Answer:
the answer is going to be district b
Answer:
Zevzda
Explanation:
Check for yourself if you think i'm wrong
Answer:
#include <iostream>
using namespace std;
int main(){
int a, b, c;
cout<<"Enter three integers: ";
cin>>a>>b>>c;
if(a<=b && a<=c){
cout<<"Smallest: "<<a; }
else if(b<=a && b<=c){
cout<<"Smallest: "<<b; }
else{
cout<<"Smallest: "<<c; }
return 0;
}
Explanation:
This line declares three integer variables a, b and c
int a, b, c;
This line prompts the user for three integer inputs
cout<<"Enter three integers: ";
This line gets the inputs
cin>>a>>b>>c;
This checks if the first is the smallest
if(a<=b && a<=c){
If yes, it prints the first as the smallest
cout<<"Smallest: "<<a; }
This checks if the second is the smallest
else if(b<=a && b<=c){
If yes, it prints the second as the smallest
cout<<"Smallest: "<<b; }
If the above conditions are not true, then the third number is printed as the smallest
<em> else{</em>
<em> cout<<"Smallest: "<<c;</em>
<em> }</em>
<em />
Hi, you haven't provided the programing language, therefore, we will use python but you can extend it to any programing language by reading the code and the explanation.
Answer:
n1 = int(input("First numeber: "))
n2 = int(input("Second numeber: "))
for i in range(5):
r1 = n1%10
r2 = n2%10
print(r1+r2)
n1 = n1//10
n2 = n2//10
Explanation:
- First, we ask for the user input n1 and n2
- We create a for-loop to calculate the sum of each place-value of two numbers
- We obtain the last number in n1 by using the mod operator (%) an the number ten this way we can always take the last value, we make the same for n2
- Then we print the result of adding the last two numbers (place value)
- Finally, we get rid of the last value and overwrite n1 and n2 to continue with the process
Answer:
4 basic components
Explanation:
Networks are comprised of four basic components: hardware, software, protocols and the connection medium. All data networks are comprised of these elements, and cannot function without them.