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the way we're seeing things, Answers might be posted at mid-night, so keep coming online and check here regularly, don't share this post with your friends pls.






MATHEMATICS OBJ 100% VERIFIED :
1-10: ADADDDCEBE
11-20: CCDCAACECA
21-30: DCDAECAEED
31-40: EAADBEEADC
41-50: BBDCCBCECD
51-60: CADBAECDEC

We are sorry for the OBJ flop,



/ means DIVISION
* means Multiplication
tita You Should Know that.
rais means Raise to power or
^ means Raise to power
sqr rut/square rut means Square root
—–(1) means Equation 1
yr means Year



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
=================

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
=======================



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
=================


(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
========================




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=144-24r+25
r^2=169-24r
r^2+24r-169=0
r^2+24r=169
r^2+24r+14^2=169+14^2
(r+14)^2=169+196
(r+14)^2=365
(r+14=sqr365
r+14=19.105
r=19.105-14
r=5.105
r=5.1cm
8aii)
circumfrenece of a circle=2pie R
C=2x22/2*(5.1)^2
C=1144.44/7
C=163.4914cm
C=163.5cm
8b)
y2-y1/x2-x1=y-y1/x-x1
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


(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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