Answer: Its A Im pretty sure
Step-by-step explanation:
The coefficient matrix is build with its rows representing each equation, and its columns representing each variable.
So, you may write the matrix as
![\left[\begin{array}{cc}\text{x-coefficient, 1st equation}&\text{y-coefficient, 1st equation}\\\text{x-coefficient, 2nd equation}&\text{y-coefficient, 2nd equation} \end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Ctext%7Bx-coefficient%2C%201st%20equation%7D%26%5Ctext%7By-coefficient%2C%201st%20equation%7D%5C%5C%5Ctext%7Bx-coefficient%2C%202nd%20equation%7D%26%5Ctext%7By-coefficient%2C%202nd%20equation%7D%20%5Cend%7Barray%7D%5Cright%5D%20%20)
which means
![\left[\begin{array}{cc}4&-3\\8&-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-3%5C%5C8%26-3%5Cend%7Barray%7D%5Cright%5D%20%20)
The determinant is computed subtracting diagonals:
![\left | \left[ \begin{array}{cc}a&b\\c&d\end{array}\right]\right | = ad-bc](https://tex.z-dn.net/?f=%20%5Cleft%20%7C%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5Cright%20%7C%20%3D%20ad-bc%20)
So, we have
![\left | \left[\begin{array}{cc}4&-3\\8&-3\end{array}\right] \right | = 4(-3) - 8(-3) = -4(-3) = 12](https://tex.z-dn.net/?f=%20%5Cleft%20%7C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-3%5C%5C8%26-3%5Cend%7Barray%7D%5Cright%5D%20%5Cright%20%7C%20%3D%204%28-3%29%20-%208%28-3%29%20%3D%20-4%28-3%29%20%3D%2012%20%20)
1. Add all of the numbers together
2. Divide the answer by how many different numbers there are (so 10)
The mean to the nearest tenth is
356.8
Y= y1
replace the y1 with the ordinate ( the y-part) of the point you are going thru
example a horizontal line going through the point (7,12) would be y=12
In
order to solve for a nth term in an arithmetic sequence, we use the formula
written as:<span>
an = a1 + (n-1)d
where an is the nth term, a1 is the first value
in the sequence, n is the term position and d is the common difference.
</span><span>THIRD
</span><span>A3=4+(3-1)(-5)
A3 = -6
A(3)=-2(3-1)(-5)
A3 = 20
</span><span>
FIFTH
</span>A5=4+(5-1)(-5)
A5 = -16
A(5)=-2(5-1)(-5)
A5 = 40<span>
TENTH
</span>A10=4+(10-1)(-5)
A10 = -41
A(10)=-2(10-1)(-5)
A10 = 90