Answer:
- Define your variable and write your equation: 36 + b + (3b-28) = 180°.
- measure of the second angle: 43°.
- measure of the third angle: 101°
Step-by-step explanation:
a = 36
c = 3b - 28
a + b + c = 180°
a = measure of the first angle
b = measure of the second angle
c = measure of the third angle
then:
1) 36 + b + (3b-28) = 180 ⇒ )Define your variable. Write your equation and solve)
b + 3b + 36 - 28 = 180
4b + 8 = 180
4b = 180 - 8
4b = 172
b = 172/4
2) b = 43° ⇒ measure of the second angle
c = 3b - 28
c = 3*43 - 28
c = 129 - 28
3) c = 101° ⇒ measure of the third angle
Check:
36° + 43° + 101° = 180°
Answer:
Yes, a minimum phase continuous time system is also minimum phase when converted into discrete time system using bilinear transformation.
Step-by-step explanation:
Bilinear Transform:
In digital signal processing, the bilinear transform is used to convert continuous time system into discrete time system representation.
Minimum-Phase:
We know that a system is considered to be minimum phase if the zeros are situated in the left half of the s-plane in continuous time system. In the same way, a system is minimum phase when its zeros are inside the unit circle of z-plane in discrete time system.
The bilinear transform is used to map the left half of the s-plane to the interior of the unit circle in the z-plane preserving the stability and minimum phase property of the system. Therefore, a minimum phase continuous time system is also minimum phase when converted into discrete time system using bilinear transformation.
Xy = 50
x + y = 15
x = 15 - y
xy = 50
(5 - y)y = 50
5y - y^2 = 50
y^2 - 5y + 50 = 0
(y - 5)(y - 10) = 0
y - 5 = 0 or y - 10 = 0
y = 5 or y = 10
x + y = 15
For y = 5:
x + 5 = 15
x = 10
For y = 10
x + 10 = 15
x = 5
Answer: the numbers are 5 and 10.
Answer:
g(-4) = -1
g(-1) = -1
g(1) = 3
Explanation:
If you are given a function that is defined by a system of equations associated with certain intervals of x, just find which interval makes x true, and then substitute x into the equation of that interval.
For example, given g(-4), this is an expression which is asking for the value of the equation when x = -4. So -4 is not ≥ 2, so ¼x - 1 will not be used. -4 is also not ≤ -1 and ≤ 2, so -(x - 1)² + 3 will not be used either. So in turn, we will just use -1 which is always -1 so g(-4) will just be -1, right because there is no x variable in -1 so it will always be the same.
Using the same idea as before g(-1) is g(x) when x = -1 so -1 will not be a solution because -1 is not less than -1 (< -1). -1 is not ≥ 2 either so we will be using the second equation because -1 is part of the interval -1≤x≤2 (it is a solution to this inequality), therefore -(x - 1)² + 3 will be used.
As x = -1, -(x - 1)² + 3 = -(-1 - 1)² + 3 = -(-2)² + 3 = -4 + 3 = -1.
It is a coincidence that g(-1) = -1.
Now for g(1), where g(x) has an input of 1 or the value of the function where x = 1, we will not use the first equation because x = 1 → x < -1 → 1 < -1 [this is false because 1 is never less than -1], so we will not use -1.
We will use -(x - 1)² + 3 again because 1 is not ≥ 2, 1≥2 [this is also false]. And -1 ≤ 1 < 2 [This is a true statement]. Therefore g(1) = -(1 - 1)² + 3 = -(0)² + 3 = 3
