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olya-2409 [2.1K]
3 years ago
12

If f(x) = (x+1)^-1 and g(x) = x-2, what is the domain of f(x) divided by g(x)

Mathematics
1 answer:
Setler [38]3 years ago
6 0
F(x)=(x+1)^-1

f(x)=1/(x+1)  so 

f(x)/g(x) is

1/[(x+1)(x-2)]

Since division by zero approaches infinity, it is not a real value, and is said to be undefined.  Division by zero is "not allowed", which just really means that it does not have a real value.  So in this case x cannot equal -1 or 2 so the domain is:

All real numbers except -1 or 2...which can be expressed as:

(-oo,-1),(-1,2),(2,+oo) 
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Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r
erastova [34]

Answer:

The probability is \frac{56!}{64!}

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, 64 \choose 8 .

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function  f : A \rightarrow A , with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}

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2 years ago
Write 5^3 x √5 as a single power of 5
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Type a= 8x60

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