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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