1a)
= 3 4/9 ÷ (5 1/3 - 2 3/4) 5 9/10
= 31/9 ÷ Lcm (3 4-9/12 ) 5 9/10
= 31/9 ÷ 3 -5/12 5 9/10
= 31/9 ÷ 2 7/12 5 9/10
= 31/9 ÷ 31/12 5 9/10
= 31/9 x 12/31 5 9/10
= 4/3 59/10
Find the Lcm
= 40 177/30
= 217/30
= 7 7/30
1b)
A { 2,3,4 }
B { 1,3,5 }
= Possibility are { 1,2 }, {2,3}, {3,1},{3,3},{3,5},{4,1},
{4,3},and {4,5}
= Prob! Of sum greater than 3 = 8/9
= Prob of sum less than 7 = 5/9
= Prob of sum greater than 3 and less than 7 = 8/9
x 5/9
= 40/81
2a)
4 + 3/4 (x+2) < eqaul to 3/8x + 1
Multiply through by 8
= 8(4) + 3(2)(x+2) < eqaul to 3x + 8(1)
= 32 + 6(x+2) < eqaul to 3x + 8
= 32 + 6x + 12 < eqaul to 3x + 8
= 6x - 3x < eqaul to 8 - 44
= 3x < eqaul to -36
= x < eqaul to -36/3
= x < eqaul to -12
2b) Area of the shaded portion = total area - area
of the removed square
484 = 20(20+x) - x^2
484 = 400 + 20x - x^2
X^2 - 20x + 84 = 0
X^2 - 6x - 14x + 84 = 0
X(x-6) - 14 ( x-6 ) = 0
(X-14) , (x-6) = 0
X=14 or 6
Ie x= 6cm or 14cm
3a) The ratio of interior angle to. Exterior angle =
6 : 2
interior angle = 5/7 x 180/1
= 900/7
= 128 4/7
Exterior angle = 2/7 x 180/1
= 360/7
= 51 3/7
Number of side = 360/exterior angle
= 360 / 51 3/7 = 6.9999 approx!! 7
The number of side of the polygon is 7
9a)
a= -8
a + 6d : a + 8d
5:8
a+6d/a+8d = 5/8
8a + 48d = 5a + 40d
8a - 5a = 40d - 48d
3a = -8d
3 (- = -8d
-24 = -8d
d= 24/8
d = 3
10a)
2tan60 + cos 30/sin 60 = 2 ( square root 3/1 + square root 3/2 )/ square root 3/2
Open the bracket
= 2 square root 3/1 + square root. 3/2 / square root. 3/2
= 4 square root 3 + square root 3/2 x 2/square root 3
= 8 square root 3 + 2 square root 3/2square root 3
= 2 ( 4square root 3 + square root 3 ) / 2square root 3
= 4 square root 3 + square root 3 / square root 3 x square root 3 / square root 3
= 4(3) + 3 / 3
= 15/3
= 5
10b)
Using < ACE
Tan Θ = opp/ adj
Tan 41 degree = CE / 1050
CE = 1050 (tan 41)
CE = 1050 ( 0.8693 )
CE = 912.751m
Using. < BCD
Tan 36/1 = CD/1050
CD = 1050 (tan 36)
CD = 762.8697m
(I) Height of control tower = CE - DE
= 912.751 - 762.8697
= 149.8813 Approx! 150m
(II) Using < ACE
Cos Θ = adj/hyp
Cos 41 = 1050/AC
AC = 1050/cos 41
AC = 1391.2636 aprox! 1391m
The shortest distance = AC
11a)
H = mt/d(m+p) find M
= cross multiply
= mt = hd(m+p)
= mt = hdm + hdp
= mt - hdm = hdp
= m(t-hd) = hdp
= m = hdp/t-hd
or
= m = hdp/t-dh
11b)
From Angle WXM
< WXM = 90 degree ( angle at the center is twice <
at circumference )
< WMX = 180degree - (90 - 48)degrees (sum of <
in a triangles)
< WMX = 42
< WMX + < XMZ - 180 ( < on a straight lines)
< XMZ = 180 - 42 = 138
< WYZ + < XMZ = 180
< WYZ = 42 degree
11c)
Operation Table
* | 1 | 3 | 5 | 6
1 | 4 | 6 | 1 | 2
3 | 6 | 1 | 3 | 4
5 | 1 | 3 | 5 | 6
6 | 2 | 4 | 6 | 0
I= {5}
II= { }
(12)
Vol of. The reservoir = vol of cone +
vol of hem = 1/3 pi r^2 h + 2/3 pi r^3
where r =x 333 1/3 = 1/3 pi x^2 6x +
2/3 pi x^3 1000/3 = 2 pi x^3 + 2/3 pi
x^3 1000/3 = 2 2/3 pi x^3 1000/3 ÷
8/3 = pi x^3 1000/3 x 3/8 = pi x^3
7000/176 = x^3 X^3 = 39.77 X=
3sqare root 39.77 X = 3.41 Radius = 3.41m
(12bI)
Vol of hemsphere = 2/3 pi r^3 Vol =
2/3 x 22/7 x ( 3.41)^3 V = 2/3 x 22/7
x 39.77 V = 1749.88/21 V = 83.33m
(12bII)
Total surface area of the reservoir =
curved surface area of the cone +
curve surface area of the hempsere
TSA = 22/7 x R x L + 2 x 22/7 x r^2
L^2 = (20.46)^2 + (3.41)^2 L^2 =
418.61 + 11.63 L^2 = 430.24 L =
square root ( 430.24) L= 20.74 TSA =
1555.91/7 + 511.54/7 = 2067.55/7 = 295.36m.
Thank me later.

thanks

send Mtn airtime of 200 or transfer to this number to get maths obj immediately 08138903001

osmond1995:
send Mtn airtime of 200 or transfer to this number to get maths obj immediately 08138903001

get obj for free at http://naijagistclub.pun.bz