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
Height of a cone-shaped attic:
h = 18 ft,
d = 18 ft,
r = 18: 2 = 9 ft
V = r² π h / 3 =
= 9² π · 18 / 3 = 81 π · 6 = 486 π ft³
Answer:
The volume of the space in the attic is 486 π ft³.
1. angles 2 and 5 are supplementary, reason : it's given
2. 3 and 2 are congruent reason: they are vertical
3. 3 and 5 are supplementary reason : substitution prop. (i think so)
4. I don't know this one but are their any options?
9=3^2
3^(2x)=(3^2)^(3x-4)
3^(2x)=3^(6x-8)
2x=6x-8
0=4x-8
8=4x
2=x