Answer: y<x-1
Step-by-step explanation:
So first we have to find the blue line. y=mx+b is the equation. For slope it is 0--1 which is 1 and 1-0 which is 1 so the slope is just one or x now we need to find the y intercept which is just -1. So y<(is less than)x-1! Hope this helps :)
Answer:
a. a = 1, b = -5, c = -14
b. a = 1, b = -6, c = 9
c. a = -1, b = -1, c = -3
d. a = 1, b = 0, c = -1
e. a = 1, b = 0, c = -3
Step-by-step explanation:
a. x-ints at 7 and -2
this means that our quadratic equation must factor to:

FOIL:

Simplify:

a = 1, b = -5, c = -14
b. one x-int at 3
this means that the equation will factor to:

FOIL:

Simplify:

a = 1, b = -6, c = 9
c. no x-int and negative y must be less than 0
This means that our vertex must be below the x-axis and our parabola must point down
There are many equations for this, but one could be:

a = -1, b = -1, c = -3
d. one positive x-int, one negative x-int
We can use any x-intercepts, so let's just use -1 and 1
The equation will factor to:

This is a perfect square
FOIL:

a = 1, b = 0, c = -1
e. x-int at 
our equation will factor to:

This is also a perfect square
FOIL and you will get:

a = 1, b = 0, c = -3
Answer:

Step-by-step explanation:

Answer:
55 1/50 or 2751/50
Step-by-step explanation:
Convert both expressions to improper form
Make their denominators the same
Multiply the numerators
Simplify the fraction if needed
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.