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\[3\sqrt{5c}*7\sqrt{15c^2}\]
then \[21\sqrt{75c^3}\]
are you solving for c? or simplifying?
then \[21*5\sqrt{3c ^{3}}\]
simplifying
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Now what am I suppose to do?
\(\large c^3 =c^2 c\)
so u can pull out a \(c\) put of squareroot ?
so then it would be \[105c \sqrt{3c^2}\]
Like that?
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not quite
\(\large 21*5\sqrt{3c^3}\) \(\large 21*5\sqrt{3c^2c}\) \(\large 21*5*c\sqrt{3c}\) \(\large 105c\sqrt{3c}\)
when u pull out, c^2 becomes c
oh
and thats it right?
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yes, \(\large \sqrt{x^2} = x\)
yup thats it
Thank you
yw
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