Answer:
<h3>Defination of Software :- </h3>
- software, instructions that tell a computer what to do. Software comprises the entire set of programs, procedures, and routines associated with the operation of a computer system. The term was coined to differentiate these instructions from hardware.
- The physical components of a computer system.
Explanation:
<h3>Hope this helps you dear ✌️</h3><h2>Carry on learning !! </h2>
A.A shot.
hope this helps!!
<h3>UNDER THE MENU OPTION INSERT.</h3>
Lets first work out how many different codes would be needed to represent everything. 26 for lowercase, 26 for uppercase, and 10 for 0-9. Total, that makes 62 needed codes.
If we start with 0, we need to go up to 61 to represent all the characters. Thus, we can convert 61 to binary and count the number of digits needed to represent that as the last number in the set and that will tell us how many digits are needed.
61 in binary is 111101, so we need 6 digits to represent that number. The answer is B.
Answer:
- public class Main {
-
- public static void main (String [] args) {
-
- for(int i = 2; i < 10000; i++){
- if(isPrime1(i)){
- System.out.print(i + " ");
- }
- }
-
- System.out.println();
-
- for(int i = 2; i < 10000; i++){
- if(isPrime2(i)){
- System.out.print(i + " ");
- }
- }
- }
-
- public static boolean isPrime1(int n){
-
- for(int i=2; i <= n/2; i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
-
- public static boolean isPrime2(int n){
-
- for(int i=2; i <= Math.sqrt(n); i++){
- if(n % i == 0){
- return false;
- }
- }
-
- return true;
- }
- }
<u></u>
Explanation:
Firstly, create the first version of method to identify a prime number, isPrime1. This version set the limit of the for loop as n/2. The for loop will iterate through the number from 2 till input n / 2 and check if n is divisible by current value of i. If so, return false to show this is not a prime number (Line 22 - 26). Otherwise it return true to indicate this is a prime number.
In the main program, we call the isPrime1 method by passing the i-index value as an argument within a for-loop that will iterate through the number 2 - 10000 (exclusive). If the method return true, print the current i value). (Line 5 - 9)
The most direct way to ensure all the prime numbers below 10000 are found, is to check the prime status from number 2 - 9999 which is amount to 9998 of numbers.
Next we create a second version of method to check prime, isPrime2 (Line 31 - 40). This version differs from the first version by only changing the for loop condition to i <= square root of n (Line 33). In the main program, we create another for loop and repeatedly call the second version of method (Line 13 - 17). We also get the same output as in the previous version.
