Answer:
im not 100% sure but to me it looks like it could be 225
There are 59 integer solutions
Such questions are best solved by writing cases and calculating the total number of cases. So beginning with
1) x = -3. The possible combinations are as follows:-
-3 2 13
-3 3 12
-3 4 11
-3 5 10
-3 6 9
-3 7 8
-3 8 7
-3 9 6
-3 10 5
-3 11 4
10 combinations
2) x = -2
-2 2 12
through
-2 11 3
10 combinations
3) x = -1
-1 2 11
through
-1 10 3
9 combinations
4) x = 0
0 2 10
through
0 9 3
8 combinations
as we can see from the pattern at x =1 we get 7 combinations, at x =2 we get 6 combinations, at x=3 we get 5 combinations and at x =4 we get 5 combinations.
Thus total number of combinations 4+5+6+7+8+9+10+10 = 59 integer solution.
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Given that R(ABCDE) is in Boyce-Codd normal form.
And AB is the only key for R.
Definition
A relational nontrivial Schema R is in BCNF if FD (X-A) holds in R, Super key of R. whenever then X is
a
Given that AB is the only key for R.
ABC E (Yes).
check if ABC is a Super key. AB is a key, ABC is A B C E is in BONE a super key.
2) ACE B
(NO). no Check if ACE As there is ACE is not a Super key? AB in Super key. ACE.
ACE B
is
Boyce-Codd Normal Form not in BENE (NO)
3) ACDE → B (NO)
check if is a super key. ACDE
As ACDE there is not any AB Tn ACDE. a super key.
ACDEB is not in BCNF.
4) BS → C → (NO)
As there is no AB in BC ~. B(→ not in BCNF
BC is not a super key.
5) ABDE (Yes).
Since AB is a key.
ABO TS a super key.
.. ABDE → E is in BCNF
Let R(ABCDE) be a relation in Boyce-Codd Normal Form (BCNF). If AB is the only key for R, identify each of these FDs from the following list. Answer Yes or No and explain your answer to receive points.
1. ABC E
2. ACE B
3. ACDE B
4. BC C
5. ABD E
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