Maths-Obj
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Maths-Theory.
13a)
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13b)
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3a)
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3b)
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3c)
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6ai)
The profit y = X²/8 + 5x
y = GHc20,000.00
Hence 20,000 = X²/8 + 5x
160,000 = X² + 40x
X² + 40x – 160,000 = 0
Since X is in thousands
X² + 40x – 160 = 0
6aii)
Using quadratic formula
X = -b±√b² – 4ac/2a
Where; a = 1, b = 40 & c = -160
X = -40±√40² – 4(1)(-160)/2(1)
X= -40 ±√1600 + 640/2
X = -40 ±√2240/2
X = -40 ± 47.32/2
X = -40±47.32/2
= 7.32/2
X = 3.66
X ≈ 4
6b)
Draw the diagram
Using ΔTOP
tan 28 = H/OP
OP = H/tan28
Then for ΔROP
tantita = H/2/OP
OP = H/2/tantita
Hence H/tan 28 = H/2/tantita
tantita = H/2 × tan 28/H
tantita = tan28/2
Hence Tita = 28/2 = 14°
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2a)
Draw the triangle
Using Pythagoras rule
a² = b² + c²
5² = 3² + c²
25 = 9 + c²
c² = 25 – 9
c² = 16
C = √16 = 4, Cos X = 4/5
5(cosx)² – 3 = 5((4/5))² – 3
= 5(4/5 × 4/5) – 3
= 16/5 – 3/1 L. C. M = 5
= 16 – 15/5 = 1/5.
2b)
Volume of a pyramid
=1/3b×L×h
Volume of a cone = 1/3πr²h
Vp = volume of pyramid
Vc = volume of cone
Hp = height of pyramid
Hc = height of cone
Vp = Vc; Hp = Hc
Vp ± Vc
1/3bLh = 1/3πr²h
42 × 11 = 22/7r²
21 = 1/7 × r²
r² = 21 × 7
r² = 147
r = √147
r = 12 3/25cm
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12a)
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12b)
using sinθ rule
i) 307/sin78 = 285/sinθ
sinθ=285sin78/307
sinθ=0.908
θ=sin^-1 0.908
=650 to the nearest whole number
the bearing of X from Z = 360-65=295degree
12bii)
|YZ|/sin37 = 307/sin78
|YZ| = 307sin37/sin78
=189m to the nearest whole number
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5a)
5 – X > 1 9 + X >_ 8
5 – 1 > X X >_ 8 – 9
4 > X X >_ -1
Range is
-1_< X < 4 OR 4 >X >_ -1
5b)
PQR + PSR = 180(supplementary angles of a cyclic quad)
PQR + 56 = 180
PQR = 180 – 56
PQR = 124°
Next, join P to R
QRP = QPR(base angles of an isosceles)
PQR + 2QRP = 180(Sum of angles in a triangle)
124 + 2QRP = 180
2QRP = 56°
QRP = 56/2 = 28°
PRS = 90°(angle in a semi – circle)
= 28 + 90
= 118°
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7a)
√^2(2x^3-4x+6)dx
2x^4/4-4x^2/2+6x
X4/2-2x^2+6x
(2)^4/2-2(2)^2+6(2)
16/2-8+12=8-8+12=12
x^4/2x^2+6x
(1)^4/2-2(1)^2+6(1)
1/2-2/1+6/1=10-14/+12/2
=13-14/2=9/2
=12/1-9/2=24-9/2
=15/2=7^1/2
U
Edited Use This 7b)
7b)
Given; P^-1 = (-1 1)
(4 -3)
P = (p-1)^-1 = C^T/|p^-¹|
=(-3 -1)
(-4 -1)/3 – 4
=(-3 -1)
(-4 -1)/-1
=(3 1)
(4 1)
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8ai)
Diagram
V1=4.158liters
R=21cm
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8aii)
V = 1/3 Ah, = x r²
V = 1/3 xr²h
V = 4.158 liters
V = 4.158cm³, V = 1/3 xr²h
4158 = 1/3 x 22/7 x 21 x 21 x h
4158 x 3 = 22 x 63h
h = 4158/21 x 22 = 9cm
h = 9cm
8b)
d = 28cm, r = d/2 = 28/2 = 14cm
V2 = 1/3 xr²h
V2 = 1/3 x 22/7 x 14 x 14 x 9
V2 = 1/2 x 22/1 x 2 X 14 x 3
= 1848cm³
V2 = 1.845 liters
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4a)
(2y+x) + (6y-2x+1) + 4y = 28 …(i)
6y-2x + 1 = 4y… (II)
2y+ x +6y – 2x + 1 +4 = 28
12y + x +6 -2x + 1 + 4 = 28
12y – x + 1 = 28
12y – x = 29… (III)
6y-2x + 1 = 4y, 6y-2x – 4y = 1
2y – 2x = -1… (iv)
24y – 2x = 54
2y – 2x = 1
22y/2w = 55/22 y = 2.5
12y – x = 27
12 (2.5) – x = 27
30 x 27 x=3
4b)
2y + x = 2 (2.5) + 3 = 5+3 = 8cm
6y – 2x + 1 = (2.5)-2 (3) + 1
= 15-6 + 1 = 10cm
4y = 4(2.5) = 10cm
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1a)
sn=n/2(2a+(n-1)d)
n=10, sn=130
130=10/2(2a+(10-1)d)
130=5(2a+ad)
130=10a+45d———(eq1)
Tn=a+(n-1)d
T5=3*a=3a,n=5
3a=a+(5-1)d
3a=a+4d
3a-a-4d=0
2a-4d=0———-(eq2)
Solve equation 1 and 2 simultaneously
10a+45d=130*1
2a-4d=0*5
10a+45d=130
10a-20d=0
65d/65=130/65
d=2
1b)
a=2d
a=2*2
a=4
1c)
Tn=a+(n-1)d
28=4+(n-1)^2
28=4+2n-2
28=2+2n
28-2=2n
26/2=2n/2
n=13
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Completed
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