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2 years ago

Which value is a solution to the inequality x < 4

Mathematics

1 answer:

denpristay [2]2 years ago

5 0

Answer:

3, 2, 1, 0, literally anything below 4

Step-by-step explanation:

The "mouth" < or > eats the bigger number, if 4 is bigger than x, x must be smaller than 4

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What is the following sum? Assume x greater-than-or-equal-to 0 and y greater-than-or-equal-to 0

Free_Kalibri [48]

Answer:

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

Hence, ption B is true.

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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1 year ago

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Pepsi [2]

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