Which expression is equivalent to the following function?

Mathematics

2 answers:

nevsk [136]3 years ago

6 0

<u>Answer </u>

B. f(x) = - (y/x)⁵

<u>Explanation</u>
A negative power means the reciprocal of that number raised to the positive value .
For example, a⁻ⁿ = 1/aⁿ

So, for the question above.
f(x) = -(x/y)⁻⁵
= - 1/(x/y)⁵
= - (y/x)⁵

mylen [45]3 years ago

4 0

The answer for the problem shown in the figure attached, is the option B, which is:
B) f(x)=(-y/x)^5
The explanatio for the exercise is shown below:
1. You have the following function given in the problem above:
f(x)=-(x/y)^-5
2. The negative exponent -5, indicates that the x must be in the denominator and the y must be in the numerator, and the exponent must be positive as following:
f(x)=(-y/x)^5

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What is the answer for 4x+y=0

leva [86]

Y=-4x because all you have to do is move the variable to the right by adding its opposite to both sides. So 4x+y-4x=0-4x. Then eliminate the opposites y=0-4x

Your welcome

7 0

3 years ago

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

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