What is the product of4 2/3 and 11 ,1/4

1.) false

2.) true

3.) k=.7777777

4.) 15 3/4-m=8 1/4

5.) m=7 1/2

Answer:20 dollars

Step-by-step explanation:

Answer:

20 dollars:)

Step-by-step explanation:

Answer:

applying for a credit card

paying taxes

opening a checking account

Step-by-step explanation:

Hope this will help yal'l

Final answer:

The optimal dimensions of a soda can to minimize surface area given a specified volume are achieved with the ratio h = 2r. This changes to h/r = 8/π or 2.55 when considering rectangles, and h/r = 4√3/π or 2.21 when considering hexagons. Real-world soda cans typically use dimensions somewhat between these models.

Explanation:

Given the volume V of a cylindrical can with height h and radius r, we can find the dimensions that minimize the surface area. The volume of a cylinder is given by the formula V = πr2h. To minimize the surface area, which is given by 2πrh + 2πr2, we should find the derivative of this with respect to r and set it equal to 0, which gives us h = 2r.

When considering waste materials from cylinders cut from a sheet of metal, this modifies the equation for the lateral surface area. If the disks for the ends of the cans leave waste, and we consider that each disk is cut from a square of side length 2r, the optimal dimensions change to h/r = 8/π ≈ 2.55.

If the ends are cut from a tiling of hexagons, the area of a hexagon can be written as A = 3√3r2/2. This minimizes the waste material and results in the optimal ratio h/r = 4√3/π ≈ 2.21.

Comparing these models to actual can dimensions, we find that real-world can dimensions usually fall somewhere between the cylindrical and the hexagonal model, leaning slightly more toward the cylindrical.

Learn more about optimal dimensions:

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Answer:

a) To minimize the total surface area for a given volume, V, we need to consider the formula for the lateral surface area of a cylinder:

Now, set this derivative equal to zero:

2πr = 0

This implies that r can be any positive value, and it doesn't affect the minimization of TSA. Therefore, we can choose any value for r. However, it's more convenient to work with h = 2r because it simplifies the calculations and is often used in the design of cans.

b) When considering the manufacturing process with minimal waste, the total material needed is minimized when the ratio h/r = 8/π ≈ 2.55. This is because when h/r is equal to 8/π, you can fit the largest number of disks with a diameter of 2r onto a square sheet of metal with side lengths of 2r, minimizing waste.

c) If the disks for the lids and bases of the cans are cut from a tiling of hexagons, the optimal ratio is h/r = 4√3/π ≈ 2.21. Hexagons are more efficient than squares for packing circles, which represents the lids and bases of the cans. This results in less waste material.

d) To compare these theoretical models to actual aluminum cans from the supermarket, you would need to measure the dimensions of real cans and calculate their ratios of h/r. While many aluminum cans are close to the idealized cylinder shape, real-world manufacturing considerations can lead to variations. For instance, cans might have slightly different ratios of h/r based on manufacturing efficiencies, branding, and design choices. Additionally, the exact shapes and dimensions of cans may vary among different brands and beverage types.

It's also important to note that real cans might have additional features such as ridges, embossing, or variations in the shape of the top and bottom, which can affect their overall dimensions and shape. Therefore, it's essential to consider these factors when comparing theoretical models to actual cans.

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