dP Answer parts (a) and (b) using the inhibited growth model - dt - = k(L - P). Let L = 6000, and assume that P passes through the point (17,460). Assume that P(0) = 0. a) Find the particular solution P. b) When will P(t) = 0.9L? . . . a) P(t) = (Type an exact answer in terms of e. Use integers or decimals for any numbers in the expression. Round to five decimal places as needed.)dP Use the limited growth model = kP(L - P) to answer parts (a) through (c). dt a) Let L = 5000, k = 0.0002, and Po = 160. Find the particular solution P. b) Find t when P = 2820. c) Find the point of inflection. a) P(t) = (Type an exact answer in terms of e.)Find the general solution of the differential equation using an integration factor. y' + 5y = 2 . . . y(X) =A ceramic bowl leaves a kiln with a temperature of 240 C and is placed in a room with a constant temperature of 25 C. After 45 min, the bowl's temperature is 208 C. a) Write an equation that gives the bowl's temperature T after t minutes. b) When will the bowl's temperature be 120 C? a) First, set up a differential equation that represents the rate at which the temperature of the bowl changes. Choose the correct answer below. dT O A. dt - = - k(T -25), where k is a constant greater than 0, and T is temperature dT O B. dt = - kT - 25, where k is a constant greater than 0, and T is temperature dT O C. dt .= -25(T - k), where k is a constant greater than 0, and T is temperature dT O D. dt = -25(k - T), where k is a constant greater than 0, and T is temperature