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Elena-2011 [213]
2 years ago
9

Determine the value of X if (211) base x = (152) base 8, how to do this?

Computers and Technology
1 answer:
Alex2 years ago
7 0
 we translate the following statement given in terms of logarithms. 211 base x can be expressed into log 211 / log x while 152 base 8 can be expressed int o log 152 over log 8. In this case,
log 211 / log x = log 152 / log 8log x = 0.962x = 10^0.962 = 9.1632
You might be interested in
Given 4 integers, output their product and their average, using integer arithmetic.
solmaris [256]

Answer:

see explaination

Explanation:

Part 1:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

int avg=0, pro=1;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

avg = (num1+num2+num3+num4)/4;

pro = num1*num2*num3*num4;

System.out.println(pro+" "+avg);

}

}

------------------------------------------------------------------

Part 2:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

double avg=0, pro=1; //using double to store floating point numbers.

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

avg = (num1+num2+num3+num4)/4.0; //if avg is declared as a float, then use 4.0f

pro = num1*num2*num3*num4;

System.out.println((int)pro+" "+(int)avg); //using type conversion only integer part

System.out.printf("%.3f %.3f\n",pro,avg);// \n is for newline

}

}

8 0
2 years ago
In this assignment you'll write a program that encrypts the alphabetic letters in a file using the Vigenère cipher. Your program
solniwko [45]

Answer:

C code is given below

Explanation:

// Vigenere cipher

#include <ctype.h>

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

/**

* Reading key file.

*/

char *readFile(char *fileName) {

   FILE *file = fopen(fileName, "r");

   char *code;

   size_t n = 0;

   int c;

   if (file == NULL) return NULL; //could not open file

   code = (char *)malloc(513);

   while ((c = fgetc(file)) != EOF) {

      if( !isalpha(c) )

          continue;

      if( isupper(c) )

          c = tolower(c);

      code[n++] = (char)c;

   }

   code[n] = '\0';

  fclose(file);

   return code;

}

int main(int argc, char ** argv){  

   // Check if correct # of arguments given

   if (argc != 3) {

       printf("Wrong number of arguments. Please try again.\n");

       return 1;

   }

 

  // try to read the key file

  char *key = readFile(argv[1]);

  if( !key ) {

      printf( "Invalid file %s\n", argv[1] );

      return 1;

  }

 

  char *data = readFile(argv[2]);

  if( !data ) {

      printf("Invalid file %s\n", argv[2] );

      return 1;

  }

 

   // Store key as string and get length

   int kLen = strlen(key);

  int dataLen = strlen( data );

 

  printf("%s\n", key );

  printf("%s\n", data );

 

  int paddingLength = dataLen % kLen;

  if( kLen > dataLen ) {

      paddingLength = kLen - dataLen;

  }

  for( int i = 0; i < paddingLength && dataLen + paddingLength <= 512; i++ ) {

      data[ dataLen + i ] = 'x';

  }

 

  dataLen += paddingLength;

 

   // Loop through text

   for (int i = 0, j = 0, n = dataLen; i < n; i++) {          

       // Get key for this letter

       int letterKey = tolower(key[j % kLen]) - 'a';

     

       // Keep case of letter

       if (isupper(data[i])) {

           // Get modulo number and add to appropriate case

           printf("%c", 'A' + (data[i] - 'A' + letterKey) % 26);

         

           // Only increment j when used

           j++;

       }

       else if (islower(data[i])) {

           printf("%c", 'a' + (data[i] - 'a' + letterKey) % 26);

           j++;

       }

       else {

           // return unchanged

           printf("%c", data[i]);

       }

      if( (i+1) % 80 == 0 ) {

          printf("\n");

      }

   }

 

   printf("\n");

 

   return 0;

}

4 0
2 years ago
Read 2 more answers
Which collaboration technology is becoming more and more popular within organizations because it provides a means for forming ad
Leona [35]

Answer:

d) Social networking sites

Explanation:

-Intranet is a network that is created by an organization to share information, tools and different services inside the company.

-Wikis are websites used to share content and knowledge and different users can modify the information.

-VoIP is a technology that allows you to make calls over the internet.

-Social networking sites are online platforms that allow people to connect with organizations and other people, create communities online and share different types of information.

-Unified communications is a system that has different communication methods through a single application.

According to this, the collaboration technology that is becoming more and more popular within organizations because it provides a means for forming ad hoc groups, networking and locating potential business allies is social networking sites.

8 0
2 years ago
Write a program trapezoid.cpp that prints an upside-down trapezoid of given width and height. However, if the input height is im
MatroZZZ [7]

Answer:

#include <iostream>

using namespace std;

int main()

{    

   int rows, width, height, spaces, stars; // declare values

   cout << "enter width" << endl;

   cin >> width;

   cout << "enter height" << endl;

   cin >> height;

   for (int row = 0; row < height; ++row) {

       for (int col = height + row; col > 0; --col) {

           if (height % 6 == 1) {  

               cout << "Impossible shape!" << endl;

               return 0;

           }

           cout << " ";

       }

       for (int col = 0; col < (width - 2 * row); ++col) {

           cout << "*";

           spaces += 1;

           stars -= 2;

       }

       cout << endl;

3 0
2 years ago
Write a program that reads the contents of a file named text.txt and determines the following: The number of uppercase letters i
murzikaleks [220]

Answer:

Explanation:

//Cpp program to count Uppercase, Lowercase, Digits count from file

#include<iostream>

#include<fstream>

using namespace std;

int main()

{

  //ifstream For reading file purpose only

ifstream file("text.txt");

 

  //cheks whether given file is present or not

  if(!file){

      cout<<"File not Found";

      return 0;

  }

  //variables for all three counts and character in file

char character;

int uppercaseCount=0, lowercaseCount=0, digitsCount=0;

  //loop for reading and checkile file character by character

while(file){

      //get a single character

file.get(character);

 

      //checks uppercase character

if((character >= 'A') && (character <= 'Z'))

uppercaseCount++;

     

      //checks lowercase character

 else if((character >= 'a') && (character <= 'z'))

lowercaseCount++;

     //checks lowercase character

      else if((character >= '0') && (character <= '9'))

      digitsCount++;

}

cout<<"\nUppercase characters: "<<uppercaseCount;

cout<<"\nLowercase characters: "<<lowercaseCount;

cout<<"\nDigits: "<<digitsCount;

return 0;

}

/*

content of File text.txt

C++

10.15: Character Analysis Write a program that reads the contents of a file named text.txt and determines the following: The number of uppercase letters in the file The number of lowercase letters in the file The number of digits in the file Prompts And Output Labels. There are no prompts-- nothing is read from standard in, just from the file text.txt. Each of the numbers calculated is displayed on a separate line on standard output, preceded by the following prompts (respectively): "Uppercase characters: ", "Lowercase characters: ", "Digits: ". Input Validation. None.

Output :-

Uppercase characters: 19

Lowercase characters: 434

Digits: 4

*/

3 0
2 years ago
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