Question 1171812
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Assuming x > 0 and y > 0, we can say


{{{sqrt((45x^3y^4)/(16x^2))}}}


{{{(sqrt(45x^3y^4))/(sqrt(16x^2))}}} Use the rule that {{{sqrt(a/b) = sqrt(a)/sqrt(b)}}}


{{{(sqrt(9*5*x^2*x*y^4))/(sqrt((4x)^2))}}}


{{{(sqrt((9x^2y^4)*(5x)))/(4x)}}}


{{{(sqrt((3xy^2)^2*5x))/(4x)}}}


{{{(sqrt((3xy^2)^2)*sqrt(5x))/(4x)}}} Use the rule that {{{sqrt(a*b) = sqrt(a)*sqrt(b)}}}


{{{(3xy^2*sqrt(5x))/(4x)}}}


{{{(3y^2*sqrt(5x))/(4)}}}


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Answer:  {{{(3y^2*sqrt(5x))/(4)}}}


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