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Thread: algebra fraction

  1. #1
    Member realintegerz's Avatar
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    algebra fraction

    <br /> <br />
pi^2/-s^3-2 = - (pi^2/-s^3-2) <br /> <br /> <br />
    <br />
                  = pi^2/ -s^3+2<br />
    am i using the subtraction correctly?
    Last edited by realintegerz; Sep 21st 2008 at 09:41 PM.
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  2. #2
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    Hello,

    I don't know what you mean by /- but I assume you left out some parentheses.

    pi^2/(-s^3-2)= \frac{pi^2}{-s^3-2}=-\frac{pi^2}{s^3+2}.
    pi^2/(-s^3)-2= \frac{pi^2}{-s^3}-2=-\frac{pi^2}{s^3}-2.
    -(pi^2/(-s^3-2))= -\frac{pi^2}{-s^3-2}=\frac{pi^2}{s^3+2}.
    -(pi^2/(-s^3)-2))= -\left(\frac{pi^2}{-s^3}-2\right)=\frac{pi^2}{s^3}+2.

    You wrote pi^2/-s^3-2= pi^2/ -s^3+2 which seems obviously wrong.

    Bye.
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  3. #3
    Super Member 11rdc11's Avatar
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    I'm not sure how your problem looks like. Is it



    \frac{\pi^2}{-s^3-2}

    or

    \frac{\pi^2}{-s^3} - 2
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  4. #4
    Member realintegerz's Avatar
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    ok i changed it...didnt know it would come out like that on here..

    the 2nd line is the right side

    im just checking if i distributed the negative right or wrong
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  5. #5
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    Hello,

    -(A+B)=-A-B, -(A-B)=-A+B,-(-A-B)=A+B,
    -\frac{A}{B}=\frac{A}{-B}=\frac{-A}{B}.
    Thus,
    -\frac{A}{B+C}=\frac{A}{-(B+C)}=\frac{A}{-B-C},
    -\frac{A}{B-C}=\frac{A}{-(B-C)}=\frac{A}{-B+C},
    -\frac{A}{-B-C}=\frac{A}{-(-B-C)}=\frac{A}{B+C}.

    I hope it helps. I wrote all the possibilities in the previous post.

    Bye.
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  6. #6
    Member realintegerz's Avatar
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    is that correct ??
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  7. #7
    Super Member 11rdc11's Avatar
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    If you saying the left hand side equals the bottom right hand side then nope.

    You can see this is not true by plugging in a value for s. Try plugging in 1 and you will notice

    \frac{\pi^2}{-(1)^3 - 2}~\text{does not equal}~ \frac{\pi^2}{-(1)^3+2}
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  8. #8
    Member realintegerz's Avatar
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    so i distributed the negative correctly or no?

    i just need to know if i did or not...because im checking if functions are even, odd, or neither...

    the function is h(s) = [ pi^2 over s^3 - 2 ]
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  9. #9
    Super Member 11rdc11's Avatar
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    Testing to see if it even

    y = \frac{\pi^2}{-s^3-2}

    Replace s with -s

    y = \frac{\pi^2}{-(-s)^3-2}

    which equals


    y = \frac{\pi^2}{s^3-2}

    so it is not the same as the original equation so it is not even.

    To see if it odd replace y with -y and s with -s

    -y = \frac{\pi^2}{-(-s)^3-2}

    -y = \frac{\pi^2}{s^3-2}

    y = \frac{\pi^2}{-s^3+2}

    which is not the same as the original function so it is neither
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  10. #10
    Member realintegerz's Avatar
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    the original function itself h(s) does not have -s^3 - 2...

    its s^3 - 2
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  11. #11
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by realintegerz View Post


    is that correct ??
    Wait I thought the left hand side equation was the original equation?

    So it is

    \frac{\pi^2}{s^3-2}
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  12. #12
    Member realintegerz's Avatar
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    yep..

    like i said before

    the function h(s) = pi^2/s^3-2

    i put -s into it, and it isnt the same, so its not even...

    and the picture is the work to see if its odd...
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  13. #13
    Member realintegerz's Avatar
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    h(s) = pi^2/s^3 - 2
    h(-s) = pi^2/(-s)^3 - 2
    = pi^2/-s^3 - 2

    so h(s) doesnt = h(-s)

    then i tried this

    trying to see if h(-s) = - h(s)
    pi^2/-s^3 - 2 = - ( pi^2/ s^3 - 2)
    = pi^2/-s^3 + 2



    thats all my work
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  14. #14
    Super Member 11rdc11's Avatar
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    Ok sorry I misread

    y = \frac{\pi^2}{s^3-2}

    To see if it odd replace y with -y and s with -s

    -y = \frac{\pi^2}{(-s)^3-2}

    -y = \frac{\pi^2}{-s^3-2}

    y = \frac{\pi^2}{s^3+2}

    So nope it is not odd either
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