WAEC 2016 MAY/JUNE FURTHER MATH POSTED FOR FREE

By 11:12 Fri, 13 May 2016 Comments


2016 WAEC MAY/JUNE FURTHER MATH

(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

(6a)

(1+4+k+k+4+11)/5=k+1

(20+2k)/5=k+1

20+2k=5(k+1)

20+2k=5k+5

20-5=5k-2k

15=3k

k=15/3

k=5

(6b)

1,4,5,9and 11

TABULATE

x:1,4,5,9,11

x-x^-:-4,-1,0,4,6

(x-x^-)^2:16,1,0,16,36

E(x-x^-)^2=69

Hence=E(x-x^-)^2/n

=69/5

=13.8

(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

(2)

(5,2)(-4,k)(2,1)

(y3-y2)/(x3-x2)=(y2-y1)/(x2-x1)

(1-k)/2-(-4)=(k-2)/(-4-5)

(1-k)/(2+4)=(k-2)/-9

(1-k)/6=(k-2)/-9

-9(1-k)=6(k-2)

-9+9k=6k-12

9k-6k=-12+9

3k=-3

k=-1

(3a)

If f(x+2)=6x^2+5x-8)

To find f(5)

Therefore f(x+2)=f(5)

where x+2=5

x=5-2

x=3

therefore f(5)=6(3)^2+5(3)-8

=6(9)+15-8

=54+7

=61

(3b)

(7root2+3root3)/(4root2-2roo3)*(4root2+2root3)/(4root2+2root3)

(24*2+14root6+12root6+6*3)/(16*2+8root6-8root6-4*3)

(48+26root6+18)/(32-12)

=(66+26root6)/20

=66/20+(26root6/20)

=33/10+(13root6/10)

=3.3+1.3root6

(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

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