Answer:
Explanation:
area of a triangle = base * height
let us take height to be 'x' then, base will be '2x'
since area = 512 cm
512=(x)(2x)
( 2x)²=512
x²=512/2
=256
therfore 'x' = 16
now height='x' = 16
and, base= '2x' = 2* 16 = 32
Answer:
newtons first law ; an object will stay at rest until a force is added onto it. Since the student knocked over the first domino that force the student added continues on until the row ends
Explanation:
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)
may you please right the choices for us to pick and thank you