which expression is equivalent to $sqrt{\frac{128x^{5}y^{6}}{2x^{7}y^{5}}}$? assume $x > 0$ and $y >…

Explanation:

Step1: Simplify the fraction inside the square - root

First, simplify the coefficient and use the rule of exponents $\frac{a^m}{a^n}=a^{m - n}$. For the coefficient: $\frac{128}{2}=64$. For the $x$ - terms: $\frac{x^5}{x^7}=x^{5 - 7}=x^{-2}=\frac{1}{x^2}$. For the $y$ - terms: $\frac{y^6}{y^5}=y^{6 - 5}=y$. So, $\frac{128x^5y^6}{2x^7y^5}=64\frac{y}{x^2}$.

Step2: Take the square - root

$\sqrt{\frac{128x^5y^6}{2x^7y^5}}=\sqrt{64\frac{y}{x^2}}$. Using the property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ ($a = 64$, $b=\frac{y}{x^2}$), we get $\sqrt{64}\cdot\sqrt{\frac{y}{x^2}}$. Since $\sqrt{64}=8$ and $\sqrt{\frac{y}{x^2}}=\frac{\sqrt{y}}{x}$ (because $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ and $\sqrt{x^2}=x$ for $x>0$), the result is $\frac{8\sqrt{y}}{x}$.

Answer:

$\frac{8\sqrt{y}}{x}$