Answer:
Step-by-step explanation:
D)
Hope that helps!
Answer:
you are correct, they are supplementary
Step-by-step explanation:
Answer:
Let the integers are p and q.
In case both integers are positive or negative, you add the numbers up and apply their sign to the sum.
<u>Examples:</u>
- 10 + 88 = 98
- -20 + (-15) = -35
In case one of the integers is negative and one positive.
Use absolute value in this case.
Subtract the numbers and apply the sign of the number with the greater absolute value.
<u>Examples:</u>
- - 10 + 15 = |15 - 10| = |5| = 5
- - 15 + 10 = - |15 - 10| = - |5| = -5
Answer:
Matrix multiplication is not conmutative
Step-by-step explanation:
The matrix multiplication can be performed if the number of columns of the first matrix is equal to the number of rows of the second matrix
Let A with dimension mxn and B with dimension nxp represent two matrix
The multiplication of A by B is a matrix C with dimension mxp, but the multiplication of B by A is can't be calculated because the number of columns of B is not the number of rows of A. Therefore, you can notice that is not conmutative in general.
But even if the multiplication of AB and BA is defined (For example if A and B are squared matrix of 2x2) the multiplication is not necessary conmutative.
The matrix multiplication result is a matrix which entries are given by dot product of the corresponding row of the first matrix and the corresponding column of the second matrix:
![A=\left[\begin{array}{ccc}a11&a12\\a21&a22\end{array}\right]\\B= \left[\begin{array}{ccc}b11&b12\\b21&b22\end{array}\right]\\AB = \left[\begin{array}{ccc}a11b11+a12b21&a11b12+a12b22\\a21b11+a22b21&a21b12+a22b22\end{array}\right]\\\\BA=\left[\begin{array}{ccc}b11a11+b12a21&b11a12+b12a22\\b21a11+b22ba21&b21a12+b22a22\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da11%26a12%5C%5Ca21%26a22%5Cend%7Barray%7D%5Cright%5D%5C%5CB%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db11%26b12%5C%5Cb21%26b22%5Cend%7Barray%7D%5Cright%5D%5C%5CAB%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da11b11%2Ba12b21%26a11b12%2Ba12b22%5C%5Ca21b11%2Ba22b21%26a21b12%2Ba22b22%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CBA%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db11a11%2Bb12a21%26b11a12%2Bb12a22%5C%5Cb21a11%2Bb22ba21%26b21a12%2Bb22a22%5Cend%7Barray%7D%5Cright%5D)
Notice that in general, the result is not the same. It could be the same for very specific values of the elements of each matrix.
Substitute
, so that

Then the resulting ODE in
is separable, with

On the left, we can split into partial fractions:

Integrating both sides gives




Now solve for
:

