Answer:
Malleability and ductility
Explanation:
Metals are also easily distinguished from non-metals and other materials in their ductility (ability to be drawn into a wire) and malleability (the ability to be beaten into a. sheet).
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Answer:
Full
Explanation:
According to my research on IT tasks, I can say that based on the information provided within the question this type of software installation is called a Full Software Installation. This type of installation is usually done when installing a brand new operating system, which you then need to install all the necessary software that is commonly needed.
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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;
- }
- }
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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.
