\begin{tabular}{|c|c|c|c|}
\hline
[tex]$x$[/tex] & 3 & 7 \\
\hline
5 & 7 & 4 & [tex]$\div$[/tex] \\
\hline
& 8 & 7 & 2 \\
\hline
\end{tabular}



Answer :

Sure, let's solve the problem step by step using the information provided in the table.

First, let’s identify the structure and relevant values from the table, and determine what we need to find:

[tex]\[ \begin{array}{|c|c|c|c|} \hline x & 3 & 7 \\ \hline 5 & 7 & 4 & \div \\ \hline & 8 & 7 & 2 \\ \hline \end{array} \][/tex]

We need to find the value of [tex]\(x\)[/tex] in the first row and first column. Let’s follow the steps to achieve this:

1. Notice the values in the last row: [tex]\(8, 7, 2\)[/tex]
2. These values might be related through the specified division operation.

We are given that in the second row, the symbol [tex]\(\div\)[/tex] is present in the last column, suggesting that the division operation is related to finding [tex]\(x\)[/tex].

We assume the relationship involves the values in the table in some form of multiplication or division involving the elements in the last row because the division operation is implied:

[tex]\[ 8 \times 7 \times 2 \][/tex]

Next, we determine how these values relate to [tex]\(x\)[/tex]:

[tex]\[ \frac{8 \times 7 \times 2}{\div \} \][/tex]

Notice the division symbol ([tex]\(\div\)[/tex]) might indicate that these values should be divided by something, presumably the value in the mentioned division column of the second row.

Therefore, [tex]\(x\)[/tex] can be calculated as the division of the product of the values in the last row by this divisor:

[tex]\[ \text{Product of the values in the last row} = 8 \times 7 \times 2 = 112 \][/tex]

Given that the value immediately related to the division in the second row is:

[tex]\[ \div = 4 \][/tex]

So dividing the product by the value we get:

[tex]\[ x = \frac{112}{4} \][/tex]

Finally:

[tex]\[ x = 28 \][/tex]

Thus, the value of [tex]\(x\)[/tex] is [tex]\(28\)[/tex].

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