Given the code: int x, y; cout << "Enter two integers: "; cin >> x >> y; Write a few lines of code that will print all numbers that are factors of x and not factors of y. Example Output 1 Enter two integers: 12 30 4 12 Example Output 2 Enter two integers: 15 24 5 15
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- Given that m = 0, what is the output of this code #include <stdio.h> int main (void) { int m; printf ("enter an integer number between 0 and 5\n"); scanf("%d",&m); switch (m) {case 0: printf(" the number is low\n"); break; case 1: printf("the number is ok\n"); case 2: printf("the number is high\n"); } }Driving is expensive. Write a program with a car's miles/gallon and gas dollars/gallon (both doubles) as input, and output the gas cost for 20 miles, 75 miles, and 500 miles. Output each floating-point value with two digits after the decimal point, which can be achieved by executingcout << fixed << setprecision(2); once before all other cout statements. Ex: If the input is: 20.0 3.1599 the output is: 3.16 11.85 79.00Given that x = 0, what is the output of this code #include <stdio.h> int main (void) { int x; printf ("enter an integer number between 0 and 2\n"); scanf("%d",&x); switch (x) {case 0: printf(" the number is low\n"); break; case 1: printf("the number is ok\n"); case 2: printf("the number is high\n"); } }
- int input; cout << "enter input " << endl; cin >> input; int i = input / 2; int d; for (i >= 2; i--;) { d = input % i; if (d == 2 || i == 1) cout << "not prime" << endl; else cout << "prime" << endl; } This code outputs prime numbers as both "prime" and "not prime" but I just want it to say "prime". Please help me with this problem using c++.Suppose j, sum, and num are int variables, and the input is : 25 30 51 3 - 2 What is the output of the following code? int num, j, sum = 0; cout << "Please input five numbers here: "; cin >> num; for (j = 1; j <= 4; j++) { sum = sum + num; cin >> num; } cout << "\nThe sum is: " << sum << endl;1. Write the code to check if y is between 80 and 100 (including 80 and not including 100)? Answer: 2. Given that x = 2, what is the output of this code #include <stdio.h> int main (void) { int x; printf ("enter an integer number between 0 and 2\n"); scanf("%d",&x); switch (x) {case '0': printf(" the number is low\n"); break; case '1': printf("the number is ok\n"); case '2': printf("the number is high\n"); } } Answer: WRITE C REPL
- int a = 4, b;b = a-- + 2;What is the value of b in the above code? a. 7 b. 6 c. Error d. 5Print "user_num1 is negative." if user_num1 is less than 0. End with newline. Assign user_num2 with 1 if user_num2 is greater than 12. Otherwise, print "user_num2 is less than or equal to 12.". End with newline. My code: user_num1 = int(input())user_num2 = int(input()) if user_num1 < 0: print("user_num1 is negative.") if user_num2 > 12: user_num2 == 1 else: print("user_num2 is less than or equal to 12.") print('user_num2 is', user_num2) When the input is 0 and 13 my output for user_num2 is 13 and it should be 1. I don't know what I'm doing wrong.What will be printed out as a result of the following code?int x = 5, y = 7;System.out.println(++x);x += y++;System.out.println(x + "," + y); a. 66,8 b. 613,8 c. 513,8 d. 512,7
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