What are the equation on these

1. Determine the amplitude of y = -16 cos (x)
2. Determine the amplitude of = -sin (x)
3. Determine the domain of = -5 cos (x)
4. Determine the range of y = -10 sin (x)
5. Determine the period of y = -5 cos (2ðx)
6. Determine the phase shift of y = -5 cos (3x -2)
7. Solve the follow equation for the indicated value of x
y= -sin (x); for x = ð
8. Solve the following epuation for the indicated value of x
y= - sin (x); for x = - 6ð
9. What is the equation of the graph in your handout?
10. What is the equation of the graph in your handout?

1. To determine the amplitude of y = -16 cos(x), we need to look at the coefficient in front of the cosine function. In this case, the coefficient is -16, which represents the amplitude. Therefore, the amplitude is 16.

2. To determine the amplitude of y = -sin(x), we need to look at the coefficient in front of the sine function. In this case, there is no coefficient, so we assume it to be 1. Therefore, the amplitude is 1.

3. To determine the domain of y = -5 cos(x), we need to consider the possible values of x. Since the cosine function is defined for all real numbers, there are no restrictions on the domain. Therefore, the domain is (-∞, ∞).

4. To determine the range of y = -10 sin(x), we need to consider the possible values of y. The sine function has a range between -1 and 1, inclusive. So, when we multiply by -10 in front of the sine function, we get the range of y to be between -10 and 10, inclusive. Therefore, the range is [-10, 10].

5. To determine the period of y = -5 cos(2πx), we need to look at the coefficient in front of x, which is 2π. The period of the cosine function is given by 2π divided by the coefficient. Therefore, the period is 2π / (2π) = 1.

6. To determine the phase shift of y = -5 cos(3x - 2), we need to compare the input to the cosine function (3x - 2) to the standard expression of the cosine function, which is just x. In this case, the phase shift is given by setting 3x - 2 equal to 0 and solving for x. Therefore, the phase shift is x = 2/3.

7. To solve y = -sin(x) for x = π, we substitute π into the equation. Therefore, y = -sin(π) = 0.

8. To solve y = -sin(x) for x = -6π, we substitute -6π into the equation. Therefore, y = -sin(-6π) = 0.

9. The equation of the graph in the handout is not provided in the question.

10. The equation of the graph in the handout is not provided in the question.

1. To determine the amplitude of the function y = -16 cos (x), you can use the general form of a cosine function, which is y = A cos (Bx + C), where A is the amplitude. In this case, the amplitude is 16 because the coefficient in front of the cosine function is -16.

2. To determine the amplitude of the function y = -sin (x), you can use the general form of a sine function, which is y = A sin (Bx + C). In this case, the amplitude is 1 because there is no coefficient in front of the sine function. By default, the amplitude of a sine function is 1.

3. To determine the domain of the function y = -5 cos (x), you need to identify all the possible values that x can take. In trigonometric functions, the domain is usually all real numbers. Therefore, the domain of this function is (-∞, ∞).

4. To determine the range of the function y = -10 sin (x), you need to identify all the possible values that y can take. In a sine function, the range is typically between -1 and 1. However, since the coefficient in front of the sine function is -10, the range of this function is (-10, 10).

5. To determine the period of the function y = -5 cos (2πx), you can use the formula for the period of a cosine function, which is T = 2π/B. In this case, B is 2π, so the period is 2π/(2π) = 1.

6. To determine the phase shift of the function y = -5 cos (3x - 2), you need to set the argument of the cosine function (3x - 2) equal to zero and solve for x. In this case, the phase shift is 2/3 units to the right.

7. To solve the equation y = -sin (x) for x = π, you substitute π into the equation and solve for y. In this case, y = -sin (π) = 0.

8. To solve the equation y = -sin (x) for x = -6π, you substitute -6π into the equation and solve for y. In this case, y = -sin (-6π) = 0.

9. Since you mentioned a graph in your handout without providing any specific equation, I am unable to provide the equation for that particular graph. However, if you provide me with more information regarding the graph, such as its characteristics or any points on the graph, I can help derive the equation.

10. Similarly, without any specific information or context about the graph in your handout, I am unable to provide the equation for that graph. If you can provide more details about the graph or any specific points or features, I can help determine the equation.