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)
Answer:
see explanation
Step-by-step explanation:
Given
2x² + 7x = 15 ( subtract 15 from both sides )
2x² + 7x - 15 = 0 ← in standard form
To factorise the left side
Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term
product = 2 × - 15 = - 30 and sum = + 7
The factors are + 10 and - 3
Use these factors to split the x- term
2x² + 10x - 3x - 15 ( factor the first/second and third/fourth terms )
2x(x + 5) - 3(x + 5) ← factor out (x + 5) from each term
(x + 5)(2x - 3) ← in factored form
Answer:
.87 I think
Step-by-step explanation:
I believe you have to evaluate the inverse tangent first then do the sine.
tan^-1 34/19 is about 1.06
sin 1.06 is about .87
<span>B) 2 3/4 , 2.34, 9/4 hope this helps
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A number line. like where u plot the points (usually numbers) to compare them.
