Answer:
1/3 is the answer.
Step-by-step explanation:
Tanya prepared 4 different letters to be sent to 4 different addresses.
To solve this we can do the following:
The probability that the 1st letter is in the right envelope is = 
The probability that the 2nd letter is in the wrong envelope is = 
The probability that the 3rd letter is in the wrong envelope is = 
The probability that the 4th letter is in the wrong envelope is = 1
So, the answer becomes:
= 
As we need 4 correct letters in the envelope, we will multiply by 4:

The domain is all the numbers of which the inputs can be put. Here, we have a graph, which simplifies things. We just need to look at the x-values of the graph, or the horizontal line. The graph extends both negatively and positively. Thus since the graph extends both ways (right and left) forever, the answer would be D, or all real numbers.
I would rather use a ruler but it depends on what you are doing.
Answer:
The last choice.
Step-by-step explanation:

The value of k must be restricted from being one that makes the original denominator zero. Hence k ≠ -1 or 5.
Answer:
A:
C -20 -15 -5
F(C) -4 5 23
Step-by-step explanation: