SIMPLIFIED EQUATIONS HELP, PLEASE.

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?

A)

d =240/a

B) 2 loads of gravel

Step-by-step explanation:

Step 1:

Given one load of gravel is 240 ft2.

We have depth = 240/area

Hence the inverse variation model is d =240/a

Step 2 :

The length and breadth of the playground is 40 ft and 24 ft

The depth of the playground is 6 in = 6/12 ft

Hence the volume of the given playground is 40 * 24 * 6/12 = 480 cu ft.

There is  240 ft2 of gravel in one load

So we need 480 cu ft of gravel =  480/240 = 2 loads of gravel.

828. all there is to it!

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