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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

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

4)

(x^2+5x+1)sqroot(2x^3+mx^2+nx+11)=(2x-5)

remainder:30x+16

(x^2+5x+1)(2x-5)

=2x^3+10x^2+2x-5x^2-25x-5

=2x^3+10x^2-5x^2-25x-5

=2x^3+5x^2-23x+30x+16-5

=2x^3+5x^2+7x+11

Therefore m=5, n=7

—————————————————-

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/s

——————————————-

(11a)

Given:

f(x)={(4x-x^2)dx

f(x)=2x^2 – x^3/3 + K

f(3)=2(3)^2 – (3)^2/3 + K =21

18 – 9 + K=11

9+K=21

K=21-9

K=12

Therefore

f(x)= -x^3 + 2x^2 + 12

——————————————

 
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
=================

11a)

Kp2=72

K!/(k-2)!=72

K(k-1)(K-2)!/(K-2)!=72

K^2-K=72

K^2-K-72=0

READ ALSO  Financial Accounting (Essay & Objective) Waec 2017 Free Expo Answers

K^2-9k+8k-72=0

K(K-9)+8(k-9)=0

(K+8)(K-9)=0

k=-8,K=9

We consider positive value of K=9

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 = sqr 45 . 25 = 6 . 73

 

11b)

 

The equation 2cos^2tita-5costita=3

Let cos tita=x

2x^2-5x=3

using quadratic formular

a=2,b=-5,c=-3

5+_root(25+24)/4

=5+_root(49)/4

=(5+_7)/4

=(5+7)4=3 or (5-7)/4=-2/4=-1/2

since x cos tita

cos tita=-0.5

tita=cos^-1(-0.5)

tita=120degrees

——————

 

 

15a)

 

at max height

V=0m/s

g=10m/s^2

V^2=u^2-2gh

0^2=30^2-2 10H

0=900-20H

20H=900

H=900/20

H=45m

 

15b)

 

time taken to get to max height

V=u-gt

0=30-10t

10t=30

t=30/10

t=3secs

Timetaken to return=2t

=2 3=6secs

 

15c)

 

H=40m

H=ut-1/2gt^2=40

30t-1/210t^2=40

30t-5t^2=40

5t^2-30t+40=0

t^2-2t-4t+8=0

(t^2-2t)-(4t+8)=0

t(t-2)-4(t-2)=0

(t-2)(t-4)=0

t-2=0 or t-4=0

t=2secs or t=4secs

(14ai)
SKETCH THE DIAGRAM

(14aii)
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
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

 
 
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