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Alex Ar [27]
3 years ago
8

The region which satisfies all of the constraints in graphical linear programming is called the:

Computers and Technology
1 answer:
liraira [26]3 years ago
6 0
<span>The region which satisfies all of the constraints in graphical linear programming is called the feasible solution space. Correct answer: E

</span>The feasible solution space includes values for the decision variables that satisfies all of the constraints in an optimization problem. The optimization problem can be an <span>inequality, an equality, and integer constraint.</span>

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Is there an ethically acceptable reason to study and use the various attack methods described in this chapter? Can you think of
avanturin [10]

Answer:

Please see below  

Explanation:

Yes, there indeed is ethical justification for hacking certain computer systems. Since computer scientists are required to keep the system secure from external threats, so they make use of it when testing the network for potential loopholes that could make it vulnerable. It is beneficial in that it can help manifest the weaknesses present in the system, which can then be corrected for.

6 0
4 years ago
Write a loop that sets new scores to old scores shifted once left, with element 0 copied to the end. ex: if old scores = {10, 20
Inga [223]

#include <iostream>
using namespace std;

int main() {
   const int SCORES_SIZE = 4;
   int oldScores[SCORES_SIZE];
   int newScores[SCORES_SIZE];
   int i = 0;

   oldScores[0] = 10;
   oldScores[1] = 20;
   oldScores[2] = 30;
   oldScores[3] = 40;

   /* Your solution goes here */

   for (i = 0; i < SCORES_SIZE; ++i) {
      cout << newScores[i] <<" ";
   }
   cout << endl;

   return 0;
}

3 0
3 years ago
Read 2 more answers
Ria has modified the hard disk that hosts the operating system by using the fdisk command. The fdisk command indicates that the
pickupchik [31]

Answer:

a and d

Explanation:

A. Run the parted command

D. Reboot the system

These two actions will make it easier to manually reload the fdisk

8 0
3 years ago
Write two alternate functions specified below, each of which simply triples the variable count defined in main. These two functi
Vedmedyk [2.9K]

Answer:

Following are the code to this question:

#include <iostream>//header file

using namespace std;

int triplebyValue(int count)//defining a method triplebyValue

{

int x=count*3;//defining a variable x that multiply by 3 in the count  

return x;//return value of x

}

void triplebyReference(int& count)//defining a method triplebyReference that hold count variable as a reference in parameter

{

count*=3;//multipling a value 3 in the count variable

}

int main()//main method

{

int count;//defining integer variable

count=triplebyValue(3);//use count to call triplebyValue method

cout<<"After call by value, count= "<<count<<endl;//print count value with message

triplebyReference(count);//calling a method triplebyReference

cout<<"After call by reference, count= "<<count<<endl;//print count value with message

return 0;

}

Output:

After call by value, count= 9

After call by reference, count= 27

Explanation:

In this code two methods "triplebyValue and triplebyReference" are declared, which accepts a count variable as a parameter, and in both multiply the count value by 3 and return its value.

Inside the main method, an integer variable "count" is declared that first calls the "triplebyValue" method and holds its value into count variable and in the next, the method value is a pass in another method that is "triplebyReference", and use print method to print both methods value with a message.

7 0
3 years ago
Program to count how many even and odd number are in the range from M to N using C++ programming language
NemiM [27]

Answer:

#include<stdio.h>

#include <iostream>

using namespace std;

 

int main(){

   int number, min, max;

   cout << "Enter the minimum range: ";

   cin >> min;

  cout << "Enter the maximum range: ";

   cin >> max;

   cout << "Odd numbers in given range are: ";

   for(number = min; number <= max; number++)

        if(number % 2 !=0)

            cout << number<< " ";

   cout<< "Even numbers in given range are: ";

   for(number = min; number <= max; number++)

        if(number % 2 ==0)

            cout << number << " ";

 

   return 0;

}

Explanation:

First of all, we take declare two variables. one as the lowest number of the range and the other as the upper limit of the range (in this case: <em>min</em> and <em>max</em>). We declare another variable (<em>number</em>) and store in it the lowest number of the range (<em>min</em>). To check whether the number currently stored as the value of the variable (<em>number</em>) is even, we take in account the remainder of that number divided by 2. If the remainder does not equals 0, we print that number as odd. We again check the remainder by dividing the number by 2. If the remainder does equals 0, we print that number as even.

8 0
3 years ago
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