Question 14

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waecmaths question: 

Make s the subject of the relation: $p=s+\frac{s{{m}^{2}}}{nr}$

Option A: 

$s=\frac{mrp}{nr+{{m}^{2}}}$

Option B: 

$s=\frac{nr+{{m}^{2}}}{mrp}$

Option C: 

$s=\frac{nrp}{mr+{{m}^{2}}}$

Option D: 

$s=\frac{nrp}{nr+{{m}^{2}}}$

waecmaths solution: 

$\begin{align}  & p=s+\frac{s{{m}^{2}}}{nr} \\ & p=\frac{snr+s{{m}^{2}}}{nr} \\ & pnr=s(nr+{{m}^{2}}) \\ & \frac{nrp}{nr+{{m}^{2}}}=s \\\end{align}$

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