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Julli [10]
3 years ago
11

Which one of these tasks is part of the pre-production phase of game development?

Computers and Technology
2 answers:
Illusion [34]3 years ago
5 0

Answer:

b

Explanation:

Alexxx [7]3 years ago
3 0
[B], developing the art style guide and production plan.

It wouldn't be [A], because patches are released to consumers of the game, to fix bugs and add new content, which won't be done until post-production.

It wouldn't be [C] either, as it is also post-production, because you are sending the game to produced, packaged and shipped, meaning the game has already been pretty much fully developed.
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Plato :
ddd [48]

Answer:

filler content

Explanation:

Isabel must use some sort of filler content while creating a wireframe that she is up with. For the banner, she can make use of the placeholder, and for the text, she can make use of the lorem ipsum. Also, if the wireframe is low fidelity then only she can use above. And if the wireframe is high fidelity then she should use the original banner and text.

3 0
3 years ago
Read 2 more answers
I'm using assembly language. This is my assignment. I kinda confused about how to write code for this assignment. Can anyone exp
Brut [27]

Answer:

oid changeCase (char char_array[], int array_size ) {

__asm{

   mov eax, char_array;    

   mov edi, 0;

readArray:

   cmp edi, array_size;

   jge exit;

   mov ebx, edi;          

   shl ebx, 2;

   mov cl, [eax + ebx];    

check:

   //working on it

   cmp cl, 0x41;      

   jl next_indx;

   cmp cl, 0x7A;      

   jg next_indx;

   cmp cl, 'a';

   jl convert_down;

   jge convert_up;

convert_down:

   or cl, 0x20;        //make it lowercase

   jmp write;

convert_up:

   and cl, 0x20;      

   jmp write;

write:

   mov byte ptr [eax + ebx], cl    

next_indx:

   inc edi;

exit:

   cmp edi, array_size;

   jl readArray;

mov char_array, eax;

}

}

Explanation:

  • Move char_array to eax as it is base image .
  • Use ebx as offset .
  • Use ecx as the storage register .
  • check if cl is <= than ASCII value 65 (A) .
6 0
3 years ago
State whether True / False:<br>The pause option stops the recording temporarily.*<br>True<br>False​
defon

Answer:

true

Explanation:

when you pause a recording it will temporarily stop but, when you hit record again it will continue where you left off

7 0
3 years ago
Determine the number of character comparisons made by the brute-force algorithm in searching for the pattern GANDHI in the text
leonid [27]

Answer:

Total number of character comparison = 43

Explanation:

Using the Brute force algorithm

The string of n characters is known as text, and the string of m characters is known as the pattern.

From the given information:

The text (n)=THERE_IS_MORE_TO_LIFE_THAN_INCREASING_ITS_SPEED

The pattern (m) = GANDHI

The total no of characters that we have in the text = 47

The total number of characters in pattern = 6

For a brute force algorithm;

Since; the first character of the pattern does not exist in the text, then the number of trials made can be attempted can be expressed as = n – m + 1

= 47 – 6 + 1

= 47 – 5

= 42

Thus; the algorithm will attempt the trial 42 times.

Now, for loop in the algorithm to run 42 times, the G in the pattern will have to align against the for T in the text, and in the last case, it will be aligned against the last space.

On each attempted trial, the algorithm will make one unsuccessful comparison.

However, at the trial at which the G in the pattern Is aligned with the G in the text, there will be two successful comparisons.

Hence, we can calculate the total number of character comparison as follows:

Total number of character comparison = \mathbf{\bigg ( ( 42 -  (no. \ of \  failed \ comparison) ) \times 1 + (1 \times ( Two \ successful \  comparisons) ) \bigg ) }

Total number of character comparison = ( (( 42 – 1) × 1 ) + ( 1 × 2) )

Total number of character comparison = 41 + 2

Total number of character comparison = 43

3 0
3 years ago
Which part holds the "brains" of the computer?
Rus_ich [418]
The (CPU) holds the “brains” of the computer
7 0
3 years ago
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