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The fish population in a local lake is 10,000. The population increase by half every year. Which function represents the fish population after x years?

1. f(x) = 1.5(10,000)^x
2. 10,000(-1.5)^x
3. None
4. f(x) = 10,000(0.5)^x
5. f(x) = 10,000 (1.5)^x

User Davidcelis
by
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1 Answer

3 votes

Answer:

5. f(x) = 10,000 (1.5)^x

Explanation:

We would have to multiply the original amount by 1.50^x because the initial amount would be 1, and 50% increase would be .5 so 1.5 and you raise it to the number of years to show the total increase.

Let's test it.

Initial:

10,000

After 1 year

10,000 + (.5*10000)

10,000 + 5000 = 15,000

After 2 years

15,000 + (.5*15000)

15,000 + 7500 = 22,500

Let's try our equation.

f(x) = 10,000 (1.5)^x

x = 2

10,000(1.5)^2

10,000(2.25) = 22,500

The same!

User Abhishek Dey
by
8.7k points

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