## anonymous one year ago This question has something to do with quotient rule and simplify. Write your answer in positive exponents.

1. anonymous

3k^2l^-8/6k^7l^-2

2. anonymous

|dw:1442358587704:dw| this?

3. zepdrix

I think it's this:$\large\rm \frac{3k^{2}\ell^{-8}}{6k^{7}\ell^{-2}}$

4. anonymous

correct

5. anonymous

now how do i do this

6. anonymous

please explain all steps because i will be offline. thank you so much for having the time to help me solve this

7. anonymous

Okay.. I know you can simplify that because of the 3/6 and the -8/-2. They both can be simplified. I think you can also simplify the k2 and k7... but let's see. Three and six is easy, you cancel out the 3 and divide 6 by 3, you end up with 1 and 2 right? Move on to the next part. k2 and k7, I believe you can cancel out the k2 and take k2 off of k7 and that leaves you with just k5 in the denominator.. -8 and - 2.. they can divide. Cancel out the -2 with -2 and divide -8 by -2. That leaves you with l^4 in the numerator. Does this sound about right..?

8. anonymous

yes i does sound right im checking right now

9. zepdrix

Woooops! Subtraction Rule of Exponents:$\large\rm \color{royalblue}{\frac{x^a}{x^b}=x^{a-b}}$

10. zepdrix

$\large\rm \frac{\ell^{-8}}{\ell^{-2}}=\ell^{-8-(-2)}$

11. anonymous

wow thank you so much that is absolutly right

12. zepdrix

k5 sounds right though :) 2-7 = -5 leaving it in the denominator.

13. anonymous

so it is 1/2k^5l^6

14. anonymous

From the looks of it it's just l^4 over k5..

15. zepdrix

$\large\rm =\frac{1}{2k^5\ell^{6}}$Ya this sounds right :)

16. zepdrix

Rheaaaa >.< I'm confused why you applied subtraction to the 2 and 7, but division to the -8 and -2 ... hmm weird.

17. anonymous

Oh wait. Wait goodness lol I just noticed that mistake thank you for pointing it out.