If a varies inversely as b squared​, and aequals25 when bequalsfour fifths

​, find a when bequalsfive fourths
.

1 answer

If \( a \) varies inversely as \( b^2 \), we can express this relationship mathematically as:

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

where \( k \) is a constant.

From the information given, when \( b = \frac{4}{5} \), \( a = 25 \). We can use this to find the constant \( k \).

Substituting the values into the equation:

\[ 25 = \frac{k}{\left(\frac{4}{5}\right)^2} \]

Calculating \( \left(\frac{4}{5}\right)^2 \):

\[ \left(\frac{4}{5}\right)^2 = \frac{16}{25} \]

Now substituting this back into the equation:

\[ 25 = \frac{k}{\frac{16}{25}} \]

Rearranging gives:

\[ 25 \cdot \frac{16}{25} = k \]

This simplifies to:

\[ k = 16 \]

We now have the constant \( k \). The relationship between \( a \) and \( b \) is now:

\[ a = \frac{16}{b^2} \]

Next, we need to find \( a \) when \( b = \frac{5}{4} \):

\[ a = \frac{16}{\left(\frac{5}{4}\right)^2} \]

Calculating \( \left(\frac{5}{4}\right)^2 \):

\[ \left(\frac{5}{4}\right)^2 = \frac{25}{16} \]

Substituting this back into the equation for \( a \):

\[ a = \frac{16}{\frac{25}{16}} = 16 \cdot \frac{16}{25} = \frac{256}{25} \]

Thus, the value of \( a \) when \( b = \frac{5}{4} \) is:

\[ \boxed{\frac{256}{25}} \]