[3(6x) - 5][2(6x)m + 1]
[3(6x) + 5][2(6x) - 1]
[(6x) - 5][3(6x) + 1]
[2(6x) + 5][3(6x) - 1]
Correct answer is B
62x + 1 + 7(6x) - 5
Let 6x = y....(i)
6yx + 7y - 5 = 0
(3y + 5)(2y - 1) = 0.....(i)
Subt. for y from eqn. (i) in eqn. (ii)
= [3(6x) + 5][2(6x) - 1]
-1, 1
-1, 2
1, -2
2, -2
Correct answer is C
x2 + 2(m + 1)x + m + 3 from (A + B)2 = a2 + 2AB + b2
2B = \(\frac{2(m + 1)}{2}\) = m + 1 .......(i)
Also B2 = m + 3..........(ii)
Subst. for B in equation ..........(iii)
= (m + 1)2 = m + 3 = m2 + 2m + 1 = m + 3
= m2 + 2m - m + 1 - 3 = 0
m2 + m - 2 = 0
(m + 2)(m - 1) = 0 or m - 1 = 0
m = -2 or 1
Make y the subject of the formula Z = x\(^2\) + \(\frac{1}{y^3}\)
y = \(\frac{1}{(Z - x^2)^3}\)
y = \(\frac{1}{(Z + x^2)^{\frac{1}{3}}}\)
y = \(\frac{1}{(Z - x^2)^{\frac{1}{3}}}\)
y = \(\frac{1}{\sqrt[3]{Z} - \sqrt[3]{x^2}}\)
Correct answer is C
Z = x\(^2\) + \(\frac{1}{y^3}\)
Z - x\(^2\) = \(\frac{1}{y^3}\)
y\(^3\) = \(\frac{1}{Z - x^2}\)
y = \(\sqrt[3]{\frac{1}{Z - x^2}}\)
∴ y = \(\frac{1}{\sqrt[3]{Z - x^2}}\)
y = \(\frac{1}{(Z - x^2)^{\frac{1}{3}}}\)
P = 98R2
PR2 = 98
P = \(\frac{1}{98R2
P = \(\frac{PR2}{98}\)
Correct answer is B
P = \(\frac{1}{v}\) and vR2 = P = \(\frac{k}{v}\)......(i)
and v KR2 .......(ii)
(where k is constant)
Subst. for v in equation (i) = p = \(\frac{1^2}{KR}\).....(ii)
when r = 7, p = 2
2 = \(\frac{k}{7^2}\)
k = 2 x 49
= 98
Subt. foe k in ....(iii)
P = \(\frac{98}{R^2}\)
PR2 = 98
350k
200k
150k
50k
Correct answer is B
Q = 1.5 + 0.5n gives the cost 1(in Naira) of feeding n people for a week. Extra cost of feeding one additional person = n = 1 Subt. for n in the formula Q = 1.5 + 0.5(1) = 15 + 0.5 Q = N2 = 200k