VERIFIED MATH OBJ:
1-10: ADADDDCEBE
11-20: CCDCAACECA
21-30: DCDAECAEED
31-40: EAADBEEADC
41-50: BBDCCBCECD
51-60: CADBAECDEC
(1a)
Tabulate: Draw both vertical and horizontal
lines across the digits. I drew only verticals cos
it's txt messaging.
x|1|2|3|4
1|1|2|3|4
2|2|4|0|2
3|3|0|3|0
4|4|2|0|4
(1b)
S.I=PRT
P=N15,000, R=10%, T=3yrs
S.I=#15,000, R=10%, T=3yrss
S.I=15,000* 10/100 *3
S.I=N4,500
A=P+S.I
A=N15,000+N4,500
A=#19,500
============================
3)
Let x represent Donald's age
and let y represent his son's age
x=5y—(1)
(x-4(y-4)=448—(11)
substitute for x in (11)
5y^2-20y-4y+16=448
5y^2-24y+16-448=0
5y^2-24y-432=0
y=-b+_sqroot(b^2-4ac)/2a
a=5, b=-24 c=-432
y=24+_sqroot[(-24^2)-4*5*-432]/2*5
y=24+_sqroot(576+8640)/10
y=24+_sqroot(9216)/10
y=24+_96/10
y=24+96/10 or 24-96/10
y=120/10 or -72/10
y=12 or -7.2
since age cannot be negative
y=12years
but x=5y
x=5*12
x=60years
================================
(4a)
Convert 30000 litres to metres
= 30000/1000 metres
= 30metres
Depth of fuel = h * 7.5*4.2 = 30m^3
= 31.5hm^3 = 30m^3
= h = 30m^3/31.5m
= h = 0.9m
(4b)
Depth of the tank = l*b*h
where l= 7.5, b = 4.2, h = 1.2
=(7.5 * 4.2 * 1.2)m^3
=37.8m^3
Convert metres to litres = 37.8*
1000litres
= 37800litres
Litres of fuel needed to fill the
tank=37800litres/30800litres
=7800litres
================================
(5a)
sector for building project
=48000/144000*360=120degree
sector for education =
32,000/144000*360=80degree
sector for saving = 19200/144000*360=48degree
sector for maintenance =
12000/144000*360=30degree
sector for miscellaneous =
7200/144000*360=18degree
sector for food items = 360-(120+80+48+30+18)
=360-296
=64degree
THEN DRAW A PIE CHART WITH THE ANSWERS
YOU GOT ABOVE.
(5b)
Food = 48000 + 32000 + 19200 +
1200 +7200 + x = 144000
Food => 118400 + x = 144000
Food => x = 144000 — 118400
Food = x = #25,600
=================================
(6)
3sqroot[(41.02*sqroot0.7124)/
(42.87*0.207*0.0404)]
Tabulate
No|log
41.02|16130
*sqroot0.7124|(1^-.8527)/2
|(2^-+1.8527)/2
|1^-9264
41.02*sqroot0.7124|1.6130
|+1^-.9264
|=1.5394
42.87|1.6322
*0.207|+1^-.3160
|0.9482
*0.0404|+2^-.6064
|=1^-.5546
|1.5394
|1^-.5546
|=(1.9848)/3
antilog4.588|0.6616
Answer=4.588
===============================
(7a)
3^2n+1 – 4(3^n+1)+9=0
3^2-3 – 4(3^n -3)+9=0
(3^n)^2-3 – 4(3^n -3)+9=0
let 3^n = p
p^2 -3 – 4(p-3)+9=0
3p^2/3 – 12p/3 + 9/3 = 0
p^2 – 4p + 3 = 0
p^2 – 3p – p + 3 = 0
p^2p(p-3) – 1(p-3) = 0
(p-1)(p-3) = 0
p-1 = 0 or p-3 = 0
p = 1 or 3
Recall 3^n = p
when p=1
3^n = 3^0
n = 0
when p = 3
3^n = 3^1
n = 1
(7b)
log(x^2+4) = 2+logx – log^20
log(x^2+4) = log^100 = log^x – log^20
(x^2+4) = log(xx)
x^2+4 = 5x
x^2-5x+4 = 0
x^2-4x – x +4 = 0
x(x-4) – 1(x-4) = 0
(x-1)(x-4) = 0
x-1 = 0 or x-4 = 0
x = 1 or 4
=============
(8ai)
In <> ABC
Cos B=(a^2+c^2-b^2)/2ac
Cos B=(13^3+13^2-10^2)/(2*13*13)
Cos B=(338-100)/338
Cos B=238/338
Cos B=0.7041
B=cos^-0.7041
B=45.2degrees
therefore =90.4degrees
In <>AOC
By sine rule
r/sine44.8=10/sine90.4
r=10sin44.8/sin90.4
r=7.05cm
(8aii)
Circumference =2*pie*R
=2*22/7*7.05
=44.31
=44.3cm(1d.p)
(8b)
p(-1,2) , (2,6)
(y-y1)/(y2-y1)=(x-x1)/(x2-x1)
(y-2)/(6-2)=(x+1)/(2+1)
(y-2)/4=(x+1)/3
3(y-2)=4(x+1)
3y-6=4x+4
3y-4x=10
==========================
(9a)
Let the lens digit x and unit digit be y,
therefore x-y=5 —(1)
3xy-(10x+y)=14 —-(2)
3xy-10x-y=14 —–(3)
frm eq(1); x=5+y — (4)
therefore, 5(5+y)(y)-10(5+y)-y=14
(15+3y)y-50-10y-y=14
3y^2+4y-50-14=0
3y^2+4y-64=0
3y^2 -12y + 16y-64=0
3y(y-4)+16(y-4)=0
(3y+16)(y-4)=0
y=-16/3 or 4
therefore from eqn(1);
x+4=5
x=5+4=9
the number is 94
(9b)
(3-2x)/ 4 + (2x-3)/3
(3(3-2x)+4(2x-3))/12
(9-6x+8x-12)/12
=(2x-3)/12
============================
(10a)
y=(2x^2 + 3)^5
let U=2x^2 + 3
Y=u^5
du/dx = 4x
dy/du = 5u^4
dy/du = (2x^2 + 3)^4
dy/dx = du/dx dy/du
dy/dx = 4x.5(2x^2 + 3)^4
dy/dx = 20x(2x^2 + 3)^4
(10b)
y=3x^2 + 2x +5
dy/dx =6x + 2
dy/dx =6(3) +2
dy/dx =18+2
dy/dx =20
(10c)
R-W=Wv^2/gx
Wv^2=gx(R-W)
Wv^2=gRx-Wgx
Wv^2+Wgx=gRx
W(v^2 + gx) =gRx
W=gRx/V^2 + gx
R=2, g=10, x=3/2, V=3
W= 10*2*3/2/3^2 + 10*3/5
W=30/9+15
W=30/24
W=5/4
(12a)
TABULATE
class interval:50-54,55-59,60-64,65-69,70-74
F:5,5,15,12,3=40
x:52,57,62,67,72
x:260,285,930,804,216=2495
x-x^-:-10.4,-5.4,-0.4,4.6,9.6
(x-x^-)^2:108.16,29.16,0.16,23.16,92.16
F(x-x^-)^2:540.8,145.8,2.4,253.92,276.48=1219.4
(12bi)
Mean(x^-)=Efx/Ef=2495/40
=62.4
(12bii)
Variance=Ef(x-x^-)^2/Ef=1219.4/40
=30.485
(12biii)
Standard deviation=sqroot[Ef(x-x^-)^2/Ef
=Sqroot1219.4/40
=sqroot30.485
=5.52
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Michael Ishola Sanni

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