SOLUTION: (2x^2/3y^5)^-3

Algebra ->  Systems-of-equations -> SOLUTION: (2x^2/3y^5)^-3      Log On


   

Question 983780: (2x^2/3y^5)^-3
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(2x^2/3y^5)^-3
-----
The negative in the exponent means "invert"
-----
So you get:
= [(3y^5)/(2x^2)]^3
--------
= [27y^15/8x^6]
---------
Cheers,
Stan H.
-----------