Friday, 24 March 2017

Further Mathematics Answers.... OBJ and Theory..

Advertise with Us Here


OBJ LOADING




NIGERIA ANSWERS

1a)
g(x)=y
y=x+6
x=y-6
g^- f(x-6)
=4-5(x-6)/2=4-5x+30/2
=34-5x/2
1b)
coodinate=(x1+x2/2 ,y1+y2/2)
=(7-2/2,7-5/2)=(5/2,2/2)
=(5/2,1)


2)
x + 1 /3 x ^ 2 – x – 2
3x ^ 2- x – 2 / – 6 (- 3 2 )
3x ^ 2- 3 x + 2 x – 2
3x ( x – 1 ) + 2( x – 1 )
(3 x +2 ) ( x – 1 )
x + 1 /( 3x + 2 ) (x – 1)= A/ 3 x + 2 +B / x – 1
x + 1 /( 3x + 2 ) (x – 1)= A( x – 1 ) + B (3 x +2 ) /
(3 x +2 ) (x – 1 )
x + 1 = A( x – 1 ) + B (3 x +2 )
3x + 2 =0
x = – 2/ 3


5a)
pr(age)=4/5
pr(fully)=3/4
pr(must)=2/3
pr(age not admitted)=1-4/5
=1/5
pr(fully not admitted)=1-3/4
=1/4
pr(must not admitted)=1-2/3
=1/3
Therefore pr(none admitted)=1/5*1/4*1/3
=1/60

5b)
pr(only age and fully gained admission)=4/5*3/4*1/3
=1/5

7)
m1=3
u1=8m/s
m2=?
u2=5m/s
v=6m/s
m1u1+m1u2=(m1+m2)v
3*8+m2*5=(3+m2)6
24+5m2=18+6m2
24-18=6m2-5m2
m2=6

(7b)
m2u2-m1u1=V(m1+m2)
6*5-3*8=V(3+6)
30-24=9v
9v=6
v=6/9
v=0.67m/

10a)
i ) ( x ^2-1 ) ( x +2 )= 0
( x -1 ) ( x +1 ) ( x +2 )
x =1, or – 1 or -2
ii ) 2 x -3/( x -1) ( x + 1) ( +2)
=A /x -1+B /x + 1+C /x +2
2x – 3=A ( x +1) ( x +2 ) +B ( x -1) ( x +2)
+C ( x -1) ( x + 1)
let x + 1=0, x =- 1
2( -1 ) -3=B ( – 1-1) ( – 1+2)
-5/2 =-2B /- 2 B =5/2
let x – 1 = 0 x = 1
2( 1) – 3=A ( 1 +1) ( 1+2 )
-1= CA , A = -1/6
Let x +2=0 x =- 2
2( -2 ) -3=C ( -2- 1) ( -2 +1)
-7= 3C , C =-7 /3


(14ai)
SKETCH THE DIAGRAM
14ii)
Using lami’s theory
T1/sin60=T2/sin30
48N/sin60=T2/sin30
48N/0.8660=T2/0.5
0.5(48)/0.8660=T2(0.8660)/0.8660
T2=24/0.8660
T2=27.7N
(14b)
Using the equation of motion
H=U^2/2g
H=(20)^2/2*10
=20*20/20
H=20m
Timetaken to reach the maximum height
S=Ut+1/2at^2
20=0+1/2(100)t^2
20/5=5t^2/5
t^2=4
t=sqroot4
t=2S


Question 11a
Given
F(x)=§(4x-x^2)dx
F(x)=2x^2-x^3/3+k
F(3)=2(3)^2-(3)^3/3+k=21
18-9+k=21
K=21-9
K=12
Therefore
F(x)=x³/3+2x²+12

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

KEEP REFRESHING FOR THE ANSWERS
MORE LOADING…

GHANA ANSWERS

2)
x+1/3x^2-x-2
3x^2-x-2/-6(-3 2)
3x^2-3x+2x-2
3x(x-1)+2(x-1)
(3x+2)(x-1)
x+1/(3x+2)(x-1)=A/3x+2+B/x-1
x+1/(3x+2)(x-1)=A(x-1)+B(3x+2)/(3x+2)(x-1)
x+1=A(x-1)+B(3x+2)
3x+2=0
x=-2/3
=================

8a)
60,56,70,63,50,72,65,60
mean=£x/n=60 + 56 + 70 + 63 + 50 + 72 + 65 + 60/8
mean=62
8b)
variance=£(x-x^-)^2/n
=(62-60)^2 + (62-56)^2 + (62-70)^2 + (62-63)^2 + (62-50)^2 + (62-72)^2 + (62-63)^2 + (62-60)^2/8
=362/8=45.25
SD=sqr variane =sqr45.25=6.73


(9a)
1/1-cos tita + 1/1+cos tita
=1+cos tita + 1-cos tita//(1-cos tita) (1+cos tita)
= 2/1+cos tita – cos tita – cos^2 tita
= 2/1-cos^2 tita
Recall that :
Cos^2 tita + sin^2 tita = 1
.:. Cos^2 tita = 1-sin^2 tita
.:. 1/1-cos^2 tita + 1/1+cos tita
= 2/1-(1-sin^2 tita)

(9b)
At stationary points,
dy/dx=0.
y=x^0(x-3)
Let u=x^2,v=x-3.
du/dx=2x dv/dx=1.
dy/dx= Udv/dx + Vdu/dx
dy/dx=x^2(1)+(x-3)(2x)
.:. dy/dx=x^2+2x^2-6x
dy/dx=3x^2-6x
At stationary point,
dy/dx=0..
.:.3x^2-6x=0
Equation of line=> 3x^2-6x=0
==================

10b)
X1 Y2
(3, 1)
r=sqr(x2-x1)^2+(y2-y1)^2
r=sqr(3+3)^2+(1-1)^2
r=sqr6^2+0=sqr36=6
the equatuon of a circle
(x-a)^2+(y-b)^2=r^2
(x-(-3))^2+(y-1)^2=6^2
(x+3)^2+(y-1)^2=36
x^2+6x*9+y^2-2y+1=36
x^2+y^2+6x-2y+9+1-36=0
x^2+y^2+6x-2y-26=0

=================
©Ay2much Blog

No comments:

Post a Comment