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 \(\boxed{4}\).