Answer:16:18
Step-by-step explanation:
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
X=5.2
Hope this was helpful
The distance from A to B is the radius, as point A is the center of the circle, and B is a point on the circle.
AB is 5 inches, so the radius will be 5 inches.
D is another point on the circle, so AD would also be considered a radius. This means that the distance from A to D would be 5 inches.
2 is common factor
2x=2(x)
4=2(2)
ab+ac=a(b+c)
2(x)+2(2)=2(x+2)