<h2>Solution :</h2>
The given problem is solved in Python.
def ComputeSquare():
side = float(input('Enter the side of the square: '))
if side > 0:
perimeter = 4 * side
area = side * side
print('Perimeter of the square is:',
perimeter, 'unit.')
print('Area of the square is:', area, 'square unit.')
<h3> else:</h3>
print('Invalid input.')
ComputeSquare()
<h2>Explanation :-</h2>
- In this program, we create a function ComputeSquare() to calculate the perimeter and area of a square.
- The function asks the user to enter the side of the square. The side is then stored in a variable.
- Now, we check whether the side is greater than 0 or not using if-else statement.
- If the condition is true, the perimeter and area is calculated which is displayed using print() statement.
- If the condition is false, the else blocks executes printing the error message.
<h3>Refer to the attachment for output.</h3>
Step 2: Multiply total number of pixels by the bit depth of the detector (16 bit, 14 bit etc.) to get the total number of bits of data. Step 3: Dividing the total number of bits by 8 equals the file size in bytes. Step 4: Divide the number of bytes by 1024 to get the file size in kilobytes.
It is actually podcast! i took the quiz as well :)
Answer:
import java.io.*;
import java.util.Scanner;
class divide {
public static void main (String[] args) {
Scanner num=new Scanner(System.in);//scanner object.
int userNum=num.nextInt();
while(userNum>1)//while loop.
{
userNum/=2;//dividing the userNum.
System.out.print(userNum+" ");//printing the userNum.
}
}
}
Input:-
40
Output:-
20 10 5 2 1
Input:-
2
Output:-
1
Input:-
0
Output:-
No Output
Input:-
-1
Output:-
No Output.
Explanation:
In the program While loop is used.In the while loop it divides the userNum by 2 in each iteration and prints the value of userNum.The inputs and corresponding outputs are written in the answer.
Answer:
O(n²)
Explanation:
The worse case time complexity of insertion sort using binary search for positioning of data would be O(n²).
This is due to the fact that there are quite a number of series of swapping operations that are needed to handle each insertion.
