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I need some help simplifying 'e'.

 

Thanks for any help!

Guest Oct 4, 2018

Best Answer 

 #1
avatar+3148 
+2

\(\Large \dfrac{a^{2n-1}\times b^3 \times c^{1-n}}{a^{n-3}\times b^{2-n}\times c^{2-2n}} = \\ \\ \Large a^{(2n-1)-(n-3))}\times b^{3-(2-n)}\times c^{(1-n)-(2-2n)} = \\ \\ \Large a^{n+2}\times b^{n+1}\times c^{n-1}\)

Rom  Oct 4, 2018
 #1
avatar+3148 
+2
Best Answer

\(\Large \dfrac{a^{2n-1}\times b^3 \times c^{1-n}}{a^{n-3}\times b^{2-n}\times c^{2-2n}} = \\ \\ \Large a^{(2n-1)-(n-3))}\times b^{3-(2-n)}\times c^{(1-n)-(2-2n)} = \\ \\ \Large a^{n+2}\times b^{n+1}\times c^{n-1}\)

Rom  Oct 4, 2018
 #2
avatar+41 
0

Ok. 'e' is Euler's Constant, or approximately 2.718281828459045235360...

Euler's constant can be defined by terms of an infinite sum in the form of \(e=\sum_{n=0}^{}{\frac{1}{n!}}\)

creepercraft97T3  Oct 4, 2018
 #3
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0

no thats lord farquaad from shrek

Guest Oct 4, 2018

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