Answer:
Good luck
Explanation:
Laptop and iPad
laptop because you can open many pages and tabs to help you multitask your goals
and iPad because most free business would be on an iPad and apps are cheap and would help you focus/help you work harder
Cleaning up a system helps clear up the space on the drives. It may also clear up processor usage, ram usage if you uninstall programs that automatically started when the system booted. You may also delete some unwanted programs in the process.
Answer:
19.95
Explanation:
Power ratio (in db) = 10 * log [P(out)/P(in)]
Here, P(in) and P(out) correspond to input and output power in Watts respectively.
This implies, 13 db = 10 * log [P(out)/P(in)]
Rearranging, log [P(out)/P(in)] = (13/10)
Or, P(out)/P(in) = 10 ^ 1.3 = 19.95
But, P(out)/P(in) corresponds to the power ratio of output and input power.
So the required power ratio corresponds to 19.95
Answer:
import java.util.*;
public class Main
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter the first: ");
int first = input.nextInt();
System.out.print("Enter the second: ");
int second = input.nextInt();
input.nextLine();
System.out.print("Enter your name: ");
String name = input.nextLine();
int difference = Math.abs(second - first);
System.out.println(name + ", the difference is " + difference);
}
}
Explanation:
*The code is in Java.
Create a Scanner object to get input from the user
Ask the user to enter the first, second, and name
Calculate the difference, subtract the second from the first and then get the absolute value of the result by using Math.abs() method
Print the name and the difference as in required format
Answer:
Complete Python code with step by step comments for explanation are given below.
Python Code:
# creating a function named scrabble_number that will take num as input argument
def scrabble_number(num):
# print the original number
print("The original number is: ",num)
# we can implement the required logic by using python built-in functions join() and zip()
scrambled = ''.join([char[1]+char[0] for char in zip(num[::2], num[1::2])])
# print the scrambled number
print("The scrambled number is: " + str(scrambled))
Driver code:
scrabble_number('123456')
Output:
The original number is: 123456
The scrambled number is: 214365
