Question

Help pleaseeee! ASAP!

Answer 1

Answer : The correct option is, (C) $\sqrt{3y}(3+2y^2)$

Step-by-step explanation :

The given expression is:

$3\sqrt{3y}-\sqrt{27y^5}+y^2\sqrt{75y}$

First we have to break into factors.

$3\sqrt{3y}-\sqrt{3^3y^5}+y^2\sqrt{5^2\times 3y}$

Now we have to make the power in even numbers.

$3\sqrt{3y}-\sqrt{3^2y^4(3y)}+y^2\sqrt{5^2\times 3y}$

$3\sqrt{3y}-3y^2\sqrt{3y}+5y^2\sqrt{3y}$

Now we have to take common $\sqrt{3y}$ from the given expression, we get:

$\sqrt{3y}(3-3y^2+5y^2)$

Now we have to add like terms, we get:

$\sqrt{3y}(3+2y^2)$

Therefore, the correct option is, (C) $\sqrt{3y}(3+2y^2)$

Answer 2

Answer:

C

Step-by-step explanation:

First we simplify what we can in the expressions.  3sqrt(3y) can't be so we leave that.  We want to try and get everythig to have sqrt(3y) in it though.

sqrt(27y^5) = sqrt((3^3)y^2y^3) = sqrt(3^2y^4*3y) = 3y^2*sqrt(3y)

y^2*sqrt(75y) = y^2*sqrt(5^2*3y) = 5y^2*sqrt(3y)

So now we have 3sqrt(3y) - 3y^2*sqrt(3y) + 5y^2*sqrt(3y)

If we factor out sqrt(3y) we get (3-3y^2+5y^2)sqrt(3y) = (3+2y^2)sqrt(3y) So that's C

Let me know if you don't get how I did something though and I'll be happy to explain it.