#### crappiefisher26

##### New member

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x^2y^-2*x^5yz^5*9x^-2y^-4

(9)(x^2*x^5*x^-2)(y^-2*y^-4)(z^5)

9x^2+5-2 y^-2-4 z^5

= 9x^5y^-6z^5

9y^-6z^5/x^5 thats my answer.

heres the correct answer

9x^5z^5/y^5

- Thread starter crappiefisher26
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x^2y^-2*x^5yz^5*9x^-2y^-4

(9)(x^2*x^5*x^-2)(y^-2*y^-4)(z^5)

9x^2+5-2 y^-2-4 z^5

= 9x^5y^-6z^5

9y^-6z^5/x^5 thats my answer.

heres the correct answer

9x^5z^5/y^5

- Joined
- Jan 29, 2005

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\left( {x^2 y^{ - 2} } \right)\left( {x^5 y^1z^5 } \right)\left( {9x^{ - 2} y^{ - 4} } \right) = 9\left( {x^{2 + 5 - 2} } \right)\left( {y^{ - 2 + 1 - 4} } \right)\left( {z^5 } \right) = 9x^5 y^{ - 5} z^5 =\frac{9 x^5 z^5}{y^{5}}\)

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This is the same type but im not sure how you would set it up.

(x^2/y^3z)*(xy/z^4)*(y^2z^3/x^9)

(x^2/y^3z)*(xy/z^4)*(y^2z^3/x^9)

Bring the denominators to the top by changing the sign of the power;crappiefisher26 said:This is the same type but im not sure how you would set it up.

(x^2/y^3z)*(xy/z^4)*(y^2z^3/x^9)

like x^2 / (y^3 z^1) becomes x^2 y^(-3) x^(-1);

then complete as pka showed you.

Your 1st term (x^2/y^3z) requires another set of brackets: (x^2/(y^3z))

As example:

20 / 5 * 2 = 4 * 2 = 8

20 / (5 * 2) = 20 / 10 = 2 : capish :?: :shock:

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- Jul 23, 2006

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im confused now.

sorry, is there any way to explain it more thoroughly.

Thanks.

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Man crappiefisher26 you are having far too much trouble with all of this.crappiefisher26 said:so i would set the problem up like this correct

im confused now. sorry, is there any way to explain it more thoroughly.

It just is not that hard. Do you have a teacher/instructor?

Do you have someone you can go to and have a good sit-down talk to?

You really need to do that before you find yourself totally lost.

If you don’t do that, you may find yourself beyond repair.