[98serious answers | i have to show "my" work "a high school is installing a brand new regulation soccer field measuring 100 yards by 60 yards. in order to meet zoning regulaitons, the field must ahve a 6-inch layer of gravel under the topsoil to provide adequate drainage. the gravel supplier charges either $12.50 per cubic yard of gravel or a flat rate of $95 for a full truck carrying 9 cubic yards of gravel."

The volume of gravel is therefore 6000 square yards x the depth of the gravel. As a yard is 36 inches, the depth (6 inches) is 1/6 of a yard. Therefore the volume of gravel required is 
6000 x 1/6 cubic yards = 1000 cubic yards. 

The gravel in a full truck carrying 9 cubic yards is cheaper than if 9 individual cubic yards was purchased (Truck = $95, separate = 9 x $12.50 = $112.50). Therefore the school must purchase as many full trucks as possible. 
It needs 1000 cubic yards of gravel. Divide this by 9 = 111 full trucks 999 cubic yards plus 1 separate cubic yard. 
Therefore the minimum cost of the gravel is $95 x 111 trucks + 1 x $12.50 
= $10545 + $12.50 = $10557.50 

I hope this helps didint really know whaa to solve for. 

The volume of the gravel would equal 100 times 60 which is 6000!

However, since one yard’s depth is 6 inches out of 36 inches, we multiply the 6000 by 1/6 which would equal 1000 yd^3!

So, the required gravel volume would be 1000 yd^3 (cubic yards)

The gravel in a full truck carrying 9 cubic yards would cost less than 9 separate trucks!

As the school needs 1000 cubic yards of gravel, we divide this by 9 which would equal 111 full trucks of gravel containing 999 cubic yards of gravel plus 1 separate cubic yard!

We do the math :
$95 (full truck) times 111 trucks + = $10,545 thousand
Plus $12.50 for the one separate one cubic yard of gravel = $ 10,557.50

So, the minimum cost of gravel would be
$10,557.5 thousands

Hope this helps :)