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Mandarinka [93]
2 years ago
7

// Problem 7: inputMatrix (10 points)// Ask the user for each element of the 3X3 matrix and store the elements in"matrixA[][]"//

Display the matrix in the following form:// matrixA =// 1 2 3// 4 5 6// 7 8 9// The user may input any inetgers for matrix elements, not necessarily same asexample above.void inputMatrix(int matrixA[3][3]){}
Computers and Technology
1 answer:
Firdavs [7]2 years ago
4 0

Answer:

Following are the program for the above question:

Explanation:

#include <stdio.h> //header file

void inputMatrix(int matrixA[3][3]) //inputmatrix function to take the inputs and display the outputs.

{

  int i,j; //declare the symbols.

//loop to take the inputs.

 for(i=0;i<3;i++)

  for(j=0;j<3;j++)

   scanf("%d",&matrixA[i][j]);

//loop to print the output

 for(i=0;i<3;i++)

 {

  for(j=0;j<3;j++)

   printf("%2d",matrixA[i][j]);

  printf("\t");

 }

}

int main() //main function

{

  int matrixA[3][3]; //matrix declaration.

inputMatrix(matrixA); //calling function.

   return 0;

}

Output:

  • The above code take the 9 inputs from the user and display the output as described question format.

Code Explanation:

  • There is one user-defined function that takes the double dimension array as the argument.
  • Then it takes the 9 value for the double dimension array.
  • Then it displays the value to the user in that format which is defined by the question.
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8 0
2 years ago
Counting calculus students. About A university offers 3 calculus classes: Math 2A, 2B and 2C. In both parts, you are given data
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Answer:

Part (a) n(A∩B∩C) = 4

Part (b) n(A∪B∪C) = 54

Explanation:

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n(A∩B) = no. of students who took both Math 2A and 2B

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∩ represents Intersection and ∪ represents Union

Part (a)

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(A∩C) - n(B∩C) + n(A∩B∩C)  

Where n(A∩B∩C) represents the number of students who took all three classes and n(A∪B∪C) represents the total number of students in group A

157 = 51 + 80 + 70 - 15 - 20 - 13 + n(A∩B∩C)

Re-arranging the equation to solve for n(A∩B∩C) since we want to find out those students who took all three classes

n(A∩B∩C) = 157 - 51 - 80 - 70 + 15 + 20 + 13

n(A∩B∩C) = 4

So there are 4 students in group A who took all three classes

Part (b)

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(A∩C) - n(B∩C) + n(A∩B∩C)

This time we are given n(A∩B∩C) students who took all three classes and want to find n(A∪B∪C) that is total number of students

n(A∪B∪C) = 28 + 28 + 25 - 11 - 9 - 10 + 3

n(A∪B∪C) = 54

So there are total 54 students in group B

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garri49 [273]
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