a. Only a Hamiltonian path
One such path is
1 → 2 → 0 → 4 → 3
which satisfies the requirement that each vertex is visited exactly once.
There is no Hamiltonian circuit, however, since it is impossible for any Hamiltonian path on this graph to visit vertex 0 exactly once.
Answer:
where is the graph if you can put it i will give you the answer
Step-by-step explanation:
Negative since the answer would be -13
Short Answer A
Comment
It's a good thing that the domain is confined or the graph is. That function is undefined if x < - 3
At exactly x = - 3, f(x) = 0 and that's your starting point.
So look what happens to this graph. If x = 0, f(0) = y = 3. So we are starting to see that as x get's larger so does f(x). The graph tells us that x = 0 is bigger than x = - 3.
Let's keep on plugging things in.
As x increases to 5, f(5) = 5. x = 5 is larger than x = 0, and f(5) > 3.
One more and then we'll start drawing conclusions. If x = 9 then f(9) = y = 6.
x = 9 is larger than x = 5. f(9) = 6 is just larger than f(5) which is 5
OK I think we should be ready to look at answers. There's nothing there that makes the answer anything but a. Let's find out what the problem is with the rest of the choices.
B
The problem with B is that as x increases, f(x) does not decrease. We didn't find one example of that. So B is wrong.
C
C has exactly the same problem as B.
D
The second statement in D is incorrect. As x increases f(x) never decreases. No example showed that.
The answer is A <<<< Answer.
2(-11)-19
-22-19
-41
D is the answer
