SR (600y^23)

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- anonymous

SR (600y^23)

- schrodinger

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- anonymous

\[\sqrt{600y ^{23}}\]
is the actual equation

- bahrom7893

\[(600y^{23})^{1/2}\] = \[10 (y^{22/2})\sqrt{6*y} \]

- bahrom7893

\[10y^{11}\sqrt{68y}\]

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## More answers

- bahrom7893

woops sorry:
\[10y^{11}\sqrt{6*y}\]

- bahrom7893

Please click on become a fan if I helped, I really want to get to the next level!! Thanks =)

- anonymous

any questions?

- anonymous

yeah I am still lost as can be...

- anonymous

your in alegebra 2 right? im might be in the same chapter as you.

- anonymous

essentially it is algebra 2

- anonymous

okay soo first see in you can find a sqrt of 600.

- anonymous

it comes out as a decimal so I know that I have to simplify first and to me it would make the most sense to have it be \[\sqrt{100}\sqrt{6}\]

- anonymous

being as 100 is a perfect square

- anonymous

what my teacher taught me was to write it like Sqrt(100x6) then break those down to more simple numbers. Sqrt(50x2x2x3) [50x2=100] and 2x3=6

- anonymous

Do you guys want to go through the problem

- anonymous

?

- anonymous

i am going throught the problem...

- anonymous

the sqrt will eventually look like this (5x5x2x2x2x3)

- anonymous

you have 2 5's and 2 2's. take both and multiple 5 and 2 and put their product on the outside of the squareroot.

- anonymous

your equation should look like this. 10sqrt(6y^23) [6=product of remianing numbers in root]

- anonymous

for just figure out how many times 2 can go into 23 any remaing numbers remain in the root. answer = \[10y^{11}\sqrt{6y}\]

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