Question

which expression is equivalent to \\(\sqrt{\frac{25x^9y^3}{64x^6y^{11}}}\\)? assume \\(x > 0\\) and \\(y > 0\\).\\(\frac{8y^4\sqrt{x}}{5x}\\)\\(\frac{8y^2\sqrt{x}}{5}\\)\\(\frac{5\sqrt{x}}{8y^2}\\)\\(\frac{5x\sqrt{x}}{8y^4}\\)

Explanation:

Step1: Simplify the fraction inside the square root

First, simplify the fraction \(\frac{25x^{9}y^{3}}{64x^{6}y^{11}}\) using the quotient rule for exponents \( \frac{a^m}{a^n}=a^{m - n} \). For the coefficients, \(\frac{25}{64}\) remains as is. For \(x\): \(x^{9-6}=x^{3}\), and for \(y\): \(y^{3 - 11}=y^{-8}=\frac{1}{y^{8}}\). So the fraction becomes \(\frac{25x^{3}}{64y^{8}}\). Now the expression is \(\sqrt{\frac{25x^{3}}{64y^{8}}}\).

Step2: Apply the square root to the numerator and denominator

Using the property \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\) (for \(a\geq0,b > 0\)), we get \(\frac{\sqrt{25x^{3}}}{\sqrt{64y^{8}}}\).

Step3: Simplify the square roots

Simplify each square root: \(\sqrt{25}=5\), \(\sqrt{64}=8\), \(\sqrt{y^{8}} = y^{4}\) (since \(y>0\)), and \(\sqrt{x^{3}}=\sqrt{x^{2}\cdot x}=x\sqrt{x}\) (since \(x > 0\)). Substituting these in, we have \(\frac{5\cdot x\sqrt{x}}{8\cdot y^{4}}\).

Answer:

\(\frac{5x\sqrt{x}}{8y^{4}}\) (which corresponds to the last option, likely labeled as D or the fourth option, e.g., D. \(\frac{5x\sqrt{x}}{8y^{4}}\))