Answer with Step-by-step explanation:
We are given S be any set which is countable and nonempty.
We have to prove that their exist a surjection g:N
Surjection: It is also called onto function .When cardinality of domain set is greater than or equal to cardinality of range set then the function is onto
Cardinality of natural numbers set =( Aleph naught)
There are two cases
1.S is finite nonempty set
2.S is countably infinite set
1.When S is finite set and nonempty set
Then cardinality of set S is any constant number which is less than the cardinality of set of natura number
Therefore, their exist a surjection from N to S.
2.When S is countably infinite set and cardinality with aleph naught
Then cardinality of set S is equal to cardinality of set of natural .Therefore, their exist a surjection from N to S.
Hence, proved
Answer:
A: g(x)
Step-by-step explanation:
The equation is technically in vertex form "y = a(x – h)² + k", where the vertex is (h, k).
For the domain and range to be as stated in the problem, your "vertex" needs to be (2, 3), and so you want y = a(x – 2)² + 3 which is answer choice A
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Answer:
coefficients- 3 2 -1
constant terms- -1
like terms- 3r & 2r
Step-by-step explanation:
:)