A "true proportion" is when you know for a fact the equations are equivalent. You can easily multiply or divide without doing any work.
From 2:30-3:30 is 1hr and from 3:30-4:00 is 30 min from 4:00-4:15 is 15 minutes
Add up these time you get 1hr and 45min
Answer:
Slope: 2
y = 2x + 50
Step-by-Step Solution:
Slope: rise/run = 40/20 = 4/2 = 2
y = mx + b
m= slope = 2
b= y-intercept = 50
y = 2x + 50
Answer:
14.5
Step-by-step explanation:
I got 14.5 i hope this helps
![\begin{cases}a_1=-2\\a_2=2\\a_{n+1}=-2a_n+8a_{n-1}&\text{for }n\ge2\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Da_1%3D-2%5C%5Ca_2%3D2%5C%5Ca_%7Bn%2B1%7D%3D-2a_n%2B8a_%7Bn-1%7D%26%5Ctext%7Bfor%20%7Dn%5Cge2%5Cend%7Bcases%7D)
From the given starting values, we have
![\begin{bmatrix}a_3\\a_2\end{bmatrix}=\begin{bmatrix}-2&8\\1&0\end{bmatrix}\begin{bmatrix}a_2\\a_1\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7Da_3%5C%5Ca_2%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D-2%268%5C%5C1%260%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7Da_2%5C%5Ca_1%5Cend%7Bbmatrix%7D)
and more generally, the
![(n+1,n)](https://tex.z-dn.net/?f=%28n%2B1%2Cn%29)
-th term pair is captured by
![\begin{bmatrix}a_{n+1}\\a_n\end{bmatrix}=\begin{bmatrix}-2&8\\1&0\end{bmatrix}\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7Da_%7Bn%2B1%7D%5C%5Ca_n%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D-2%268%5C%5C1%260%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7Da_n%5C%5Ca_%7Bn-1%7D%5Cend%7Bbmatrix%7D)
so that
![\mathbf M=\begin{bmatrix}-2&8\\1&0\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathbf%20M%3D%5Cbegin%7Bbmatrix%7D-2%268%5C%5C1%260%5Cend%7Bbmatrix%7D)
We notice that
![\begin{bmatrix}a_{n+1}\\a_n\end{bmatrix}=\mathbf M\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\mathbf M^2\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\cdots=\mathbf M^{n-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7Da_%7Bn%2B1%7D%5C%5Ca_n%5Cend%7Bbmatrix%7D%3D%5Cmathbf%20M%5Cbegin%7Bbmatrix%7Da_n%5C%5Ca_%7Bn-1%7D%5Cend%7Bbmatrix%7D%3D%5Cmathbf%20M%5E2%5Cbegin%7Bbmatrix%7Da_%7Bn-1%7D%5C%5Ca_%7Bn-2%7D%5Cend%7Bbmatrix%7D%3D%5Ccdots%3D%5Cmathbf%20M%5E%7Bn-1%7D%5Cbegin%7Bbmatrix%7Da_2%5C%5Ca_1%5Cend%7Bbmatrix%7D)
which is to say that the explicit formula for
![a_n](https://tex.z-dn.net/?f=a_n)
is given by the dot product of the second row of
![\mathbf M^{n-1}](https://tex.z-dn.net/?f=%5Cmathbf%20M%5E%7Bn-1%7D)
with
![\begin{bmatrix}a_2&a_1\end{bmatrix}^\top](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7Da_2%26a_1%5Cend%7Bbmatrix%7D%5E%5Ctop)
.
First we diagonalize
![\mathbf M](https://tex.z-dn.net/?f=%5Cmathbf%20M)
. It has eigenvalues
![\lambda](https://tex.z-dn.net/?f=%5Clambda)
, where
![\det(\mathbf M-\lambda\mathbf I)=\begin{vmatrix}-2-\lambda&8\\1&-\lambda\end{vmatrix}=0\implies\lambda_1=2,\lambda_2=-4](https://tex.z-dn.net/?f=%5Cdet%28%5Cmathbf%20M-%5Clambda%5Cmathbf%20I%29%3D%5Cbegin%7Bvmatrix%7D-2-%5Clambda%268%5C%5C1%26-%5Clambda%5Cend%7Bvmatrix%7D%3D0%5Cimplies%5Clambda_1%3D2%2C%5Clambda_2%3D-4)
with corresponding eigenvectors
![\vec\eta](https://tex.z-dn.net/?f=%5Cvec%5Ceta)
, where
![(\mathbf M-2\mathbf I)\vec\eta_1=\mathbf0\implies\vec\eta_1=\begin{bmatrix}2\\1\end{bmatrix}](https://tex.z-dn.net/?f=%28%5Cmathbf%20M-2%5Cmathbf%20I%29%5Cvec%5Ceta_1%3D%5Cmathbf0%5Cimplies%5Cvec%5Ceta_1%3D%5Cbegin%7Bbmatrix%7D2%5C%5C1%5Cend%7Bbmatrix%7D)
![(\mathbf M+4\mathbf I)\vec\eta_2=\mathbf0\implies\vec\eta_2=\begin{bmatrix}-4\\1\end{bmatrix}](https://tex.z-dn.net/?f=%28%5Cmathbf%20M%2B4%5Cmathbf%20I%29%5Cvec%5Ceta_2%3D%5Cmathbf0%5Cimplies%5Cvec%5Ceta_2%3D%5Cbegin%7Bbmatrix%7D-4%5C%5C1%5Cend%7Bbmatrix%7D)
Then we have
![\mathbf M=\mathbf S^{-1}\mathbf{DS}](https://tex.z-dn.net/?f=%5Cmathbf%20M%3D%5Cmathbf%20S%5E%7B-1%7D%5Cmathbf%7BDS%7D)
where
![\mathbf S](https://tex.z-dn.net/?f=%5Cmathbf%20S)
is the matrix whose columns are
![\vec\eta_1](https://tex.z-dn.net/?f=%5Cvec%5Ceta_1)
and
![\vec\eta_2](https://tex.z-dn.net/?f=%5Cvec%5Ceta_2)
and
![\mathbf D](https://tex.z-dn.net/?f=%5Cmathbf%20D)
is the diagonal matrix whose non-zero entries are the eigenvalues of
![\mathbf M](https://tex.z-dn.net/?f=%5Cmathbf%20M)
. So
![\mathbf M=\begin{bmatrix}2&-4\\1&1\end{bmatrix}^{-1}\begin{bmatrix}2&0\\0&-4\end{bmatrix}\begin{bmatrix}2&-4\\1&1\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathbf%20M%3D%5Cbegin%7Bbmatrix%7D2%26-4%5C%5C1%261%5Cend%7Bbmatrix%7D%5E%7B-1%7D%5Cbegin%7Bbmatrix%7D2%260%5C%5C0%26-4%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7D2%26-4%5C%5C1%261%5Cend%7Bbmatrix%7D)
which gives
![\mathbf M^{n-1}=\begin{bmatrix}2&-4\\1&1\end{bmatrix}^{-1}\begin{bmatrix}2^{n-1}&0\\0&(-4)^{n-1}\end{bmatrix}\begin{bmatrix}2&-4\\1&1\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathbf%20%0AM%5E%7Bn-1%7D%3D%5Cbegin%7Bbmatrix%7D2%26-4%5C%5C1%261%5Cend%7Bbmatrix%7D%5E%7B-1%7D%5Cbegin%7Bbmatrix%7D2%5E%7Bn-1%7D%260%5C%5C0%26%28-4%29%5E%7Bn-1%7D%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7D2%26-4%5C%5C1%261%5Cend%7Bbmatrix%7D)
From here, you would just need to determine the bottom row of the final matrix product.