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Simplify completely √27y²
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\[3 y \sqrt{3}\]
3|y|sqrt(3). You need the absolute value around the y.
\[\sqrt{9*3}*\sqrt{y^2}\] =\[3y \sqrt{3}\]
i dont see the choice/
3\[3y \sqrt{3}\]
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and yeah there should be an absolute value sign!
is it 3y power of 4√3
If the problem is sqrt(27y^2) then the answer equals sqrt(27)sqrt(y^2) = sqrt(9*3)|y| = 3|y|sqrt(3).
and sqrt(x^2) = |x| = +-x in other words if you graph sqrt(x^2) it equals a straight line in both the positive and negative directions (like an aboslute value graph)
____ Simplify completely √27y²
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Yes, then the answer you seek is 3, times absolute value of y, times square root of 3.
you guys are the best! :)
|dw:1329798162343:dw|
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