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
Hence,
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
(7)
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
Please Note the below symbols
^ means raise to power
* means multiplication
/ means division
(1)
| x-3 -4 3 |
| 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
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(7)
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

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