1. **State the problem:** We are given the sequence $35, \frac{35}{3}, \frac{8}{9}, \frac{17}{27}, \ldots$ and asked to find: (a) the common difference or common ratio, (b) the type of function (linear or exponential), (c) the equation of the corresponding continuous function. 2. **Find the common ratio or difference:** Calculate the ratio between consecutive terms: $$\frac{\frac{35}{3}}{35} = \frac{35}{3} \times \frac{1}{35} = \frac{1}{3}$$ $$\frac{\frac{8}{9}}{\frac{35}{3}} = \frac{8}{9} \times \frac{3}{35} = \frac{24}{315} = \frac{8}{105}$$ Since these are not equal, check the problem's given terms carefully. The user wrote $11 \frac{2}{3}$ which is $\frac{35}{3}$, and $1 \frac{8}{27}$ which is $\frac{35}{27}$ (not $\frac{17}{27}$). So the sequence is: $$35, \frac{35}{3}, \frac{35}{9}, \frac{35}{27}, \ldots$$ Check ratio: $$\frac{\frac{35}{3}}{35} = \frac{1}{3}$$ $$\frac{\frac{35}{9}}{\frac{35}{3}} = \frac{35}{9} \times \frac{3}{35} = \frac{1}{3}$$ $$\frac{\frac{35}{27}}{\frac{35}{9}} = \frac{35}{27} \times \frac{9}{35} = \frac{1}{3}$$ So the common ratio is $\frac{1}{3}$, not $\frac{2}{3}$. 3. **Type of function:** Since there is a constant ratio, the function is exponential. 4. **Equation of the continuous function:** The first term $f(0) = 35$, and the ratio per step is $\frac{1}{3}$. So the function is: $$f(x) = 35 \left(\frac{1}{3}\right)^x$$ **Summary:** (a) The common ratio is $\frac{1}{3}$. (b) The function is exponential because there is a common ratio. (c) The equation is $f(x) = 35 \left(\frac{1}{3}\right)^x$.