Question

How do you write `0.3666...` as a fraction? The `6` repeating.

Answer 1

`18333/50000`

Explanation:

Put the number over 1( make denominator 1)
`0.36666/1`
and multiply by 10 for every number after the decimal point.
as there is 5 numbers you have to multiply by 100000
=`(0.36666*100000)/(1*100000)`

=`36666/100000`
and divide by the greatest factor, which is 2

=`(36666/2)/(100000/2)`

=`18333/50000` final answer.

Answer 2

See a solution process below:

Explanation:

First, we can write:

`x = 0.3bar6`

Next, we can multiply each side by `10` giving:

`10x = 3.bar6`

Then we can subtract each side of the first equation from each side of the second equation giving:

`10x - x = 3.bar6 - 0.3bar6`

We can now solve for `x` as follows:

`10x - 1x = (3.6 + 0.0bar6) - (0.3 + 0.0bar6)`

`(10 - 1)x = 3.6 + 0.0bar6 - 0.3 - 0.0bar6`

`9x = (3.6 - 0.3) + (0.0bar6 - 0.0bar6)`

`9x = 3.3 + 0`

`9x = 3.3`

`(9x)/color(red)(9) = 3.3/color(red)(9)`

`(color(red)(cancel(color(black)(9)))x)/cancel(color(red)(9)) = (3 xx 1.1)/color(red)(3 xx 3)`

`x = (color(red)(cancel(color(black)(3))) xx 1.1)/color(red)(color(black)(cancel(color(red)(3))) xx 3)`

`x = 1.1/3`

`x = 10/10 xx 1.1/3`

`x = 11/30`

Answer 3

`0.366...=11/30`

Explanation:

There's nice algebraic trick to get rid of repeating tail:

`x=0.366...`

`10x=3.666...`

`10x-x=3.6cancel(66...)-0.3cancel(66...)`

`9x=3.6-0.3=3.3`

`90x=33`

`x=33/90=11/30`

If there's `n` digits in repeating element, then multiply by `10^n`

`x=0.36565...`

`100x=36.56565...`

`100x-x=36.5cancel(6565...)-0.3cancel(6565...)`

`99x=36.5-0.3=36.2`

`990x=362`

`x=362/990=181/495`