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prisoha [69]
3 years ago
10

Which of the following is equivalent??

Mathematics
1 answer:
Katen [24]3 years ago
7 0
(-3 7following is equivalent
__.
4
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(\frac{15}{2},\frac{5}{2})

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How to write 84 divided into sevenths
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84 x (1/7) or simply 84/7
In the end, the answer would be 12. 
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A ball is dropped from 45 feet above the ground. With each bounce, its max height is 80% of its previous height. What is the bal
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Find all solutions of each equation on the interval 0 ≤ x < 2π.
Korvikt [17]

Answer:

x = 0 or x = \pi.

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}.

\displaystyle \sec{x} = \frac{1}{\cos{x}}.

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, \sin^{2}{x} = 1 - \cos^{2}{x}. Therefore, for the square of tangents,

\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}.

This equation will thus become:

\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2.

To simplify the calculations, replace all \cos^{2}{x} with another variable. For example, let u = \cos^{2}{x}. Keep in mind that 0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1.

\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2.

\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2.

Solve this equation for u:

\displaystyle \frac{u^{2} + 1}{u^{2}} = 2.

u^{2} + 1 = 2 u^{2}.

u^{2} = 1.

Given that 0 \le u \le 1, u = 1 is the only possible solution.

\cos^{2}{x} = 1,

x = k \pi, where k\in \mathbb{Z} (i.e., k is an integer.)

Given that 0 \le x < 2\pi,

0 \le k.

k = 0 or k = 1. Accordingly,

x = 0 or x = \pi.

8 0
3 years ago
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