Help Dividing this Problem?

[Hanajima]

I got this answer because

a^5 / a^6 = 1/ a
b / b^11 = 1/ b^10
and the square root of 72 is 6 square root of 2

I tried to get rid of the radical in the denominator and ended up getting the answer you see in the screenshot.

 

[lookagain]

I got this answer because

a^5 / a^6 = 1/ a
b / b^11 = 1/ b^10
and the square root of 72 is 6 square root of 2

I tried to get rid of the radical in the denominator and ended up getting the answer you see in the screenshot.

The screenshot is not showing up for me.

 

[Hanajima]

I got this answer because

a^5 / a^6 = 1/ a
b / b^11 = 1/ b^10
and the square root of 72 is 6 square root of 2

I tried to get rid of the radical in the denominator and ended up getting the answer you see in the screenshot.

The screenshot is not showing up for me.

Sorry about that!

Here it is:



Uploaded with ImageShack.us

 

[lookagain]

I got this answer because

a^5 / a^6 = 1/ a
b / b^11 = 1/ b^10
and the square root of 72 is 6 square root of 2

I tried to get rid of the radical in the denominator and ended up getting the answer you see in the screenshot.

\(\displaystyle 6\sqrt{\frac{2}{7ab^{10}}} \ = \\)


\(\displaystyle \frac{6}{b^5}\sqrt{\frac{2}{7a}} \ = \\)


\(\displaystyle \frac{6}{b^5}\sqrt{\frac{2}{7a}} \cdot \frac{\sqrt{7a}}{\sqrt{7a}}\)


Can you finish it from here?

 

[Hanajima]

I got this answer because

a^5 / a^6 = 1/ a
b / b^11 = 1/ b^10
and the square root of 72 is 6 square root of 2

I tried to get rid of the radical in the denominator and ended up getting the answer you see in the screenshot.

\(\displaystyle 6\sqrt{\frac{2}{7ab^{10}}} \ = \\)


\(\displaystyle \frac{6}{b^5}\sqrt{\frac{2}{7a}} \ = \\)


\(\displaystyle \frac{6}{b^5}\sqrt{\frac{2}{7a}} \cdot \frac{\sqrt{7a}}{\sqrt{7a}}\)


Can you finish it from here?

I got 6 square root of 14a over 7ab^5

Thank you so much! :]