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

• Aug 29th 2012, 04:14 PM
EJdive43
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!
• Aug 29th 2012, 09:40 PM
Prove It
Re: How would you solve this -(-3)^2-7(1/2)/5^3?
\displaystyle \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*}
• Aug 30th 2012, 12:12 PM
EJdive43
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?
• Aug 30th 2012, 03:58 PM
Wilmer
Re: How would you solve this -(-3)^2-7(1/2)/5^3?
(a / x) / (b / y) = (a / x) * (y / b)
• Aug 30th 2012, 07:33 PM
Prove It
Re: How would you solve this -(-3)^2-7(1/2)/5^3?
Quote:

Originally Posted by EJdive43
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.