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Exercise (b) For what values of P is the population decreasing? Step 1 N is decreasing when N' is negative Step 2 To find where N is decreasing, we must decide where N' = 1.7N 7N(1-. Assuming that N> 0 always, then we need 1 - Exercise (c) What are the equilibrium solutions? negative Therefore, N' is negative whenever NE (5200,00) Step 1 This occurs when N = and en Equilibrium solutions occur when 0 = N' = 1.7 Submit Skip (you cannot come back) N 5200 N * = 1.7N(1- 5200 N 5200 < 0. - (0, 5200) N 5200 your answer using interval is negative. (5200,00) (Enter your answer using interval notation.) = 0. So the equilibrium solutions are N = 0 X. (Enter answers as a comma-separated list.)

          Exercise (b) For what values of P is the population decreasing? Step 1 N is decreasing when N' is negative Step 2 To find where N is decreasing, we must decide where N' = 1.7N 7N(1-. Assuming that N> 0 always, then we need 1 - Exercise (c) What are the equilibrium solutions? negative Therefore, N' is negative whenever NE (5200,00) Step 1 This occurs when N = and en Equilibrium solutions occur when 0 = N' = 1.7 Submit Skip (you cannot come back) N 5200 N * = 1.7N(1- 5200 N 5200 < 0. - (0, 5200) N 5200 your answer using interval is negative. (5200,00) (Enter your answer using interval notation.) = 0. So the equilibrium solutions are N = 0 X. (Enter answers as a comma-separated list.)

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