2 − y ÷ 2x − 2y × 2
−
2xy + y 2
x
x − xy
(a + 1)(a + 1)
(a + 2)(a + 1)
×
3 − 3x 2
2x + 2y
2
a2 + 2a + 1
a2 + 3a + 2
=
2
8.
31
÷
=
r( p − q)( p + q)
× r((pp−−qq))
( p − q)( p − q)
=
1
p( p − q)
=
(x − y )(x + y )
x(x − y )
=
2(x + y)
3(x − y)
×
2(x − y)
3(x + y )
×
x(x + y)
(x − y)(x − y)
3p 2 + 3pq
3
×
3
3p( p + q)
2
2
4x 2 − 4x + 1
12. 8x 2 − 8 ÷ 4x −24x + 1 × −4x 3 + 4x
16x − 4
16x − 4
13.
=
8(x 2 − 1)
4(4x 2 − 1)
=
8(x − 1)(x + 1)
4(2x − 1)(2x + 1)
=
2
−x
4(x 2 − 1)
(2x − 1)(2x − 1)
×
2
a2 − a − 6
−2 + 4a + a2
2
a −a−6
a2 + 4a − 2
=
(a − 3)(a + 2)
(a2 + 4a − 2)
=
a+2
a −1
×
(x + 12)(x − 1)
(x − 8)(x − 2)
=
x −1
x−2
(2x − 1)(2x − 1)
−4x(x − 1)(x + 1)
2
2
14. x 2 − 3x − 10 ÷ x +25x + 6
x −9
x − x − 20
=
(x − 5)(x + 2)
(x − 5)(x + 4)
(a2 + 4a − 2)
(a − 3)(a − 1)
=
x −3
x+4
a + 4a − 2
a2 − 4a + 3
2
x 2 − 7x − 8
15. x 2 + 11x − 12 × 2
=
×
2
×
x − 10x + 16
(2x − 1)(2x − 1)
−4x(x 2 − 1)
4(2x − 1)(2x + 1)
4(2x − 1)(2x − 1)
a2 − 4a + 3
a2 + 4a − 2
÷
=
×
×
x + 13x + 12
×
(x − 8)(x + 1)
(x + 12)(x + 1)
×
(x − 3)(x + 3)
(x + 3)(x + 2)
2
2
2
2
a2 − 2ab + b 2
16. a − b 2 × a2 + ab − 2b 2 ÷
2
ab + 2b
a − 2ab − 3b
=
(a − b)(a + b)
b(a + 2b)
=
a
b
×
(a + 2b)(a − b)
(a − 3b)(a + b)
a − 3ab
×
a(a − 3b)
(a − b)(a − b)
32
Module 2 • Factorisation, HCF, LCM and algebraic fractions
Activity 2.9
1.
3
7
+
4
7
=
7
7
=
=1
=
2.
3
8
=
3.
4
x
=
5.
x – 2x
2
3x
24 − 3x 2 − 4x
6x
–
a −1
3a2
–
4.
=
2b 2(a − 1) − a2(2a + 1)
6a2b 2
=
2ab 2 − 2b 2 − 2a3 − a2
6a2b 2
7.
2
3
x+y + x+y
= x3 ++ 2y
= x +5 y
9.
6x − 1
3
–
3−x
4
y
6x
=
2a + 1
6b 2
–
x −5
6
2 + 1
3
4
3(3) − 8(2) + 6(1)
24
9 − 16 + 6
24
1
– 24
–
–
x
y
y 2 − 6x 2
6xy
6.
x −2 – x+3
5
2
2(x − 2) − 5(x + 3)
=
10
2
x
−
4
−
5x − 15
=
10
= −3x10− 19
8.
4 – 5
a−b
a−b
4
−
5
= a−b
−1
= a−b
10. 3a3a− 1 – a2−a1 + 12
=
4(6x − 1) − 3(3 − x) − 2(x − 5)
12
=
2(3a − 1) − 3(a − 1) + 3a
6a
=
24x − 4 − 9 + 3x − 2x + 10
12
=
6a − 2 − 3a + 3 + 3a
6a
=
25x − 3
12
=
6a + 1
6a
11. a −2 2 – 2(a4− 1) + 2a8− 5
=
=
=
4(a − 2) − 2.2(a − 1) + (2a − 5)
8
4a − 8 − 4a + 4 + 2a − 5
8
2a − 9
8
2
5
13. 3pq2 + 32 – 12pq
8p q
12. 1 + x – 2x2− 1
=
=
2 + 2x − 2x + 1
2
3 or 1 1
2
2
14. x +33 + 14
2x
=
8p(2) + 3q(3) − 2pq(5)
24 p 2q 2
=
x + 3 + 14(2x 3)
2x 3
=
16p + 9q − 10pq
24 p 2q 2
=
x + 3 + 28x 3
2x 3
N2 Mathematics Lecturer Guide|Hands-On
3
16. a −2 4 + 13 (a – 2) – a −4 2
2
15. x − y + x − y – 1
=
3
(x − y)
=
3 + 2 − (x − y)
(x − y)
5−x+ y
x−y
=
+
2
(x − y)
=
–1
=
=
6(a − 4) + 4(a − 2) − 3(a − 2)
12
6a − 24 + 4a − 8 − 3a + 6
12
7a − 26
12
Activity 2.10
1.
3 – 1
x +1
x +1
3
−
1
= x +1
= x 2+ 1
2.
2 + 4
x−2
x+2
2(x + 2) + 4(x − 2)
= (x − 2)(x + 2)
= 2(xx +−42)(+ x4x+ −2)8
2(3x − 2)
6x − 4
= (x − 2)(x + 2) or (x − 2)(x + 2)
3.
a
a−3
4.
2
x2 − 4
5.
–
a−4
a
1
x+2
=
a2 − (a − 3)(a − 4)
a(a − 3)
=
2
(x − 2)(x + 2)
=
a2 − a2 + 7a − 12
a(a − 3)
=
2−x+2
(x − 2)(x + 2)
=
7a − 12
a(a − 3)
=
−x + 4
(x − 2)(x + 2)
x −3
x−4
x +1
x+2
+
6.
2
x 2 − 49
–
(x − 3)(x + 2) + (x + 1)(x − 4)
(x − 4)(x + 2)
=
2
(x − 7)(x + 7)
=
x 2 − x − 6 + x 2 − 3x − 4
(x − 4)(x + 2)
=
2x − (x − 7)
x(x + 7)(x − 7)
=
2x 2 − 4x − 10
(x − 4)(x + 2)
=
2x − x + 7
x(x + 7)(x − 7)
=
1
x +1
–
2
x +1
–
3
(x + 1)2
8.
1
(x + 2)
–
1
x 2 + 7x
=
=
7.
–
–
1
x(x + 7)
( x + 7)
x(x + 7)(x − 7)
1
x(x − 7)
4
x2 + x
+
1
x − x2
–
3
x
=
x + 1 − 2(x + 1) − 3
(x + 1)2
=
4
x(x + 1)
=
x + 1 − 2x − 2 − 3
(x + 1)2
=
4(1 − x) + (x + 1) − 3(x + 1)(1 − x)
x(x + 1)(1 − x)
=
−x − 4
(x + 1)2
=
4 − 4x + x + 1 − 3(x − x 2 + 1 − x)
x(x + 1)(1 − x)
=
5 − 3x − 3x + 3x 2 − 3 + 3x
x(x + 1)(1 − x)
=
3x 2 − 3x + 2
x(x + 1)(1 − x)
+
1
x(1 − x)
–
3
x
33
34
Module 2 • Factorisation, HCF, LCM and algebraic fractions
9.
x+3
x 2 − 7x + 12
x −3
x2 − 9
=
(x + 3)
– (x −(x3)(−x3)+ 3)
(x − 4)(x − 3)
(x + 3)(x + 3) − (x − 3)(x − 4)
(x − 4)(x − 3)(x + 3)
3
– 2 2
(x + 1)
x3 + x2
3
= x 2(x + 1) – (x + 1)(2 x + 1)
3(x + 1) − 2x 2
= x 2(x + 1)(x + 1)
=
x 2 + 6x + 9 − x 2 + 7x − 12
(x − 4)(x − 3)(x + 3)
=
=
13x − 3
(x − 4)(x − 3)(x + 3)
=
–
10.
2x
12. 3x 4+ y – 3x 5− y + 9x 2 − y 2
2
11. a −3 3 + 2 a
– a +2 2
a −a−6
a2
(a − 3)(a + 2)
=
3
(a − 3)
=
3(a + 2) + a2 − 2(a − 3)
(a − 3)(a + 2)
=
3a + 6 + a2 − 2a + 6
(a − 3)(a + 2)
=
a2 + a + 12
(a − 3)(a + 2)
+
–
13. a 5− 3 – a +1 2 – 2 a + 7
a −a−6
=
5
(a − 3)
=
1
(a + 2)
3x + 3 − 2x 2
x 2(x + 1)2
2
(a + 2)
14.
(a + 7)
(a − 3)(a + 2)
=
4
(3x + y)
=
4(3x − y) − 5(3x + y) + 2x
(3x − y)(3x + y)
=
12x − 4 y − 15x − 5y + 2x
(3x − y)(3x + y)
=
−x − 9y
(3x − y)(3x + y)
3
4x 2 − 9
–
–
5
(3x − y)
+
1
4x 2 − 12x + 9
=
3
(2x − 3)(2x + 3)
5(a + 2) − (a − 3) − (a + 7)
(a − 3)(a + 2)
=
3(2x − 3) − (2x + 3)
(2x − 3)(2x + 3)(2x − 3)
=
5a + 10 − a + 3 − a − 7
(a − 3)(a + 2)
=
=
3a + 6
(a − 3)(a + 2)
=
=
3(a + 2)
(a − 3)(a + 2)
6x − 9 − 2x − 3
(2x − 3)2(2x + 3)
4x − 12
(2x − 3)2(2x + 3)
=
3
a−3
–
–
15. 3x 9+ 9 – 2 2x + 8
2x + 14x + 24
2(x + 8)
2(x 2 + 7x + 12)
=
9
3(x + 3)
=
3
(x + 3)
=
3(x + 4) − (x + 8)
(x + 3)(x + 4)
=
3x + 12 − x − 8
(x + 3)(x + 4)
=
2x + 4
(x + 3)(x + 4)
–
–
(x + 8)
(x + 4)(x + 3)
16.
2xy
x2 − y2
+
2x
(3x − y)(3x + y)
x
x+y
–
–
1
(2x − 3)(2x − 3)
y
x−y
+1
x
(x + y)
–
y
(x − y)
=
2xy
(x − y)(x + y)
=
2xy + x(x − y) − y(x + y) + (x + y)(x − y)
(x + y)(x − y)
=
2xy + x 2 − xy − xy − y 2 + x 2 − y 2
(x + y)(x − y)
=
2x 2 − 2y 2
(x + y)(x − y)
=
2(x − y)(x + y)
(x + y)(x − y)
=2
+
+1
N2 Mathematics Lecturer Guide|Hands-On
17.
19.
3
x 2 + 2x − 3
2
x 2 − 3x + 2
–
18.
x +1
x 2 − 16
–
x −1
(x − 4)(x − 3)
(x − 1)
(x − 4)(x − 3)
=
x +1
(x − 4)(x + 4)
3(x − 2) − 2(x + 3)
(x + 3)(x − 1)(x − 2)
=
(x + 1)(x − 3) − (x − 1)(x + 4)
(x − 4)(x + 4)(x − 3)
=
3x − 6 − 2x − 6
(x + 3)(x − 1)(x − 2)
=
x 2 − 2x − 3 − x 2 − 3x + 4
(x − 4)(x + 4)(x − 3)
=
x − 12
(x + 3)(x − 1)(x − 2)
=
−5x + 1
(x − 4)(x + 4)(x − 3)
=
3
(x + 3)(x − 1)
=
x
(x + y)2
+
–
2
(x − 2)(x − 1)
y
x2 − y2
–
x2 + y2
1
x−y
y
(x − y)(x + y)
–
20. x + y – 3x − 3y
=
x
(x + y)2
=
x(x − y) + y(x + y) − (x + y)2
(x + y)2(x − y)
=
x
1
+
y
1
–
(x − y )(x + y)
3(x − y )
=
x 2 − xy + xy + y 2 − x 2 − 2xy − y 2
(x + y)2(x − y)
=
+
y
1
–
x+y
3
=
−2xy
(x + y)2(x − y)
x
1
=
3x + 3y − x − y
3
2x + 2y
or 2(x 3+ y)
3
+
–
1
(x − y)
=x+y–
=
2
21. 3x 2 − 4x + 1 + 2 5
– x 2− 1
6x + x − 1
5
(3x − 1)(2x + 1)
=
2
(3x − 1)(x − 1)
=
2(2x + 1) + 5(x − 1) − 2(3x − 1)(2x + 1)
(3x − 1)(x − 1)(2x + 1)
=
4x + 2 + 5x − 5 − 12x 2 − 2x + 2
(3x − 1)(x − 1)(2x + 1)
=
−12x 2 + 7x − 1
(3x − 1)(x − 1)(2x + 1)
+
–
2
(x − 1)
x2 − y2
3(x − y)
35
36
Module 2 • Factorisation, HCF, LCM and algebraic fractions
Module 2: Summative assessment answers
Question 1
1.1 6 – 352
x
=
6x 2 − 35
x2
(2)
1.2 x2(a – b) – 2x(a – b) – b + a
= x2(a – b) – 2x(a – b) + (a – b)
= (a – b)(x2 – 2x + 1)
= (a – b)(x – 1)2
(3)
1.3 81 – x4
= (9 – x2)(9 + x2)
= (3 – x)(3 + x)(9 + x2) (3)
1.4 4a – 3b + 12 – ab
= (4a + 12) + (–3b – ab)
= 4(a + 3) + b(–3 – a)
= 4(a + 3) – b(a + 3)
= (a + 3)(4 – b) (4)
1.5 3x2 + 3x – 36
= 3(x2 + x – 12)
= 3(x + 4)(x – 3) (3)
1.6 5x2 – 17x + 6
= (5x – 2)(x – 3) (2)
[17]
Question 2
2.1 25x2 + 125x = 25x(x + 5)
x2 + 10x + 25 = (x + 5)(x + 5)
x2 – 6x + 5x – 30 = (x2 – 6x) + (5x – 30)
= x(x – 6) + 5(x – 6)
= (x – 6)(x + 5)
LCM = 25x(x + 5)(x + 5)(x – 6) or 25x(x + 5)2(x – 6)
HCF = (x + 5) (5)
2.2 x2 + xy = x(x + y)
x2 – y2 = (x – y)(x + y)
(x + y)2 = (x + y)(x + y)
HCF = x + y
LCM = x(x + y)(x – y)(x + y) or x(x + y)2(x – y) (5)
[10]
N2 Mathematics Lecturer Guide|Hands-On
3.1
3.2
3.3
3.4
1
a2bc
–
2a
b 2c 2
5
ac
+
=
bc − 2a(a2) + 5ab 2c
a2b 2c 2
=
bc − 2a3 + 5ab 2c
a2b 2c 2
x2
x −4
2
÷
(2)
x 2 − 3x
x 2 − 5x + 6
=
x2
(x − 2)(x + 2)
=
x x+2
(x − 3)(x − 2)
x(x − 3)
×
(5)
3x 2
x −x −6
3x
x −3
–
2
–
1
x+2
=
3x 2
(x − 3)(x + 2)
=
3x 2 − 3x(x + 2) − (x − 3)
(x − 3)(x + 2)
=
3x 2 − 3x 2 − 6x − x + 3
(x − 3)(x + 2)
=
−7x + 3
(x − 3)(x + 2) 25 − 25y 2
3y 2 − 3
=
=
5
÷
–
3x
(x − 3)
×
1
(x + 2)
–
(5)
9y 2 − 6y + 1
3y 2 − 3
25( y 2 − 1)
3( y 2 − 1)
5
y
37
×
3( y 2 − 1)
(3y − 1)2
(3y − 1)2
−5y 3 + 5y
×
(3y − 1)2
−5 y( y 2 + 1)
(5)
3.5 2a + 32 – a +1 2 + 22a
a −4
(a + 2)
=
2a + 3
(a + 2)(a + 2)
=
(2a + 3)(a − 2) − (a + 2)(a − 2) + 2a(a + 2)
(a + 2)(a + 2)(a − 2)
=
2a2 − 4a + 3a − 6 − a2 + 4 + 2a2 + 4a
(a + 2)2(a − 2)
=
3a2 + 3a − 2
(a + 2)2(a − 2)
1
(a + 2)
–
+
2a
(a − 2)(a + 2)
(6)
[23]
Total [50]
MODULE
3
Equations, word problems and
manipulation of technical formulae
Activity 3.1
1.
q – 1 = 5q + 3q – 8
q – 1 = 8q – 8
q – 8q = –8 + 1
–7q = –7
q=
3.
2.
−7
−7
18y – 24 = –3(2 – 5y)
18y – 24 = –6 + 15y
18y – 15y = –6 + 24
3y = 18
y=
=1
–4(1 – 6b) = 24b – 22
–4 + 24b = 24b – 22
24b – 24b = –22 + 4
0b = –18
b has no solution.
4.
y=6
3(n – 4) = 7(n – 2) + 2n
3n – 12 = 7n – 14 + 2n
3n – 12 = 9n – 14
3n – 9n = –14 + 12
–6n = –2
n=
n=
5.
–(1 + 7x) –6(–7 – x) = 36
–1 – 7x + 42 + 6x = 36
41 – x = 36
–x = 36 – 41
–x = –5
x=
6.
−5
−1
3y − 2 = 10
3y – 2 = 100
3y = 100 + 2
3y = 102
y=
102
3
y = 34
−2
−6
1 or
3
0,333
–5(1 – 5x) + 5(–8x – 2) = –4x – 8x
–5 + 25x – 40x – 10 = –12x
–15 – 15x = –12x
–15x + 12x = 15
–3x = 15
x=
x=5
7.
18
3
8.
15
−3
x = –5
–3x + 2(–1 – 3x) = 4 – 3(2 – x)
–3x – 2 – 6x = 4 – 6 + 3x
–9x – 2 = –2 + 3x
–9x – 3x = –2 + 2
–12x = 0
x=
0
−12
x=0
N2 Mathematics Lecturer Guide|Hands-On
9.
39
10. 3(2 – 2a) – 7(2 – a) = –(a + 4) + 3
–5 = – 40 − 3x
25 = 40 – 3x
3x = 40 – 25
3x = 15
x=5
6 – 6a – 14 + 7a = –a – 4 + 3
–8 + a = –a – 1
a + a = –1 + 8
2a = 7
a=
7
2
a = 3 12 or 3,5
4
11. x3 + 2x
=7
6 + 4 = 14x
10 = 14x
10 =
x = 14
5
7
or 0,714
Activity 3.2
1. a)
c)
e)
x2 – 13x + 12 = 0
(x – 12)(x – 1) = 0
∴ x – 12 = 0 or x – 1 = 0
∴ x = 12 or x = 1
2
2
2x + 7x + 3 = 0
d)
p + 7p – 30 = 0
(2x + 1)(x + 3) = 0
(p + 10)(p – 3) = 0
∴ 2x + 1 = 0 or x + 3 = 0 ∴ p + 10 = 0 or p – 3 = 0
∴ 2x = –1 or x = –3 ∴ p = –10 or p = 3
x = – 12
(x – 6)(x – 4) = 0
x – 6 = 0 or x – 4 = 0
∴ x = 6 or x = 4
12b2 – 75 = 0
3(4b2 – 25) = 0
4b2 – 25 = 0
(2b – 5)(2b + 5) = 0
∴ 2b – 5 = 0 or 2b + 5 = 0
2b = 5
or 2b = –5
5
2
b=
b = 2,5
f)
b)
b = – 52
b = –2,5
2a2 + 6a + 14 = 0
a2 + 8a + 7 = 0
(a + 7)(a + 1) = 0
∴ a + 7 = 0 or a + 1 = 0
∴ a = –7 or a = –1
g)
x2 +10x = –25
x2 + 10x + 25 = 0
(x + 5)(x + 5) = 0
∴ x + 5 = 0 or x + 5 = 0
∴ x = –5
40
Module 3 • Factorisation, HCF, LCM and algebraic fractions
h)
s2 = 3s + 28
s2 – 3s – 28 = 0
(s – 7)(s + 4) = 0
∴ s – 7 = 0 or s + 4 = 0
∴ s = 7 or s = –4
j)
3x2 + 45 = 24x
3x2 – 24x + 45 = 0
x2 – 8x + 15 = 0
(x – 5)(x – 3) = 0
∴ x – 5 = 0 or x – 3 = 0
∴ x = 5 or x = 3
2.
x2 – 16x + 14 = 0
a)
x=
=
=
=
=
=
i)
x2 = 36
x2 – 36 = 0
(x – 6)(x + 6) = 0
∴ x – 6 = 0 or x + 6 = 0
∴ x = 6 or x = –6
−b ± b 2 − 4ac
2a
−(−16) ± (−16)2 − 4(1)(14)
2(1)
16 ± 256 − 56
2
16 ± 200
2
16 ± 14,142
2
16 + 14,142 or 16 − 14,142
2
2
= 15,071
b) x(x – 9) –2 = 0
x2 – 9x – 2 = 0
or 0,929
2
−b ± b − 4ac
2a
−(−9) ± (−9)2 − 4(1)(−2)
2(1)
x=
=
=
=
=
=
= 9,217 or –0,217
9 ± 81 + 8
2
9 ± 89
2
9 ± 9,443
2
9 + 9,443 or 9 − 9,443
2
2
c)
9x2 + 3x + 1 = 0
x=
=
=
=
−b ± b 2 − 4ac
2a
−(3) ± (3)2 − 4(9)(1)
2(9)
−3 ± 9 − 36
18
−3 ± −27
18
(imaginary roots)
N2 Mathematics Lecturer Guide|Hands-On
d)
2x2 + 9x = 3
2x2 + 9x – 3 = 0
x=
=
=
=
=
=
e)
=
=
=
x=
x=
−9 ± 81 − 24
4
−9 ± 57
4
−9 ± 7,55
4
−9 + 7,55 or −9 − 7,55
4
4
= –0,363 or –4,137
– 8 = –3x
2
6x + 3x – 8 = 0
x=
f)
−b ± b 2 − 4ac
2a
−(9) ± (9)2 − 4(2)(3)
2(2)
6x2
41
−b ± b 2 − 4ac
2a
−(3) ± (3)2 − 4(6)(−8)
2(6)
−3 ± 9 + 192
12
−3 ± 201
12
−3 ± 14,177
12
−3 + 14,177 or
12
x = 0,931
16x2 – 9 = 0
x=
=
x=
−3 − 14,177
12
or x = –1,431
2
−b ± b − 4ac
2a
g)
3x2 – 3x – 6 = 0
x2 – x – 2 = 0
−(0) ± (0)2 − 4(16)(−9)
2(16)
x=
0 ± 576
32
±24
32
+24 or x = −24
32
32
3
or x = – 43
4
=
=
=
=
∴x=
=
=
∴x=
x=
x = 0,75 or x = –0,75
−b ± b 2 − 4ac
2a
−(−1) ± (−1)2 − 4(1)(−2)
2(1)
1 ± 1+ 8
2
1± 9
2
1± 3
2
1 + 3 or x
2
x=2
=
1−3
2
or x = –1
42
Module 3 • Factorisation, HCF, LCM and algebraic fractions
h)
7x2 = 2x + 6
7x2 – 2x – 6 = 0
x=
=
=
=
=
∴x=
−b ± b 2 − 4ac
2a
−(−2) ± (−2)2 − 4(7)(−6)
2(7)
2 ± 4 + 168
14
2 ± 172
14
2 ± 13,115
14
2 + 13,115 or 2 − 13,115
14
14
x = 1,08
or x = –0,794
Activity 3.3
1.
x – y = 1 ..................... 1
x + 2y = –5.................... 2
1 – 2 –3y = 6
y = −63
2.
y = –2
Substitute into 1:
x – (–2) = 1
x+2=1
x=1–2
x = –1
∴ x = –1 and y = –2 or (–1; –2)
3.
y + x = 5.................. 1
3y + 2x = 13............... 2
1 × 2: 2y + 2x = 10............... 3
2: 3y + 2x = 13............... 2
3 – 2: –y = –3
y = −−31
y=3
Substitute into 1:
3+x=5
x=5–3
x=2
∴ (2; 3) or x = 2 and y = 3
3x + 4y = 3.......................... 1
–2x – 4y = 2.......................... 2
1+2x=5
Substitute into 1:
3(5) + 4y = 3
15 + 4y = 3
4y = 3 – 15
4y = –12
y=
−12
4
y = –3
∴ x = 5 and y = –3 or (5; –3)
4.
3x – 4y = 4................. 1
–2x + 8y = 2................. 2
1 × 2: 6x – 8y = 8................. 3
2: –2x + 8y = 2................. 2
3 + 2: 4x = 10
x=
10
4
x = 2,50
Substitute into 1:
3(2,50) – 4y = 4
7,50 – 4y = 4
–4y = 4 – 7,50
–4y = –3,50
y=
−3,50
−4
= 0,875
x = 2,5 and y = 0,875
N2 Mathematics Lecturer Guide|Hands-On
5.
2a – 3b = –9............... 1
3a – 4b = –11............. 2
1 × 3: 6a – 9b = –27............. 3
2 × 2: 6a – 8b = –22............. 4
3 – 4: –b = –5
b = −−51
6.
3x + 3y = 15........... 1
8x + 5y = 13........... 2
1 × 5: 15x + 15y = 75........... 3
2 × 3: 24x + 15y = 39........... 4
3 – 4: –9x = 36
x = −369
= –4
Substitute into 1:
3(–4) + 3y = 15
–12 + 3y = 15
3y = 15 + 12
3y = 27
y=9
∴ x = –4 and y = 9 or (–4; 9)
4x + y = 0........................ 1
–2x – y = 2........................ 2
1 + 2: 2x = 2
x=1
Substitute into 1:
4(1) + y = 0
4+y=0
y=0–4
y = –4
∴ x = 1 and y = –4 or (1; –4)
8.
8a + 3b = 7............... 1
20a + 9b = 13............. 2
1 × 3: 24a + 9b = 21............. 3
20a + 9b = 13............. 2
3 – 2: 4a = 8
=5
Substitute into 1:
2a – 3(5) = –9
2a – 15 = –9
2a = –9 + 15
2a = 6
a = 62
=3
∴ a = 3 and b = 5 or (3; 5)
7.
a=
8
4
a=2
Substitute into 1:
8(2) + 3b = 7
16 + 3b = 7
3b = 7 – 16
3b = –9
b=
−9
3
b = –3
∴ a = 2 and b = –3 or (2; –3)
43
44
Module 3 • Factorisation, HCF, LCM and algebraic fractions
Activity 3.4
1.
Let the first number be x and the second number be y.
4x + 3y = 22.................................. 1
2(x – y) = 4 ................................... 2
1: 4x + 3y = 22.................................. 1
2: 2x – 2y = 4 ................................... 2
4x + 3y = 22.................................. 1
×
2:
4x – 4y = 8 ................................... 3
2
1 – 3: 7y = 14
y=2
Substitute into 1:
4x + 3(2) = 22
4x + 6 = 22
4x = 22 – 6
4x = 16
∴ x = 16
=4
4
The numbers are 4 and 2.
2.
Let Thabo be x years old and his sister y years old.
x + y = 24 . ................................................. 1
Three years ago Thabo was x – 3 years old and his sister was y – 3 years old.
∴ x – 3 = 2(y – 3)
x – 3 = 2y – 6
x – 2y = –6 + 3
x – 2y = –3........................................ 2
x + y = 24 ....................................... 1
x – 2y = –3........................................ 2
–
1 2: 3y = 27
y=9
Substitute into 1:
x + 9 = 24
x = 24 – 9
x = 15
Thabo is 15 years old and his sister is 9 years old.
N2 Mathematics Lecturer Guide|Hands-On
3.
Let the price of an apple be x and the price of a banana be y.
∴ 3x + 5y = 24,10 . ...................... 1
5x + 3y = 27,90 . ...................... 2
×
5:
15x
+ 25y = 120,50 ..................... 3
1
2 × 3: 15x + 9y = 83,70 ....................... 4
3 – 4: 16y = 36,80
y = 36,80
16
y = R2,30
Substitute into 1:
3x + 5(2,30) = 24,10
3x + 11,50 = 24,10
3x = 24,10 – 11,50
3x = 12,60
x = 12,60
3
x = 4,20
An apple costs R4,20 and a banana costs R2,30.
4.
Let the number of adults be x and the number of children be y.
∴ x + y = 1 000 . .................... 1
95x + 30y = 74 200 . .................. 2
1 × 30: 30x + 30y = 30 000 ................... 3
95x + 30y = 74 200 . .................. 2
–
3 2: –65x = –44 200
x=
−44 200
−65
= 680
680 adults attended the concert
5.
Let the speed Sipho was rowing be x km/h and the speed of the current be
y km/h.
When he rows downstream, he rows at (x + y) km/h.
Distance = speed × time
∴ (x + y)3 = 39
3x + 3y = 39
or x + y = 13
• Divide both sides by 3.
45
46
Module 3 • Factorisation, HCF, LCM and algebraic fractions
When Thabo rows upstream, he rows at (x – y) km/h.
∴ (x – y)3 = 9
3x – 3y = 9
or x – y = 3
• Divide both sides by 3.
x + y = 13........................................ 1
x – y = 3.......................................... 2
1 + 2: 2x = 16
x = 8 km/h
6.
Substitute into 1:
8 + y = 13
y = 13 – 8
= 5 km/h
Sipho rows at 8 km/h. The speed of the current is 5 km/h.
Let the athlete’s running speed be x km/h and his walking speed be y km/h.
∴ 3x + y = 39.................................. 1
4x + 2y = 56.................................. 2
×
2:
6x
+ 2y = 78.................................. 3
1
4x + 2y = 56.................................. 2
3 – 2: 2x = 22
x = 11 km/h
Substitute into 1:
3(11) + y = 39
33 + y = 39
y = 39 – 33
= 6 km/h
The athlete runs at 11 km/h and walks at 6 km/h.
7.
Let the price of the door be Rx and the price of the window be Ry.
7x + 12y = 12 100 ..................... 1
10x + 18y = 17 740 ..................... 2
1 × 3: 21x + 36y = 36 300 ..................... 3
2 × 2: 20x + 36y = 35 480 ..................... 4
3 – 4: x = 820
Substitute into 1:
7(820) + 12y = 12 100
5 740 + 12y = 12 100
12y = 12 100 – 5 740
12y = 6 360
y=
6 360
12
= 530
The door costs R820 and the window costs R530.
N2 Mathematics Lecturer Guide|Hands-On
8.
A
B
x km/h
distance: 2,5x
y km/h
distance: 2,5y
∴ 2,5x + 2,5y = 125 km .......................... 1
or x + y = 50:
A
x km/h = 3x distance
B
y km/h = 3y distance
3x – 3y = 18 ....................................... 2
or x – y = 6 .............................................
∴ x + y = 50 ....................................... 1
x – y = 6 ......................................... 2
+
1 2: 2x = 56
x = 28
Substitute into 1:
28 + y = 50
y = 50 – 28
= 22
One cycles at 28 km/h, and the other cycles at 22 km/h.
Activity 3.5
E = 12 mV2 ................. (V)
1.
2.
2E = mV2
mV2 = 2E
V2 =
D=
x2
h=
2E
m
V=±
3.
πr(1 + rh) = A ................. (h)
πr + πr2h = A
πr2h = A – πr
A − πr
πr 2
2E
m
x2
4h
x2
+ h . .......... (x)
1
r1
+
1
r2
=
1
R
................. (R)
r2R + r1R = r1r2
R(r2 + r1) = r1r2
4h2
4hD = +
+ 4h2 = 4hD
x2 = 4hD – 4h2
x = ± 4hD − 4h
4.
2
R=
r1r2
r2 + r1
47
48
Module 3 • Factorisation, HCF, LCM and algebraic fractions
5.
C
5
=
F − 32
9 . ....................... (F)
2s = 2ut + at2
at2 + 2ut = 2s
at2 = 2s – 2ut
9C = 5(F – 32)
9C
5
= F – 32
F=
7.
9C
5
+ 32
l
g
T = 2π
T
2π
2
a=
................... (g)
8.
l
g
=
1 2πT 2
=
l
g
T2
4π 2
=
l
g
s = ut + 12 at2........ (a)
6.
2
2
WL = I − R . (R)
(WL)2 = I2 – R2
2
R + (WL)2 = I2
R2 = I2 – (WL)2
R = ± I2 − (WL)2
or R = ± I2 − W 2L2
gT2 = 4π2l
g=
9.
4π 2l
T2
5w(T − w)
12a
D=
D2 =
.......... (T)
10.
5w(T − w)
12a
12aD2 = 5w(T – w)
12aD2
5w
T=
F=
t=
12aD2
5w
12aD2
5w
V=
13.
R − ri
rx
+w
mV 2
gr . ..................... (V)
grF = mV2
mV2 = grF
V2 =
R = r(i + xt) ............ (t)
R = ri + rxt
rxt + ri = R
rxt = R – ri
=T–w
T–w=
11.
2s − 2ut
t2
grF
m
t=
12.
t2 = b2 – 4ac
t2 + 4ac = b2
4ac = b2 – t2
a=
grF
m
S=
n
2
[2a + (n – 1)d)............ (d)
2S = n[2a + (n – 1)d]
2S = 2an + n(n – 1)d
2an + n(n – 1)d = 2S
n(n – 1)d = 2S – 2an
d=
2S − 2an
n(n − 1)
b 2 − 4ac . ...... (a)
b2 − t 2
4c
N2 Mathematics Lecturer Guide|Hands-On
F=
14.
1
........... (C)
2π LC
1
LC
2πF =
(2πF)2 =
x2 =
s(s − a)
bc
bcx2 = s(s – a)
bcx2 = s2 – sa
bcx2 + sa = s2
sa = s2 – bcx2
1
LC
LC(2πF)2 = 1
C=
s(s − a) ........... (a)
bc
x=
15.
1
L(2πF)2
a=
s 2 − bcx 2
s
Activity 3.6
1.
x2
a2
+
y2
b2
2.
=1
b2x2 + a2y2 = a2b2
a2y2 = a2b2 – b2x2
y2 =
a2b 2 − b 2x 2
a2
y=
a2b 2 − b 2x 2
a2
=±
(−2)2(5)2 − (5)2(1)2
(−2)2
=±
4(25) − (25)
4
5t = 4x − p
(5t)2 = 4x – p
4x – p = 25t2
4x = 25t2 + p
x=
25t 2 + p
4
x=
25(3)2 + (4)
4
= 57,25
= ±4,33
3.
I=
nE
R + nr
I(R + nr) = nE
IR + Inr = nE
Inr – nE = –IR
n(Ir – E) = –IR
n=
−IR
Ir − E
n=
−2,3(7,2)
(2,3)(4,6) − 20
= 1,758
4.
V=
V2 =
g(wH − mN)
2
g(wH − mN)
2
2V2 = g(wH – mN)
g=
=
2V 2
wH − mN
2(10,1)2
(3,2)(4,6) − (2,5)(1,4)
= 18,184
49
50
Module 3 • Factorisation, HCF, LCM and algebraic fractions
P = IV +
5.
V2
R
V2 2
P − IV
2
6.
2
2
T
L = 1 2π
2 –g
=R
2
= 1 34,6
2π 2 – 12
2
2
R = 1 42,3 −10,2
(5,6)(10,2) 2
= 18,324
= 49,284
H=
7.
T
2π
T
L + g = 1 2π
2
R
=
2π L + g = T
L+ g =
= P – IV
V2
P − IV
1
V2
R
a2 + h2
4
H2 =
a2
4
H2 – h2 =
a2
4
8.
2πF =
+ h2
(2πF)2 =
4(H2 – h2) = a2
a=
F=
LC =
4(H2 − h 2)
or
2
=2 H −h
2
a = 2 12,62 − 3,4 2
= 24,265
1
2π LC
1
LC
1
LC
1
(2πF)2
C=
1
L(2πF)2
C=
1
1,64(2π1,37)2
= 0,008
N2 Mathematics Lecturer Guide|Hands-On
51
Module 3: Summative assessment answers
Question 1
1.1
y + 2(y – 2) – 3(5 – y) = 4y – 5
y + 2y – 4 – 15 + 3y = 4y – 5
6y – 19 = 4y – 5
6y – 4y = –5 + 19
2y = 14
y = 7
1.2 2x –
1
4 (3x
2x –
3x
4
– 4) =
1
2x
(3)
–5
1
+ 1 = 2x – 5
8x – 3x + 4 = 2x – 20
5x + 4 = 2x – 20
5x – 2x = –20 – 4
3x = –24
x=
−24
3
= –8
(3)
[6]
Question 2
2.1
6x2 – 13x + 5 = 0
(3x – 5)(2x – 1) = 0
3x – 5 = 0 or 2x – 1 = 0
3x = 5 or 2x = 1
x = 53
x = 12 2.2
6x2 + 16x = 6
6x2 + 16x – 6 = 0
3x2 + 8x – 3 = 0
(3x – 1)(x + 3) = 0
3x – 1 = 0 or x + 3 = 0
3x = 1
x = –3
x = 13 (3)
(4)
2.3 (2x – 1)(x + 4) = –7
2x2 + 8x – x – 4 = –7
2x2 + 7x + 3 = 0
(2x + 1)(x + 3) = 0
2x + 1 = 0 or x + 3 = 0
2x = –1
x = –3
x = – 12 (4)
52
Module 3 • Factorisation, HCF, LCM and algebraic fractions
2.4 5x2 – 3x – 2 = 0
x=
−b ± b 2 − 4ac
2a
=
3 ± (−3)2 − 4(5)(−2)
2(5)
=
3 ± 9 + 40
10
3 ± 49
10
3±7
10
10
−4
10 or 10
=
=
=
= 1 or –0,4
2.5
12x2
(4)
12x2 – 13x = 10
– 13x – 10 = 0
x=
=
=
=
−b ± b 2 − 4ac
2a
−(−13) ± (−13)2 − 4(12)(−10)
2(12)
13 ± 169 + 480
24
13 ± 649
24
13 ± 25,047
24
=
= 1,603 or –0,52
(4)
[23
Question 3
3.1
3.2
x + y = 9 . .............................
x – 2y = –3 ............................
1 – 2: 3y = 12
y=4
Substitute into 1:
x+4=9
x = 5
1
2
3y – 6x = –3 ..................
–2y + 2x = –2 ..................
1: 3y – 6x = –3 ..................
2 × 3 –6y + 6x = –6 ..................
1 + 3: –3y = –9
y=3
Substitute into 1:
3(3) – 6x = –3
9 – 6x = –3
–6x = –12
x = 2
1
2
1
3
(3)
(4)
[7]
N2 Mathematics Lecturer Guide|Hands-On
53
Question 4
Let petrol cost x and let oil cost y.
∴ 9x + 3y = R186,30 ........................... 1
14x + 5y = R297,30 ........................... 2
1 × 5: 45x + 15y = R931,50 ........................... 3
2 × 3: 42x + 15y = R891,90 ........................... 4
3 – 4: 3x = R39,60
x = R13,20
Substitute into 1:
9(R13,20) + 3y = R186,30
R118,80 + 3y = R186,30
3y = R67,50
y = R22,50
4.1 Petrol costs R13,20/ℓ.
4.2 Oil costs R22,50/can.
[4]
Question 5
r=
R(E − V)
V
rV = R(E – V)
rV = RE – RV
RV + rV = RE
V(R + r) = RE
V=
RE
R+r
[4]
Question 6
I=
6.1
E
R+r
I(R + r) = E
IR + Ir = E
Ir = E – IR
r=
E − IR
I
(4)
6.2 r = E −I IR
=
38 − (1,4)(36)
1,4
= –8,857
(2)
[6]
Total [50]
MODULE
4 Algebraic graphs
Activity 4.1
1. a)
c)
2.
f(x) = –x – 3
f(–2) = –(–2) – 3 = 2 – 3 = –1
f(a) = –a – 3
b)
f(0) = 0 – 3 = –3
d)
f(x – 2) = –(x – 2) –3
= –x + 2 – 3
= –x – 1
f(x) = 23 x – 1
a)
f(–2) = 23 (–2) – 1 = – 43 – 1 = – 73 = –2 13
f(–1) = 23 (–1) – 1 = – 23 – 1 = – 53 = –1 23
f(0) = 23 (0) – 1 = –1
f(1) = 23 (1) – 1 =
f(2) = 23 (2) – 1 =
b)
x
f(x)
c)
2
3
4
3
– 1 = – 13
–1=
1
3
–2
–1
0
1
2
–2 13
–1 23
–1
– 13
1
3
(–2; –2 13 ) (–1; –1 23 ) (0; –1) (1; – 13 ) (2; 13 )
y
1
d)
y =23 x – 1
–2
–1
0
–1
–2
–3
e)
x = 0 where y = –1
1
2 x
55
N2 Mathematics Lecturer Guide|Hands-On
3.
a)
2y = –x + 6
∴ y = – 12 x + 3
f(x) = – 12 x + 3
f(–3) = – 12 (–3) + 3 =
f(–2) = – 12 (–2) + 3 = 1 + 3 = 4
f(–1) = – 12 (–1) + 3 =
f(0) = – 12 (0) + 3 = 3
f(1) = – 12 (1) + 3 = – 12 + 3 = 2 12
f(2) = – 12 (2) + 3 = –1 + 3 = 2
f(3) = – 12 (3) + 3 = – 32 + 3 = 1 12
x
f(x)
3
2
1
2
+ 3 = 4 12
+ 3 = 3 12
–3
–2
–1
0
1
2
3
4 12
4
3 12
3
2 12
2
1 12
y
5
4
3
y = – 12 x + 3
2
1
–3
–2
–1
0
–1
b)
y = 2 when x = 2
1
2
3
x
56
Module 4 • Algebraic graphs
4.
y=x
f(–1) = –1
f(0) = 0
f(1) = 1
y = –x
f(–1) = –(–1) – 1
f(0) = 0
f(1) = –1
x
y=x
y = –x
–1
–1
1
0
0
0
1
1
–1
y
3
y = –x
y=x
2
1
–3
–2
0
–1
1
2
3
x
–1
–2
–3
5.
y = 3x
f(–2) = 3(–2) = –6
f(–1) = 3(–1) = –3
f(0) = 0
f(1) = 3(1) = 3
f(2) = 3(2) = 6
x
y = 3x
y = –2x –1
y = –2x – 1
f(–2) = –2(–2) – 1 = 4 – 1 = 3
f(–1) = –2(–1) – 1 = 2 – 1 = 1
f(0) = –2(0) – 1 = –1
f(1) = –2(1) – 1 = –2 – 1 = –3
f(2) = –2(2) – 1 = –4 – 1 = –5
–2
–6
3
–1
–3
1
0
0
–1
1
3
–3
2
6
–5
N2 Mathematics Lecturer Guide|Hands-On
y
6
y = –2x – 1
y = 3x
5
4
3
2
1
–3
–2
–1
0
–1
–2
–3
–4
–5
–6
Activity 4.2
1.
2.
a)
m = – 92 ; c = –2
b)
m = 2; c = 6
c)
m = – 73 ; c = 7
d)
m = 0; c = –4
e)
m = – 42 = – 12 ; c = 0
f)
m = 52 ; c = 0
m=
y 2 − y1
x 2 − x1
=
−3 − 3
1 − (−2)
=
−6
3
= –2
1
2
3 x
57
Module 4 • Algebraic graphs
3.
m=
=
4.
5.
2 − (−2)
3 − (−2)
4
5
a)
m=
3
2
b)
3
c)
x = –2
d)
Straight line
e)
y = 32 x + 3
a)
(4; 0)
b)
(0; 2)
c)
m = – 42 = – 12
d)
y = – 12 x + 2
Activity 4.3
Gradient
y-intercept
a)
2
0
b)
2
– 23
c)
undefined
no y-intercept
d)
1
2
3
4
e)
0
3
f)
8
16
g)
–1
0
h)
1
–6
i)
– 23
4
3
j)
– 13
None
2.
y
–4
–3
–1
1
x=4
x=0
–2
x = 1 13
1.
x = –3
58
2
3
4 x
N2 Mathematics Lecturer Guide|Hands-On
3.
y
y=3
3
2
1
y=0
y=
–1
x
– 12
–2
–3
–4
y = –5
–5
Activity 4.4
1.
y
a)
4
3
2
y = –2x + 3
1
–2
–1
0
1
3 x
2
–1
–2
y
b)
y
c)
2
2
y=x
1
–2
–1
0
1
2
1
x
–2
–1
0
–1
–1
–2
–2
1
2
x
y = –x
59
60
Module 4 • Algebraic graphs
d)
y
y = – 41x + 3
4
3
2
1
–4
–3
–2
–1
0
1
2
3
–1
y
e)
3
y = 12 x +
3
4
2
1
–3
–2
–1
3
4
0
1
–1
y
f)
5
4
3
2
1
–2
–1
0
–1
–2
–3
–4
–5
1
2 x
2
3
x
4 x
N2 Mathematics Lecturer Guide|Hands-On
2.
y = –3x + 1
a) –3
b) 1
y
c)
y = –3x + 1
3
2
1
–3
–2
–1
0
1
2
3
x
–1
–2
–3
d)
(0; 1)
e)
1
3
1;
3
1
f)
02
Activity 4.5
1.
y + 4x = 2
x-intercept: Let y = 0
0 + 4x = 2
∴ x = 12
y-intercept: Let x = 0
y + 4(0) = 2
∴y=2
y
3
2
1
–2
–1
0
1
2
1
2
3
–1
–2
y = –4x + 2
x
61
62
Module 4 • Algebraic graphs
2.
y = – 13 x + 2
x + 2y = 3
x-intercept:
x-intercept:
– 13 x
x + 2(0) = 3
∴x=3
0=
1x
3
+2
=2
∴x=6
y-intercept:
y = – 13 (0) + 2
∴y=2
y-intercept:
0 + 2y = 3
∴y=
3
2
or 1 12
y
3
2
y = – 13x + 2
1
–4
–3
–2
0
–1
1
2
3
4
5
6
–1
–2
–3
3.
x
2
–
y
4
=1
x-intercept:
x
2
=1
∴x=2
y-intercept: – 4y = 1
∴ y = –4
∴ (2; 0) and (0; –4)
∴ C is the correct answer
4.
x = –2y + 4
x-intercept: x = 4
y-intercept: 2y = 4
∴y=2
y = – 12x +
3
2
x
N2 Mathematics Lecturer Guide|Hands-On
y
3
2
y = – 12x + 2
1
–4
–3
–2
–1
0
1
2
3
4
x
–1
–2
–3
5.
y = –3x
x-intercept: 0 = –3x
∴x=0
y-intercept: y = 0
y
(–1)
3
2
(+3)
1
–2
0
–1
1
2
x
–1
(–3)
–2
–3
y = –3x
(+)
Activity 4.6
1. a)
y – y1 = m(x – x1)
y – (–2) = 3[x – (–3)]
y + 2 = 3[x + 3]
y + 2 = 3x + 9
y = 3x + 7
•
(–3; –2)
(x1; y1)
63
Module 4 • Algebraic graphs
b)
x-intercept:
0 = 3x + 7
3x = –7
x = –2 13
y
7
6
y-intercept:
y=7
5
4
y=3
x+7
64
3
2
1
–4
–3
–2
–1
0
–1
–2
2.
x
−3
+
y
6
=1
x-intercept:
x
−3
=1
∴ x = –3
y-intercept:
y
6
=1
∴y=6
y
6
x
y
4
–3+ 6 = 1
2
–3
–2
–1
0
–2
3.
a)
y – y1 = m(x – x1)
y –(–2) = – 12 (x – 2)
y + 2 = – 12 x + 1
y = – 12 x – 1
1
2
3
x
1
2 x
N2 Mathematics Lecturer Guide|Hands-On
y
b)
–2
0
–1
1
–1
2
x
y = –12 x – 1
–2
4. a)
2y = x + 4
x-intercept: 2(0) = x + 4
x = –4
y-intercept: 2y = 4
y=2
y – 3x – 2 = 0
x-intercept: 0 – 3x – 2 = 0
3x = –2
x = – 23
y-intercept: y – 3(0) – 2 = 0
y=2
y
3
2
2y = x + 4
1
–4
–3
–2
–1
0
1
2
1
2
x
–1
y – 3x – 2 = 0
–2
–3
y
1
5.
–4
–3
–2
–1
0
3
–1
–2
–3
–4
x
4
+ −4y = 1
4
x
65
66
Module 4 • Algebraic graphs
Activity 4.7
1. a)
y = x2 – 9
f(–3) = (–3)2 – 9 = 0
f(–2) = (–2)2 – 9 = –5
f(–1) = (–1)2 – 9 = –8
f(0) = (0)2 – 9 = –9
f(1) = (1)2 – 9 = –8
f(2) = (2)2 – 9 = –5
f(3) = (3)2 – 9 = 0
x
y
–3
0
–2
–5
–1
–8
0
–9
1
–8
y
1
–4
–3
–2
–1
0
1
2
3
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
y = x2 – 9
4 x
2
–5
3
0
N2 Mathematics Lecturer Guide|Hands-On
b)
y = –x2
f(–3) = –(–3)2 = –9
f(–2) = –(–2)2 = –4
f(–1) = –(–1)2 = –1
f(0) = –(0)2 = 0
f(1) = –(1)2 = –1
f(2) = –(2)2 = –4
f(3) = –(3)2 = –9
x
y
–3
–9
–2
–4
–1
–1
0
0
1
–1
y
1
–4
–3
–2
–1
0
1
2
3
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
y = –x2
4 x
2
–4
3
–9
67
68
Module 4 • Algebraic graphs
c)
y = x2 – 2x – 3
f(–2) = (–2)2 – 2(–2) – 3 = 4 + 4 – 3 = 5
f(–1) = (–1)2 – 2(–1) – 3 = 1 + 2 – 3 = 0
f(0) = –3
f(1) = (1)2 – 2(1) – 3 = 1 – 2 – 3 = –4
f(2) = (2)2 – 2(2) – 3 = 4 – 4 – 3 = –3
f(3) = (3)2 – 2(3) – 3 = 9 – 6 – 3 = 0
f(4) = (4)2 – 2(4) – 3 = 16 – 8 – 3 = 5
x
y
–2
5
–1
0
0
–3
1
–4
2
–3
y
5
y = x2 – 2x – 3
4
3
2
1
–3
–2
–1
0
1
2
3
4
–1
–2
–3
–4
• x = 0 and x = 2 where y = –3
2. a)
f(x) = –x2 + 2x + 3
f(–2) = –(–2)2 + 2(–2) + 3 = –4 – 4 + 3 = –5
f(–1) = –(–1)2 + 2(–1) + 3 = –1 – 2 + 3 = 0
f(0) = –(0)2 + 2(0) + 3 = 3
f(1) = –(1)2 + 2(1) + 3 = –1 + 2 + 3 = 4
f(2) = –(2)2 + 2(2) + 3 = –4 + 4 + 3 = 3
f(3) = –(3)2 + 2(3) + 3 = –9 + 6 + 3 = 0
f(4) = –(4)2 + 2(4) + 3 = –16 + 8 + 3 = –5
x
3
0
4
5
N2 Mathematics Lecturer Guide|Hands-On
x
f(x)
–2
–5
–1
0
0
3
1
4
2
3
3
0
4
–5
y
4
3
2
1
–3
–2
0
–1
1
2
3
5 x
4
–1
–2
–3
y = –x2 + 2x + 3
–4
–5
–6
b)
3.
x = –1 and x = 3
2y = x2
a)
y = 12 x2
f(x) = 12 x2
f(–2) = 12 (–2)2 = 2
f(–1) = 12 (–1)2 =
f(0) = 12 (0)2 = 0
f(1) = 12 (1)2 =
f(2) = 12 (4) = 2
1
2
1
2
x
–2
–1
0
1
2
y
2
1
2
0
1
2
2
69
70
Module 4 • Algebraic graphs
y
3
y = 12 x2
2
1
–3
–2
0
–1
1
x
2
–1
• y = 2 where x = –2
b)
y = x2 + 4x + 3
f(x) = x2 + 4x + 3
f(–4) = (–4)2 + 4(–4) + 3 = 16 – 16 + 3 = 3
f(–3) = (–3)2 + 4(–3) + 3 = 9 – 12 + 3 = 0
f(–2) = (–2)2 + 4(–2) + 3 = 4 – 8 + 3 = –1
f(–1) = (–1)2 + 4(–1) + 3 = 1 – 4 + 3 = 0
f(0) = (0) + 4(0) + 3 = 3
f(1) = (1)2 + 4(1) + 3 = 1 + 4 + 3 = 8
x
y
–4
3
–3
0
–2
–1
–1
0
y
y = x2 + 4x + 3
8
7
6
5
4
3
2
1
–5
–4
–3
–2
–1
0
–1
• x = –2 where y = – 1
1
2
3 x
0
3
1
8
71
N2 Mathematics Lecturer Guide|Hands-On
c)
2y + x2 + 4 = 0
2y = –x2 – 4
y = – 12 x2 – 2
f(–3) = – 12 (–3)2 – 2 = – 12 (9) – 2 = –6 12
f(–2) = – 12 (–2)2 – 2 = – 12 (4) – 2 = –4
f(–1) = – 12 (–1)2 – 2 = –2 12
f(0) = – 12 (0)2 – 2 = –2
f(1) = – 12 (1)2 – 2 = –2 12
f(2) = – 12 (2)2 – 2 = –4
f(3) = – 12 (3)2 – 2 = –6 12
x
–3
–2
–1
0
1
2
3
y
–6 12
–4
–2 12
–2
–2 12
–4
–6 12
y
2
1
–4
–3
–2
0
–1
1
2
3
4 x
–1
–2
–3
–4
–5
–6
–7
•
y = –2 where x = 0
y = – 12 x2 – 2
72
Module 4 • Algebraic graphs
d)
y = –(x – 1)2
y = –(x – 1) (x – 1)
y = –(x2 – 2x + 1)
y = –x2 + 2x – 1
f(x) = –x2 + 2x – 1
f(–2) = –(–2)2 + 2(–2) – 1 = –4 – 4 – 1 = –9
f(–1) = –(–1)2 + 2(–1) – 1 = –1 – 2 – 1 = –4
f(0) = –(0)2 + 2(0) – 1 = –1
f(1) = –(1)2 + 2(1) – 1 = –1 + 2 – 1 = 0
f(2) = –(2)2 + 2(2) – 1 = –4 + 4 – 1 = –1
f(3) = –(3)2 + 2(3) – 1 = –9 + 6 – 1 = –4
x
y
–2
–9
–1
–4
0
–1
1
0
2
–1
y
2
1
–4
–3
–2
–1
0
1
2
3
–1
–2
–3
–4
–5
–6
–7
–8
–9
• y = –4 where x = 3
y = –x2 + 2x – 1
4
x
3
–4
N2 Mathematics Lecturer Guide|Hands-On
Activity 4.8
1. a)
c)
2.
(i)
(ii)
(iii)
(iv)
(i)
(ii)
(iii)
(iv)
(–3; 0) and (3; 0)
(0; 9)
x=0
(0; 9)
None
(0; 4)
x=0
(0; 4)
y = (x – 4)2
y = (x – 4)(x – 4)
y = x2 – 8x + 16
Axis of symmetry: x = – 2ba
)
= – (2−8
(1)
=
8
2
∴x= 4
Turning point: y = x2 – 8x + 16
y = (4)2 – 8(4) + 16
y = 16 – 32 + 16
y=0
∴ TP: (4; 0)
f(x) = –x2 + 4
a) x-intercept: y = 0
0 = –x2 + 4
x2 = 4
∴ x = ±2
3.
b)
c)
y-intercept: x = 0
y = –(0)2 + 4
∴y=4
x = – 2ba
= – 2(−01)
∴x=0
d)
TP: y = –(0)2 + 4
y=4
∴ TP: (0; 4)
(i)
(ii)
(iii)
(iv)
d) (i)
(ii)
(iii)
(iv)
b)
(–3; 0)(1; 0)
(0; –6)
x = –1
(–1; –8)
(–2; 0)
(0; 6)
x = –2
(–2; 0)
73
74
Module 4 • Algebraic graphs
e)
y
5
4
3
f(x) = –x2 + 4
2
1
–4
–3
–2
0
–1
–1
–2
–3
–4
f)
x = –2 and x = 2
y = x2 – x – 1
a) roots: y = 0
0 = x2 – x – 1
4.
−b ± b 2 − 4ac
2a
x=
=
−(−1) ± (−1)2 − 4(1)(−1)
2(1)
=
1± 1+ 4
2
=
1± 5
2
∴x=
1+ 5
2
b)
= 1,6
y = –1
or x =
or
1− 5
2
= –0,6
1
2
3
4 x
N2 Mathematics Lecturer Guide|Hands-On
c)
x=
−b
2a
=
−(−1)
2(1)
=
1
2
d)
TP: y = x2 – x – 1
2
= 1 12 2 – 1 12 2 – 1
=
=
=
=
∴ TP: 1
1 – 1 –
4
2
1− 2− 4
4
5
–4
–1 14
1 ; –1 1
2
4
1
2
e)
y
4
3
2
1
–3
–2
–1
0
–1
–2
–3
f)
y = –1 14
g)
x = –0,6 and x = 1,6
h)
y = –1
1
2
( 12 ; – 1 14 )
3
x
75
76
Module 4 • Algebraic graphs
y + x2 – 2x – 3 = 0
y = –x2 + 2x + 3
a) x-intercept: y = 0
0 = –x2 + 2x + 3
0 = x2 – 2x – 3
0 = (x – 3)(x + 1)
∴ x = 3 or x = –1
5.
b)
c)
d)
y-intercept: x = 0
y=3
x = – 2ba
=
− (2)
2( − 1)
∴x=1
TP: y = –x2 + 2x + 3
= –(1)2 + 2(1) + 3
= –1 + 2 + 3
=4
∴ TP: (1; 4)
e)
y
5
4
3
2
1
–4
–3
–2
–1
0
1
2
3
–1
–2
–3
f)
g)
h)
(–1; 0) and (3; 0)
(0; 3)
Maximum
y = –x2 + 2x + 3
4 x
N2 Mathematics Lecturer Guide|Hands-On
6.
y = x2 – 4x – 2
a)
x-intercepts: y = 0
0 = x2 – 4x – 2
−b ± b 2 − 4ac
2a
x=
=
−(−4) ± (−4)2 − 4(1)(−2)
2(1)
=
4 ± 16 + 8
2
=
4 ± 24
2
∴ x = 4,4 or x = –0,4
b)
y = –2
c)
x=
=
d)
−b
2a
−(−4)
2(1)
∴x=2
TP: y = x2 – 4x – 2
= (2)2 – 4(2) – 2
=4–8–2
= –6
∴ TP: (2; –6)
e)
y
2
y = x2 – 4x – 2
–3
–2
–1
1
0
–1
–2
–3
–4
–5
–6
f)
y = –6
1
2
3
4
5 x
77
78
Module 4 • Algebraic graphs
Activity 4.9
1.
a)
x-intercept
y-intercept
y = 2x – 1
0 = 2x – 1
2x = 1
∴ x = 12
y = –x + 5
0 = –x + 5
∴x=5
y = –1
y=5
y
5
y = 2x – 1
4
(2; 3)
3
2
1
–2
0
–1
1
2
3
4
–1
5
6
x
y = –x + 5
–2
Solution is (2; 3)
y = 2x – 3
b)
x-intercept
0 = 2x – 3
2x = 3
∴x=
y-intercept
3
2
y = –2x – 7
0 = –2x – 7
2x = –7
x = – 72
or x = 1 12
∴ x = –3 12
y = –3
y = –7
N2 Mathematics Lecturer Guide|Hands-On
y
1
y = 2x – 7
–5
–4
–3
–2
–1
0
y = 2x – 3
1
2
3
4 x
–1
–2
–3
–4
(–1; –5)
–5
–6
–7
Solution is (–1; –5)
x + y = 10
x = 10
y = 10
c)
x-intercept
y-intercept
x–y=6
x=6
y = –6
y
10
8
6
y=x–6
4
(8; 2)
2
–4
–2
0
–2
–4
–6
Solution: (8; 2)
2
4
6
8
10 x
y = –x + 10
79
80
Module 4 • Algebraic graphs
y=x–2
0=x–2
∴x=2
d)
x-intercept
2x + y = 7
2x = 7
x=
7
2
∴ x = 3 12
y-intercept
y = –2
y=7
y
7
6
5
4
3
y=x–2
2
(3; 1)
1
–3
–2
–1
0
1
2
3
–1
4
x
y = –2x + 7
–2
–3
Solution is (3; 1)
e)
x-intercept
y-intercept
y = 2x – 6
0 = 2x – 6
2x = 6
∴x=3
y = –6
y = –x
x=0
y=0
N2 Mathematics Lecturer Guide|Hands-On
y
2
y = 2x – 6
1
–3
–2
–1
0
1
2
4 x
3
–1
(2; –2)
–2
–3
y = –x
–4
–5
–6
f)
Solution: (2; –2)
y
3
2
x
4
1
–3
–2
–1
0
1
2
–1
3
y
+ −4 = 1
4
5 x
y = –1
(3; –1)
–2
–3
–4
2.
Solution: (3; –1)
a)
y = 23 x + 1
x-intercept
0 = 23 x + 1
2x
3
= –1
3y = –x – 6
0 = –x – 6
x = –6
x = – 32 or –1 12
y-intercept
y=1
y = –2
81
82
Module 4 • Algebraic graphs
y
3
2
y = 23 x + 1
1
–6
–5
–4
–3
–2
(–3; –1)
0
–1
1
2
3
x
–1
–2
y = –13 x – 2
–3
b)
Solution: (–3; –1)
y = 23 x + 1 ................................ 1
3y = – x – 6
∴ y = – 13 x – 2 .............................. 2
2x
3
+ 1 = – 13 x – 2 ......................... 1 = 2
2x + 3 = –x – 6
3x = –9
x = –3
Substitute x = –3 into 1
∴ y = 23 (–3) + 1
= –2 + 1
∴ y = –1
Solution: (–3; –1)
Activity 4.10
y – 3x = –1
y = x2 + 2x – 3
x-intercept: –3x = –1
x = 13
x-intercept: 0 = (x + 3)(x – 1)
∴ x = –3; x = 1
y-intercept: y = –1
y-intercept: y = –3
Symmetry axis: x =
1.
a)
−2
2(1)
∴ x = –1
TP: y = (–1)2 + 2(–1) – 3
=1–2–3
= –4
∴ TP: (–1; –4)
N2 Mathematics Lecturer Guide|Hands-On
y
(2; 5)
5
y = x2 + 2x – 3
4
3
2
1
–4
–3
–2
–1
0
1
2
3
x
–1
–2
–3
(–1; –4)
y = 3x – 1
–4
Solutions are: (–1; –4) and (2; 5)
b) f(x) = –x + 2
g(x) = x2 – 4
x-intercept: x = 2
x-intercepts: 0 = x2 – 4
x2 = 4
x = ±2
y-intercept: y = 2
y-intercept: y = –4
Symmetry axis: x = – 2(1)
0
TP: y = –4
∴ TP: (0; –4)
x=0
83
84
Module 4 • Algebraic graphs
y = –x2 + 2x + 3
a) x-intercepts: 0 = –x2 + 2x + 3
0 = x2 – 2x – 3
0 = (x – 3)(x + 1)
x = 3; x = –1
2.
b)
y-intercept: y = 3
c)
Symmetry: x = – 2(−1)
2
∴x=1
y
5
(–3; 5)
4
3
g(x) = x2 – 4
2
1
(2; 0)
–4
–3
–2
0
–1
1
2
3
4 x
–1
–2
–3
–4
–5
Solutions are: (2; 0) and (–3; 5)
d)
TP: y = –(1)2 + 2(1) + 3
= –1 + 2 + 3
∴y=4
∴ TP: (1; 4)
f(x) = –x + 2
N2 Mathematics Lecturer Guide|Hands-On
e)
y = 2x + 2
x-intercept: 2x = –2
x = –1
y-intercept: y = 2
y
5
y = 2x + 2
(1; 4)
4
3
2
1
(–1; 0)
–4
–3
–2
–1
0
1
2
3
–1
–2
–3
f)
3.
Solutions: (–1; 0) and(1; 4)
y – x2 + x + 6 = 0
∴ y = x2 – x – 6
a)
)
Symmetry: x = – (2−1
(1)
∴x=
b)
c)
x-intercept: 0 = x2 – x – 6
0 = (x – 3)(x + 2)
x = 3; x = –2
2
TP: y = 1 12 2 – 1 12 2 – 6
=
=
=
=
∴ TP: 1
d)
1
2
1 – 1 –
4
2
1 − 2 − 24
4
25
–4
–6 14
1 ; –6 1
2
4
6
2
y-intercept: y = –6
y = –x2 + 2x + 3
4
x
85
86
Module 4 • Algebraic graphs
e)
2y – 4x = –12
x-intercept: –4x = –12
x=3
y-intercept: 2y = –12
y = –6
y
3
2
y = x2 – x – 6
y = 2x – 6
1
(3; 0)
–4
–3
–2
–1
0
1
2
3
4
5
x
–1
–2
–3
–4
–5
(0; –6)
–6
–7
f)
4.
x
−2
(12 ; –6 14 )
Solutions: (3; 0) and (0; –6)
+
y
4
y = –x2 + 4x + 12
=1
x-intercept:
y-intercept:
x
−2
y
4
x = –2
x-intercept: 0 = –x2 + 4x + 12
0 = x2 – 4x – 12
0 = (x – 6)(x + 2)
x = 6; x = –2
=1
y-intercept: y = 12
=1
y=4
Symmetry axis: x = – 2(−41)
∴x=2
TP: y = –(2)2 + 4(2) + 12
= –4 + 8 + 12
= 16
∴ TP: (2; 16)
N2 Mathematics Lecturer Guide|Hands-On
y
18
16
x
−2
14
+ x4 = 1
(4; 12)
12
10
8
6
4
2
(–2; 0)
–6
–4
0
–2
2
4
6
–2
–4
y = –x2 + 4x + 12
–6
Solutions are: (–2; 0) and (4; 12)
5.
y = 24 + 10x – x2
y = –x2 + 10x + 24
a)
b)
c)
d)
x-intercept: 0 = –x2 + 10x + 24
0 = x2 – 10x – 24
0 = (x – 12)(x + 2)
∴ x = 12; x = –2
y-intercept: y = 24
Symmetry axis: x = – 2(10
−1)
∴x=5
TP: y = –(5)2 + 10(5) + 24
= –25 + 50 + 24
y = 49
TP: (5; 49)
8 x
87
88
Module 4 • Algebraic graphs
e)
y = 2x + 4
x-intercept: 0 = 2x + 4
2x = –4
∴ x = –2
y-intercept: y = 4
y
(5; 49)
50
y2 = –x2 + 10x + 24
45
40
35
y = 2x + 4
30
25
(10; 24)
20
15
10
5
(–2; 0)
–8
–6
–4
0
–2
2
4
–5
f)
6.
Solutions are: (–2; 0) and (10; 24)
y = x2 – 9
y = –5
x-intercept: 0 = x2 – 9
x2 = ± 3
y-intercept: y = –9
0
Axis of symmetry: x = – 2(1)
x=0
TP: y = (0)2 – 9
y = –9
∴ TP: (0; –9)
6
8
10
12
14 x
N2 Mathematics Lecturer Guide|Hands-On
y
y = x2 – 9
2
1
–4
–3
–2
–1
0
1
2
3
4 x
–1
–2
–3
–4
(–2; –5)
–5
–6
–7
–8
–9
Solutions: (–2; –5) and (2; –5)
(2; –5)
y = –5
89
90
Module 4 • Algebraic graphs
Module 4: Summative assessment answers
Equation
3y – 6x + 9 = 0
1y–x=0
2
y = mx + c
y = 2x – 3
y = 2x
m
2
2
c
–3
0
1.1.3
y
3
y = 6x – 2
6
–2
1.1.4
y=4
y=4
0
4
1.1
1.1.1
1.1.2
– 2x = – 23
(12 × 12 ) (6)
1.2 y = –x2 + 9
f(–3) = –(–3)2 + 9 = 0
f(–2) = –(–2)2 + 9 = 5
f(–1) = –(–1)2 + 9 = 8
f(0) = –(0)2 + 9 = 9
f(1) = –(1)2 + 9 = 8
f(2) = –(2)2 + 9 = 5
f(3) = –(3)2 + 9 = 0
x
y
–3
0
–2
5
–1
8
0
9
1
8
2
5
3
0
(5)
y
9
8
7
6
5
4
y = –x2 + 9
3
2
1
–4
–3
–2
–1
0
–1
1
2
3
4
x
[11]
N2 Mathematics Lecturer Guide|Hands-On
2.1
y
4
x=2
3
y=x
x=0
y = –x
91
2
1
y=0
–4
–3
–2
–1
0
1
2
3
4
x
–1
–2
y = –3
–3
–4
(6)
2.2 2.1.1
x
5
+
y
−5
=1
y
1
–3
–2
–1
0
1
2
3
4
5 x
–1
–2
–3
–4
x
5
+ −5y = 1
–5
–6
(2)
2.2.2 (0; –5)
2.2.3 (5; 0)
(1)
(1)
[10]
92
Module 4 • Algebraic graphs
2. 2.1
m = –2
y – y1 = m(x – x1)
y –(–2) = –2(x – 1)
y + 2 = –2x + 2
y = –2x
2.2
y
(x1; y1) = (1; –2)
(3)
3
2
1
–3
–2
0
–1
1
2
3
x
–1
–2
3.
y = –2x
–3
(2)
y + x2 + 2x – 3 = 0
∴ y = –x2 – 2x + 3
3.1.1
(−2)
x = – 2(−1)
∴ x = –1
(1)
3.1.2 TP: y = –1(–1)2 – 2(–1) + 3
y = –1 + 2 + 3
∴ y =4 and TP: (–1; 4)
(2)
3.1.3 Let y = 0:
0 = –x2 – 2x + 3
0 = x2 + 2x –3
0 = (x + 3)(x – 1)
∴ x = –3; x = 1
(3)
3.1.4 y-intercept: x = 0
y=3
(1)
N2 Mathematics Lecturer Guide|Hands-On
y
5
4
y=x–1
3
2
1
–5
–4
–3
–2
–1
0
(1; 0)
1
2
3
4
5
x
–1
–2
–3
y = –x2 – 2x + 3
–4
(–4; –5)
–5
3.2 y – x = –1
x-intercept: y = 0
y-intercept: x = 0
0 – x = –1
y = –1
∴x=1
Solution: (–4; –5) and (1; 0)
4.1 y – 2x = 4
y = 2x + 4
–y + 9 = 3x
y = –3x + 9
x-intercept: 0 = 2x + 4
2x = –4
x = –2
x-intercept: 0 = –3x + 9
3x = 9
x=3
y-intercept: y = 4
y-intercept: y = 9
93
94
Module 4 • Algebraic graphs
y
9
y = 2x + 4
8
7
6
(1; 6)
5
4
3
y = –3x + 9
2
1
–4
–3
–2
0
–1
1
2
3
4 x
–1
The point of intersection is (1; 6)
4.2 y = x2
x-intercept: 0
y-intercept: 0
axis of symmetry: x = 0
TP: (0; 0)
(6)
x+y–2=0
x-intercept: x = 2
y-intercept: y = 2
N2 Mathematics Lecturer Guide|Hands-On
y
95
y = x2
5
4
(–2; 4)
3
2
(1; 1)
1
–4
–3
–2
–1
0
1
2
3
x
4
–1
–2
y = –x + 2
The points of intersection are (1; 1) and (–2; 4)
(7)
[13]
Total: 50
MODULE
5
Measuring of angles, angular and
peripheral velocity and sectors of circles
Activity 5.1
1.
a)
c)
e)
g)
i)
65° = (65 × 60)'
= 3 900'
135,21° = 135° + (0,21 × 60)'
= 135° + (12,6)'
= 135°13'
0,25 rev = (0,25 × 360)°
= 90°
320° =
320
360
= 0,889 rev
26
55°26' = 55° + 1 60
2
55° + (0,433)
=
= 55,433°
=
= 0,154 rev
k)
283° =
283
360
= 0,786 rev
= 92,4°
a) 0,16 rev = (0,16 × 360)°
= 57,6°
= 57° + (0,6 × 60)'
= 57°36'
c) 30,259° = 30° + (0,259 × 60)'
= 30°16'
e)
f)
0,56 rev = (0,56 × 360°)
= 201,6°
= 201° + (0,6 × 60)'
= 201°36'
h)
20,41° =
j)
55,433
360
24
m) 92°24’ = 92° + 1 60
2
2.
0,4° = (0,4 × 60)'
= 24'
41
d) 35°41' = 35° + 1 60
2
=
35,68°
b)
82,5° = 82° + (0,5 × 60)
= 82°30'
20,41
360
= 0,057 rev
0,17 rev = (0,17 × 360)°
= 61,2°
= 61° + (0,2 × 60)'
= 61°12'
54,31° = 54° + (0,31 × 60)'
= 54° + (18,6)
= 54°19'
n) 256,14° = 256° + (0,14 × 60)
= 256°8'
l)
b)
251,34° = 251° + (0,34 × 60)'
= 251°20'
d)
0,34 = (0,34 × 360)°
= 122,4°
= 122° + (0,4 × 60)'
= 122°24'
N2 Mathematics Lecturer Guide|Hands-On
Activity 5.2.1
1.
54° =
54
57,3
rad
2.
215° =
= 0,942 rad
3.
75,12° =
= 3,752 rad
75,12
57,3
55°19' = 55° + 1 19
60 2
= 55,32°
=
55,32
57,3
360° = 57,3
or 360° = 2π rad
= 6,283 rad
= 6,283 rad
6.
42
352°42' = 352° + 1 60
2
= 352,7°
rad
=
= 0,965 rad
7.
360
4.
= 1,311 rad
5.
215
57,3
180
57,3
180° =
or 180° = π rad
= 3,141 rad
= 3,141 rad
8.
9.
π
2
90° =
rad or 90° = rad
= 1,571 rad
= 1,571 rad
11. 0,88 rev = (0,88 × 360)°
= 316,8°
=
316,8
57,3
rad
= 6,155 rad
0,28 rev = (0,28 × 360)°
= 100,8°
=
90
57,3
352,7
57,3
100,8
57,3
rad
= 1,759 rad
185,51
10. 185,51° = 57,3
= 3,238 rad
12. 75°59' = 75° + 1 59
60 2
= 75,983°
rad
=
= 5,529 rad
75,983
57,3
rad
= 1,326 rad
Activity 5.2.2
1.
3.
5.
7.
4 rad = 4 × 57,3°
= 229,2°
= 229° + (0,2 × 60)'
= 229°12'
0,41 rad = 0,41 × 57,3°
= 23,493°
= 23° + (0,493 × 60)'
= 23°30'
2π rad = 360°
0,25 rev = (0,25 × 360)°
= 90°
= π2 rad or 1,571 rad
2.
4.
6.
8.
2,491 rad = 2,491 × 57,3°
= 142,734°
= 143° + (0,734 × 60)'
= 143°44'
π
2
rad = 90°
π rad = 180°
0,92 rad = 0,92 × 57,3°
= 52,716°
= 52° + (0,716 × 60)'
= 52°43'
97
98
Module 5 • Measuring of angles, angular and peripheral velocity and sectors of circles
9.
0,516 rev = (0,516 × 360)°
= 185,76°
= 185° + (0,76 × 60)'
= 185°46'
11. 25π rad = 25π × 57,3°
= 72,005°
= 72° + (0,005 × 60)'
= 72°0'
10. 2,6 rad = 2,6 × 57,3°
= 148,98°
= 148° + (0,98 × 60)'
= 148°59'
12. 1 rad = 57,3°
Activity 5.3
1. i)
2.
Given: d = 450 mm
∴ r = 225 mm
n = 920 r/m
= 15,333 r/sec
w = 2πn
= 2(π)(15,333)
= 96,340 rads/sec
ii)
V = πDn
= π(450)(15,333)
= 21 676,518 m/s
Given: ω = 15 rad/min
35 revolutions = 70π rads
= 219,911 rads
time =
219,911
15
min
= 14,661 min
3.
Given: D = 700 mm
n = 800 r/min ⇒ 13,333 r/sec
i) ω = 2πn
= 2(π)(13,333)
= 83,774 rad/s
r = 350 mm
iii) V = 29,321 m/s
∴ in 4 min = 240 sec it will cover
240 × 29,32 m = 7 037,04 m
4.
ii)
V = πDn
= π(700) 13,333
= 29 320,798 mm/s
= 29,32 m/s
Given: V = 34,5 km/h = 9,583 m/s d = 2,25 m
i) V = πDn
n=
n=
ii)
V
πD
9,583
π(2,25)
= 1,356 r/sec
= 81,343 r/min
n = 1,356 r/sec
N2 Mathematics Lecturer Guide|Hands-On
5.
Given: ω = 7,42 rad in 2,5 sec = 2,968 rad/s
t=
=
θ
t
θ
5 rev
2,968
=
10π
2,968
i)
ii)
6.
ω=
= 10,585 sec
Given: d = 900 mm = 0,9 m ∴ r = 0,45 m
V = ωr
= 2,968 (0,45)
= 1,336 m/sec
Given: ω = 85,342 rad/s
2π
85,342
2π
i)
n=
=
= 13,583 r/sec
= 814,98 r/min
ii)
D = 500 mm = 0,5 m
•
ω = 2πn
r/sec
if calculator is switched off
= 814,956 r/min
V = Dπn
= 0,5 × π × 13,583
= 21,336 m/s
θ
iii) ω = t
θ=
ωt
= 85,342 × 2,5 × 60
= 12 801,3 radians
iv) rev =
7.
8.
12 801,3
2π
= 2 037,39 revolutions
Given: n = 640 r/min
= 10,667 r/sec
i) V = Dπn
= (0,180) π (10,667)
= 6,032 m/s
Given: d = 1,5 m
i) V = ωr
= 80(0,75) m/s
= 60 m/s
r = 0,75 m
D = 180 mm = 0,180 m
ii)
ω = 2πn
= 2π 10,667
= 67,023 rad/sec
w = 80 rad/s
ii) ω = 2πn
2π
80
2π
n=
=
= 12,732 r/s
99
100
Module 5 • Measuring of angles, angular and peripheral velocity and sectors of circles
iii) θ = ωt
iv) V = 60 m/s
3 min = 180 sec
∴ distance = V × t
= 12,732(50)
= 636,6 rads
= 60 × 180
= 10 800 m
9.
d = 600 mm = 0,6 m
V = 74 km/h
i)
V = 74 km/h = 20,556 m/s
ii)
n=
=
V
πD
20,556
π(0,6)
= 10,905 r/s
iv) V = 20,556 m/s ω =
iii) V = ωr
V
r
20,556
0,3
ω=
=
= 68,52 rad/s
v)
20,556
(0,3)
∴ω=
= 68,52 rad/s
θ = angular displacement
= 68,52 × 14
= 959,28 rads
distance = V × t
= 20,556 × 35
= 719,46 m
10. Given: r = 4,25 km
t = 4 min for one revolution ∴ in 1 min it only completes 0,25 rev
∴ d = 8,5 km ∴ n = 0,25 r/min or n = 15 r/h (60 min = 1 hr)
V
ii) ω = r
i) V = πDn
n = 15 r/h
= π(8,5)(15)
= 400,553
4,25
= 400,553 km/h
= 94,248 rad/h
= 0,026 rad/sec
Activity 5.4
1.
Given: d = 30 cm
r = 15 cm
i)
31,5° = 31,5 ×
θ = 31,5°
= 0,55 rads
π
180
= 0,55 rads
iii) l = rθ
= 15(0,55)
= 8,25 cm
V
r
ii)
A = 12 θr2
= 12 (0,55)(15)2
= 61,875 cm2
θ = ωt
N2 Mathematics Lecturer Guide|Hands-On 101
2.
Given l = 14,4 cm
r = 5,7 cm
θ=
=
l
r
14,4
5,7
= 2,526 rads
180
π
= 2,526 ×
= 144,729° if calculator is switched off.
= 144,747° if calculator is not switched off.
3.
r = 350 mm
θ = 24° ×
θ = 24°
π
180
= 0,419 rads
i) l = θr
= 0,419 × 350
= 146,53 mm
4.
Given: l = 7,2 cm
l
r
7,2
2,3
i)
θ=
=
= 3,13 rads
180
π
= 3,13 ×
= 179,336°
Given: A = 360 cm2
i)
A=
1
2 rl
2A = rl
r=
=
6.
2A
l
2(360)
24
= 30 cm
∴ d = 60 cm
A = 12 θr2
= 12 (0,419)(350)2
= 25 663,75 mm2
r = 2,3 cm
ii)
5.
ii)
A = 12 θr2
= 12 (3,13)(2,3)2
= 8,279 cm2
l = 24 cm
ii)
θ=
l
r
24
30
=
= 0,8 rads × 180
π
= 45,837°
= 45° + (0,837 × 60)
= 45° 50,22'
Given: l = 1,7 m
θ = 48°
π
= 48 × 180
= 0,838 rads
l
θ
1,7
0,838
i)
r=
=
= 2,029 m
102
Module 5 • Measuring of angles, angular and peripheral velocity and sectors of circles
ii)
A = 12 θr2
= 12 (0,838)(2,029)2
= 1,725 m2
7.
A = 12 rl
or
= 12 (2,029)(1,7)
= 1,725 m2
r = 7,5 cm
Given: A = 26,4 cm2
i)
A = 12 θr
2A = θr
θ=
=
= 0,939 rads
0,939 × 180
π
= 53,801°
=53° 48' 3,6''
ii)
2A
r2
2(26,4)
(7,5)2
A = 12 rl
2A = rl
iii) l = 2A
r
=
2(26,4)
7,5
= 7,04 cm
Given: θ = 52°
π
i)
θ = 52 × 180
8.
= 0,908 rads
= 12 (12)2(0,908)
iii) r =
cm2
2A
l
•
2(26,3)
10,2 =
= 5,157 m 1
iv) A = 2 r2θ
2A = r2θ
A = 12 r2θ
= 65,376
A = 12 r2θ
ii)
θ=
=
= 2,726 rads
A = 12 r2θ
\ A = 12 r2 rl
\ A = 12 rl
\ 2A
l =r
1 2 as θ = rl
• θ = 50°
=
1 l 2
2 θ θ =
1 l2
2 θ2
1 2
= 50 ×
π
180
× θ = 0,873 rads
2
l
2θ
(12)2
2(0,873)
=
=
= 82,474 cm2
2A
r2
2(52,4)
(6,2)2
N2 Mathematics Lecturer Guide|Hands-On 103
Activity 5.5
1.
Given:
D=
2
x
4h
475 =
x2
360
x2
360
D = 475 mm
+h
x2
4(90)
+ 90
= 475 – 90
h = 90 mm
or x =
=
4hD − 4h 2
4(90)(475) − 4(90)2
= 372,290 mm
= 385
x2 = 385 × 360
= 138 600
x = 138 600
= 372,290 mm
2.
Given: x = 56 mm
D=
=
2
x
4h
h = 12 mm
+h
2
(56)
4(12)
+ 12
= 77,333 mm
3.
Given: D = 305 mm
D=
305
305
x = 79 mm
2
x
4h + h
2
= 79
4h + h
= 6 241
h +h
305h = 6 241 + h2
h2 – 305h + 6 241 = 0
h=
−b ± b 2 − 4ac
2a
=
−(−305) ± (−305)2 − 4(1)(6 241)
2(1)
=
305 ± 68 061
2
=
305 ± 260,885
2
= 282,943 mm or 22,057 mm
4.
Given: x = 270 mm
D=
=
2
x
4h
+h
2
270
4(80)
+ 80
= 227,8125 + 80
= 307,813 mm
h = 80 mm
104
Module 5 • Measuring of angles, angular and peripheral velocity and sectors of circles
5.
Given: r = 105 mm
D = 210 mm
D=
x2
4h
210 =
x2
280
+h
x2
4(70)
h = 70 mm
or x =
+ 70
= 210 – 70 = 140
=
4hD − 4h 2
4(70)(210) − 4(70)2
= 197,99 mm
x2 = 140 × 280
= 39 200
x = 39 200
= 197,99 mm
6.
x = 980 mm
= 0,980 m
D=2m
D=
2=
2=
2=
x2
4h + h
(0,980)2
+
4h
0,96
4h + h
0,24
h +h
h
2h = 0,24 + h2
h2 – 2h + 0,24 = 0
h=
=
=
=
=
−b ± b 2 − 4ac
2a
−(−2) ± (−2)2 − 4(1)(0,24)
2(1)
2 ± 4 − 0,96
2
2 ± 3,04
2
2 ± 1,744
2
= 1,872 m or 0,128 m
7.
Given: h = 6 cm
D=
=
2
x
4h + h
(15)2
4(6) + 6
= 15,375 cm
x = 150 mm = 15 cm
N2 Mathematics Lecturer Guide|Hands-On 105
8.
Given: x = 30 cm
2
D = x4h + h
)2
34 = (30
4h + h
34 = 900
4h + h
34 =
225
h
D = 34 cm
+h
34h = 225 + h2
h2 – 34h + 225 = 0
−b ± b 2 − 4ac
2a
h=
=
−(−34) ± (−34)2 − 4(1)(225)
2(1)
=
34 ± 1 156 − 900
2
=
34 ± 256
2
34 ± 16
2
50
18
2 or 2
=
=
= 25 cm or 9 cm
9.
Given: AC = 110 cm
OC = OA (radii) = r
AC2 = OC2 + OA2 (Pythag)
(110)2 = r2 + r2
(110)2 = 2r2
(110)2
2
= r2
6 050 = r2
∴ r = 6 050
= 77,782
∴ D = 2r
= 155,564 cm
10. Given: D = 40 mm
D=
40 =
40 =
2
x
36
x2
2
x
4h
+h
2
x
4(9)
2
x
36
+9
+9
= 31
31 × 36
=
= 1 116
x = 1 116
= 33,407 mm
h = 9 mm
106
Module 5 • Measuring of angles, angular and peripheral velocity and sectors of circles
Module 5: Summative assessment answers
Question 1
1.1 35,7° = 35° + (0,7 × 60)'
= 35°42'
51
1.2 88°51' = 88° + 1 60
2
= 88,85°
=
88,85
57,3
rad
= 1,551 rad
1.3 65°35' = 65° + 1 35
60 2
= 65,583°
=
65,583
360
= 0,182 revolutions
1.4 0,57 rev = (0,57 × 360)°
= 205,2°
= 205° + (0,2 × 60)'
= 205°12'
1.5 0,588 rad = 0,588 × 57,3°
= 33,692°
= 33° + (0,692 × 60)'
= 33°42'
2.
Given: D = 30 cm
θ = 32,4°= 0,565 rads
r = 15 cm
2.1
θ = 32,4°
π
= 32,4 × 180
= 0,565 rads
2.2
2.3
A = 12 r2θ
= 12 (15)2(0,565)
= 63,563 cm2
l = rθ
= (15)(0,565)
= 8,475 cm
N2 Mathematics Lecturer Guide|Hands-On 107
3.
Given: D = 800 mm= 0,8 m
n = 600 r/min =10 r/sec
3.1
ω = 2πn
= 2π(10)
= 62,832 rad/s
3.2
V = πDn
= π(0,8)(10)
= 25,133 m/s
3.3
V = 25,133 m/s
∴ 4 min = 240 sec
∴ distance = V × t
= 25,133 × 240
= 6 031,92 m
4.
Given: x = 20 cm D = 54 cm
x2
4h + h
)2
= (20
4h + h
= 400
h +h
D=
54
54
54h = 400 + h2
h2 – 54h + 400 = 0
h=
−b ± b 2 − 4ac
2a
=
−(−54) ± (−54)2 − 4(1)(400)
2(1)
=
54 ± 1 316
2
=
54 ± 36,277
2
= 45,138 cm or 8,862 cm
5.
Given: l = 10,5 cm
θ=
=
l
r
10,5
4,2
r = 4,2 cm
= 2,5 rads
= 2,5 ×
180
π
= 143,239°
6.
Given: V = 45,3 km/h = 12,583 m/s
6.1
n=
V
πD
= π12,583
(1,25)
= 3,204 r/s
D = 1,25 m
108
Module 5 • Measuring of angles, angular and peripheral velocity and sectors of circles
∴ 60 sec in 1 min
= 3,204 × 60
= 192,24 r/min
6.2
7.
Given: ω = 8,62 rads in 1,5 sec
∴ ω = 5,747 rads/s
7.1
ω = 2πn
n=
2π
= 5,747
2π
= 0,915 r/s
∴
= 4,372 seconds
Given: D = 800 mm = 0,8 m
V = πDn
= π(0,8)(0,915)
= 2,2996
= 2,3 m/s
7.2
8.
4
0,915
Given: D = 50 mm
D=
2
x
4h
+h
2
50 =
x
4(12)
50 =
x2
48
2
x
48
h = 12 mm
+ 12
+ 12
= 50 – 12 = 38
x2 = 38 × 48
= 1 824
x = 1 824
= 42,708 mm
9.
Given: D = 400 mm= 0,4 m
9.1
n=
2π
62,832
2π
=
= 10 r/s
∴ 600 r/min
9.2
V = πDn
= π(0,4)(10)
= 12,566 m/s
ω = 62,832 rad/sec
N2 Mathematics Lecturer Guide|Hands-On 109
9.3
Given: ω = 62,382 rad/sec
∴ 2,5 min = 150 sec
∴ 62,382 × 150 sec
= 9 357,3 rads
9.4
9 357,3 rads
= 9357,3
2π rev
= 1 489,2605 rev
10. Given: x = 30 cm h = 5 cm
D=
D=
x2
4h
+h
2
30
4(5)
+5
= 50 cm
MODULE
6 Trigonometry
Activity 6.1
B
1.
2m
,
23
13,5 m
C
A
opp
hyp
13,5
23,2
i)
sin A =
=
= 0,582
ii)
tan B =
opp
adj
AC
BC
18,868
13,5
•
AC2 = AB2 – BC2 (Pythagoras)
=
=
= 1,398
AC =
hyp
AB
iii) cosec B = opp
= AC
23,2
18,868
=
= 1,23
adj
iv) cot A = opp = AC
BC
18,868
13,5
=
= 1,398
v)
cos A =
adj
hyp
=
AC
AB
18,868
23,2
=
= 0,813
hyp
AB
vi) sec B = adj = BC
23,2
13,5
=
= 1,719
= 23,22 – 13,52
= 355,99
355,99
= 18,868
N2 Mathematics Lecturer Guide|Hands-On 111
2.
i)
tan C =
AB
BC
A
∴ BC tan C = AB
BC =
=
= 5,174
ii)
AB
tan C
7,2
tan 54,3°
sin C =
7,2
AB
AC
AC sin C = AB
AC =
=
= 8,866
B
8,4
D
54,3°
AB
sin C
7,2
sin 54,3°
AB
iii) sin AD̂B = AD
=
7,2
8,4
= 0,857
AD̂B = sin–10,857
= 58,981°
hyp
To find BD use Pythagoras
8,4
4,327
•
iv) sec ∠ADB = adj = AD
BD
=
v)
BD2 = AD2 – AB2
= 1,941
= (8,4)2 – (7,2)2
= 18,72
BD = 18,72
= 4,327
cot ∠BAC =
adj
opp
=
7,2
5,174
=
= 1,392
hyp
vi) cosec C = opp = AC
AB
8,866
7,2
=
= 1,231
opp
vii) tan ∠BAD = adj
=
BD
AB
=
4,327
7,2
= 0,601
AB
BC
C
112
Module 6 • Trigonometry
hyp
adj
i)
sec A =
=
= 1,118
3.
ii)
cot B =
=
AC
AD
100
89,45
adj
opp
=
BD
DC
22,35
44,7
=
= 0,5
opp
iii) sin ∠DCB = hyp
BD
BC
22,35
50
=
=
= 0,447
hyp
iv) cosec ∠DCA = opp
=
=
= 1,118
tan A =
opp
adj
=
BC
AC
=
50
100
= 0,5
v)
AC
AD
100
89,45
adj
vi) cos B = hyp
=
BD
BC
=
22,35
50
= 0,447
hyp
vii) cosec ∠DCB = opp
50
22,35
=
= 2,237
adj
viii) cot ∠DAC = opp
89,45
44,7
=
= 2,001
N2 Mathematics Lecturer Guide|Hands-On 113
Activity 6.2
1.
sin 323,4°
= –0,596
2.
cos 174°36'
= cos 174,6°
= –0,996
3.
tan 240,6°
= 1,775
4.
cosec 174,8°
=
1
sin174,8°
= 11,034
5.
sec 237°47'
6.
= sec 237,783°
=
cot 158,2°
=
1
cos 237,783°
1
tan158,2°
= –2,5
= –1,876
7.
9.
cos 330°15'
8.
cosec 299,4°
1
sin 299,4°
= cos 330,25°
=
= 0,868
= –1,148
tan 170°4'
10. sin 240,6°
= tan 170,067°
= –0,871
= –0,175
11. sec 137,8°
=
12. cot 310°25'
1
cos137,8°
= cot 310,417°
= –1,35
=
1
tan 310,417°
= –0,852
Activity 6.3
1.
(hypotenuse) = 22 + 52
= 29
hypotenuse = 29
i)
∴ cos θ =
5
29
= 0,928
iii) cosec θ = 229
= 2,693
v)
sec θ =
29
5
= 1,077
2
θ
ii)
5
sin θ =
2
29
= 0,371
iv) cot θ = 52
= 2,5
114
Module 6 • Trigonometry
2.
i)
sin θ =
=
ii)
opp
hyp
•
Given: Hypotenuse = 13
−10,954
13 x=7
13
∴ y = – 132 − 7 2 = – 120
= –10,954 (neg because of position)
= –0,843
tan θ =
y
opp
x = adj
−10,954
7
=
= –1,565
hyp
iv) sec θ = adj
hyp
iii) cosec θ = opp
13
−10,594
=
= –1,187
v)
cot θ =
adj
opp
7
−10,954
=
13
7
=
= 1,857
= –0,639
from Pythagoras x = –5
3.
–5
–12
θ
13
hyp
adj
=
hyp
x
opp
adj
y
x
−12
−5
i)
sec θ =
=
13
−5
=
= –2,6
=
= 2,4
hyp
iii) cosec θ = opp
ii)
tan θ =
adj
iv) cos θ = hyp
=
hyp
y
=
x
hyp
=
13
−12
=
−5
13
= –1,083
= –0,385
4.
7
θ
8
If tan θ = – 13
then θ lies in 2nd or 4th quadrant
8
ref angle = tan–11 13
2
= 31,608°
∴ θ = 180 – 31,608° if it lies in second quadrant
= 148,392° or
θ = 360 – 31,608° if it lies in 4th quadrant
= 328,392°
N2 Mathematics Lecturer Guide|Hands-On 115
5.
If sec θ = –2,15
then cos θ =
1
−2,15
•
1
sec θ
= cos θ
= –0,465
When cos is neg then
θ lies in 2nd or 3rd quadrant
reference angle = cos–1(0,465)
= 62,29°
∴ θ = 180° – 62,29°
= 117,17°
or θ = 180 + 62,29°
= 242,29°
6.
cot θ = 4,2
1
cot θ
=
1
4,2
tan θ = 0,238
if tan θ is pos then θ lies in the first or the 3rd quadrant
reference angle = tan–1 0,238
= 13,392°
∴ θ = 13,392° if it lies in first quadrant
or θ = 180 + 13,392°
= 193,392° if it lies in second quadrant
7.
CB2 = (–7)2 + (–4)2
= 49 + 16
= 65
CB = 65
= 8,062
opp
hyp
–4
B
A
–7
C(–4; –7)
=
BC
AC
sec BĈA =
hyp
adj
=
i)
sin BÂC =
=
−7
8,062
=
= –0,868
= –1,152
ii)
adj
8,062
−7
adj
AB
iv) cos BÂC = hyp = CA
iii) cot AĈB = opp = BC
AB
−4
8,062
=
−7
−4
=
= 1,75
= –0,496
v)
cosec ∠CÂB =
hyp
opp
=
AC
BC
AC
BC
opp
AB
vi) tan BĈA = adj = BC
−4
−7
=
8,062
−7
=
= –1,152
= 0,571
116
Module 6 • Trigonometry
8.
sin B = 0,8 =
∴
8
10
A
10
8
B
C
BC2 = AB2 – AC2
= 100 – 64
= 36
BC = 36 = 6
opp
adj
i)
∴ tan B =
=
=
= 1,333
8
6
AC
BC
ii)
cos B =
=
=
adj
hyp
BC
AB
6
10
= 0,6
iv) cosec B – cot B
hyp
iii) sec B = adj
AB
BC
10
6
=
=
= 1,667
9.
A
–12
θ
B
=
=
=
10
6
8 – 8
10 − 6
= 84
8
1
2
(–12)2 + BC2 = 132
144 + BC2 = 169
BC2 = 169 – 144
13
C
hyp
adj ∴ AC = 13
and AB = –12
i)
cos θ tan θ
=
=
=
adj
hyp
−12
13
−5
13
•
•
opp
adj
−5
−12
cot B =
• AB2 + BC2 = AC2
–5
sec θ =
hyp
opp
adj
opp
cosec B =
= 25
∴ BC2 = ± 25 = –5
but because of the position it is –5
ii)
1
cosec θ
= sin θ
=
opp
hyp
=
−5
13
N2 Mathematics Lecturer Guide|Hands-On 117
10. 2 cosec θ = –4,38
∴ cosec θ = –2,19
1
cosec θ
=
1
−2,19
sin θ = –0,457
sin θ is negative ∴ θ lies in the third or the fourth quadrant
reference angle = sin–1(0,457)
= 27,194°
∴ θ if it lies in the 3rd quadrant
180 + 27,194° = 207,194°
and if it lies in the 4th quadrant
= 360° – 27,194°
= 332,806°
Activity 6.4
1.
strin
g
20 m
17°
sin 17° =
20
length of string
length of string × sin 17° = 20
length of string =
20
sin 17°
= 68,406 m
2.
A
15°
high?
32,5 m
15°
B
∠AĈB = 15°
AB
∴ sin 15° = AC
AB = AC sin 15°
= 32,5 × sin 15°
= 8,412 m
The bluff is 8,412 m high
C
118
Module 6 • Trigonometry
3.
A
124 m
60°
B
Point ?
tan 60° =
Tower
C
AC
BC
BC tan 60° = AC
BC =
AC
tan 60°
=
124
tan 60°
= 71,591 m
4.
D
A
46°
30°
110 m
tan 30° =
C
B
CB
AB
tan 46° =
BD
AB
CB = AB tan 30°
DB = AB tan 46°
= 110 tan 30° = 110 tan 46°
= 63,509 m = 113,908 m
∴ DC = BD – CB
= 113,908 – 63,509 = 50,399 m
The tower must be raised 50,399 m.
5.
C
Cliff A
Tower
B
height of cliff
tan 46° =
AB
BD
30°
45 m
46°
D
height of tower CD
tan 30° =
CD
BD
AB = BD tan 46°
CD = BD tan 30°
= 45 tan 46° = 45 tan 30°
height of cliff = 46,599 m
height of tower = 25,981 m
N2 Mathematics Lecturer Guide|Hands-On 119
6.
A
70°
50°
B
Cliff
3,5 m
C
sea level
50°
E
E
3,5 m
D
tan 50° =
ED
CD
CD tan 50° = ED
CD =
=
ED
tan 50°
3,5
tan 50°
= 2,937 m
∴ The ship is 2,937 m from the foot of the cliff.
tan 70° =
AB
BE
AB = BE tan 70°
= 2,937 tan 70° (CD = BE)
= 8,069 m
∴ The height of the cliff is AB + BC = 8,069 + 3,5 = 11,569 m
7.
A
D
x
B
35°
E
11 m
20°
11 – x
C
Let BE = x then EC = 11 – x
tan 35° =
height (AB)
x
but
x tan 35° = height (AB)
(height) AB = height (DC)
x tan 35° = (11 – x) tan 20°
0,7x = (11 – x) 0,364
0,7x = 4,004 – 0,364x
0,7x + 0,364x = 4,004
1,064x = 4,004
x=
4,004
1,064
x = 3,763 m
tan 35° =
AB
x
∴ Height AB = x tan 35° = 3,763 m × 0,7
Trees are = 2,635 m high
tan 20° =
height (DC)
11 − x
(11 – x) tan 20° = height (DC)
Module 6 • Trigonometry
8.
B
C
38° 30°
25 m
A
D
CD
8.1 tan 30° = AD
AD tan 30° = CD
25 tan 30° = CD
14,43 m = CD height of office building
BD
8.2 tan 38° = AD
AD tan 38° = BD
25 tan 38° = BD
19,532 m = BD
But height of flagpole = BD – CD
= 19,532 – 14,43 m
= 5,098 m
The flagpole is 5,098 m tall.
A
9.
50,96
120
Simon
140 m
C
B
The angle of depression is the same as the angle AĈB.
tan ACB =
tan ∠ACB
AB
CD
= 50,96
140
= 0,364
∠ACB = tan–1(0,364)
= 20,002°
N2 Mathematics Lecturer Guide|Hands-On 121
10.
A
1,5 m
39°
B
sin 39° =
C
AB
AC
AC sin 39° = AB
AC =
AB
sin 39°
=
1,5
sin 39°
= 2,384
∴ The tree was 1,5 + 2,384 = 3,884 m high
11.
A
E
46°
30°
80 m
46°
B
C
30°
D
Angle EÂD = AD̂B = 30° (alt angles)
EÂC = AĈB = 46° (alt angles)
∴ tan 46° =
AB
BC
BC tan 46° = AB
BC =
=
AB
tan 46°
80
tan 46°
= 77,255 m
tan 30° =
AB
BD
BD tan 30° = AB
BD =
=
AB
tan 30°
80
tan 30°
= 138,564 m
The distance between the cars CD = BD – BC
= 138,564 – 77,255 m
= 61,309 m
Module 6 • Trigonometry
12.
A
Tom
B
Sam
140 m
122
46°
60°
C
tan 46° =
AC
BC BC tan 46° = AC
BC =
=
D
AC
tan 46° 140
tan 46° 13.
A
70 m
shadow
192,307 m
tan AB̂C =
AC
BC
tan AB̂C =
70
192,307
tan AB̂C = 0,364
AB̂C = tan– (0,364)
= 20,002°
AC
CD
CD tan 60° = AC
= 135,196 m
Distance between Tom and Sam
BC + CD
135,196 + 80,829
= 216,025 m
B
tan 60° =
C
CD =
AC
tan 60°
=
140
tan 60°
= 80,829 m
N2 Mathematics Lecturer Guide|Hands-On 123
14.
observation point
B
14444244443
A
18°
31°
C
150 m
E
height
D
In 3BCD
tan 31° =
CD
BC
=
150
BC
(BE = CD = 150)
BC tan 31° = 150
BC =
150
tan 31
= 249,642 m
In 3ABC
tan 18° =
∴ Height
AC
BC
= AC + CD
AC = BC tan 18°
= 249,642 × tan 18°
= 81,114 m
= 81,114 + 150
= 231,114 m
the cloud is 231,114 m above the lake
15.
B
(not drawn to scale)
A
54°
1,8 m
D
67°
x
Let DC = x
BC
C
tan 54° =
BC
(1,8 + x)
tan 67° = x
(1,8 + x) tan 54° = BC
x tan 67° = BC
(1,8 + x)(1,376) = BC
2,356x = BC 2,477 + 1,376x = BC
∴ 2,356x = 2,477 + 1,376x
2,356x – 1,376x = 2,477
0,98x = 2,477
x=
2,477
0,98
The wendy house is 2,528 m high.
124
Module 6 • Trigonometry
Activity 6.5
1.
x
0°
30°
60°
90°
y = sin x
0
0,5
0,866
1
y = 1,5 sin x
0
0,75
1,3
1,5
y = sin x – 1
–1 –0,5 –0,134
0
120° 150° 180° 210°
240° 270° 300°
330° 360°
0,866
0,5
0
–0,5 –0,866 –1 –0,866 –0,5
0
1,3
0,75
0
–0,75 –1,3 –1,5 –1,3 –0,75
0
–0,134 –0,5 –1
–1,5 –1,866 –2 –1,866 –1,5
–1
y
2
y = 1,5 sin x
1,5
y = sin x
1
0,5
0
30°
60°
90°
120°
150°
180°
210°
240°
270°
300°
330°
360°
–0,5
y = sin x – 1
–1
–1,5
–2
2.
x
0°
30°
60°
90°
120°
150°
180°
y = cos x
1
0,866
0,5
0
–0,5
–0,866
–1
y = cos x – 2
–1
–1,134
–1,5
–2
–2,5
–2,866
–3
y = 1,5 cos x
1,5
1,3
0,75
0
–0,75
–1,3
–1,5
x
N2 Mathematics Lecturer Guide|Hands-On 125
y
1,5
1
0,5
0
30°
60°
90°
120°
150°
180°
x
–0,5
y = cos x
–1
y = 1,5 cos x
–1,5
–2
–2,5
y = cos x – 2
–3
3.
x
0°
30°
60°
90°
y = cos x + 1
2
1,866
1,5
1
120° 150° 180° 210° 240° 270° 300° 330° 360°
0,5
0,134
0
0,134
0,5
1
1,5
1,866
2
y
2,5
2
1,5
1
(i)
(ii)
0,5
0
30°
60°
90°
120°
150°
180°
210°
240°
270°
–0,5
i)
If cos x = 12 = 0,5
Then cos x + 1 = 0,5 + 1
∴ cos x + 1 = 1,5
Therefore from the graph y = 1,5 when x = 60° and 300°
300°
330°
360°
x
126
Module 6 • Trigonometry
If cos x = 0
Then cos x + 1 = 0 + 1
∴ cos x + 1 = 1
Therefore from the graph y = 1 when x = 90°and 270°
ii)
4.
x
0°
30°
60°
90°
120°
150°
180°
210°
240°
270°
y = 2 sin x
0
1
1,732
2
1,732
1
0
–1
–1,732
–2
y = cos x – 1
0
–0,134
0,5
–1
–1,5
–1,866
–2
–1,866
–1,5
–1
210°
240°
270°
y
y = 2 sin x
2
1,5
1
0,5
0
–0,5
–1
–1,5
(i)
(iii)
(ii)
30°
60°
90°
120°
150°
180°
y = cos x – 1
(i)
–2
i)
If cos x – 1 = 2 sin x it means where do they cross
At x = 0° and for x = ± 230°
ii)
If sin x = –0,5
Then 2 sin x = –1. From the graph that happens when x = 210°
iii)
If 2 cos x = 1
Then cos x = 0,5
Then cos x – 1 = 0,5 – 1
∴ cos x – 1 = –0,5
From the graph this happens when x = 60°
x
N2 Mathematics Lecturer Guide|Hands-On 127
5.
x
0°
30°
60°
90°
y = 3 sin x
0
1,5
2,6
3
2,6
1,5
0
–1,5
–2,6
–3
–2,6
–1,5
0
y = 2 cos x
2
1,73
1
0
–1
–1,73
–2
–1,73
–1
0
1
1,73
2
y
120° 150° 180° 210° 240° 270° 300° 330° 360°
y = 3 sin x
3
2,5
2
(i)
1,5
1
(iii)
y = 2 cos x
(ii)
0,5
0
30°
60°
90°
120°
150°
180°
210°
240°
–0,5
–1
–1,5
(i)
–2
–2,5
–3
i)
If 2 cos x – 3 sin x = 0
Then 2 cos x = 3 sin x
Then x = ± 33° or x = ± 213°
ii)
If 2 cos x – 1 = 0
Then 2 cos x = 1
This occurs when x = 60° or when x = 300°
iii)
If 2 sin x = 1
Then sin x = 0,5
and 3 sin x = 3 × 0,5 = 1,5
This occurs when x = 30° or when x = 150°
270°
300°
330°
360°
x
128
Module 6 • Trigonometry
Module 6: Summative assessment answers
1.
tan A =
1
4
1
A
4
hypotenuse = 12 + 4 2
= 17
1.1 sin A = 117
1.2 cosec A =
2.
17
2.1 sec 45° = 1
1
cos 45°
=
= 1,414
2.2 cot 116°
1
tan 116°
=
= –0,488
2.3 12 tan 40°42' + 3 cosec 30°45'
3
sin 30,75°
= 0,5 tan 40,7° +
= 0,43 + 5,867
= 6,297 or 6,298 (calculator not switched off)
3.
sin B = 0,6 =
10
B
6
10
6
∴ missing sides (adjacent) = 100 − 36 = 8
3.1 tan B = 68 = 43
8
3.2 cos B = 10
= 0,8
3.3
(sin B)2 + (cos B)2
= (0,6)2 + (0,8)2
= 0,36 + 0,64
=1
5
3.4 cosec (90° – B) = 10
8 or 4
N2 Mathematics Lecturer Guide|Hands-On 129
4.
4.1 tan θ ⋅ cos θ
=
=
−3
−4
– 53
×
From Pythagoras
−4
5
Missing side =
=
2
5 (−4) –3
θ
5
=–
3
(negative because of position)
1
4.2 cosec
θ = sin θ
–4
2
−3
5
5.
33°24'
41 m
tan 33°24' =
33°24'
height
41 m
6. 6.1 tan AĈB = 15
4
tan AĈB = 3,75
AĈB = tan–(3,75)
= 75,069°
A
2,5 m
15 m
6.2 AD2 = AB2 + BD2
AD2 = (15)2 + (7)2
= 274
AD = 16,553 m long
7.
14243
height = 41 tan 33,4°
= 27,03 m
B
4m
C
3m
D
x
0°
30°
60°
90°
120°
150°
180°
y = cos x
1
0,866
0,5
0
–0.5
–0,866
–1
y = sin x + 1
1
1,5
1,866
2
1,866
1,5
1
130
Module 6 • Trigonometry
y
2
1,5
1
0,5
0
30°
60°
90°
120°
150°
180° x
–0,5
–1
8.
y
3
x
0°
30°
60°
90°
y = 2 sin x
0
1
1,73
2
1,73
y = 3 cos x
3
2,6
1,5
0
–1,5 –2,6
120° 150° 180° 210° 240° 270° 300° 330° 360°
1
0
–3
–1
1,73
–2,6 –1,5
–2 –1,73 –1
0
0
3
1,5
2,6
y = 3 cos x
2,5
2
y = 2 sin x
(i)
(ii)
1,5
1
0,5
0
30°
60°
90°
120°
150°
180°
210°
240°
–0,5
(iii) –1
–1,5
–2
–2,5
–3
(i)
270°
300°
330°
360°
x
N2 Mathematics Lecturer Guide|Hands-On 131
8.1 3 cos x – 2 sin x = 0
3 cos x = 2 sin x
From graph x = ± 56° or x = ±236°
8.2 3 cos x – 1,5 = 0
3 cos x = 1,5
From graph x = 60° or x = 300°
8.3 sin x = 0,5
2 sin x = 2 × 0,5 = 1
From graph x = 30° or x = 150°
MODULE
7 Mensuration
Activity 1.7
1.
i)
A = 2πrh
ii)
= 2π1 24
(30)
2 2
= 2 261,947 cm2
iii) Vcylinder = πr2h
2.
i)
= 13 571,68 cm3
Asphere = 4πr2
i)
= 2π(12)(30) + 2π(12)2
= 3 166,725 cm2
ii)
9 2
= 254,469 cm2
= 4π1 2 2
3.
= π(12)2(30)
A = 2πrh + 2πr2
Acone = πr2 + πr h 2 + r 2
Vsphere = 43 πr3
= 43 π(4,5)3
= 381,704 cm3
ii)
2
= π1 10
+ π(5) 182 + 52 2 2
= 371,989 cm2
4. Vone metre block = l × b × h
= 45 × 36 × 120
Vcone = 13 πr2h
= 13 π(5)2(18)
= 471,239 cm3
V of 4 blocks = 4 × 45 × 36 × 120
= 777 600 mm3
Vcylinder = πr2h
2
777 600 = π1 120
2 2 h
777 600
π(60)2
=h
68,755 mm = h
5.
i)
Vsphere = 43 πr3
3
= 43 π1 600
2 2
= 113 097 335,5
6.
ii)
V = 13 πr2h
= 4π(300)2
= 1 130 973,355 mm2
12 cm
12 2(15)
2
2
= 565,487 cm3
15 cm
= 13 π1
mm3
Asphere = 4πr2
N2 Mathematics Lecturer Guide|Hands-On 133
Vsphere = 43 πr3
7.
75 000 =
3(75 000)
4π
4 πr3
3
=
r3
17 904,931 =
r3
3 17 904,931
=r
26,161 mm = r
A = 4πr2
= 4π(26,161)2
= 8 600,398 mm2
9.
Vsphere =
=
8.
Acylinder = 2πrh
4 362,8 = 2πr(38)
4 362,8
2π(38)
=r
18,273 cm = r
Vcylinder = πr2h
= π(18,273)2(38)
= 39 861,458 cm3
4 πr3
3
4 π(90)3
3
= 3 053 628,059 mm3
Vcube = s3
3 053 628,059 = s3
3 3 053 628,059 = s
145,079 mm = side
10. Vcylinder = πr2h
= π(0,4)2(1,2)
= 0,603
• D = 0,8 m
m3
1 m3 = 1 000 ℓ
= 0,603 × 1 000
= 603 ℓ
11. Vcone = 13 πr2h
• (603,186 ℓ if calculator not switched off)
12. Vcylinder = πr2h
2
= 13 π1 180
(250)
2 2
= 2 120 575,041 mm3
Vsphere = 43 πr3
2 120 575,041 = 43 πr3
3 × 2 120 575,041
4π
= r3
506 250 = r3
3
506 250 = r
79,699 mm = r
2
= π1 72 2 (40)
= 1 539,38 cm3
V15 sinkers = 1 539,38
Vsinker = 102,625 cm3
Vsinker = 13 πr2h
102,625 = 13 πr2(5)
3 × 102,625
π ×5
= r2
19,6 = r2
19,6 = r
4,427 = r
4,427 × 2 = D
8,854 cm = D
134
Module 7 • Mensuration
13. a)
Acircle sector = 12 rl
= 12 (1,6)(2,5)
b) i)
ii)
= 2 m2
Acone = πrl
2 = πr(1,6)
2
π(1,6)
Slant height of cone is the radius of circle sector.
=r
0,398 m = r
or Arc length = 2πr
2,5 = 2πr
r=
2,5
2π
= 0,398 m
Acone = πr h 2 + r 2
2 = π(0,398) h 2 + (0,398)2
2
π(0,398)
=
h 2 + (0,398)2
1,6 =
h 2 + (0,398)2
2,56 = h2 + (0,398)2
2,56 – (0,398)2 = h2
2,402 = h2
2,402 = h
1,55 m = h
or
h2 + r2 = l2
2
h + (0,398)2 = (1,6)2
h2 = 2,402
h = 1,55 m
Vcone = 13 πr2h
iii)
= 13 π(0,398)2(1,55)
= 0,257 m3
N2 Mathematics Lecturer Guide|Hands-On 135
14. i)
Vcone = 13 πr2h
ii)
2
= 4π(83,679)2
= 2 454 369,261 mm3
= 87 991,927 mm2
= 13 π1 250
(150)
2 2
Vsphere = 43 πr3
2 454 369,261 = 43 πr3
3 × 2 454 369,261
4π
= r3
585 937,5 = r3
Asphere = 4πr2
3 585 937,5
=r
83,679 = r
2 × 83,679 = D
167,358 mm = D
15. Vsphere = 43 πr3
=
4
3
3
π1 0,6
2 2
= 0,113 m3
Vcylinder = πr2h
2
0,113 = π1 0,6
h
2 2
0,113
π(0,3)2
=h
0,4 m = h
16. Acone = πr h 2 + r 2
8,2
2π
=r
1,305 m = r
Acone = πr h 2 + r 2
= π(1,305) (2)2 + (1,305)2
= 9,791 m2
or A = πrs
1,48 = πr(1,7)
r=
2m
Base diameter = 2πr
8,2 = 2πr
1,48
π(1,7)
= 0,277 m
2πr = 8,2 m
Module 7 • Mensuration
17. Acircle sector = 12 rl
l
1,48 = 12 (1,7)l
=l
1,48 m2
1,741 = 2πr
1,741
2π
h
m
1,741 m = l
1,7
2 × 1,48
1,7
1,7
m
136
r
=r
2πr
0,277 m = r
by means of Pythagoras: 1,72 = h2 + r2
1,72 = h2 + (0,277)2
1,72 – (0,277)2 = h2
2,813 = h2
2,813 = h
1,677 m = h
18. The top part of the shape is 12 cylinder.
V 12 cylinder = 12 (πr2h)
=
1
2
(π( 24
)2(40))
2
= 9 047,787 cm3
The bottom part is a rectangular block.
Vblock = l × b × h
= 62 × 24 × 22
= 32 736 cm3
Total volume = V12
cylinder
•
Height = 34 – radius
+ Vblock
= 9 047,787 + 32 736
= 41 783,787
Vsphere = 43 πr3
41 783,787 = 43 πr3
3 × 41 783,787
4×π
= r3
9 975,144 = r3
3
9 975,144 = r
21,526 cm = r
Activity 7.2
1.
A = 1 first + last2 ordinate + rest2 × distance
= 1 120 +2 118 + 137 + 148 + 147 + 154 + 142 + 136 + 1272 × 20
= 22 200 mm2
N2 Mathematics Lecturer Guide|Hands-On 137
2.
A = 1 first + last2 ordinate + rest2 × distance
= 1 0 +2 0 + 4 + 8 + 6 + 10 + 14 + 22 × 3
= 132 mm2
3.
A = 1 first + last2 ordinate + rest2 × distance
= 1 6 +2 8 + 7 + 5 + 72 × 2
Distance diff in x = 2
•
= 26 × 2
= 52 units2
4.
i)
A = (sum of mid ordinates) × distance
405 = 1 first + last2 ordinate + rest2 × distance
= 1 1,2 +2 3,7 + 3 + 2,8 + 2,8 + 3,6 + 3,1 + 4,22 × distance
= 21,95 × distance
405
21,95
18,451 cm = distance
ii)
Straight side length
= 7 × distance
= 7 × 18,451
= 129,157 cm
5.
= distance
•
There are 8 pts ∴ 7 intervals
A = 1 first + last2 ordinate + rest2 × distance
= 1 0 + 222,5 + 4,8 + 9,2 + 14,5 + 12 + 18,8 + 192 × 5
= 447,75 units2
6.
A = 1 first + last2 ordinate + rest2 × distance
8 905,5 mm2 = 1 212 +2 200 + 308 + 156 + 300 + 250 + 198 + 275 + 2862 × distance
8 905,5
1 979
= 1 979 × distance
= distance
4,5 mm = distance
7.
A = 1 first + last2 ordinate + rest2 × distance
= 1 5 +2 0 + 6 + 9 + 10 + 12 + 152 × 2,5
= 136,25 units2
8.
2 km2 = 2 000 000 m2
A = 1 first + last2 ordinate + rest2 × distance
2 000 000 = 1 102 +2 170 + 140 + 150 + 174 + 186 + 210 + 190 + 1682 × distance
= 1 354 × distance
2 000 000
1 354
= distance
1 477,105 m = distance = width of strip
Module 7 • Mensuration
Module 7: Summative assessment answers
1.
A = 4πr2
2
= 4π1 142 2
= 615,752 cm2
2.
i)
Vcone = 13 πr2h
h=
= 13 π(75)2(206,821)
mm3
ii)
= 1 218 276,252
Acone = πrl
= π(75)(220)
= 51 836,279 mm2
3. i)
Vpipe = Vcylinder
= πr2h
ii)
4.
•
=
= 229 022,104 cm3
Apipe = Acurved surface cylinder
= 2πrh
= 2π(9)(900)
= 50 893,801 cm2
= 5,089 m2
= 206,821
h
r=
D
2
=
18
2
cm
= 9 cm
9 m = 900 cm
Vcone = 13 πr2h
Vcylinder = πr2h
= 153 938,04 mm3
Vcylinder = πr2h
= π(7)2(22)
= 3 386,637 cm3
5. i)
ii)
(220)2 − (75)2
75
π(9)2(900)
= π(35)2(40)
6.
220
138
153 938,04 = 13 π(70)2h
3 × 153 938,04
π(70)2
=h
30 mm = h
•
220 mm = 22 cm
Acylinder = 2πrh + πr2
= 2π(7)(22) + π(7)2
= 1 121,549 cm2
Vbig ball = 43 πr3
3
= 43 π1 180
2 2
= 3 053 628,059 mm3
No of balls =
=
Vbig ball
Vsmall ball
3 053 628,059
1 767,146
= 1 728,00
= 1 728 balls
Vsmall ball = 43 πr3
3
= 43 π1 15
22
= 1 767,146 mm3
N2 Mathematics Lecturer Guide|Hands-On 139
(slant side)2 = h2 + r2
= (10)2 + (4)2
• r=
= 116
slant side = 116
= 10,77 cm
ii)
Acone = πrl + πr2
= π(4)(10,77) + π(4)2
= 185,605 cm2
D
2
= 4 cm
iii) Vcone = 13 πr2h
= 13 π(4)2(10)
= 167,552 cm3
8.
A = 1 first + last2 ordinate + rest2 × distance
= 1 1 +2 8 + 3 + 4 + 7 + 102 × 2
= 57 units2
9.
D = 20 mm = 2 cm
radius =
Vsphere =
=
D = 1 cm
2
4 πr3
3
4 π(1)
3
= 4,189 cm3
3 500 balls = 3 500 × 4,189
= 14 661,5 cm3
mass = 14 661,5 × 7 800
= 114 359 700 kg
10. A = 1 first + last2 ordinate + rest2 × distance
= 1 400 +2 400 + 450 + 470 + 480 + 500 + 4302 × 30
= 81 900 m2
=
81 900
10 000
= 8,19 ha
11. A = 1 first + last2 ordinate + rest2 × distance
6 048 = 1 32 +2 36 + 47 + 40 + 53 + 68 + 54 + 442 × distance
= 340 × distance
6 048
340
= distance
17,788 mm = distance
10 cm
7. i)
4 cm
8 cm
140
Module 7 • Mensuration
003