Can I please get some help

Mathematics

1 answer:

Nitella [24]4 years ago

5 0

Answer:

$f(3) = 3\\g(2) = -3\\g(\frac{1}{2}) = 1\\f(2) = \frac{2}{3}\\g(-2) = 0\\f(\pi) = 2\\$

Step-by-step explanation:

The functions are given f and g using coordinates.

Whenever we will ask for f(a), we look for "a" in the x coordinate of the function f and find the corresponding value. THAT IS THE ANSWER.

If we ask for g(b), we look for "b" in the x coordinate of the function g and find the corresponding value. THAT IS THE ANSWER.

So,

$f(3) = 3\\g(2) = -3\\g(\frac{1}{2}) = 1\\f(2) = \frac{2}{3}\\g(-2) = 0\\f(\pi) = 2\\$

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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 $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .