Simulate the predator-prey equations. a. p (t ) =  p (t ) −  p (t ) r (t ) r (t ) = − r (t ) +  p (t ) r (t ) hint: i used the values alp=0.04; gam=0.08; bet=0.002; del=0.0004. b. add controls u, v to the predator-prey equations. p (t ) =  p (t ) −  p (t ) r (t ) + u (t ) r(t)= −r(t)+ p(t)r(t)+v(t) the controls could be predator or prey animals added or subtracted from the population (at some rate). suppose, for example, suppose prey animals are added at a fixed rate over a fixed period u(t)=u00 t1tt2  0 elsewhere investigate the impact of this control on the evolution of the population. c. find if possible, feedback control laws that can move the non-trivial equilibrium from its uncontrolled value to a target value. d. use the simulator to check the stability of the new (controlled) equilibrium.

salary

step-by-step explanation:

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

two? xd

step-by-step explanation:

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0