Question

HELP WILL GIVE ALOT OF POINT AND WILL GIVE BRAINLYIEST

Answer 1

Answer:  C. About 6 days

Step-by-step explanation:

Here the function that shows the population of fish after x weeks,

$P(x) = 6^{bx}$

Where b is any unknown,

If b = 3,

Then, the function is,

$P(x) = 6^{3x}$

Which is a exponentially increasing function,

That having y-intercept = (0,1)

And, horizontal asymptote,

y = 0

End behavior of the function:

As $x\rightarrow \infty$ , $y\rightarrow \infty$

As $x\rightarrow -\infty$ , $y\rightarrow 0$

Thus, by the above information we can graph the given relation.

Now, For at least 100 fish in the pound,

$6^{3x}\geq 100$

By taking log on both sides,

$3x log(6) \geq log(100)$

$3x\geq \frac{log(100)}{log(6)}$

$3x\geq 2.57019441788$

$x\geq 0.85673147262\approx 0.8567$

Thus, after 0.8567 weeks (approx) the fish on the pound will be at least 100.

1 week = 7 days,

0.8567 weeks = 5.9969 days ≈  6 days

Hence, after 6 days (approx) the fish on the pound will be at least 100.