Answer:
isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.
Step-by-step explanation:
Let
denote a set of elements.
would denote the set of all ordered pairs of elements of
.
For example, with
,
and
are both members of
. However,
because the pairs are ordered.
A relation
on
is a subset of
. For any two elements
,
if and only if the ordered pair
is in
.
A relation
on set
is an equivalence relation if it satisfies the following:
- Reflexivity: for any
, the relation
needs to ensure that
(that is:
.)
- Symmetry: for any
,
if and only if
. In other words, either both
and
are in
, or neither is in
.
- Transitivity: for any
, if
and
, then
. In other words, if
and
are both in
, then
also needs to be in
.
The relation
(on
) in this question is indeed reflexive.
,
, and
(one pair for each element of
) are all elements of
.
isn't symmetric.
but
(the pairs in
are all ordered.) In other words,
isn't equivalent to
under
even though
.
Neither is
transitive.
and
. However,
. In other words, under relation
,
and
does not imply
.
Answer:
For example, slide ∠ 1 down the transversal and it will coincide with ∠2. are equal in measure. If two parallel lines are cut by a transversal, the corresponding angles are congruent. If two lines are cut by a transversal and the corresponding angles are congruent, the lines are parallel.
Step-by-step explanation:
I am not going to give you the answer, but I am going to tell you how to do it. You first put 18 over 1, and multiply that by one half. (You multiply across.)
Answer:
In what article are you most likely to find the sentence “The coma streamed out until it formed a tail of dust and light”?
“Comets in the Sky”
“How to Punctuate Sentences”
“Water Under the Bridge”
“Medical Histories”
Step-by-step explanation: