Question

SR (600y^23)

Answer 1

\[\sqrt{600y ^{23}}\]

is the actual equation

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

any questions?

Answer 7

yeah I am still lost as can be...

Answer 8

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

Answer 9

essentially it is algebra 2

Answer 10

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

Answer 11

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 \[\sqrt{100}\sqrt{6}\]

Answer 12

being as 100 is a perfect square

Answer 13

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

Answer 14

Do you guys want to go through the problem

Answer 15

?

Answer 16

i am going throught the problem...

Answer 17

the sqrt will eventually look like this (5x5x2x2x2x3)

Answer 18

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.

Answer 19

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

Answer 20

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