Answer:
Proved
Step-by-step explanation:
To prove that every point in the open interval (0,1) is an interior point of S
This we can prove by contradiction method.
Let, if possible c be a point in the interval which is not an interior point.
Then c has a neighbourhood which contains atleast one point not in (0,1)
Let d be the point which is in neighbourhood of c but not in S(0,1)
Then the points between c and d would be either in (0,1) or not in (0,1)
If out of all points say d1,d2..... we find that dn is a point which is in (0,1) and dn+1 is not in (0,1) however large n is.
Then we find that dn is a boundary point of S
But since S is an open interval there is no boundary point hence we get a contradiction. Our assumption was wrong.
Every point of S=(0, 1) is an interior point of S.
<span>D: a line passing through the points (2, –6) and (4, –16)</span>
Answer:
16
Step-by-step explanation:
4*4
Answer:
Yes
Step-by-step explanation:
To be a function, the graph must pass the vertical line test.
The vertical line test is simple. Imagine you draw an infinite number of vertical lines. In all of these lines, it should never hit the function twice in the same x value.
For example, if it hits (4, 2) and (4, 7), the graph is not a function because there are two points with the same x value.
