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R(A,B,C,D) F={A->C, D->B}
using Armstrong's axioms how do I prove that AD->B?
what I currently have is: using augmentation if D->B and A is any subset of U, then AD->AB
but I'm not confident on that answer
Here is a simple proof:
1. D -> B by hypothesis
2. AD -> AB by augmentation of 1
3. AB -> B by reflexivity
4. AD -> B by transitivity of 2 and 3
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Q: Write an assembly code that sorts a given 10-element array in ascending order.
For example:
from A: .word 9,8,7,6,5,4,3,2,1,0
to A=0,1,2,3,4,5,6,7,8,9
https://cpulator.01xz.net/?sys=mipsr5
MIPS32r5
this link I want all steps to answers my question please .
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How to make sparse array c language without condition i !=0 ??
I tried i!=0 condition but teacher doesn't want that
Teacher wants another path ??
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I am trying to retrieve the last 10 mentions of the text that's in cell B8.
For example, if B8 is "Tom Brady", I want to retrieve the last 10 results for Passing Yards (DB:M) that mention Tom Brady.
I got this formula but it displays the first 15 it seems.
=QUERY(DB!B:AZ,"select M where C = '"&B8&"' order by B desc limit 15")
https://docs.google.com/spreadsheets/d/12b9dxeSt4_F9hdgaqDzggNW-fbMyXesWaqT0JjqMVhk/edit?usp=sharing
Any help would be appreciated!
use:
=INDEX(SORT(SORTN(FILTER({D5:D, ROW(D5:D)}, D5:D<>""), 10, 0, 2, 0), 2, 1),,1)
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I want to display the only the highlighted values (They are 9 digits). Can we do this using RegExp?
You can use regexp_substr(column1, '[0-9]{9}'). Below is an example:
select regexp_substr(column1, '[0-9]{9}') nine_digits
from (values
(';**260488570**;1;13.25;20339=22.99')
,('1293=::info::0,;**297100755**;1;2.86;20339=4.49')
,('1293=::info::0,;**338010030**;3;6.71;20339=2.69')
,('1293=::info::0,;**260142941**;1;2.38;20339=4.59')
,('1293=::info::0,;**370039059**;1;2.86;20339=3.79')
);
Result:
COL2
260488570
297100755
338010030
260142941
370039059
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I understand the usage and purpose of Array#inject but why is it called inject? I don't understand what's being injected where.
I prefer to think of inject as "injecting" an operation among the items inside the given array and returning the final result of the calculation.
(1..5).inject(:+) #=> 15
In my example, it takes the number 1 to 5 and "injects" a sum operation among them, resulting in 1 + 2 + 3 + 4 + 5 = 15.
Also, it is aliased by reduce, as stated with details in https://ruby-doc.org/core-2.4.1/Enumerable.html#method-i-inject.