word problem, math formula? |
word problem, math formula? |
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#1
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Senior Member ![]() ![]() ![]() ![]() Group: Member Posts: 151 Joined: Dec 2004 Member No: 74,827 ![]() |
A city has planner placed a square wading pool in the center of a park. The pool surrounded by a fence, 20 meters wide. The tiles for the floor of the wading pool cover 121 square meters. How wide is the wooden deck surrounding the wading pool?
Takheela plans to shop a doll house in a box that is 30 inches along each edge. The doll house measures 24 inches wide, 20 inches long, and 28 inches high. How much foam packing material (in cubic inches) will Takheela need? What is the approximate side length of the cubic box that holds only the packaging material? I dont remeber what formulas to use..does anyone know |
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#2
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![]() in a matter of time ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Staff Alumni Posts: 7,151 Joined: Aug 2005 Member No: 191,357 ![]() |
Okay...well. This is what I understand from your first question:
![]() ------------------------------------ For the second question, find the volume of the box first, then subtract the volume of the doll house from it. V of Box = 30 x 30 x 30 V of Box = 27000 cubic inches V of Doll House = 24 x 20 x 28 V of Doll House = 13440 cubic inches V of Box - V of Doll House = Volume of Packing Material 27000 - 13440 = 13560 cubic inches Therefore, you would need 13560 cubic inches of packing material. ............ Okay, now for the second part of the question. Take the dollhouse out of the equation now. What you're left with is 13560 cubic inches of packing material. ![]() You know that the dimensions of the packaging material would be 30 * 30 * x, because it's in the 30 * 30 * 30 box. V of Packaging Material = 30*30*x You know what the volume is, it is 13560 cubic inches. So, 13560 = 30*30*x x = 13560/(30*30) x = 13560/900 x = 15 1/15 or 15.066666repeating. Therefore, the approximate side length of the cubic box that only holdsthe packaging material would be 15.1 inches. |
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