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Doominabox
Level 20 Comments: Peasant
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Last status update:

Date Signed Up:
10/08/2010
Last Login:
2/01/2013
Stats
Content Thumbs:
44
total, 116
, 72
Comment Thumbs:
71
total, 645
, 574
Content Level Progress:
81.35% (48/59)
Level 0 Content: Untouched account → Level 1 Content: New Here
Comment Level Progress:
0% (0/1)
Level 20 Comments: Peasant → Level 21 Comments: Peasant
Subscribers:
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Content Views:
2923
Total Comments Made:
321
FJ Points:
127
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MLP/Night at the...
niiiice
latest user's comments
#789
 just to watch him die?
01/02/2013 on
What's one secret that...
0
#27
 /watch?v=1IKbtBbl3EA
11/19/2012 on
too soon?
0
#595
 Obviously a lie, god
11/19/2012 on
10%
0
#166
 obviously im not going to accept your theory, and you mine, so…
11/18/2012 on
TODAY
0
#164
 the eqation wrote should have said either 9=10x.9.. …
[+]
(2 replies)
11/18/2012 on
TODAY
0
#165

GeorgeBush
(11/18/2012)
[]
That still doesn't mean that in the end .999... doesn't = 1, because it does. I can't explain it as well as my teacher could, but it's basic mathematic theory that they, and all other infinitely repeating integers with .999... equals the same number rounded up. 46.999... = 47, etc
#166

Doominabox
(11/18/2012)
[]
obviously im not going to accept your theory, and you mine, so lets both just stop talking
#161
 what you're not getting is that x≠.99999.... because she cha…
[+]
(4 replies)
11/18/2012 on
TODAY
0
#163

GeorgeBush
(11/18/2012)
[]
She didn't
change
what x equaled, she changed the equation x was used in.
In math you
always
do to one side of the equation what you do to the other, as I'm sure you know. She simply multiplied the starting equation by 10 on each side.
(1) x = .999... multiplied by 10 is
10x = 9.999... , the value of x did not change.
If you then divide the equation 10x = 9.999... (in which you claim x no longer equals .999...)
by
10 you get
(1) x = .999... again, meaning that in the two equations x remains constant.
#164

Doominabox
(11/18/2012)
[]
the eqation wrote should have said either
9=10x.9..
or 9x=9x
just because it was writen like that, doesn't mean its true
X=5
50X=50 (times it by 10)
49X=49 (i subtracted one from each side)
X=1 (divided by 49)
and presto 5=1
#165

GeorgeBush
(11/18/2012)
[]
That still doesn't mean that in the end .999... doesn't = 1, because it does. I can't explain it as well as my teacher could, but it's basic mathematic theory that they, and all other infinitely repeating integers with .999... equals the same number rounded up. 46.999... = 47, etc
#166

Doominabox
(11/18/2012)
[]
obviously im not going to accept your theory, and you mine, so lets both just stop talking
#159
 https://www. [url deleted] ?feature=player_embedded &v…
11/18/2012 on
TODAY
0
#158
 Comment deleted
11/18/2012 on
TODAY
0
#157
 exactly thats what she did wrong
[+]
(6 replies)
11/18/2012 on
TODAY
0
#160

GeorgeBush
(11/18/2012)
[]
No, she did nothing wrong. Every step of the equation had the same process done unto both sides. Subtracting x and subtracting .999... is the same processes, using the same amount, meaning that it is a perfectly valid. They
said
x = .999... meaning subtracting .999.... and subtracting x makes
no difference
as they are the
same number
. The
reason
they subtracted x on one side and .999... on one side is because one side (left) already had x as a variable, and the right side did not.
#161

Doominabox
(11/18/2012)
[]
what you're not getting is that x≠.99999.... because she changed it after the first step
you would not subtract x from the other side the actual equation would be
9=10x.99999....
wich is not the same as
9=9x
#163

GeorgeBush
(11/18/2012)
[]
She didn't
change
what x equaled, she changed the equation x was used in.
In math you
always
do to one side of the equation what you do to the other, as I'm sure you know. She simply multiplied the starting equation by 10 on each side.
(1) x = .999... multiplied by 10 is
10x = 9.999... , the value of x did not change.
If you then divide the equation 10x = 9.999... (in which you claim x no longer equals .999...)
by
10 you get
(1) x = .999... again, meaning that in the two equations x remains constant.
#164

Doominabox
(11/18/2012)
[]
the eqation wrote should have said either
9=10x.9..
or 9x=9x
just because it was writen like that, doesn't mean its true
X=5
50X=50 (times it by 10)
49X=49 (i subtracted one from each side)
X=1 (divided by 49)
and presto 5=1
#165

GeorgeBush
(11/18/2012)
[]
That still doesn't mean that in the end .999... doesn't = 1, because it does. I can't explain it as well as my teacher could, but it's basic mathematic theory that they, and all other infinitely repeating integers with .999... equals the same number rounded up. 46.999... = 47, etc
#166

Doominabox
(11/18/2012)
[]
obviously im not going to accept your theory, and you mine, so lets both just stop talking
#156
 also, you have to do exactly the same on both sides, so when t…
11/18/2012 on
TODAY
0
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