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.
- m<1 = 145 | Supplementary
- m<3 = 35 | Vertical
- m<4 = 145 | Supplementary
- m<5 = 145 | (With Angle 4) If parallel then alternate interior angles congruent
- m<6 = 35 | (With Angle 5) Supplementary
- m<7 = 35 | (With Angle 6) Vertical
- m<8 = 145 | (With Angle 7) Supplementary
Hope it helps <3
(If it does, maybe brainliest :) Need one more for rank up)
Answer:
3126
Step-by-step explanation:
Because it is so much bigger than the other numbers.
