Consider the universal relation r = {a, b, c, d, e, f, g, h, i, j} and the set of functional dependencies f = { {a, b } → {c}, {a}→{d, e}, {b}→{f}, {f}→{g, h}, {d}→{i, j} }. what is the key for r? decompose r into 2nf and then 3nf relations

relation of FDs are {a, b, d, h}. From these all but {h}

seem plausibly part of keys. Calculating the closures of likely candidates gives:

{a, b} + = {a b, c, i}

{b, d} + = {b, d, e, f}

{a, d} + = {a, d, g, h, i, j}

{a, b, d} + = {a, b, c, d, e, f, g, h, i, j}  

decompose r into 2nf and then 3nf relations shown in attachment.                    

Step- by-Step Explanation

Explanation is given in the attachment.

360 different combinations

the correct answer option is (2x-3)(2x+3).

step-by-step explanation:

we are given the following expression and we are to factorize it:

we are to find such factors that when multiplied they give a result of 36 (9 x 4 = 36) and when added they give a result of 0.

the two such factors of 36 are negative six and postive six.

factorizing the given expression to get:

therefore, the correct factorization of is (2x-3)(2x+3).