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oh got it !!! yes his test series is worst.
I mentioned RBR but this particular name is...
@Musa why are u not mentioning *** test series...
I took made easy,RBR ,Gate overflow test...
yes...right!! i also took ME and Applied before...
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Happy Birthday Sir Arjun
Ode of Gratitude Bringing learning under one roof Of theorems and mathematical proofs Born on this auspicious day To you Sir Ode of Gratitude we pay The contribution is beyond measure We gain insight into the vast treasure Birthdays are a special ... hopes revived True learning in form of GO arrived Once again wishing a Happiest Birthday from all Learners to Sir Arjun.
(C) is the answer. Because of AB $\to$ C and C $\to$ A, we cannot have A, B and C together in any BCNF relation- in relation ABC, C is not a super key and C$\to$ A exists violating BCNF condition. So, we cannot preserve AB $\to$ C dependency in any decomposition of ABCD. ... and B the respective keys For (B) we can have AB, BC, CD, A, B and C the respective keys For (D) we can have ABCD, A is key
Nov 6, 2018
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