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Alecsey [184]
3 years ago
14

Please solve this question quick! The anwser is not 19 or 12

Mathematics
1 answer:
Viktor [21]3 years ago
8 0

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 17


Thank you, please give Brainliest! :)

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liubo4ka [24]

Answer:

There is only one real zero and it is located at x = 1.359

Step-by-step explanation:

After the 4th iteration the solution was repeating the first 3 decimal places.  The formula for Newton's Method is

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If our function is

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f'(x)=5x^4+1

I graphed this on my calculator to see where the zero(s) looked like they might be, and saw there was only one real one, somewhere between 1 and 2.  I started with my first guess being x = 1.

When I plugged in a 1 for x, I got a zero of 5/3.  

Plugging in 5/3 and completing the process again gave me 997/687

Plugging in 997/687 and completing the process again gave me 1.36976

Plugging in 1.36976 and completing the process again gave me 1.359454

Plugging in 1.359454 and completing the process again gave me 1.359304

Since we are looking for accuracy to 3 decimal places, there was no need to go further.

Checking the zeros on the calculator graphing program gave me a zero of 1.3593041 which is exactly the same as my 5th iteration!

Newton's Method is absolutely amazing!!!

5 0
4 years ago
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