Let's approach this question step by step by filling in the missing [tex]\( y \)[/tex] values in the table for the given [tex]\( x \)[/tex] values, using the equation [tex]\( y = 15x - 2 \)[/tex].
### Step-by-Step Solution
1. Given Values:
The equation provided is [tex]\( y = 15x - 2 \)[/tex].
The table has the following [tex]\( x \)[/tex] values with some [tex]\( y \)[/tex] values missing:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-3 & -65 \\
\hline
0 & -2 \\
\hline
3 & \\
\hline
6 & \\
\hline
\end{array}
\][/tex]
2. Finding [tex]\( y \)[/tex] for [tex]\( x = 3 \)[/tex]:
[tex]\[
y = 15(3) - 2
\][/tex]
Calculate:
[tex]\[
y = 45 - 2 = 43
\][/tex]
So, when [tex]\( x = 3 \)[/tex], [tex]\( y = 43 \)[/tex].
3. Finding [tex]\( y \)[/tex] for [tex]\( x = 6 \)[/tex]:
[tex]\[
y = 15(6) - 2
\][/tex]
Calculate:
[tex]\[
y = 90 - 2 = 88
\][/tex]
So, when [tex]\( x = 6 \)[/tex], [tex]\( y = 88 \)[/tex].
4. Completed Table:
Now, let's fill in the [tex]\( y \)[/tex] values in the table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-3 & -65 \\
\hline
0 & -2 \\
\hline
3 & 43 \\
\hline
6 & 88 \\
\hline
\end{array}
\][/tex]
### Determining if Each Input Has Exactly One Output
- In the given equation [tex]\( y = 15x - 2 \)[/tex], for every unique value of [tex]\( x \)[/tex], there is a unique corresponding value of [tex]\( y \)[/tex].
- This is a linear equation, and in such equations, each input (value of [tex]\( x \)[/tex]) will always map to exactly one output (value of [tex]\( y \)[/tex]).
### Conclusion:
Therefore, each input in the table does have exactly one output.
To answer the question:
> Would each input have exactly one output?
Yes.
