simplify (3rs)^(-2) / (3^2 r^2 s^(-4))

[jshaziza]

Would someone help me with these two equations?

(3rs) to negative 2 exponent divided by 3 to the second r to the second and s to the -4

12k to the -2(k to the -3) to the -4 divide by 6k to the 5

thx in advance.

 

[stapel]

These are not equations, since they contain no "equals". They can not be "solved". However, expressions can be "simplified". Is that what you were supposed to do?

I will guess that you mean the following:

. . . . .1) [(3rs)^-2] / [3² r² s^-4]

. . . . .2) [(12k)^-2 (k^-3)^-4] / [(6k)⁵]

Please reply with correction or confirmation, along with the instructions and a clear listing of the steps you have tried so far.

Thank you.

Eliz.

 

[jshaziza]

To Stapel

Hey Stapel, you are right I am supposed to simplify the two problems, and the two problems you wrote are also correct.

for #1 my problem is how far do I simplify it? I can either make (3rs)^-2 into 3^-2
r^-2 and s^-2 divided by the bottome row my final answer or simplify it further to 1^-4 r^-4 and 1^2. And also though, I am not allowed to have negative exponents in my final answer.

for #2 I am also having the same problem.

So if you can give me some help I would appreciate it thx

 

[skeeter]

\(\displaystyle \L \frac{(3rs)^{-2}}{3^2r^2s^{-4}} =\)

\(\displaystyle \L \frac{s^4}{(3rs)^23^2r^2} =\)

\(\displaystyle \L \frac{s^4}{3^4r^4s^2} =\)

\(\displaystyle \L \frac{s^2}{3^4r^4}\)

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

\(\displaystyle \L \frac{(12k)^{-2}(k^{-3})^{-4}}{(6k)^5} =\)

\(\displaystyle \L \frac{k^{12}}{(12k)^2(6k)^5}=\)

\(\displaystyle \L \frac{k^{12}}{12^2 \cdot 6^5k^7} =\)

\(\displaystyle \L \frac{k^5}{12^2 \cdot 6^5}\)