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:
v = 0
Step-by-step explanation:
12 = 12 + v / 8
0 = 1/8v
v = 0
Step-by-step explanation and answer:
For this you just need to plug in x for y = f(1/5x)
1/5(-4) = -4/5 or -0.8
1/5(-1) = -1/5 or -0.2
1/5(0) = 0
1/5(3) = 3/5 or 0.6
1/5(6) = 6/5 or 1.2 or 1 1/5
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Answer:
Solution
To solve the problem first we have simplify the improper fraction.
⇒ 33/4 ÷ 4 pounds.
= (4 × 3 + 3) /4 ÷ 4 pounds.
= 15/4 ÷ 4 pounds.
= 15/4 × 1/4 pounds.
= 15/16 pounds.
Now we know that, 1 pound = 16 ounces.
Therefore, 15/16 pounds .
= 15/16 × 16 ounces.
= 15/1
= 15
= 15 ounces.
Answer: 15 ounces.
