If a varies inversely as b^2 and a =48 when b=4, find a when b=7

1 answer

If \( a \) varies inversely as \( b^2 \), we can write the relationship as:

\[ a = \frac{k}{b^2} \]

where \( k \) is a constant.

Given that \( a = 48 \) when \( b = 4 \), we can substitute these values into the equation to find \( k \):

\[ 48 = \frac{k}{4^2} \]

This simplifies to:

\[ 48 = \frac{k}{16} \]

Now, multiply both sides by \( 16 \):

\[ k = 48 \times 16 \]

Calculating \( 48 \times 16 \):

\[ k = 768 \]

Now we have the relationship:

\[ a = \frac{768}{b^2} \]

Next, we need to find \( a \) when \( b = 7 \):

\[ a = \frac{768}{7^2} \]

Calculating \( 7^2 \):

\[ 7^2 = 49 \]

Now substitute this back into the equation:

\[ a = \frac{768}{49} \]

Now, dividing \( 768 \) by \( 49 \):

\[ a \approx 15.6734693877551 \]

Thus, the value of \( a \) when \( b = 7 \) is approximately:

\[ a \approx 15.67 \]

To summarize:

\[ \text{When } b = 7, , a \approx 15.67. \]