Question

the number of fish, $f$, in skippers pond at the beginning of each year can be modeled by the equation $f(x)=3(2^{x})$, where $x$ represents the number of years after the beginning of the year 2000. for example, $x = 0$ represents the beginning of the year 2000, $x = 1$ represents the beginning of the year 2001, and so forth.sketch and label a graph that models the number of fish in skippers pond from 2000 to 2005.answer this question on paper.part baccording to the model, how many fish were in skippers pond at the beginning of the year 2006?answer this question on paper.part caccording to the model, what year will be the first time skippers pond has more than 1000 fish at the beginning of the year?answer this question on paper.

Explanation:

Step1: Define x for 2006

x = 2006 - 2000 = 6

Step2: Calculate fish count for 2006

$f(6) = 3(2^6) = 3 \times 64 = 192$

Step3: Set up inequality for >1000 fish

$3(2^x) > 1000$

Step4: Isolate exponential term

$2^x > \frac{1000}{3} \approx 333.33$

Step5: Solve for x using log

$x > \log_2(333.33) = \frac{\ln(333.33)}{\ln(2)} \approx \frac{5.81}{0.693} \approx 8.38$

Step6: Find corresponding year

x=9 corresponds to 2000+9=2009

Answer:

Part B: 192

Part C: 2009

(Note: For Part A, the graph is an exponential curve starting at (0, 3) (2000), passing through (1,6) (2001), (2,12) (2002), (3,24) (2003), (4,48) (2004), (5,96) (2005), with the x-axis labeled "Years after 2000 (x)" and y-axis labeled "Number of fish (f(x)")