Please help me with these problems. Work and answer is appreciated! :) thanks.
1. Find on the interval theta is greater than or equal to 0 and less than 2pi; then find the general solution:
5tantheta+1=-2.89
16cos^2theta-6=3
6sin^2theta+sintheta=2
costheta=.84
2tanx=7
2.Find all solutions to 5cos(2pi x)=3 on the interval of x is greater or equal to 0 and less than or equal to 3.
3.Find # of solutions to cos(4x)=.724 on the interval of theta is greater or equal to 0 and less than 2pi
4.The angle of elevation from an airport to plane is 5 degrees. The altitude of the plane is 25000 ft. What is the distance from the plane to the airport?
5.A satellite launched into space reaches its farthest pt north of the equator 30 min after the launch. It reaches its farthest south of equator, distance of approx 2000 km, 90 min after that. If it takes the satellite 3 hours to orbit the earth, find an equation to model the satellites distance from the equator. (t=time)
PLEASE HELP URGENT
It is not the function of this webpage to simply do the homework or assignments for students.
It sure looks like that is what you want.
These look like routine questions for a standard Calculus course.
I will do #2 the way I would do it, follow this procedure to do the rest.
5cos(2pi x)=3
cos(2pi x)=3/5 = .6
so the angle (2pi x) is either in quadrants I or IV according to the CAST rule
set your calculator to radians
then (2pi x) = .9273 OR (2pi x) = 2pi-.9273 = 5.358
then x = .14758 or x = .8524
Now the period of your original function is 2pi/2pi = 1
so you can get addional answers by adding the period of 1 to any of your answers.
So in the interval from 0 to 3
x = .14758, 1.14758, 2.14758, .8524
(I tested all of them, they work)
I don't get percentage.
Of course, I'm here to help you! Let's go through each problem step by step.
1.
a) For the equation 5tanθ + 1 = -2.89, we'll start by isolating the tangent term. Subtracting 1 from both sides gives us 5tanθ = -3.89. Then, we divide by 5 to get tanθ = -3.89/5. To find the values of θ, we can use the inverse tangent function (tan^(-1)) on both sides. Therefore, θ = tan^(-1)(-3.89/5).
b) For the equation 16cos^2θ - 6 = 3, we'll isolate the cosine term. Adding 6 to both sides gives us 16cos^2θ = 9. Dividing by 16 gives us cos^2θ = 9/16. Taking the square root of both sides, we get cosθ = ±√(9/16). Since cosine is positive in the first and fourth quadrants, we have cosθ = ±3/4. To find the values of θ, we can use the inverse cosine function (cos^(-1)) on both sides. Therefore, θ = cos^(-1)(3/4) and θ = cos^(-1)(-3/4).
c) For the equation 6sin^2θ + sinθ = 2, we'll isolate the sine term. Subtracting 2 from both sides, we get 6sin^2θ + sinθ - 2 = 0. This is a quadratic equation in terms of sinθ. We can factor it or use the quadratic formula to solve for sinθ. The solutions to this equation will give us the values of θ.
d) For the equation costheta = 0.84, we can simply use the inverse cosine function (cos^(-1)) to find θ. Therefore, θ = cos^(-1)(0.84).
e) For the equation 2tanx = 7, we'll isolate the tangent term. Dividing both sides by 2 gives us tanx = 7/2. To find the values of x, we can use the inverse tangent function (tan^(-1)). Therefore, x = tan^(-1)(7/2).
2. For the equation 5cos(2πx) = 3, we can start by isolating the cosine term. Dividing by 5 gives us cos(2πx) = 3/5. To find the values of x, we need to take the inverse cosine function (cos^(-1)) on both sides. Therefore, 2πx = cos^(-1)(3/5). Solving for x, we have x = (cos^(-1)(3/5))/(2π). Since the interval for x is from 0 to 3, we need to check if there are any additional solutions within this interval.
3. The equation cos(4x) = 0.724 has infinitely many solutions. To find the number of solutions on the interval 0 ≤ θ < 2π, we need to determine the number of times the graph of y = cos(4x) intersects the line y = 0. The number of intersections is equal to the number of solutions.
4. To find the distance from the plane to the airport, we can use trigonometry. Given that the angle of elevation is 5 degrees and the altitude of the plane is 25000 ft, we can use the tangent function. tan(angle of elevation) = opposite/adjacent. In this case, the opposite side is the altitude of the plane, which is 25000 ft. Let's call the distance from the plane to the airport "x". Therefore, tan(5 degrees) = 25000/x. Now, we can solve for x by rearranging the equation to x = 25000 / tan(5 degrees).
5. To find an equation to model the satellite's distance from the equator, we can use the concept of periodic functions. Given that the satellite takes 3 hours to orbit the Earth and reaches its farthest north point 30 minutes after launch, we can start with the equation d(t) = a + bcos(ct + d). Here, a represents the average distance from the equator, b represents the amplitude or maximum distance from the equator, c represents the period, and d represents the phase shift.
Since the satellite reaches the farthest north point 30 minutes after launch, its phase shift will be -pi/6 (since 90 minutes is equal to pi/2, and 30 minutes is equal to pi/6). The time period is 3 hours, which is equal to 2pi. The amplitude is given as 2000km. Therefore, the equation to model the satellite's distance from the equator is d(t) = a + 2000cos((2pi/3)(t - pi/6)), where t represents time.
I hope this helps! Let me know if you have any further questions or need more clarification.