Assuming you are playing 7 card stud, 2h 2c 2d 6d 10d Qc Ah will win the pot with 3 of a kind with the Queen and Ace as kickers.
The other hands, (Two pair (5's and K's) are second) and the last hand (High card ace) is the worse hand.
None of these hands qualify for a "Lo" hand if you are playing Hi/Lo 8 or better. So the pot is only won by the first hand, 3 of a kind.
10h
If you mean: 10h+6-5h+3 then it is 5h+9 when simplified
10h
h/(-3 - 7) = 10h/-10 = 10h = -100
The simplified expression of 10h+6-5h+3 is 5h+9
8h - 10h = 3h + 25-2h = 3h + 25-5h = 25-h = 5h = -5
today, July 10h
Two pair will win, in Texas Holdem' hands ranking two pair is higher than three of a kind. The reverse is for all other types of poker. This is not even close to correct, whoever typed this is not smart or simply does not know poker rules. The following is the official poker hand rankings for Texas holdem poker and for that matter, all poker! FROM BEST TO WORST ROYAL FLUSH (A,K,Q,J,10 - ALL THE SAME SUIT....NOTE THAT NO SUIT OUTRANKS ANY OTHER, IN TEXAS HOLDEM THERE CAN ONLY BE ONE ROYAL FLUSH PER HAND) STRAIGHT FLUSH (ANY 5 CARDS IN A ROW OF THE SAME SUIT, EG: 2,3,4,5,6 ALL SUITED OR 5,6,7,8,9 ALL SUITED...NOTE THAT A ROYAL FLUSH IS THE HIGHEST RANKING STRAIGHT FLUSH) FOUR OF A KIND (ENOUGH SAID A,A,A,A .....10,10,10,10 ETC......NOTE THAT IF THERE IS A FOUR OF A KIND ON THE BOARD IN TEXAS HOLDEM THEN THE PERSON WITH THE HIGHEST KICKER WILL WIN) FULL HOUSE (BASICALLY A PAIR AND THREE OF A KIND AT THE SAME TIME...EG: K,K,K,Q,Q) FLUSH (5 CARDS OF THE SAME SUIT THAT DONT FALL INTO THE STRAIGHT FLUSH CATEGORY...EG: KH,QH,10H,5H,2H) STRAIGHT (5 UNSUITED CARDS IN A ROW ...EG: KH,QH,JS,10C,9D) THREE OF A KIND (TRIPS) K,K,K OR A,A,A OR 3,3,3 ETC TWO PAIR AA,KK OR 10,10,2,2 ETC ONE PAIR (A,A) HIGH CARD (ACE HIGH ETC)
You do not need the distributive property for to do that!
10h 14m
Solve this please in a few minutes
10h 00m 00s, −20° 00′ 00″