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Answer:
#include <iostream>
using namespace std;
int main()
{
int i = 45;
for ( i = 45; i <=165; i = i + 6)
cout << i << endl;
}
Explanation:
I corrected your code and highlighted the mistakes. Even though you wrote the correct algorithm, your code did not compile because of the typos you made.
Remember, C++ is a case-sensitive language. That means, "For" is not same as "for".
Generally, variables and keywords are written in lower case. Of course, there are exceptions, such as constant variables are all written in uppercase letter and class names start with an uppercase letter.
Solution:
It is important because, it has to do operation on so without, data structures and algorithms.
It performs these operation:
1)Take an input
2) Process it
3) Give back the output.
The input can be in any form, for ex while searching for directions on google maps, the starting point and the destination as input to google maps, while logging in to any social sites, We have to give our email and password as input and so on.
Similarly, in the third step, the computer application gives us output in some form or the other.
To make this process efficient, we need to optimize all the three steps.
Answer:
Metrical task system algorithm is the online algorithm that is used for organizing the online problems like k-server issues, paging issues etc.This task system works in the form of metrics to decrease the complete cost experienced due to processing of the operation and analyzing the competition.
It can be used for the online shopping project for the analyzing the comparison between the performance on basis of online and offline trends and then optimization can take place according to the the results.
Answer:
z = a.c' + a.b.d' + b.c'.d'
Explanation:
The truth table for this question is provided in the attachment to this question.
N.B - a' = not a!
The rows with output of 1 come from the following relations: 01 > 00, 10 > 00, 10 > 01, 11 > 00, 11 > 01, 11 > 10
This means that the Boolean expression is a sum of all the rows with output of 1.
z = a'bc'd' + ab'c'd' + ab'c'd + abc'd' + abc'd + abcd'
On simplification,
z = bc'd' + ab'c' + ac'd' + ac'd + abc' + abd'
z = ac' + abd' + bc'd'
Hope this helps!
