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)
The options to fill in the gaps for the Long-run macroeconomic equilibrium are:
- equals
- intersects as a point on
<h3>What is macroeconomic equilibrium?</h3>
Macroeconomic equilibrium is an equilibrium situation that occurs when the quantities of the real GDP demanded equals the real GDP supplied at the point of intersection of the AD curve and the AS curve.
Long-run macroeconomic equilibrium occurs when aggregate demand equals short-run aggregate supply and they intersects as a point on the long-run supply curve.
In conclusion, Long-run macroeconomic equilibrium results in equilibrium of demand and supply of GDP.
Learn more about Long-run macroeconomic equilibrium at: brainly.com/question/27606663
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Answer:
280
Explanation:
For ease of comparison $20/hr < $12/hr+ commission, let's do it by the hour first.
Hourly pay of the food truck is $12, every burger she sells, she gets 10% of the $12 burger, which is $1.2. Therefore all we need to figure out is 1.2x +$12 > $20. Minus both sides by 12, now all we have is down to the bare minimum: 1.2*X = $8
8 divided by 1.2 is 6.66 (infinite), but since she can't sell 0.6 of a burger, we need to round up, which is 7, keep in mind we are still at hourly. Now we know that she has to sell at least 7 burgers every hour she's working for the food truck, it's a 40-hour work week, 7*40 is 280.
Check:
Food truck: $12+ 12*0.1*7 = $20.4/hr $20.4 > $20
Best of luck for your SAT!
one year as a full-time student........ hope this helps you