2 years ago

(If this question is easy, I am just horrible with percentages, lol :p)

Mathematics

1 answer:

Sati [7]2 years ago

5 0

55% of 20 is 11. the answer is 20

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$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}$

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Step-by-step explanation:

Given the expression

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}$

solving the expression

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}$

$\sqrt{x^2y^3}=xy\sqrt{y}$

$2\sqrt{x^3y^4}=2xy^2\sqrt{x}$

so the expression becomes

$\:\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=xy\sqrt{y}+2xy^2\sqrt{x}+xy\sqrt{y}$

Group like terms

$=2xy^2\sqrt{x}+xy\sqrt{y}+xy\sqrt{y}$

Add similar elements

$=2xy^2\sqrt{x}+2xy\sqrt{y}$

Therefore, we conclude that:

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}$

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