Answer:
k = 30, ![y(t) = C_1e^{5t}+C_2e^{6t}](https://tex.z-dn.net/?f=y%28t%29%20%3D%20C_1e%5E%7B5t%7D%2BC_2e%5E%7B6t%7D)
Step-by-step explanation:
Since
is a solution, then it must satisfy the differential equation. So, we calculate the derivatives and replace the value in the equation. We have that
![\frac{d^2y}{dt^2} = 25 e^{5t},\frac{dy}{dt} = 5e^{5t}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2y%7D%7Bdt%5E2%7D%20%3D%2025%20e%5E%7B5t%7D%2C%5Cfrac%7Bdy%7D%7Bdt%7D%20%3D%205e%5E%7B5t%7D)
Then, replacing the derivatives in the equation we have:
![25e^{5t}-11(5)e^{5t}+ke^{5t}=0 e^{5t}(25-55+k) =0](https://tex.z-dn.net/?f=25e%5E%7B5t%7D-11%285%29e%5E%7B5t%7D%2Bke%5E%7B5t%7D%3D0%20e%5E%7B5t%7D%2825-55%2Bk%29%20%3D0)
Since
is a positive function, we have that
.
Now, consider a general solution
, then, by calculating the derivatives and replacing them in the equation, we get
![Ae^{rt}(r^2-11r+30)=0](https://tex.z-dn.net/?f=Ae%5E%7Brt%7D%28r%5E2-11r%2B30%29%3D0)
We already know that r=5 is a solution of the equation, then we can divide the polynomial by the factor (r-5) to the get the other solution. If we do so, we get that (r-6)=0. So the other solution is r=6.
Therefore, the general solution is
![y(t) = C_1e^{5t}+C_2e^{6t}](https://tex.z-dn.net/?f=y%28t%29%20%3D%20C_1e%5E%7B5t%7D%2BC_2e%5E%7B6t%7D)
12500 cause 5000 goes into 12500 twice and it would be 24$ for 10000 sqft and then another 6 so 30$ cause 2500 is half of 5000
Answer:C
Step-by-step explanation:
3.1 x 10^(-4) x 8.5 x 10^(-2)
3.1 x 8.5 x 10^(-4) x 10^(-2)
26.35 x 10^(-4-2)
26.35 x 10^(-6)
2.635 x 10^1 x 10^(-6)
2.635 x 10^(1-6)
2.635 x 10^(-5)
Answer:
Its 6 bill because people think its flashy
Step-by-step explanation: