3 years ago

Please help me so my mom doesn't sell my kitten.

Mathematics

2 answers:

Kryger [21]3 years ago

6 0

Answer:

false I think

Step-by-step explanation:

because a term in an algebraic expression that does not contain any variables and therefore is constant.

nevsk [136]3 years ago

5 0

Answer:

I think it is true but I'm not sure about it

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To check you multiply:

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2 years ago

Which of these limits evaluate to 0?

Vikentia [17]

<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

$\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\$

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Part II

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\$

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Part III

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