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

Question

​, find a when bequalsseven halves
.

Answer 1

Since $a$ varies inversely as $b^2$, we can express this relationship as:

$$a=\frac{k}{b^2}$$

where $k$ is a constant.

We know that $a=25$ when $b=\frac{4}{5}$. Using this information, we can solve for $k$:

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

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

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

Now, substituting this back into the equation gives:

$$25=\frac{k}{\frac{16}{25}}$$

To isolate $k$, we multiply both sides by $\frac{16}{25}$:

$$k=25\cdot \frac{16}{25}=16$$

Now that we have the value of $k$, we can use it to find $a$ when $b=\frac{7}{2}$:

$$a=\frac{16}{{\left(\frac{7}{2}\right)}^2}$$

Calculating ${\left(\frac{7}{2}\right)}^2$:

$${\left(\frac{7}{2}\right)}^2=\frac{49}{4}$$

Now substituting this back into the equation for $a$:

$$a=\frac{16}{\frac{49}{4}}=16\cdot \frac{4}{49}=\frac{64}{49}$$

Thus, the value of $a$ when $b=\frac{7}{2}$ is:

$$\boxed{{\displaystyle \frac{64}{49}}}$$