int sum = 0;
int number;
for (int i=0; i<10; i++) {
printf("Enter an integer number: ");
scanf_s("%d", &number);
sum += number;
}
printf("The sum is %d", sum);
Answer:
The definition including its context of this discussion is outlined in the following clarification portion.
Explanation:
- An array seems to be a storage research method to collect a set number of constant weight. Whenever the array was indeed generated, the length including its array has been determined. Its period is set after development.
- Depending mostly on array description, the array may contain data structures and even some objects in something like a class. For primitive data types, the individual values become located at contiguous locations throughout the store.
Scenario:
- If we're to preserve a summary, we could use the appropriate array size.
- An individual can however preserve an object list.
Answer:
False
Explanation:
Chloroplasts = photosynthesis
mitochondria= cellular respiration
Answer:
A. Acceptable use policies
Explanation:
These specific actions that the user may perform are all described in the acceptable use policy. The company usually requires that the user read and sign this policy before being granted a network ID into the system itself. By signing this agreement the individual user agrees to perform ONLY the actions listed in the agreement and nothing else and are restricted from unauthorized areas.
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
</span>
As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
