]Eigenvectors are found by the equation

implying that

. We then can write:
And:
Gives us the characteristic polynomial:

So, solving for each eigenvector subspace:
![\left [ \begin{array}{cc} 4 & 2 \\ 5 & 1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} -x \\ -y \end{array} \right ]](https://tex.z-dn.net/?f=%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%204%20%26%202%20%5C%5C%205%20%26%201%20%5Cend%7Barray%7D%20%5Cright%20%5D%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bc%7D%20x%20%5C%5C%20y%20%5Cend%7Barray%7D%20%5Cright%20%5D%20%3D%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bc%7D%20-x%20%5C%5C%20-y%20%5Cend%7Barray%7D%20%5Cright%20%5D%20)
Gives us the system of equations:
Producing the subspace along the line

We can see then that 3 is the answer.
Answer:
A
Step-by-step explanation:
B and D have to be supplementary otherwise it wouldn't be a parallelogram
Answer:
cot theta = -√3
Step-by-step explanation:
we are given that cos theta = - √3/2 within the range 180° < theta < 270° and
since cos theta = adj/hyp
adj = - √3
hyp = 2
Get the opposite using the pythagoras theorem
hyp^2 = opp^2 + adj^2
2² = opp² + (-√3)²
4 = opp² + 3
opp² = 4-3
opp² = 1
opp = 1
tan theta = opp/adj
tan theta = - 1/√3
Recall that cot theta = 1/tan theta
cot theta = 1/(-1/√3)
cot theta = -√3
1/15=144/x
Cross multiply
X=2,160
Gymnasium=2,160 ft2