Question #95e73

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Question #95e73

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Answer 1 · 2017-03-09T18:56:53

$\frac{\sqrt{2}}{2}$ $sqrt(2)/2$

Explanation:

$\sqrt{{(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2}}$ $sqrt((1/2)^2+(1/2)^2)$

First find out what ${(\frac{1}{2})}^{2}$ $(1/2)^2$ is:

${(\frac{1}{2})}^{2} = \frac{1}{4}$ $(1/2)^2 = 1/4$

So, we now know that the expression is now $\sqrt{\frac{2}{4}}$ $sqrt(2/4)$ which is the same as $\sqrt{\frac{1}{2}}$ $sqrt(1/2)$ .

Then, we rationalize the denominator as we cannot have a root as
the denominator:

$\sqrt{1} = 1$ $sqrt(1) = 1$ , so the expression is now $\frac{1}{\sqrt{2}}$ $1 /sqrt(2)$

Multiply both the numerator and the denominator by $\sqrt{2}$ $sqrt(2)$

Denominator $= \sqrt{2} \cdot \sqrt{2} = 2$ $= sqrt(2) * sqrt(2) = 2$

Numerator $= 1 \cdot \sqrt{2} = \sqrt{2}$ $= 1 * sqrt(2) = sqrt(2 )$

Hence,

$= \frac{\sqrt{2}}{2}$ $= sqrt(2)/2$

Answer 2 · 2017-03-09T19:30:24

The expression is equivalent to $\frac{\sqrt{2}}{2}$ $sqrt(2)/2$ .

Explanation:

The square of $\frac{1}{2}$ $1/2$ is $\frac{1}{4}$ $1/4$ because $(\frac{1}{2}) (\frac{1}{2}) = \frac{1}{2 \cdot 2} = \frac{1}{4}$ $(1/2)(1/2) = 1/(2*2) = 1/4$ .

Therefore,

$\sqrt{\frac{1}{4} + \frac{1}{4}}$ $sqrt(1/4 + 1/4)$

$\sqrt{\frac{2}{4}}$ $sqrt(2/4)$

$\sqrt{\frac{1}{2}}$ $sqrt(1/2)$

We can separate the radicals.

$\frac{\sqrt{1}}{\sqrt{2}}$ $sqrt(1)/sqrt(2)$

$\frac{1}{\sqrt{2}}$ $1/sqrt(2)$

I would recommend you rationalize the denominators.

$\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$ $1/sqrt(2) * sqrt(2)/sqrt(2)$

$\frac{\sqrt{2}}{2}$ $sqrt(2)/2$

Hopefully this helps!

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Question #95e73

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