Answer:
(y+2)(y+4) would be the answer. :)
It looks like this is a system of linear ODEs given in matrix form,
![x' = \begin{bmatrix}10&-1\\5&8\end{bmatrix} x](https://tex.z-dn.net/?f=x%27%20%3D%20%5Cbegin%7Bbmatrix%7D10%26-1%5C%5C5%268%5Cend%7Bbmatrix%7D%20x)
with initial condition x(0) = (-6, 8)ᵀ.
Compute the eigenvalues and -vectors of the coefficient matrix:
![\det\begin{bmatrix}10-\lambda&-1\\5&8-\lambda\end{bmatrix} = (10-\lambda)(8-\lambda) + 5 = 0 \implies \lambda^2-18\lambda+85=0 \implies \lambda = 9\pm2i](https://tex.z-dn.net/?f=%5Cdet%5Cbegin%7Bbmatrix%7D10-%5Clambda%26-1%5C%5C5%268-%5Clambda%5Cend%7Bbmatrix%7D%20%3D%20%2810-%5Clambda%29%288-%5Clambda%29%20%2B%205%20%3D%200%20%5Cimplies%20%5Clambda%5E2-18%5Clambda%2B85%3D0%20%5Cimplies%20%5Clambda%20%3D%209%5Cpm2i)
Let v be the eigenvector corresponding to λ = 9 + 2i. Then
![\begin{bmatrix}10-\lambda&-1\\5&8-\lambda\end{bmatrix}v = 0 \implies \begin{bmatrix}1-2i&-1\\5&-1-2i\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D10-%5Clambda%26-1%5C%5C5%268-%5Clambda%5Cend%7Bbmatrix%7Dv%20%3D%200%20%5Cimplies%20%5Cbegin%7Bbmatrix%7D1-2i%26-1%5C%5C5%26-1-2i%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7Dv_1%5C%5Cv_2%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D0%5C%5C0%5Cend%7Bbmatrix%7D)
or equivalently,
![\begin{cases}(1-2i)v_1-v_2=0 \\ 5v_1-(1+2i)v_2=0\end{cases} \implies 5v_1 - (1+2i)v_2 = 0](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%281-2i%29v_1-v_2%3D0%20%5C%5C%205v_1-%281%2B2i%29v_2%3D0%5Cend%7Bcases%7D%20%5Cimplies%205v_1%20-%20%281%2B2i%29v_2%20%3D%200)
Let
; then
, so that
![\begin{bmatrix}10&-1\\5&8\end{bmatrix}\begin{bmatrix}1\\1-2i\end{bmatrix} = (9+2i)\begin{bmatrix}1\\1-2i\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D10%26-1%5C%5C5%268%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7D1%5C%5C1-2i%5Cend%7Bbmatrix%7D%20%3D%20%289%2B2i%29%5Cbegin%7Bbmatrix%7D1%5C%5C1-2i%5Cend%7Bbmatrix%7D)
and we get the other eigenvalue/-vector pair by taking the complex conjugate,
![\begin{bmatrix}10&-1\\5&8\end{bmatrix}\begin{bmatrix}1\\1+2i\end{bmatrix} = (9-2i)\begin{bmatrix}1\\1+2i\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D10%26-1%5C%5C5%268%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7D1%5C%5C1%2B2i%5Cend%7Bbmatrix%7D%20%3D%20%289-2i%29%5Cbegin%7Bbmatrix%7D1%5C%5C1%2B2i%5Cend%7Bbmatrix%7D)
Then the characteristic solution to the system is
![x = C_1 e^{(9+2i)t} \begin{bmatrix}1\\1-2i\end{bmatrix} + C_2 e^{(9-2i)t} \begin{bmatrix}1\\1+2i\end{bmatrix}](https://tex.z-dn.net/?f=x%20%3D%20C_1%20e%5E%7B%289%2B2i%29t%7D%20%5Cbegin%7Bbmatrix%7D1%5C%5C1-2i%5Cend%7Bbmatrix%7D%20%2B%20C_2%20e%5E%7B%289-2i%29t%7D%20%5Cbegin%7Bbmatrix%7D1%5C%5C1%2B2i%5Cend%7Bbmatrix%7D)
From the given condition, we have
![\displaystyle \begin{bmatrix}-6\\8\end{bmatrix} = C_1 \begin{bmatrix}1\\1-2i\end{bmatrix} + C_2 \begin{bmatrix}1\\1+2i\end{bmatrix} \implies C_1 = -3-\frac i2, C_2=-3+\frac i2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Bbmatrix%7D-6%5C%5C8%5Cend%7Bbmatrix%7D%20%3D%20C_1%20%5Cbegin%7Bbmatrix%7D1%5C%5C1-2i%5Cend%7Bbmatrix%7D%20%2B%20C_2%20%5Cbegin%7Bbmatrix%7D1%5C%5C1%2B2i%5Cend%7Bbmatrix%7D%20%5Cimplies%20C_1%20%3D%20-3-%5Cfrac%20i2%2C%20C_2%3D-3%2B%5Cfrac%20i2)
and so the particular solution to the IVP is
![\displaystyle \boxed{x = -\left(3+\frac i2\right) e^{(9+2i)t} \begin{bmatrix}1\\1-2i\end{bmatrix} - \left(3-\frac i2\right) e^{(9-2i)t} \begin{bmatrix}1\\1+2i\end{bmatrix}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cboxed%7Bx%20%3D%20-%5Cleft%283%2B%5Cfrac%20i2%5Cright%29%20e%5E%7B%289%2B2i%29t%7D%20%5Cbegin%7Bbmatrix%7D1%5C%5C1-2i%5Cend%7Bbmatrix%7D%20-%20%5Cleft%283-%5Cfrac%20i2%5Cright%29%20e%5E%7B%289-2i%29t%7D%20%5Cbegin%7Bbmatrix%7D1%5C%5C1%2B2i%5Cend%7Bbmatrix%7D%7D)
which you could go on to rewrite using Euler's formula,
![e^{(a+bi)t} = e^{at} (\cos(bt) + i \sin(bt))](https://tex.z-dn.net/?f=e%5E%7B%28a%2Bbi%29t%7D%20%3D%20e%5E%7Bat%7D%20%28%5Ccos%28bt%29%20%2B%20i%20%5Csin%28bt%29%29)
Answer:
a large dance star
Explanation:
because it is bigger and heavier
D.
If you are also asking this from a personal standpoint, my advice is that the dual enrollment courses are much easier. However, this may not be the best option depending on which college it is based through (Dual credit is NOT universal, and there are some colleges, especially the ones that are harder to get into, that will not accept transfer credit from many colleges.) My advice would be, if you don't know where you are going to college yet, to take the AP and study a lot for that test. Or, if you know where you are going, to take advantage of CLEP tests, which are easier and strictly pass/fail subject tests. Most colleges would have the AP/Transfer Credit/CLEP requirements on their websites if you do a little digging.