y=3x^3-5X^2+6 bUT IT WON'T RUN. pLEASE HELP...

Question

#include<stdio.h>

int main()

{

int x;

setbuf(stdout,NULL);

printf("Enter x: \n");

scanf("%d\n",&x);

printf("You entered x: %d", x);

printf("y=%d\n",(3*x*x*x)-(5*x*x)+6);

return 0;

}

Answer 1

#include<stdio.h>
#include<math.h>

int main()

{

int x;
printf("Enter x: \n");

scanf("%d",&x);

printf("You entered x: %d", x);

printf("\ny= %d",(3*x*x*x)-(5*x*x)+6);

return 0;

}

Answer 2

#include<stdio.h>

int main()

{

int x;

setbuf(stdout,NULL);

printf("Enter x: \n");

scanf("%d",&x);

printf("You entered x: %d", x);

printf("y=%d\n",(3*x*x*x)-(5*x*x)+6);

return 0;

}

dont write '\n' in scanf function

Answer 3

the value of the expression is y=3*x^3-5*x^2+6 and setbuf function call is not required

Answer 4

Remove \n in scanf since that is stopping scanf from terminating properly

Answer 5

Bro you just add math.h library

or cmath in it should be true when you use power function or ^

Answer 6

#include<stdio.h>

int main()

{
int x;
printf("Enter x: \n");
scanf("%d",&x);
printf("You entered x: %d", x);
float f=3*(x*x*x)-(5*(x*x))+6;
printf("y=%f\n",f);
return 0;
}

Answer 7

#include<stdio.h>

int main()

{

int x;

printf("Enter x: \n");

scanf("%d",&x);

printf("You entered x: %d\n", x);

printf("y=%d\n",(3*x*x*x)-(5*x*x)+6);

return 0;

}