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2 years ago

The radius of a circle is 10 m. Find its area to the nearest whole number.

Mathematics

1 answer:

Bess [88]2 years ago

7 0

<h3>given:</h3>

radius= 10 m

the area of the given circle.

<h3>solution:</h3>

$a = \pi {r}^{2}$

$a = \pi \times {10}^{2}$

$a = 314.15927 \: {m}^{2}$

$a = 314 \: {m}^{2}$

therefore, the area of the given circle is 314 square meters. 

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2 years ago

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charle [14.2K]

Answer:

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Step-by-step explanation:

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3 years ago

Help math sucks THESE MAKE MY BRAIN DIE

Flura [38]

Answer:

A ∩ B = {1, 3, 5}

A - B = {2, 4}

Step-by-step explanation:

The given problem regards sets and set notation, a set can simply be defined as a collection of values. One is given the following information:

A = {1, 2, 3, 4, 5}

B = {1, 3, 5, 6, 9}

One is asked to find the following:

A ∩ B,

A - B

1. Solving problem 1

A ∩ B,

The symbol (∩) in set notation refers to the intersection between the two sets. It essentially asks one to find all of the terms that two sets have in common. The given sets (A) and (B) have the values ({1, 3, 5}) in common thus, the following statement can be made,

A ∩ B = {1, 3, 5}

2. Solving problem 2

A - B

Subtracting two sets is essentially taking one set, and removing the values that are shared in common with the other set. Sets (A) and (B) have the following values in common ({1, 3, 5}). Thus, when doing (A - B), one will omit the values ({1, 3, 5}) from set (A).

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2 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

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

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2 years ago

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3 years ago