Answer :
1. As the graph is exponential the sequence is geometric.
2. - Find the ratio r betwen each k(x) value and the previous one:
[tex]\begin{gathered} \frac{27}{81}=\frac{1}{3} \\ \\ \frac{9}{27}=\frac{1}{3} \\ \\ \frac{3}{9}=\frac{1}{3} \\ \\ \frac{1}{3}=\frac{1}{3} \end{gathered}[/tex]Use the next formula to get the recursive formula:
[tex]\begin{gathered} k_1=\text{first term} \\ k(x)=r\cdot k_{x-1} \end{gathered}[/tex]Being k1 the first term (81)
r the ratio you find above: (1/3)
k(x-1) is the prevoius term
You get the next recursive formula:[tex]\begin{gathered} k_1=81 \\ k(x)=\frac{1}{3}k_{x-1} \end{gathered}[/tex]3. As x is the number of cuts, it cannot be a negative number. And it needs to be an inerger number (as the number of cuts cannot be a decimal number)
Then:
[tex]x\ge0;x\in Z[/tex]x can be any interger possitive number (example: 6,7,8) also, the number of cuts cannot be a very large number as you cannot cut an object in infinitely parts.
Then, x can be a possitive interger and cannot be a negative number, a decimal, or a very large possitive interger