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allochka39001 [22]

3 years ago

6

How do i solve and graph inequalities

1 answer:

Darina [25.2K]3 years ago

7 0

I hope this helps you

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where is the graph

Step-by-step explanation:

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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

$\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\$

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3 years ago

Simplify this expression

Gnesinka [82]

It’s A!! hope that helped

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What expression is equivalent to 4 to the negative 3 power

Elina [12.6K]

Answer:

1/64

Step-by-step explanation:

4^(-3)

1/4³

1/64

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