MAths-Obj
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MATHS THOERY
1a)
Tabulate
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)
I = PRT/100,
p=N15000
R=10% and I=3years
A = P+ I where I =
15000*10*3/100=N4500
A=4500+15000 =N19500
2a)
using sine rule
b/sin20 = 6/sin30
bsin30 = 6sin120
b 6sin120/sin30
b = 6x0.2511/0.4540
b = 5.7063/0.4540
b = 12.57 ≠ 12.6cm
2bi)
the diagram is euivalent triangles.
where
|AX|/|BC| = |BY|/|AC| = |XY|/|
YC|
XY = 9, BY = 7
YC = 18-7=11
9/11 = 7/|AC|
9|AC| = 77
|AC| = 77/9
|AC| = 8cm
2bii)
XY/AB = BY/AC
9/|AB| = 7/8.6
|AB| = 9x8.6/7
|AB| = 11cm
3)
let the son age be x
man=5x
son=x
4yrs ago;the man age = 5x - 4
the son age = x - 4
the product of their ages
(5x - 4)(x - 4) =448
4a)
volume of fuel = cross-sectional area
of X depth of fuel rectangular tank
30,000litres = 7.5*4.2*d m^3
but; 1000litres =1m^3
therefore;30(M^3) = 7.5*4.2*d(M^3)
30=31.5d
====> d = 30/31.5 = 0.95(2d.p)
4b)
to fill the tank/volume of fuel
needed = 7.5*4.2*1.2 = 37.8m^3 =
37,800
litres addition fuel = 37,800-30,000 =
7,800
litres therefore, 7,800
more litres would be needed
================
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
5b)
amount spent=144000-
[48,000+32000+19200+12000+7200]
=144000-118400 =N25600
===============
7a)
3²ⁿ+¹ — 4(3ⁿ+¹) + 9 = 0
3²ⁿ × 3 — 4(3ⁿ× 3¹) + 9 = 0
(3ⁿ)² × 3 — 4(30ⁿ× ) + 9 = 0
Let 3ⁿ= x
3x² — 4 × 3 × x + 9 = 0
3x² — 12x + 9 = 0
Divide all by 3
3x²/3 — 12x/3 + 9/3 = 0
x² — 4x + 3 = 0
x² — 3x — x + 3 = 0
x(x—3) -1(x—3) = 0
(x—3)(x—3) = 0
x—3 = 0 or x—1 = 0
x = 3, x = 1
Substitute x = 3
3ⁿ = 3 or 3ⁿ = 1
3ⁿ = 3¹ or 3ⁿ= 3°
n = 1 or n= 0
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
8a )
| AD | ^2= 13^2- 5^2
| AD | ^2= 169- 25
| AD | ^2= 144
AD= sqr144
AD= 12CM
| AD | =12 -r
r ^2=( 12- r )^2 - 5^2
r ^2=( 12- r )(12 -r ) +25
r ^2=144 -24 r +25
r ^2=169 -24 r
r ^2+24 r -169 =0
r ^2+24 r =169
r ^2+24 r +14^2= 169+14 ^2
(r +14 )^2= 169+196
(r +14 )^2= 365
(r +14 =sqr 365
r +14= 19. 105
r =19. 105- 14
r =5. 105
r =5. 1cm
8aii )
circumfrenece of a circle = 2pie R
C =2x 22 /2*(5. 1)^2
C =1144 . 44/7
C =163. 4914 cm
C =163. 5cm
8b)
y 2- y 1/x 2 -x 1 =y - y 1/x - x 1
6- 2/2- (- 1)=y -2/x -( -1)
4/2+1 = y -2 /x +1
4/3=y -2/x +1
3(y -2)= 4( x +1)
3y - 6=4x +4
3y - 4x =4 +6
3y - 4x =10
y =4x /3+10/3
==================================
9a )
let Xy represent the two digit number
x - y =5 -- -- - (i )
3xy - ( 10x +y )= 14
3xy - 10 x - y =14 -- - -( ii )
from eqn (i )
x =5+ y
3y ( 5+y ) -10 (5+y )- y =14
15y + 3y ^2 - 50 - 10 y - y =14
3y ^2 + 4 y - 50 = 14
3y ^2 + 4 y - 50 - 14 = 0
3y ^2 + 4 y - 64 =0
3y ^2 + 12 y + 16y - 64 =0
(3y ^2 - 12 y ) (+16 y - 64 )=0
by
(y - 4) +16( y -4 )=0
(y - 4)= 0
ii )
(3y +16) (y - 4)= 0
3y +16 =0 or y - 4=0
3y =- 16 or y =4
y =- 16/3 or y =4
when y =4
x =5+ y
x =5+ 4
x =9
the no is 94
9b)
3- 2x /4 + 2x -3/3
=3( 3-2 x )+4 (2x - 3) /12
=9- 6x +8 x -12 /12
=2x - 2/12
==================================
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
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