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Tamiku [17]
3 years ago
12

The number of operations executed by algorithms A and B is 40n2 and 2n3, respectively. Determine n0 such that A is better than B

for n ≥ n0.
Computers and Technology
1 answer:
Tom [10]3 years ago
8 0

Answer:

Given that:

A= 40n^2

B = 2n^3

By given scenario:

40n^2=2n^3

dividing both sides by 2

20n^2=n^3

dividing both sides by n^2 we get

20 = n

Now putting n=20 in algorithms A and B:

A=40n^2

= 40 (20)^2

= 40 * (400)

A= 16000

B= 2n^3

= 2 (20)^3

= 2(8000)

B= 16000

Now as A and B got same on n = 20, then as given:

n0 <20 for n =20

Let us take n0 = 19, it will prove A is better than B.

We can also match the respective graphs of algorithms of A and B to see which one leads and which one lags, before they cross at n= 20.

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In a switch statement, if a break statement is missing, _______________. Select one: a. the default case is automatically execut
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Answer:

c. execution falls through the next branch until a break statement is reached

Explanation:

In a switch statement, if break is missing, then program continues to the next cases until it finds a break statement.

Consider the following example!

Example:

#include <iostream>

using namespace std;

int main()

{

  int choice = 1;

  switch (choice)

  {

      case 1:

      printf("This is case 1, No break\n");

      case 2:

      printf("This is case 2, No break\n");

      case 3:

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  }

  return 0;

}

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This is case 2, No break

This is case 3, with break

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