Answer:
Given that
N(t)=2t³+3t²-5t+1000
By differentiating with respect to t
[tex]\frac{dN}{dt}=6t^2+6t-5[/tex]
a)
At t= 2 year
[tex]\frac{dN}{dt}=6t^2+6t-5[/tex]
[tex]\frac{dN}{dt}=6\times 2^2+6\times 2-5[/tex]
[tex]\frac{dN}{dt}=31\ turtles/yr[/tex]
b)
at t= 8 yr
[tex]\frac{dN}{dt}=6t^2+6t-5[/tex]
[tex]\frac{dN}{dt}=6\times 8^2+6\times 8-5[/tex]
[tex]\frac{dN}{dt}=427\ turtles/yr[/tex]
c)
At t= 10 year
N(t)=2t³+3t²-5t+1000
N(t)=2 x 8³+3 x 8²-5 x 8+1000
N(t)=2,176 turtles