To solve for \( C \) when \( A = 20 \) and \( B = 15 \) given that \( A \) varies directly with \( B \) and \( C \), we can use the information provided to find a proportionality constant.
From the information given, we know that:
\[ A = k \cdot B \cdot C \]
where \( k \) is a constant.
Using the values provided: when \( A = 6 \), \( B = 4 \), and \( C = 9 \),
\[ 6 = k \cdot 4 \cdot 9 \]
Calculate \( k \):
\[ 6 = k \cdot 36 \] \[ k = \frac{6}{36} = \frac{1}{6} \]
Now, we can use this constant \( k \) to find \( C \) when \( A = 20 \) and \( B = 15 \).
Using the equation:
\[ 20 = \frac{1}{6} \cdot 15 \cdot C \]
Multiply both sides by 6:
\[ 120 = 15 \cdot C \]
Now, solve for \( C \):
\[ C = \frac{120}{15} = 8 \]
Therefore, when \( A = 20 \) and \( B = 15 \), the value of \( C \) is \( \boxed{8} \).
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