Answer:
<h2>2 and three-fourths </h2>
Step-by-step explanation:
Given the expression
, the equivalent expression can be gotten as shown;

2 and three-fourth therefore gives the required expression
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 time of a commercial airplane is 280 minutes
Step-by-step explanation:
Let
x -----> the speed of a commercial airplane
y ----> the speed of a jet plane
t -----> the time that a jet airplane takes from Vancouver to Regina
we know that
The speed is equal to divide the distance by the time
y=2x ----> equation A
<u><em>The speed of a commercial airplane is equal to</em></u>
x=1,730/(t+140) ----> equation B
<u><em>The speed of a jet airplane is equal to</em></u>
y=1,730/t -----> equation C
substitute equation B and equation C in equation A
1,730/t=2(1,730/(t+140))
Solve for t
1/t=(2/(t+140))
t+140=2t
2t-t=140
t=140 minutes
The time of a commercial airplane is
t+140=140+140=280 minutes
Answer:
a) 10/3
b) hyperbola
c) x = ± 6/5
Step-by-step explanation:
a) A conic section with a focus at the origin, a directrix of x = ±p where p is a positive real number and positive eccentricity (e) has a polar equation:

Given the conic equation: 
We have to make it to be in the form
:

Comparing with 
e = 10/3 = 3.3333, p = 6/5
b) since the eccentricity = 3.33 > 1, it is a hyperbola
c) The equation of the directrix is x = ±p = ± 6/5
plz mark brainly
plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
plz?