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1) (4xy³)-² (5x-³y^4)³
2) y-2 / y^5-3y²-20
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1) i have 2000/x^11y^6
2) i have y^4-3y^2+10
Thanks in advance.
1)
Last edited by krassi_holmz (2005-12-30 10:17:57)
IPBLE: Increasing Performance By Lowering Expectations.
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2) It is not monomial It cannot be simplified:
IPBLE: Increasing Performance By Lowering Expectations.
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Hi
1) 1/(4xy^3)^2 * 125x^-9y^12
= (125y^6)/16x^11
2) Let us first divide (y^5-3y²-20) by (y-2)
The quotient will be y^4 + 2y^3 + 4y^2 +5y+10
which means that
(y-2)(y^4 + 2y^3 + 4y^2 +5y+10) = y^5-3y²-20
Therefore,
y-2 / y^5-3y²-20
=y-2/ (y-2)(y^4 + 2y^3 + 4y^2 +5y+10)
(y-2) in the numerator and denominator gets cancelled and the answer will be
= 1/(y^4 + 2y^3 + 4y^2 +5y+10)
Last edited by deepu (2005-12-30 10:40:26)
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Yes, Deepu, you're right.
IPBLE: Increasing Performance By Lowering Expectations.
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Thanks, i got them now.
Hi krassi Holmz
Thanks for helping
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Hi. Happy New Year!!!
IPBLE: Increasing Performance By Lowering Expectations.
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