waecmaths question:
Given that ${{p}^{\tfrac{1}{3}}}=\frac{\sqrt[3]{q}}{r}$, make q the subject of the equation
Option A:
$q=p\sqrt{r}$
Option B:
$q={{p}^{3}}r$
Option C:
$q=p{{r}^{3}}$
Option D:
$q=p{{r}^{\tfrac{1}{3}}}$
waecmaths solution:
$\begin{align} & {{p}^{\tfrac{1}{3}}}=\frac{\sqrt[3]{q}}{r} \\ & \text{Multiply both sides by }r \\ & \sqrt[3]{q}={{p}^{\tfrac{1}{3}}}r \\ & \text{Cube both sides} \\ & q=p{{r}^{3}} \\\end{align}$
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