Answer:
Please see the graph below
Step-by-step explanation:
The table is just selecting numbers for x and then solving for what y would be. Once we have the ordered pairs (x,y) We can graph the points and see the line.
( c^2 )^3 = 64 <=>( c^2 )^3 = 4^3 <=> c^2 = 4 <=> c = 2 or c = -2.
You could start on 34 on the number line and jump back 28 times to get your answer.
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;
}
