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anonymous
 5 years ago
I am a little lost on this
20x^840x^6+40x^2/5x^2
anonymous
 5 years ago
I am a little lost on this 20x^840x^6+40x^2/5x^2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can divide 5x^2 from both the top and the bottom. Divide each coefficient in the top by 5, and subtract 2 from each exponent. (x^m / x^n = x^(mn))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so would this be the answer 4x^68x^4+8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then how would you do this one 30x^3y^2/6x^7y^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would it be this 5X^10y^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just do the same thing except for both x and y. You can do it in steps. First divide the top and bottom by 6. Then divide the top and bottom by x^7, then divide the top and bottom by y^2. When you are done you might want to the negative exponent to the denominator.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, you are close, but you are adding the exponents. You need to subtract when you are dividing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that would make them negative correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^m / x^n = x^(mn) Also, x^(m) = 1/x^m

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, x^3/x^7 is x^(37) = x^(4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well then I have stupid question how do you make it where the exponents are nonnegative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^(m) = 1/x^m you can turn x^(4) to 1/(x^4) A negative exponent just means that it is 1 over to what it would be if the exponent were positive. http://tutorial.math.lamar.edu/Classes/Alg/IntegerExponents.aspx
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