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IrinaVladis [17]
2 years ago
14

! (( i > ( f + 1)) es verdadero o falso segun la logica de programación.

Computers and Technology
1 answer:
Ostrovityanka [42]2 years ago
3 0

Answer:

yeajadaj

Explanation:

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7 0
4 years ago
Read 2 more answers
a) (20 points) Define the method inDogish recursively such that it returns true if the word is in dog-ish and false if it is not
tia_tia [17]

Answer:

/ Main.java

class Main {

     public static void main(String[] args) {

           //System.out.println(inDogish("aplderogad"));

           //System.out.println(inXish("aplderogad", "dog"));

     }

     // returns true if the word is in dog-ish

     // returns false if word is not in dog-ish

     public static boolean inDogish(String word) {

           /**

           * Note: we can use the inXish() method to complete this method in a

           * single line if you want. just write 'return inXish(word,"dog");' and

           * you are done.

           */

           // if word is null or empty, returning false

           if (word == null || word.length() == 0) {

                 return false;

           }

           // converting word to lower case

           word = word.toLowerCase();

           // if 'd' does not exist on the word, returning false

           if (!dogishHelper(word, 'd')) {

                 return false;

           }

           // removing the characters upto first 'd' in word

           word = word.substring(word.indexOf('d') + 1);

           // if 'o' does not exist on the word, returning false

           if (!dogishHelper(word, 'o')) {

                 return false;

           }

           // removing the characters upto first 'o' in word

           word = word.substring(word.indexOf('o') + 1);

           // if 'g' does not exist on the word, returning false, otherwise

           // returning true

           if (!dogishHelper(word, 'g')) {

                 return false;

           } else {

                 return true;

           }

     }

     // necessary to implement inDogish recursively.

     // returns true if letter is in word, else not.

     public static boolean dogishHelper(String word, char letter) {

           // if word is empty, returning -1

           if (word == null || word.length() == 0) {

                 return false;

           }

           // if first character in word is letter, returning 0

           if (word.charAt(0) == letter) {

                 return true;

           }

           return dogishHelper(word.substring(1), letter);

     }

     // the best thing about the above helper method is that we can use it in

     // inXish method also

     // a generalized version of the inDogish method

     public static boolean inXish(String word, String x) {

           // if x is empty, returning true (base case, which means all the letters

           // in x are in word in that order)

           if (x.length() == 0) {

                 return true;

           }

           // getting first character of x

           char first = x.charAt(0);

           // if first character does not exist on the word, returning false

           if (!dogishHelper(word, first)) {

                 return false;

           }

           // calling and returning inXish recursively, passing the word after

           // removing characters upto first occurrance of char first, and x after

           // removing first character from it

           return inXish(word.substring(word.indexOf(first) + 1), x.substring(1));

     }

}

Explanation:

7 0
4 years ago
On a piano, a key has a frequency, say f0. Each higher key (black or white) has a frequency of f0 * rn, where n is the distance
lesantik [10]

Answer:

In C:

#include <stdio.h>

#include <math.h>

int main(){

   float f0,r,temp;

   r = pow(2.0,1.0/12);

   printf("f0: ");    scanf("%f", &f0);

   temp = f0;

   for(int i = 0; i<=4;i++){

       f0 = f0 * pow(r,i);

       printf("%.2lf ", f0);

       f0 = temp;    }

   return 0;

}

Explanation:

This declares f0, r and temp as float

   float f0,r,temp;

This initializes r to 2^(1/12)

   r = pow(2.0,1.0/12);

This prompts the user for f0

   printf("f0: ");    scanf("%f", &f0);

This saves f0 in temp

   temp = f0;

This iterates the number of keys from 0 to 4

   for(int i = 0; i<=4;i++){

This calculates each key

       f0 = f0 * pow(r,i);

This prints the key

       printf("%.2lf ", f0);

This gets the initial value of f0

       f0 = temp;    }

   return 0;

8 0
3 years ago
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