I can't figure this one out on my own either
$$\frac{(3x^{3/2}y^3)} {(x^2y^{-1/2})}^{-2}$$
I am a little confused on all the small rules at play here but I know that a negative exponent will flip a fraction so I square the top and then flip it. But before that I should work in the parentheses first since that is the order of operataions. I am not sure how to cancel out each of the numbers though I am a little confused what a negative fraction in the numerator does to a larger positive exponent in the denominator.
Best Answer
You have: $$\begin{align*} \frac{(3x^{3/2}y^3)^{-2}}{(x^2y^{-1/2})} &= \frac{1}{(3x^{3/2}y^3)^2}\times \frac{1}{x^2} \times y^{1/2}\\ &= \frac{1}{(3^2)(x^{3/2})^2(y^3)^2}\times \frac{1}{x^2}\times\frac{y^{1/2}}{1}\\ &= \frac{1}{9x^3y^6}\times\frac{1}{x^2}\times \frac{y^{1/2}}{1}\\ &= \frac{1}{9x^3x^2}\times\frac{y^{1/2}}{y^6}\\ &= \frac{1}{9x^5}\times \frac{1}{y^{11/2}}\\ &= \frac{1}{9x^5y^{11/2}}. \end{align*}$$