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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