Answer:
2.81
Step-by-step explanation:
.81 has a bar over it.
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:
Radius = 2.90 cm
Step-by-step explanation:

220 = 3.14 × {r}^{2} × 8.3
220 = 26.062 × {r}^{2}
220/26.062 = {r}^{2}
{r}^{2} = 8.44 ( approx )
r = √8.44
r = 2.90 cm. (approx)
Hope this helps .....
Please mark as Brainliest!!!!
Answer:
The bus is late 4 times in March.
Step-by-step explanation:
Solving this problem is very easy, using this formula:
20(0.2) = times late in March
(The 0.2 is the decimal form of 20%)
This is equal to 4, the times in March when the bus was late.