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Papessa [141]
3 years ago
15

Im so stuck bruh helpppppppppp

Computers and Technology
1 answer:
goblinko [34]3 years ago
5 0

Answer:

24 is the answer hope that helped!

•w•

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A weakness of a system, process, or architecture that could lead to compromised information or unauthorized access is known as a
Anna11 [10]

Answer: vulnerability

Explanation:

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A tooltip is a _____________ that allows the programmer to create a small popup message that displays when the user places the m
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I believe a message
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Nested if-else structures can contain many blocks of code. How many of those blocks of code might be executed?
dezoksy [38]

Answer:            

A nested if else structure is defined as using one if-else inside another one. It is used when you want a take a decision by checking several conditions. Nested if can contain many block of codes with each block executing if the certain condition is met for each of that block.

The number of blocks of code, that might be executed depends on the condition that is intended to be met, in order to take a certain decision based on that condition. Let suppose if the desired condition is met in the very first block of code then only one block of code is executed. For example consider the following code:

if(number<5)

{  if(number==1)  {

  cout<<"number is 1"<<number; }

 else

 {       cout<<"number is greater than 1"<<number; } }

       else

      {         cout<<"number is greater than 5";}

Suppose the number entered is 1. So the first nested if condition is true and number is 1 will be displayed. So none of the else blocks has to be executed.

Number of code blocks that might be executed also relies on the number of block present in the code.

The execution also depends on the number of blocks that are true or false. If you want to execute a block for true condition, nested if else looks through every block until the block with the true desired condition is reached but if you want false condition to be checked to display the block of code when condition gets false then it will look through all the blocks until it gets the block with that false part.

For example consider the following code:

if ( number > 0)

   {        cout << "its a positive number: " << number;    }

   else if (number < 0)

   {         cout<<"its a negative integer: " << number;    }

   else     {

            cout << "its a 0";   }

Suppose the number is 0 and the program has to sequentially checks for each condition. It will check number>0 which is not what we need, then it will check else if statement for number<0 which is again not desired and finally it will display the block "its a 0" which matches the desired condition.

7 0
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Marys mom had four kids: April, May, June. Who is the fourth child?
Rainbow [258]

Answer:

July

Explanation:

Because they have a order April than May then June then July is after June so it has to be the right answer.

7 0
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Phone numbers and PIN codes can be easier to remember when you find words that spell out the numberon a standard phone pad. For
Alik [6]

Answer:

Kindly note that, you're to replace "at" with shift 2 as the brainly text editor can't take the symbol

Explanation:

Here is the completed code for the problem in the question above.

// PinWordEnumerator.java

import java.util.Scanner;

public class PinWordEnumerator {

               public static void main(String[] args) {

                               Scanner scanner = new Scanner(System.in);

                               System.out.print("Enter pin number: ");

                               String num = scanner.nextLine();

                               System.out.println();

                               System.out.printf("The keypad encodings for %s are:%n", num);

                               enumerateWords(num);

               }

               /**

               * method to return a character array containing all possible keypad

               * encodings for a digit in a standard phone

               */

               static char[] getKeysForButton(int n) {

                               switch (n) {

                               case 0:

                                               return new char[] { ' ' }; //returning only space

                               case 1:

                                               return new char[] { '.' }; //returning only dot

                               case 2:

                                               //button 2 contains A, B and C keys

                                               return new char[] { 'A', 'B', 'C' };

                               case 3:

                                               return new char[] { 'D', 'E', 'F' };

                               case 4:

                                               return new char[] { 'G', 'H', 'I' };

                               case 5:

                                               return new char[] { 'J', 'K', 'L' };

                               case 6:

                                               return new char[] { 'M', 'N', 'O' };

                               case 7:

                                               return new char[] { 'P', 'Q', 'R', 'S' };

                               case 8:

                                               return new char[] { 'T', 'U', 'V' };

                               case 9:

                                               return new char[] { 'W', 'X', 'Y', 'Z' };

                               }

                               return null;

               }

               /**

               * method to enumerate words

               *

               * "at"param num

               *            - String containing pin numbers, assuming it has only numbers

               */

               static void enumerateWords(String num) {

                               /**

                               * calling the recursive method to perform the enumeration

                               */

                               if (num != null)

                                               enumerateWords(num, "");

               }

               /**

               * the main method which performs the recursion

               *

               * "at"param num

               *            - current number

               * "at"param text

               *            - text containing converted spellings

               */

               static void enumerateWords(String num, String text) {

                               if (num.length() == 0) {

                                               // base case, displaying the text

                                               System.out.println(text);

                               } else {

                                               // finding the digit at 0th position

                                               int digit = num.charAt(0) - '0';

                                               // finding possible phone keys for this digit

                                               char letters[] = getKeysForButton(digit);

                                               if (letters != null) {

                                                               // looping through all possible keys

                                                               for (int i = 0; i < letters.length; i++) {

                                                                               /**

                                                                               * appending the current letter to the text and calling the

                                                                               * recursive method also neglecting the first letter of

                                                                               * current 'num' string

                                                                               */

                                                                               enumerateWords(num.substring(1), text + letters[i]);

                                                               }

                                               }

                               }

               }

}

/*OUTPUT*/

Enter pin number: 5282

The keypad encodings for 5282 are:

JATA

JATB

JATC

JAUA

JAUB

JAUC

JAVA

JAVB

JAVC

JBTA

JBTB

JBTC

JBUA

JBUB

JBUC

JBVA

JBVB

JBVC

JCTA

JCTB

JCTC

JCUA

JCUB

JCUC

JCVA

JCVB

JCVC

KATA

KATB

KATC

KAUA

KAUB

KAUC

KAVA

KAVB

KAVC

KBTA

KBTB

KBTC

KBUA

KBUB

KBUC

KBVA

KBVB

KBVC

KCTA

KCTB

KCTC

KCUA

KCUB

KCUC

KCVA

KCVB

KCVC

LATA

LATB

LATC

LAUA

LAUB

LAUC

LAVA

LAVB

LAVC

LBTA

LBTB

LBTC

LBUA

LBUB

LBUC

LBVA

LBVB

LBVC

LCTA

LCTB

LCTC

LCUA

LCUB

LCUC

LCVA

LCVB

LCVC

7 0
3 years ago
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