High School

Water hyacinth is an invasive plant species found in many lakes that typically grows at a rate of [tex]$7 \%$[/tex] per day. As part of a study, a scientist introduces a 150-gram sample of water hyacinth into a testing pool.

Which of the following functions gives the amount of water hyacinth in the testing pool [tex]$t$[/tex] weeks after the sample is introduced? (Note: 1 week is 7 days.)

A. [tex]f(t)=150\left(1+0.07^{(1 / 7)}\right)^t[/tex]

B. [tex]g(t)=150\left(1.07^{(1 / 7)}\right)^t[/tex]

C. [tex]h(t)=150\left(1+0.07^{(7)}\right)^t[/tex]

D. [tex]k(t)=150\left(1.07^{(7)}\right)^t[/tex]

Answer :

To solve this problem, we need to determine the weekly growth rate of the water hyacinth, which grows at a rate of 7% per day, and then write a function that represents the amount of water hyacinth in the pool weeks after it is introduced.

### Step-by-step Solution

1. Understanding the Daily Growth Rate:
- The water hyacinth grows at a rate of 7% per day. This means that each day, the quantity of water hyacinth is multiplied by 1.07 (since 100% + 7% = 107% = 1.07).

2. Converting to Weekly Growth Rate:
- Since we want to know the function in terms of weeks, and there are 7 days in a week, we need to find the growth rate over 7 days.
- We do this by raising the daily growth factor (1.07) to the power of 7, which is mathematically expressed as [tex]\(1.07^7\)[/tex].

3. Equation for Weekly Growth Rate:
- Calculating this, we find that the weekly growth factor is approximately 1.6058. This means that after one week, the water hyacinth will grow by a factor of about 1.6058.

4. Creating the Function for [tex]\(t\)[/tex] Weeks:
- If [tex]\(t\)[/tex] represents the number of weeks, the amount of water hyacinth after [tex]\(t\)[/tex] weeks can be represented by multiplying the initial amount (150 grams) by the weekly growth factor raised to the power of [tex]\(t\)[/tex].
- Therefore, the function is:
[tex]\[ k(t) = 150 \times (1.07^7)^t \][/tex]

5. Select the Correct Option:
- By examining the choices, the function [tex]\( k(t) = 150(1.07^7)^t \)[/tex] corresponds to option (D).

Hence, option (D) is the correct choice as it properly accounts for the weekly exponential growth based on a daily growth rate of 7%.