Question

Help please

Answer 1

@xkylex

Answer 2

oh, this one. just a sec...

Answer 3

Btw, are you in honors algebra 2 connects academy like me?

Answer 4

Why yes I am

Answer 5

Do you mind a quote?
If so

"sqrt(250x^16) / sqrt(2x)
You have to find the square root of each. They do not cancel eachother out. So, you really do have to find the sqrt of 250, the sqrt of x^16, and the sqrt of 2x
A trick to remember here is when finding a root of something that is "to the power of," you subtract the root value (2) from the powers.
I suggest the factoring method 250=5*5*5*2
So the top is sqrt[ (5^3)(2)(x^16) ], which is (5^(3-2)) (sqrt2)(x^(16-2))
This is simplified to (5)(sqrt2)(x^14).
Your problem is now [5sqrt(2)x^14]/sqrt(2x)"

Answer 6

Ok, gimme a sec to comprehend all this lol.

Answer 7

I gtg soon, il be back in about 2 hours at most. Please hold out until then

Answer 8

hmm, okay, thanks

Answer 9

If you're still here, I would like to know what to do next.

Answer 10

hint: \[\Large \sqrt{x^{16}}=\sqrt{x^2x^2\cdots x^2}=\sqrt{x^2}\sqrt{x^2}\cdots \sqrt{x^2}\]

Answer 11

Ok I figured the problem out, thanks!:)