Answer:
Remember that a set is a base for some topology on if satisfy the following properties:
1.
2. For , If then exist such that .
Now, for we verify the above properties:
1. It's clear that
2. Let . Without loss of generality suppose that . Then , this implies that and and .
Then, B satisfy the two properties. This show that B is a basis for a topology in
#4 vertex form
and
vertex (h,k)
Step-by-step explanation:
vertex form
-2(x-3)^2+32
vertex
(h,k)
(3,32)
Answer c: 2 and 3
Of course Q is on plane PQR. If S is as well, then QS is as well.