Answer:
#include<iostream>//library inclusion
using namespace std;
int main()
{
int userInput;
do//start of do while loop
{
cout << "Enter a number less than a 100" << endl;
cin >> userInput;
if (userInput < 100) //condition
{
cout << "YOu entered less than a hundred: " << userInput << endl;
}
else
{
cout << "your number is greater than 100" << endl;
}
} while (userInput > 100);//condition for do while
return 0;//termination of int main
}
Explanation:
The program has been commented for you. The do-while loop enters the first loop regardless of the condition. Then after the first iteration, it checks for the condition. If the condition is being met, it will iterate through, again. Otherwise it will break out of the loop and land on the "return 0;" line. Which also happens to be the termination of the program in this case. The if-else condition is used for the user to see when prompted.
Answer:
The answer would be A: Random
Explanation:
The random() function returns a generated random number (a pseudorandom number)
A cloud-first strategy helps this client by providing a data-centric approach that offers more speed, agility, and security. Thus, the correct option is B.
<h3>What is a Cloud-first strategy?</h3>
A cloud-first strategy may be defined as an operational strategy that allows the users to reposition all or most of their infrastructure to cloud-computing platforms.
A cloud-first strategy helps this client in numerous ways. It provides a data-centric approach that offers more speed, agility, and security to its customers or clients in a facilitated manner.
Therefore, the correct option for this question is B.
To learn more about the Cloud-first strategy, refer to the link:
brainly.com/question/24719487
#SPJ1
(A) the same, by applying styles the same formats are being applied each time.
The recursive function would work like this: the n-th odd number is 2n-1. With each iteration, we return the sum of 2n-1 and the sum of the first n-1 odd numbers. The break case is when we have the sum of the first odd number, which is 1, and we return 1.
int recursiveOddSum(int n) {
if(2n-1==1) return 1;
return (2n-1) + recursiveOddSum(n-1);
}
To prove the correctness of this algorithm by induction, we start from the base case as usual:
by definition of the break case, and 1 is indeed the sum of the first odd number (it is a degenerate sum of only one term).
Now we can assume that 
returns indeed the sum of the first n-1 odd numbers, and we have to proof that 
returns the sum of the first n odd numbers. By the recursive logic, we have
and by induction, is the sum of the first n-1 odd numbers, and 2n-1 is the n-th odd number. So, is the sum of the first n odd numbers, as required:
