That's cool, I guess. What was the question?
RDO.
RDO uses the lower-level DAO and ODBC for direct access to databases.
Answer:
C.
Explanation:
A line graph can show how both the movie and the novel are compared to each other. It can give a visual of both mediums of the story rather than one or the other. Hope this helped :)
Answer:
The answers are explained below
Explanation:
1) Identify the potential classes in this problem domain be list all the nouns
class Customer
class Acco unt
2) Refine the list to include only the necessary class names for this problem
the class customer is not necessary to solve the problem itself, therefore the only class could be the account class
3) Identify the responsibilities of the class or classes.
The responsibilities of the class account will be
* determination of the type of account--> Acc ount . type(char)
* deposit money into the account --> Acc ount . de posit(float)
* withdraw money into the account --> Acc ount . with draw(float)
* show balance of the account --> Acc ount . bal ance()
* generate interest --> Acc ount . int erest()
Please join the words together. I used spaces due to regulations
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.
