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
![x(t) = \left[\begin{array}{ccc}0&1\\-1&0\end{array}\right] x(t)](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C-1%260%5Cend%7Barray%7D%5Cright%5D%20x%28t%29)
[] signs mean that the number is included in the domain and () mean that the number is not included. Now,
1. [-1,5)
Because x is greater than or equal to -1 (included) but less than 5 (excluded
2. [-5,13)
When we plug in x = -1, we get the minimum value of -5, which is included because its x-value was included. The, we plug in x = 5 to find the range's maximum and this is excluded because its x-value was excluded.
3. 218
x = 6 lies in the domain of the first function so we will use it to evaluate the value of x = 6. We do this by simply substituting x into the function.
4. 12
x = 9 is included in the domain of the second function and the value of the second function at all points is 12.
5. 103
x = 10 is included in the domain of the second function so we will use it to evaluate the value of x = 10. This is done by substituting x into the equation.
Answer:
x<4
Step-by-step explanation:
-2x+20<2x+4
−2x+20−2x<2x+4−2x
−4x+20<4
−4x+20−20<4−20
−4x<−16
−4x/-4<-16/-4
x<4 its a open circle
<em>372₁₀ = 2442₅</em>
- Step-by-step explanation:
<em>372 : 5 = 74 r 2</em>
<em>74 : 5 = 14 r 4</em>
<em>14 : 5 = 2 r 4</em>
<em>2 : 5 = 0 r 2 ↑</em>
<em>write down the remainders in the reverse order</em>
<em />
<em>372₁₀ = 2442₅</em>
