SOLUTION: The fish population in a certain lake rises and falls according to the formula. F=1000(30+15t-t2). Here F is the number of fish at a time t , where t is measured in y

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Question 339926: The fish population in a certain lake rises and falls according to
the formula.
F=1000(30+15t-t2).
Here F is the number of fish at a time t , where t is measured in years
since January 1, 2002. Use this information to answer the following
questions.

1. The fish population will again be the same as on January 1, 2002
on the date January 1,_________(that is____________ years later).

2. All the fish in the lake will die after 'approximately'
t=_________________years (the answer might be a decimal numbers)

Answer by jrfrunner(365) About Me  (Show Source):
You can put this solution on YOUR website!
F=1000%2830%2B15t-t%5E2%29
==
1. The fish population will again be the same as on January 1, 2002
January 1,2002 means t=0
when t=0, F=30,000
so you want to know when F=30000=1000%2830%2B15t-t%5E2%29
which means 30000%2F1000=30%2B15t-t%5E2
0=15t-t%5E2=%2815-t%29t
t=0 or t=15
answer t=15 years from Jan 1, 2002 or January 1, 2017
----
2. All the fish in the lake will die after 'approximately'
You basically want to know t, for F=0
F=1000%2830%2B15t-t%5E2%29
0=1000%2830%2B15t-t%5E2%29
0=30%2B15t-t%5E2
use quadratic equation with a=-1, b=15, c=30
t=%28-15-sqrt%2815%5E2-4%28-1%2930%29%29%2F%282%28-1%29%29=16.79 or t=%28-15%2Bsqrt%2815%5E2-4%28-1%2930%29%29%2F%282%28-1%29%29=-1.79
answer: 16.79 years after january 1, 2002 or roughly Sept 14,2018