Answer:
ez
Step-by-step explanation:
start at the point you were given, and use the slope. remember rise over run.
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>.
Answer:
hey did you ever found the answer because i need help with this question too
Step-by-step explanation:
Answer:
3, 2, 1, 0, literally anything below 4
Step-by-step explanation:
The "mouth" < or > eats the bigger number, if 4 is bigger than x, x must be smaller than 4
Isolate the w
w/3 = 5
Note the equal sign. What you do to one side, you do to the other.
Multiply 3 to both sides
w/3(3) = 5(3)
w = 5(3)
Multiply
w = 15
15 is your answer for w
hope this helps
