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need help dividing algebra terms

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45 TIMES Xto the fourth power TIMES Y to the fourth power divided by 15 TIMES Xto the second power TIMES Y to third power
other way 15 on top
\[45x^{4}y ^{4} \over15x ^{2}y ^{3}\]

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Other answers:

like this?
yes sir
you mean\[15x ^{2}y ^{3}\over45x ^{4}y ^{4}\]
yes precal has it
=\[=\left( 15 \over 45 \right)\left( x ^{2} \over x ^{4}\right)\left( y ^{3}\over y ^{4} \right)\]
Can you simplify it now?
ok 1/3 and x2power
i dont know the last one
precal is right pfenn1 is dummmm
y to the 1 power?
remember \[\left( 1 \over x ^{a} \right)=x ^{-a}\]and\[x ^{a}x ^{b}=x ^{a+b}\]
|dw:1336094932499:dw| that is how it should look
they got u pfenn1
@cesar2462, don't really appreciate the attitude. There are several ways to look at this problem and sometimes different ways can help someone understand it better. We are trying to help people learn, not just get the answer.
@pfenn1 i like the idea how you separate it into sections can you do it again with another problem
-56 TIMES A 4thpower TIMES X 4thpower over -8 TIMES A 3rdpower TIMES X
\[-56A ^{4}x ^{4}\over -8A ^{3}x\]
-56A^4X^4/-8A^3X that what you mean?
Sorry, I had problems with my computer.\[\left( -56 \over-8 \right)\left( A ^{4}\over A^{3} \right)\left( x ^{4}\over x\right)=\]
thank you everyone but this medal goes @pfenn1
@cesar2462 The attitude is not necessary. If you have nothing nice to say, say nothing at all.
i agree^

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