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 January 20th, 2016, 07:17 PM #11 Member   Joined: Sep 2015 From: Philadelphia Posts: 36 Thanks: 3 I'll be here for hours typing my work but my final answer was a(squared) b( to the 11th power) over 2 c(squared) a
 January 20th, 2016, 07:31 PM #12 Member   Joined: Sep 2015 From: Philadelphia Posts: 36 Thanks: 3 I just figured out my mistake. It was in my subtraction. It's amazing how one little thing can affect the whole problem. Thanks again
January 21st, 2016, 06:18 AM   #13
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From: Ottawa Ontario, Canada

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Quote:
 Originally Posted by hhmathgeek I'll be here for hours typing my work but my final answer was a(squared) b( to the 11th power) over 2 c(squared) a
So you got:
(a^2 * b^11) / (2 * c^2 * a)

That's not really incorrect: just needs one further simplification:
(a * b^11) / (2 * c^2)
...which you say you got...so all's well

 January 21st, 2016, 08:16 PM #14 Senior Member   Joined: May 2015 From: Varanasi Posts: 110 Thanks: 5 Math Focus: Calculus Well i think that is quiet simple just subtract the exponents of same variable below fraction from the exponents of variable above fraction. Thats the central idea.

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