The population of a community of foxes is observed to fluctuate on a 10-year cycle due to variations in the availability of prey. When population measurements began (t=0, the population was 35 foxes. The growth rate in units of foxes>>year was observed to be P′(t)=5+10sinπt/5  a. What is the population 15 years later? 35 years later?
b. Find the population P(t) at any time t≥0.

Population 15 years later  P(15)  =  100 + 50/π

Population 35 years later P(35)  =  200 + 50/π

Population any t ≥ 0    P(t) = 35 + 50 /π + 5*t + 10*cos(π*t/5)  

Step-by-step explanation:

P´(t)  =  5  + 10*sinπt/5       ⇒    dP/dt =  5  + 10*sinπt/5    

dP = ( 5  + 10*sinπt/5  ) *dt

P(t) = ∫  ( 5  + 10*sinπt/5 ) dt

P(t)  =  5*t  +  10 * ∫ sinπ*t/5* dt

P(t)  =  5*t  -  10*5/π *cos πt/5 + k

To determine k    t = 0   P(t) = 35

P(0) = 5*0  - 50/π (1) + k

35 = - 50/π  + k

k  =  35  + 50/π     and

P(t)  =  5*t  + 10*cos(π*t/5) + 35 + 50/π

b)P(t) = 35 + 50 /π + 5*t + 10*cos(π*t/5)    (1)

a) Population 15 years later

P(15)  =  35 + 50/π  + 5*15 - 10

P(15)  =  100 + 50/π

Again from equation  (1)

P(35)  =  35 + 50 /π + 5*35 + 10*cos(35*π/5)

P(35)  =  35 + 50/π  + 175 + 10*cos (7*π )

P(35)  =  210 + 50/π  - 10

P(35)  =  200 + 50/π

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)