Answer:
Your answer is initial-response field kit
Explanation:
Idk but I use to know because I took computer classes
Answer:
Explanation:
class TimeToSleep() {
main bunch of stuff (string argos) {
int age;
string comedy;
bool guess;
public bool imTheMasterMethod() {
guess = true;
while (guess) {
comedy = "You forgot to put the flowchart, you meathead!";
println(comedy);
scan("%d", &age);
if (age < 18) {
println("Go to sleep kiddo!");
}
}
}
}
}
Answer:
2^11
Explanation:
Physical Memory Size = 32 KB = 32 x 2^10 B
Virtual Address space = 216 B
Page size is always equal to frame size.
Page size = 16 B. Therefore, Frame size = 16 B
If there is a restriction, the number of bits is calculated like this:
number of page entries = 2^[log2(physical memory size) - log2(n bit machine)]
where
physical memory size = 32KB which is the restriction
n bit machine = frame size = 16
Hence, we have page entries = 2^[log2(32*2^10) - log2(16)] = 2ˆ[15 - 4 ] = 2ˆ11
The recursive function would work like this: the n-th odd number is 2n-1. With each iteration, we return the sum of 2n-1 and the sum of the first n-1 odd numbers. The break case is when we have the sum of the first odd number, which is 1, and we return 1.
int recursiveOddSum(int n) {
if(2n-1==1) return 1;
return (2n-1) + recursiveOddSum(n-1);
}
To prove the correctness of this algorithm by induction, we start from the base case as usual:
by definition of the break case, and 1 is indeed the sum of the first odd number (it is a degenerate sum of only one term).
Now we can assume that 
returns indeed the sum of the first n-1 odd numbers, and we have to proof that 
returns the sum of the first n odd numbers. By the recursive logic, we have
and by induction, is the sum of the first n-1 odd numbers, and 2n-1 is the n-th odd number. So, is the sum of the first n odd numbers, as required:
