SOLUTION: Hi....I am having a really hard time with word problems. I am not sure how to make this into an equation. Please help!!! A bug population initially has 10^5 bugs, but is decreas

Question 163678: Hi....I am having a really hard time with word problems. I am not sure how to make this into an equation. Please help!!!
A bug population initially has 10^5 bugs, but is decreasing at a steady rate. In 30 days there are only 3.325 x 10^4 bugs. When was there 1.1 x 10^4 bugs? To the nearest day, when will there be no bugs?
Thanks,
Kathy Gafford
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A bug population initially has 10^5 bugs, but is decreasing at a steady rate.
In 30 days there are only 3.325 x 10^4 bugs. When was there 1.1 x 10^4 bugs?
:
Find the reduction in bugs per day (use a calc):
:
= 2225 bugs per day less
:
Let t = no. of days
:
10^5 - 2225t = 1.1(10^4)
:
Simplify divide equation by 10000 (10^4)
10 - .2225t = 1.1
:
10 - 1.1 = .2225t
:
8.9 = .2225t
t =
t = 40 days, bugs reduced to 3.325(10^4)
:
To the nearest day, when will there be no bugs?
Same equation except
10^5 - 2225t = 0
:
See if you can solve this now:
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