*waec 2018 further maths answers is out*

*(6)*

It follow that:

P(n) = 1/3 and p(k)¹ = 1 – 1/3 = 2/3

P(T) = 1/5 and P(T)¹ = 1- 1/5 = 4/5

It follow that:

P(n) = 1/3 and p(k)¹ = 1 – 1/3 = 2/3

P(T) = 1/5 and P(T)¹ = 1- 1/5 = 4/5

*Hence,*

probability that only one if the them will be solve the questions will be:

probability that only one if the them will be solve the questions will be:

*= (1/3 × 4/5) + (1/5 × 2/3)*

= 4/15 + 2/15

= 6/15

= 2/5

= 4/15 + 2/15

= 6/15

= 2/5

*(7)*

m = 3i – 2j ; n = 2i + 3j ; p = i + 6j

m = 3i – 2j ; n = 2i + 3j ; p = i + 6j

*Therefore:*

4(3i – 2j) +2(2i +3j) -3(-i + 6j)

12i – 8j + 4i + 6j + 3i – 18j

12i + 4i + 3i – 8j + 6j – 18j

19i = 20j

[4/9, 09:47] Mr_Cyborg: 2018 WAEC MAY/JUNE FURTHER MATHS OBJ AND THEORY

4(3i – 2j) +2(2i +3j) -3(-i + 6j)

12i – 8j + 4i + 6j + 3i – 18j

12i + 4i + 3i – 8j + 6j – 18j

19i = 20j

[4/9, 09:47] Mr_Cyborg: 2018 WAEC MAY/JUNE FURTHER MATHS OBJ AND THEORY

*Please Note the below symbols*

*^ means raise to power*

* means multiplication

/ means division

* means multiplication

/ means division

*(1)*

*| x-3 -4 3 |*

| 5 2 2 | = -24

| 2 -4 6-x |

| 5 2 2 | = -24

| 2 -4 6-x |

*(x-3)[2(6-x) +8]+4[5(6-x) -4]+3(-20-4) =-24*

(x-3)[-2x+20]+4[-5x+26]+3(-24)= -24

-2x^2+26x-60-20x+104-72= -24

-2x^2+6x-4=0

X^2-3x+2=0

X^2-2x-x+2=0

X(x-2)-1(x-2)=0

(x-1)(x-2)=0

X-1=0 OR x-2=0

X=1 OR x=2

(x-3)[-2x+20]+4[-5x+26]+3(-24)= -24

-2x^2+26x-60-20x+104-72= -24

-2x^2+6x-4=0

X^2-3x+2=0

X^2-2x-x+2=0

X(x-2)-1(x-2)=0

(x-1)(x-2)=0

X-1=0 OR x-2=0

X=1 OR x=2

*=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=*

*(7)*

m = 3i – 2j ; n = 2i + 3j ; p = i + 6j

m = 3i – 2j ; n = 2i + 3j ; p = i + 6j

*Therefore:*

4(3i – 2j) +2(2i +3j) -3(-i + 6j)

12i – 8j + 4i + 6j + 3i – 18j

12i + 4i + 3i – 8j + 6j – 18j

19i = 20j

4(3i – 2j) +2(2i +3j) -3(-i + 6j)

12i – 8j + 4i + 6j + 3i – 18j

12i + 4i + 3i – 8j + 6j – 18j

19i = 20j

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