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anonymous
 5 years ago
(x^(2)y^(2))/(x^(1)y^(1)) please?
anonymous
 5 years ago
(x^(2)y^(2))/(x^(1)y^(1)) please?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{x^{2}y^{2} \over x^{1}y^{1}}?\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x^{2}y^{2}}{x^{1}y^{1}}\] right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its just a matter of flipping all the ^ exponents to their reciprocals and re writeing it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0what part are you stuck at?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just factor the numerator: \[={(x^{1}y^{1})(x^{1}+y^{1}) \over x^{1}y^{1}}=x^{1}+y^{1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x^{2}y^{2}}{x^{1}y^{1}}=\frac{\frac{1}{x^2}\frac{1}{y^2}}{\frac{1}{x}\frac{1}{y}}\] Answar's ways method is the correct snappy way to do it but if you want to see why it is true, multiply numerator and denominator by \[x^2y^2\] to clear the fractions, or subtract and divide and you will get \[\frac{yx}{xy}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow, thanks satelite73! how do i give a medal for your answer? you had the clearest solution of all! made me really understand the problem well. :) THANK YOU SO MUCH TO THE OTHERS AS WELL! <3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By the way satellite, it's Anwar :) .. My way is simpler if you focus :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you're right Anwar! it is! i guess i just have to be more focused!
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