All categories

3 years ago

A. [-1, 1] b. [0, pi] c. [-pi/2, pi/2] d. -♾, ♾]

Mathematics

1 answer:

meriva3 years ago

6 0

f(x) = sin(x)

Domain (–∞, ∞)

Range [–1, 1]

y = arcsine(x)

y = f–¹(x)

Domain [–1, 1]

Range [–π/2, π/2]

You might be interested in

Only answer if you know this if i see a link its a report ∞√5±x+5√15x+∞=

fenix001 [56]

Answer:

the exact form is 2/5. the decimal form is 0.4 ig

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A dance instructor chose four of his 10 students to be on stage for a performance. If order does not matter, in how many differe

Vikki [24]

This is a combination in which you choose 4 from 10.
The formula is
combinations = 10! / 4! * (10-4)!
combinations = 10! / 4! * 6!
combinations = 10 * 9 * 8 * 7 * 6! / 4! * 6!
combinations = 10 * 9 * 8 * 7 / 4 * 3 * 2
combinations = 10 * 3 * 7
combinations = 210
Source:
http://www.1728.org/combinat.htm

4 0

3 years ago

Read 2 more answers

A person on a runway sees a plane approaching. The angle of elevation from the runway to the plane is 11.1° . The altitude of th

Gnoma [55]

Answer:

The horizontal distance from the plane to the person on the runway is 20408.16 ft.

Step-by-step explanation:

Consider the figure below,

Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner. The angle of elevation from the runway to the plane is 11.1°

BC is the horizontal distance from the plane to the person on the runway.

We have to find distance BC,

Using trigonometric ratio,

$\tan\theta=\frac{Perpendicular}{base}$

Here, $\theta=11.1^{\circ}$ ,Perpendicular AB = 4000

$\tan\theta=\frac{AB}{BC}$

$\tan 11.1^{\circ} =\frac{4000}{BC}$

Solving for BC, we get,

$BC=\frac{4000}{\tan 11.1^{\circ} }$

$BC=\frac{4000}{0.196}$ (approx)

$BC=20408.16$ (approx)

Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft

8 0

3 years ago

How many one-half cubes with dimensions of 1/2 x1x1 fit in a unit cube? 1 2 4 6

Leni [432]

I got 2 but this could be wrong. Hope this helps ;D

4 0

3 years ago

Read 2 more answers

Find the points on the ellipse 4x2 + y2 = 4 that are farthest away from the point (−1, 0).

Hitman42 [59]

The two points that are most distant from (-1,0) are exactly (1/3, 4sqrt(2)/3) and (1/3, -4sqrt(2)/3) approximately (0.3333333, 1.885618) and (0.3333333, -1.885618) Rewriting to express Y as a function of X, we get 4x^2 + y^2 = 4 y^2 = 4 - 4x^2 y = +/- sqrt(4 - 4x^2) So that indicates that the range of values for X is -1 to 1. Also the range of values for Y is from -2 to 2. Additionally, the ellipse is centered upon the origin and is symmetrical to both the X and Y axis. So let's just look at the positive Y values and upon finding the maximum distance, simply reflect that point across the X axis. So y = sqrt(4-4x^2) distance is sqrt((x + 1)^2 + sqrt(4-4x^2)^2) =sqrt(x^2 + 2x + 1 + 4 - 4x^2) =sqrt(-3x^2 + 2x + 5) And to simplify things, the maximum distance will also have the maximum squared distance, so square the equation, giving -3x^2 + 2x + 5 Now the maximum will happen where the first derivative is equal to 0, so calculate the first derivative. d = -3x^2 + 2x + 5 d' = -6x + 2 And set d' to 0 and solve for x, so 0 = -6x + 2 -2 = -6x 1/3 = x So the furthest point will be where X = 1/3. Calculate those points using (1) above. y = +/- sqrt(4 - 4x^2) y = +/- sqrt(4 - 4(1/3)^2) y = +/- sqrt(4 - 4(1/9)) y = +/- sqrt(4 - 4/9) y = +/- sqrt(3 5/9) y = +/- sqrt(32)/sqrt(9) y = +/- 4sqrt(2)/3 y is approximately +/- 1.885618

7 0

4 years ago