Question

he population of fish in a pond grew over a five-year period according to the function P of t equals 20 left-parenthesis 2.50 right-parenthesis Superscript StartFraction t over 5 EndFraction Baseline, where t represents time in years.



Move the options to the spaces to rewrite P of t in an equivalent form that reveals the annual growth rate of the fish.

Response area with 2 blank spaces
P of t equals
Blank space 2 empty
left-parenthesis
Blank space 3 empty
right-parenthesis Superscript t Baseline
Answer options with 7 options.

Answer 1

To rewrite \( P(t) = 20 \left(2.50\right)^{\frac{t}{5}} \) in a form that reveals the annual growth rate, we can re-express the term \( \left(2.50\right)^{\frac{t}{5}} \).

First, we can express \( 2.50 \) as follows:

\[
2.50 = (2.50)^{1} = (2.50)^{\frac{5}{5}} = \left((2.50)^{\frac{1}{5}}\right)^{t}
\]

Thus,

\[
\left(2.50\right)^{\frac{t}{5}} = \left((2.50)^{\frac{1}{5}}\right)^{t}
\]

Now we can rewrite \( P(t) \) as:

\[
P(t) = 20 \cdot \left((2.50)^{\frac{1}{5}}\right)^{t}
\]

In order to fill in the blanks, we can identify:

1. The first blank (blank space 2) will contain the coefficient (20).
2. The second blank (blank space 3) will contain \( (2.50)^{\frac{1}{5}} \).

Therefore, filling in the response area, we get:

\[
P(t) = 20 \left( (2.50)^{\frac{1}{5}} \right)^{t}
\]

So the final answer is:

- Blank space 2: 20
- Blank space 3: \( (2.50)^{\frac{1}{5}} \)