$\dfrac{1}{\sqrt{6}} + \dfrac{1}{\sqrt{6}}$ equals
$\dfrac{1}{\sqrt{6}} + \dfrac{1}{\sqrt{6}}$ equals
- $\dfrac{\sqrt{6}}{3}$
- $\sqrt{6}$
- $\dfrac{2}{6}$
- $\dfrac{2}{\sqrt{6}}$
Solution
$\dfrac{2}{\sqrt{6}} \cdot \dfrac{\sqrt{6}}{\sqrt{6}} = \dfrac{2\sqrt{6}}{6} = \dfrac{\sqrt{6}}{3}$.
Asked in: MH-SSC-9
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