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DENIUS [597]
3 years ago
9

Help pleaseeee! ASAP!

Mathematics
2 answers:
yaroslaw [1]3 years ago
5 0

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)

Arturiano [62]3 years ago
3 0

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.  

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