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Daniel [21]
2 years ago
10

Technology that transfers page images composed inside a computer directly to printing plates is called:

Computers and Technology
1 answer:
AlekseyPX2 years ago
6 0

Technology that transfers page images composed inside a computer directly to printing plates is called Computer-to-plate.

<h3>What is Computer-to-plate?</h3>

Computer-to-plate (CTP) is known to be a kind of modern imaging technology. It is known to be used in the area of modern printing processes.

This technology is one that often forms an image suing a Desktop Publishing (DTP) application which is the output to a printing plate. In the CTP procedure, the plates are said to be made from a variety of materials that is from aluminum to polyester.

Learn more about Technology from

brainly.com/question/25110079

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Any websites online to make $50 daily?​
andreyandreev [35.5K]

Answer:

hmmm research some and find out because there are alot

7 0
3 years ago
ENG103 DISCUSSION BOARD 5
ehidna [41]
Plagiarism is a crime, flat out. It is unethical and immoral to copy, verbatim, another persons work and claim it as your own, especially for scholastic uses. In order to avoid plagiarism, it is important that we give accurate references and paraphrase, rather that copy and paste information. It is important to consider the widespread effect of plagiarism. It effects all involved: The student, the owner of the information, the instructor, and even the site from which the information was stolen. Use the proper channels and be honest when completing assignments goes a long way. We must do what we can to prevent plagiarism, starting with ourselves and our own research. <span />
5 0
4 years ago
2- (8 point) Write a program using the instructions below. Assume that integers are stored in 4 bytes. a) Define an array of typ
lakkis [162]

Answer:

a)  

int apples [5] = {2, 4, 6, 8, 10};

b)

int *aPtr   //this is the pointer to int

Another way to attach a pointer to a an int variable that already exists:

int * aPtr;

int var;

aPtr = &var;

c)

for (int i = 0; i < size; i++){

       cout << values[i] << endl;    }

d)  

   aPtr = values;

   aPtr = &values[0];    

both  the statements are equivalent

e)

If its referring to the part d) then the address is:

cout<<aPtr;

f)

     for (int i = 0; i < size; ++i) {

            cout<<*(vPtr + i)<<endl;    }

g)

   cout << (aPtr + 3) << endl;  // address referenced by aPtr + 3

   cout << *(aPtr + 3) << endl; // value stored at that location

This value stored at location is 8

h)

    aPtr = &apples[4];

    aPtr -= 4;

    cout<<aPtr<<endl;

    cout<<*aPtr<<endl;  

Explanation:        

a)

int apples [5] = {2, 4, 6, 8, 10};

In this statement the array names is apples, the size of the array is specified in square brackets. so the size is 5. The type of array apples is int this means it can store integer elements. The values or elements of the array apples are even integers from 2 to 10. So the elements of array are:

apples[0] = 2

apples[1] = 4

apples[2] = 6

apples[3] = 8

apples[4] = 10

b)

In this statement int *aPtr  

The int* here is used to make the pointer aPtr points to integer object. Data type the pointer is pointing to is int. The asterisk symbol used with in makes this variable aPtr a pointer.

If there already exists an int type variable i.e. var and we want the pointer to point to that variable then declare an int type pointer aPtr and aPtr = &var; assigns the address of variable var to aPtr.

int * aPtr;

int var;

aPtr = &var;

c)

The complete program is:

int size= 5;

int values[size] = {2,4,6,8,10};

for (int i = 0; i < size; i++){

       cout << values[i] << endl; }

The size of array is 5. The name of array is values. The elements of array are 2,4,6,8,10.

To print each element of the values array using array subscript notation, the variable i is initialized to 0, because array index starts at 0. The cout statement inside body of loop prints the element at 0-th index i.e. the first element of values array at first iteration. Then i is incremented by 1 each time the loop iterates, and this loop continues to execute until the value of i get greater of equal to the size i.e. 5 of values array.

The output is:

2

4

6

8

10

d)

aPtr = values;

This statement assigns the first element in values array to pointer aPtr. Here values is the address of the first element of the array.

aPtr = &values[0];    

In this statement &values[0] is the starting address of the array values to which is assigned to aPtr. Note that the values[0] is the first element of the array values.

e)

Since &values[0] is the starting address of the array values to which is assigned to aPtr. So this address is the physical address of the starting of the array. If referring to the part d) then use this statement to print physical address is aPtr pointing to

cout<<aPtr;

This is basically the starting address of the array values to which is assigned to aPtr.

The output:

0x7fff697e1810                

f)

i variable represents offset and corresponds directly to the array index.

name of the pointer i.e. vPtr references the array

So the statement (vPtr + i) means pointer vPtr that references to array values plus the offset i array index that is to be referenced. This statement gives the address of i-th element of values array. In order to get the value of the i-th element of values array, dereference operator * is used.  It returns an ith value equivalent to the address the vPtr + i is pointing to. So the output is:

2

3

6

8

10

g)

values[0] is stored at 1002500

aPtr + 3 refers to values[3],

An integer is 4 bytes long,

So the address that is referenced by aPtr + 3 is

1002500 + 3 * 4 = 1002512

values[3] is basically the element of values array at 3rd index which is the 4th element of the array so the value stored at that referred location  is 8.

h)

Given that aPtr points to apples[4], so the address stored in aPtr is

1002500 + 4 * 4 = 1002516

aPtr -= 4  is equivalent to aPtr = aPtr - 4

The above statement decrements aPtr by 4 elements of apples array, so the new value is:

1002516 - 4 * 4 = 1002500

This is the address of first element of apples array i.e 2.

Now

cout<<aPtr<<endl; statement prints the address  referenced by aPtr -= 4 which is 1002500  

cout<<*aPtr<<endl;  statement prints the value is stored at that location which is 2.

6 0
4 years ago
APCPS B U10L12 exam (Connexus)
il63 [147K]

I don't know how to that i don't even know how to do my work

7 0
2 years ago
Negative numbers are encoded using the __________ technique. a. two’s complement b. floating point c. ASCII d. Unicode
OLEGan [10]

Negative numbers are encoded using the two’s complement technique. Two's complement is used to encode negative numbers.

Option  A is correct .

<h3>What method does the in data type use to store negative numbers?</h3>

Most implementations you're likely to encounter store negative signed integers in a form known as two's complement. Another important way of storing negative signed numbers is called one's complement. The one's complement of an N-bit number x is basically defined as x with all bits inverted.

<h3>What is encoding method?</h3>

An encoding technique is the application of established industry conventions to a coded character set to create a coded character scheme. Such rules determine the number of bits required to store the numeric representation of a given character and its code position in the encoding.

Learn more about two complement technique encoding :

brainly.com/question/26668609

#SPJ4

5 0
1 year ago
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