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    Is this right? It's in the mark scheme...



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    (Original post by MsFahima)
    Is this right? It's in the mark scheme...



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    Yup
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    (Original post by MsFahima)
    Is this right? It's in the mark scheme...



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    :yep:

    The x at the top and one x at the bottom cancel with each other, so you're just left with 8/x !
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    (Original post by James A)
    :yep:

    The x at the top and one x at the bottom cancel with each other, so you're just left with 8/x !
    But.. x^1 /x^2 would it not equal x^-1 ?
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    It becomes 8x/x^2 which cancels to become 8/x
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    (Original post by MsFahima)
    But.. x^1 /x^2 would it not equal x^-1 ?
    x^{-1}=\dfrac{1}{x}
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    (Original post by MsFahima)
    But.. x^1 /x^2 would it not equal x^-1 ?
    Yes. \dfrac8x = 8x^{-1}
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    (Original post by MsFahima)
    But.. x^1 /x^2 would it not equal x^-1 ?
    8x^{-1} = \dfrac{8}{x}

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    (Original post by alow)
    Yes. \dfrac8x = 8x^{-1}

    Thanks!
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    (Original post by Robbie242)
    x^{-1}=\dfrac{1}{x}
    (Original post by Bashur)
    It becomes 8x/x^2 which cancels to become 8/x

    Thanks!
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    \frac{8}{x^2}*\dfrac{x}{1}=\frac  {8x}{x^2} Which then cancel out to leave you with \frac{8}{x}
 
 
 
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