SOLUTION: I really need some help. the square root of 3x^4y^3 divided the sqaure root of 9xy^5 The only thing I thought to do was to put the problems vertical instead of horizontal and then
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Question 142299: I really need some help. the square root of 3x^4y^3 divided the sqaure root of 9xy^5 The only thing I thought to do was to put the problems vertical instead of horizontal and then get the square root of 9. Im not sure what the next step is. Found 2 solutions by stanbon, rapaljer:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! square root of 3x^4y^3 divided the sqaure root of 9xy^5
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= sqrt[3x^4y^3/9xy^5]
Multiply numerator and denominator by xy so the denominator becomes
a perfect square:
= sqrt[3x^5y^4/9x^2y^6]
Simplify the denominator:
= sqrt[3x^5y^4] / 3xy^3
Simplify the numerator:
= [x^2y^2/3xy^3] sqrt(3x)
Reduce the fraction:
= [x/3y]*sqrt(3x)
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Cheers,
Stan H.
Next, reduce the fraction, but I recommend NOT reducing the 3 and the 9, since you prefer to have a perfect square of 9 in the denominator. Do it this way:
The perfect square denominator is great. Take this square root:
Simplify the numerator by breaking it down into two square roots, placing all perfect squares in the first square root sign.
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