Question

PLZ HELP ASAP

2. One load of gravel contains 240 ft2 of gravel. The area A that the gravel will cover is inversely proportional to the depth of d to which the gravel is spread.
a. Write an inverse variation model for the relationship between the area and depth for one load of gravel.
b. A designer plans a playground with gravel 6 in. deep over the entire play area. If the play area is a rectangle 40 ft. wide and 24 ft long, how many loads of gravel will be needed? (4 points)

I understand everything else but this question.

Answer 1

A. y=
`(z*240ft\\\\^(3) )/(x)`

B. z= 2

Explanation:

A.

The format for the inverse variation formula is y =

This is a bit different compared to the format of directly variation formula y= kx

In this case, one load of gravel have
`240 ft^(3)`volume. The area is inversely proportional with depth, so formula for the volume is Area x Depth (V= A x D). The inverse variation formula will be: A= V/D

If the number of loads=z, each load is
`240 ft^(3)`, area = y and depth = x , then the model will be:

y=
`(z*\\k)/(x)`

y=
`(z*240ft\\\\^(3) )/(x)`

B.

The play area is a rectangle 40 ft wide and 24 ft long, then the area will be: area = 40ft * 24ft= 960
`ft^(2)`

We have area(y= 960
`ft^(2)`) and depth(x= 6 in- 0.5 ft) , now we can find out the number of loads needed(z) by using the equation.

y=
`(240ft*z\\\\^(3) )/(x)`

960
`ft^(2)`=
`(z*240ft\\\\^(3) )/(x)`

`480 ft^(2) * z`= 960
`ft^(2)`

z=
`(960ft^(2) )/(480ft^(2))`= 2