Question 20

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waecmaths question: 

Simplify $2\sqrt{3}-\frac{6}{\sqrt{3}}+\frac{3}{\sqrt{27}}$

Option A: 

1

Option B: 

$\tfrac{1}{3}\sqrt{3}$

Option C: 

$2\sqrt{3}-5\tfrac{2}{3}$

Option D: 

$6\sqrt{3}-17$

waecmaths solution: 

$\begin{align}  & 2\sqrt{3}-\frac{6}{\sqrt{3}}+\frac{3}{\sqrt{27}}=2\sqrt{3}-\frac{6}{\sqrt{3}}+\frac{3}{3\sqrt{3}} \\ & =2\sqrt{3}-\frac{6}{\sqrt{3}}+\frac{1}{\sqrt{3}} \\ & =\frac{6-6+1}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\ & =\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \end{align}$

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