imgur: the simple image sharer

My attempt:

imgur: the simple image sharer

I'm not sure if I was right about dividing .3 in the beginning from the radical.

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- May 12th 2013, 11:05 AMMathnood768Radicals #1
imgur: the simple image sharer

My attempt:

imgur: the simple image sharer

I'm not sure if I was right about dividing .3 in the beginning from the radical. - May 12th 2013, 02:10 PMShakarriRe: Radicals #1
You didn't need to divide by 0.3. You also divided 6.3 by 0.3 as well as dividing $\displaystyle \sqrt{48}$ by $\displaystyle \sqrt{0.3}$ so in total you divided by $\displaystyle 0.3\sqrt{0.3}$. If you want to do this you also have to multiply by $\displaystyle 0.3\sqrt{0.3}$ to effectively multiply by 1 so that you don't change the value of it, to balance things.

Also there is no need bring everything inside the square root so that it is $\displaystyle \sqrt{70560}$ because you'll end up bringing it all outside the square root in the end.

You can simplify it like this

$\displaystyle 6.3\sqrt{48}= 6.3\sqrt{6\times8}=6.3\sqrt{2\times3\times2\times2 \times2}= 6.3\times 2\times 2\sqrt{3}=25.2\sqrt{3}$

Somehow your answer still came out right - May 13th 2013, 06:41 PMMathnood768Re: Radicals #1
Thank you!