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)
If we know your Pythagorean Triples we can immediately recognize that the last choice is a right triangle:
8² + 15² = 17²
If you don't know your Pythagorean Triples, it's worth learning the first few off the list because teachers use them in problems all the time. But for now let's just exhaustively check the Pythagorean Theorem for each triangle. We don't have to multiply everything out; we can analyze the common factors. If two have a common factor that the third one doesn't have, there's no way for the Pythagorean Theorem to add up.
Clearly 5²+15² is a multiple of 5 but 18² isn't so that one isn't a right triangle.
6²+12² is a multiple of 6, 16² isn't a multiple of 6, not an RT.
15²-5² is a multiple of 5, 13² isn't, no joy.
8²+15² = 64 + 225 = 289 = 17² -- that's a real right triangle, a valid Pythagorean Triple.
Answer:
g(f(x)) = 3.15x
Step-by-step explanation:
To find the number of Japanese yen equivalent to x russian rubles, we need to put one function into another.
If we take f(x) and put in into g(x), we will get Japanese yen in terms of rubles. Thus,
g(f(x)) = 90 (0.035x)
g(f(x)) = 3.15x
THis is the composite function which represents the number of Japanese yen equivalent to x Russian Rubles.
Answer:
2369.86
Step-by-step explanation:
f(36)=3700(1−0.06) ^36/5
2369.857704098921
Answer:
-3+(-5)
Checking our answer:
Adding this does indeed give -8
