Entered 0.3*5-3*[y/(100+2*t)] 4.1577 At least one of the answers above is NOT correct. Answer Preview =...

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Entered 0.3*5-3*[y/(100+2*t)] 4.1577 At least one of the answers above is NOT correct. Answer Preview = 0.3.5 - 3 2. How much salt is in the tank after 40 minutes? Answer (in kilograms): 4.1577 Y 100 + 2t 4.1577 Result correct A very large tank initially contains 100L of pure water. Starting at time t = 0 a solution with a salt concentration of 0.3kg/L is added at a rate of 5L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min. Answer the following questions. incorrect 1. Let y(t) be the amount of salt (in kilograms) in the tank after t minutes. What differential equation does y satisfy? Use the variable y for y(t). dy dt Answer (in kilograms per minute): 0.3*5-3(y/(100+2t)) Entered 0.3*5-3*[y/(100+2*t)] 4.1577 At least one of the answers above is NOT correct. Answer Preview = 0.3.5 - 3 2. How much salt is in the tank after 40 minutes? Answer (in kilograms): 4.1577 Y 100 + 2t 4.1577 Result correct A very large tank initially contains 100L of pure water. Starting at time t = 0 a solution with a salt concentration of 0.3kg/L is added at a rate of 5L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min. Answer the following questions. incorrect 1. Let y(t) be the amount of salt (in kilograms) in the tank after t minutes. What differential equation does y satisfy? Use the variable y for y(t). dy dt Answer (in kilograms per minute): 0.3*5-3(y/(100+2t))