Answer:
B. 5=5
Step-by-step explanation:
Answer:
b=23 degrees
Step-by-step explanation:
a was 66 degrees so b=23 degrees 180-90-66=23
The line y = mx + c has a slope of m.
Parallel line:
A parallel line will also have a slope of m, and should be of the form
y = mx + k
Because the line passes through the point (a,b), therefore
ma + k = b
k = b - ma
The equation of a parallel line is
y = mx + b - ma
= m(x-a) + b
Perpendicular line:
A perpendicular line will have a slope of -1/m, because the product of slopes should be -1.
The equation for a perpendicular line is of the form
y = -(1/m)x + k
Because the line passes through (a,b), therefore
-(1/m)a + k = b
k = b + a/m
The equation for a perpendicular line is
y = -(x/m) + b + a/m
= -(1/m)(x-a) + b
Answer:
Parallel line:

Perpendicular line:
Step-by-step explanation:
3 to 8= 5
8 to 17= 9
17 to 30= 13
30 to 47= 17
47 to 68= 21
68 to 93= 25
93 to 122= 29
122 to 155= 33
155 to 192= 37
difference is 4
solve by

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.