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dem82 [27]
3 years ago
7

Vanessa is a set designer. She is working on a period film set in Egypt. She has selected several items for a scene in a palace.

She visits a library and a museum in Cairo to check the authenticity of the items. What responsibility is Vanessa undertaking?
Computers and Technology
2 answers:
BARSIC [14]3 years ago
4 0

C - researching

PLATO

Leona [35]3 years ago
3 0

Answer:

the responsibility of cultural and historical accuracy checking

Explanation:

Vanessa seems to be undertaking the responsibility of cultural and historical accuracy checking. By visiting the library and museum, Vanessa is researching historical data in order to make sure that the items that she has selected match the same type of items that Individuals living in Ancient Egypt times would use as well. This provides both historical and cultural accuracy to the film by representing The citizens of Ancient Egypt as they were in that time.

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Let T be the statement: The sum of any two rational numbers is rational. Then T is true, but the following "proof is incorrect.
ikadub [295]

Answer:

the mistake is in the first statement.

Explanation:

Now lets us put the statement into consideration:

"The sum of any two rational numbers is irrational"

The negation is: " there exists a pair of rational numbers whose sum is irrational". (Existence of at least one of such a pair).

The negation is not "the sum of any two rational numbers is irrational"

Therefore the mistake is in the first statement and it is due to incorrect negation of the proof.

4 0
3 years ago
. Reorder the following efficiencies from smallest to largest:
alexdok [17]

Answer:

The order of the efficiencies is as following:-

10,000 < 2n < nlog(n) < n5 < n!.

Explanation:

10,000 is constant time whatever will be the size of the problem the efficiency will remain the same.

2n this efficiency is linear it will grow proportionally as the size of the problem increases.

nlog(n) this efficiency is is a bit greater than 2n though it will grow faster than 2n but slower than n2 as the size of the problem increases.

n5 this efficiency is very poor.It is growing very rapidly as the size of the problem increases.

n! is the worst efficiency of them all.

n!=n*(n-1)*(n-2)*(n-3)*(n-4)*.......2*1.

It will grow beanstalk in jack and the beanstalk.

3 0
3 years ago
Do you believe that OOP should be phased out and we should start working on some alternative(s)?
katen-ka-za [31]

Answer:

The answer to this question can be defined as below:

Explanation:

There are many new directions for solutions for any given problem in today's computer age, in which the "OOP's" is one of the philosophies of preference.  

  • If the issue for each of the several cases can be degraded in various classes, that reveal complex relationships.  
  • In this process the OO analysis is acceptable with other concepts, these issues usually produce difficult outcomes, that's why I don't believe in this modeling.
5 0
3 years ago
"How has graphic design changed through time?"
noname [10]

The main purpose behind graphic designing is to communicate, The different colors, shapes and patterns used in graphics aim to deliver a meaningful message, which sometimes is hidden. Back in the time, graphic designs usually portrayed what was going on at that time, for instance wars, with larger icons and catchy slogans. Over the time, the graphic designs have evolved where psychedelic, geometrical and monochromatic designs have gained more popularity, with different typographic styles to follow. Moreover, since the world has been taken over by technology in a storm, we get to see some drastic changes in the photography styles, use of color balances with minimal designs and subtle tones.

8 0
3 years ago
palindrome is a string that reads the same forwards as backwards. Using only a xed number of stacks, and a xed number of int and
bixtya [17]

Solution :

check_palindrome$(string)$

   lower_$case$_string$=$ string$. to$_lower()

   Let stack = new Stack()

   Let queue = new Queue();

   for each character c in lower_case_string:

       stack.push(c);

       queue.enqueue(c);

   let isPalindrome = true;

   while queue is not empty {

       if (queue.remove().equals(stack.pop())) {

           continue;

       } else {

           isPalindrome=false;

           break while loop;

       }

   }

   return isPalindrome

Input = aabb

output = true

input =abcd

output = false

6 0
3 years ago
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