Consider all parabolas:
1.
When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.
2.
When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.
3.
When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.
4.
When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.
5.
When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.
6.
When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.
Answer:
-3r + 15 ---> answer
Step-by-step explanation:
r < 5
You are going to multiply both sides with 3. The reason being is that 3 is a positive number and the equality sign will not change if you use +3.
3r < 15
Now, subtract 15 from both sides, you will get this:
3r < 15
-15 -15
-------------
3r — 15 < 0
Lastly, using the Modulus function, we are going to add a negative sign to the content of our previous step because it's already negative.
So, -3r + 15 is the final solution if r < 5 in the given equation of l3r-15l
Maximum Material Condition or for short, MMC, is a feature of size symbol that describes the condition of a feature or part where the maximum amount of material (volume/size) exists within its dimensional tolerance.
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.
