Define
![{x} = \left[\begin{array}{ccc}x_{1}\\x_{2}\end{array}\right]](https://tex.z-dn.net/?f=%7Bx%7D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_%7B1%7D%5C%5Cx_%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20)
Then
x₁ = cos(t) x₁(0) + sin(t) x₂(0)
x₂ = -sin(t) x₁(0) + cos(t) x₂(0)
Differentiate to obtain
x₁' = -sin(t) x₁(0) + cos(t) x₂(0)
x₂' = -cos(t) x₁(0) - sin(t) x₂(0)
That is,
![\dot{x} = \left[\begin{array}{ccc}-sin(t)&cos(t)\\-cos(t)&-sin(t)\end{array}\right] x(0)](https://tex.z-dn.net/?f=%5Cdot%7Bx%7D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-sin%28t%29%26cos%28t%29%5C%5C-cos%28t%29%26-sin%28t%29%5Cend%7Barray%7D%5Cright%5D%20x%280%29)
Note that
![\left[\begin{array}{ccc}0&1\\-1&09\end{array}\right] \left[\begin{array}{ccc}cos(t)&sin(t)\\-sin(t)&cos(t)\end{array}\right] = \left[\begin{array}{ccc}-sin(t)&cos(t)\\-cos(t)&-sin(t)\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C-1%2609%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%28t%29%26sin%28t%29%5C%5C-sin%28t%29%26cos%28t%29%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-sin%28t%29%26cos%28t%29%5C%5C-cos%28t%29%26-sin%28t%29%5Cend%7Barray%7D%5Cright%5D%20)
Therefore
<h2>
Answer:</h2>
<h3>(0,14) and (2,11)</h3>
<h2>
Step-by-step explanation:</h2>
points are on the graph of the line:-
⠀
we get,
⠀
⠀
In,(0,14)
If we put the value ,x=0
y=
So,we get one point,(0,14)✔
In,(2,11)
If we put ,x=2
y=
so,we get another point (2,11)✔
in other given ponts{(1,12),(4,9),(8,4)},It is impossible. ⠀ ⠀
1) Solve one of the equations for either variable.
2) Substitute the expression from Step 1 into the other equation.
3) Solve the resulting equation.
4) Substitute the solution in Step 3 into one of the original equations to find the other variable.
5) Write the solution as an ordered pair.
Answer: Choice D
b greater-than 3 and StartFraction 2 over 15 EndFraction
In other words,
b > 3 & 2/15
or

========================================================
Explanation:
Let's convert the mixed number 2 & 3/5 into an improper fraction.
We'll use the rule
a & b/c = (a*c + b)/c
In this case, a = 2, b = 3, c = 5
So,
a & b/c = (a*c + b)/c
2 & 3/5 = (2*5 + 3)/5
2 & 3/5 = (10 + 3)/5
2 & 3/5 = 13/5
The inequality
is the same as 
---------------------
Let's multiply both sides by 15 to clear out the fractions

---------------------
Now isolate the variable b

Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how
47/15 = 3 remainder 2
The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.
1/4 is the answer is is a lower fraction