Question #c6b3b

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Question #c6b3b

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Answer 1 · 2017-08-22T04:36:22

$\frac{31}{15}$ $31/15$

Explanation:

So you want $6$ $6$ to be recurring. This is shown as $2.0 . 6 . 6$ $2.0dot6dot6$ .

We set this equal to $x$ $x$ .

$x = 2.0 . 6 . 6$ $x=2.0dot6dot6$

Multiply by $10$ $10$

$10 x = 20 . . 6 . 6$ $10x=20.dot6dot6$

Multiply by $10$ $10$ again

$100 x = 206 . . 6 . 6$ $100x=206.dot6dot6$

Now we know what $10 x$ $10x$ and $100 x$ $100x$ is equal to. Then, subtracting $10 x$ $10x$ from $100 x$ $100x$ will get rid of the $. . 6 . 6$ $.dot6dot6$ (Annoying part)

$90 x = 186$ $90x=186$

Divide by $90$ $90$

$x = \frac{31}{15}$ $x=31/15$

Now since we set $x = 2.0 . 6 . 6$ $x=2.0dot6dot6$ ,

$2.0 . 6 . 6 = \frac{31}{15}$ $2.0dot6dot6=31/15$

And bob's your uncle.
Hope this helped! ;)

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Question #c6b3b

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