SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given equation for explicit formula
![\begin{gathered} a_n=a_1\cdot r^(n-1) \\ where\text{ }a_1\text{ is the initial count\lparen first term\rparen} \\ r\text{ is the common ratio} \\ n\text{ is the number of years} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/rrkmr1cld7t0g2ayc5jvd5on0cx6us231x.png)
STEP 2: Write the given details
![\begin{gathered} a_1=9000 \\ r=1+(69)/(100)=1.69\text{ since it is a growth rate} \\ \\ Hence,the\text{ equation is given as:} \\ a_x=9000(1.69)^(x-1) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/fv9zh8yec2mj2t1o7nkb89g62yllrmfxce.png)
STEP 3: Get the explicit equation for f(n)
n = x
Substitute n for x in the equation in step 2.
Therefore, the explicit equation is given as:
![f(n)=9000\cdot(1.69)^(n-1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/dlz1nz2uzw5kcftul1apkuzxg1dp1sft9a.png)
STEP 4: Answer part B
To get how many lionfish in the bay after 6 years
![\begin{gathered} From\text{ equation above,} \\ n=6 \\ f(6)=9000\cdot(1.69)^(6-1) \\ f(6)=9000\cdot1.69^5 \\ f(6)=9000\cdot13.78584918 \\ f(6)=124072.6427 \\ f(6)\approx124073 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/opl6w6g5kl4orkoqg4gotncxcqukr7q7b4.png)
Hence, there will be approximately 124073 lionfish
STEP 5: Get the recursive formula
1400 lionfish was removed per year, this gives an equation defined below:
Recursive formula is given as
![a_n=r(a_(n-1))](https://img.qammunity.org/2023/formulas/mathematics/college/btu83tkjdkdu6o0swzpiik1d048bc5tvbd.png)
Since we know that the difference each year is 1400, this gives the equation below:
![a_n-1400](https://img.qammunity.org/2023/formulas/mathematics/high-school/37vo282zcpo0a619a737jnw55qa36s92b8.png)
By substitution, the recursive formula will be given by:
Since 1400 is removed each year, we have:
![\begin{gathered} f(n)=a_(n-1)-1400n \\ f(n)=9000\cdot(1.69)^(n-1)-1400n \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/vcrqto39tx5hr2wo1o0ne3tw19wfz3j0bo.png)