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

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

Step-by-step 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!

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb

answer: 2x + 2 − 2 = 10 − 2; subtraction property of equality

step-by-step explanation: