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anonymous
 one year ago
how the heck do I do this?!?!?!
(2xy5x)^2
anonymous
 one year ago
how the heck do I do this?!?!?! (2xy5x)^2

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what are you trying to do?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im trying to foil this expression, but I dont know what to do with the xy and the x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oooh okay so what you want to do is write that same equation out twice so it will be (2xy5x)(2xy5x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and what we basically want to do is kind distribute

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 2xy times 2xy = 4x^2y^2 so now do that to the 5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 2xy(5) which is 10xy so so far we have 4x^2y^210xy

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0Well, can you do this? \((ab)^2\) If so, then it's just same process. You can let \(a = 2xy\) and \(b = 5x\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we want to do the 5 (2xy) which gives 10xy so we now have 4x^2y^210xy10xy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do I multiply xy by x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0after foiling 2xy to the other 2xy, I have to multiply the 2xy by the 5x, and thats where I get stuck.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh you just do it dont stop for nothing 2xy(5x) you do 5 times 2 which gives you 10 how many x's are there? 2 so 10x^2 and now how many y's? 1 so it becomes 10x^2y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so when I foil xy and x, I have to combine the x's and leave the y?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0If by "leaving" y, you mean do nothing, then yes. You have \(xy\cdot x ~~~=~~~ x\cdot x\cdot y = x^2y\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes and the same would go if it was y y y x z it would be y^3xz....your basically counting the variables
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