After I looked up some quick information, your car loses 15-25% each year. After 5 years, it goes to only 37%. So, a car that would cost 20000 would be valued at around 7k after 5 years. So, in the first 5 years, it lost 14k. Also, a 1999 Corolla goes for 2.5k, and that's nearly 20 years. So, I'm just going to go with, based off these numbers, probably 25-30 years...if it still works by then.
Also, if you get in a heavy accident the next day, making the car unusable, you rip off your friend. The first scenario only applies if the engine works well and no accidents were on the vehicle either.
Answer: 0.0655
Step-by-step explanation: if you divide 6.55 by 100, you I’ll get 0.0655 .
Answer:
18
Step-by-step explanation:
9+9=18 both sides of the rectangle is 9 so the top is longer which means 18
Answer:-56
Step-by-step explanation:The product is your answer
Answer:
Step-by-step explanation:
The complete question is
Water flows into a tank according to the rate F(t)= (t+6)/(1+t), and at the same time empties out at the rate E(t)= (ln(t+2))/(t+1), with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest galllon, is in the tank at time t=10 minutes.
Let C(t) be the amount of water in the tank at time t. We now that the rate of change of the tank is given by
![\frac{dC}{dt}=[\tex]rate at which water flows in- rate at which water flows out. Then [tex]\frac{dC}{dt}=\frac{t+6}{t+1}-\frac{\ln(t+2)}{(t+1)}[\tex]so, the desired expression is obtained by integrating with respect to t. This leads us to [tex]C(t) = \int \frac{t+1}{t+1}+ \frac{5}{t+1} - \frac{\ln(t+2)}{(t+1)} dt=t+ 5 \ln (|t+1|)-\int \frac{\ln(t+2)}{(t+1)} dt +C](https://tex.z-dn.net/?f=%5Cfrac%7BdC%7D%7Bdt%7D%3D%5B%5Ctex%5Drate%20at%20which%20water%20flows%20in-%20rate%20at%20which%20water%20flows%20out.%20%3C%2Fp%3E%3Cp%3EThen%20%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cfrac%7BdC%7D%7Bdt%7D%3D%5Cfrac%7Bt%2B6%7D%7Bt%2B1%7D-%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3Eso%2C%20the%20desired%20expression%20is%20obtained%20by%20integrating%20with%20respect%20to%20t.%20%3C%2Fp%3E%3Cp%3EThis%20leads%20us%20to%20%3C%2Fp%3E%3Cp%3E%5Btex%5DC%28t%29%20%20%3D%20%5Cint%20%5Cfrac%7Bt%2B1%7D%7Bt%2B1%7D%2B%20%5Cfrac%7B5%7D%7Bt%2B1%7D%20-%20%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%20dt%3C%2Fp%3E%3Cp%3E%3Dt%2B%205%20%5Cln%20%28%7Ct%2B1%7C%29-%5Cint%20%5Cfrac%7B%5Cln%28t%2B2%29%7D%7B%28t%2B1%29%7D%20dt%20%2BC)
Unfortunately, the integral 
cannot be expressed using fundamental functions. So, the problem cannot have an specific function (if you are willing to know the complete answer, the integral of this function uses the polylogarithm function with n=2).
Since you want the exact amount of water at time, you need to give C a value, that is, you need to know an initial condition for the problem. This means, you need to know the amount of water in the tank at time 0
