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77julia77 [94]
3 years ago
7

Hey can someone read My I.P address? My teacher want's us to learn port forwarding but I can't find my I.P. because its a school

computer
Computers and Technology
1 answer:
Alborosie3 years ago
7 0
Obviously you have access to any sort of device. 
click on your internet and click more info: 
your IP address will be stated clearly, for example as 666<span>.666.6.666</span>
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Answer:

The main physical components of cloud infrastructure are networking equipment, servers and data storage

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2 years ago
Which tool allows multiple users to dynamically update a document at the same time? Question 3 options: Google Docs OneDrive Exc
GenaCL600 [577]

Answer:

Excel

Explanation:

7 0
3 years ago
Write a do-while loop that continues to prompt a user to enter a number less than 100, until the entered number is actually less
Ad libitum [116K]

Answer:

#include<iostream>//library inclusion

using namespace std;

int main()

{

int userInput;

do//start of do while loop

{

 cout << "Enter a number less than a 100" << endl;

 cin >> userInput;

 if (userInput < 100) //condition

 {

  cout << "YOu entered less than a hundred: " << userInput << endl;

 }

 else

 {

  cout << "your number is greater than 100" << endl;

 }

} while (userInput > 100);//condition for do while

return 0;//termination of int main

}

Explanation:

The program has been commented for you. The do-while loop enters the first loop regardless of the condition. Then after the first iteration, it checks for the condition. If the condition is being met, it will iterate through, again. Otherwise it will break out of the loop and land on the "return 0;" line. Which also happens to be the termination of the program in this case. The if-else condition is used for the user to see when prompted.

6 0
3 years ago
Pass the two arrays to a method that will display the integer value(s), if any, that appear in both arrays (note that the two ar
vampirchik [111]

Answer:

The solution code is written in Java

  1.    public static void checkCommonValues(int arr1[], int arr2[]){
  2.        if(arr1.length < arr2.length){
  3.            for(int i = 0; i < arr1.length; i++){
  4.                for(int j = 0; j < arr2.length; j++){
  5.                    if(arr1[i] == arr2[j]){
  6.                        System.out.print(arr1[i] + " ");
  7.                    }
  8.                }
  9.            }
  10.        }
  11.        else{
  12.            for(int i = 0; i < arr2.length; i++){
  13.                for(int j = 0; j < arr1.length; j++){
  14.                    if(arr2[i] == arr1[j]){
  15.                        System.out.print(arr2[i] + " ");
  16.                    }
  17.                }
  18.            }
  19.        }
  20.    }

Explanation:

The key idea of this method is to repeated get a value from the shorter array to check against the all the values from a longer array. If any comparison result in True, the program shall display the integer.

Based on this idea, an if-else condition is defined (Line 2). Outer loop will traverse through the shorter array (Line 3, 12) and the inner loop will traverse the longer array (Line 4, 13). Within the inner loop, there is another if condition to check if the current value is equal to any value in the longer array, if so, print the common value (Line 5-7, 14-16).

5 0
3 years ago
Select all that apply.
True [87]

Answer:

                                          Option    A, B & C   are correct answers .

Explanation:

   

4 0
3 years ago
Read 2 more answers
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