any1 know how to do decimals to fracts with repeating?? if so please hlp?

Question

fo example 0.6 repeating

Answer 1

0.6666... is the same as 6/9 = 2/3

Take the repeating portion and put it over an equivalent number of 9s.

0.414141... is the same as 41/99
0.531531531... is the same as 531/999 which can be reduced to 59/111

If there are no repeating decimals, you just put it over 1 with as many zeros.

0.512 = 512/1000 = 64/125

Now if you have both a non-repeating and a repeating, you combine the two methods.

0.51241414141... = 512/1000 + 41/99000

Now add the fractions and reduce, if you can.

(99*512 + 41)/99000 = 50729/99000

Answer 2

.6 Repeating Is 2/3. Some Calculaters Have A Function That Is An Arrow Pointing To A F Or D Which Means Convert To A Decimal Or Fraction. And To Check If .666666... Is 2/3 Just Divide 2 By 3. And If You're Still Not Sure, Then... I Don't Know...

Answer 3

0.6666... = 6 x 0.1111... = 6 x ( 1 / 9 ) = 6 / 9 = 2 / 3

because 9 x 0.11111... = 0.99999... = 1

Now for a hard one:

0.294117647058823529411764705882352....

= 2941176470588235 x 0.00000000000000010000000000000001 ...

= 2941176470588235 / 9999999999999999

= ( 3^2 × 5 × 11 × 7^3 × 101 × 137 × 5882353) /
( 3^2 × 11 × 17 × 7^3 × 101 × 137 × 5882353 )

= 5 / 17

Can you now see the general rule

Example: 0.571428571428...

Write the repeating part as an integer, and divide by 999...999 with the number of 9's being the same as the number of digits that repeat:

= 571428 / 999999

Factor both numerator and denominator and cancel:

( 2^2 × 3^3 × 11 × 13 × 37 ) / ( 3^3 × 7 × 11 × 13 × 37 )

= 2^2 / 7 = 4 / 7

Alternatively, find the largest comon factor of the numerator annd denominator;

gcd of 571428 and 999999 = 142857

and divide:

( 571428 / 142857 ) / ( 999999 / 142857 )

= 4 / 7

Answer 4

This is as easy as pie
You just have to do the following

1) first assign the value to a variable.In your case the value is 0.666666666666666......to infinity
so let x(variable)=.66666666666666....

2)This is the most imp step of all. Multiply both sides by a multiple of ten having as many zeros as the number of repeating digits. In your case only 6 is repeating so only 1 zero
=>10x=6.6666666666666.......to infinity

3) Now you have 2 equations:
10x=6.666666666666666666.... (1)
x=0.666666666666666666.... (2)
Subtract the two
=>9x=6 (as all digits after the decimal get cancelled, this is because of infinite number of 6)
=>x=2/3

Another eg:
let a repeating decimal be 3.142142142142142......

follow the steps provided:
1)let x=3.142142142......

2)Repeating digits are 142 i.e. 3 digits
so, multiply by 1000(3 zeros)
=>1000x=3142.142142142....
3) 2 equations are:
1000x=3142.142142....... (1)
x= 3.142142....... (2)
subtract the 2
=>999x=3139
=>x=3139/999

Answer 5

Let x=0.6666666_ _ _ --1)
Since only one digit i.e.6 is repeating
therefore multiply both sides of 1) by 10
10x=6.66666_ _ _ --2)
Subtract 1) from 2)
9x=6
x=6/9=2/3
Hence 0.666666_ _ _=2/3

Answer 6

any number reapeating like that is always itself over 9..try it out to ee for yourself

Answer 7

use your graphics display calculator

Answer 8

2.22222xxx...

is

22.2222xxxx...../10

etc.