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## Topic review (newest first)

krassi_holmz
2006-01-01 00:43:05

Hi. Happy New Year!!!

deepu
2006-01-01 00:36:58

Hi krassi Holmz

Thanks for helping

njkidd
2005-12-31 12:08:38

Thanks, i got them now.

krassi_holmz
2005-12-31 10:35:33

Yes, Deepu, you're right.

deepu
2005-12-31 09:37:08

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)

krassi_holmz
2005-12-31 09:21:21

2) It is not monomial It cannot be simplified:

krassi_holmz
2005-12-31 09:16:02

1)

njkidd
2005-12-31 08:54:12

1) (4xy³)-² • (5x-³y^4)³

2) y-2  /  y^5-3y²-20

------------------------------

1) i have 2000/x^11y^6

2) i have y^4-3y^2+10

Thanks in advance.

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