Respuesta :
Answer:
1.) What is the circumference, in inches, of a donut?
The circumference of a circle is given by the formula `C = πd`, where `C` is the circumference, `π` is a mathematical constant approximately equal to 3.14, and `d` is the diameter.
Given that the diameter of each donut is 3.5 inches, the circumference is:
C = πd = 3.14 * 3.5 in = 11 in
Therefore, the circumference of a donut is 11 inches.
2.) There are 144 square inches in 1 foot. How many square inches are in 2.5 square feet?
There are 144 square inches in 1 square foot, so there are 144 * 2.5 = 360 square inches in 2.5 square feet.
3.) To make sure there is enough space for the donuts, Dave wants to add 1/2 inch to the minimum length, width, and height of the box. Including the additional space, what should be the length, width, and height of the new box, in inches?
The minimum length of the box should be equal to the circumference of a donut, which is 11 inches. Adding 1/2 inch to this gives a length of 11.5 inches.
The minimum width of the box should be equal to the diameter of a donut, which is 3.5 inches. Adding 1/2 inch to this gives a width of 4 inches.
The minimum height of the box should be equal to the height of a donut, which is 1.5 inches. Adding 1/2 inch to this gives a height of 2 inches.
Therefore, the length, width, and height of the new box, including the additional space, should be 11.5 inches, 4 inches, and 2 inches, respectively.
4.) Based on your response to question 3, what is the area of the bottom of the new box? How do the length and width of the new box meet Dave’s requirements?
The area of the bottom of the new box is equal to the length times the width. Given that the length is 11.5 inches and the width is 4 inches, the area of the bottom of the new box is:
Area = length * width = 11.5 in * 4 in = 46 in^2
The length and width of the new box meet Dave's requirements because they are both greater than or equal to the minimum length and width required to fit the donuts. The minimum length required is the circumference of a donut, which is 11 inches. The minimum width required is the diameter of a donut, which is 3.5 inches. The length of the new box is 11.5 inches, which is greater than the minimum length of 11 inches. The width of the new box is 4 inches, which is greater than the minimum width of 3.5 inches. Therefore, the length and width of the new box meet Dave's requirements.
5.) Using the dimensions from question 3, what is the surface area, in square inches, of your new box?
The surface area of the new box is equal to the sum of the areas of all six sides. The bottom and top of the box are both rectangles with an area of 46 square inches. The two sides of the box are both rectangles with an area of 11.5 * 2 = 23 square inches. The two ends of the box are both rectangles with an area of 4 * 2 = 8 square inches. Therefore, the surface area of the new box is:
Surface area = 2 * (46 in^2 + 23 in^2 + 8 in^2) = 154 in^2
6.) It costs $15.00 for 25 of the original boxes. Dave wants the cost of the new box to be less than or equal to the cost of the original box. Does the new box cost less than the original box? Include all necessary work to support your answer.
The cost of one original box is $15.00 / 25 = $0.60.
The cost of the new box is $0.20 per square foot of cardboard. The surface area of the new box is 154 square inches, which is equal to 154 / 144 = 1.07 square feet. Therefore, the cost of the new box is:
Cost = $0.20 / ft^2 * 1.07 ft^2 = $0.214
Since the cost of the new box ($0.214) is less than the cost of the original box ($0.60), the new box costs less than the original box.
