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√20x * √3x^3
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any answers for this one?
yes
|dw:1368843714251:dw|
you got part backwards
6x^2 square root 3
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\[\sqrt{60}=\sqrt{4\times 15}=2\sqrt{15}\] so a \(2\) comes out and a \(15\) stays in
the \(x^2\) part is right, because \(\sqrt{x^4}=x^2\)
18x
final answer should be \[2x^2\sqrt{15}\]
not an option
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really? is the question \[\sqrt{20x}\times \sqrt{3x^3}\]?
\[\sqrt{20x}*\sqrt{3x^3}\]\[\sqrt{60x^4}\]\[2x^2\sqrt{15}\]
30*
because if so i am sticking with the answer of \(2x^2\sqrt{15}\)
ahhhh
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my sincerest apologies..... :(
I thought you multiplied...
\[\sqrt{30x}\sqrt{3x^2}\] right?
let me scroll up to see your answer again
|dw:1368844042803:dw|
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