Answer:
It would be the first one
Step-by-step explanation:
because there are 3 positive and 2 negative
Any line parallel to this line can be written as y=-2/3x+c and passes through (9,6).thus 6=-2/3(9)+c.c-6=6.c=12.equation is y=-2/3x+12.
Answer: D
Step-by-step explanation: I said D because it's bigger than every other thing
Answer:
The general solution to this differential equation is ![P(t) = \frac{Ke^{kt} + h}{k}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20%5Cfrac%7BKe%5E%7Bkt%7D%20%2B%20h%7D%7Bk%7D)
Step-by-step explanation:
We are given the following differential equation:
![\frac{dP}{dt} = kP - h](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%20%3D%20kP%20-%20h)
Solving by separation of variables:
![\frac{dP}{kP-h} = dt](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7BkP-h%7D%20%3D%20dt)
Integrating both sides:
![\int \frac{dP}{kP-h} = \int dt](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7BdP%7D%7BkP-h%7D%20%3D%20%5Cint%20dt)
On the left side, by substitution, u = kP - h, du = kDp, Dp = du/k. Then
![\frac{1}{k} \ln{kP-h} = t + K](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bk%7D%20%5Cln%7BkP-h%7D%20%3D%20t%20%2B%20K)
In which K is the constant of integration.
![\ln{kP-h} = kt + K](https://tex.z-dn.net/?f=%5Cln%7BkP-h%7D%20%3D%20kt%20%2B%20K)
![e^{\ln{kP-h}} = e^{kt + K}](https://tex.z-dn.net/?f=e%5E%7B%5Cln%7BkP-h%7D%7D%20%3D%20e%5E%7Bkt%20%2B%20K%7D)
![kP - h = Ke^{kt}](https://tex.z-dn.net/?f=kP%20-%20h%20%3D%20Ke%5E%7Bkt%7D)
![kP = Ke^{kt} + h](https://tex.z-dn.net/?f=kP%20%3D%20Ke%5E%7Bkt%7D%20%2B%20h)
![P(t) = \frac{Ke^{kt} + h}{k}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20%5Cfrac%7BKe%5E%7Bkt%7D%20%2B%20h%7D%7Bk%7D)
The general solution to this differential equation is ![P(t) = \frac{Ke^{kt} + h}{k}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20%5Cfrac%7BKe%5E%7Bkt%7D%20%2B%20h%7D%7Bk%7D)