A rare species of insect was discovered in the amazon rainforest. To protect the species, environmentalists declare theinsect endangered and transplant the insect into a protected area.The population P of the insect t months after being transplantedis P(t)=(50(1+0.5t))/((2+0.01t)) (a) How many insects were discovered? In other words, what wasthe population when t=0? (b) What will the population be after 5 years? (c)Determine the horizontal asymptote of P(t).What is thelargest population that the protected area can sustain?

Makena Preston 2022-08-01 Answered
A rare species of insect was discovered in the amazon rainforest. To protect the species, environmentalists declare theinsect endangered and transplant the insect into a protected area.The population P of the insect t months after being transplantedis
P ( t ) = 50 ( 1 + 0.5 t ) ( 2 + 0.01 t )
(a) How many insects were discovered? In other words, what wasthe population when t=0?
(b) What will the population be after 5 years?
(c)Determine the horizontal asymptote of P(t).What is thelargest population that the protected area can sustain?
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Answers (1)

grocbyntza
Answered 2022-08-02 Author has 25 answers
a.)
P(0) is the population at t=0 so substitute into the equation and solve
P ( 0 ) = 50 ( 1 + .5 ( 0 ) ) 2 + .01 ( 0 ) = 25
b.)
because t is in months convert years to months and substitute (60months)
P ( 0 ) = 50 ( 1 + .5 ( 60 ) ) 2 + .01 ( 60 ) = 596.1
c.)
You basically want to know the value when T equals . This will result in the ratio of the coefficents of the variable which will be 25/.01 which is 2500. So 2500 is the max this population can handle.
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