Please help in really stuck

Mathematics

1 answer:

sattari [20]2 years ago

6 0

$\int\limits {\pi y^2} \, dy$

y = 2 + x^6

y - 2 = x^6

x = (y - 2)^1/6

$\pi \int\limits^3_2 {(y - 2)^{1/6}} \, dx$

Well, solving this integral

$\pi \int\limits^3_2 {(y - 2)^{1/6}} \, dx$ = 2,69279

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X + 23 = 40 what does "x" equal??? PLz explain

vladimir1956 [14]

Hey there!

40 - 23 = 17

x = 17

Hope this helps
Have a great day (:

6 0

3 years ago

Read 2 more answers

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