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simplify square root of 72x^3y^4
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how does 3x*9y*\[\sqrt{8x}\]sound
\[6\sqrt{2} x^{3/2}y^{2}\]
6xy^2 root2x
bracket can you show how you did that please, we need to show the work
root (a*b*c) = root(a) * root(b) * root(c) so\[ \sqrt{72x ^{3}}y ^{4} = \sqrt{72}*\sqrt{x ^{3}}*\sqrt{y ^{4}}\]
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and 72=2*2*2*3*3=2^2*3^2*2
\[\sqrt{x ^{3}} = (x ^{3})^{1/2} = x ^{3/2}\]
\[\sqrt{y ^{4}} = (y ^{4})^{1/2} = y ^{4/2} = y ^{2}\]
so\[\sqrt{72}x ^{3}y ^{4}=6\sqrt{2}*x ^{3/2}*y ^{2}\]
Hope this clear??? If yes, pls click the GOOD ANSWER button for me :)
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