# Radicals #2

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• May 12th 2013, 12:07 PM
Mathnood768
Radicals #2
imgur: the simple image sharer

My attempt:

imgur: the simple image sharer

I was really close, that 8 is suppose to be a 12, could I have gone somewhere in the factoring tree? I'm not sure....
• May 12th 2013, 03:29 PM
Shakarri
Re: Radicals #2
For mercury you got to $^3\sqrt{193600}= ^3\sqrt{3025\times 2^6}= ^3\sqrt{3025} \times ^3\sqrt{2^{6}}$

Remember you are taking the cube root so $^3\sqrt{2^{6}}=2^2$
• May 12th 2013, 06:17 PM
topsquark
Re: Radicals #2
Quote:

Originally Posted by Shakarri
For mercury you got to $^3\sqrt{193600}= ^3\sqrt{3025\times 2^6}= ^3\sqrt{3025} \times ^3\sqrt{2^{6}}$

Remember you are taking the cube root so $^3\sqrt{2^{6}}=2^2$

Just an FYI about the LaTeX code:
$$\sqrt[3]{2^6}$$

gives
$\sqrt[3]{2^6}$

-Dan
• May 13th 2013, 07:44 PM
Mathnood768
Re: Radicals #2
Thank you!