Questions: y varies directly as x and inversely as the square of z. y=32 when x=100 and z=5. Find y when x=3 and z=9.

Transcript text: $y$ varies directly as $x$ and inversely as the square of $z$. $y=32$ when $x=100$ and $z=5$. Find $y$ when $x=3$ and $z=9$.

Solution

Solution Steps

To solve this problem, we need to use the concept of direct and inverse variation. The relationship given is that $ y $ varies directly as $ x $ and inversely as the square of $ z $. This can be expressed as $ y = k \cdot \frac{x}{z^2} $, where $ k $ is a constant. First, we will find the value of $ k $ using the given values $ y = 32 $, $ x = 100 $, and $ z = 5 $. Then, we will use this constant to find the new value of $ y $ when $ x = 3 $ and $ z = 9 $.

Step 1: Determine the Constant $ k $

Given the relationship $ y = k \cdot \frac{x}{z^2} $, we can find the constant $ k $ using the values $ y = 32 $, $ x = 100 $, and $ z = 5 $.

\[ k = \frac{y \cdot z^2}{x} = \frac{32 \cdot 5^2}{100} = \frac{32 \cdot 25}{100} = \frac{800}{100} = 8.0 \]

Step 2: Calculate $ y $ for New Values

Now, we will use the constant $ k $ to find $ y $ when $ x = 3 $ and $ z = 9 $.

\[ y = k \cdot \frac{x}{z^2} = 8.0 \cdot \frac{3}{9^2} = 8.0 \cdot \frac{3}{81} = 8.0 \cdot \frac{1}{27} = \frac{8.0}{27} \approx 0.2963 \]