Question #fd348

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Question #fd348

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Answer 1 · 2016-03-18T01:00:25

$\frac{1 - (- \frac{1}{3}) = \frac{4}{3}}{2} = \frac{2}{3}$ $(1 - (-1/3) = 4/3)/2 = 2/3$ and $\sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$ $sqrt(2/3) = sqrt2/sqrt3 = sqrt6/3$

Explanation:

Take the $- (- \frac{1}{3})$ $- ( -1/3)$ and turn it into $\frac{1}{3}$ $1/3$

Then, turn the $1 \frac{1}{3}$ $1 1/3$ into $\frac{4}{3}$ $4/3$

When you divide $\frac{4}{3}$ $4/3$ by $2$ $2$ it gets you $\frac{2}{3}$ $2/3$

Next, using that one square root identity, $\sqrt{\frac{2}{3}}$ $sqrt(2/3)$ becomes $\frac{\sqrt{2}}{\sqrt{3}}$ $sqrt2/sqrt3$

Finally, remove the radical on bottom by multiplying by $\frac{\sqrt{3}}{\sqrt{3}}$ $sqrt3/sqrt3$ to turn $\frac{\sqrt{2}}{\sqrt{3}}$ $sqrt2/sqrt3$ into $\frac{\sqrt{6}}{3}$ $sqrt6/3$ .

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Question #fd348

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