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
C= πr²
C= 3.14×8²
C=3.14×64
C=200.96
Answer:
1. -370
2. 70
3. 10.75
4. 6700
5. 37.26
Step-by-step explanation:
1. Estimating the solution means round the numbers to the nearest ten, then subtract the rounded numbers.
2. Same thing as #1
3. Using long division, it should be quite easy. If you need more evaluation, comment and I'll answer.
4. To the nearest hundred, you write the closest exact hundred. example: 8334 would be rounded to 8300.
5. To the nearest hundredth, you write the closest exact hundredth. example: 65.734 would be rounded to 65.73
Hope this helped, good luck!
Answer:
a. -64
b. 115
think its good not sure tho
