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The town experiences quadratic growth of the form P=8t squared +Po, where t is the time in years from 2010.?
A) Insert the value Po =1000 into the equation, use t^2 to type t-squared...Answer
B) Fill in the following Year(t)......Population(p)
t=0 Po=1000
(2010)
t=1 Po= ???
(2011)
t=2 Po= ???
(2012)
t=3 Po ???
(2013)
t=6 Po ???
(2016)
C) Use your equation from part a to approximate how many years it will take for the population to reach 7000, round to nearest whole year.
D) Graph this function in MS Excel by plotting the points found in your chart in part b. Label your axes with time on the x-axis and population on the y-axis.
2 Answers
- 1 decade agoFavorite Answer
Po represents a constant. It's value is the initial population and won't change throughout the entire problem.
a) so from the problem, Po = 1000, for all values of t and P
and your equation becomes
P = 8t^2 + 1000
b) just fill in the values
t=0 Po=1000
P = 8(0)^2 + 1000 = 1000
t=1 Po will not change, it's still 1000
P = 8*(1)^2 + 1000 = 1008
t=2
P = 8*(2)^2 + 1000 = 8*4 + 1000 = 1032
t=3
P = 8*(3)^2 + 1000 = 8*9 + 1000 = 1072
t=6
P = 8*(6)^2 + 1000 = 8*36 + 1000 = 1288
c) they give you the final population P
so you need to solve for the number of years t
7000 = 8t^2 + 1000
6000 = 8t^2
750 = t^2
t = √750
either use your calculator if you are allowed or
√750 = √25*√30 = 5√30
d) just plug in the points in excel and let P = y-axis on the chart and t = x axis
- fontagneLv 45 years ago
For section a: you want the values of P no longer Po. Po is inhabitants at t=0. So Po = 1500. The equation is: P = 8t^2 + 1500. Plug the values of t into the equation. The solutions are the ensuing cost of inhabitants length (P). Eg: whilst t=2, P = 8(2)^2 + 1500 etc. For section b: you need to come across t whilst P = 7000. Eqn is: P = 8*t^2 + Po 7000 = 8t^2 + 1500 Rearrange this with t on the left. sparkling up for t. around t to the closest comprehensive quantity. it incredibly is your answer.
