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Given that \( y \) varies directly with \( x \) in the table below, what is the value of \( y \) if the value of \( x \) is 7?

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
x & 2 & 4 & 6 & 10 \\
\hline
y & 12 & 24 & 36 & 60 \\
\hline
\end{tabular}
\][/tex]

A. 37
B. 42
C. 48
D. 54


Sagot :

Sure, let's solve the problem step-by-step.

We are given that \( y \) varies directly with \( x \). This means that the relationship between \( x \) and \( y \) can be described by the equation:
[tex]\[ y = kx \][/tex]
where \( k \) is the constant of variation.

Given the table:
[tex]\[ \begin{array}{|c|c|c|c|c|} \hline x & 2 & 4 & 6 & 10 \\ \hline y & 12 & 24 & 36 & 60 \\ \hline \end{array} \][/tex]

First, let's find the constant of variation \( k \) using one of the pairs \((x, y)\) from the table. For example, using \( x = 2 \) and \( y = 12 \):
[tex]\[ 12 = k \cdot 2 \][/tex]
[tex]\[ k = \frac{12}{2} = 6 \][/tex]

To confirm that this \( k \) value is consistent, we can check it with the other pairs:
[tex]\[ \text{For } x = 4, y = 24: \quad k = \frac{24}{4} = 6 \][/tex]
[tex]\[ \text{For } x = 6, y = 36: \quad k = \frac{36}{6} = 6 \][/tex]
[tex]\[ \text{For } x = 10, y = 60: \quad k = \frac{60}{10} = 6 \][/tex]

The constant \( k = 6 \) is consistent across all the given pairs.

Now, we need to find the value of \( y \) when \( x = 7 \):
[tex]\[ y = kx \][/tex]
[tex]\[ y = 6 \cdot 7 \][/tex]
[tex]\[ y = 42 \][/tex]

Therefore, the value of [tex]\( y \)[/tex] when [tex]\( x = 7 \)[/tex] is [tex]\(\boxed{42}\)[/tex].
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