Question 34

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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: 

$\begin{align}  & T=\sqrt{\frac{Tk-H}{k-H}} \\ & \text{Square both sides} \\ & {{T}^{2}}=\frac{TK-H}{K-H} \\ & {{T}^{2}}(K-H)=TK-H \\ & {{T}^{2}}K-{{T}^{2}}H=TK-H \\ & K({{T}^{2}}-T)=H({{T}^{2}}-1) \\ & K=\frac{H({{T}^{2}}-1)}{{{T}^{2}}-T} \\\end{align}$

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