Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

simplify? question in comments

OpenStudy (anonymous):

|dw:1382659015194:dw|

OpenStudy (anonymous):

is it (144a)^3 or 144(a^3) ?

OpenStudy (anonymous):

144(a^3)

OpenStudy (anonymous):

k, give me a second...

OpenStudy (anonymous):

i suck..

OpenStudy (anonymous):

it's all good haha i couldn't get it either! my answer doesn't make sense

OpenStudy (anonymous):

I'm not sure which direction to take, if I should change it into exponents or keep it as radicals

OpenStudy (mathlegend):

@DemolisionWolf might I ask what did you get so far?

OpenStudy (anonymous):

i broke it down like this: (2/5)*sqrt(11a)sqrt(4) + sqrt(36a)sqrt(4) - (1/2)sqrt(11a) so now I have a sqrt(4) and a sqrt(11).... maybe this is more like a polynomial?

OpenStudy (mathlegend):

I basically have... \[\frac{ 4 }{ 5 }\sqrt{11a} + 12a \sqrt{a} - \frac{ \sqrt{11a} }{ 2 }\]

OpenStudy (mathlegend):

But I am not that sure of this...

OpenStudy (anonymous):

@MathLegend @DemolisionWolf the textbook says: (3/10)(sqroot 11a) +(12a sqroot a) , a is greater than or equal to 0

OpenStudy (mathlegend):

as the answer?

OpenStudy (anonymous):

yep \[\frac{ 3 }{ 10}\sqrt{11a}+12a \sqrt{a}, a \ge0\]

OpenStudy (mathlegend):

okay I played around with the numbers... and I got two parts correct... so let me show you how I did the two parts and see if we can figure out the next part.

OpenStudy (mathlegend):

\[\frac{ 2 }{ 5 }\sqrt{44a}\]

OpenStudy (mathlegend):

The highest perfect square that goes into 44 is... what?

OpenStudy (anonymous):

I got it figure out now, if any one cares

OpenStudy (mathlegend):

Yeah, I got the whole thing too but he is offline. :(

OpenStudy (anonymous):

oh really haha *high five us*

OpenStudy (anonymous):

|dw:1382661738251: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!