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The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.
Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).
(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get
P(t)=40
(50(1+0.05t))/(2+0.01t)=40
Cross multiply both sides, we get
50(1+0.05t)=40(2+0.01t)
Apply the distributive property a(b+c)=ab+ac, we get
50+2.5t=80+0.4t
Subtract 0.4t and 50 from both sides, we get
50+2.5t-0.4t-50=80+0.4t-0.4t-50
2.1t=30
Divide both sides with 2.1, we get
t=14.29 months
(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get
(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.
Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.
Learn more about limiting factor from here brainly.com/question/18415071.
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I think that equation is supposed to look like this:
The reason I say that is because this is an exponential decay problem, with a starting amount of 50 pieces of candy.
If that's the case, then it would look like this with a t value of 5:
and
so the amount of candy left after 5 minutes is 16.384 or 16
The reason I say that is because this is an exponential decay problem, with a starting amount of 50 pieces of candy.
If that's the case, then it would look like this with a t value of 5:
and
so the amount of candy left after 5 minutes is 16.384 or 16