so, you just use the x's from the table and plug them into the equation to find the y.
y=(1)+9
y=10
y=(2)+9
y=11
y=(3)+9
y=12
y=(4)+9
y=13
i hope this helps :)
Answer:
80% of 20 is 16.
Step-by-step explanation:
Answer:
Plot points at (0,1) and (-3,3) and draw a line going through both points.
Step-by-step explanation:
Let's start by graphing the y intercept.
y=mx+b
m is the slope. b is the y intercept. Since the equation is y=-2/3x+1, we can conclude the y intercept is 1. We graph a point at (0,1).
If you didn't know the y intercept is where the line intercepts the y-axis.
Now, from the point (0,1) we go up 2 and to the left 3 as it is a negative slope. We reach (-3,3). Plot a point there. Then draw a line going through both points. There's your line!
Answer:
Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.
<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that 
. Thus, A↔A.
<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that 
. In this equality we can perform a right multiplication by 
and obtain 
. Then, in the obtained equality we perform a left multiplication by P and get 
. If we write 
and 
we have 
. Thus, B↔A.
<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have and from B↔C we have 
. Now, if we substitute the last equality into the first one we get
.
Recall that if P and Q are invertible, then QP is invertible and 
. So, if we denote R=QP we obtained that
. Hence, A↔C.
Therefore, the relation is an <em>equivalence relation</em>.
