Respuesta :

semsee45

Answer:

[tex]10 \sqrt{5} \ x^2y^3\sqrt{y}[/tex]

Step-by-step explanation:

simplify [tex]2\sqrt{125x^4y^7}[/tex]

Apply radical rule [tex]\sqrt{ab} =\sqrt{a} \sqrt{b}[/tex]:

[tex]\implies 2\sqrt{125}\sqrt{x^4}\sqrt{y^7}[/tex]

[tex]\implies 2 \cdot 5 \sqrt{5}\sqrt{x^4}\sqrt{y^7}[/tex]

[tex]\implies 10 \sqrt{5}\sqrt{x^4}\sqrt{y^7}[/tex]

Apply radical rule [tex]\sqrt[n]{a^m} =a^{\frac{m}{n}}[/tex]:

[tex]\implies 10 \sqrt{5} \ x^{\frac42}y^{\frac72}[/tex]

[tex]\implies 10 \sqrt{5} \ x^2y^3\sqrt{y}[/tex]

animalactivist

Answer:

[tex]7x^{2}\sqrt{5y^{7} }[/tex]

Step-by-step explanation:

First, we'll start by simplifying [tex]\sqrt{125}[/tex].  2 factors of 125 are 5 and 25, and since 25 is a perfect square, that can be taken out of the radicand and added to the 2 that's already outside of it, creating [tex]7\sqrt{5x^{4}y^{7}}[/tex].

x to the power of 4 can be simplified to x squared and taken out of the radicand along with that 7, so:

[tex]7x^{2}\sqrt{5y^{7} }[/tex]

y to the power of 7 is a bit trickier to solve, so I left it like this, because I think it's the simplest version, but if this isn't an option then simplify further by working on the y.  

I hope this helps :)  

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