What is the area of this figure?

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

2 years ago

Read 2 more answers

Solve for x and y. 4x-3y=16 -2x+4y=-2 DUE 2/7/17!

Trava [24]

$\left \{ {{4x-3y=16\:(I)} \atop {-2x+4y=-2\:(II)}} \right.$

Multiply the 2nd equation by (-4). This will eliminate the x's when you add the two new equations together.
 $\left \{ {{4x-3y=16\:\:\:\:\:\:\:\:\:\:} \atop {-2x+4y=-2\:*(-4)}} \right.$
$\left \{ {{\diagup\!\!\!\!4x-3y=16} \atop {-\diagup\!\!\!\!4x+8y=-4}} \right.$
--------------------
$5y = 12$
$\boxed{\boxed{y = \frac{12}{5}}}\end{array}}\qquad\quad\checkmark$

Let's replace the value found in the first equation:
$4x-3y=16\:(I)$
$4x-3*( \frac{12}{5}) =16$
$4x - \frac{36}{5} = 16$
least common multiple (5)
$\frac{20x}{\diagup\!\!\!\!5} - \frac{36}{\diagup\!\!\!\!5} = \frac{80}{\diagup\!\!\!\!5}$
$20x - 36 = 80$
$20x = 80 + 36$
$20x = 116$
$x = \frac{116}{20} \frac{\div4}{\div4} \to\: \boxed{\boxed{x = \frac{29}{5} }}\end{array}}\qquad\quad\checkmark$

4 0

3 years ago

Can you please help me....Thank you!

PIT_PIT [208]

Answer:

#1

mean=7

median=8

range=9

mode=3

#2

mean=15

mode=17

range=14

median=14

#3

mean=11

mode=none

range=10

median=4.5

#4

mean=6

mode=4

range=13

median=4

Step-by-step explanation:

there you go, I think that's right but it's been awhile since I've had to do mode mean median and range

4 0

2 years ago

mike bought a 24 pack of mountain dew for $8.40. write and solve an equation to calculate how much each indivdual can soda cost

vekshin1

Answer:

$0.31

Step-by-step explanation:

You need to take the amount of money each pack costs and divide it by the number of cans to find the unit rate. Hope this helps!!

6 0

2 years ago

Question 2 Unsaved

m_a_m_a [10]

The answer is Fourth degree binomial.
The exponent is 4 (fourth degree) and it has two terms (binomial).

5 0

3 years ago