A varies BC when B=4 and C=9,A=6. find A when B=3 and C=10

1 answer

To determine how \( A \) varies with \( B \) and \( C \), we can represent the relationship as:

\[ A = k \cdot \frac{B^m}{C^n} \]

for some constants \( k, m, n \). Given the values when \( B = 4 \) and \( C = 9 \), we know:

\[ 6 = k \cdot \frac{4^m}{9^n} \]

Next, we need to find the value of \( A \) when \( B = 3 \) and \( C = 10 \):

\[ A = k \cdot \frac{3^m}{10^n} \]

To solve for \( A \) with the new values, we first need to find \( k, m, \) and \( n \) based on the relationship. However, the problem does not provide enough information to uniquely determine \( k, m, \) and \( n \).

If we make a common assumption, such as assuming \( m = 1 \) and \( n = 1 \) (implying a direct variation), we would rewrite the relationship as:

\[ A = k \cdot \frac{B}{C} \]

Using this assumption for the initial condition \( (B = 4, C = 9, A = 6) \):

\[ 6 = k \cdot \frac{4}{9} \implies k = 6 \cdot \frac{9}{4} = \frac{54}{4} = 13.5 \]

Thus, the relationship becomes:

\[ A = 13.5 \cdot \frac{B}{C} \]

Now substituting \( B = 3 \) and \( C = 10 \):

\[ A = 13.5 \cdot \frac{3}{10} = 13.5 \cdot 0.3 = 4.05 \]

So, \( A \) when \( B = 3 \) and \( C = 10 \) is \( 4.05 \).

If the variation is different (i.e., different values for \( m \) and \( n \)), you would need additional information to determine those.