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Mathematics 16 Online
OpenStudy (anonymous):

how would I turn .29292929 into a fraction. multiply by 100... then what?

OpenStudy (unklerhaukus):

oh wait you mean \[0.\dot2\dot9\]

OpenStudy (anonymous):

.29 repeating yes

OpenStudy (unklerhaukus):

yeah multiply by 100 is right, then take away the original,, you'll have 99 (original )=29

OpenStudy (anonymous):

what do you mean? take away the original?

OpenStudy (sburchette):

You can use a geometric series and then find the fraction that it converges to.

OpenStudy (unklerhaukus):

so \[x=0.\dot2\dot9\] \[100x=29.\dot2\dot9\] \[99x=29\] \[x=29/99\] \[0.\dot2\dot9=29/99\]

OpenStudy (anonymous):

why'd you subtract 1 though?

OpenStudy (anonymous):

anyone?

OpenStudy (anonymous):

@gd0x simply to get rid of nasty repeating parts

OpenStudy (anonymous):

@gd0x when you do calculation 100x-x we get rid of repeating parts from the decimal

OpenStudy (anonymous):

ohhhhh. got it got it. ty

OpenStudy (unklerhaukus):

we are subtracting x

OpenStudy (sburchette):

.292929... = .29+.0029+.000029+...=(29/100)+(29/10000)+(29/1000000)+... \[\sum_{n=1}^{\infty}(29/100^n)\]\[29\sum_{n=1}^{\infty}(1/100)^n\]This is a geometric series whose common ratio is (1/100) which is less than 1, thus it converges at \[a_1/(1-r)\]\[29[(1/100)/(1-(1/100)]\]\[29[(1/100)/(99/100)]\]\[29(1/99)\]\[29/99\]

OpenStudy (anonymous):

@SBurchette your method is awesome but i remember i used to solve these questions in the 6th or 7th standard & at that level Geometric Progression will go out of the beyond.

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