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Math Help - Is this right?

  1. #1
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    Is this right?

    Can someone tell me if I did the following question right and if not what I did wrong?

    1 - 1
    x + h x
    ______________
    h

    Answer: 1 - h
    x +h
    Last edited by JessicaHoaglund; November 6th 2007 at 12:53 PM. Reason: fix problem
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JessicaHoaglund View Post
    Can someone tell me if I did the following question right and if not what I did wrong?

    1 - 1
    x + h x
    ______________
    h

    Answer: 1 - h
    x +h
    no, that is incorrect i'm afraid, and fix your formatting. if you can't use LaTex, type fractions like this: (numerator)/(denominator)

    so, for example, you would type your (incorrect) answer like this: (1 - h)/(x + h)
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  3. #3
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    Okay. Sorry I am not trying to make this difficult I just signed up for this an hour ago so I was not aware how to put the answers in correctly. Thank you.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JessicaHoaglund View Post
    Okay. Sorry I am not trying to make this difficult I just signed up for this an hour ago so I was not aware how to put the answers in correctly. Thank you.
    that's fine. did you try again? what's your answer now? begin by combining the fractions in the numerator. do you know how to do that?
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  5. #5
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    Nope. I am honestly really terrible at this!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JessicaHoaglund View Post
    Nope. I am honestly really terrible at this!
    \frac {\frac 1{x + h} - \frac 1x}h = \frac {\frac {x - x - h}{x(x + h)}}h

    = \frac {-h}{xh(x + h)}

    = \frac {-1}{x(x + h)}

    any questions?

    post number 5 here may help
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  7. #7
    Super Member angel.white's Avatar
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    Okay, so you have a fraction over a fraction, such as

    \frac{\mbox{    }\frac{a}{b}\mbox{    }}{\frac{c}{d}}

    This is basically saying \frac{a}{b} \div \frac{c}{d}

    Whenever you divide by a fraction, you are multiplying by it's inverse, so you just rewrite it:

    \frac{a}{b} \times \frac{d}{c}

    Which is then very easy:

    \frac{ad}{bc}

    ---------

    So lets say
    \frac{\mbox{    }\frac{1}{2}\mbox{    }}{\frac{3}{4}}

    Then write as division:
    \frac{1}{2} \div \frac{3}{4}

    Then write as multiplying by the inverse:
    \frac{1}{2} \times \frac{4}{3}

    Multiply numerator, multiply denominator
    \frac{1\times4}{2\times3}

    Simplify
    \frac{4}{6}

    Simplify
    \frac{2}{3}
    -----

    So I can't read your equation, but try again using that rule.
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  8. #8
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    I guess I just don't understand the first step:

    How did you get

    (x-x-h)/x(x+h)

    sorry!
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JessicaHoaglund View Post
    I guess I just don't understand the first step:

    How did you get

    (x-x-h)/x(x+h)

    sorry!
    in the post i directed you to, we saw that we can add/subtract fractions in the following way

    \frac ab \pm \frac cd = \frac {ad \pm bc}{bd}

    thus, \frac 1{x + h} - \frac 1x = \frac {x - (x + h)}{x(x + h)} = \frac {x - x - h}{x(x + h)} (since here, a = 1, b = x + h, c = 1 and d = x)
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