Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

(x*3)^2 * (x*5)^4

OpenStudy (anonymous):

Use: \[\large (x^a)^b = x^{a \times b}\] \[\large x^a \times x^b = x^{a+b}\]

OpenStudy (anonymous):

so would it be x^6+^20?

hartnn (hartnn):

where did u get + from?? it would be x^6*x^20 then use the 2nd formula given by waterineyes

OpenStudy (anonymous):

\[\large x^{6+20}\]

OpenStudy (anonymous):

you can add 6 + 20 here..

OpenStudy (anonymous):

thats what i meant... i cant use the small script so i added the 6 and the 20

OpenStudy (anonymous):

So finally what did you get ??

OpenStudy (anonymous):

x^26

hartnn (hartnn):

correct :)

OpenStudy (anonymous):

thanks guys! i appreciate that!

OpenStudy (anonymous):

Sorry for late.. yes \(x^[26}\) is correct..

OpenStudy (anonymous):

\(x^{26}\) is correct..

OpenStudy (anonymous):

Welcome dear..

OpenStudy (anonymous):

Thanks @hartnn for being on time here...

hartnn (hartnn):

no need of thanks...i was just looking around...u actually solved it....

OpenStudy (anonymous):

f^2*g^7

OpenStudy (anonymous):

you cannot solve it further as f and g are two different variables..

OpenStudy (anonymous):

they have me over here doing these ridiculous math problems at 2 am!

OpenStudy (anonymous):

So you will leave it as such: \[f^2 \cdot g^7\]

OpenStudy (anonymous):

-2x^8*-2x^7 (i got 4x^15 for my final answer is that correct?

hartnn (hartnn):

yup :)

OpenStudy (anonymous):

ok.. i had to make sure! thanks!

OpenStudy (anonymous):

Yep..

OpenStudy (anonymous):

4^3*4^7*4^4 it wants me to simplify and i got 4^14 for my final answer do you think i worked it out right?

OpenStudy (anonymous):

Yeah it looks good to me..

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

Welcome..

OpenStudy (anonymous):

8/s^-9

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!