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lozanna [386]
3 years ago
6

Which feature of a blog helps to store and retrieve older posts?

Computers and Technology
2 answers:
Rashid [163]3 years ago
8 0
Could it be archive posts? I'm not sure, but I believe it's archive posts.
Alexeev081 [22]3 years ago
8 0

archive pages.. just took test

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Based on this information, what is their gross income for six months? Roderick
alexira [117]
Multiply what he makes a month by six
8 0
3 years ago
A portrait shows a person's face entirely in silhouette, shadowing the subject's features completely.
Sonja [21]

Answer:

From behind

Explanation:

This is because, a silhouette can only be formed by an object if the light is coming from behind the object. For example, if a shadow of an object is to be formed, the light shone on the object must be coming from behind the object.

Similarly, for a complete silhouette of the person's face, the light must be coming from behind the person's face for it to form a shadow and then a silhouette.

So the light must come from behind.

6 0
4 years ago
How to sync music from my phone to groove music
Svetllana [295]
<span><span>Make sure your device supports OneDrive music. Music will not show up on Windows 8, Windows Phone 8, Zune, or Windows Media Player. To find out whether your device supports OneDrive content, See System requirements above.</span><span>Make sure your music is in your OneDrive Music folder.</span>Make sure your music files are one of the supported file types: MP3, M4A, or WMA unprotected. Protected files added to your OneDrive Music folder will not show up on other devices.<span>Check to see that you're not over the 50,000 track limit in your Music collection. If you have more than 50K files in your OneDrive music folder, only the first 50K files will show.
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4 0
3 years ago
Write a program for Horizon Phones, a provider of cellular phone service. Prompt a user for maximum monthly values for talk minu
Anarel [89]

Answer:

mport java.util.Scanner;

public class CellPhoneService

{

    public static void main (String args[])

    {

         Scanner sc=new Scanner(System.in);

         System.out.println("Enter the Talk minutes");

         int talkmin=sc.nextInt();

         System.out.println("Enter the text message");

         int textmsg=sc.nextInt();

         System.out.println("Enter the Gigabyte Data");

         int gbdata=sc.nextInt();

         if(talkmin<500 && textmsg==0 && gbdata==0)

         System.out.println("PLAN A IS RECOMMENDED FOR THIS USER at 49$ per month");

         else if(talkmin<500 && textmsg>0 && gbdata==0)

         {

              System.out.println("PLAN B IS RECOMMENDED FOR THIS USER at 55$ permonth");

         }

         else if(talkmin>=500 && textmsg<100 && gbdata==0)

         {

              System.out.println("PLAN C IS RECOMMENDED FOR THIS USER at 61$ permonth");

         }

         else if(talkmin>=500 && textmsg>=100 && gbdata==0)

         {

              System.out.println("PLAN D IS RECOMMENDED FOR THIS USER at 70$ permonth");

         }

         else if(gbdata>0 && gbdata<2)

         {

              System.out.println("PLAN E IS RECOMMENDED FOR THIS USER at 79$ permonth");

         }

         else if(gbdata>=2)

         {

              System.out.println("PLAN F IS RECOMMENDED FOR THIS USER at 87$ permonth");

         }

    }

}

Explanation:

  • Take the talk minute and text message as input from user.  
  • Apply the conditional statement to check if talk minute is less than 500 along with text message and GB data having a value 0.
  • At last check if GB data is greater than or equal to 2, then print the relevant response.
3 0
4 years ago
In this lab you will write a program that simulates a mouse in a maze. The maze will have one exit location. The mouse will star
sergij07 [2.7K]

Answer:

/* C/C++ program to solve Rat in a Maze problem using  

backtracking */

#include <stdio.h>  

// Maze size  

#define N 4  

bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N]);  

/* A utility function to print solution matrix sol[N][N] */

void printSolution(int sol[N][N])  

{  

for (int i = 0; i < N; i++) {  

 for (int j = 0; j < N; j++)  

  printf(" %d ", sol[i][j]);  

 printf("\n");  

}  

}  

/* A utility function to check if x, y is valid index for N*N maze */

bool isSafe(int maze[N][N], int x, int y)  

{  

// if (x, y outside maze) return false  

if (x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1)  

 return true;  

return false;  

}  

/* This function solves the Maze problem using Backtracking. It mainly  

uses solveMazeUtil() to solve the problem. It returns false if no  

path is possible, otherwise return true and prints the path in the  

form of 1s. Please note that there may be more than one solutions,  

this function prints one of the feasible solutions.*/

bool solveMaze(int maze[N][N])  

{  

int sol[N][N] = { { 0, 0, 0, 0 },  

    { 0, 0, 0, 0 },  

    { 0, 0, 0, 0 },  

    { 0, 0, 0, 0 } };  

if (solveMazeUtil(maze, 0, 0, sol) == false) {  

 printf("Solution doesn't exist");  

 return false;  

}  

printSolution(sol);  

return true;  

}  

/* A recursive utility function to solve Maze problem */

bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N])  

{  

// if (x, y is goal) return true  

if (x == N - 1 && y == N - 1) {  

 sol[x][y] = 1;  

 return true;  

}  

// Check if maze[x][y] is valid  

if (isSafe(maze, x, y) == true) {  

 // mark x, y as part of solution path  

 sol[x][y] = 1;  

 /* Move forward in x direction */

 if (solveMazeUtil(maze, x + 1, y, sol) == true)  

  return true;  

 /* If moving in x direction doesn't give solution then  

 Move down in y direction */

 if (solveMazeUtil(maze, x, y + 1, sol) == true)  

  return true;  

 /* If none of the above movements work then BACKTRACK:  

  unmark x, y as part of solution path */

 sol[x][y] = 0;  

 return false;  

}  

return false;  

}  

// driver program to test above function  

int main()  

{  

int maze[N][N] = { { 1, 0, 0, 0 },  

    { 1, 1, 0, 1 },  

    { 0, 1, 0, 0 },  

    { 1, 1, 1, 1 } };  

solveMaze(maze);  

return 0;  

}  

3 0
3 years ago
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