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storchak [24]
3 years ago
10

JavaScript is the same thing as Java?

Computers and Technology
2 answers:
Mila [183]3 years ago
5 0

Not really, because Java is an OOP programming language and JavaScript is an OOP scripting language.

Hope I helped!

~Mshcmindy

stepan [7]3 years ago
4 0

<em>Key differences between Java and JavaScript: Java is an OOP programming language while Java Script is an OOP scripting language. Java creates applications that run in a virtual machine or browser while JavaScript code is run on a browser only. Java code needs to be compiled while JavaScript code are all in text.</em>

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If you see an advertisement for a 3TB portable drive, this is most likely a reference to the device having a capacity of three
Anika [276]
Hello there.

Question: <span>If you see an advertisement for a 3TB portable drive, this is most likely a reference to the device having a capacity of three _____.

Answer: It is 3 terabytes. .

Hope This Helps You!
Good Luck Studying ^-^</span>
4 0
3 years ago
An array is sorted (in ascending order) if each element of the array is less than or equal to the next element .
Aleks04 [339]

Answer:

// The program below checks if an array is sorted or not

// Program is written in C++ Programming Language.

// Comments are used for explanatory purpose

//Program Starts here

#include<iostream>

using namespace std;

//Function to check if the array is sorted starts here

int isSorted(int arr[], int count)

{

// Check if arrays has one or no elements

if (count == 1 || count == 0) {

return 1;

}

else

{

// Check first two elements

if(arr[0] >= arr[1])

{

return 0; // Not sorted

}

else

{ // Check other elements

int check = 0;

for(int I = 1; I<count; I++){

if (arr[I-1] > arr[I]) { // Equal number are allowed

check++; // Not sorted

}

}

}

if(check==0)

return isSorted(arr, count - 1); //Sorted

else

return 0; // Not sorted

}

// Main Method starts here

int main()

{

int count;

cin<<count;

int arr[count];

for(int I = 1; I<=count; I++)

{

cin>>arr[I-1];

}

int n = sizeof(arr) / sizeof(arr[0]);

if (isSorted(arr, n))

cout << "Array is sorted";

else

cout << "Array is not sorted";

}

8 0
4 years ago
Which of the following protocols support the encryption and decryption of e-mail messages?
Katena32 [7]
It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway, here is the answer. The protocol that supports the encryption and decryption of e-mail messages is this: <span>Secure Multipurpose Internet Mail Extensions (S/MIME) and Pretty Good Privacy (PGP). Hope this helps.</span>
7 0
3 years ago
A type of specialty processor devoted exclusively to protecting your privacy.
drek231 [11]
 Cryptoprocessor would be it.
8 0
3 years ago
Read 2 more answers
Write a static method called bothStart that allows the user to input two Strings and returns the String that is the longest subs
marishachu [46]

Answer:

  1.    public static String bothStart(String text1, String text2){
  2.        String s = "";
  3.        if(text1.length() > text2.length()) {
  4.            for (int i = 0; i < text2.length(); i++) {
  5.                if (text1.charAt(i) == text2.charAt(i)) {
  6.                    s += text1.charAt(i);
  7.                }else{
  8.                    break;
  9.                }
  10.            }
  11.            return s;
  12.        }else{
  13.            for (int i = 0; i < text1.length(); i++) {
  14.                if (text1.charAt(i) == text2.charAt(i)) {
  15.                    s += text1.charAt(i);
  16.                }else{
  17.                    break;
  18.                }
  19.            }
  20.            return s;
  21.        }
  22.    }

Explanation:

Let's start with creating a static method <em>bothStart()</em> with two String type parameters, <em>text1 </em>&<em> text2</em> (Line 1).  

<em />

Create a String type variable, <em>s,</em> which will hold the value of the longest substring that both inputs start with the same character (Line 2).

There are two possible situation here: either <em>text1 </em>longer than<em> text2 </em>or vice versa. Hence, we need to create if-else statements to handle these two position conditions (Line 4 & Line 13).

If the length of<em> text1</em> is longer than <em>text2</em>, the for-loop should only traverse both of strings up to the length of the <em>text2 </em>(Line 5). Within the for-loop, we can use<em> charAt()</em> method to extract individual character from the<em> text1</em> & <em>text2 </em>and compare with each other (Line 15). If they are matched, the character should be joined with the string s (Line 16). If not, break the loop.

The program logic from (Line 14 - 20) is similar to the code segment above (Line 4 -12) except for-loop traverse up to the length of <em>text1 .</em>

<em />

At the end, return the s as output (Line 21).

5 0
3 years ago
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