To solve this problem, we need to determine the values of [tex]\( y \)[/tex] based on the given values of [tex]\( x \)[/tex], given that [tex]\( y \)[/tex] is directly proportional to [tex]\( x \)[/tex]. This means we can express [tex]\( y \)[/tex] as [tex]\( y = kx \)[/tex] where [tex]\( k \)[/tex] is some constant of proportionality.
Let's start with the table provided:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & 4 & 20 & 24 & 36 & 44 \\
\hline
y & & & 6 & 9 & 11 \\
\hline
\end{array}
\][/tex]
Step 1: Identify a known pair of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
In this table, we know that when [tex]\( x = 24 \)[/tex], [tex]\( y = 6 \)[/tex].
Step 2: Use the known pair to find the constant of proportionality [tex]\( k \)[/tex].
[tex]\[
y = kx \implies 6 = k \cdot 24 \implies k = \frac{6}{24} = 0.25
\][/tex]
Step 3: Use the constant [tex]\( k \)[/tex] to find the missing values of [tex]\( y \)[/tex].
For [tex]\( x = 4 \)[/tex]:
[tex]\[
y = k \cdot x \implies y = 0.25 \cdot 4 = 1.0
\][/tex]
For [tex]\( x = 20 \)[/tex]:
[tex]\[
y = k \cdot x \implies y = 0.25 \cdot 20 = 5.0
\][/tex]
Replace the values in the table:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & 4 & 20 & 24 & 36 & 44 \\
\hline
y & 1.0 & 5.0 & 6 & 9 & 11 \\
\hline
\end{array}
\][/tex]
So, the completed table is:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & 4 & 20 & 24 & 36 & 44 \\
\hline
y & 1.0 & 5.0 & 6 & 9 & 11 \\
\hline
\end{array}
\][/tex]
This completes the necessary calculations for the given table.
