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Mathematics 15 Online
OpenStudy (anonymous):

Convert the repeating decimal to a/b form, where a,b are integers and b≠0

OpenStudy (mathmate):

Love this question, wonder why it's not taught in most schools.

OpenStudy (tyteen4a03):

There is a general way of doing this. Let's try to convert 0.898989... to the rational form (the correct term for a/b). Firstly, we will assume that x is the number with recurring digits, that's 0.898989... in this case. Since the repeating part is the part we need to get rid of, we multiply x by the number of digits the repeating bit has (in this case, 2). Hence we yield 100x = 89.898989 (notice that the recurring digits are still there) Now we subtract x from 100x, getting 99x = 89. (What happened during this process is that the recurring part was cancelled out from the subtraction) Now do simple algebra and you get x = 89/99. Make sure at the end you mention the relation between 0.898989... and 89/99, for clarity.

OpenStudy (anonymous):

|dw:1346954646388:dw|Oh, I am sorry I forgot to give you the rest of the problem, here it goes;Convert the repeating decimal to a/b form, where a,b are integers and b≠0

OpenStudy (mathmate):

Use what tyteen4a03 suggested: |dw:1346954781756:dw|

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