Answer:
1. X=95
2. M=60
3.R=70
4.Q=75
5.A=75
6.B=105
7. ∠ACB = ∠BCA
8. ∠DEG = 75
9.∠FOE = 64 ∠COF = 116
10. ∠COE = 137 ∠DOE= 43
11. ∠WXZ = 149
12. ∠RSZ = 99.5
Step-by-step explanation:
Answer:
The amplitude of the oscillation in this oscillating system, after reaching the steady-state, is 3.984.
Step-by-step explanation:
Assume you know a and b values, which you do not actually need to know to solve the question.
The given general solution of the equation is made up of three terms, namely:
- A constant value (equal to 4),
- a cosine term: 3.984cos(2nt-B)
- a cosine, exponential term: 2.81exp(-8.58t)cos(at + o)
This means that, after a certain time (for large values of time "t"), when the system reaches its <u>steady-state</u>, the following will happen to these parts, respectively:
- The constant value will remain as 4. It shall not cancel, but it shall not provide any oscillation either, and, in turn, no amplitude nor frequency.
- The second term will always be a cosine, since it is a periodic function. Its amplitud is "3.984" and its frequency is "2n" radians/second. Its phase is actually "B".
- This third term is not a periodic functon. It is made up of a periodic function multiplied by an exponential function, whose exponent is negative (for any positive value of time variable "t"). This exponential function approaches to zero when its exponent approaches to minus infinity. So, after a certain time -or, in other words, once the steady-state is eventually reached- the product will be a delimited function (cosine, whose absolute value is always than "1") multiplied by zero. That is, this third term, as a whole, approaches to zero for large, infinite values of time "t".
All in all, once the steady-state is reached, the solution shall remain as:
x = 4+ 3.984cos(2nt-B).
The only oscillation would be that of the cosine term, and its amplitude will be 3.984, an actual value given by the question itself.
Well using the slope-intercept the answer is or should be 1 positive
