Question 14

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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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