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Math Help - test

  1. #1
    Senior Member
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    test

    \sqrt[n]{\frac{a}{b}}\:\:=\:\:\frac{ {\sqrt[n]{a}} {\sqrt[n]{b}}   }\;\;\mbox{such that a and b}>\:0\:
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  2. #2
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    Lexington, MA (USA)
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    Hello, ^_^Engineer_Adam^_^

    You have too many braces . . .


    You typed:

    [math ] \sqrt[n]{\frac{a}{b}}\:\:=\:\:\frac{ {\sqrt[n]{a}} {\sqrt[n]{b}} }\;\;\mbox{such that a and b}>\:0\:[/tex]


    and produced this:

    \sqrt[n]{\frac{a}{b}}\:\:=\:\:\frac{ {\sqrt[n]{a}} {\sqrt[n]{b}}   }\;\;\mbox{such that a and b}>\:0\:



    Eliminate the red braces above:

    [math ] \sqrt[n]{\frac{a}{b}}\:\:=\:\:\frac {\sqrt[n]{a}} {\sqrt[n]{b}}\;\;\mbox{such that a and b}>\:0\:[/tex]


    which produces:

     \sqrt[n]{\frac{a}{b}}\:\:=\:\:\frac {\sqrt[n]{a}} {\sqrt[n]{b}} \;\;\mbox{such that a and b}>\:0\:

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  3. #3
    Junior Member
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    Sep 2009
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    just another test

    [tex]j-i > 2^n \rightarrow A(i,j) \leq n[\math]


    [tex]x^2\sqrt{x}[\math]
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  4. #4
    Super Member

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    Hello, Madspeter!

    [tex]j-i > 2^n \rightarrow A(i,j) \leq n[\math]

    [tex]x^2\sqrt{x}[\math]

    You must close LaTeX with: .[/math]


    Then you will get:

    j-i > 2^n \rightarrow A(i,j) \leq n

    x^2\sqrt{x}

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  5. #5
    Junior Member
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    thanks a lot. )
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