How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
Given that t = \(2 ^{-x}\), find \(2 ^{x + 1}\) in terms of t.
\(\frac{2}{t}\)
\(\frac{t}{2}\)
\(\frac{1}{2t}\)
t
Correct answer is A
t = \(2^{-x} = \frac{1}{2^{x}}\)
\(\implies 2^{x} =\frac{1}{t}\)
\(2^{x+1} = 2^{x} \times 2^{1}\)
= \(\frac{1}{t} \times 2 = \frac{2}{t}\)
p \(\iff\) \(\sim\)q
p \(\iff\) q
\(\sim\)p \(\iff\) \(\sim\)q
q \(\iff\) \(\sim\)p
Correct answer is D
No explanation has been provided for this answer.
Make x the subject of the relation d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\)
x = \(\frac{6 + 12}{d^2 + y}\)
x = \(\frac{12}{d^2 - y}\)
x = \(\frac{12}{y} - 2d^2\)
x = \(\frac{12}{2d^2 + y}\)
Correct answer is D
d = \(\sqrt{\frac{6}{x} - \frac{y}{2}}\)
\(d^2 = \frac{6}{x} - \frac{y}{2}\)
\(2xd^2 = 12 - xy\)
\(2xd^2 + xy = 12\)
x = \(\frac{12}{2d^2 + y}\)
The roots of a quadratic equation are \(\frac{-1}{2}\) and \(\frac{2}{3}\). Find the equation.
\(6x^2 - x + 2 = 0\)
\(6x^2 - x - 2 = 0\)
\(6x^2 + x - 2 = 0\)
\(6x^2 + x + 2 = 0\)
Correct answer is B
(x + \(\frac{1}{2}\)) (n - \(\frac{2}{3}\))
\(x^2 - \frac{2}{3^x} + \frac{x}{2} - \frac{1}{3}\)
\(6x^2 - 4n + 3n - 2 = 0\)
\(6x^2 - x - 2 = 0\)
Find the 6th term of the sequence \(\frac{2}{3} \frac{7}{15} \frac{4}{15}\),...
-\(\frac{1}{3}\)
-\(\frac{1}{5}\)
-\(\frac{1}{15}\)
\(\frac{1}{9}\)
Correct answer is A
a = \(\frac{2}{3}\), d = \(\frac{7}{15}\) - \(\frac{2}{3}\)
= 7 - 10
= \(\frac{-3}{15}\)
d = - \(\frac{-1}{5}\)
T6 = a + 5d
= \(\frac{2}{3}\) + 5(\(\frac{-1}{5}\)
= \(\frac{2}{3}\) - 1
= \(\frac{2 - 3}{3}\)
= \(\frac{-1}{3}\)