Simultaneous equations can be solved using inverse matrix operation.
The complete steps of Jacob's solution are:
![\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%5E%7B-1%7D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cfrac%7B1%7D%7B14%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-1%5C%5C2%264%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-22%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-22%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cfrac%7B1%7D%7B14%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D28%26-84%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-6%5Cend%7Barray%7D%5Cright%5D)
We have:


Calculate the determinant of ![\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D)



So, the inverse matrix becomes
![A = \frac{1}{14}\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B1%7D%7B14%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D)
Replace the first column with
to calculate the value of x
![x = \frac{1}{14}\left[\begin{array}{cc}2&1\\-22&3\end{array}\right]](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B14%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%261%5C%5C-22%263%5Cend%7Barray%7D%5Cright%5D)
So, we have:




Replace the second column with
to calculate the value of y
![y = \frac{1}{14}\left[\begin{array}{cc}4&2\\-2&-22\end{array}\right]](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B14%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%262%5C%5C-2%26-22%5Cend%7Barray%7D%5Cright%5D)
So, we have:




Hence, the complete process is:
![\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%5E%7B-1%7D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cfrac%7B1%7D%7B14%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-1%5C%5C2%264%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-22%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%261%5C%5C-2%263%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-22%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cfrac%7B1%7D%7B14%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D28%26-84%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%26-6%5Cend%7Barray%7D%5Cright%5D)
Read more about matrices at:
brainly.com/question/11367104
Answer:
The measure of Arc EH is 123°.
Step-by-step explanation:
Consider the diagram below.
It is provided that ∠EDH ≅ ∠EDG.
This implies that: ∠EDH = ∠EDG
The arc measure is same as the measure of the central angle.
That is:
arc FE = ∠EDF = 57°
arc FG = ∠FDG = 66°
Compute the measure of angle ∠EDH as follows:
arc EH = ∠EDH
=∠EDG
= ∠EDF + ∠FDG
= 57° + 66°
= 123°
Thus, the measure of Arc EH is 123°.
I believe the correct answer is D - 25 centimeters based on my calculations
Answer:
508.8 seconds
Step-by-step explanation:
The most accurate determination mathematically is to assume that Lola will maintain an average of 5.3 seconds per signature as she signs all 96 invitations.
Therefore, multiply the time it takes her to sign each invitation (5.3 seconds) by the total number of invitations there are (96 invitations) to get the projected total amount of time that it will take Lola to sign all 96 invitations: