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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