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d1i1m1o1n [39]
2 years ago
11

Write a code segment that displays the values of the integers x, y, and z on a single line, such that each value is right-justif

ied with a field width of 6.
Computers and Technology
1 answer:
Elena L [17]2 years ago
3 0

Answer:

x = int(input ("enter first number: "))

y = int(input ("enter second number: "))

z = int(input ("enter third number: "))

print('%6d %6d %6d' %(x,y,z))

Explanation:

Using python programming language we receive three integers variables (x,y,z) then using string formatting (%6) which specifies that the output should be right justified with a width of 6, the values are printed out.

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Els
galina1969 [7]

Answer:

ds2d2d2d2dd

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3 years ago
Which of the following is an example of a complex formula?
myrzilka [38]

Answer:

=A1<=A14

Explanation:

complex formula in excel is which contain more than 1 mathematical operators . An order of mathematical operations are important to understand

there are different type of operators

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  • Comparison operators
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  • Operators reference

here Comparison operator is an example of complex formula . Comparison operator returns TRUE/FALSE it is use to compare two values

= Equal to

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3 years ago
Which of the following is an image that enhances comunaction? Illustration, logo, maps, or visual cue
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The answer is A.  Illustration

6 0
3 years ago
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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