roots and radicals roots and radicals....

3sqrt0.008y^3 x^6

So we have

3√(.008* y³ x⁶) =  3(.2)y lx³l √(.2 y) = (.6 lx³l y)√(.2 y)

(Note, Nataszaa, that .2³  = .008).......hey, I even included the absolute value bars this time...did you notice???

Haha! Thanks!

I hope you have it right! (you always do but these roots have recently been wrong for some reason???)

my instructor says that its incomplete CPHill?

The only other thing I see, Nataszaa, is that the ".2" inside is  "2/10'  = "1/5' and we would have a radical in the denominator..........

If so...let's see what we can do

(.6 lx³l y)√(.2 y)  =  (.6 lx³l y)√[(2 y)/ 10] = (.6 lx³l y)√(y/5)

We can multiply the numerator and denominator inside the square root by 5 and get

(.6 lx³l y)√[(5 y)/25] = [(.6 lx³l y)/5]√(5 y) = (.12 lx³l y)√(5 y)

See if this is correct!!!

thank you!

Was that the issue???

"after you take the cubed root, you remove the radical. dont forget the x's. the cubed root of x^6 is x^3"