Answer:
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.
Answer:
B and C
Step-by-step explanation:
Rearrange each equation into slope- intercept form
A
3x - 2y = 4 ( subtract 3x from both sides )
- 2y = - 3x + 4 ( divide all terms by - 2 )
y = x - 2 ← not equivalent
B
2x - 3y = 12 ( subtract 2x from both sides )
- 3y = - 2x + 12 ( divide all terms by - 2 )
y = x - 4 ← equivalent
C
- 4(2x - 3y ) = - 48 ( divide both sides by - 4 )
2x - 3y = 12 ← same as B ⇒ equivalent
D
2(x + 6) = 3y
3y = 2x + 12 ( divide all terms by 3 )
y = x + 4 ← not equivalent
E
2x - 3y = 4 ( subtract 2x from both sides )
- 3y = - 2x + 4 ( divide all terms by - 3 )
y = x -
← not equivalent