Answer:
Step-by-step explanation:
As the statement is ‘‘if and only if’’ we need to prove two implications
is surjective implies there exists a function 
such that 
.- If there exists a function such that , then is surjective
Let us start by the first implication.
Our hypothesis is that the function is surjective. From this we know that for every 
there exist, at least, one 
such that 
.
Now, define the sets 
. Notice that the set 
is the pre-image of the element 
. Also, from the fact that 
is a function we deduce that 
, and because the sets are no empty.
From each set choose only one element 
, and notice that 
.
So, we can define the function 
as 
. It is no difficult to conclude that 
. With this we have that 
, and the prove is complete.
Now, let us prove the second implication.
We have that there exists a function such that .
Take an element , then 
. Now, write 
and notice that . Also, with this we have that 
.
So, for every element we have found that an element (recall that ) such that , which is equivalent to the fact that is surjective. Therefore, the prove is complete.
Answer:
9
Step-by-step explanation:
This is because 21/7 is 3. So, you use the scale factor of 3 to divide 27/3 to get x which is 9.
False it is not it just said that if B is 100 then A is 35
Answer:
cot∅ = (-2√30)/7.
Step-by-step explanation:
Given the value of csc∅ = -13/7 and ∅ is in quad III.
We know y = r sin∅ and r > 0. So csc∅ = r/y = -13/7 = 13/(-7).
It means y = -7, r = 13.
We know x² + y² = r².
x² = r² - y²
x² = (13)² - (-7)² = 169 - 49 = 120.
x = √120 = 2√30.
we know cot∅ = x/y = (2√30)/(-7) = (-2√30)/7.
Hence, cot∅ = (-2√30)/7.
