Answer:
Using Matlab code for Fourier series to calculate for the function, see the attached
Step-by-step explanation:
Go through the picture step by step.
Problem 1
x = measure of angle N
2x = measure of angle M, twice as large as N
3(2x) = 6x = measure of angle O, three times as large as M
The three angles add to 180 which is true of any triangle.
M+N+O = 180
x+2x+6x = 180
9x = 180
x = 180/9
x = 20 is the measure of angle N
Use this x value to find that 2x = 2*20 = 40 and 6x = 6*20 = 120 to represent the measures of angles M and O in that order.
<h3>Answers:</h3>
- Angle M = 40 degrees
- Angle N = 20 degrees
- Angle O = 120 degrees
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Problem 2
n = number of sides
S = sum of the interior angles of a polygon with n sides
S = 180(n-2)
2700 = 180(n-2)
n-2 = 2700/180
n-2 = 15
n = 15+2
n = 17
<h3>Answer: 17 sides</h3>
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Problem 3
x = smaller acute angle
3x = larger acute angle, three times as large
For any right triangle, the two acute angles always add to 90.
x+3x = 90
4x = 90
x = 90/4
x = 22.5
This leads to 3x = 3*22.5 = 67.5
<h3>Answers:</h3>
- Smaller acute angle = 22.5 degrees
- Larger acute angle = 67.5 degrees
f(3) is equal to -101.
<u>Step-by-step explanation</u>:
- The sequence is followed by f(0)=4
- The given expression is f(n+1)= -3f(n)+1
Put n=0,
f(0+1)= -3f(0)+1
f(1)= -3(4)+1
f(1)= -12+1 = -11
Put n=1,
f(1+1)= -3f(1)+1
f(2)= -3(-11)+1
f(2)= 33+1 = 34
Put n=2,
f(2+1)= -3f(2)+1
f(3)= -3(34)+1
f(3)= -102+1
f(3)= -101
