The book does not explain this one, for some reason:
(-3m^2y^ -1) (2m^ -3 y^ -1)
need help with exponent problem!?
2008-10-09 2:38 pm
回答 (5)
2008-10-09 2:53 pm
✔ 最佳答案
(-3m^2 y^-1)(2m^-3 y^-1) : original expressionMultiply constants, and add exponents
= -6m^-1 y^-2
Without negative exponents, this will be :
-6 / (m y^2)
To make a negative exponent positive, you cross the fraction bar with it -- top goes to bottom and bottom goes to top. These were both on top, so they go to the bottom.
2008-10-09 10:53 pm
(-3m^2y^-1)(2m^-3y^-1)
= (-3)(2)(m^2)(m^-3)(y^-1)(y^-1)
= [-6][m^(2 - 3)][y^(-1 - 1)]
= [-6][m^-1][y^-2]
= [-6][1/m][1/y^2]
= -6/my^2
= (-3)(2)(m^2)(m^-3)(y^-1)(y^-1)
= [-6][m^(2 - 3)][y^(-1 - 1)]
= [-6][m^-1][y^-2]
= [-6][1/m][1/y^2]
= -6/my^2
2008-10-09 9:56 pm
just use FOIL
multiply the F.irst terms
-3m^2(2m^-3)=-6m^-1
then the O.uter terms
-3m^2(y^-1)=-3m^2(y^-1)
then the I.nner terms
2y^-1(2m^-3)=2y^-1(2m^-3)
then the L.ast terms of each equation
2y^-1(y^-1)=2y^-2
then you put all your answers together
-6m^-1+-3m^2(y^-1)+2y^-1(2m^-3)+2y^-2
and thats your answer
multiply the F.irst terms
-3m^2(2m^-3)=-6m^-1
then the O.uter terms
-3m^2(y^-1)=-3m^2(y^-1)
then the I.nner terms
2y^-1(2m^-3)=2y^-1(2m^-3)
then the L.ast terms of each equation
2y^-1(y^-1)=2y^-2
then you put all your answers together
-6m^-1+-3m^2(y^-1)+2y^-1(2m^-3)+2y^-2
and thats your answer
參考: =)
2008-10-09 9:54 pm
Simply add or subtract exponents of the same variable.
(-3m^2y^ -1) (2m^ -3 y^ -1) = (-3m^2) (2m^ -3) (y^ -1)( y^ -1)
= -6 m^-1 y^-2)
(-3m^2y^ -1) (2m^ -3 y^ -1) = (-3m^2) (2m^ -3) (y^ -1)( y^ -1)
= -6 m^-1 y^-2)
2008-10-09 9:51 pm
Multiply:
-6 m^2
---------
-m^3y^2
= -6 / my^2
-6 m^2
---------
-m^3y^2
= -6 / my^2
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