MATHS OBJ:
1-10: DCBBDCCADB
11-20: ACBACCDADA
21-30: DAAAABDBBB
31-40: ABCDABCDCA
41-50: DBABCBABBB
*COMPLETED...!
1-10: DCBBDCCADB
11-20: ACBACCDADA
21-30: DAAAABDBBB
31-40: ABCDABCDCA
41-50: DBABCBABBB
*COMPLETED...!
ANSWER NUMBER 4
ANSWER NUMBER ONE(1)
ANSWER NUMBER 8
ANSWER NUMBER 7
ANSWER NUMBER
ANSWER NUMBER 6
(7a)
2
S(2x³ - 4x + 6)dx
1
= 2x³+¹/3+1 - 4x¹+¹/1+1 + 6x]2
1
=2x^4/4 - 4x²/2 +6x]2, 1
= x^4/2 - 2x² + 6x]2, 1
=(2^4/2 - 2(2²) + 6(2)) - (1^4/2 - 2(1)² + 6(1))
=(8 - 8 + 12) - (1/2 - 2 + 6)
=12 - 4½
= 7½
(7b)
Given; P^-1 = (-1 1)
(4 -3)
P = (p-1)^-1 = C^T/|p^-¹|
=(-3 -1)
(-4 -1)/3 - 4
=(-3 -1)
(-4 -1)/-1
=(3 1)
(4 1)
(6ai)
The profit y = X²/8 + 5x
y = GHc20,000.00
Hence 20,000 = X²/8 + 5x
160,000 = X² + 40x
X² + 40x - 160,000 = 0
Since X is in thousands
X² + 40x - 160 = 0
(6aii)
Using quadratic formula
X = -b±√b² - 4ac/2a
Where; a = 1, b = 40 & c = -160
X = -40±√40² - 4(1)(-160)/2(1)
X= -40 ±√1600 + 640/2
X = -40 ±√2240/2
X = -40 ± 47.32/2
X = -40±47.32/2
= 7.32/2
X = 3.66
X ≈ 4
(6b)
Draw the diagram
Using ΔTOP
tan 28 = H/OP
OP = H/tan28
Then for ΔROP
tantita = H/2/OP
OP = H/2/tantita
Hence H/tan 28 = H/2/tantita
tantita = H/2 × tan 28/H
tantita = tan28/2
Hence Tita = 28/2 = 14•
(4a)
(2y+x) + (6y-2x+1) + 4y = 28 ...(i)
6y-2x + 1 = 4y... (II)
2y+ x +6y - 2x + 1 +4 = 28
12y + x +6 -2x + 1 + 4 = 28
12y - x + 1 = 28
12y - x = 29... (III)
6y-2x + 1 = 4y, 6y-2x - 4y = 1
2y - 2x = -1... (iv)
24y - 2x = 54
2y - 2x = 1
22y/2w = 55/22 y = 2.5
12y - x = 27
12 (2.5) - x = 27
30 x 27 x=3
(b)
2y + x = 2 (2.5) + 3 = 5+3 = 8cm
6y - 2x + 1 = (2.5)-2 (3) + 1
= 15-6 + 1 = 10cm
4y = 4(2.5) = 10cm
Question 1
A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x: 3 < x < 9} are subsets of the universal-set
U = {2, 3, 4, 5, 6, 7, 8, 9}. Find
(a) A n(B’nC’);
(b) (AuB) n(BuC).
Mathematics waec gce 2017 ObservationPlease ensure to use curly brackets to enclose the elements of the sets.
Furthermore always list the elements of set C hence, this is for you to be able to find its complement.
Do it this way C i.e. C = {4, 5,6, 7, 8}, obtain the compliments of the sets Band C thus B’ = {4,5,6, 8}, C’ = {2, 3, 9,}.
Using these sets, the following procedures were to be followed:
(a)(B’ nC’) = { } Hence An (B’ n C’) = { }. Don’t be like Some candidates know to write like this { 0 } instead of { } or 0.
(b) (A u B) = { 2, 3,4,6, 7, 8, 9 }, (BuC) = {2, 3, 4,5,6, 7, 8, 9 } Therefore { Au B } n (BuC) = { 2, 3, 4, 6, 7, 8, 9 }.
Question 2
(a) The angle of depression of a boat from the mid-point of a vertical cliff is 35°. If the boat is 120 m from the foot of the cliff, calculate the height of the cliff.
(b) Towns P and Q are x km apart. Two motorists set out at the same time from P to Q at steady speeds of 60 km/h and 80 km/h. The faster motorist got to Q 30 minutes earlier than the other. Find the value of x.
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