the square root of 9x^3/5y^2 show your work. (Do not forget to rationalize the denominator)

To find the square root of the algebraic expression `9x^3 / 5y^2`, we'll break the problem into two parts: finding the square root of the numerator and finding the square root of the denominator separately.

1. Square root of `9x^3`: Since `9` is a perfect square and `x^3` can be written as `x^2*x`, the square root of `9x^3` can be found using the square root of each factor. √(9x^3) = √(9) * √(x^2*x) = (3) * (x * √x) = 3x√x

2. Square root of `5y^2`: `5` is not a perfect square, but `y^2` is. So we can only take the `y` outside of the square root. √(5y^2) = √(5) * √(y^2) = √(5) * y = y√5

Finally, we combine the square roots of the numerator and the denominator: √(9x^3 / 5y^2) = (3x√x) / (y√5)

Now, to rationalize the denominator, we have to get rid of the square root in the denominator. To do this, we'll multiply both the numerator and denominator by the square root that’s in the denominator (which is √5, in this case).

(3x√x) / (y√5) * (√5/√5) = (3x√x * √5) / (y * 5)

Now let's simplify it: = (3x√(5x^2)) / (5y)

This is the fully simplified and rationalized form of the square root of the expression `9x^3 / 5y^2`.