What is the cosine equation of the function shown? -6 -10 -11 -12 Enter your answer by filling in the boxes. Enter any phase shift as its smallest multiple from the fundamental period. f (x) = cos
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- 1. Graphs of the position functions of two particles are shown, where t is measured in seconds.When is each particle speeding up? When is it slowing down? Explain.A mass suspended from a spring is pulled down a distance of 2 ft from its rest position, as shown in the figure. The mass is released at time t = 0 and allowed to oscillate. If the mass returns to this position after 2 s, find an equation that describes its motion. y =47. A sinusoidal function below is used to model the motion of a pendulum over time, where y is the distance, in centimetres, from its rest position and t is the time, in seconds. Suppose the period of the motion doubles. If all other factors in the equation remain unchanged, what is the new equation?
- Find the Phase shift at y = cos x of the following functions and show your complete solution. Answer a-d. I already provided the answers I just need the SOLUTION. Thank you? a. –3 cos 6xb. – 2 cos 5/4 xc. 2 cos (x – 2π)d. 3 cos (x + π/6) Answer: a. 0 b. 0 c. 2π d. π/6 Note: 0 ≤ x ≤ 2π2.6 Can you show me the steps to solve this problem? Directions: Find dy/dx by implicit differentiation. 17) √xy = 1 + x2y9.- A strong gust of wind strikes a tall building, causing it to sway back and forth in damped harmonic motion. The frequency of the oscillation is 0.5 cycle per second, and the damping constant is c = 0.3. Find an equation of the form y = ke−ctsin(ωt) that describes the motion of the building. (Assume that k = 1, and take t = 0 to be the instant when the gust of wind strikes the building. Assume y begins increasing at t = 0.)
- solve the nonhomogenous equation y''' - 6y' = -cos(x)(a) Sketch the direction field for the differential equation y ' = x sq . + y sq . -1 (b) Use part (a) to sketch the solution curve that passes through the origin.Solve this calclus Find the complementary function yc, the particular solution yp and the general solution of the nonhomogeneous equation y'' + y'−6y = e-t+ tYou may consider using the formula
- Find the order one Differential equation of (x-2)y'=y+2(x-2)^3solution of the equation zxx + zyy =e-xcos(y) which tends to zero as x-> ∞ and has the value cos(y) when x = 0.Can you explain how to find the period and the phase shift with there being two points? (-9,0) and (1,0) That part is throwing me off guard. The equation I made was f(x)= 2sin(pi/5(x-0)+2) I did the point-slope formula between the two points to get 0 for the phase shift but I'm sure if that's what I'm supposed to do