Question

The total weight of a rock depends on its size and is proportional to its density. In this context, density is the weight per cubic inch. Let w denote the weight of the rock in pounds, s the size of the rock in cubic inches, and d the density of the rock in pounds per cubic inch. If a 48-cubic-inch rock weighs w pounds, write an equation that shows the proportional relation.

Answer 1

Answer:

$d = \frac{w}{48}$

If  $\frac{1}{48}$ is constant, say k

Then,

$d = kw$

∴ d ∝ w

Hence, weight is proportional to the density

Step-by-step explanation:

From the question,

Let w denote the weight of the rock in pounds

s denote the size of the rock in cubic inches and

d denote the density of the rock in pounds per cubic inch.

First, we will write the equation connecting w, s, and d.

We get

$density (pounds/inch^{3} ) = \frac{weight(pounds)}{size (inch^{3}) }$

That is,

$d = \frac{w}{s}$

Now, given a 48-cubic-inch rock with weight w pounds, to show the proportional relation between the weight and the density, we will write

$d = \frac{w}{48}$

If  $\frac{1}{48}$ is constant, say k

Then,

$d = kw$

∴ d ∝ w

Hence, density is proportional to the weight OR weight is proportional to the density