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gizmo_the_mogwai [7]
3 years ago
9

The ____ category of apps makes the computer easier for blind people to use.

Computers and Technology
1 answer:
Schach [20]3 years ago
4 0

The accessibility category of apps makes the computer easier for blind people to use. In general the accessibility apps are apps that help people with disabilities use a particular piece of hardware. For example there is an app designed to help blind people use their devices by paring them with a voluntary non-blind people trough audio-video connections.





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Gina is upgrading your computer with a new processor. She installed the processor into your motherboard and adds the cooling sys
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Answer:

The answer is option d.

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Given a positive real number, print its fractional part.
givi [52]
<span>The modf() function will do this for you:

double x, y, d;
x = -14.876;
y = modf(x, &d);
printf("Fractional part = %lf\n", y);

</span>
8 0
4 years ago
1st row has 3 possible answers software program, web page, and web browser
Maurinko [17]

Answer:

1st row: software program

2nd row: a user's computer

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7 0
3 years ago
Create a structure representing a student. The member variables should include student name, student ID, and four test grades fo
Neporo4naja [7]

Answer:

Output:

Name: Brainly

ID:0001

Write the 4 tests grades of the student separated by space :10 9 8 10

Brainly

0001

10 9 8 10

Average :9.66667

'1' to continue '0' to exit :

Explanation:

#include<iostream>

#include<string>

using namespace std;

//variables declaration

struct Student {

  string id; //string declaration ID

  string name; // string declaration name

  int grades[4]; //array of 4 because it is 4 grades

};

//definition of the function get information

void inputData(Student &s){

 

   

   cout << "Name:" ;

   getline(cin,s.name);

   cout << "ID:";

   cin >> s.id;

   cout << "Write the 4 tests grades of the student separated by space :";

   for (int i = 0; i<4; i++)

       cin >> s.grades[i];

}

//definition of the function of average

double inputAvg(Student s){

   double summation;

   int temporary;

   double average;

   for (int i = 0; i<4; i++){  

      for (int j = i; j<4; j++){

         if (s.grades[j] > s.grades[i]){

             temporary = s.grades[i];

             s.grades[i] = s.grades[j];

             s.grades[j] = temporary;

         }

      }

    }

    summation = 0;

    for (int i = 0; i<3; i++){

        summation = summation + s.grades[i];  

    }

    average = summation/3;

    return average;

}

void disp(Student *s){

   cout << s->name << endl;

   cout << s->id << endl;

   for (int i = 0; i<4; i++)

       cout << s->grades[i] << " ";

   cout << endl;

   cout << "Average :" << inputAvg(*s) << endl;

   

}

int main(){

  Student st;

  int ch;

  while(true){

      inputData(st);

      disp(&st);

      cout << " '1' to continue '0' to exit :";

      cin >> ch;

      if (ch == 0)

         break;      

  }

  return 0;

}

6 0
3 years ago
"Write an iterative function iterPower(base, exp) that calculates the exponential baseexp by simply using successive multiplicat
sp2606 [1]

Answer:

I am writing the function using Python. Let me know if you want the program in some other programming language.            

def iterPower(base, exp):    

    baseexp = 1

    while exp > 0:

        baseexp = baseexp*base

        exp= exp - 1

    return baseexp

base = 3

exp = 2

print(iterPower(base, exp))

Explanation:

  • The function name is iterPower which takes two parameters base and exp. base variable here is the number which is being multiplied and this number is multiplied exponential times which is specified in exp variable.
  • baseexp is a variable that stores the result and then returns the result of successive multiplication.
  • while loop body keeps executing until the value of exp is greater than 0. So it will keep doing successive multiplication of the base, exp times until value of exp becomes 0.
  • The baseexp keeps storing the multiplication of the base and exp keeps decrements by 1 at each iteration until it becomes 0 which will break the loop and the result of successive multiplication stored in baseexp will be displayed in the output.
  • Here we gave the value of 3 to base and 2 to exp and then print(iterPower(base, exp)) statement calls the iterPower function which calculates the exponential of these given values.
  • Lets see how each iteration works:
  • 1st iteration

baseexp = 1

exp>0 True because exp=2 which is greater than 0

baseexp = baseexp*base

               = 1*3 = 3

So baseexp = 3

exp = exp - 1

      = 2 - 1 = 1    

exp = 1

  • 2nd iteration

baseexp = 3

exp>0 True because exp=1 which is greater than 0

baseexp = baseexp*base

               = 3*3 = 9

So baseexp = 9

exp = exp - 1

      = 1-1 = 0    

exp = 0

  • Here the loop will break now when it reaches third iteration because value of exp is 0 and the loop condition evaluates to false now.
  • return baseexp statement will return the value stored in baseexp which is 9
  • So the output of the above program is 9.
5 0
3 years ago
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