Zeroes only appear in the division

Question

The result of division is only zero, which is wrong?

#include <stdio.h>
void main()
{   
    double Nota1, Nota2, MediaPond;
    printf("Escreva a nota 1: ");
    scanf("%f", &Nota1);
    printf("\nEscreva a nota 2: ");
    scanf( "%f", &Nota2 );
    MediaPond = ((Nota1 * 2) + (Nota2 * 3)) / 5;
    printf("\nA media ponderada e: ");
    printf("%.2f", &MediaPond);
    system("pause");
}

This is the return of the program

Write the note 1: 2

Write note 2: 5

The weighted average is: 0.00

Answer 1

There are other errors, but not what you are describing. You need to print in the format of lf for double and it makes no sense to have the address of a variable printed with &, nor need to create this variable, using the normal patterns of how to code:

#include <stdio.h>

int main() {   
    double nota1, nota2;
    printf("Escreva a nota 1: ");
    scanf("%lf", &nota1);
    printf("\nEscreva a nota 2: ");
    scanf("%lf", &nota2);
    printf("\nA media ponderada e: ");
    printf("%.2lf", (nota1 * 2 + nota2 * 3) / 5);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

Answer 2

The main error is in reading the variables declared as double. When the function was used scanfyou have stored these variables using the %f,the correct in this case would be using %lf. Example :

#include <stdio.h>
#include <stdlib.h>

int main() {

double n1,n2;

printf("Nota 1: ");
scanf("%lf",&n1);

printf("Nota 2: ");
scanf("%lf",&n2);

printf("Media : %.2lf\n",(n1 + n2) / 2);

system("pause");
return 0;

}