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Making a Table of Values for a Linear Equation

[tex]\[
\begin{tabular}{|c|c|}
\hline
Time $(t)$ & Elevation $(e)$ \\
\hline
-2 & $a$ \\
\hline
3.5 & $b$ \\
\hline
30 & $c$ \\
\hline
\end{tabular}
\][/tex]

Rory is staying in a cabin on a hill 300 feet above sea level. She walks down the hill to the water's edge. The equation of her average change in elevation over time is [tex]\( e = 300 - 10t \)[/tex], where [tex]\( t \)[/tex] is the time in minutes since she left the cabin, and [tex]\( e \)[/tex] is her elevation with regard to sea level. Which values are viable points, and what are their values in the table relating [tex]\( t \)[/tex] and [tex]\( e \)[/tex]?

[tex]\[
\begin{array}{l}
a = \square \checkmark \\
b = \bigvee 265 \vee \\
c = \bigvee 0 \quad \checkmark
\end{array}
\][/tex]

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Answer :

GinnyAnswer
Let's examine the relationship between time [tex]\( t \)[/tex] and elevation [tex]\( e \)[/tex] using the given equation [tex]\( e = 300 - 10t \)[/tex].

1. For [tex]\( t = -2 \)[/tex]:
[tex]\[ e = 300 - 10(-2) \][/tex]
Substituting [tex]\( t = -2 \)[/tex] into the equation:
[tex]\[ e = 300 + 20 = 320 \][/tex]
Therefore, [tex]\( a = 320 \)[/tex].

2. For [tex]\( t = 3.5 \)[/tex]:
[tex]\[ e = 300 - 10(3.5) \][/tex]
Substituting [tex]\( t = 3.5 \)[/tex] into the equation:
[tex]\[ e = 300 - 35 = 265 \][/tex]
Thus, [tex]\( b = 265 \)[/tex] which is a viable value.

3. For [tex]\( t = 30 \)[/tex]:
[tex]\[ e = 300 - 10(30) \][/tex]
Substituting [tex]\( t = 30 \)[/tex] into the equation:
[tex]\[ e = 300 - 300 = 0 \][/tex]
Therefore, [tex]\( c = 0 \)[/tex].

So, the values in the table of values for a linear equation are:

\begin{tabular}{|c|c|}
\hline
Time [tex]$(t)$[/tex] & Elevation [tex]$(e)$[/tex] \\
\hline
-2 & 320 \checkmark \\
\hline
3.5 & 265 \checkmark \\
\hline
30 & 0 \checkmark \\
\hline
\end{tabular}

Thus, the viable points and their values in the table are:
[tex]\[ \begin{array}{l} a = 320 \checkmark \\ b = 265 \checkmark \\ c = 0 \quad \checkmark \\ \end{array} \][/tex]

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