Answer:
[tex]d = \frac{w}{48}[/tex]
If [tex]\frac{1}{48} \\[/tex] is constant, say k
Then,
[tex]d = kw[/tex]
∴ 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
[tex]density (pounds/inch^{3} ) = \frac{weight(pounds)}{size (inch^{3}) }[/tex]
That is,
[tex]d = \frac{w}{s}[/tex]
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
[tex]d = \frac{w}{48}[/tex]
If [tex]\frac{1}{48} \\[/tex] is constant, say k
Then,
[tex]d = kw[/tex]
∴ d ∝ w
Hence, density is proportional to the weight OR weight is proportional to the density
