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.

In this case, one load of gravel have [tex]240 ft^{3}[/tex]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 [tex]240 ft^{3}[/tex], area = y and depth = x , then the model will be:

y= [tex]\frac{z*\\k}{x}[/tex]

y= [tex]\frac{z*240ft\\\\^{3} }{x}[/tex]

B.

The play area is a rectangle 40 ft wide and 24 ft long, then the area will be: area = 40ft * 24ft= 960 [tex]ft^{2}[/tex]

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

y= [tex]\frac{240ft*z\\\\^{3} }{x}[/tex]

960 [tex]ft^{2}[/tex]= [tex]\frac{z*240ft\\\\^{3} }{x}[/tex]

[tex]480 ft^{2} * z[/tex]= 960 [tex]ft^{2}[/tex]

z= [tex]\frac{960ft^{2} }{480ft^{2}}[/tex]= 2