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valina [46]
4 years ago
8

Which tasks can be completed within the File tab?

Computers and Technology
1 answer:
elena-14-01-66 [18.8K]4 years ago
7 0
<span><span>The file is an information stored on a computer under a single name. A folder contains several files.A file can be a Word document, a PowerPoint presentation, a PDF document,....</span>The file tab displays</span><span> all the things you can do with a file referred as file management tasks, like (opening, saving, printing, sharing file, publishing). </span>


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Assume that d is a double variable. Write an if statement that assigns d to the int variable i if the value in d is not larger t
mamaluj [8]

Answer:

That's because the value has reached the size limit of the int data type. ... you should use long rather than int , because long can store much larger numbers than int . If ... In other words, a float or double variable can't accurately represent 0.1 . ... If you're using Java to measure the size of your house, you'd need an electron ...

Explanation:

3 0
3 years ago
Consider the partially-filled array named a. What does the following loop do? (cin is a Scanner object)
Jlenok [28]

Answer:

3. Reads up to 3 values and places them in the array in the unused space.

Explanation:

cin.NextInt() reads the next value from the Scanner.

The while is just while the size is lesser than the capacity and the value is positive. It may read up to 3 values, and puts at the position size; Size starts at 3, that is, the first index which the value is 0, and goes up to 5. So it is the unused positions.

The correct answer is:

3. Reads up to 3 values and places them in the array in the unused space.

3 0
3 years ago
Keeping memos on your checks is important because they
Anastasy [175]

Answer: they can determine how you get paid, so if you get a memo about schedules, and you work hourly, you keep keep that memo to show that you are following direction in hours, which result in getting paid more/less.

8 0
3 years ago
Data Structure in C++
agasfer [191]

The code .cpp is available bellow

#include<iostream>

using namespace std;

//declaring variables

void merge(int* ip, int sz, int* opt, bool opt_asc); //merging

int* mergesort(int* ip, int sz);

void mergesort(int *ip, int sz, int* opt, bool opt_asc);

void merge(int* ip, int sz, int* opt, bool opt_asc)

{

  int s1 = 0;

  int mid_sz = sz / 2;

  int s2 = mid_sz;

  int e2 = sz;

  int s3 = 0;

  int end3 = sz;

  int i, j;

   

  if (opt_asc==true)

  {

      i = s1;

      j = e2 - 1;

      while (i < mid_sz && j >= s2)

      {

          if (*(ip + i) > *(ip + j))

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j--;

          }

          else if (*(ip + i) <= *(ip + j))

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i++;

          }

      }

      if (i != mid_sz)

      {

          while (i < mid_sz)

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i++;

          }

      }

      if (j >= s2)

      {

          while (j >= s2)

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j--;

          }

      }

  }

  else

  {

      i = mid_sz - 1;

      j = s2;

      while (i >= s1 && j <e2)

      {

          if (*(ip + i) > *(ip + j))

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i--;

          }

          else if (*(ip + i) <= *(ip + j))

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j++;

          }

      }

      if (i >= s1)

      {

          while (i >= s1)

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i--;

          }

      }

      if (j != e2)

      {

          while (j < e2)

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j++;

          }

      }

  }

   

  for (i = 0; i < sz; i++)

      *(ip + i) = *(opt + i);

}

int* mergesort(int* ip, int sz)

{

  int* opt = new int[sz];

   

  mergesort(ip, sz, opt, true);

  return opt;

}

void mergesort(int *ip, int sz, int* opt, bool opt_asc)

{

  if (sz > 1)

  {

      int q = sz / 2;

      mergesort(ip, sz / 2, opt, true);

      mergesort(ip + sz / 2, sz - sz / 2, opt + sz / 2, false);

      merge(ip, sz, opt, opt_asc);

  }

}

int main()

{

  int arr1[12] = { 5, 6, 9, 8,25,36, 3, 2, 5, 16, 87, 12 };

  int arr2[14] = { 2, 3, 4, 5, 1, 20,15,30, 2, 3, 4, 6, 9,12 };

  int arr3[10] = { 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 };

  int *opt;

  cout << "Arays after sorting:\n";

  cout << "Array 1 : ";

  opt = mergesort(arr1, 12);

  for (int i = 0; i < 12; i++)

      cout << opt[i] << " ";

  cout << endl;

  cout << "Array 2 : ";

  opt = mergesort(arr2, 14);

  for (int i = 0; i < 14; i++)

      cout << opt[i] << " ";

  cout << endl;

  cout << "Array 3 : ";

  opt = mergesort(arr3, 10);

  for (int i = 0; i < 10; i++)

      cout << opt[i] << " ";

  cout << endl;

  return 0;

}

4 0
4 years ago
Suppose that f is a function with a prototype like this:
GrogVix [38]

Answer:

C. node*.

Explanation:

A linked list is a linear data structure where which contains elements stored at non contiguous memory location.They are linked to each other.Every node in a linked list is a pointer.Node consists of a data element and a pointer of node type which contains the address of the next node.

So to connect all the nodes we need the node to be a pointer.That's how we can connect them by address.

8 0
3 years ago
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