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Sindrei [870]
3 years ago
8

Wendy has to capture a sculpture in the middle of the park at night. She is not able to get the entire sculpture into focus in o

ne image. She decides to take multiple sharp images of different angles of the sculpture and then merge the images into one sharp complete picture of the sculpture. What is this technique called? A. depth mapping B. depth of focus C. focus stacking D. focus sharing
Computers and Technology
1 answer:
dsp733 years ago
6 0

Answer:

Focus Stacking

Explanation:

F-stop stacking which is also referred to as image stacking is a powerful technique that improves the quality of an image by stacking images taken at different f-stops in order to improve corner sharpness and overcome blurriness. Once photographs are taken at different f-stops (focus stops), a final complete composite picture is created using only the sharpest portions of the photograph.

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If the value of score1 is 350 and the value of score2 is 210, what will be the value of result after the code segment is execute
MAVERICK [17]

  The value of result of the  code segment is executed is known to be 4.

<h3>Why is the value of the code segment so?</h3>

  When the result of  is not  executed because the condition is said to be false and also when there is a false condition is, the else statement will be said to be true

  Therefore,   result = result + 2; -> result is brought up by 2 to bring about 4 and as such, the value of result of the  code segment is executed is known to be 4.

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2 years ago
You have received a "no boot device found" notification upon booting your system. what does this mean, and what can you do to tr
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Basically no boot device could be 1 of three things listed below.

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CALL US ☏1≠855☼241☼6569 Aol mail not working on windows 10 ||123||**USA***
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Explanation:

3 0
3 years ago
Locker doors There are n lockers in a hallway, numbered sequentially from 1 to n. Initially, all the locker doors are closed. Yo
kow [346]

Answer:

// here is code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int n,no_open=0;

   cout<<"enter the number of lockers:";

   // read the number of lockers

   cin>>n;

   // initialize all lockers with 0, 0 for locked and 1 for open

   int lock[n]={};

   // toggle the locks

   // in each pass toggle every ith lock

   // if open close it and vice versa

   for(int i=1;i<=n;i++)

   {

       for(int a=0;a<n;a++)

       {

           if((a+1)%i==0)

           {

               if(lock[a]==0)

               lock[a]=1;

               else if(lock[a]==1)

               lock[a]=0;

           }

       }

   }

   cout<<"After last pass status of all locks:"<<endl;

   // print the status of all locks

   for(int x=0;x<n;x++)

   {

       if(lock[x]==0)

       {

           cout<<"lock "<<x+1<<" is close."<<endl;

       }

       else if(lock[x]==1)

       {

           cout<<"lock "<<x+1<<" is open."<<endl;

           // count the open locks

           no_open++;

       }

   }

   // print the open locks

   cout<<"total open locks are :"<<no_open<<endl;

return 0;

}

Explanation:

First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.

Output:

enter the number of lockers:9

After last pass status of all locks:

lock 1 is open.

lock 2 is close.

lock 3 is close.

lock 4 is open.

lock 5 is close.

lock 6 is close.

lock 7 is close.

lock 8 is close.

lock 9 is open.

total open locks are :3

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<span>C. pretending to be someone else when asking for information</span>
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