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 solutions
Step-by-step explanation:
Since the two terms have the same base, we are able to use the rule for subtracting logarithms:
Therefore, the equation can be written as:
By using the definition of a logarithm we can say that:
When plugging this solution in, you find that the term has x-6 evaluate to a number less than 0. This is not included in the domain of log functions, so is not a valid solution. This means that there are no solutions.
Where are the triangles so I can compare to find the triangle congruence?
Answer:
heyyyyyheyyyyyyyyyyyyyyyyyyyyyyyyyyheyyyyyyyyyyyyyyyyyyyyyyyyyyy
D that's the answer dont worry Ik I got u
Answer:
a = 28.14 (Estimated)
Step-by-step explanation:
7a + 50 = 247
=> 7a = 197
=> a = 28.1428571429 (Using Calculator)
=> a = 28.14 (Estimated)
Hoped this helped.