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Which table has a constant of proportionality between [tex]\( y \)[/tex] and [tex]\( x \)[/tex] of [tex]\( \frac{8}{5} \)[/tex]?

A.
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
4 & \frac{32}{5} \\
10 & 16 \\
11 & \frac{88}{5} \\
\hline
\end{array}
\][/tex]

B.
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
2 & \frac{14}{5} \\
6 & \frac{42}{5} \\
20 & 28 \\
\hline
\end{array}
\][/tex]

C.
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
5 & \frac{20}{3} \\
9 & 12 \\
\hline
\end{array}
\][/tex]

Sagot :

GinnyAnswer
To determine which table shows a constant of proportionality of [tex]\( \frac{8}{5} \)[/tex] between [tex]\( y \)[/tex] and [tex]\( x \)[/tex], we will examine each table and check whether the ratio [tex]\( \frac{y}{x} \)[/tex] is consistently [tex]\( \frac{8}{5} \)[/tex] across all the given data points.

### Table (A)
[tex]\[ \begin{array}{|cc|} \hline x & y \\ \hline 4 & \frac{32}{5} \\ 10 & 16 \\ 11 & \frac{88}{5} \\ \hline \end{array} \][/tex]

1. Calculate [tex]\( \frac{y}{x} \)[/tex] for each row:
- For [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{32}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{32}{5}}{4} = \frac{32}{5} \times \frac{1}{4} = \frac{32}{20} = \frac{8}{5} \][/tex]
- For [tex]\( x = 10 \)[/tex], [tex]\( y = 16 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{16}{10} = \frac{8}{5} \][/tex]
- For [tex]\( x = 11 \)[/tex], [tex]\( y = \frac{88}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{88}{5}}{11} = \frac{88}{5} \times \frac{1}{11} = \frac{88}{55} = \frac{8}{5} \][/tex]

All ratios are [tex]\( \frac{8}{5} \)[/tex].

### Table (B)
[tex]\[ \begin{array}{|cc|} \hline x & y \\ \hline 2 & \frac{14}{5} \\ 6 & \frac{42}{5} \\ 20 & 28 \\ \hline \end{array} \][/tex]

1. Calculate [tex]\( \frac{y}{x} \)[/tex] for each row:
- For [tex]\( x = 2 \)[/tex], [tex]\( y = \frac{14}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{14}{5}}{2} = \frac{14}{5} \times \frac{1}{2} = \frac{14}{10} = \frac{7}{5} \][/tex]
- For [tex]\( x = 6 \)[/tex], [tex]\( y = \frac{42}{5} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{42}{5}}{6} = \frac{42}{5} \times \frac{1}{6} = \frac{42}{30} = \frac{7}{5} \][/tex]
- For [tex]\( x = 20 \)[/tex], [tex]\( y = 28 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{28}{20} = \frac{7}{5} \][/tex]

All ratios are [tex]\( \frac{7}{5} \)[/tex], not [tex]\( \frac{8}{5} \)[/tex].

### Table (C)
[tex]\[ \begin{array}{|cc|} \hline x & y \\ \hline 5 & \frac{20}{3} \\ 9 & 12 \\ \hline \end{array} \][/tex]

1. Calculate [tex]\( \frac{y}{x} \)[/tex] for each row:
- For [tex]\( x = 5 \)[/tex], [tex]\( y = \frac{20}{3} \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{20}{3}}{5} = \frac{20}{3} \times \frac{1}{5} = \frac{20}{15} = \frac{4}{3} \][/tex]
- For [tex]\( x = 9 \)[/tex], [tex]\( y = 12 \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{12}{9} = \frac{4}{3} \][/tex]

All ratios are [tex]\( \frac{4}{3} \)[/tex], not [tex]\( \frac{8}{5} \)[/tex].

Based on this analysis, Table (A) is the one that shows a constant of proportionality of [tex]\( \frac{8}{5} \)[/tex] between [tex]\( y \)[/tex] and [tex]\( x \)[/tex].

Thus, the answer is:
(A)
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