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SCORPION-xisa [38]
2 years ago
12

The computerization of the layout and paste-up process further complicated printing, as did the digitizing of photographs.

Computers and Technology
1 answer:
andrey2020 [161]2 years ago
7 0

Answer:

The computerization of the layout and paste-up process further complicated printing, as did the digitizing of photographs.

b. False

Explanation:

It is false as computerization has simplified, as in many other instances paste up or laying out methods instead, because the printing steps are now fewer and strightforwarder thanks to the desktop standardized creative page-design publishing programs.

Previously, printing laid on the offset lithography process, which first needed a completed page or "camera-ready page", known as mechanicals or mechanical art firts, to be them sent to be photographed using a specialized camera to get a same-size film negative for each printing plate designed.

Paste up also depended on a several steps process called phototypesetting. First you needed to obtain what was called a "cold type" printed on photographic paper shaped in long columns of text which turn ot to be a single column in a scroll of narrow (3-inch or 4-inch) printouts, as deep as the text was. There were crafted people, known as paste-up artist, layout artist, mechanical artist, production artist, or compositor who, after receiving the types from the typesetter, trimmed them carefully into sections and sequenced them along the necessary columns, so he did with headlines and all the other designed typographic to get the final page design to be printed.   The result were series of strips that were sticked out using rubber cement as a partial glue on the back applied using a brush or passing them through a machine that would apply an adhesive type of wax, so thatthe strips were removable if they needed to be changed on a stiff white paper-board, the artist managed to set them in publication's margins and columns, using a pencil, non-photographic blue ink or a light cyan color which the orthochromatic film would ignore in the printing plates of offset lithography.

Digital photography with its better image exposure, contrast, color correction image output as well, has developed fast from traditional darkroom methods to continuous-tone digital printers and quality computer processing of superior images obtained from a digital camera or from scanned conventional film directly to digital color printers, compared to conventional darkroom prints, when evaluated for resolution, image quality, costs, chemical-free digital photography.

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Explanation:

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Which comparison operator is valid for greater than or equal to?<br><br> &gt;<br> &gt;=<br> =&gt;
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Assume you have a button control named btndisplaylist. Which is the default name for an event procedure that will be executed wh
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8 0
1 year ago
1. How many bits would you need to address a 2M × 32 memory if:
Dominik [7]

Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0
2 years ago
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