## simplifying [(-2 x^0)/(y^2 z^{-3})]^{-2} x^{-3} y^4 z

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
EdSewersPark
Posts: 22
Joined: Mon May 23, 2011 8:30 pm
Contact:

### simplifying [(-2 x^0)/(y^2 z^{-3})]^{-2} x^{-3} y^4 z

$(\frac{-2x^0}{y^2z^-3})^-2$ $(x^-3y^4z)$

I know the correct answer, but I can't get it right.

EdSewersPark
Posts: 22
Joined: Mon May 23, 2011 8:30 pm
Contact:

### Re: simplifying [(-2 x^0)/(y^2 z^{-3})]^{-2} x^{-3} y^4 z

I am learning the scripts...here it is correctly typed: $(\frac{-2x^0}{y^2z^{-3}})^{-2}$$x^{-3}y^4z$

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
I am learning the scripts...here it is correctly typed: $(\frac{-2x^0}{y^2z^{-3}})^{-2}$$x^{-3}y^4z$
What are your steps so far? (If you need to refresh, you can try here.)

EdSewersPark
Posts: 22
Joined: Mon May 23, 2011 8:30 pm
Contact:

### Re: simplifying [(-2 x^0)/(y^2 z^{-3})]^{-2} x^{-3} y^4 z

Is this correct?

First I flipped the inside and changed the negative exponent (on the outside) to a positive: $(\frac{y^2z^{-3}}{-2x^0})^2$

Then I squared the inside: $(\frac{y^4z^{-6}}{4})$

I put the second part into a fraction: $(\frac{x^{-3}y^4z}{1})$ and flipped the "x": $(\frac{y^4z}{x^3})$

Then multiplied across: $(\frac{y^8z^{-5}}{4x^3})$ and brought the "z" down: $\frac{y^8}{4x^3z^5}$

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
Looks good to me!