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zaharov [31]
3 years ago
10

Adjusting the ______ adjusts the difference in appearance between light and dark areas of the photo.​

Computers and Technology
2 answers:
kupik [55]3 years ago
6 0

Answer: contrast

Explanation:

Adjustments made to the contrast makes images more brighter where light areas becomes lighter and dark areas becomes darker and visible. When contrast is decreased, it makes the picture look flat.

I hope this helps, please mark as brainliest answer.

Svetach [21]3 years ago
4 0

Answer:

contrast

Explanation:

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Which tag will you use if you want your web page to look the same as your source code?
Stolb23 [73]

Answer:

The <samp> element allows you to output sample code as output.

The <pre> element also provides you the choice to output exactly what written in the source code

Explanation:

The two tags affords you the chance of outputting exactly what you have in the source code as your outputs.

6 0
3 years ago
Write a C++ program to count even and odd numbers in array. The array size is 50. The array elements will be entered by the user
vlabodo [156]

Answer:

The program in C++ is as follows:

#include <iostream>

using namespace std;

int main(){

   int numbers[50];

   int evekount = 0, odkount = 0;

   for(int i = 0; i<50;i++){

       cin>>numbers[i];

       if(numbers[i]%2==0){            evekount++;        }

       else{            odkount++;        }

   }

   cout<<"Even Count: "<<evekount<<endl;

   cout<<"Odd Count: "<<odkount<<endl;

   return 0;

}

Explanation:

This declares the integer array of number

   int numbers[50];

This initializes the even count and odd count to 0

   int evekount = 0, odkount = 0;

This iterates from 1 to 50

   for(int i = 0; i<50;i++){

This gets input for the array

       cin>>numbers[i];

This checks for even

<em>        if(numbers[i]%2==0){            evekount++;        }</em>

This checks for odd

<em>        else{            odkount++;        }</em>

   }

This prints the even count

   cout<<"Even Count: "<<evekount<<endl;

This prints the odd count

   cout<<"Odd Count: "<<odkount<<endl;

3 0
3 years ago
How many possible password of length four to eight symbols can be formed using English alphabets both upper and lower case (A-Z
Fynjy0 [20]

Answer:

In a password, symbol/characters can be repeated. first calculate the total

symbols which can be used in a password.

So there are total 26(A-Z),26(a-z),10(0-9) and 2(_,$) symbols.

that is equal to 26+26+10+2=64.

Total number of password of length 4:

here at each place can filled in total number of symbols i.e 64 way for each

place.Then total number of possible password is:

64*64*64*64=16777216

Total number of password of length 5:

here at each place can filled in total number of symbols i.e 64 way for each

place.Then total number of possible password is:

64*64*64*64*64=1073741824

Similarly,

Total number of password of length 6:

64*64*64*64*64*64=68719476736

Total number of password of length 7:

64*64*64*64*64*64*64=4398046511104

Total number of password of length 8:

64*64*64*64*64*64*64=281474976710656

Hence the total number of password possible is:285,942,833,217,536

7 0
3 years ago
Which of these YA (young adult movies is not base on a book is it A wrinkle in time, or Midnight sun,or Love,simon.
ExtremeBDS [4]
I think it’s midnight sun
4 0
3 years ago
Using Module operator, write a java program to print odd number between 0 and 1000​
jolli1 [7]

Answer:

class OddNumber

{

  public static void main(String args[])  

       {

         int n = 1000;  //Store 1000 in Variable n typed integer

         System.out.print("Odd Numbers from 1 to 1000 are:");  // Print headline of output window

         for (int i = 1; i <= n; i++)  //For loop to go through each number till end

          {

            if (i % 2 != 0)  //check if number is even or odd.Not divisible by 2 without reminder means it is odd number

             {

               System.out.print(i + " "); //print odd numbers

             }

          }

       }

}              

5 0
3 years ago
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