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  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,984 Thanks: 995  Quote:
(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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