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Illusion [34]
3 years ago
12

Which of these is outside the scope of an art director's responsibility?

Computers and Technology
2 answers:
frez [133]3 years ago
7 0

Answer:

Establishing Tight deadlines for a rush job.

Explanation:

The time taken for a rush job has nothing to do with the art director because it has nothing to do with art. Also rush jobs already call for tight deadlines so technically the deadline is already established.

Morgarella [4.7K]3 years ago
4 0

Answer: telling the animators that an entire scene must be reworked

Explanation:

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Implementations <br> in c++ programming when finding lcm and gcd in c++​
vazorg [7]

Answer:

#include<iostream>

using namespace std;

int lcm(int m, int n) {

  int a;

  a = (m > n) ? m: n;

  while (true) {

     if (a % m == 0 && a % n == 0)

        return a;

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int gcd(int m, int n) {

  int r = 0, a, b;

  a = (m > n) ? m : n;

  b = (m < n) ? m : n;

  r = b;

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     r = a % b;

     a = b;

     b = r;

  }

  return r;

}

int main(int argc, char **argv) {

  cout << "Enter the two numbers: ";

  int m, n;

  cin >> m >> n;

  cout << "The LCM of two numbers is: " << lcm(m, n) << endl;

  cout << "The GCD of two numbers is: " << gcd(m, n) << endl;  

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Explanation:

7 0
3 years ago
The relationship between social media and the Internet is complex. Individual Internet behavior involves a myriad of factors tha
Artemon [7]

Answer:

Explanation:

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3 years ago
An array is sorted (in ascending order) if each element of the array is less than or equal to the next element .
Aleks04 [339]

Answer:

// The program below checks if an array is sorted or not

// Program is written in C++ Programming Language.

// Comments are used for explanatory purpose

//Program Starts here

#include<iostream>

using namespace std;

//Function to check if the array is sorted starts here

int isSorted(int arr[], int count)

{

// Check if arrays has one or no elements

if (count == 1 || count == 0) {

return 1;

}

else

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}

else

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}

// Main Method starts here

int main()

{

int count;

cin<<count;

int arr[count];

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{

cin>>arr[I-1];

}

int n = sizeof(arr) / sizeof(arr[0]);

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}

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