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3 years ago

PLEASE HEELLLLP WILL MARK BRAINLIEST

Mathematics

2 answers:

Mrrafil [7]3 years ago

6 0

Answer:

312km

Step-by-step explanation:

1. Multiply. 4800 x 7% = 3744

2. Divide. 3744 / 12 = 312km sq.

So the area after 12 years is 312 km sq.

Gwar [14]3 years ago

4 0

Answer:

2,009.2

Step-by-step explanation:

4,800×(1−7%)^(12)

2,009.2

hope it helps

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SOLVE THIS PROBLEM ASAP PLS

deff fn [24]

Answer:

See attached

Step-by-step explanation:

-> Also see attached

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

7 0

3 years ago

Evaluate the function f(x)=x^2+6x+2 at the given values of the independent variable and simplify.

____ [38]

Answer:

a) 18

b)x^2+10x+18

c)x^2 -6x+2

Step-by-step explanation:

This is a case of plugging in the value into f(x).

a) f(-8)= -8^2 + 6(-8) +2

f(-8)= 64 + (-48) +2

f(-8)=64 + (-46)

f(-8)=18

b) f(x+2)= (x+2)^2+6(x+2)+2

So here I'll take a break to explain what's going on, because x+2 is a binomial meaning two terms and it is being squared I have to multiply the whole thing by itself. Meaning: (x+2) x (x+2), this is also known as foiling!! and for the next part its distributing 6 into x and 2.

f(x+2)= x^2+4x+4+6x+12+2

I'll reorder it

f(x+2)= x^2+4x+6x+12+2+4

f(x+2)= x^2+10x+18

c) f(-x)= -x^2+6(-x) +2

f(-x)= x^2 -6x+2

6 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

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