Answer: 4i
Step-by-step explanation:
Supposse that the distance from the point 
to the point 
is equal to the distance from to the point 
. Then, by the formula of the distnace we must have
cancel the square root and the 
's, and then expand the parenthesis to obtain
then, simplifying we obtain
therfore we must have
this means that the points satisfying the propertie must have first component equal to 5. So we can give a lot of examples of such points: 
. The set of this points give us a straight line and the points (3,0) and (7,0) are symmetric with respect to this line.
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.
Step-by-step explanation:
-8.0-(-0.4+0.6)÷2
= -0.8-(0.2)÷2
= -0.8 -0.2 ÷2
= -0.9
use Bodmas method
Answer:
reflection then translation
Step-by-step explanation:
you reflected of the y axis and then you translated 2 left and 5 up which brings you to 3
