by binary132 December 15, 2004
by DylanPaterson December 26, 2019
A firearm chambered in 9mm. Properly used, it refers to the 9x19mm cartridge, aslo known as 9mm Parabelum or 9mm Luger. Though several other calibers are 9mm in diameter, they should not correctly be refered to as "nines".
9x17 AKA 9K AKA 9mm Kurtz AKA 9mm short is properly refered to in ghetto slang as a .380, referring to .380ACP the proper American nomenclature.
9x18 AKA 9mm Makarov is not commonly found in North America, and is therefore unavailable to most criminals.
9x21, is a more powerful variation of the 9x19 cartridge, and the two are not compatible in autopistols.
9x23, is a powerful cartridge, but is seldom found in cheap pistols, and therefore out of reach for most criminals.
9x17 AKA 9K AKA 9mm Kurtz AKA 9mm short is properly refered to in ghetto slang as a .380, referring to .380ACP the proper American nomenclature.
9x18 AKA 9mm Makarov is not commonly found in North America, and is therefore unavailable to most criminals.
9x21, is a more powerful variation of the 9x19 cartridge, and the two are not compatible in autopistols.
9x23, is a powerful cartridge, but is seldom found in cheap pistols, and therefore out of reach for most criminals.
by .357ROB October 1, 2003
A "smiley" similar to :P, but meant to indicate something appetizing or drool-inducing. The 9 forms both the smile and the tongue of the face. Often used in chatrooms.
by pwrnapper March 16, 2009
by Answer 101 May 29, 2019
A number equal to 1 to anyone who understands math.
Topic should not even be a debate, however, it is understandibly quite hard to accept by some.
Topic should not even be a debate, however, it is understandibly quite hard to accept by some.
1.) All sequential repeating decimals can be expressed as the repeating series divided by a 9 per each digit.
For example, 3/7 = .428571~
Thus, 428571/999999 = .428571~
Therefore, 428571/999999 = 3/7
When we apply this rule to .9~:
9 is the repeating sequence. We put it over an identical number of 9's, which is one 9, thus giving us 9/9, which is 1.
2.) All rational numbers can be expressed as a/b with a and b as integers. .9~ is a rational number. Thusly, it can be expressed as a/b. 9.~'s a/b expression is 1/1. To all those who disagree, I challenge you to find another possible way that .9~ can be expressed as a/b.
3.) For any two numbers, there is an infinite number of other numbers that fit between them.
Example:
1.08, and 1.09.
1.081, 1.0801, 1.08001...
There are no numbers that can fit between .9~ and 1. thus, they must be equal.
Demonstrations:
1.)
Define x as .9~
x = .9~
10x = 9.9~
10x - x = 9
9x = 9
x = 1
Thus, .9~ = x = 1
2.)
1/3 = .3~
1/3 X 3 = 1
.3~ X 3 = .9~
If .3~ and 1/3 are equal, identical operations on them result in an equal product. Thus, .9~ = 1.
ARGUMENTS:
".9~ is not equal to 1, it gets closer to 1 with each 9 added but never reaches it"
Reply: .9~ is a number. This means it has value. The value of .9~ doesn't "get closer" to anything. It is a number, and has value.
"If .9~ equals 1, then doesn't 1.9~ equal 2, 2.9~ equal 3, 3.9~ = 4, and so forth?"
Reply: Yes.
"There is no point at which as certain number of 9's makes .9~ equal 1."
Reply:
Of course there isn't. If there were a point at which a certain number of nines "made" .9~ equal 1, then it would not be .9~ that equaled 1, just the decimal with enough nines to reach that point.
A "certain number of nines" directly contradicts the concept of an infinite string of nines. Approaching the question from this angle is entirely illogical.
For example, 3/7 = .428571~
Thus, 428571/999999 = .428571~
Therefore, 428571/999999 = 3/7
When we apply this rule to .9~:
9 is the repeating sequence. We put it over an identical number of 9's, which is one 9, thus giving us 9/9, which is 1.
2.) All rational numbers can be expressed as a/b with a and b as integers. .9~ is a rational number. Thusly, it can be expressed as a/b. 9.~'s a/b expression is 1/1. To all those who disagree, I challenge you to find another possible way that .9~ can be expressed as a/b.
3.) For any two numbers, there is an infinite number of other numbers that fit between them.
Example:
1.08, and 1.09.
1.081, 1.0801, 1.08001...
There are no numbers that can fit between .9~ and 1. thus, they must be equal.
Demonstrations:
1.)
Define x as .9~
x = .9~
10x = 9.9~
10x - x = 9
9x = 9
x = 1
Thus, .9~ = x = 1
2.)
1/3 = .3~
1/3 X 3 = 1
.3~ X 3 = .9~
If .3~ and 1/3 are equal, identical operations on them result in an equal product. Thus, .9~ = 1.
ARGUMENTS:
".9~ is not equal to 1, it gets closer to 1 with each 9 added but never reaches it"
Reply: .9~ is a number. This means it has value. The value of .9~ doesn't "get closer" to anything. It is a number, and has value.
"If .9~ equals 1, then doesn't 1.9~ equal 2, 2.9~ equal 3, 3.9~ = 4, and so forth?"
Reply: Yes.
"There is no point at which as certain number of 9's makes .9~ equal 1."
Reply:
Of course there isn't. If there were a point at which a certain number of nines "made" .9~ equal 1, then it would not be .9~ that equaled 1, just the decimal with enough nines to reach that point.
A "certain number of nines" directly contradicts the concept of an infinite string of nines. Approaching the question from this angle is entirely illogical.
by YHHAWNFTPSHI May 4, 2009