Answer:
Codes for each of the problems are explained below
Explanation:
PROBLEM 1 IN C++:
#include<iostream>
using namespace std;
//fib function that calculate nth integer of the fibonacci sequence.
void fib(int n){
// l and r inital fibonacci values for n=1 and n=2;
int l=1,r=1,c;
//if n==1 or n==2 then print 1.
if(n==1 || n==2){
cout << 1;
return;
}
//for loop runs n-2 times and calculates nth integer of fibonacci sequence.
for(int i=0;i<n-2;i++){
c=l+r;
l=r;
r=c;
cout << "(" << i << "," << c << ") ";
}
//prints nth integer of the fibonacci sequence stored in c.
cout << "\n" << c;
}
int main(){
int n; //declared variable n
cin >> n; //inputs n to find nth integer of the fibonacci sequence.
fib(n);//calls function fib to calculate and print fibonacci number.
}
PROBLEM 2 IN PYTHON:
def fib(n):
print("fib({})".format(n), end=' ')
if n <= 1:
return n
else:
return fib(n - 1) + fib(n - 2)
if __name__ == '__main__':
n = int(input())
result = fib(n)
print()
print(result)
Answer:
2.44 mV
Explanation:
This question has to be one of analog quantization size questions and as such, we use the formula
Q = (V₂ - V₁) / 2^n
Where
n = 12
V₂ = higher voltage, 5 V
V₁ = lower voltage, -5 V
Q = is the change in voltage were looking for
On applying the formula and substitutiting the values we have
Q = (5 - -5) / 2^12
Q = 10 / 4096
Q = 0.00244 V, or we say, 2.44 mV
...simplify devices, reducing weight and the chance of failure.
