The problem states that $a$ varies directly as the square of $b$ and inversely as $c$. Given $a=2$ when $b=4$ and $c=24$, we need to find: a. The value of the constant of proportionality $k$. b. The value of $a$ when $b=9$ and $c=27$. c. The values of $b$ when $a=8$ and $c=6$.

1. Problem Description

The problem states that $a$ varies directly as the square of $b$ and inversely as $c$. Given $a=2$ when $b=4$ and $c=24$, we need to find:

a. The value of the constant of proportionality $k$.

b. The value of $a$ when $b=9$ and $c=27$.

c. The values of $b$ when $a=8$ and $c=6$.

2. Solution Steps

a. Since $a$ varies directly as the square of $b$ and inversely as $c$, we can write the relationship as:

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

We are given that $a=2$, $b=4$, and $c=24$. Substituting these values into the equation, we get:

$2 = k \frac{4^2}{24}$

$2 = k \frac{16}{24}$

$2 = k \frac{2}{3}$

To find $k$, multiply both sides by $\frac{3}{2}$:

$k = 2 \cdot \frac{3}{2} = 3$

b. Now that we know $k=3$, we can use the same formula to find $a$ when $b=9$ and $c=27$:

$a = 3 \frac{b^2}{c}$

$a = 3 \frac{9^2}{27}$

$a = 3 \frac{81}{27}$

$a = 3 \cdot 3 = 9$

c. We need to find the values of $b$ when $a=8$ and $c=6$. Using the formula $a = k \frac{b^2}{c}$ with $k=3$, we have:

$8 = 3 \frac{b^2}{6}$

Multiply both sides by 6:

$48 = 3b^2$

Divide both sides by 3:

$b^2 = 16$

Take the square root of both sides:

$b = \pm 4$

3. Final Answer

a. Value of $k$: $3$

b. Value of $a$ when $b=9$ and $c=27$: $9$

c. Values of $b$ when $a=8$ and $c=6$: $4$, $-4$