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

(2a)exponent 3

OpenStudy (anonymous):

\[8a^3\]

OpenStudy (anonymous):

\[(2a)^3=2^3a^3=8a^3\]

OpenStudy (anonymous):

what did you do to get 8?

OpenStudy (anonymous):

2^3 means 2*2*2=8

OpenStudy (anonymous):

8A^2

OpenStudy (anonymous):

oh, duh! i knew that.

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

will you help me with just one more question?

OpenStudy (anonymous):

sure thing

OpenStudy (anonymous):

(x^2 y^2)^4xy^5

OpenStudy (anonymous):

Is that \[(x^2y^2)^4xy^5\]

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[x^{2*4+1}y^{2*4+5}\] \[=x^9y^13\]

OpenStudy (anonymous):

^^^^AGREE

OpenStudy (anonymous):

That's supposed to be 9 and 13 by the way, it's a bit messy sorry.

OpenStudy (anonymous):

If so then the 4 is applied to both factors inside the parentheses first\[(x^2)^4(y^2)^4xy^5\]then you have\[x^8y^8xy^5= x ^{9}y ^{13}\]

OpenStudy (anonymous):

okay, so my answer would be x^9 y^13?

OpenStudy (anonymous):

Exactly

OpenStudy (anonymous):

thankyou so much!!!!

OpenStudy (anonymous):

i tried this one on my own and i was just wondering if you could tell me if i did it right..... the question is: 12x^8/3x^6 and i got 15x^14

OpenStudy (anonymous):

NO MEDALS FOR ME :(

OpenStudy (anonymous):

thats confusing

OpenStudy (anonymous):

If the 3x^6 is in the denominator, then you have an error, x^8/x^6=x^2 it should be in the numerator. 12/3 would reduce to 4 so the answer would be 4x^2.

OpenStudy (anonymous):

There was no 15, I accidentally pulled it from your answer.

OpenStudy (anonymous):

The meaning of x^8 is that there are 8 factors of x in the numerator; the meaning of x^6 is that there are 6 factors of x in the denominator. The 6 factors of x in the deniminator cancel 6 of the 8 factors of x in the numerator leaving 2 factors of x in the numerator.

OpenStudy (anonymous):

|dw:1315266143663:dw|

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!