Let's work through this step-by-step to determine the outputs for each input [tex]\( x \)[/tex] given the rule [tex]\( 60 \div 2x \)[/tex]:
1. For [tex]\( x = 0 \)[/tex]:
The rule is [tex]\( 60 \div 2x \)[/tex]. Plugging in [tex]\( x = 0 \)[/tex], we get:
[tex]\[
60 \div 2 \cdot 0
\][/tex]
This involves division by zero, which is undefined. Therefore, the output cannot be determined and is represented as [tex]\( \text{None} \)[/tex] (undefined).
2. For [tex]\( x = 1 \)[/tex]:
The rule is [tex]\( 60 \div 2x \)[/tex]. Plugging in [tex]\( x = 1 \)[/tex], we get:
[tex]\[
60 \div 2 \cdot 1 = 60 \div 2 = 30
\][/tex]
So, the output for [tex]\( x = 1 \)[/tex] is 30.0.
3. For [tex]\( x = 2 \)[/tex]:
The rule is [tex]\( 60 \div 2x \)[/tex]. Plugging in [tex]\( x = 2 \)[/tex], we get:
[tex]\[
60 \div 2 \cdot 2 = 60 \div 4 = 15
\][/tex]
So, the output for [tex]\( x = 2 \)[/tex] is 15.0, which is already provided in the table.
4. For [tex]\( x = 3 \)[/tex]:
The rule is [tex]\( 60 \div 2x \)[/tex]. Plugging in [tex]\( x = 3 \)[/tex], we get:
[tex]\[
60 \div 2 \cdot 3 = 60 \div 6 = 10
\][/tex]
So, the output for [tex]\( x = 3 \)[/tex] is 10.0.
Given all these calculations, we can complete the table as follows:
[tex]\[
\begin{tabular}{|l|l|l|l|l|}
\hline
Input $(x)$ & 0 & 1 & 2 & 3 \\
\hline
Output & \text{None} & 30.0 & 15.0 & 10.0 \\
\hline
\end{tabular}
\][/tex]
