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Mathematics 19 Online
OpenStudy (ageta):

Free medal!!!

OpenStudy (ageta):

√84y^7

OpenStudy (ageta):

@sleepyjess @confluxepic @TheSmartOne @ageta

OpenStudy (ageta):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

The problem is this \[\Large \sqrt{84y^7}\] correct?

OpenStudy (ageta):

yea it is the problem

OpenStudy (ageta):

we have to simplify it

jimthompson5910 (jim_thompson5910):

what are some ways to factor 84? which factors are perfect squares?

OpenStudy (ageta):

4 and 21

jimthompson5910 (jim_thompson5910):

good

jimthompson5910 (jim_thompson5910):

with the y^7, we can break it up into y^6*y agreed?

OpenStudy (ageta):

yep

jimthompson5910 (jim_thompson5910):

So based on all that, we can say \[\Large \sqrt{84y^7}\] \[\Large \sqrt{4*21*y^6*y}\] \[\Large \sqrt{4y^6*21y}\] \[\Large \sqrt{4y^6}*\sqrt{21y}\]

jimthompson5910 (jim_thompson5910):

what comes next?

OpenStudy (ageta):

2

jimthompson5910 (jim_thompson5910):

square root of 4 is 2, yes how about the y^6 part?

OpenStudy (ageta):

divide it by two

OpenStudy (ageta):

i think

jimthompson5910 (jim_thompson5910):

yes 6/2 = 3 is the new exponent

jimthompson5910 (jim_thompson5910):

So \[\Large \sqrt{4y^6}*\sqrt{21y}\] turns into \[\Large 2y^3*\sqrt{21y}\]

jimthompson5910 (jim_thompson5910):

and that's as simplified as it gets

OpenStudy (ageta):

oh i get it now

OpenStudy (ageta):

thank you

jimthompson5910 (jim_thompson5910):

no problem

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