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16u^4v^3
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\[16u^4v^3 \] this?
yea
whats there to solve then?
the square root of 16u^4v^3
\[(16u^4v^3 )^\frac{ 1 }{ 2 }\] will give you \[4u^2v^\frac{ 3 }{ 2 } \]
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where did you get the 1/2
square root can also be written as 'to the power 1/2' ;)
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