Answer:
a) n<1 and n>5
b) 0 < n < -4
c) n > 2 and n < -2
Step-by-step explanation:
The signal is given by x[n] = 0 for n < -1 and n > 3
The problem asks us to determine the values of n for which it's guaranteed to be zero.
a) x[n-2]
We know that n -2 must be less than -1 or greater than 3.
Therefore we're going to write down our inequalities and solve for n

Therefore for n<1 and n>5 x [n-2] will be zero
b) x [n+ 3]
Similarly, n + 3 must be less than -1 or greater than 3

Therefore for n< -4 and n>0, in other words, for 0 < n < -4 x[n-2] will be zero
c)x [-n + 1]
Similarly, -n+1 must be less than -1 or greater than 3

Therefore, for n > 2 and n < -2 x[-n+1] will be zero
This sine function has the form y = f(x) = a sin (bx + c), where |a| is the amplitude, b is the frequency, and c determines the phase shift.
The period of this sine curve is 2 pi/b, or (in this problem) 2 pi/(1/2).
The phase shift of this curve is -c/b, or -(pi/4)/(1/2). Simplify this.
The amplitude is |-3|, or just 3.
Draw a set of coordinate axes. Draw light horizontal lines through y=3 and y=-3. These are the max and min. values of this sine function.
Plot the phase shift on your x-axis.
Your sine function will begin at (phase shift value, 0) and continue for one period. Add the period to the phase shift value. Draw a sine curve beginning and ending at these two x-values.
I encourage you to share your calculations and to sketch the graph described here. Then I could give you more specific feedback on your work. Good luck!
Step-by-step explanation:
here is the answer attached
Answer:
For first question answer is D.0.8
For second question answer is D.ΔMGR is similar to ΔNST.
Step-by-step explanation:
Here triangle FUN is dilated to get triangle PET
So, let's take F and P.
F(-2,4) and P(-1.6, 3.2)
Let's find the scale factor by dividing the coordinates.
Scale factor =
Using y coordinates of the same points
Scale factor =
So, scale factor is 0.8
Now, for the second problem
<G= <S
<R= <T
<M=<N
So, the corresponding angles are congruent when two triangles similar.
So, we can say ΔMGR is similar to ΔNST.