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o-na [289]
3 years ago
5

What is a Windows feature that allows you to temporarily store text?

Computers and Technology
1 answer:
nata0808 [166]3 years ago
3 0
This would be copy and paste.

When you select a text, you can use your keys CTRL and C to copy that same selected text. After this, you can use the paste feature by selecting CTRL and V. Doing this, your selected text will be copied and pasted to a certain location. 
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Assume you are given a boolean variable named isNegative and a 2-dimensional array of ints that has been created and assigned to
Anon25 [30]

Answer:

#include <iostream>

using namespace std;

int main(){

   int rows, cols;

   bool isNegatives;

   cout<<"Rows: ";

   cin>>rows;

   cout<<"Columns: ";

   cin>>cols;

   int a2d[rows][cols];

   for(int i =0;i<rows;i++){

   for(int j =0;j<cols;j++){

       cin>>a2d[i][j];}    }

   int negatives, others = 0;

   for(int i =0;i<rows;i++){

   for(int j =0;j<cols;j++){

       if(a2d[i][j]<0){

           negatives++;}

       else{

           others++;}}    }

   if(negatives>others){

       isNegatives = true;}

   else{

       isNegatives = false;}

   cout<<isNegatives;  

   return 0;

}

Explanation:

For clarity and better understanding of the question, I answered the question from the scratch.

This line declares number of rows and columns

   int rows, cols;

This line declares the Boolean variable

   bool isNegatives;

This line prompts user for rows

   cout<<"Rows: ";

This line gets the number of rows

   cin>>rows;

This line prompts user for columns

   cout<<"Columns: ";

This line gets the number of columns

   cin>>cols;

This line declares the array

   int a2d[rows][cols];

This line gets user input for the array

<em>    for(int i =0;i<rows;i++){</em>

<em>    for(int j =0;j<cols;j++){</em>

<em>        cin>>a2d[i][j];}    }</em>

This line declares and initializes number of negative and others to 0

   int negatives, others = 0;

The following iteration counts the number of negatives and also count the number of non negative (i.e. others)

<em>    for(int i =0;i<rows;i++){</em>

<em>    for(int j =0;j<cols;j++){</em>

<em>        if(a2d[i][j]<0){</em>

<em>            negatives++;}</em>

<em>        else{</em>

<em>            others++;}}    }</em>

This checks of number of negatives is greater than others

   if(negatives>others){

If yes, it assigns true to isNegatives

       isNegatives = true;}

   else{

If otherwise, it assigns false to isNegatives

       isNegatives = false;}

This prints the value of isNegatives

   cout<<isNegatives;  

<em>See attachment</em>

Download cpp
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Answer: your document will be inaccessible

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Answer:

D) x=8

Explanation:

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