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mylen [45]
4 years ago
8

Why are specification for food processing tool,equipmentand untensils necessary?​

Computers and Technology
1 answer:
anygoal [31]4 years ago
6 0

Answer:

Aluminum is the best for all-around use. It is the most popular, lightweight, attractive and less expensive. It requires care to keep it shiny and clean. Much more, it gives even heat distribution no matter what heat temperature you have. It is available in sheet or cast aluminum. Since it is a soft metal, the lighter gauges will dent and scratch easily, making the utensil unusable. Aluminum turns dark when used with alkalis, such as potatoes, beets, carrots and other vegetables. Acid vegetables like tomatoes will brighten it.

Stainless Steel is the most popular material used for tools and equipment, but is more expensive. It is easier to clean and shine and will not wear out as soon as aluminum. Choose those with copper, aluminum or laminated steel bottoms to spread heat and keep the pot from getting heat dark spots. Stainless steel utensils maybe bought in many gauges, from light to heavy.

Glass is good for baking but not practical on top or surface cooking. Great care is needed to make sure for long shelf life.

Cast Iron is sturdy but must be kept seasoned to avoid rust. Salad oil with no salt or shortening can be rub inside and out and dry. Wash with soap (not detergent) before using.

Ceramic and heat-proof glass is used especially for baking dishes, casseroles, and measuring cups. Glass and ceramic conduct the heat slowly and evenly. Many of these baking dishes are decorated and can go from stove or oven to the dining table.

Teflon is a special coating applied to the inside of some aluminum or steel pots and pans. It helps food from not sticking to the pan. It is easier to wash and clean, however, take care not to scratch the Teflon coating with sharp instrument such as knife or fork. Use

wooden or plastic spatula to turn or mix food inside.

Explanation:

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Small programs called _______ are used to communicate with Paris Friel devices such as monitors printers portable storage device
algol13

the answer is D drivers

7 0
3 years ago
Match each of the following terms to its function:_________
Crazy boy [7]

Answer:

I = B

II = E

III = A

IV = D

V = C

7 0
3 years ago
I have six nuts and six bolts. Exactly one nut goes with each bolt. The nuts are all different sizes, but it’s hard to compare t
juin [17]

Answer:

Explanation:

In order to arrange the corresponding nuts and bolts in order using quicksort algorithm, we need to first create two arrays , one for nuts and another for bolts namely nutsArr[6] and boltsArr[6]. Now, using one of the bolts as pivot, we can rearrange the nuts in the nuts array such that the nuts on left side of the element chosen (i.e, the ith element indexed as nutArr[i]) are smaller than the nut at ith position and nuts to the right side of nutsArr[i] are larger than the nut at position "I". We implement this strategy recursively to sort the nuts array. The reason that we need to use bolts for sorting nuts is that nuts are not comparable among themselves and bolts are not comparable among themselves(as mentioned in the question)

The pseudocode for the given problem goes as follows:

// method to quick sort the elements in the two arrays

quickSort(nutsArr[start...end], boltsArr[start...end]): if start < end: // choose a nut from nutsArr at random randElement = nutsArr[random(start, end+1)] // partition the boltsArr using the randElement random pivot pivot = partition(boltsArr[start...end], randElement) // partition nutsArr around the bolt at the pivot position partition(nutsArr[start...end], boltsArr[pivot]) // call quickSort by passing first partition quickSort(nutsArr[start...pivot-1], boltsArr[start...pivot-1]) // call quickSort by passing second partition quickSort(nutsArr[pivot+1...end], boltsArr[pivot+1...end])

// method to partition the array passed as parameter, it also takes pivot as parameter

partition(character array, integer start, integer end, character pivot)

{

       integer i = start;

loop from j = start to j < end

       {

check if array[j] < pivot

{

swap (array[i],array[j])

               increase i by 1;

           }

 else check if array[j] = pivot

{

               swap (array[end],array[j])

               decrease i by 1;

           }

       }

swap (array[i] , array[end])

       return partition index i;

}

7 0
4 years ago
Most printers are plug and play compatible and must be manually configured when they are plugged into the system.
nikdorinn [45]

Answer:

False

Explanation:

Plug and play devices are computer peripheral devices that can be used immediately, with little or no necessary configuration when plugged or connected to the computer system.

Printers are mostly electronic devices used as an output of a hard copy of a computer system application document. They are not part of the computer system (peripheral devices). They come with installation disk which a computer must install, in order to be able to use the printer.

Yes they are peripheral devices, but they are not plug and play devices since its software must be installed on the computer, to use it

8 0
4 years ago
Read 2 more answers
Write a function negateOdds that takes a list of integers and returns a list of integers with all of the odd integers negated. n
aleksley [76]

Is this computer science?

If so, then the function you would need for your code is this...

_____

if (someValue%2 != 0) {

 value *= -1;

        }

_____

//basically <u>number%2 == 0</u> means even so "!" means false so "not even" meaning "odd."

Assuming you are doing an array list (given a set value) or a for-loop with an

int someValue = Integer.parseInt(args[i]); inside (not given a set value and not restricted)

Otherwise ignore me....lol

5 0
4 years ago
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