Mathz obj
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Theory is ready

1a)
3/5(4x+3) = 3/4(3x-1)
3(4x+3)/5 = 3(3x-1)/4
12x+9/5 = 9x-3/4
5(9x-3) = 4(12x+9) 45x-15=48x+36
45x-48x=36+15
-3x=51
x=-51/3
x = -17
(1b) x/sin60 degree = 10/sin30 degree
x = 10sin60 degree/sin 30 degree = 10 *
0.866/0.5
x = 8.66/0.5
=17.32cm
Height = 17.32cm

07066531599 .

9a)
Draw the diagram
Radius = 13cm
|OM| = 5cm In triangle MOB,
|OB|^2 = |OM|^2 + |MB|^2
13^2 = 5^ + |MB|^2
|MB|^2 = 13^2 - 5^2
|MB|^2 = 169-25
|MB|^2 = 144 |MB| = root 144
|MB| = 12cm
Hence, the length of the chord = 2|MB|
=2*12cm
=24cm

9b) Green balls = 6
Blue balls = 10
Total balls = 16
Prob (that the balls are of different
colors)
= Prb(GB) + Prb(BG) =(6/16*10/15) + (10/16*6/15)
=(3/8*2/3) + (2/8*3/3)
=(1/4) + (1/4)
=1/4 + 1/4
=1+1/4
=2/4 =1/2

9c)
x & 1/y^2
x = k/y^2
12 = k/(1/2)^2
k = 12*1/4 k = 12*1/4 = 3
.:. x = 3/y^2
when y = 3/4
x = 3/(3/4)^2 =3/9/16
=3*16/9
=16/3 =5whole number 1/3

(12a)
y=(x^3+1)^4=U^4
Let U=x^3+1
dy/dx=du/dx*dy/du
du/dx=3x^2
dy/du=4u^3
dy/dx=3x^2*4(x^3+1)^3
=12x^2(x^3+1)^3
(12bi)
Tabulate
x-0-1-2-3-4-5-6
0-0-0-0-0-0-0-0
1-0-1-2-3-4-5-6
2-0-2-4-6-1-3-5
3-0-3-6-2-5-1-4
4-0-4-1-5-2-6-3
5-0-5-3-1-6-4-2
6-0-6-5-4-3-2-1
(12bii)
2sn=n[2a+(n-1)d]
2sn=n[2a+nd-d]
2sn-2an=dn^2-dn
2(sn-an)=d(n^2-n)
d=2(sn-an)/n^2-n
d=2(sn-an)/n(n-1)

(11ai)
First term=2
Common ratio=6/2=3
Number of terms required=6
Un=ar^n-1
U6=2*(3)^6-1
=2*3^5
=2*243
=486
(11aii)
Radius=6cm
pie=22/7
Volume=4752cm^3
V=Pie*r^2*h
4752=22/7*6^2*h
4752*7=22*36*h
33264=792h
h=33264/792
h=42cm
therefore height of the cylinder is 42cm
(11bi)
Let x=-2 or x=3/5
x-2=0 or 5x=3
(x-2)(5x-3)=0
5x^2-3x-10x+6=0
5x^2-13x+6=0
(11bii)
2x^2+5x-12=0
2x^2+8x-3x-12=0
(2x^2+8x)-(3x+12)=0
2x(x+4)-3(x+4)=0
(2x-3)(x+4)=0