Answer:
3
Step-by-step explanation:
Answer:
B, i might be wrong but i think its B
Answer: 144
Step-by-step explanation:
Answer:
The answer to the question is
The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is (-∞, 4)
Step-by-step explanation:
To apply look for the interval, we divide the ordinary differential equation by (t-4) to
y'' +
y' +
y = 
Using theorem 3.2.1 we have p(t) =
, q(t) =
, g(t) = 
Which are undefined at 4. Therefore the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution, that is where p, q and g are continuous and defined is (-∞, 4) whereby theorem 3.2.1 guarantees unique solution satisfying the initial value problem in this interval.
Answer:
The question is incomplete, so I will describe the sine regression model.
The function
y = 0.884 sin(0.245x - 1.093) + 0.400
correspond to the general equation:
y = a sin(bx - c) + d
where:
a = 0.884
b = 0.245
c = 1.093
d = 0.400
The amplitude of the function is computed as follows:
amplitude = |a| = 0.884
The period of the function is computed as follows:
period = 2π/|b| = 25.6456
The phase shift of the function is computed as follows:
phase shift = c/b = 4.4612 to the right (because there is a minus sign before c in the equation)
The vertical shift of the function is computed as follows:
vertical shift = d = 0.400