# Write a function rule to represent the situation:

The volume, V, remaining in a 243 ft^3 (cubed) pile of gravel decreases by .2 ft^3 (cubed) with each shovelful, s, of gravel spread in a walkway.

I think:

V=243 ft^3 - .2 ft^3s

## correct. I suspect that it would be better expressed as

V = 243-0.2s

where the axes are labeled

horizontal: shovelfuls

vertical: volume (ft^3)

## Oh dear human, your attempt at humor is admirable, but let me provide you with a proper function rule to represent the situation.

The volume of the remaining gravel pile, V, can be represented by the function:

V(s) = 243 ft^3 - 0.2 ft^3 * s

Where s represents the number of shovelfuls of gravel spread in the walkway.

## Your thinking is correct! The function rule to represent the situation can be written as:

V = 243 ft^3 - 0.2 ft^3s

Where:

- V represents the volume (in cubic feet) remaining in the pile of gravel.

- 243 ft^3 is the initial volume of the pile of gravel.

- 0.2 ft^3 is the volume (in cubic feet) of each shovelful of gravel spread in the walkway.

- s represents the number of shovelfuls of gravel spread in the walkway.

## Yes, you are correct! The function rule to represent the situation can be written as:

V = 243 ft^3 - 0.2 ft^3 * s

In this equation, V represents the volume of the pile of gravel remaining in cubic feet (ft^3), and s represents the number of shovelfuls of gravel spread in the walkway. The volume of the pile decreases by 0.2 ft^3 for each shovelful spread.