The
correct answer is in file attached
<span>we have
triangle RSTm∠R = 60, m∠S = 80 and m∠T=180-(80+60)=40</span>
<span>RS=4
triangle EFDm∠F = 60, m∠D = 40 and m∠E=180-(60+40)=80</span>
EF=4
Therefore
<span>The triangles
RST and
EFD are
congruents because they have two angles and the side common to them,
respectively, equal.
This is the theorem of ASA (angle-side-angle).</span>
Explication
Side common--------> RS=EF
<span>Angles RS------------- > m∠R =
60 m∠S =
80 </span>
<span>Angles EF------------- > m∠E =
80 m∠F = 60 </span>
<span>Angles RS and Angles EF are equals</span>
<span>
The segment which is congruent
to RT is FD, because angles of RT are 60 and 40, and angles of FD also are 60
and 40.</span>
<span>I can solve for x and y
A) 2x - 4y = 12
A) 2x = 4y + 12
A) x = 2y + 6
Substituting into B
B) 3* (2y +6) - 5y = 20
B) 6y + 18 -5y = 20
B) y = 2
B) x = 10
</span>
Answer: x = -2
Step-by-step explanation:
-x + 4x = x - 4
=> 3x = x - 4
=> 3x - x = -4
=> 2x = -4
=> x = -2
Answer:
x=-7, y=2
Step-by-step explanation:
You are given the matrix equation
![\left[\begin{array}{cc}2&-2\\-1&3\end{array}\right] \cdot \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}-18\\13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-2%5C%5C-1%263%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-18%5C%5C13%5Cend%7Barray%7D%5Cright%5D)
Find the inverse matrix for the matrix
![\left[\begin{array}{cc}2&-2\\-1&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-2%5C%5C-1%263%5Cend%7Barray%7D%5Cright%5D)
1. The determinant is
![\left|\left[\begin{array}{cc}2&-2\\-1&3\end{array}\right]\right|=2\cdot 3-(-1)\cdot (-2)=6-2=4](https://tex.z-dn.net/?f=%5Cleft%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-2%5C%5C-1%263%5Cend%7Barray%7D%5Cright%5D%5Cright%7C%3D2%5Ccdot%203-%28-1%29%5Ccdot%20%28-2%29%3D6-2%3D4)
2.

3. Inverse matrix is
![\dfrac{1}{4}\left[\begin{array}{cc}3&1\\2&2\end{array}\right]^T=\dfrac{1}{4}\left[\begin{array}{cc}3&2\\1&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B4%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%261%5C%5C2%262%5Cend%7Barray%7D%5Cright%5D%5ET%3D%5Cdfrac%7B1%7D%7B4%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%262%5C%5C1%262%5Cend%7Barray%7D%5Cright%5D)
So, the solution of the equation is
![\left[\begin{array}{c}x\\y\end{array}\right]=\dfrac{1}{4}\left[\begin{array}{cc}3&2\\1&2\end{array}\right]\cdot \left[\begin{array}{c}-18\\13\end{array}\right]=\\ \\=\dfrac{1}{4}\left[\begin{array}{cc}3\cdot(-18)+2\cdot 13\\1\cdot (-18)+2\cdot 13\end{array}\right]=\dfrac{1}{4}\left[\begin{array}{c}-28\\8\end{array}\right]=\left[\begin{array}{c}-7\\2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%3D%5Cdfrac%7B1%7D%7B4%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%262%5C%5C1%262%5Cend%7Barray%7D%5Cright%5D%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-18%5C%5C13%5Cend%7Barray%7D%5Cright%5D%3D%5C%5C%20%5C%5C%3D%5Cdfrac%7B1%7D%7B4%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%5Ccdot%28-18%29%2B2%5Ccdot%2013%5C%5C1%5Ccdot%20%28-18%29%2B2%5Ccdot%2013%5Cend%7Barray%7D%5Cright%5D%3D%5Cdfrac%7B1%7D%7B4%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-28%5C%5C8%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-7%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
Answer:
Step-by-step explanation:
C