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Inessa [10]
4 years ago
13

As a rising photographer who has recently been recognized for several impactful images, why might you explore copyright protecti

on? Consider and explain what it would offer and what, if any, limitations that you should be aware of. Finally, address how this differs from “exclusive rights” and which might be the most applicable in regard to the images you’ve been recognized for?
Computers and Technology
1 answer:
Korolek [52]4 years ago
5 0

<u>Photography copyright protection:</u>

In digital world people use to capture many images or medial files and publish to public.  The images or media files that capture person are responsible and have all rights and ownership of the images.

Some people just steal images or media files and publish to public and inform that as if they are the owner of the images. To protect the stealing it is bounded with law. Owner of the images has full rights and responsibility.

Normally images or medial files are shared with ownership agreement between owner and other party’s exclusive agreements before public to public access.

Exclusive rights rights determined that owner ship of the any media files is exclusive to only to the owner.

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Ryan needs to see the space available to insert content on a slide in his presentation. Which feature of a presentation program
Sav [38]

The feature that Ryan should take advantage is Layout, which is available in his presentation, to insert the content he wants, be it a graph, picture, paragraph, bullet points, video, etc. Assuming this is a Microsoft PowerPoint, Ryan can right-click the slide he wants to put the content in, where an option window would appear. He should click on the ‘Layout’ option and several types of layouts available for him to put content on would appear: Title & Content, Two Content, Comparison, and Content with Caption.

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3 years ago
A user of Augmented Reality is able to experience<br> of information on their surroundings.
Margarita [4]

Explanation:

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Ex: projecting a phone pad to your hand, and Pokemon Go,

7 0
3 years ago
Flujograma de nómina ​
balu736 [363]
Explanation there is the rest in comments

7 0
3 years ago
Create a Java program with threads that looks through a vary large array (100,000,000 elements) to find the smallest number in t
olchik [2.2K]

Answer:

See explaination

Explanation:

import java.util.Random;

public class Sample{

static class MinMax implements Runnable{

int []arr;

int start,end,min,max;

MinMax(int[]arr, int start,int end){

this.start=start;

this.end=end;

min=Integer.MAX_VALUE;

max=Integer.MIN_VALUE;

this.arr=arr;

}

atOverride

public void run() {

for(int i=start;i<=end;i++){ //search min and max form strant to end index

min=Math.min(min,arr[i]);

max=Math.max(max, arr[i]);

}

}

}

public static void main(String[] args) throws Exception{

long beginTime = System.nanoTime();

Random gen = new Random();

int n=100000000;

int[] data = new int[n]; //generate and fill random numbers

for(int i = 0; i < data.length; i++) {

data[i] = gen.nextInt()%1000000;

}

long endTime = System.nanoTime();

System.out.println("Done filling the array. That took " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 1 thread"); //1 thread

MinMax m1=new MinMax(data,0,n-1); //class object

Thread t1=new Thread(m1); //new thread

beginTime=System.nanoTime(); //start timer

t1.start(); //start thread

t1.join(0); //wait until thread finishes

endTime=System.nanoTime(); //end timer

System.out.println("Min,Max: "+m1.min+","+m1.max); //print minimum and maximum

System.out.println("Time using 1 thread " + (endTime - beginTime)/1000000000f + " seconds."); //print time taken

//-----------------------------------------

System.out.println("Using 2 thread");

m1=new MinMax(data,0,n/2);

MinMax m2=new MinMax(data,n/2+1,n-1);

t1=new Thread(m1);

Thread t2=new Thread(m2);

beginTime=System.nanoTime();

t1.start();

t2.start();

t1.join(0);

t2.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(m1.min,m2.min)+","+Math.max(m1.max,m2.max));

System.out.println("Time using 2 thread " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 3 thread");

m1=new MinMax(data,0,n/3);

m2=new MinMax(data,n/3+1,2*n/3);

MinMax m3=new MinMax(data,2*n/3+1,n-1);

t1=new Thread(m1);

t2=new Thread(m2);

Thread t3=new Thread(m3);

beginTime=System.nanoTime();

t1.start();

t2.start();

t3.start();

t1.join(0);

t2.join(0);

t3.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(Math.min(m1.min,m2.min),m3.min)+","+Math.max(Math.max(m1.max,m2.max),m3.max));

System.out.println("Time using 3 thread " + (endTime - beginTime)/1000000000f + " seconds.");

//-----------------------------------------

System.out.println("Using 4 thread");

m1=new MinMax(data,0,n/4);

m2=new MinMax(data,n/4+1,2*n/4);

m3=new MinMax(data,2*n/4+1,3*n/4);

MinMax m4=new MinMax(data,3*n/4+1,n-1);

t1=new Thread(m1);

t2=new Thread(m2);

t3=new Thread(m3);

Thread t4=new Thread(m4);

beginTime=System.nanoTime();

t1.start();

t2.start();

t3.start();

t4.start();

t1.join(0);

t2.join(0);

t3.join(0);

t4.join(0);

endTime=System.nanoTime();

System.out.println("Min,Max: "+ Math.min(Math.min(m1.min,m2.min),Math.min(m3.min,m4.min))+","+Math.max(Math.max(m1.max,m2.max),Math.max(m3.max,m4.max)));

System.out.println("Time using 4 thread " + (endTime - beginTime)/1000000000f + " seconds.");

}

}

6 0
3 years ago
Return true if the given non-negative number is a multiple of 3 or 5, but not both. Use the % "mod" operator.
Pani-rosa [81]

the following C++ function will return true if the given non - negative number is multiple of 5 or 3 else it will return false.

bool function( int x ){

// variable to check if it is multiple of both or not

int number =0;

if(3%x == 0){

number++;

}

if(5%x == 0){

number++;

}

// Now returning by deciding

if( number <=1)

return true;

else

return false

}

4 0
3 years ago
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