Answer:
#include <iostream>
using namespace std;
int main() {
int i, lo=12, hi=45, result=0; // initializing the required variables..
for(i=lo;i<=hi;i++) // implementing a for loop.
{
result+=i; // accumulating the sum of integers..
}
cout<<result<<endl; // displaying the resulting total.
return 0;
}
Output:-
969
Explanation:
This program is written in C++. Initially, the variables i, lo, hi, and result are declared, with only lo, hi, and result initialized. A for loop is employed to loop through all values from lo to hi, adding each integer to the result.
Answer:
num1 = int(input("Numerator: "))
num2 = int(input("Denominator: "))
if num1 < 1 or num2<1:
print("Input must be greater than 1")
else:
print("Quotient: "+str(num1//num2))
print("Remainder: "+str(num1%num2))
Explanation
The next two lines prompt the user for two numbers
num1 = int(input("Numerator: "))
num2 = int(input("Denominator: "))
The next if statement checks whether either or both inputs are not positive
if num1 < 1 or num2<1:
print("Input must be greater than 1")-> If true, this print statement will run
If the conditions are not met, the program prints the quotient and remainder
else:
print("Quotient: "+str(num1//num2))
print("Remainder: "+str(num1%num2))
A method used to guide traffic to various destinations based on considerations like location, congestion, or link health is termed Anycast. It ranks among the top five techniques for Internet traffic routing, alongside unicast, broadcast, multicast, and geocast.
Answer:
They forecasted yet another cold day in Seattle combined with another day of wind.
Answer:
Vendor B
Explanation:
By using letters to signify each vendor and the symbol < to indicate which is cheaper, the problem can be summarized as follows: A<C, B<C, B<D, D<A.
These inequalities show that the vendors on the left are less expensive than those on the right, therefore, any vendor appearing on the right in these expressions cannot be the cheapest, ruling out C, D, and A. Thus, the least expensive vendor is B.
