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We have found the following equation for the amount of chemical C, X(t). 150 - X = ce180kt 120 - 2X To solve for the constants, we use the initial conditions. At time t = 0, no quantity of chemical C has been produced, or X(0) = 0. This can be used to solve for c. 150 - X ce180kt %3D 120 2X 150 – 0 ce180k(0) 120 C = We also know that 10 grams of C is formed after 9 minutes, or X(9) = 10, which can be used to solve for k. (Round your final answer to 7 decimal places.) 150 – 10 5180k(9) 120 – 2(10) = e1,620k 500 In = 1,620k 500 k =

We have found the following equation for the amount of chemical C, X(t). 150 - X = ce180kt 120 - 2X To solve for the constants, we use the initial conditions. At time t = 0, no quantity of chemical C has been produced, or X(0) = 0. This can be used to solve for c. 150 - X ce180kt %3D 120 2X 150 – 0 ce180k(0) 120 C = We also know that 10 grams of C is formed after 9 minutes, or X(9) = 10, which can be used to solve for k. (Round your final answer to 7 decimal places.) 150 – 10 5180k(9) 120 – 2(10) = e1,620k 500 In = 1,620k 500 k =

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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We have found the following equation for the amount of chemical C, X(t). 150 – X = ce180kt 120 - 2X To solve for the constants, we use the initial conditions. At time t = 0, no quantity of chemical C has been produced, or X(0) = 0. This can be used to solve for c. 150 - X ce180kt %3D 120 2X 150 – 0 ce180k(0) 120 C = 4 We also know that 10 grams of C is formed after 9 minutes, or X(9) = 10, which can be used to solve for k. (Round your final answer to 7 decimal places.) 150 – 10 5,180k(9) 120 – 2(10) = e1,620k 500 In = 1,620k 500 k =
Transcribed Image Text:We have found the following equation for the amount of chemical C, X(t). 150 – X = ce180kt 120 - 2X To solve for the constants, we use the initial conditions. At time t = 0, no quantity of chemical C has been produced, or X(0) = 0. This can be used to solve for c. 150 - X ce180kt %3D 120 2X 150 – 0 ce180k(0) 120 C = 4 We also know that 10 grams of C is formed after 9 minutes, or X(9) = 10, which can be used to solve for k. (Round your final answer to 7 decimal places.) 150 – 10 5,180k(9) 120 – 2(10) = e1,620k 500 In = 1,620k 500 k =
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