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olya-2409 [2.1K]
2 years ago
6

Your program has a loop. You want to exit the loop completely if the user guesses the correct word.

Computers and Technology
2 answers:
Wewaii [24]2 years ago
8 0

Answer:

Continue

Explanation:

balandron [24]2 years ago
5 0

break

Explanation:

breaking it will stop the loop and you are free

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What do we call exceptions to the exclusive rights of copyright law?
Studentka2010 [4]

Answer:

Fair use

Explanation:

I don't know if this is correct but Its the best option.

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You have an application that you would like to run on your Windows workstation every Monday at 3:00 p.m. Which tool would you us
NikAS [45]

Answer:

Task Scheduler

Explanation:

Task Scheduler allows you to automate tasks in Windows 10

7 0
3 years ago
4) To run program short cut​
quester [9]

Answer:

Run Dialog

Open it by pressing WinKey+R. In the Run dialog, type the name of a program's .exe file and press Enter to launch it.

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6 0
3 years ago
you have a website that customers use to view product information and place orders. You would like to identify the maximum numbe
notka56 [123]

Answer:

a load tester.

Explanation:

Based on the information provided within the question it can be said that in this situation the tool that you should use is known as a load tester. This tool allows you to test your servers in order to see how it behaves under both normal conditions as well as the anticipated peak load conditions that it may be encountering in the near future. Which is exactly what you need for your website.

4 0
3 years ago
github Portfolio Balances An investor opens a new account and wants to invest in a number of assets. Each asset begins with a ba
OLEGan [10]

The following code will be used to determine the maximum amount invested

<u>Explanation:</u>

long maxValue(int n, int rounds_rows, int rounds_columns, int into rounds)

{

   // Define the variable to store

   // the maximum amount invested.

   long max = 0;

   

   // Define an array of size n,

   // to store the n investments.

   long *investments = (long*)malloc(sizeof(long)*n);

   int i=0;  

   // Initially set all

   // the investments to 0.

   for(i=0;i<n;i++)

   {

       investments[i] = 0;

   }

   i=0;

   // Run the loop to

   // perform the rounds.

   while(i<rounds_rows)

   {

       // Get the left value

       // of the current round.

       int left = rounds[i][0];

       

       // Get the right value

       // of the current round.

       int right = rounds[i][1];

       // Get the contribution

       // for the current round.

       int contribution = rounds[i][2];

       // Since the user uses 1-based

       // indexing, subtract 1 from left

       // and right as the program uses

       // 0-based indexing. That is, the

       // investments in the array start

       // from 0 and not 1.

       right = right - 1;

       int j=0;

       // Run the loop to add the

       // contribution to all the investments

       // between left and right, both inclusive.

       for(j=left; j<=right; j++)

       {

           investments[j] += contribution;

       }

       i++;

   }

   // Traverse the investments array

   // to find the maximum element.

   max = investments[0];

   for(i=1; i<n;i++)

   {

       if(investments[i]>max)

       {

           max = investments[i];

       }

   }

   // Return the

   // maximum investment.

   return max;  

}

6 0
4 years ago
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