Answer:
the answer is A- B/3 or B divided by 3
Step-by-step explanation:
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)
Look at the picture.
The base of the ladder is 6.47 feet from the house.
Answer:
its alot to explain but i will try to make it as simple as possible
Step-by-step explanation:
your first goal is to make each problem into the form ax^2+bx+c=0
number 1, 2, 7 and 8 is already done for you
now all you have to do is plug in each number in the standard form into the quadtratic formula.
basically at this point you can just use your calculator to do the rest of the work. dont forget parentheses so it doesnt get confused...
or you can perform the algebraic work.. its all just a matter of plugging in the right numbers into the quadratic formula...
cant really do the work for you since im on my phone. but yeah all you need to do step one is transform each problem into ax^2+bx+c=0 form
then step 2, plug in each number in to the quadtratic formula. from there calculate using basic algebraic rules
Answer:
a reflection over the x-axis and then a 90 degree clockwise rotation about the origin
Step-by-step explanation:
Lets suppose triangle JKL has the vertices on the points as follows:
J: (-1,0)
K: (0,0)
L: (0,1)
This gives us a triangle in the second quadrant with the 90 degrees corner on the origin. It says that this is then transformed by performing a 90 degree clockwise rotation about the origin and then a reflection over the y-axis. If we rotate it 90 degrees clockwise we end up with:
J: (0,1) , K: (0,0), L: (1,0)
Then we reflect it across the y-axis and get:
J: (0,1), K:(0,0), L: (-1,0)
Now we go through each answer and look for the one that ends up in the second quadrant;
If we do a reflection over the y-axis and then a 90 degree clockwise rotation about the origin we end up in the fourth quadrant.
If we do a reflection over the x-axis and then a 90 degree counterclockwise rotation about the origin we also end up in the fourth quadrant.
If we do a reflection over the x-axis and then a reflection over the y-axis we also end up in the fourth quadrant.
The third answer is the only one that yields a transformation which leads back to the original position.
