Sunday

April 20, 2014

April 20, 2014

Posted by **Nicole** on Tuesday, November 26, 2013 at 5:30pm.

In simpler terms it would look like this:

2x^4 X y^-4 X z^-3

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

3x^2 X y^-3 X z^4

When I tried to solve this I got 6x^2 X y^-1 X 2^-7. Am I correct? If not could you please help me on how to solve it.

- Math #2 -
**Steve**, Tuesday, November 26, 2013 at 6:04pmNope. The coefficients just divide normally: 2/3

x^4/x^2 = x^2

y^-4/y^-3 = y^-1

z^-3/z^4 = z^-7

so, you wind up with

2/3 x^2 y^-1 z^-7, or

2x^2

--------

3yz^7

**Related Questions**

Math33333 - Please help me! Problem: x+2 x-3 _________ - _________ 2x^2+5x+2 2x^...

Math - I factored 3.6x^6+2x^6-2.8x^2+3.2x as; 3.6x^7+2x^6-2.8x^2+3.2x However my...

Algebra - Below is a "complete the square problem" from a math help site on the ...

Calculus - relative extrema x^4-2x^2+5 so far I know how to find the derivative ...

Math - This is a 3rd grade math problem that my son had on homework. Can you ...

Discrete Math - I have a review problem I am having problems with. This is the ...

calculus - So I am suppose to evaulate this problem y=tan^4(2x) and I am ...

math,correction - Is this correct or no. I need help in a couple of problems can...

Rational Expression - (x+1/2x-1 - x-1/2x+1) * (2x-1/x - 2x-1/x^2) (x+1/2x-1 - x-...

Math - Trigonometry Identities problem: Prove the following; (tan^2x)(cos^2x...