Question

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

Answer 1

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:

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