-1(x^-2)y/ 5
First divide the numbers -3 and 15.
Remember you can move varibles up and down the division symbol is you change the sign of the exponent. For example x/1= 1/ x^-1
Therefore just move the variables and do the subtraction =]
2006-12-12 12:47:29
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answer #1
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answered by mini_roller 3
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To simplify, all you have to do is cancel common terms. Regarding the actual numbers, all you have to do is simplify them as you would the fraction.
First, let's start with the numbers. What does -3/15 reduce to? The answer to that is -1/5, so you'd have a (-1) on top and a 5 on the bottom while cancelling out the -3 and 15.
Now, the variables. Every x that appears on the top can cancel out an x that appears on the bottom. Similarly, every x that appears on the bottom can cancel out an x that appears on the top. In this case, we have one x on the top, so we cancel that one out and reduce the power of the x on the bottom by 1. We also cancel out the y^3 and reduce the power of y on the top by 3.
(-1)y^3
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5x^2
or
-y^3
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5x^2
2006-12-12 20:43:31
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answer #2
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answered by Puggy 7
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the answer is...
-1x(3)
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3y
the (3) is suppossed to be an exponent of 3 but im not sure how to do that. I havent done this math in a while but im almost sure thats what the answer is.
2006-12-12 20:45:03
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answer #3
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answered by softballgurl13 1
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when dividing.......subtract exponents.....
-3/15 -1/5
x^1 / x^3 = x^-2 = 1/x^2
y^4/ y^3 = y^1
thus.........-y/5x^2
2006-12-12 20:43:30
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answer #4
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answered by luvmath03 5
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I haven't taken math in 8 years, sorry. I would guess it would be:
-1y
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3x2
...but I could be wrong.
2006-12-12 20:40:06
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answer #5
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answered by T 3
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The answer is definitely
-1x˲y
______
5
or you can put it as
-1y
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5x²
2006-12-13 05:02:58
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answer #6
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answered by Fräulein Jaclyn 2
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-1y
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5x^2
divide integers, subtract exponents.
2006-12-12 20:40:42
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answer #7
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answered by keirnal_2002 1
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like this : -1y/x2.....dont know how to type exponents....thats x squared.
2006-12-12 20:44:37
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answer #8
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answered by thappygolucky 2
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-1y/5x^2
2006-12-12 20:44:08
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answer #9
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answered by Anonymous
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-y/5xx
2006-12-12 20:42:33
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answer #10
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answered by rascoe627 1
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