Which save as element allows a user to save a presentation on a computer?

Answer:

Alaric Saltzman.

The Originals.

Car crash, ran off Wickery bridge.

Explanation:

Answer:

okkk I got this TVD since forever

Explanation:

Alaric Saltzman

They are called the original vampires (Mikael, Finn, Elijah, Klaus, Kol, and Rebekah)

Elena parents died on the way back home from picking up Elena from a party. They end up driving off the Wickery Bridge. Stefan saved Elena though he was going to save Elena dad first however her dad told Stefan to get Elena.

Answer:

wide area network (WAN)

Explanation:

Answer: B. Local area

A local area network is good for connecting computer clusters

Answer:

The answer is "Choice c".

Explanation:

Please find the complete question in the attached file.

To begin with, allow its principle of the numerically solving to be recognized, three stages are concerned.

1. Topic n=1

2. Suppose n to be true

3. Display n+1 it retains

We have LHS as 2+3+ for the third step now [tex](n+1) = (2+3+.. \& n) + n+1[/tex]

We can now replace the bracket of RHS by [tex]\frac{(n+2)(n-1)}{2},[/tex]as we assumed its valid for n in step 2

if we do that we get

[tex]= \frac{(n+2)(n-1)}{2+(n+1)}\\\\= \frac{(n^2-n+2n-2+2n+2)}{2}\\\\= \frac{(n^2+3n)}{2}\\\\= \frac{n(n+3)}{2}[/tex]

Answer:

Answer choice 4

Explanation:

If Josh sends an e-mail to his... co-worker.... wouldn't that be... workplace communication?

Answer: Find the app you want to close, then go to the moveable panel and click the X.

Answer:

it depend on the software the computer has along with how much data it can hold

Explanation:

Answer:

Depends on the software code the computer has

Explanation:

It is a web site that searches anything

Answer:

It is a web site that searches anything, the guy above me was Correct EDGE 2022!

Explanation:

Answer:

see picture

Explanation:

There are several syntax errors in the program that need to be fixed:

So the actual answer would be: the compiler will report syntax errors.

Answer:

the no is20

Hello#######

Bye:

Reason: Programming, hoped this is right!

Answer:

Following are the solution to this question:

Explanation:

Its complexity of both the pseudo-code described is indeed proportional to the number of digits. So, how often number there are in this specific number is the query. Whenever a number is considered, the d digit would be between [tex]10^{(d-1)}.[/tex] inclusive exclusive [tex]10^d[/tex] That would be as, let d become the number of digits at N, and the inequalities, They can tell

[tex]10^{(d-1)} \leq N < 10^d[/tex]

We get, we take a logarithm,

[tex]d-1 \leq \log(N) < d[/tex]

The increase of 1 to the left inequality, [tex]d \leq \log(N)+1[/tex], and Combining the previous outcome, we got, [tex]\log(N) < d \leq \log(N) + 1[/tex]. That's would be to say, that number of number d by [tex]O(\log(N))[/tex] is higher and lower. Consequently, the number of transactions in the code is [tex]O(\log(N))[/tex]

We have that the number of operations that are executed in the code in terms of n  is mathematically given as

The quantity of operations done in the code is O(log(N))

Generally the equation for the   is mathematically given as

If we think about a number,

with d digits is between 10^(d-1) inclusive and 10^d exclusive.

Let d be the wide variety of digits in N

10^(d-1) <= N < 10^d

d-1 <= log(N) < d

d <= log(N) + 1,

The Inequality

log(N) < d <= log(N) + 1.

Therefore

The quantity of operations done in the code is O(log(N))

For more information on Inequality visit

https://brainly.com/question/19491153

Answer:

As long as it works like a game and not a book i would say its a game.

Explanation:

Answer:

to make a name for themselves

Explanation:

Answer:

What are the 2 options?

Explanation:

Answer:

Do you have the options?

Explanation:

Answer:

#include <iostream>//header file

#include <string>//header file

using namespace std;

string integer_to_reverse_binary(int integer_value)//defining a method integer_to_reverse_binary  

{

   string ret = "";//defining string variable

   while (integer_value > 0) //using while loop to check integer_value value is greater than 0

   {

       ret += '0' + (integer_value % 2);//adding zeros in remainder value

       integer_value /= 2;//holding quotient value

   }

   return ret;//return string value

}

string reverse_string(string input_string)//defining a method reverse_string that holds a parameter user_String  

{

   string result;//defining a string variable  

   for (int i = 0; i < input_string.length(); ++i)//use for loop to calculate value  

   {

       result += input_string[input_string.length()-i-1];//add value in result variable

   }

   return result;//result result variable value

}

int main()//defining main method  

{

   int num;//defining integer variable

   string str;//defining string variable

   cin >> num;//input num value

   str = integer_to_reverse_binary(num);//use str variable to call the integer_to_reverse_binary method

   cout << reverse_string(str) << endl;//printing the reverse_string method value

   return 0;

}

Output:

6

110

Explanation:

In this code two string method "integer_to_reverse_binary and reverse_string" is defined that holds one parameter "integer_value and input_string".

In the first method a string variable is defined, that use the while loop to check integer value is greater than 0 and add zeros in the value and return its value as a string.

In the second it reverse the string value and store into the result variable, and in the main method the "num and str" variable is defined, and in the num it takes integer value and pass into the above method and print its return value.    

Explanation:

friction drag that is your answer 12 letters

Answer:

drag

Explanation:

im not the brightest but in the sentients it should have "drag" in it but u can listen to others answers to make sure

Answer:

Defined Expression

Explanation:

This will be your answer

Answer:

The stator..?

Explanation:

Stators are the stationary parts of motors. Dunno if I'm right lol

Answer:

the answer is in the image below:

Explanation:

Answer:

A

Explanation:

Answer:

b. 0001100

Explanation:

1100

Answer:

a. True

Explanation:

The above listed information are part of the reasons why so many IT projects are abandoned by the business entities after a given period of time frame.

Answer:

first one is "int" second one is "string" and third one should be "float"

Explanation:

not sure if the first one is right but try this.

Answer:

Removes characters behind (or to the left of) the insertion point.

Explanation:

Answer:

its is 810000

Explanation:

hope this helps

Answer:

In Java:

import java.util.*;

public class Main{

public static boolean isArrayEven(int[] arrayValues, int arraySize){

   boolean val = true;

   int odd = 0;

   for(int i =0;i<arraySize;i++){

       if(arrayValues[i]%2!=0){

           odd++;        }}

   if(odd>0){

       val = false;}

   return val;

}

public static boolean isArrayOdd(int[] arrayValues, int arraySize){

   boolean val = true;

   int even = 0;

   for(int i =0;i<arraySize;i++){

       if(arrayValues[i]%2==0){

           even++;}}

   if(even>0){

       val = false;}

   return val;

}

public static void main(String[] args) {

 Scanner input = new Scanner(System.in);

 System.out.print("Array Length: ");

 int n = input.nextInt();

 int[] array = new int[n];

 System.out.print("Array Elements: ");

 for(int i = 0;i<n;i++){

     array[i] = input.nextInt();  }

boolean even = isArrayEven(array,n);

 boolean odd = isArrayOdd(array,n);  

 if(even == true && odd == false){

     System.out.print("all even");  }

 else if(even == false && odd == true){

     System.out.print("all odd");  }

 else{

     System.out.print("not even or odd");  }

}

}

Explanation:

This declares the isArrayEven method

public static boolean isArrayEven(int[] arrayValues, int arraySize){

This initializes the return value to true

   boolean val = true;

This initializes the frequency of odd numbers to 0

   int odd = 0;

This iterates through the array

   for(int i =0;i<arraySize;i++){

This checks if the current array element is odd

       if(arrayValues[i]%2!=0){

If yes, the frequency of odd numbers is incremented by 1

           odd++;        }}

If the frequency of odd is greater than 1, then the array is not an even array

   if(odd>0){

       val = false;}

This returns true or false

   return val;

}

This declares the isArrayOdd method

public static boolean isArrayOdd(int[] arrayValues, int arraySize){

This initializes the return value to true

   boolean val = true;

This initializes the frequency of even numbers to 0

   int even = 0;

This iterates through the array

   for(int i =0;i<arraySize;i++){

This checks if the current array element is even

       if(arrayValues[i]%2==0){

If yes, the frequency of even numbers is incremented by 1

           even++;}}

If the frequency of even is greater than 1, then the array is not an odd array

   if(even>0){

       val = false;}

This returns true or false

   return val;

}

The main method begins here

public static void main(String[] args) {

 Scanner input = new Scanner(System.in);

This prompts for length of array:  System.out.print("Array Length: ");

This gets input for length of array  int n = input.nextInt();

 int[] array = new int[n];

This prompts for array elements: System.out.print("Array Elements: ");

This gets input for the array

 for(int i = 0;i<n;i++){

     array[i] = input.nextInt();

 }

This calls the isArrayEven method and the result is stored in even

boolean even = isArrayEven(array,n);

This calls the isArrayOdd method and the result is stored in odd

 boolean odd = isArrayOdd(array,n);  

The following prints the result of the methods

 if(even == true && odd == false){

     System.out.print("all even");  }

 else if(even == false && odd == true){

     System.out.print("all odd");  }

 else{

     System.out.print("not even or odd");  }