Hi help rn plss this is a system of equations question

Mathematics

1 answer:

otez555 [7]3 years ago

8 0

Answer:

x=6, y=2

Step-by-step explanation:

You might be interested in

F (x) = x + 3; g(x) = 1/x What is the domain and range of (f • g)

ArbitrLikvidat [17]

The domain of g alone is {x | x ≠ 0}, and the domain of f is all reals. So the domain of (f ◦ g) is the domain of g {x | x ≠ 0}. (f ◦ g)(x) = 1/x + 3. The range of g(x) = 1/x is actually the same as its domain {y | y ≠ 0}. Adding three, the range of f ◦ g is all reals except for 3, {y | y ≠ 3} The line y = 3 is actually an asymptote (horizontal) to the graph of f ◦ g.

8 0

3 years ago

Emelio wants to solve this equation. x 5 = 35 Help him follow the steps to finish solving for x. 1. Multiplication property of e

antiseptic1488 [7]

Answer: 175

Step-by-step explanation

3 0

3 years ago

Read 2 more answers

two points are given (-1,-4) and (-3,6) use the coordinates of the two points to set up the equation for rate of change the solv

lisabon 2012 [21]

The formula to find the rate of change is y2-y1/x2-x1
6-(-4)/-3-(-1)
=10/-2
=-5

4 0

3 years ago

What's two thirds of a straight line?

Komok [63]

Well if the straight line is equal to 1 then your answer is 2/3 but if the straight line is not equal to 1 you need to know what it is equal to or you could use the equation y = (2/3)x
"x" would be the length of the straight line and "y" would be two-thirds of it

6 0

4 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

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

7 0

3 years ago