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prisoha [69]
3 years ago
5

Which of the following best explains the different between Cut and Copy? a. When you copy text you are permanently deleting it,

whereas when you cut text you are removing it, but placing it on the Clipboard for later use. b. When you copy text, it remains in its original location but it is also placed on the Clipboard. Cutting text removes it from its original location and places it on the Clipboard. c. When you cut text you are permanently deleting it, whereas when you copy text you are removing it, but placing it on the Clipboard for later use. d. When you cut text, it remains in its original location but it is also placed on the Clipboard. Copying text removes it from its original location and places it on the Clipboard.
Computers and Technology
2 answers:
Vilka [71]3 years ago
5 0
B. When you copy text it remains in its original location and places it on the clipboard. Cutting text removes it from its original location and places it on the clipboard

wolverine [178]3 years ago
5 0

Answer:

I believe the answer is B

Explanation:

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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)

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   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.

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