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Grazes

  • 3 years ago

By looking at 0.36 (repeating decimal) like an infinite geometric sequence, convert it into a fraction.

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  1. anonymous
    • 3 years ago
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    you can do this a couple ways

  2. anonymous
    • 3 years ago
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    is it \(\overline{.36}\) ?

  3. Grazes
    • 3 years ago
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    Yes.

  4. Grazes
    • 3 years ago
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    I have the formula \[S _{n}=\frac{ a _{1}-a_{1}r ^{n} }{ 1-r }\] If that helps.

  5. anonymous
    • 3 years ago
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    ok one simple method is to call \(x=\overline{.36}\) so \(100x=36\overline{.36}\) subtract and get \(100x-x=36\) so \[99x=36\] and therefore \[x=\frac{36}{99}=\frac{4}{11}\] but that might not be what you want

  6. erin512
    • 3 years ago
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    @satellite73 hey, when you get a chance can you finish helping me with that problem we started? thanks :)

  7. anonymous
    • 3 years ago
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    if you want to use the formula you wrote above, then since you are summing an infinite geometric series omit the \(-a_1r^n\) part and go with \[S=\frac{a_1}{1-r}\]

  8. Grazes
    • 3 years ago
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    Oh. I think I copied the formula down incorrectly...

  9. Grazes
    • 3 years ago
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    Cuz I was looking through my notes and wondering how I got the answers with the -a1r^n part

  10. anonymous
    • 3 years ago
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    yeah since \(r<1\) you have \(\lim_{n\to \infty}r^n=0\)

  11. Grazes
    • 3 years ago
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    Thanks.

  12. anonymous
    • 3 years ago
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    in this case you can use \[a=\frac{36}{100}\] and \[r=\frac{1}{100}\]

  13. Grazes
    • 3 years ago
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    For reference, though, does that formula equal\[S_{n}= \frac{ a_{1}(1-r ^{n)} }{ 1-r }\]

  14. anonymous
    • 3 years ago
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    you get \[S=\overline{.36}=\frac{\frac{36}{100}}{1-\frac{1}{100}}\] \[=\frac{\frac{36}{100}}{\frac{99}{100}}\] \[=\frac{36}{99}\] etc

  15. anonymous
    • 3 years ago
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    no that is a formula for a finite series infinite series is the one i wrote above

  16. Grazes
    • 3 years ago
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    Ah. Well, thanks!

  17. anonymous
    • 3 years ago
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    @erin512 repost i cannot find it

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