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notsponge [240]
3 years ago
7

Propose, two new, proudly South African ways, which you can visualize that the internet of things, can be used in at work to mak

e life better.
Computers and Technology
1 answer:
Eduardwww [97]3 years ago
8 0
1. Internet should be used as a medium of learning.

Through internet, such African people could obtain a lot of knowledge from various fields that provided by more developed countries, and use that knowledge to help the development in south africa.

2. It should be used for utilizing business.

There are a lot of new potential business that could be done because of internet, such as blogging, affiliate marketing, public endorsement through social media followers, etc.
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Create a class called Fraction. Provide a constructor that takes 2 integers. Provide methods for:
tekilochka [14]

Answer:

Explanation;

else

System.out.println("f1 and f2 are not equal");

switch (input.charAt(0)

{

case '+':

f3 = f1.add(f2);

System.out.println("f1+f2=" + f3);

break;

case '-':

f3 = f1.subtract(f2);

System.out.println("f1-f2=" + f3);

break;

case '*':

f3 = f1.multiply(f2);

System.out.println("f1*f2="+f3);

break;

case '/':

f3 = f1.divide(f2);

System.out.println("f1/f2="+f3);

break;

default:

System.out.println("Illegal command: " + input );

break;

}

}// end of while loop

} // end of main

}

 

Note ; this is the last part of the programme check the attachment from 1-5  this is the 6th .

8 0
3 years ago
What tool is used to find and organize files on a mac
Margaret [11]

Answer:

the tool is literally called <u><em>The Finder.</em></u>

Explanation:

hope this helps

4 0
3 years ago
A7DF is the hexadecimal representation for what bit pattern?
kipiarov [429]

Answer:

1010 0111 1101 1111

Explanation:

A = 10 in decimal = 1010 in binary

7 = 7 in decimal = 0111 in binary

D = 13 in decimal = 1101 in binary

F = 15 in decimal = 1111 in binary

Therefore 0xA7DF = 1010 0111 1101 1111 in binary

6 0
3 years ago
United Broke Artists (UBA) is a broker for not-so-famous artists. UBA maintains a small database to track painters, paintings, a
lions [1.4K]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the relation and the table names.

7 0
3 years ago
Read 2 more answers
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
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