anonymous 4 years ago $(2^{-1}x ^{-2}y ^{-1})^{-2}(2x ^{-3}y)^{-2}(16x ^{-3}y ^{3})^{0}/ (2x ^{-3}y ^{-5)^{2}}$

1. anonymous

$\frac{x^{13} y^{25}}{2}$

2. anonymous

Rottobey can you show me how you got that answer. since the answer is x^18 y^6/4. I'm just little lost on the equation (16x ^{-3}y ^{3})^{0) is the zero represent all the number inside the equation is zero or one itself

3. anonymous

It's fine Robtobey, I thank you for your help. :)

4. anonymous

There is a syntax error in the last term. The following is a step by step evaluation of the problem expression.$\left(\left(\frac{1}{2} \frac{1}{x^2} \frac{1}{y}\right){}^{\wedge}(-2)\right) \left(\left(\frac{2}{x^3} y\right){}^{\wedge}(-2) \right)\left(\left(\frac{16}{x^3} y^3\right){}^{\wedge}(0)\right)/\left(\frac{2}{x^3} \frac{1}{y^5}\right){}^{\wedge}(2)$$\frac{\left(4 x^4 y^2\right)\left(\frac{x^6}{4 y^2}\right) (1)}{\left(\frac{4}{x^6 y^{10}}\right)}$$\frac{x^{10}}{\frac{4}{x^6 y^{10}}}$$\frac{x^{16} y^{10}}{4}$WolframAlpha.com agrees http://www.wolframalpha.com/input/?i=simplify+((1%2F2+1%2Fx%5E2+1%2Fy)%5E(-2))+((2%2Fx%5E3+y)%5E(-2))+((16%2Fx%5E3+y%5E3)%5E(0))%2F(2%2Fx%5E3+1%2Fy%5E5)%5E(2)&t=sftb01

5. anonymous

The divisor should have been written as an equivalent to the following: ( (2 x^-3)(y^-5) )^2