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>.
Y would be 16,
y=kx
6=3k
2=k
y=2(8)
y=16
Answer:
24
Step-by-step explanation:
6×5=30
6*4=24
you have to find more then half which would be 3:6 so it's more then 15 but less then 30
Answer:
11 1/2 teaspoons
Step-by-step explanation:
1 2/4 basil= 1.5 tsp If you multiply this by two to double the recipe you will have 3 teaspoons of basil.
1/4*2= 2/4 pepper
4*2=8 parsley
After you double everything you need to add them together:
3+2/4+8
Which brings you to the final answer of 11 2/4 teaspoons or if you simplify it.... 11 1/2 teaspoons
You can go through the set and test each one out:
x=0:
2(0)+3 ≥ 7? Or 3≥7? No
x=0.6:
2(0.6)+3≥7? 4.2≥7? No.
x=2:
2(2)+3≥7? 7≥7? Yes.
x=4:
2(4)+3≥7? 11≥7? Yes.
x=6:
2(6)+3≥7? 15≥7? Yes.
The set is {2, 4, 6}.
