Question 29

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waecmaths question: 

If  $y=\frac{2\sqrt{{{x}^{2}}+m}}{3N}$ make x the subject of the formula

Option A: 

$\frac{\sqrt{9{{y}^{2}}{{N}^{2}}-2m}}{2}$

Option B: 

$\frac{\sqrt{9{{y}^{2}}{{N}^{2}}+2m}}{2}$

Option C: 

$\frac{\sqrt{9{{y}^{2}}{{N}^{2}}-4m}}{2}$

Option D: 

.$\frac{\sqrt{9{{y}^{2}}{{N}^{2}}+4m}}{2}$

waecmaths solution: 

$\begin{align}  & y=\frac{2\sqrt{{{x}^{2}}+m}}{3N} \\ & 2\sqrt{{{x}^{2}}+m}=3yN \\ & \sqrt{{{x}^{2}}+m}=\frac{3yN}{2} \\ & {{x}^{2}}+m=\frac{9{{y}^{2}}{{N}^{2}}}{4} \\ & {{x}^{2}}=\frac{9{{y}^{2}}{{N}^{2}}}{4}-m \\ & {{x}^{2}}=\frac{9{{y}^{2}}{{N}^{2}}-4m}{4} \\ & x=\sqrt{\frac{9{{y}^{2}}{{N}^{2}}-4m}{4}} \\ & x=\frac{\sqrt{9{{y}^{2}}{{N}^{2}}-4m}}{2} \\\end{align}$

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