which expression is equivalent to $left(\frac{(3xy^{-5})^{3}}{(x^{-2}y^{2})^{-4}}\right)^{-2}$? assume…

which expression is equivalent to $left(\frac{(3xy^{-5})^{3}}{(x^{-2}y^{2})^{-4}}\right)^{-2}$? assume $x\neq0,y\neq0$.\n$\frac{x^{10}y^{14}}{729}$\n$\frac{x^{22}}{18y^{46}}$\n$\frac{729}{x^{10}y^{14}}$\n$\frac{18y^{46}}{x^{22}}$

which expression is equivalent to $left(\frac{(3xy^{-5})^{3}}{(x^{-2}y^{2})^{-4}}\right)^{-2}$? assume $x\neq0,y\neq0$.\n$\frac{x^{10}y^{14}}{729}$\n$\frac{x^{22}}{18y^{46}}$\n$\frac{729}{x^{10}y^{14}}$\n$\frac{18y^{46}}{x^{22}}$

Answer

Explanation:

Step1: Simplify numerator and denominator inside the big - fraction

First, simplify ((3xy^{-5})^{3}) using the power - of - a - product rule ((ab)^n=a^n b^n) and power - of - a - power rule ((a^m)^n=a^{mn}). ((3xy^{-5})^{3}=3^{3}x^{3}y^{-15}=27x^{3}y^{-15}). Next, simplify ((x^{-2}y^{2})^{-4}) using the power - of - a - product rule and power - of - a - power rule: ((x^{-2}y^{2})^{-4}=x^{8}y^{-8}). So the fraction inside the big - fraction is (\frac{(3xy^{-5})^{3}}{(x^{-2}y^{2})^{-4}}=\frac{27x^{3}y^{-15}}{x^{8}y^{-8}}). Using the quotient rule (\frac{a^m}{a^n}=a^{m - n}), we get (\frac{27x^{3}y^{-15}}{x^{8}y^{-8}} = 27x^{3-8}y^{-15 + 8}=27x^{-5}y^{-7}).

Step2: Apply the outer negative exponent

We have ((\frac{(3xy^{-5})^{3}}{(x^{-2}y^{2})^{-4}})^{-2}=(27x^{-5}y^{-7})^{-2}). Using the power - of - a - product rule ((ab)^n=a^n b^n) and power - of - a - power rule ((a^m)^n=a^{mn}), we get ((27x^{-5}y^{-7})^{-2}=27^{-2}x^{10}y^{14}). Since (27^{-2}=\frac{1}{27^{2}}=\frac{1}{729}), the expression is (\frac{x^{10}y^{14}}{729}).

Answer:

(\frac{x^{10}y^{14}}{729})