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Alexeev081 [22]
4 years ago
5

He ____ content of a web page contains information used by search engines to help users find your website.

Computers and Technology
1 answer:
luda_lava [24]4 years ago
6 0
The header content of a web-page can contain meta tags that classify the page. Search engines use this information.
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Smileys Pizzeria has a special on cheese pizza this month. 6- inch personal pizzas are $5, 10-inch small pizzas are $8, 14-inch
ValentinkaMS [17]

Answer:

THE PROGRAM IS ANEW ONE IT IS CALLED SMILEYS CHOOSER

Explanation:

4 0
3 years ago
Write a MATLAB code for the following problem:
AysviL [449]
<span>Here is matlab that should work % cos(x) = 1 - (x^2)/2! + (x^4)/4! -(x^6)/6!+(x^8)/8!... % let y= x*x % cos(x) = sum( (-y)^n/(2n)! ) format short x= 0.3*pi; y= x*x; for N= 1:6 n= 0:N; s1= [(-y).^n./factorial(2*n) ] mac= sum(s1); cx= cos(x); str= sprintf('%d terms. series: %12.10f cos(x): %12.10f\n %12.10f',... N, mac,cx, (cx-mac)); disp(str); end;</span>
7 0
3 years ago
Consider the function definition void Demo( int&amp; intVal, float floatVal ) { intVal = intVal * 2; floatVal = float(intVal) +
Fynjy0 [20]

Answer:

myInt=40

myFloat=4.8

Explanation:

First look at the function definition .It has two arguments intVal is passed by reference while floatVal is passed by value.So the changes done on the myInt variable will be reflected on the original argument because when a variable is passed by reference the the changes are reflected on the original argument but when a variable is passed by value the function created a duplicate copy of it and work on them so changes are not reflected on the original argument.So myInt will get doubled while myFloat will remain the same.

5 0
3 years ago
What do we call exceptions to the exclusive rights of copyright law?
Studentka2010 [4]

Answer:

Fair use

Explanation:

I don't know if this is correct but Its the best option.

4 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
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