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77julia77 [94]
3 years ago
13

Give three reasons why you think Clip Art is so widely used in many different types of documents.

Computers and Technology
2 answers:
Kipish [7]3 years ago
7 0

here is the answer for ur question

Well Here Are Some Advantages To Clip Art,

There Is A Large Variety Of Clip Art To Choose From.

You Don't Need To Buy Any Special Equipment Like A Scanner.

It Takes No Artistic Skills To Produce.

Most Clip Art Is Royalty And Copyright Free.

It Doesn't Have To Be Developed From Scratch.

Hope I Helped




irinina [24]3 years ago
6 0
Well Here Are Some Advantages To Clip Art,
There Is A Large Variety Of Clip Art To Choose From.
You Don't Need To Buy Any Special Equipment Like A Scanner.
It Takes No Artistic Skills To Produce.
Most Clip Art Is Royalty And Copyright Free.
It Doesn't Have To Be Developed From Scratch.
Hope I Helped
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Write a class called MagicSquare, which contains a single method called check that accepts a two-dimensional array as an argumen
ira [324]

Answer:

public class MagicSquare {

   public static void main(String[] args) {

       int[][] square = {

               { 8, 11, 14, 1},

               {13, 2, 7,12},

               { 3, 16, 9, 6},

               {10, 5, 4, 15}

       };

       System.out.printf("The square %s a magic square. %n",

               (isMagicSquare(square) ? "is" : "is not"));

   }

   public static boolean isMagicSquare(int[][] square) {

       if(square.length != square[0].length) {

           return false;

       }

       int sum = 0;

       for(int i = 0; i < square[0].length; ++i) {

           sum += square[0][i];

       }

       int d1 = 0, d2 = 0;

       for(int i = 0; i < square.length; ++i) {

           int row_sum = 0;

           int col_sum = 0;

           for(int j = 0; j < square[0].length; ++j) {

               if(i == j) {

                   d1 += square[i][j];

               }

               if(j == square.length-i-1) {

                   d2 += square[i][j];

               }

               row_sum += square[i][j];

               col_sum += square[j][i];

           }

           if(row_sum != sum || col_sum != sum) {

               return false;

           }

       }

       return d1 == sum && d2 == sum;

   }

}

5 0
3 years ago
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