Answer:
def print_sum(a,b,c):
print(a+b+c)
one = int(input("Enter the 1st number: "))
two = int(input("Enter the 2nd number: "))
three = int(input("Enter the 3rd number: "))
print_sum(one,two,three)
Explanation:
hope this helped :)
Answer:
4000k-ohm to 10,000k-ohm
Explanation:
As we know that time constant for an RC circuit is t=RC
Putting the values of t we can get the range of varaiable resistor as;
t=RC
Putting t=2 we get the first value of the range for the variable resistor
2=R*0.500*10^-6
R=2/(0.500*10^-6)
R=4*10^6
R=4000k-ohm
Now putting t=5 we get the final value for the range of variable resistor
t=RC
5=R*0.500*10^-6
R=5/(0.500*10^-6)
R=10*10^6
R=10,000k-ohm
So variable resistance must be made to vary in the range from 4000k-ohm to 10,000k-ohm
Answer:
In C++:
#include <bits/stdc++.h>
#include <iostream>
#include <vector>
using namespace std;
int main(){
vector<int> vectItems;
cout << "Vector length: ";
int ln; cin>>ln;
int num;
for (int ikk = 0; ikk < ln; ikk++){
cin >> num;
vectItems.push_back(num);}
int small, secsmall;
small = secsmall = INT_MAX;
for (int ikk = 0; ikk < ln; ikk++){
if(vectItems[ikk] < small){
secsmall = small;
small = vectItems[ikk]; }
else if (vectItems[ikk] < secsmall && vectItems[ikk] != small) {
secsmall = vectItems[ikk];} }
cout<<small<<" "<<secsmall;
return 0;}
Explanation:
See attachment for program file where comments are used for explanation
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