1337489 is the answer lolz
9514 1404 393
Answer:
(dN)/(dt) = (0.4)/(1200)N(1200-N) -50
142 fish
Step-by-step explanation:
A) The differential equation is modified by adding a -50 fish per year constant term:
(dN)/(dt) = (0.4)/(1200)N(1200-N) -50
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B) The steady-state value of the fish population will be when N reaches the value that makes dN/dt = 0.
(0.4/1200)(N)(1200-N) -50 = 0
N(N-1200) = -(50)(1200)/0.4) . . . . rewrite so N^2 has a positive coefficient
N^2 -1200N + 600^2 = -150,000 +600^2 . . . . complete the square
(N -600)^2 = 210,000 . . . . . simplify
N = 600 + √210,000 ≈ 1058
This steady-state number of fish is ...
1200 - 1058 = 142 . . . . below the original carrying capacity
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Step-by-step explanation:
See attached picture for the answer:
