Question 34

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waecmaths question: 

If $P={{\left[ \frac{Q(R-T)}{15} \right]}^{\tfrac{1}{3}}}$ Make T the subject of the relation

Option A: 

$T=\frac{R+{{P}^{3}}}{15Q}$

Option B: 

$T=\frac{R-15{{p}^{3}}}{Q}$

Option C: 

$T=R-\frac{15{{p}^{3}}}{Q}$

Option D: 

$T=\frac{15R-Q}{{{P}^{3}}}$

waecmaths solution: 

$\begin{align}  & P={{\left[ \frac{Q(R-T)}{15} \right]}^{\tfrac{1}{3}}} \\ & \text{Cube both sides} \\ & {{P}^{3}}=\frac{Q(R-T)}{15} \\ & \text{Multiply both sides by }15 \\ & 15{{P}^{3}}=Q(R-T) \\ & \frac{15{{P}^{3}}}{Q}=R-T \\ & T=R-\frac{15{{P}^{3}}}{Q} \\\end{align}$

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