Answer:
(3,-3)
Step-by-step explanation:
If a coordinate (x, y) is dilated by a factor k, the resulting coordinate will be (kx, ky)
Given the coordinate (1, -1), if dilated by a factor of 3, the resulting coordinate will be (3(1), 3(-1)) = (3, -3)
Hence the required coordinate will be (3,-3)
Step-by-step explanation (Question 1):
<u>Step 1: (Question 1): Subtract x from both sides.</u>
5x+3=x+13
5x+3−x=x+13−x
4x+3=13
<u>Step 2: (Question 1): Subtract 3 from both sides.</u>
4x+3−3=13−3
4x=10
<u>Step 3: (Question 1): Divide both sides by 4.</u>
4x/4 = 10/4
FIRST ANSWER: x = 5/2
Step-by-step explanation (Question 2):
<u>Step 1: (Question 2): Multiply by LCM</u>
15x^2 - 6 = x
<u>Step 2: (Question 2): Solve 15x^2</u>
15x^2 - 6 = x:
x = 2/3
x = -3/5
<u>Step 3: (Question 2): Solve</u>
ANSWER: x=0.666667 or x=−0.6
See Attachment 1 for question 1 steps (FULL)
See Attachment 2 for question 2 steps (FULL)
Answer:
1) x = 5/2
2) x=0.666667 or x=−0.6
Hope this helps.
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.
Answer:
the answer is 34.
Step-by-step explanation: