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3 years ago

This describes an airplane's flight:

Mathematics

1 answer:

Maksim231197 [3]3 years ago

4 0

Answer:

The third box, the fourth box, and the second box

Step-by-step explanation:

The first segment that describes Part A is the one that is increasing because based on the information, the airplane climbs, meaning it's going up.

The second segment that describes Part B is the one that is a straight horizontal line, and this is because the airplane is flying at a constant rate, so it will be straight.

The third segment that describes Part C is the one that is decreasing (going down) slowly. This is because the airplane is down descending, which is also means going down.

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Answer:

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Step-by-step explanation:

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2 years ago

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Step-by-step explanation:

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2 years ago

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jek_recluse [69]

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3 years ago

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mojhsa [17]

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3 years ago

Can you please answer the question?

Roman55 [17]

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that one side of the equality is equal to the expression of the other side, requiring the use of algebraic and trigonometric properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$