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natka813 [3]
4 years ago
15

Building relationships during your career exploration is called

Computers and Technology
1 answer:
miskamm [114]4 years ago
7 0
Building relationships during your career exploration is called Networking.  Networking doesn't necessarily guarantee you a job, but when you're in career exploration phase, building relationships with various people may expose you to various kind of opportunities that could benefit your career paths
You might be interested in
Subscribe to (Thicc sickqueen) and like 2 vids and the claim your prize
lara [203]

Answer:

done

Explanation:

5 0
4 years ago
Create a short document titled "Guide to web searches" that Sergio can give to users of the facility he supervises. The document
Studentka2010 [4]

Answer:

Jobs Seeker's

Web Search Guide

Sergio Escobar

PERFORMING A SEARCH:

All search engines will have a search field in which you are able to enter your search terms,

criteria, keywords, or even a website.

Search terms are the keywords a search engine will use to try to find the most relevant search

results.

If you’re looking for general information, just enter the name of the topic you’re

interested in into the search field.

So, if you’re looking for work, you can use the search terms ‘employment’ and ‘jobs’.

The search engine will return you a list of web pages that include the words

‘employment’ and ‘jobs’.

IF you’re looking for more specific information, you can narrow your search results by

using more specific search terms.

So, if you wanted employment opportunities in Phoenix, AZ, you could use the search terms:

‘local’, ‘employment’, ‘jobs’ and ‘Phoenix’.

Once you’ve entered your search terms and selected the search or go button, the search

engine will provide you with a list oF search results.

Depending on your search, you could get thousands or even millions of search results. This is

because search engines will usually list every single web page on the Internet that used those

search terms somewhere on a web page.

(However, the most current and/ or relevant results should be at the top of the list.)

Searching Strategy:

Using Boolean Operators:

use AND,OR,NOT between the words

ex: LOCAL AND JOBS

JOBS NOT IN NEWYORK

Proximity Operators:

With some search engines you can use proximity operators such as OpenText's NEAR operators or Webcrawler's ADJecent or the FOLLOWED BY operator.

With each of these operators, word order is important.

For example: if you place square brackets such as [local jobs] causes a hit if they are found within 100 words of each other.

Truncation (*)

You can use truncation on most search engines.

That is, you can use the asterisk (*) operator to end a root word.

For example: searching for comput* will find computer, computing, and computer jobs.

Note: the asterisk can not be the first or second letter of a root word.

Wildcard (?)

You can find words that share some but not all characters using the question mark (?) operator.

For example: Com?er J?b will find Computer Jobs and Computer Job.

Note: the ? can not be the first character in the search.

You may also use combinations of truncation (*) and single character wildcard (?) in your searches.

Conclusion:

These are just a few of the strategies and new tools trainers can use to make working on the web more productive.

As trainers continue to use the web they will soon see the next generation of web "knowledge tools" begin to emerge.

These will include multidimensional tools that are created to manage data on the web using factors such as "virtual neighborhoods of information,

" "organic structuring," and "mental model based searching and flying mechanisms" .

These are tools which are intended to make the world wide web more manageable for the user.

Let us now go back to my original statement...that the goals of search strategies and engines should be to increase your efficiency

and effectiveness when looking for information on the web.

Only you can decide which search/knowledge management strategies and tools actually improve your productivity.

It is my hope that this article helps you with making these decisions.

8 0
4 years ago
If you were looking for a record in a very large database and you knew the ID number, which of the following commands would be t
OLga [1]

Answer:

The Correct answer is Find.

Explanation:

7 0
3 years ago
Consider the following code segment, which is intended to set the Boolean variable inRange to true if the integer value num is g
Natasha_Volkova [10]

Answer:

Option d  num = 50, min = 50, max = 50

Explanation:

Given the code segment:

  1. boolean isBigger;
  2. boolean isSmaller;
  3. boolean inRange;
  4. if (num < max)
  5.        {
  6.            isSmaller = true;
  7.        }
  8.        else {
  9.            isSmaller = false;
  10.        }
  11.        if (num > min)
  12.        {
  13.            isBigger = true;
  14.        }
  15.        else {
  16.            isBigger = false;
  17.        }
  18.        if (isBigger == isSmaller) {
  19.            inRange = true;
  20.        } else {
  21.            inRange = false;
  22.        }

If we have num = 50, min = 50 , max = 50,  the condition num < max will be evaluated to false and therefore isSmaller is set to false.

The condition num > min will be evaluated to false as well and therefore isBigger is set to false.

Since isSmaller and isBigger are both false and therefore isBigger == isSmaller will be evaluated to true and set the inRange = true. This has violated the statement that if the integer value num is greater than min value and less than max value, then only set inRange to true. This is because num = 50 is neither greater than min nor less than max, it is supposedly not in range according to the original intention of the code design.

8 0
3 years ago
Write a loop that sets new scores to old scores shifted once left, with element 0 copied to the end. ex: if old scores = {10, 20
Inga [223]

#include <iostream>
using namespace std;

int main() {
   const int SCORES_SIZE = 4;
   int oldScores[SCORES_SIZE];
   int newScores[SCORES_SIZE];
   int i = 0;

   oldScores[0] = 10;
   oldScores[1] = 20;
   oldScores[2] = 30;
   oldScores[3] = 40;

   /* Your solution goes here */

   for (i = 0; i < SCORES_SIZE; ++i) {
      cout << newScores[i] <<" ";
   }
   cout << endl;

   return 0;
}

3 0
3 years ago
Read 2 more answers
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