Answer:
Answered below
Explanation:
//Program is written in Java programming //language
Class Box{
private double length;
private double width;
private double height;
Box(double len, double wid, double hgt){
length = len;
width = wid;
height = hgt;
}
public double volumeOfBox( ){
double volume = length * width * height;
return volume;
}
public double getLength( ){
return length;
}
public double getWidth( ){
return width;
}
public double getHeight( ){
return height;
}
}
Hello, you haven't provided the programing language in which you need the code, I'll explain how to do it using Python, and you can follow the same logic to make a program in the programing language that you need.
Answer:
1. # -*- coding: utf-8 -*-
2. #Python
3. class Calculator:
4. def add(self):
5. print(a + b)
6. def sub(self):
7. print(a - b)
8. def mul(self):
9. print(a * b)
10. def div(self):
11. print(a / b)
12.
13. obj = Calculator()
14. choice = 1
15. while choice != 0:
16. a = int(input("\nEnter first number: "))
17. b = int(input("Enter first number: "))
18.
19. print("\n0. EXIT")
20. print("1. DIVISION")
21. print("2. ADDITION")
22. print("3. SUBTRACTION")
23. print("4. MULTIPLICATION")
24.
25. choice = int(input("\nEnter your choice: "))
26. if choice == 1:
27. obj.div()
28. elif choice == 2:
29. obj.add()
30. elif choice == 3:
31. obj.sub()
32. elif choice == 4:
33. obj.mul()
34. else:
35. break
Explanation:
- From lines 1 to 12 we define the object with four methods, addition, subtraction, multiplication, and division. Each method contains the operation between two variables inside a print function
- On line 13 we instantiate our class
- On line 14 we declare the variable choice that is going to store the operation that the user wants to perform
- On line 15 we declare a while loop, this is going to keep running the program until the user wants to exit
- From line 16 to 18 we ask the user to enter two numbers
- From line 19 to 24 we print the possible operation, assigning a number for each operation, this indicates to the user what number to press for what operation
- On line 25 we ask the user for the operation
- From lines 26 to 35 we check the user input an accordingly we call the corresponding method to performs the operation
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.
The correct answer is Technician A ONLY. I had this on my test, so im sure of my answer!
