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**njkidd****Guest**

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.

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

1)

*Last edited by krassi_holmz (2005-12-30 10:17:57)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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2) It is not monomial It cannot be simplified:

IPBLE: Increasing Performance By Lowering Expectations.

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**deepu****Member**- Registered: 2005-12-30
- Posts: 3

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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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

Yes, Deepu, you're right.

IPBLE: Increasing Performance By Lowering Expectations.

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**njkidd****Guest**

Thanks, i got them now.

**deepu****Member**- Registered: 2005-12-30
- Posts: 3

Hi krassi Holmz

Thanks for helping

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

Hi. Happy New Year!!!

IPBLE: Increasing Performance By Lowering Expectations.

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