The fish population in a pond with carrying capacity 1200 is modeled by the following logistic equation where N(t) denotes the number of fish at time t in years. (dN)/(dt) = (0.4)/(1200)N(1200-N)
Starting at the time at which the number of fish reached 150, the owner of the pond removed (harvested) fish at a constant rate of 50 fish per year. We take t = 0 to be the time at which the owner started to harvest fish from the pond.

A) Modify the differential equation to model the population of fish from the time it reached 150.
dN/dt =

B) How far below the original carrying capacity will the number of fish be in the long run? (Give your answer correct to the nearest whole fish.)
Number of fish=

9514 1404 393

  (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

so a parallelogram is just like a square base times height. multiply 40 and 39 to get 1560 inches.

a triangle is base times height divided by two. so multiply 7.9 and 5 and then divide it by 2 to get 19.75inches.

the kite can be split into two separate triangles and added together to get the total area. 30 x 4 /2 which equals 60. make sure to double it for the two triangles making 120.

answer: what is the question you're trying to ask?

step-by-step explanation: