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Mathematics 11 Online
OpenStudy (anonymous):

m^x+y / m^x-y

OpenStudy (jhannybean):

Is it \[\large\cfrac{m^{x+y}}{m^{x-y}}\]

OpenStudy (anonymous):

yeah

OpenStudy (jhannybean):

This would result in \[\large m^{(x+y)-(x-y)}\] Can you evaluate this?

OpenStudy (anonymous):

sorry im abit confused how to get m(x+y)-(x-Y)

OpenStudy (anonymous):

ohh nevermind I got it !:)

OpenStudy (jhannybean):

Awesome :D

OpenStudy (jhannybean):

What did you get as your answer? :)

OpenStudy (anonymous):

im not sure if this right but I distributed the m and got mx+my-mx-my

OpenStudy (jhannybean):

just distribute the negative sign to (x-y)

OpenStudy (anonymous):

m(x+y)-x+y ? is that right

OpenStudy (jhannybean):

\[\huge m^{(x+y)-(x-y)} = m^{x+y-x+y}\]

OpenStudy (jhannybean):

Yes you are right :)

OpenStudy (jhannybean):

Now simplify the exponent and combine like-terms :P

OpenStudy (anonymous):

mx^2y^2?

OpenStudy (jhannybean):

Close! \[\huge m^{2y}\]

OpenStudy (jhannybean):

The x's canceleachother out and you're left with \(\large m^{1y+1y} = m^{2y}\)

OpenStudy (anonymous):

ohh oops I thought both the x's were positive :P Thank you soso much for your help :D

OpenStudy (jhannybean):

No problem :D

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