Without using a division or multiplication operator and without using iteration , define a recursive method named product that accepts two int parameter , m and k, and calculates and returns the integer product of m times k . you can count on m> =0 and k> =0.

I will be using the language C++. Given the problem specification, there are an large variety of solving the problem, ranging from simple addition, to more complicated bit testing and selection. But since the problem isn't exactly high performance or practical, I'll use simple addition. For a recursive function, you need to create a condition that will prevent further recursion, I'll use the condition of multiplying by 0. Also, you need to define what your recursion is.
To wit, consider the following math expression 
 f(m,k) = 0 if m = 0, otherwise f(m-1,k) + k 
 If you calculate f(0,k), you'll get 0 which is exactly what 0 * k is. 
 If you calculate f(1,k), you'll get 0 + k, which is exactly what 1 * k is.   
 So here's the function   
 int product(int m, int k) 
 { 
  if (m == 0) return 0; 
  return product(m-1,k) + k;  
}

Step-by-step explanation:

This is a differential equation problem most easily solved with an exponential decay equation of the form

. We know that the initial amount of salt in the tank is 28 pounds, so

C = 28. Now we just need to find k.

The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is . Thus, the change in the concentration of salt is found in

inflow of salt - outflow of salt

Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:

Therefore,

or just

and in terms of time,

Thus, our equation is

and filling in 16 for the number of minutes in t:

y = 24.834 pounds of salt