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Math Help - HELP PLEASE! URGENT!!!

  1. #1
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    HELP PLEASE! URGENT!!!

    Here is the problem:
    .................................................. ................Fraction
    23. Write an equivalent expression by symplifying 12xyz/8y

    Please find the answer, Becuase I don't understand

    Thank You So Much
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Bmxmster View Post
    Here is the problem:
    .................................................. ................Fraction
    23. Write an equivalent expression by symplifying 12xyz/8y

    Please find the answer, Becuase I don't understand

    Thank You So Much
    \frac{12xyz}{8x}=\frac{4(3)(x)(y)(z)}{4(2)(x)}

    Cancel out the terms that are common to the numerator and denominator. That will give you the answer.

    I hope this helps

    --Chris
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Bmxmster View Post
    Here is the problem:
    .................................................. ................Fraction
    23. Write an equivalent expression by symplifying 12xyz/8y

    Please find the answer, Becuase I don't understand

    Thank You So Much
    since 12/8 = 3/2 and y/y = 1, we have: \frac {12xyz}{8y} = \frac {\not 1 \not 2^3x \not yz}{\not 8_2 \not y} = \frac {3xz}{2}

    the principle behind this is "canceling common factors, which really falls back on the rule: \frac {x^a}{x^b} = x^{a - b}

    so \frac {12xyz}{8y} = \frac {12}8 \cdot x \cdot yy \cdot z = \frac 32 \cdot x \cdot 1 \cdot z = \frac {3xz}{2} since \frac yy = \frac {y^1}{y^1} = y^{1 - 1} = y^0 = 1
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    \frac{12xyz}{8x}=\frac{4(3)(x)(y)(z)}{4(2)(x)}

    Cancel out the terms that are common to the numerator and denominator. That will give you the answer.

    I hope this helps

    --Chris
    It does somewhat but I still don't completly understand so I can't figure out the answer.

    Thanks, Niles Nimmo
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    since 12/8 = 3/2 and y/y = 1, we have: \frac {12xyz}{8y} = \frac {\not 1 \not 2^3x \not yz}{\not 8_2y} = \frac {3xz}{2}

    the principle behind this is "canceling common factors, which really falls back on the rule: \frac {x^a}{x^b} = x^{a - b}

    so \frac {12xyz}{8y} = \frac {12}8 \cdot x \cdot yy \cdot z = \frac 32 \cdot x \cdot 1 \cdot z = \frac {3xz}{2} since \frac yy = \frac {y^1}{y^1} = y^{1 - 1} = y^0 = 1

    Thank You very much

    Its exactly what I was looking for
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    \frac{12xyz}{8x}=\frac{4(3)(x)(y)(z)}{4(2)(x)}

    Cancel out the terms that are common to the numerator and denominator. That will give you the answer.

    I hope this helps

    --Chris
    Sorry, I didn't see you there, Chris
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Jhevon View Post
    Sorry, I didn't see you there, Chris
    Its ok...I think I just happened to beat you by a minute!

    ...and took the gold medal in the dash



    --Chris
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Its ok...I think I just happened to beat you by a minute!

    ...and took the gold medal in the dash



    --Chris
    Haha, oh no, you did not just go there! haha! good stuff!
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