SOLUTION: (2x^2/3y^5)^-3
Algebra.Com
Question 983780: (2x^2/3y^5)^-3
Answer by stanbon(75887) (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.
-----------
RELATED QUESTIONS
Y= -2 x
-------
3
2x + 3y = 5
(answered by Mathtut)
-5x+3y=-2... (answered by ewatrrr)
(3,_),(_,-1),2x-3y=5 (answered by bucky)
(5x^-3y^-2)(2x^3y^3) (answered by jim_thompson5910)
Solve the system
2(2x+3y)=5... (answered by lynnlo)
Factorise... (answered by Alan3354)
Factorise... (answered by Alan3354)
2-5[5-(3y-3)]= (answered by tommyt3rd)