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.
Answer no answers
Step-by-step explanation:
<h2>
Answer:</h2>
We need to determine the equation of both lines first.
- Line 1: <em>y = -2x + 3</em>
- Line 2: <em>y = -1/3x - 2</em>
Now that we know the equations, we can set up a system of equations for this graph where both equations are in standard form.
Line 1:
Line 2:
<em>Final answer:</em>
namely, what is 1/2 of 3/4, well, is just their product.
