Ask your own question, for FREE!
Mathematics 11 Online
OpenStudy (landon34):

Find the GCF of n^3 t^2 and nt^4. help needed

OpenStudy (landon34):

@whpalmer4

OpenStudy (landon34):

@just_one_last_goodbye

OpenStudy (landon34):

ok I think the answer is nt^2 but im not sure

OpenStudy (whpalmer4):

\[n^3t^2 = n*n*n*t*t\]3 ns, 2 ts \[nt^4 = n*t*t*t*t\]1 n, 4 ts The GCF will have the largest number of each factor which can be gotten from both expressions. For example, there will be 2 ts or \(t^2\) in the GCF, because both expressions have at least 2 ts

OpenStudy (landon34):

ok.

OpenStudy (landon34):

I'll write this down in my notes.

OpenStudy (whpalmer4):

Yes, \(nt^2 \) is the correct answer for the GCF of those two. How about this one: GCF of \(3xy\) and \(6x^2y\)

OpenStudy (landon34):

I need some help for the question you just gave me

OpenStudy (landon34):

hello?

OpenStudy (landon34):

I guess their doing something?

OpenStudy (landon34):

plz help me

OpenStudy (landon34):

hope, can u help?

OpenStudy (hope210):

what do you need?

OpenStudy (whpalmer4):

Sorry, OpenStudy hasn't shown me any of the things you typed until just a few seconds ago. Can you factor \[6x^2y\]like I did with \(n^3t^2\) and so on?

OpenStudy (landon34):

yeah, I can do that, whpalmer4

OpenStudy (whpalmer4):

Good. Will you?

OpenStudy (landon34):

Yeah, I can try.

OpenStudy (landon34):

Im home alone, thats why I told you guys if I didnt answer, to get help. ( im 14, im old enough to stay home, so dont worry.)

OpenStudy (landon34):

I dont get it. this stuff is confusing.

OpenStudy (landon34):

whpalmer, can you help with that question?

OpenStudy (landon34):

nobody is answering.

OpenStudy (landon34):

I guess ill go do something else, message me when you are ready to answer.

OpenStudy (whpalmer4):

Do you know what it means to factor something?

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!