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Tutor profile: Ishan A.

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Ishan A.
Undergraduate Student at Netaji Subhas Institute of Technology | Former Software Development Intern at Droom Technology
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Questions

Subject: Linear Algebra

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Question:

Question. Find the eigenvalues of the following matrix. A = (2 7 −1 −6)

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Ishan A.
Answer:

The first thing that we need to do is find the eigenvalues. That means we need the following matrix, Characteristic polynomial: A −λI = (2 7 −1 −6) - λ (1 0 0 1) = (2−λ 7 −1 −6−λ) In particular we need to determine where the determinant of this matrix is zero. det (A − λI) = (2−λ)(−6−λ) +7 = λ^2 + 4λ − 5 = (λ+5) (λ−1) = 0 So, it looks like we will have two simple eigenvalues for this matrix, λ1 = -5 and λ2 = 1.

Subject: Databases

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Question. What do you understand by database normalization? What are the different types of normalization? Is the following relation STUDENT in 1NF? If not, then how will you convert it into 1NF? STUD_NO | STUD_NAME | STUD_PHONE 1. Mark 9643637280 2. John 8076523341, 9818587812 3. Harry 7654210065

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Ishan A.
Answer:

Answer. Normalization is a process of analyzing the given relational databases based on their functional dependencies and primary keys to achieve the following properties: 1) Minimizing Redundancy 2) Minimizing the Insertion, Deletion, And Update Anomalies Different types of Normalization are: 1. First Normal Form (1NF): A relation is said to be in 1NF only when all the entities of the table contain unique or atomic values. 2. Second Normal Form (2NF): A relation is said to be in 2NF only if it is in 1NF and all the non-key attribute of the table is fully dependent on the primary key. 3. Third Normal Form (3NF): A relation is said to be in 3NF only if it is in 2NF and every non-key attribute of the table is not transitively dependent on the primary key. 4. Boyce Code Normal form (BCNF): It is the higher version of 3NF which does not have any multiple overlapping candidate keys. Relation STUDENT in is not in 1NF because of multi-valued attribute STUD_PHONE for the person named John with STUD_NO 2. Since all the entities of the table do not contain atomic values, it is not in 1NF. Its decomposition into 1NF has been shown below: STUD_NO | STUD_NAME | STUD_PHONE 1. Mark 9643637280 2. John 8076523341 3. John 9818587812 4. Harry 7654210065

Subject: C++ Programming

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Question:

#include <iostream> using namespace std; int main() { int a = 32, *ptr = &a; char ch = 'A', &cho = ch; cho += a; *ptr += ch; cout << a << ", " << ch << endl; return 0; } Question. What is the output of the C++ code mentioned above?

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Ishan A.
Answer:

Answer: 129, a Explanation: The integer variable 'a' is assigned the value 32. The 'ptr' variable is a pointer which holds the address of variable 'a'. And '*ptr' returns the value of 'a'. 'cho' is a reference variable to 'ch'. So any change made to 'cho' will be reflected to 'ch'. As such, when 'cho' is increased by 32, it adds to the ASCII value of “A”(which is 65), and this results to 97 which is the ASCII value of 'a' (from the alphabet). So this 'a' gets stored in 'ch'. As for when '*ptr' is incremented by 'ch', it gives value 97+32=129.

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