Program to evaluate the expression y=x^n.c

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

int main()
{
    double a, b, y;
    printf("Enter the value for x and n : ");
    scanf("%lf %lf", &a, &b);
    y = pow (a, b);
    printf ("The value of y in the expression y = x^n is %.2lf", y);
    return 0;
}