B. I mean it’s super easy but I hope it helps
Two angles adjacent angles that add up to 180 degrees
Answer:
With rare exceptions, cars decrease in value with age. Depending on other factors, like accidents, repairs, or other damage, the value of a car may decrease even faster. If you borrowed money to buy a car, you might owe more on your car loan than its current value. When that happens, you have negative equity in the car. Some car dealers say you won’t be responsible for the remaining balance on your old car loan when you trade in your old car. But that might not be true. Dealers sometimes just roll over the negative equity into your new car loan, so you still end up paying it.
Step-by-step explanation:
Say you want to trade in your car for a newer model.
Your loan payoff is $18,000
Your car is worth $15,000
You have negative equity of $3,000. That must be paid if you want to trade in your vehicle. If the dealer promises to pay off the $3,000, it shouldn’t be included in your new loan.
But some dealers
add that $3,000 to the loan for your new car
subtract the amount from your down payment
or do both
The answer is 20!
−100×(−20)=?
2000
Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min