Answer:
Explanation:
Considering the number are stored using binary notations.
If there are total 86 bits and 1 bit is used as sign bit. Then in total one can have 
bit combinations since a bit can be either stored as 0 or as 1.
Therefore, the largest number that can be stored will be .
To understand it in better way let's scale down the problem to 3 bits representation. The largest number that can be stored using 3 bits is 111 which in decimal form is 7 and is equal to 
.
D that’s the answer I learn that in my old school
Answer:
import java.util.*;
public class MyClass {
public static void main(String args[]) {
Scanner input = new Scanner(System.in);
System.out.print("Input a word: ");
String userinput = input.nextLine();
for(int i =0;i<userinput.length();i+=2) {
System.out.print(userinput.charAt(i));
}
}
}
Explanation:
This line prompts user for input
System.out.print("Input a word: ");
This declares a string variable named userinput and also gets input from the user
String userinput = input.nextLine();
The following iterates through every other character of userinput from the first using iteration variable i and i is incremented by 2
for(int i =0;i<userinput.length();i+=2) {
This prints characters at i-th position
System.out.print(userinput.charAt(i));
No I don’t think so personally
Or by searching up the desired content and or information on a search engine. I believe.
