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Math Help - Someone help if you get a chance

  1. #1
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    Someone help if you get a chance

    if any of you get time help me with this problem, thankssss

    http://img231.imageshack.us/img231/1427/ughmath6lp.jpg
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  2. #2
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    Quote Originally Posted by Aarxn
    if any of you get time help me with this problem, thankssss

    http://img231.imageshack.us/img231/1427/ughmath6lp.jpg
    The image is fuzzy, but I think it says  \frac{21a^2b}{7a^2b}+\left(\frac{a^2}{b^2}\right)^  0

    In review,  \frac{a^3}{a^2}=a^{(3-2)}=a^1=a and  a^0=1 when a\neq0.


     \frac{21a^2b}{7a^2b}+\left(\frac{a^2}{b^2}\right)^  0

     \frac{21a^{(2-2)}b^{(1-1)}}{7}+1

     \frac{21a^0b^0}{7}+1

     \frac{21\cdot1\cdot1}{7}+1

     \frac{21}{7}+1

     3+1

     4
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