Consider the growth of a population p(t). it starts out with p(0) = a. suppose the growth is unchecked, and hence p' = k p for some constant k. then p(t) = ae^(kt) of course populations don't grow forever. let's say there is a stable population size q that p(t) approaches as time passes. thus the speed at which the population is growing will approach zero as the population size approaches q. one way to model this is via the differential equation p' = kp(q-p), p(0) = a. the solution of this initial value problem is p(t) =

b

step-by-step explanation:

answer: evaluate the objective function at each vertex to determine which - and -values, if any, maximize or minimize the function. example: find the minimum value and maximum value of the objective function f ( x , y ) = 4 x + 5 y , subject to the following constraints

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1

27 +27 or 27times 2=54

the diffrence is 0