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;
}
Answer:
Step-by-step explanation:
It’s B
None of these couples are solutions, (11, 3)(11, 3); (−1, −6)(−1, −6); (−3, 3)(−3, 3); <span>(7, 0). Perhaps the choice of answer are insufficient. we can add (1, 48) the couple (7, 0) and (7, 0)(1, 48) is a true answer, why? because it verifies the equation.</span>
Answer:
C(t) = 20t
Step-by-step explanation:
The graph of C(t) appears to be linear
The y intercept is 0
The slope is 200/10 = 20
y = mx+b where m is the slope and b is the y intercept
y = 20x+0
y = 20x
C(t) = 20t
2, 3, 4, 5, 6
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