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Verified Answer :

kianna0028

We have: 2mn^2 √5mn^2, where √ is a cubed root.

Hence, we can do (2mn^2)^3 to get 8m^3n^6. Now we'll take that into the root:

√5mn^2*8m^3n^6 = √120m^4n^8

So, √120m^4n^8 simplifies into 2mn^2 √5mn^2.

Cricetus

Answer:

\(2\sqrt[3]{5}m^{2}n^{4}\)

Step-by-step explanation:

\(2mn^2\times\sqrt[3]{5}mn^2\)

Let us bring the term of same type together we get

\(2\times\sqrt[3]{5}\times mn^2 \times mn^2\)

\(2\sqrt[3]{5}\times mn^2\times mn^2\)

\(2\sqrt[3]{5}\times m\times m\times n^2\times n^2\)

Now here we apply the rule of exponents , which is like this

m^p\times m^q=m^{p+q}

Hence

\(2\sqrt[3]{5}\times m^{1+1}\times n^{2+2}\)

\(2\sqrt[3]{5}\times m^{2}\times n^{4}\)

\(2\sqrt[3]{5}m^{2}n^{4}\)

Hence this is our answer.

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