You're trying to find constants
![a_0,a_1,a_2](https://tex.z-dn.net/?f=a_0%2Ca_1%2Ca_2)
such that
![\hat y=a_0+a_1\hat x+a_2{\hat x}^2](https://tex.z-dn.net/?f=%5Chat%20y%3Da_0%2Ba_1%5Chat%20x%2Ba_2%7B%5Chat%20x%7D%5E2)
. Equivalently, you're looking for the least-square solution to the following matrix equation.
![\underbrace{\begin{bmatrix}1&6&6^2\\1&3&3^2\\\vdots&\vdots&\vdots\\1&9&9^2\end{bmatrix}}_{\mathbf A}\underbrace{\begin{bmatrix}a_0\\a_1\\a_2\end{bmatrix}}_{\mathbf x}=\underbrace{\begin{bmatrix}100\\110\\\vdots\\70\end{bmatrix}}_{\mathbf b}](https://tex.z-dn.net/?f=%5Cunderbrace%7B%5Cbegin%7Bbmatrix%7D1%266%266%5E2%5C%5C1%263%263%5E2%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%5C%5C1%269%269%5E2%5Cend%7Bbmatrix%7D%7D_%7B%5Cmathbf%20A%7D%5Cunderbrace%7B%5Cbegin%7Bbmatrix%7Da_0%5C%5Ca_1%5C%5Ca_2%5Cend%7Bbmatrix%7D%7D_%7B%5Cmathbf%20x%7D%3D%5Cunderbrace%7B%5Cbegin%7Bbmatrix%7D100%5C%5C110%5C%5C%5Cvdots%5C%5C70%5Cend%7Bbmatrix%7D%7D_%7B%5Cmathbf%20b%7D)
To solve
![\mathbf{Ax}=\mathbf b](https://tex.z-dn.net/?f=%5Cmathbf%7BAx%7D%3D%5Cmathbf%20b)
, multiply both sides by the transpose of
![\mathbf A](https://tex.z-dn.net/?f=%5Cmathbf%20A)
, which introduces an invertible square matrix on the LHS.
![\mathbf{Ax}=\mathbf b\implies\mathbf A^\top\mathbf{Ax}=\mathbf A^\top\mathbf b\implies\mathbf x=(\mathbf A^\top\mathbf A)^{-1}\mathbf A^\top\mathbf b](https://tex.z-dn.net/?f=%5Cmathbf%7BAx%7D%3D%5Cmathbf%20b%5Cimplies%5Cmathbf%20A%5E%5Ctop%5Cmathbf%7BAx%7D%3D%5Cmathbf%20A%5E%5Ctop%5Cmathbf%20b%5Cimplies%5Cmathbf%20x%3D%28%5Cmathbf%20A%5E%5Ctop%5Cmathbf%20A%29%5E%7B-1%7D%5Cmathbf%20A%5E%5Ctop%5Cmathbf%20b)
Computing this, you'd find that
![\mathbf x=\begin{bmatrix}a_0\\a_1\\a_2\end{bmatrix}\approx\begin{bmatrix}121.119\\-3.786\\-0.175\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathbf%20x%3D%5Cbegin%7Bbmatrix%7Da_0%5C%5Ca_1%5C%5Ca_2%5Cend%7Bbmatrix%7D%5Capprox%5Cbegin%7Bbmatrix%7D121.119%5C%5C-3.786%5C%5C-0.175%5Cend%7Bbmatrix%7D)
which means the first choice is correct.
The answer to your problem is: 122.32
Answer:
Step-by-step explanation:
We have plane 1 flying SW for 4 hours at a rate of 415 mph. The distance he covers using the d = rt formula for distance, is 415(4) = 160 miles.
We also have plane 2 flying directly east (along the x-axis) for 4 hours at 327 mph. The distance he covers using the d = rt formula for distance, is 327(4) = 1308 miles. The angle in between them at this point is 135 degrees, and what we need to find is the length of the vector connecting the 2 planes. IF this was right triangle trig that distance would be the hypotenuse and we could solve for it using Pythagorean's Theorem. BUT it is NOT a right triangle, so we have to find some other means with which to solve for that length. We will use the Law of Cosines to do this.
which simplifies a bit to
![?^2=2755600+1712864-(-3070653.624)](https://tex.z-dn.net/?f=%3F%5E2%3D2755600%2B1712864-%28-3070653.624%29)
If you add all of that together, you'll get
and you'll take the square root of that to get that the distance between the 2 planes after 4 hours is
2745 miles
A is the answer (1/16,14)
Answer:
ones, tenth, tens, hundredth
Step-by-step explanation: