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For this problem A is the amount of salt in the tank. If a tank contains 250 liters of liquid with 13 grams of salt. A mixture containing 10 grams per liter is pumped into the tank at a rate of 2 liters/minute. The mixture is well-mixed, and pumped out at a rate of 5 liters/minute. The amount of salt in the tank satisfies the differential equation 250 = 0. 5000-4987e Rewriting this as a linear differential equation we get x + X 250+ t The integrating factor is 250+ t and the solution is 13 = 20

For this problem A is the amount of salt in the tank. If a tank contains 250 liters of liquid with 13 grams of salt. A mixture containing 10 grams per liter is pumped into the tank at a rate of 2 liters/minute. The mixture is well-mixed, and pumped out at a rate of 5 liters/minute. The amount of salt in the tank satisfies the differential equation 250 = 0. 5000-4987e Rewriting this as a linear differential equation we get x + X 250+ t The integrating factor is 250+ t and the solution is 13 = 20

College Algebra
College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
ChapterP: Prerequisites
SectionP.4: Factoring Polynomials
Problem 84E: The rate of change of an autocatalytic chemical reaction is kQxkx2 where Q is the amount of the...
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For this problem A is the amount of salt in the tank. If a tank contains 250 liters of liquid with 13 grams of salt. A mixture containing 10 grams per liter is pumped into the tank at a rate of 2 liters/minute. The mixture is well-mixed, and pumped out at a rate of 5 liters/minute. The amount of salt in the tank satisfies the differential equation 250 = 0. 5000-4987e Rewriting this as a linear differential equation we get x + X 250+ t The integrating factor is 250+ t and the solution is 13 = 20
Transcribed Image Text:For this problem A is the amount of salt in the tank. If a tank contains 250 liters of liquid with 13 grams of salt. A mixture containing 10 grams per liter is pumped into the tank at a rate of 2 liters/minute. The mixture is well-mixed, and pumped out at a rate of 5 liters/minute. The amount of salt in the tank satisfies the differential equation 250 = 0. 5000-4987e Rewriting this as a linear differential equation we get x + X 250+ t The integrating factor is 250+ t and the solution is 13 = 20
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