Simplify $\frac{1}{\sqrt{2}} + \frac{3}{\sqrt{8}} + \frac{6}{\sqrt{32}}$ $1/sqrt2+3/sqrt8+6/sqrt32$ . Help, Plz?

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Simplify $\frac{1}{\sqrt{2}} + \frac{3}{\sqrt{8}} + \frac{6}{\sqrt{32}}$ $1/sqrt2+3/sqrt8+6/sqrt32$ . Help, Plz?

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Answer 1 · 2018-05-01T01:46:09

The way I would answer this is by first simplifying the bottom denominators as you need those to add. To do this I would multiply $\frac{1}{\sqrt{2}}$ $1/sqrt2$ by 16 to get $\frac{16}{\sqrt{32}}$ $16/sqrt32$ . I would multiply $\frac{3}{\sqrt{8}}$ $3/sqrt8$ by 4 to get $\frac{12}{\sqrt{32}}$ $12/sqrt32$ . This leaves you with $\frac{16}{\sqrt{32}} + \frac{12}{\sqrt{32}} + \frac{6}{\sqrt{32}}$ $16/sqrt32 + 12/sqrt32 + 6/sqrt32$ . From here we can add these to get $\frac{34}{\sqrt{32}}$ $34/sqrt32$ . We can simplify this even more by dividing by two to get $\frac{17}{\sqrt{16}}$ $17/sqrt16$ this is as simplified as this equation gets.

Answer 2 · 2018-05-01T02:06:06

$2 \sqrt{2}$ $2sqrt2$

Explanation:

First we need a common denominator. In this case, we'll use $\sqrt{32}$ $sqrt32$ .

Convert $\frac{1}{\sqrt{2}}$ $1/sqrt2$ by multiplying it by $\frac{\sqrt{16}}{\sqrt{16}}$ $sqrt16/sqrt16$

$\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{16}}{\sqrt{16}} = \frac{\sqrt{16}}{\sqrt{32}}$ $1/sqrt2 * sqrt16/sqrt16 = sqrt16/sqrt32$

We must also convert $\frac{3}{\sqrt{8}}$ $3/sqrt8$ by multiplying it by

$\frac{3}{\sqrt{8}} \cdot \frac{\sqrt{4}}{\sqrt{4}} = \frac{3 \sqrt{4}}{\sqrt{32}}$ $3/sqrt8 * sqrt4/sqrt4 = (3sqrt4)/sqrt32$

This leaves us with a simple equation:

$\frac{\sqrt{16}}{\sqrt{32}} + \frac{3 \sqrt{4}}{\sqrt{32}} + \frac{6}{\sqrt{32}}$ $sqrt16/sqrt32 + (3sqrt4)/sqrt32 + 6/sqrt32$

Now we simplify the numerators, and finish the equation.

$\frac{4}{\sqrt{32}} + \frac{6}{\sqrt{32}} + \frac{6}{\sqrt{32}} = \frac{16}{\sqrt{32}}$ $4/sqrt32 + 6/sqrt32 + 6/sqrt32 = 16/sqrt32$

We can also simplify this.

$\frac{16}{\sqrt{32}} = \frac{16}{4 \sqrt{2}} = \frac{4}{\sqrt{2}}$ $16/sqrt32 = 16/(4sqrt2) = 4/sqrt2$

If necessary, this can be rationalized.

$\frac{4}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{4 \sqrt{2}}{2} = 2 \sqrt{2}$ $4/sqrt2 * sqrt2/sqrt2 = (4sqrt2)/2 = 2sqrt2$

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Simplify $\frac{1}{\sqrt{2}} + \frac{3}{\sqrt{8}} + \frac{6}{\sqrt{32}}$ $1/sqrt2+3/sqrt8+6/sqrt32$ . Help, Plz?

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