## anonymous 5 years ago SR (600y^23)

1. anonymous

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

2. bahrom7893

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

3. bahrom7893

$10y^{11}\sqrt{68y}$

4. bahrom7893

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

5. bahrom7893

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

6. anonymous

any questions?

7. anonymous

yeah I am still lost as can be...

8. anonymous

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

9. anonymous

essentially it is algebra 2

10. anonymous

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

11. 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 $\sqrt{100}\sqrt{6}$

12. anonymous

being as 100 is a perfect square

13. 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

14. anonymous

Do you guys want to go through the problem

15. anonymous

?

16. anonymous

i am going throught the problem...

17. anonymous

the sqrt will eventually look like this (5x5x2x2x2x3)

18. 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.

19. anonymous

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

20. anonymous

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