55°
Step-by-step explanation:
110+15+x =180
x = 180 - 125
x = 55°
Answer:
![\left[\begin{array}{ccc}3&-1&\\-1&1/2\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-1%26%5C%5C-1%261%2F2%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The matrix system for the linear equations: x + 2y = 8, 2x + 6y = 9
![\left[\begin{array}{ccc}1&2&\\2&6\\\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}8\\9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26%5C%5C2%266%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D8%5C%5C9%5Cend%7Barray%7D%5Cright%5D)
To get the coefficient of x and y, the inverse of the first matrix (let the first matrix be A) must be known.
= (1 / determinant of A) x Adjoint of A
the determinant of A = (1 x 6) - (2 x 2) = 6 - 4 = 2
Adjoint of A = ![\left[\begin{array}{ccc}6&-2&\\-2&1\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%26-2%26%5C%5C-2%261%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=
= ![\left[\begin{array}{ccc}3&-1&\\-1&1/2\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-1%26%5C%5C-1%261%2F2%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Answer:
A. parallel lines
You can also use the app desmos to help you graph equations.
Step-by-step explanation:
first we notice that the line RS makes a diagonal in the circle.Since we are given ROP we can find SOP by:
180⁰-125⁰=55⁰
since they are in each others opposite point they are equal so that means that if we try the equation we did before: SOQ =ROP and QOR=SOP
<h2>
Answer:</h2>
Like you said, we need to use <u><em>substitution</em></u> to solve for y.

Now that we've solved for y, we can plug in the number we got into one of the <u><em>original</em></u> equations. (You can choose either equation, but I will use the first one.)

Now we have the solution of this system.
