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madam [21]
2 years ago
10

4²(4²) 2. 5⁴(5)÷2¹ 3. (5+3)² Solve the following expressions

Mathematics
1 answer:
omeli [17]2 years ago
3 0
  1. 4^2*4^2
  2. 4^4
  3. 256

  1. 5^4(5)/2^1
  2. 625*5/2^1
  3. 625*5/2
  4. 625*5/2
  5. 3125/2

  1. (5+3)^2
  2. 8^2
  3. 64

Hope this helps, have a BLESSED and wonderful day! :-)

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Answer:

a) Equilibrium point : [ 947, 53 ]

b) N = 947 is stable equilibrium, N = 53  is unstable equilibrium

c) N0, the population will not go extinct

Step-by-step explanation:

a)

Given that;

r = 2, k = 1000, H = 100

dN/dT = R(1 - N/k)N - H

so we substitute

dN/dt = 2( 1 - N/1000)N - 100

now for equilibrium solution, dN/dt = 0

so

2( 1 - N/1000)N - 100 = 0

((1000 - N)/1000)N = 50

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N = 1000 ± √(-1000)² - 4(1)(50000)) / 2(1)

N = 947.213 OR 52.786

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N = 947 OR 53

Therefore Equilibrium point : [ 947, 53 ]

b)

g(N) = 2( 1 - N/1000)N - 100

= 2N - N²/500 - 100

g'(N) = 2 - N/250

SO AT 947

g'(N) = g'(947) =  2 - 947/250 = -1.788 which is less than (<) 0

so N = 947 is stable equilibrium

now AT 53

g'(N) = g"(53) = 2 - 53/250 = 1.788 which is greater than (>) 0

so N = 53  is unstable equilibrium

The capacity k=1000

If the population is less than 53 then the population will become extinct but since the capacity is equal to 1000 then the population will not go extinct.

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Step-by-step explanation:

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