Answer:
Explanation:
a deli wraps its cylindrical containers of hot food items with plastic wrap. the minimum amount of plastic wrap needed to completely wrap 8 containers is: 785.4 square inches.
What is the minimum amount of plastic wrap needed?
The area of each circular end is:
A = πr^2
where r is the radius of the end. The diameter of each container is given as 5.5 inches, so the radius is half of that, or 2.75 inches. Using π = 3.14, we get:
A = 3.14 * (2.75in)^2
A = 23.68in^2 (rounded to two decimal places)
The area of the rectangular side is:
A = h * circumference
where h is the height of the container and circumference is the distance around the circular end. The circumference is equal to the diameter times π, so we have:
circumference = 5.5in * π
circumference = 17.27in (rounded to two decimal places)
Using the given height of 3 inches, we get:
A = 3in * 17.27in
A = 51.81in^2 (rounded to two decimal places)
Therefore, the total surface area of each container is:
A = 2 * 23.68in^2 + 51.81in^2
A = 98.17in^2 (rounded to two decimal places)
To wrap 8 containers, we need to multiply the surface area of each container by 8:
total surface area = 8 * 98.17in^2
total surface area = 785.36in^2 (rounded to two decimal places)
Therefore, we need a minimum of 785.36 square inches of plastic wrap to completely wrap 8 cylindrical containers of hot food items. Rounding to the nearest tenth, the answer is 785.4 square inches.