Write a C program to compute Matrix Multiplication of two matrices. Use one dimensional array to store each matrix, where each row is stored after another. Hence, the size of the array will be a product of number of rows times number of columns of that matrix. Get number of row and column from user and use variable length array to initialize the size of the two matrices as well as the resultant matrix. Check whether the two matrices can be multiplied or not. Write a getMatrix() function to generate the array elements randomly. Write a printMatrix() function to print the 1D array elements in 2D Matrix format. Also, write another function product(), which multiplies the two matrices and stores in the resultant matrix. With SEED 5, the following output is generated.
Sample Output
Enter the rows and columns of Matrix A with space in between: 3 5
Enter the rows and columns of Matrix B with space in between: 5 4
Matrix A:
8 6 4 1 6
2 9 7 7 5
1 3 1 1 2
Matrix B:
9 5 4 5
9 9 8 1
4 4 3 5
2 6 2 1
4 5 2 4
Product AxB:
168 146 106 91
161 186 125 81
50 52 37 22
In conclusion, the answer is 5x1
Please give Brainliest answer thanks! :)
we are given
Area is 1.89
now, we can use area formula
now, we can solve for c
we will get
.............Answer
Answer:
the answer is 54522
Step-by-step explanation:
hope it helps
In order to have infinitely many solutions with linear equations/functions, the two equations have to be the same;
In accordance, we can say:
(2p + 7q)x = 4x [1]
(p + 8q)y = 5y [2]
2q - p + 1 = 2 [3]
All we have to do is choose two equations and solve them simultaneously (The simplest ones for what I'm doing and hence the ones I'm going to use are [3] and [2]):
Rearrange in terms of p:
p + 8q = 5 [2]
p = 5 - 8q [2]
p + 2 = 2q + 1 [3]
p = 2q - 1 [3]
Now equate rearranged [2] and [3] and solve for q:
5 - 8q = 2q - 1
10q = 6
q = 6/10 = 3/5 = 0.6
Now, substitute q-value into rearranges equations [2] or [3] to get p:
p = 2(3/5) - 1
p = 6/5 - 1
p = 1/5 = 0.2
I believe the answer is 2(15t).
