SOLUTION: Consider this scenario: For each year t, the population of a forest of trees is represented by the function
A(t) = 116(1.023)^t.
In a neighboring forest, the population of the s
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Question 1128727: Consider this scenario: For each year t, the population of a forest of trees is represented by the function
A(t) = 116(1.023)^t.
In a neighboring forest, the population of the same type of tree is represented by the function
B(t) = 81(1.021)^t.
By how many trees does forest B over forest A after 20 years?
Can someone explain how to properly set up the equation to solve it?
Found 4 solutions by josgarithmetic, greenestamps, Cityscape16, ikleyn:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
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Consider this scenario: For each year t, the population of a forest of trees is represented by the function
A(t) = 116(1.023)^t.
In a neighboring forest, the population of the same type of tree is represented by the function
B(t) = 81(1.021)^t.
By how many trees does forest B over forest A after 20 years?
Can someone explain how to properly set up the equation to solve it?
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The question seems to be asking for the difference, but the functions don't seem to fit that. Is the question really wrong, and should instead be:
"By how many more trees will forest A have than forest B, after twenty years?" This should be .
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
The meaning of the question that is asked is uncertain, because the language makes no sense:
"By how many trees does forest B over forest A after 20 years?"
It appears we are to find how many more trees are in forest B than forest A after 20 years. But forest B initially has fewer trees than forest A, and the rate of increase for forest B is smaller -- so forest B will never have more trees than forest A.
Answer by Cityscape16(2) (Show Source): You can put this solution on YOUR website!
I went ahead and subtracted 114-81 which is 28 trees. That was the correct answer.
Answer by ikleyn(52787) (Show Source): You can put this solution on YOUR website!
.
I do not think that the post by @Cityscape is an appropriate explanation/solution.
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