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Math Advanced Math Fundamentals of Differential Equations (9th Edition)

(a) To Find: The concentration of salt in the tank after 10 minutes. The concentration of salt in the tank after 10 min will be 0.0281 kg L _ . Given: Initially, the tank filled with 500 L of water containing 5 kg of salt. A brine contains 0.2 kilogram (kg) of salt per liter (L) runs into a tank. The brine enters at the rate of 5 L/min and the mixture is kept uniform by stirring. The mixture flows out at the rate of 5 L/min. Calculation: Let A represents the number of kilograms of salt in the tank at t min after the process begins. Use the fact that, rate of increase in A = rate of input − rate of exit . That is, ( rate of increase in A ) = ( rate of salt in − rate of salt out ) d A ( t ) d t = ( 5 L min ) ( 0.2 kg L ) − ( 5 L min ) ( A ( t ) 500 kg L ) d A ( t ) d t = 1 kg min − A ( t ) 100 kg min d A d t = 1 − A 100 d A d t = 100 − A 100 Separate the variables and rewrite the equation as follows, 1 100 − A d A = 1 100 d t Integrate both sides of the equation. ∫ 1 100 − A d A = 1 100 ∫ d t − ln | 100 − A | = t 100 + C 1 ln | 100 − A | = − t 100 + C 2 100 − A = e − t 100 + C 2 Simplify further, 100 − A = e − t 100 e C 2 − A = C e − t 100 − 100 ( C is an arbitrary constant ) A ( t ) = 100 − C e − t 100 ... ( 1 ) Given that, the tank initially filled with 500 L of water containing 5 kg of salt. Substitute A ( 0 ) = 5 in the above result to find the constant C . A ( 0 ) = 100 − C e 0 5 = 100 − C C = 95 Thus, the solution becomes A ( t ) = 100 − 95 e − t 100 ... ( 2 ) Now, find the mass of salt in the tank after 10 min. Substitute t = 10 in equation (2). A ( 10 ) = 100 − 95 e − 10 100 = 100 − 95 e − 0.1 = 100 − 95 e 0.1 ≈ 14.0404452866 kg Find the concentration of salt in the tank after 10 min. Concentration = Mass of salt Volume of water in tank = 14.0404452866 500 kg L ≈ 0.0281 kg L Thus, the concentration of salt in the tank after 10 min will be 0.0281 kg L _ . (b) To Find: The concentration of salt in the tank 20 minutes after the leak develops. The concentration of salt in the tank 20 minutes after the leakage will be 0.0598 kg L _ . The additional mixture flows out at the rate of 1 L/min after 10 minutes, a leak develops in the tank. So the volume of the tank is now changing. After 10 minutes, the initial volume is V ( 10 ) = 500 L . The rate of change of volume of the tank after 10 minutes is given by, d V d t = In volume per unit time − Out volume per unit time = 5 L min − ( 5 + 1 ) L min = − 1 L min Separate the variables and rewrite the equation as follows, d V = − d t Integrate both sides of the equation. ∫ d V = − ∫ d t V = − t + C Apply the initial condition V ( 10 ) = 500 L in the above result to find the constant C . V ( 10 ) = − 10 + C 500 = − 10 + C C = 510 Thus, the volume of the tank after 10 min is V ( t ) = 510 − t . Let A represents the number of kilograms of salt in the tank at t after the leak develops. Use the fact that, rate of increase in A = rate of input − rate of exit . That is, ( rate of increase in A ) = ( rate of salt in − rate of salt out ) d A ( t ) d t = ( 5 L min ) ( 0.2 kg L ) − ( 5 L min + 1 L min ) ( A ( t ) 510 − t kg L ) d A ( t ) d t = 1 kg min − 6 A ( t ) 510 − t kg min d A d t = 1 − 6 A 510 − t d A d t + 6 510 − t A = 1 From part (a), the amount of salt after 10 minutes is A ( 10 ) = 100 − 95 e − 0.1 . Thus, the differential equation becomes d A d t + 6 510 − t A = 1 , A ( 10 ) = 100 − 95 e − 0.1 ... ( 3 ) Solve the differential equation by the method of linear equation as follows. Here, P ( t ) = 6 510 − t and Q ( t ) = 1 . Calculate the integrating factor μ ( t ) by the formula μ ( t ) = exp [ ∫ P ( t ) d t ] . μ ( t ) = exp [ ∫ ( 6 510 − t ) d t ] = exp [ − 6 ln ( 510 − t ) ] = e ln ( 510 − t ) − 6 = ( 510 − t ) − 6 = 1 ( 510 − t ) 6 Multiply the equation (3) by μ ( t ) and obtain d d t ( μ ( t ) A ) = μ ( t ) Q ( t ) . 1 ( 510 − t ) 6 ( d A d t + 6 510 − t A ) = 1 ( 510 − t ) 6 ( 1 ) 1 ( 510 − t ) 6 d y d t + 1 ( 510 − t ) 6 6 ( 510 − t ) A = 1 ( 510 − t ) 6 d d t ( A ( 510 − t ) 6 ) = 1 ( 510 − t ) 6 To obtain the general solution y ( t ) = 1 μ ( t ) [ ∫ μ ( t ) Q ( t ) d t + C ] , integrate both sides of the equation d d t ( A ( 510 − t ) 6 ) = 1 ( 510 − t ) 6 and solve for y ( t ) by dividing μ ( t ) . A ( 510 − t ) 6 = ∫ ( 510 − t ) − 6 d t A ( 510 − t ) 6 = − ∫ ( u ) − 6 d u ( Take u = 510 − t , then d u = − d t ) A ( 510 − t ) 6 = − ( u ) − 5 − 5 + C A ( 510 − t ) 6 = ( 510 − t ) − 5 5 + C ( Substitute back, u = 510 − t ) Simplify further, A = ( 510 − t ) − 5 ( 510 − t ) 6 5 + C ( 510 − t ) 6 A = 510 − t 5 + C ( 510 − t ) 6 Apply the initial condition A ( 10 ) = 100 − 95 e − 0.1 in the above result to find the constant C . A ( 10 ) = 510 − 10 5 + C ( 510 − 10 ) 6 100 − 95 e − 0.1 = 100 + C ( 500 ) 6 − 95 e − 0.1 = C ( 500 ) 6 C = − 95 e − 0.1 ( 500 ) 6 Substitute C = − 95 e − 0.1 ( 500 ) 6 in the solution A = 510 − t 5 + C ( 510 − t ) 6 and simplify further. A = 510 − t 5 − 95 e − 0.1 ( 500 ) 6 ( 510 − t ) 6 The object is to find the mass of salt in the tank 20 minutes after the leak develops. That is, t = 30 minutes . A ( 30 ) = 510 − 30 5 − 95 e − 0.1 ( 500 ) 6 ( 510 − 30 ) 6 = 480 5 − 95 e − 0.1 ( 500 ) 6 ( 480 ) 6 = 96 − 95 e − 0.1 ( 500 ) 6 ( 480 ) 6 ≈ 28.71448 kg Find the concentration of salt in the tank after 10 min. Concentration = Mass of salt Volume of water in a tank = 28.71448 510 − 30 = 28.71448 480 ≈ 0.0598 kg L Thus, the concentration of salt in the tank 20 minutes after the leakage will be 0.0598 kg L _ .

(a) To Find: The concentration of salt in the tank after 10 minutes. The concentration of salt in the tank after 10 min will be 0.0281 kg L _ . Given: Initially, the tank filled with 500 L of water containing 5 kg of salt. A brine contains 0.2 kilogram (kg) of salt per liter (L) runs into a tank. The brine enters at the rate of 5 L/min and the mixture is kept uniform by stirring. The mixture flows out at the rate of 5 L/min. Calculation: Let A represents the number of kilograms of salt in the tank at t min after the process begins. Use the fact that, rate of increase in A = rate of input − rate of exit . That is, ( rate of increase in A ) = ( rate of salt in − rate of salt out ) d A ( t ) d t = ( 5 L min ) ( 0.2 kg L ) − ( 5 L min ) ( A ( t ) 500 kg L ) d A ( t ) d t = 1 kg min − A ( t ) 100 kg min d A d t = 1 − A 100 d A d t = 100 − A 100 Separate the variables and rewrite the equation as follows, 1 100 − A d A = 1 100 d t Integrate both sides of the equation. ∫ 1 100 − A d A = 1 100 ∫ d t − ln | 100 − A | = t 100 + C 1 ln | 100 − A | = − t 100 + C 2 100 − A = e − t 100 + C 2 Simplify further, 100 − A = e − t 100 e C 2 − A = C e − t 100 − 100 ( C is an arbitrary constant ) A ( t ) = 100 − C e − t 100 ... ( 1 ) Given that, the tank initially filled with 500 L of water containing 5 kg of salt. Substitute A ( 0 ) = 5 in the above result to find the constant C . A ( 0 ) = 100 − C e 0 5 = 100 − C C = 95 Thus, the solution becomes A ( t ) = 100 − 95 e − t 100 ... ( 2 ) Now, find the mass of salt in the tank after 10 min. Substitute t = 10 in equation (2). A ( 10 ) = 100 − 95 e − 10 100 = 100 − 95 e − 0.1 = 100 − 95 e 0.1 ≈ 14.0404452866 kg Find the concentration of salt in the tank after 10 min. Concentration = Mass of salt Volume of water in tank = 14.0404452866 500 kg L ≈ 0.0281 kg L Thus, the concentration of salt in the tank after 10 min will be 0.0281 kg L _ . (b) To Find: The concentration of salt in the tank 20 minutes after the leak develops. The concentration of salt in the tank 20 minutes after the leakage will be 0.0598 kg L _ . The additional mixture flows out at the rate of 1 L/min after 10 minutes, a leak develops in the tank. So the volume of the tank is now changing. After 10 minutes, the initial volume is V ( 10 ) = 500 L . The rate of change of volume of the tank after 10 minutes is given by, d V d t = In volume per unit time − Out volume per unit time = 5 L min − ( 5 + 1 ) L min = − 1 L min Separate the variables and rewrite the equation as follows, d V = − d t Integrate both sides of the equation. ∫ d V = − ∫ d t V = − t + C Apply the initial condition V ( 10 ) = 500 L in the above result to find the constant C . V ( 10 ) = − 10 + C 500 = − 10 + C C = 510 Thus, the volume of the tank after 10 min is V ( t ) = 510 − t . Let A represents the number of kilograms of salt in the tank at t after the leak develops. Use the fact that, rate of increase in A = rate of input − rate of exit . That is, ( rate of increase in A ) = ( rate of salt in − rate of salt out ) d A ( t ) d t = ( 5 L min ) ( 0.2 kg L ) − ( 5 L min + 1 L min ) ( A ( t ) 510 − t kg L ) d A ( t ) d t = 1 kg min − 6 A ( t ) 510 − t kg min d A d t = 1 − 6 A 510 − t d A d t + 6 510 − t A = 1 From part (a), the amount of salt after 10 minutes is A ( 10 ) = 100 − 95 e − 0.1 . Thus, the differential equation becomes d A d t + 6 510 − t A = 1 , A ( 10 ) = 100 − 95 e − 0.1 ... ( 3 ) Solve the differential equation by the method of linear equation as follows. Here, P ( t ) = 6 510 − t and Q ( t ) = 1 . Calculate the integrating factor μ ( t ) by the formula μ ( t ) = exp [ ∫ P ( t ) d t ] . μ ( t ) = exp [ ∫ ( 6 510 − t ) d t ] = exp [ − 6 ln ( 510 − t ) ] = e ln ( 510 − t ) − 6 = ( 510 − t ) − 6 = 1 ( 510 − t ) 6 Multiply the equation (3) by μ ( t ) and obtain d d t ( μ ( t ) A ) = μ ( t ) Q ( t ) . 1 ( 510 − t ) 6 ( d A d t + 6 510 − t A ) = 1 ( 510 − t ) 6 ( 1 ) 1 ( 510 − t ) 6 d y d t + 1 ( 510 − t ) 6 6 ( 510 − t ) A = 1 ( 510 − t ) 6 d d t ( A ( 510 − t ) 6 ) = 1 ( 510 − t ) 6 To obtain the general solution y ( t ) = 1 μ ( t ) [ ∫ μ ( t ) Q ( t ) d t + C ] , integrate both sides of the equation d d t ( A ( 510 − t ) 6 ) = 1 ( 510 − t ) 6 and solve for y ( t ) by dividing μ ( t ) . A ( 510 − t ) 6 = ∫ ( 510 − t ) − 6 d t A ( 510 − t ) 6 = − ∫ ( u ) − 6 d u ( Take u = 510 − t , then d u = − d t ) A ( 510 − t ) 6 = − ( u ) − 5 − 5 + C A ( 510 − t ) 6 = ( 510 − t ) − 5 5 + C ( Substitute back, u = 510 − t ) Simplify further, A = ( 510 − t ) − 5 ( 510 − t ) 6 5 + C ( 510 − t ) 6 A = 510 − t 5 + C ( 510 − t ) 6 Apply the initial condition A ( 10 ) = 100 − 95 e − 0.1 in the above result to find the constant C . A ( 10 ) = 510 − 10 5 + C ( 510 − 10 ) 6 100 − 95 e − 0.1 = 100 + C ( 500 ) 6 − 95 e − 0.1 = C ( 500 ) 6 C = − 95 e − 0.1 ( 500 ) 6 Substitute C = − 95 e − 0.1 ( 500 ) 6 in the solution A = 510 − t 5 + C ( 510 − t ) 6 and simplify further. A = 510 − t 5 − 95 e − 0.1 ( 500 ) 6 ( 510 − t ) 6 The object is to find the mass of salt in the tank 20 minutes after the leak develops. That is, t = 30 minutes . A ( 30 ) = 510 − 30 5 − 95 e − 0.1 ( 500 ) 6 ( 510 − 30 ) 6 = 480 5 − 95 e − 0.1 ( 500 ) 6 ( 480 ) 6 = 96 − 95 e − 0.1 ( 500 ) 6 ( 480 ) 6 ≈ 28.71448 kg Find the concentration of salt in the tank after 10 min. Concentration = Mass of salt Volume of water in a tank = 28.71448 510 − 30 = 28.71448 480 ≈ 0.0598 kg L Thus, the concentration of salt in the tank 20 minutes after the leakage will be 0.0598 kg L _ .

Solution Summary: The author calculates the concentration of salt in the tank after 10 minutes. The brine enters at the rate of 5 L/min and the mixture is kept uniform by stirring.

Fundamentals of Differential Equations (9th Edition)

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Fundamentals of Differential Equations (9th Edition)

Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 16, determine whether the given...Ch. 2.2 - In Problems 716, solve the equation. 7. xdydx=1y3 Ch. 2.2 - In Problems 716, solve the equation. 8. dxdt=3xt2 Ch. 2.2 - In Problems 716, solve the equation. 9....Ch. 2.2 - In Problems 716, solve the equation. 10....

Ch. 2.2 - In Problems 716, solve the equation. 11....Ch. 2.2 - In Problems 716, solve the equation. 12....Ch. 2.2 - In Problems 716, solve the equation. 13....Ch. 2.2 - In Problems 716, solve the equation. 14. dxdtx3=x Ch. 2.2 - In Problems 716, solve the equation. 15....Ch. 2.2 - In Problems 716, solve the equation. 16. y1 dy +...Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 23E Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - In Problems 1726, solve the initial value problem....Ch. 2.2 - Prob. 27E Ch. 2.2 - Sketch the solution to the initial value problem...Ch. 2.2 - Uniqueness Questions. In Chapter 1 we indicated...Ch. 2.2 - As stated in this section, the separation of...Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.2 - Mixing. Suppose a brine containing 0.3 kilogram...Ch. 2.2 - Newtons Law of Cooling. According to Newtons law...Ch. 2.2 - Prob. 35E Ch. 2.2 - Prob. 36E Ch. 2.2 - Compound Interest. If P(t) is the amount of...Ch. 2.2 - Free Fall. In Section 2.1, we discussed a model...Ch. 2.2 - Grand Prix Race. Driver A had been leading...Ch. 2.2 - Prob. 40E Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 16, determine whether the given...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - In Problems 716, obtain the general solution to...Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - In Problems 1722, solve the initial value problem....Ch. 2.3 - Radioactive Decay. In Example 2 assume that the...Ch. 2.3 - Prob. 24E Ch. 2.3 - (a) Using definite integration, show that the...Ch. 2.3 - Prob. 26E Ch. 2.3 - Constant Multiples of Solutions. (a) Show that y =...Ch. 2.3 - Prob. 29E Ch. 2.3 - Bernoulli Equations. The equation (18) dydx+2y=xy2...Ch. 2.3 - Prob. 31E Ch. 2.3 - Prob. 32E Ch. 2.3 - Prob. 33E Ch. 2.3 - Prob. 34E Ch. 2.3 - Prob. 35E Ch. 2.3 - Prob. 36E Ch. 2.3 - Prob. 37E Ch. 2.3 - Prob. 38E Ch. 2.3 - Prob. 39E Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - Prob. 3E Ch. 2.4 - Prob. 4E Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - In Problems 18, classify the equation as...Ch. 2.4 - Prob. 9E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 15E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 17E Ch. 2.4 - In Problems 920, determine whether the equation is...Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 25E Ch. 2.4 - In Problems 2126, solve the initial value problem....Ch. 2.4 - Prob. 27E Ch. 2.4 - For each of the following equations, find the most...Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Orthogonal Trajectories. A geometric problem...Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.5 - Prob. 1E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 3E Ch. 2.5 - Prob. 4E Ch. 2.5 - In Problems 16, identify the equation as...Ch. 2.5 - Prob. 6E Ch. 2.5 - Prob. 7E Ch. 2.5 - Prob. 8E Ch. 2.5 - Prob. 9E Ch. 2.5 - Prob. 10E Ch. 2.5 - Prob. 11E Ch. 2.5 - Prob. 12E Ch. 2.5 - Prob. 13E Ch. 2.5 - Prob. 14E Ch. 2.5 - Prob. 15E Ch. 2.5 - Prob. 16E Ch. 2.5 - Prob. 17E Ch. 2.5 - Prob. 18E Ch. 2.5 - Prob. 19E Ch. 2.5 - Verify that when the linear differential equation...Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 2E Ch. 2.6 - Prob. 3E Ch. 2.6 - Prob. 4E Ch. 2.6 - Prob. 5E Ch. 2.6 - Prob. 6E Ch. 2.6 - In Problems 18, identify (do not solve) the...Ch. 2.6 - Prob. 8E Ch. 2.6 - Use the method discussed under Homogeneous...Ch. 2.6 - Prob. 10E Ch. 2.6 - Prob. 11E Ch. 2.6 - Prob. 12E Ch. 2.6 - Prob. 13E Ch. 2.6 - Prob. 14E Ch. 2.6 - Prob. 15E Ch. 2.6 - Prob. 16E Ch. 2.6 - Prob. 17E Ch. 2.6 - Prob. 18E Ch. 2.6 - Prob. 19E Ch. 2.6 - Prob. 20E Ch. 2.6 - Prob. 21E Ch. 2.6 - Prob. 22E Ch. 2.6 - Use the method discussed under Bernoulli Equations...Ch. 2.6 - Prob. 24E Ch. 2.6 - Prob. 25E Ch. 2.6 - Prob. 26E Ch. 2.6 - Prob. 27E Ch. 2.6 - Prob. 28E Ch. 2.6 - Use the method discussed under Equations with...Ch. 2.6 - Prob. 30E Ch. 2.6 - Prob. 31E Ch. 2.6 - Prob. 32E Ch. 2.6 - Prob. 33E Ch. 2.6 - Prob. 34E Ch. 2.6 - Prob. 35E Ch. 2.6 - In Problems 3340, solve the equation given in: 36....Ch. 2.6 - Prob. 37E Ch. 2.6 - Prob. 38E Ch. 2.6 - Prob. 39E Ch. 2.6 - Prob. 40E Ch. 2.6 - Prob. 41E Ch. 2.6 - Prob. 42E Ch. 2.6 - Prob. 43E Ch. 2.6 - Show that equation (13) reduces to an equation of...Ch. 2.6 - Prob. 45E Ch. 2.6 - Prob. 46E Ch. 2.6 - Prob. 47E Ch. 2.6 - Prob. 48E Ch. 2 - In Problems 130, solve the equation. 1....Ch. 2 - Prob. 2RP Ch. 2 - Prob. 3RP Ch. 2 - Prob. 4RP Ch. 2 - Prob. 5RP Ch. 2 - In Problems 130, solve the equation. 6. 2xy3 dx ...Ch. 2 - In Problems 130, solve the equation. 7. t3y2 dt +...Ch. 2 - Prob. 8RP Ch. 2 - In Problems 130, solve the equation. 9. (x2 + y2)...Ch. 2 - Prob. 10RP Ch. 2 - Prob. 11RP Ch. 2 - Prob. 12RP Ch. 2 - Prob. 13RP Ch. 2 - Prob. 14RP Ch. 2 - Prob. 15RP Ch. 2 - Prob. 16RP Ch. 2 - Prob. 17RP Ch. 2 - Prob. 18RP Ch. 2 - Prob. 19RP Ch. 2 - Prob. 20RP Ch. 2 - Prob. 21RP Ch. 2 - Prob. 22RP Ch. 2 - Prob. 23RP Ch. 2 - In Problems 130, solve the equation. 24. (y/x +...Ch. 2 - Prob. 25RP Ch. 2 - Prob. 26RP Ch. 2 - Prob. 27RP Ch. 2 - Prob. 28RP Ch. 2 - Prob. 29RP Ch. 2 - Prob. 30RP Ch. 2 - Prob. 31RP Ch. 2 - Prob. 32RP Ch. 2 - Prob. 33RP Ch. 2 - Prob. 34RP Ch. 2 - Prob. 35RP Ch. 2 - Prob. 36RP Ch. 2 - Prob. 37RP Ch. 2 - Prob. 38RP Ch. 2 - Prob. 39RP Ch. 2 - Prob. 40RP Ch. 2 - Prob. 41RP

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