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Math Help - Proportions

  1. #1
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    Urgent..

    help mi wif tis..
    Attached Thumbnails Attached Thumbnails Proportions-scan0010.jpg  
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by xiaoz
    help mi wif tis..
    your equation involves three variables and you are given two of them in every part. Simply plug in the values and you will get the answer.

    Keep Smiling
    Malay
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  3. #3
    MHF Contributor Quick's Avatar
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    To find k we merely need to solve for k when y=90 and x=2

    y=\frac{k}{x^2}

    90=\frac{k}{2^2}

    90=\frac{k}{4} multiply by 4

    360=k


    3(b) simple...

    y=\frac{k}{x^2}

    y=\frac{360}{(-10)^2}

    y=\frac{360}{100}

    y=\frac{36}{10}

    3(c) first, solve for x

    y=\frac{k}{x^2} reciprocate

    \frac{1}{y}=\frac{x^2}{k} multiply by k

    \frac{k}{y}=x^2 find the square root for both sides

    \sqrt[\pm]{\frac{k}{y}}=x now substitute 40 for y

    \sqrt[\pm]{\frac{360}{40}}=x solve...

    \sqrt[\pm]{9}=x solve...

    \pm3=x
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  4. #4
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    Zzz.. so diff.. ur question.. my question more easy.. XD
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  5. #5
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by xiaoz
    Zzz.. so diff.. ur question.. my question more easy.. XD
    I assume you mean the question in my signature. The only person (that I know of) to actually solve it is PH, although it's grown since then...
    Last edited by Quick; July 10th 2006 at 12:58 PM.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick
    I assume you mean the question in my signature. The only person (that I know of) to actually solve it is PH, although it's grown since then...
    I solved the signature a long time ago, I just haven't gotten the point of it yet.

    -Dan
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  7. #7
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by topsquark
    I solved the signature a long time ago, I just haven't gotten the point of it yet.

    -Dan
    Originally the answer was Q_ui_ck=co^2l whitch can be written as Q_ui_ck=cool now the answer is different by 1 number...
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  8. #8
    Member classicstrings's Avatar
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    Wait is the answer still cool? If so that is cool.
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  9. #9
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by classicstrings
    Wait is the answer still cool? If so that is cool.
    The answer is Q_ui_ck=2cool

     Q_u=\sqrt{\frac{100C}{144l^2o}}=\frac{10\sqrt{C}}{  12l\sqrt{o}}

    i_c=\sqrt[3]{8o^9l^3}=2o^3l

    k=6l\sqrt{\frac{C}{25o}}=\frac{6l\sqrt{C}}{5\sqrt{  o}}

    Q_ui_ck=\frac{10\sqrt{C}}{12l\sqrt{o}}\times2o^3l<br />
\times \frac{6l\sqrt{C}}{5\sqrt{o}}

    =\frac{120o^3l^2\sqrt{C}\sqrt{C}}{60l\sqrt{o}\sqrt  {o}}

    =\frac{120o^3l^2C}{60lo}

    =\frac{\not{60}\times2\not{o}\times o^2\not{l}\times lC}{\not{60}\not{l}\not{o}}

    2o^2lC\quad\rightarrow\quad2Co^2l\quad\rightarrow<br />
\quad <br />
 2Cool
    Last edited by Quick; July 15th 2006 at 05:09 PM.
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  10. #10
    Member classicstrings's Avatar
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    Ah ha...

    Is there a thread which shows how to work that out? Thanks I'm interested.
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  11. #11
    MHF Contributor Quick's Avatar
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    Cool

    I edited it into my last post.

    THIS IS MY 200TH POST! (and it says quick is 2cool)


    Quick is:

    The 11th highest poster!
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  12. #12
    Member classicstrings's Avatar
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    Quote Originally Posted by Quick
    The answer is Q_ui_ck=2cool

     Q_u=\sqrt{\frac{100C}{144l^2o}}=\frac{10\sqrt{C}}{  12l\sqrt{o}}

    i_c=\sqrt[3]{8o^9l^3}=2o^3l

    k=\frac{\sqrt{\frac{C}{25o}}}{6l}=\frac{6l\sqrt{C}  }{5\sqrt{o}}

    Q_ui_ck=\frac{10\sqrt{C}}{12l\sqrt{o}}\times2o^3l<br />
\times \frac{6l\sqrt{C}}{5\sqrt{o}}

    =\frac{120o^3l^2\sqrt{C}\sqrt{C}}{60l\sqrt{o}\sqrt  {o}}

    =\frac{120o^3l^2C}{60lo}

    =\frac{\not{60}\times2\not{o}\times o^2\not{l}\times lC}{\not{60}\not{l}\not{o}}

    2o^2lC\quad\rightarrow\quad2Co^2l\quad\rightarrow<br />
\quad <br />
 2Cool
    Wow that's very nice. I'm going to show it to my friends.
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick
    The answer is Q_ui_ck=2cool

     Q_u=\sqrt{\frac{100C}{144l^2o}}=\frac{10\sqrt{C}}{  12l\sqrt{o}}

    i_c=\sqrt[3]{8o^9l^3}=2o^3l

    k=\frac{\sqrt{\frac{C}{25o}}}{6l}=\frac{6l\sqrt{C}  }{5\sqrt{o}}

    Q_ui_ck=\frac{10\sqrt{C}}{12l\sqrt{o}}\times2o^3l<br />
\times \frac{6l\sqrt{C}}{5\sqrt{o}}

    =\frac{120o^3l^2\sqrt{C}\sqrt{C}}{60l\sqrt{o}\sqrt  {o}}

    =\frac{120o^3l^2C}{60lo}

    =\frac{\not{60}\times2\not{o}\times o^2\not{l}\times lC}{\not{60}\not{l}\not{o}}

    2o^2lC\quad\rightarrow\quad2Co^2l\quad\rightarrow<br />
\quad <br />
 2Cool
    NOW it makes sense. The previous formula didn't.

    -Dan
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