1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Monica [59]
3 years ago
5

Jerry wants to save his company money. He decides to move to open source software for his word processing needs. Jerry then down

loads ________.
Computers and Technology
1 answer:
Schach [20]3 years ago
6 0

Answer:

The answer is Libreoffice

Explanation:

Jerry wants to save his company money. He decides to move to open source software for his word processing needs. Jerry then downloads Libreoffice

LibreOffice is a powerful office suite – its clean interface and feature-rich tools help you unleash your creativity and enhance your productivity.

LibreOffice includes several applications that make it the most powerful Free and Open Source office suite on the market.

You might be interested in
Build three classes that conform to the following interfaces. Use arrays in creating your classes (e.g., do not use the built-in
daser333 [38]

Explanation:

public class ArrayList {

private Object[] array = new Object[1];

/**

 * Places new element at location

 * @param c, element to be inserted

 * @param i, location it is to be placed

 */

public void insert(char c, int i) {

 if(i>this.size()){

  System.out.println("Index "+i +" outside of list size; max: " +this.size());

  System.exit(0);

 }

 int count = 0;

 try {

  Object[] other = new Object[this.array.length+1];

  switch(i){

     case 0:{

      other[0] = c;

      for(int j=1;j<this.array.length;j++)

       other[j] = this.array[j-1];

       

      this.array = other;

      break;

     }

     default: {

      for(int j=0;j<i;j++)

       other[j] = this.array[j];

      other[i] = c;

      for(int k=i+1;k<this.array.length;k++)

       other[k] = this.array[k-1];

      this.array = other;

      break;

     }

     }

 }

 catch(ArrayIndexOutOfBoundsException exception) {

     largerArray();

     if(++count == 2){

      System.out.println("Something went wrong.");

      System.exit(0);

     }  

 }

}

 

/**

 * Places new element at location

 * @param object, element to be inserted

 * @param index, location it is to be placed

 */

void insert(Object object, int index){

 if(index>this.size()){

  System.out.println("Index "+index +" outside of list size; max: " +this.size());

  System.exit(0);

 }

 int count = 0;

 try {

  Object[] other = new Object[this.array.length+1];

  switch(index){

     case 0:{

      other[0] = object;

      for(int j=1;j<this.array.length;j++)

       other[j] = this.array[j-1];

       

      this.array = other;

      break;

     }

     default: {

      for(int j=0;j<index;j++)

       other[j] = this.array[j];

      other[index] = object;

      for(int k=index+1;k<this.array.length;k++)

       other[k] = this.array[k];

      this.array = other;

      break;

     }

     }

 }

 catch(ArrayIndexOutOfBoundsException exception) {

     largerArray();

     if(++count == 2){

      System.out.println("Something went wrong.");

      System.exit(0);

     }  

 }

}

 

/**

 * Removes element at index

 * @param index, location to remove

 * @return temp, object removed

 */

Object remove(int index){

 if(index>this.size()){

  System.out.println("Index "+index +" outside of list size; max: " +this.size());

  System.exit(0);

 }

 Object temp = this.array[index];

 int count = 0;

 try {

  Object[] other = new Object[this.array.length-1];

  switch(index){

     case 0:{

      other[0] = this.array[1];

      for(int j=1;j<this.size();j++)

       other[j] = this.array[j+1];

       

      this.array = other;

      break;

     }

     default: {

      for(int j=0;j<index;j++)

       other[j] = this.array[j];

      for(int k=index;k<this.size();k++)

       other[k] = this.array[k+1];

      this.array = other;

      break;

     }

     }

 }

 catch(ArrayIndexOutOfBoundsException exception) {

     largerArray();

     if(++count == 2){

      System.out.println("Something went wrong.");

      System.exit(0);

     }  

 }

 return temp;

}

 

/**

 * Copies the array to a new array twice the size.

 */

public void largerArray(){

 Object[] other = new Object[this.array.length*2];

 System.arraycopy(this.array, 0, other, 0, this.array.length);

 this.array = other;

}

 

/**

 * @return true if there are null elements

 */

public boolean isEmpty() {

 boolean bool = true;

 for (Object element : this.array) {

  if(element!=null){

   bool = false;

   break;

  }

  else{

   bool = true;

   break;

 }

 }

 return bool;

}

 

/**

 * Determines the real length of the array

 * @return length of non-null elements

 */

public int size(){

 int count = 0;

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

  if(this.array[i]==null)

   continue;

  else

   count++;

 }

 return count;

}

 

/**

 * Converts array to string.

 * @return the Array as a string

 */

public String toString(){

 String string = "";

 for(int i=0;i<this.size();i++){

  if(i<this.size()-1)

   string += this.array[i]+", ";

  else

   string += this.array[i];

 }

 return string;

}

 

/**

 * Similar to get, imput the character, retrieve index.

 * @param object, what we're looking for

 * @return index

 */

int indexOf(Object object){

 int index = -1;

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

  if(this.array[i]==(object))

   index = i;

  else continue;

 }

 return index;

}

 

/**

 * @param object, Arraylist tyoe

 * @return true, if the elements of the arrays are equivalent.

 */

public boolean equals(ArrayList object){

 boolean bool = false;

 int thisLength = this.array.length;

 int objectLength = object.array.length;

 if(thisLength > objectLength){

  for(int i=0;i<this.array.length;i++)

   if(this.array[i] == object.array[i])

    bool = true;

   else{

    bool = false;

    break;

   }

 }

 else{

  for(int i=0;i<object.array.length;i++)

   if(this.array[i] == object.array[i])

    bool = true;

   else{

    bool = false;

    break;

   }

 }

 return bool;

}

 

/**

 * Getter to check what is at an index

 * @param index, location

 * @return the character at this position

 */

Object get(int index){

 return this.array[index];

}

}

6 0
3 years ago
In addition to MLA, what are some other widely used style guides? Check all that apply.
STALIN [3.7K]

Answer:

In addition to MLA, <u>American psychological association, Associated Press and Chicago manual of style</u> are widely using referencing techniques,

Explanation:

<u>American Psychological Association(APA)</u>

APA is the largest organization of scholars, they have their own referencing style guidelines that is used by these scholars.

<u>Associated Press</u>

It is also an american based agency that have their own referencing style for citation of research work named as Associated press.

<u>Chicago Manual of Style</u>

It is also an american referencing style used for reference.

8 0
3 years ago
Which of the following could result from heat being transferred to a substance?
lbvjy [14]

Answer: i dont see the following sorry cant help ya

Explanation:

8 0
3 years ago
Blank text has a darker apperance than normal texts
sammy [17]
That's cool, I guess. What was the question?
4 0
3 years ago
a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv
Leokris [45]

Using the knowledge in computational language in C++ it is possible to write a code that  cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

<em>#include <bits/stdc++.h></em>

<em>using namespace std;</em>

<em>// chracter to digit mapping, and the inverse</em>

<em>// (if you want better performance: use array instead of unordered_map)</em>

<em>unordered_map<char, int> c2i;</em>

<em>unordered_map<int, char> i2c;</em>

<em>int ans = 0;</em>

<em>// limit: length of result</em>

<em>int limit = 0;</em>

<em>// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]  </em>

<em>bool helper(vector<string>& words, string& result, int digit, int widx, int sum) { </em>

<em>    if (digit == limit) {</em>

<em>        ans += (sum == 0);</em>

<em>        return sum == 0;</em>

<em>    }</em>

<em>    // if summation at digit position complete, validate it with result[digit].</em>

<em>    if (widx == words.size()) {</em>

<em>        if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {</em>

<em>            if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result</em>

<em>                return false;</em>

<em>            c2i[result[digit]] = sum % 10;</em>

<em>            i2c[sum%10] = result[digit];</em>

<em>            bool tmp = helper(words, result, digit+1, 0, sum/10);</em>

<em>            c2i.erase(result[digit]);</em>

<em>            i2c.erase(sum%10);</em>

<em>            ans += tmp;</em>

<em>            return tmp;</em>

<em>        } else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){</em>

<em>            if (digit + 1 == limit && 0 == c2i[result[digit]]) {</em>

<em>                return false;</em>

<em>            }</em>

<em>            return helper(words, result, digit+1, 0, sum/10);</em>

<em>        } else {</em>

<em>            return false;</em>

<em>        }</em>

<em>    }</em>

<em>    // if word[widx] length less than digit, ignore and go to next word</em>

<em>    if (digit >= words[widx].length()) {</em>

<em>        return helper(words, result, digit, widx+1, sum);</em>

<em>    }</em>

<em>    // if word[widx][digit] already mapped to a value</em>

<em>    if (c2i.count(words[widx][digit])) {</em>

<em>        if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0) </em>

<em>            return false;</em>

<em>        return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);</em>

<em>    }</em>

<em>    // if word[widx][digit] not mapped to a value yet</em>

<em>    for (int i = 0; i < 10; i++) {</em>

<em>        if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;</em>

<em>        if (i2c.count(i)) continue;</em>

<em>        c2i[words[widx][digit]] = i;</em>

<em>        i2c[i] = words[widx][digit];</em>

<em>        bool tmp = helper(words, result, digit, widx+1, sum+i);</em>

<em>        c2i.erase(words[widx][digit]);</em>

<em>        i2c.erase(i);</em>

<em>    }</em>

<em>    return false;</em>

<em>}</em>

<em>void isSolvable(vector<string>& words, string result) {</em>

<em>    limit = result.length();</em>

<em>    for (auto &w: words) </em>

<em>        if (w.length() > limit) </em>

<em>            return;</em>

<em>    for (auto&w:words) </em>

<em>        reverse(w.begin(), w.end());</em>

<em>    reverse(result.begin(), result.end());</em>

<em>    int aa = helper(words, result, 0, 0, 0);</em>

<em>}</em>

<em />

<em>int main()</em>

<em>{</em>

<em>    ans = 0;</em>

<em>    vector<string> words={"GREEN" , "BLUE"} ;</em>

<em>    string result = "BLACK";</em>

<em>    isSolvable(words, result);</em>

<em>    cout << ans << "\n";</em>

<em>    return 0;</em>

<em>}</em>

See more about C++ code at brainly.com/question/19705654

#SPJ1

3 0
2 years ago
Other questions:
  • Connected contacts require___ to open.
    11·1 answer
  • Identify the mobile device deployment option that gives the user flexibility to select a device, but allows the company to contr
    5·1 answer
  • The biggest factor in determining the price of a mortgage is:
    14·1 answer
  • When a user runs an application, what transfers from a storage device to memory?
    8·1 answer
  • _____ refers to displaying information for the user's view.
    14·2 answers
  • Which of the following best define grit
    13·1 answer
  • A coffee shop is considering accepting orders and payments through their phone app and have decided to use public key encryption
    10·1 answer
  • What output is generated by this for loop?
    6·1 answer
  • Question #2: How would you demonstrate professionalism in a video call with a teacher? Edmentum Digital world Please Help!!
    8·1 answer
  • Your company is building a new architecture to support its data-centric business focus. You are responsible for setting up the n
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!