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lorasvet [3.4K]
3 years ago
5

Your iron works has been contracted to design and build a 500-ft3 square-based, open-top, rectangular steel holding tank for a p

aper company. The tank is to be made by welding half-inch-thick stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weight as little as possible. What dimensions do you tell the shop to use?
Mathematics
1 answer:
frez [133]3 years ago
4 0
500 cubic feet of 1/2-inch thick stainless steel plates will weigh the same no matter what dimensions are use for area and height of tank.

A 5 foot by 5 foot area 20 feet high (500 cubic feet) will weigh just as much as a  10 foot by 10 foot area 5 feet high (also 500 cubic feet) since you are using the same amount of material for the whole tank. Only possible change in weight of tank is the use of welding material and how evenly and good the welder is at welding stainless steel.
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What is the decimal value of 3 quarters and 2 dimes?
Radda [10]

Answer:

0.95

Step-by-step explanation:

Quarter = 0.25

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0.25+0.25+0.25= 0.75

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Gina went to a restaurant for lunch. On her bill the sales tax equals $2.34. If the tax is 8.5%, what was the subtotal (without
bagirrra123 [75]

Answer:

$2.5389

Step-by-step explanation:

Given data

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Let us compute the cost of the tax

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3 years ago
Please Answer!!!!
Morgarella [4.7K]

Answer:

A). F = \frac{9C}{5}+32

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Step-by-step explanation:

A). Formula to convert the temperature from degree centigrade to °F is,

    C = \frac{5}{9}(F-32)

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    9 × C = 9\times \frac{5}{9}(F-32)

    9C = 5(F - 32)

    \frac{9C}{5}=\frac{5(F-32)}{5}

    \frac{9C}{5}=(F-32)

    F = \frac{9C}{5}+32

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    F = \frac{9\times 190}{5}+32

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