waecmaths question:
Make k the subject of the relation $T=\sqrt{\frac{Tk-H}{k-H}}$
Option A:
$k=\frac{H({{T}^{2}}-1)}{{{T}^{2}}-T}$
Option B:
$k=\frac{HT}{{{(T-1)}^{2}}}$
Option C:
$k=\frac{H({{T}^{2}}+1)}{T}$
Option D:
$k=\frac{H(T-1)}{T}$
waecmaths solution:
$\begin{align} & T=\sqrt{\frac{Tk-H}{k-H}} \\ & \text{Square both sides} \\ & {{T}^{2}}=\frac{TK-H}{K-H} \\ & {{T}^{2}}(K-H)=TK-H \\ & {{T}^{2}}K-{{T}^{2}}H=TK-H \\ & K({{T}^{2}}-T)=H({{T}^{2}}-1) \\ & K=\frac{H({{T}^{2}}-1)}{{{T}^{2}}-T} \\\end{align}$
maths year:
maths topics: