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@Directrix
substitute A=36pi cm^2 into this equation
\[c=2\sqrt{A~\pi}\]
ok
\[C=2\sqrt{36\pi(\pi)}\]
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i got A
yes :)
ok cx
for #2 is x = 2 the extraneous root?
Let's see.
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For the second question... you could check by replacing x with the given options and verifying it works by seeing if both sides are the same thing
I agree with B)2 as the exraneous root. How about you, @MARC_ |dw:1448699140539:dw|
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