The following tables show relationships between two quantities: one that varies directly, and one that varies inversely. In each table, the independent variable is in the first column.
Table 1
\begin{tabular}{|c|c|}
\hline$t$ & $h$ \\
\hline 10 & 380 \\
\hline 20 & 190 \\
\hline 40 & 95 \\
\hline 50 & 76 \\
\hline
\end{tabular}
Table 2
\begin{tabular}{|c|c|}
\hline$w$ & $M$ \\
\hline 3 & 12 \\
\hline 9 & 36 \\
\hline 12 & 48 \\
\hline 15 & 60 \\
\hline
\end{tabular}

Step 1 ：Observe the values in Table 1. As the value of \(t\) increases, the value of \(h\) decreases. This suggests an inverse variation.

Step 2 ：Confirm the inverse variation by checking if the product of \(t\) and \(h\) is constant for all rows in Table 1.

Step 3 ：Calculate the products for each row in Table 1: \(10 \times 380 = 3800\), \(20 \times 190 = 3800\), \(40 \times 95 = 3800\), \(50 \times 76 = 3800\).

Step 4 ：The products for all rows in Table 1 are the same, confirming that \(t\) and \(h\) vary inversely. The constant of variation is 3800.

Step 5 ：Observe the values in Table 2. As the value of \(w\) increases, the value of \(M\) also increases. This suggests a direct variation.

Step 6 ：Confirm the direct variation by checking if the ratio of \(M\) to \(w\) is constant for all rows in Table 2.

Step 7 ：Calculate the ratios for each row in Table 2: \(\frac{12}{3} = 4.0\), \(\frac{36}{9} = 4.0\), \(\frac{48}{12} = 4.0\), \(\frac{60}{15} = 4.0\).

Step 8 ：The ratios for all rows in Table 2 are the same, confirming that \(w\) and \(M\) vary directly. The constant of variation is 4.

Step 9 ：Final Answer: For Table 1, \(t\) and \(h\) vary inversely with a constant of variation of \(\boxed{3800}\). For Table 2, \(w\) and \(M\) vary directly with a constant of variation of \(\boxed{4}\).