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Math Help - How would you solve this -(-3)^2-7(1/2)/5^3?

  1. #1
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    How would you solve this -(-3)^2-7(1/2)/5^3?

    I got to the part where you have to divide (-25/2)/(5^3) and i'm stuck!
    Last edited by EJdive43; August 29th 2012 at 04:55 PM.
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    Re: How would you solve this -(-3)^2-7(1/2)/5^3?

    \displaystyle \begin{align*} \frac{-\frac{25}{2}}{5^3} &= -\frac{25}{2} \div 5^3 \\ &= -\frac{25}{2} \times \frac{1}{5^3} \\ &= -\frac{25}{2 \cdot 5^3} \\ &= -\frac{5^2}{2 \cdot 5^3} \\ &= -\frac{1}{2 \cdot 5} \\ &= -\frac{1}{10} \end{align*}
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    Re: How would you solve this -(-3)^2-7(1/2)/5^3?

    Oh, I flipped the numoator. So you only take the receptacle of the demoninator?
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    Re: How would you solve this -(-3)^2-7(1/2)/5^3?

    (a / x) / (b / y) = (a / x) * (y / b)
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    Re: How would you solve this -(-3)^2-7(1/2)/5^3?

    Quote Originally Posted by EJdive43 View Post
    Oh, I flipped the numoator. So you only take the receptacle of the demoninator?
    I always remember "Stay, Change, Flip". Keep the first fraction as is, change the divide to a times, and flip the second fraction.
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