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NISA [10]
3 years ago
9

What are three ways science and technology are related and what are three ways they are different?

Computers and Technology
1 answer:
photoshop1234 [79]3 years ago
3 0

Answer:

Science contributes to technology in at least six ways: (1) new knowledge which serves as a direct source of ideas for new technological possibilities; (2) source of tools and techniques for more efficient engineering design and a knowledge base for evaluation of feasibility of designs; (3) research instrumentation,

Explanation:

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Create a class Str that subclasses str. Add a method to the subclass that checks if the string does not start with a given strin
Stolb23 [73]

Answer:

#include <iostream>

using namespace std;

class Str{  ///baseclass

   public :

   string super_str;

   string getStr()

   {

       return super_str;

   }

   void setStr(string String)

   {

       super_str=String;

   }

};

class str : public Str{   //inheriting Str publicly

   public :

       string sub_str;

       string getstr()

       {

           return sub_str;

       }

       void setstr(string String)

       {

           sub_str=String;

       }

       bool notstartswith()

       {

           int n=sub_str.length();   //to find length of substr

           bool flag=false;

           for(int i=0;i<n;i++)     //Loop to check beginning of Str

           {

               if(super_str[i]!=sub_str[i])

               {

                   flag=true;

                   break;

               }

           }

           return flag;

       }

};

int main()

{

   str s;    //object of subclass

   s.setStr("Helloworld");  

   s.setstr("Hey");

   if(s.notstartswith()==1)     //checking if str is substring of Str

       cout<<"Str does not start with str";

   else

       cout<<"Str starts with str";

   return 0;

}

OUTPUT :

Str does not start with str

Explanation:

Above program is implemented the way as mentioned. for loop is being used to check the beginning of the str starts with substring or not.

3 0
4 years ago
Satırlar ve sütunların kesişmesiyle oluşan kutular aşağıdakilerden hangisidir
Kryger [21]

Answer:

Huh English please......

7 0
3 years ago
A spiked collar that extends horizontally for up to 3 feet from the pole is an example of what kind of technology? and that no u
slega [8]

This technology is known as an anti-climb collar.

Spiked collars are used to protect equipment mounted on poles like CCTV or in areas where you can easily climb a pole over a security fence. This technology offers a very simple and practical solution to protecting poles against thieves and vandals.

3 0
3 years ago
2. Write a program to calculate CGPA of a student. The program should be able to get input from the user of the grade for each c
Irina18 [472]
The easy way to do it is to let an array accept all the inputs and add all the inputs together and divide the sum by how many the input was. 

Example : five + five + five + five + three + two = 25
Now divide twenty five : twenty five / six = 4.1

Letter A is the closest answer.
7 0
3 years ago
Recall that within the ArrayBoundedQueue the front variable and the rear variable hold the indices of the elements array where t
Citrus2011 [14]

Answer:

int n = elements.length;

if(rear < n) {

rear = (rear + 1) % n;

elements[rear] = element;

rear = rear + 1;

}

Explanation:

Options are not made available; However, I'll answer the question base on the following assumptions.

Assumptions

Array name = elements

The front and the rear variable hold the current indices elements

Element to enqueue is "element" without the quotes

To enqueue means to add an element to an already existing queue or to a new queue.

But first, the queue needs to be checked if it can accept a new element or not; in other words, if it's full or not

To do this, we check: if rear < the length of the queue

If this statement is true then it means the array can accept a new element; the new element will be stored at elements[rear] and the rear will be icremented by 1 rear by 1

Else if the statement is false, then an overflow is said to have occurred and the element will not be added.

Going by the above explanation, we have

int n = elements.length;

if(rear < n) {

rear = (rear + 1) % n;

elements[rear] = element;

rear = rear + 1;

}

Additional explanation:

The first line calculates the length of the queue

The if condition on line 2 tests if the array can still accept an element or not

Let's assume this statement is true, then we move to liine 3

Line 3 determines the new position of rear.

Let's assume that n = 6 and current rear is 4.

Line 3 will produce the following result;

rear = (rear + 1) % n;

rear = (4 + 1)% 6

rear = 5%6

rear = 5.

So, the new element will be added at the 5th index

Then line 4 will be equivalent to:

elements[rear] = element;

elements[5] = element;

Meaning that the new element will be enqueued at the 5th index.

Lastly, rear is incremented by 1 to denote the new position where new element can be added.

6 0
4 years ago
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