Answer:
1 5 10 10 5 1
Step-by-step explanation:
The complete question has been attached as an image.
Looking at the triangle, we see a pattern. The first level we have 1, in the next level we have 1 1, the next evel we have 1 2 1, the next level we have 1 3 3 1 and so on. From here, we see that to get the numbers of the next level, we have to write 1 as the first number, then add 1 to the next number after it in the previous level to get the second number in the next level then add the second number of the previous level to the next number beside it to get the third number in the next level and so on until you get to the last number before 1 in the previous level, add that number to 1 to get the second to the last number in the next level and finally put 1 as the last number in the next level. Now, we have
1 4 6 4 1
1 5 10 10 5 1
And that is the required set of numbers.
Answer:
p = 2
Step-by-step explanation:
First, simplify:
3p = 6
Then, divide 3 on both sides:
p = 2
Hope this helps ^^
Do a proportion
60/100= x/300
60 x 300= 18,000
18,000 divided by 100= 180
Answer is 180
Or you could have done .60 x 300 and got 180.
Answer: 30 cards
Step-by-step explanation:
To find the greatest possible number of cards in each pile such that there is the same number of cards in each pile we need to find the greatest common factor of 750 and 660.
Using Euclid division method , we have
Hence, the greatest common factor of 750 and 660 = 30
Thus, there will be 30 cards in each pile.
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
Learn more about Boyce-Codd Normal Form at
brainly.com/question/14299832
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