Answer:
![\left[\begin{array}{ccc}3&-5 &|12\\4&-2 &|15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%20%20%26%7C12%5C%5C4%26-2%20%20%26%7C15%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
When making a matrix of two equations with the variables x and y, the result will be a matrix with three columns:
- a column for the values of x in each equation
- a column for the values of y in each equation
- a column for the independent values of each equation
since our system of equations is:

we can see that the value for x in the first equation is 3 and in the second equation is 4, thus the first column will have the numbers 3 and 4:
![\left[\begin{array}{ccc}3&&\\4&&\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26%26%5C%5C4%26%26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now for the values of y we hvae -5 in the first equation and -2 in the second equation, we update the matrix with another column with the values of -5 and -2:
![\left[\begin{array}{ccc}3&-5&\\4&-2&\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%26%5C%5C4%26-2%26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Finally, the last column is the independent values of each equation (or the results) in the first equation that number is 12 and in the second equation is 15, thus the matrix is:
![\left[\begin{array}{ccc}3&-5&12\\4&-2&15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%2612%5C%5C4%26-2%2615%5C%5C%5Cend%7Barray%7D%5Cright%5D)
usually there is a line separating the columns for the values of x and y, and the independent values:
![\left[\begin{array}{ccc}3&-5 &|12\\4&-2 &|15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%20%20%26%7C12%5C%5C4%26-2%20%20%26%7C15%5C%5C%5Cend%7Barray%7D%5Cright%5D)
this is the matrix of the system of equations
Twelve people join hands for a circle dance.In how many ways can they do this? Suppose six of these people are men, and the other six are women. In how many ways can they join hands for a circle dance, assuming they alternate in gender around the circle
Answer:
86400 ways
Step-by-step explanation:
Since the circle can be rotated, the number of ways to arrange a distinct number of n objects in a circle will be (n−1)!.
Now, if we rotate the circle with the six women, we will see that there are 5! ways with which they can be placed in the circle.
After picking the places for the women, we will now fill each gap between two women with a man.
We have 6 men. Thus, number of ways to arrange the men is 6!
Thus,number of ways they can join hands for a circle dance, assuming they alternate in gender around the circle = 5! × 6! = 86400 ways
Answer:
D) exponents are in decreasing order
Step-by-step explanation:
Explanation is in the file
tinyurl.com/wpazsebu