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Jobisdone [24]
3 years ago
14

1 Point

Computers and Technology
1 answer:
jekas [21]3 years ago
8 0

Answer: Its A

Explanation:

For creative writing

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fomenos

Answer:

Hi There was small mistake. It is working fine for me. When you run from command line - use LoopPattern, not looppatern

import java.util.Scanner;

public class Looppattern {

  public static void main(String[] args) {

      Scanner sc = new Scanner(System.in);

      System.out.println("Enter how man levels you need: ");

      int levels = sc.nextInt();

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      System.out.println("\n---------------Pattern B-----------------\n");

      int r = levels;

      for (int p = 1; p <= levels; p++) {

          for (int k = 1; k <= r; k++) {

              System.out.print(k);

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      System.out.println("\n---------------Pattern C-----------------\n");

      for (int p = 1; p <= levels; p++) { // here incresing

          for (int k = p; k > 0; k--) { // and here decreasing pattern to

                                          // achieve our required pattern

              System.out.print(k);

          }

          System.out.println();

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      System.out.println("\n---------------Pattern D-----------------\n");

      r = levels;

      for (int p = 1; p <= levels; p++) {

          for (int k = 1; k <= r; k++) {

              System.out.print(k);

          }

          r--; // decreasing levels

          System.out.println();

      }

  }

}

Explanation:

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attraction and attention

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In a java class, variables can be present both inside and outside of methods. question 9 options:
Tems11 [23]
The correct answer is: True

Explanation:

Consider the following program:
<span>class Main {
int a = 10;
public static void main(String[] args) {
add();
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}<span>
}


The output of the above program is 30. As you can see the variable a is declared outside the method, whereas variable b is declared inside the method.</span></span>
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