jessie13
  • jessie13
Will Medal Answer ASAP!!!!! 1: What is the GCF of the numerator and denominator in the following fraction? 21m^2n/28mn^2 7m^2n^2 7mn 21m^3n^3 mn
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
so whats the question?
jessie13
  • jessie13
im putting it up right now!
anonymous
  • anonymous
alrighty then.

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jessie13
  • jessie13
@paki
jessie13
  • jessie13
@geny55 @Photon336
Photon336
  • Photon336
So this is what we have right @jessie13 \[\frac{ 21m ^{2}n }{ 28mn ^{2} }\]
Photon336
  • Photon336
so greatest common factor, is the largest number that can go into both the numerator and denominator
jessie13
  • jessie13
1 dont get it!
Photon336
  • Photon336
Let's break this up into two parts
Photon336
  • Photon336
\[\frac{ 21 }{ 28 } * (\frac{ m ^{2}n }{ mn ^{2} })\]
Photon336
  • Photon336
do you know how to subtract powers?
Photon336
  • Photon336
whenever we divide exponents we subtract the exponents. so we will do that first \[\frac{ m ^{2}n}{ mn ^{2} } = Numerator[ m ^{2}-m ^{1} = m] denominator [n ^{1}-n ^{2}] = n ^{-1}\] we simplify the variables first. \[\frac{ m }{ n }\]
Photon336
  • Photon336
@jessie13 so now this is what we have we need to find the greatest common factor so it has to be the largest number that goes into both 21 and 28 \[\frac{ 21 }{ 28 } *(\frac{ m }{ n })\]
jessie13
  • jessie13
wait the gcf is 7
Photon336
  • Photon336
@jessie13 yep :) here's why. and that's why I told you to do it that way yeah so |dw:1444067072198:dw|
Photon336
  • Photon336
let me rewrite this
jessie13
  • jessie13
k
Photon336
  • Photon336
\[\frac{ 3 m}{ 4 n} * *\frac{ 7 mn}{ 7 mn } = \frac{ 21 }{ 28 } \frac{ m*m = m ^{2}n }{n*n = n ^{2}m }\]
Photon336
  • Photon336
\[(7mn) = GCF\] because if we multiply this by 3m/4n we get what we started with @jessie13 look at this tell me if this makes sense to you
jessie13
  • jessie13
okay!
jessie13
  • jessie13
ya it makes more sense thanks!
Photon336
  • Photon336
can you show me why this makes sense @jessie13 ?

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