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Angelina_Jolie [31]
3 years ago
6

In the dental industry, light-activated adhesives have been used for decades to help adhere braces to teeth as well as being use

d in other dental procedures. These adhesives are one-part translucent polymers that cure and harden when exposed to specific light spectrums. Many of these took up to 30 seconds to cure. New technology brings LED blue light curing which uses blue LED with UV light to cure and harden a light-activated adhesive in 3-5 seconds. Which result below would NOT be an advantage of this new technology? A) There is a higher up-front expense. B) Could be used in other areas such as sealing cracked pipes. C) No filters or cooling fan required due to quick curing and low heat emission. D) There is a major time savings when using the faster high–intensity curing lights.
Computers and Technology
2 answers:
iogann1982 [59]3 years ago
8 0

Answer:

a

Explanation:

There is a higher up-front expense since they have so many advantages over old designs and techniques.

Bas_tet [7]3 years ago
7 0

In the dental industry, light-activated adhesives have been used for decades to help adhere braces to teeth as well as being used in other dental procedures. These adhesives are one-part translucent polymers that cure and harden when exposed to specific light spectrum. Many of these took up to 30 seconds to cure. New technology brings LED blue light curing which uses blue LED with UV light to cure and harden a light-activated adhesive in 3-5 seconds. Which result below would NOT be an advantage of this new technology?<u> A)There is a higher up-front expense.</u> B)Could be used in other areas such as sealing cracked pipes. C)No filters or cooling fan required due to quick curing and low heat emission. D)There is a major time savings when using the faster high–intensity curing lights.

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Svet_ta [14]
Hi,

I changed your program using some of the concepts you were trying to use. Hopefully you can see how it works:

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using namespace std;

int main()
{
   short T;
   cin >> T;
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   string str[100];

   for(int i=0; i<T; i++)
   {
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   for (int i = 0; i < T; i++)
   {
      stringstream ss(str[i]);
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      while (ss >> tmp)
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          if (!tmp.empty())
          {
              transform(begin(tmp), end(tmp), std::begin(tmp), ::tolower);
              tmp[0] = toupper(tmp[0]);
           }
           v.push_back(tmp);
        }

        if (v.size() == 1)
        {
           cout << v[0] << endl;
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        else if (v.size() == 2)
        {
           cout << v[0][0] << ". "  << v[1] << endl;
        }
        else
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            cout << v[0][0] << ". " << v[1][0] << ". " << v[2] << endl;
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3 years ago
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Rasek [7]
This is true. I hope I was helpful!
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7 0
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Answer:

Ans1.

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