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By 12:41 Tue, 21 Jun 2016 Comments


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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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