Question:
$\left(\frac{-2}{5}\right)^{7} \div\left(\frac{-2}{5}\right)^{5}$ is equal to
(a) $\frac{4}{25}$
(b) $\frac{-4}{25}$
(c) $\left(\frac{-2}{5}\right)^{12}$
(d) $\frac{25}{4}$
Solution:
(a) 4/25
We have:
$\left(\frac{-2}{5}\right)^{7} \div\left(\frac{-2}{5}\right)^{5}=\left(\frac{-2}{5}\right)^{7-5}$
$=\left(\frac{-2}{5}\right)^{2}$
$=\frac{(-2)^{2}}{5^{2}}$
$=\frac{4}{25}$
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