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
= \(\frac{2 - 3}{3}\)
= \(\frac{-1}{3}\)