waecmaths question:
Make b the subject of the relation$lb=\tfrac{1}{2}(a+b)h$
Option A:
$\frac{ah}{2l-h}$
Option B:
$\frac{2l-h}{al}$
Option C:
$\frac{al}{2l-h}$
Option D:
$\frac{al}{2-h}$
waecmaths solution:
$\begin{align} & lb=\tfrac{1}{2}(a+b)h \\ & 2lb=ah+bh \\ & 2lb-bh=ah \\ & b(2l-h)=ah \\ & b=\frac{ah}{2l-h} \\\end{align}$
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