What is |2x-16|=10 ?

2 answers:

8 0

Well, first off, let's just temporarily forget about the absolute value. All we have is 2x - 16 = 10. First we solve this.

2x-16=10

Combine like terms

2x=26

Isolate the variable through division

x=13.

Now we remember those absolute value brackets. When we have an absolute value sign, most of the time it is going to mean that we include the answer you got, and the opposite of that number as well. So in this case, x doesn't just equal 13, but it also equals -13.

x=13,-13

Hope this helped!

-TTL

antoniya [11.8K]2 years ago

4 0

x=13 or x=3

hope this helps

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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

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