i download dy. At which time?
i iz dirasuk hantu just now, the hantu iz MJ Chat
i iz dirasuk hantu just now, the hantu iz MJ Chat
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Jul 15 2009, 12:54 AM
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All Stars
18,672 posts Joined: Jan 2003 From: Penang |
i download dy. At which time?
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Jul 15 2009, 12:54 AM
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Junior Member
139 posts Joined: Jun 2009 |
~lol~ The english not so bad la
I think u need to go temple / mosque / hindu temple/bomoh to pray or get urself clean |
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Jul 15 2009, 12:54 AM
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Junior Member
33 posts Joined: Jan 2009 |
2.54 ± 0.06 Mly (778 ± 17 kpc)
~2.6 × 1010 L☉ ⟨a1, a2, …, an⟩ (787 ± 18, 770 ± 40, 772 ± 44, 783 ± 25) = ((787 + 770 + 772 + 783) / 4) ± ((182 + 402 + 442 + 252)0.5 / 4) = 778 ± 17 [a,] = {x ∈ ℝ ∣ a ≤ x ≤ b} [a,] = {x ∈ ℝ ∣ a ≤ x ≤ b}![]() ![]() ℝ* = ℝ \ {0} |
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Jul 15 2009, 12:54 AM
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Senior Member
3,812 posts Joined: Dec 2008 From: Eden |
kau ni macam-macam betol.....
aku kena rasuk dgn jin miskin, so sekarang ni aku miskin!!! |
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Jul 15 2009, 12:55 AM
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Senior Member
1,072 posts Joined: Jan 2008 From: MY |
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Jul 15 2009, 12:56 AM
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Senior Member
1,542 posts Joined: Jul 2005 From: cheeseland |
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Jul 15 2009, 12:56 AM
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Junior Member
152 posts Joined: Jun 2008 From: Maria Ozawa House |
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Jul 15 2009, 12:57 AM
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Junior Member
99 posts Joined: Jul 2008 From: Came from the future Joined : November 2020 |
i iz dirasuk hantu betina just now.
i went toilet i pee sitting down and wash my hands and off the lights when im done. |
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Jul 15 2009, 12:58 AM
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All Stars
18,672 posts Joined: Jan 2003 From: Penang |
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Jul 15 2009, 12:58 AM
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Junior Member
33 posts Joined: Jan 2009 |
2.54 ± 0.06 Mly (778 ± 17 kpc)
~2.6 × 1010 L☉ ⟨a1, a2, …, an⟩ (787 ± 18, 770 ± 40, 772 ± 44, 783 ± 25) = ((787 + 770 + 772 + 783) / 4) ± ((182 + 402 + 442 + 252)0.5 / 4) = 778 ± 17 [a,] = {x ∈ ℝ ∣ a ≤ x ≤ b} [a,] = {x ∈ ℝ ∣ a ≤ x ≤ b}![]() ![]() ![]() ℝ* = ℝ \ {0} = 123456.7 = 1.234567 * 10^5 101.7654 = 1.017654 * 10^2 = 0.001017654 * 10^5 Hence: 123456.7 + 101.7654 = (1.234567 * 10^5) + (1.017654 * 10^2) = (1.234567 * 10^5) + (0.001017654 * 10^5) = (1.234567 + 0.001017654) * 10^5 = 1.235584654 * 10^5 |
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Jul 15 2009, 12:58 AM
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All Stars
18,672 posts Joined: Jan 2003 From: Penang |
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Jul 15 2009, 12:58 AM
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Senior Member
1,279 posts Joined: Jul 2008 From: behind you... |
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Jul 15 2009, 12:59 AM
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All Stars
18,672 posts Joined: Jan 2003 From: Penang |
LOL... u change canon to mj song...
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Jul 15 2009, 12:59 AM
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Junior Member
136 posts Joined: Nov 2008 From: The heaven above you. |
Possesed by the dupes of /k/.
Login /k/ and you are fine. |
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Jul 15 2009, 12:59 AM
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Junior Member
152 posts Joined: Jun 2008 From: Maria Ozawa House |
QUOTE(enjoice @ Jul 15 2009, 12:58 AM) 2.54 ± 0.06 Mly (778 ± 17 kpc) bro those are simple math~2.6 × 1010 L☉ ⟨a1, a2, …, an⟩ (787 ± 18, 770 ± 40, 772 ± 44, 783 ± 25) = ((787 + 770 + 772 + 783) / 4) ± ((182 + 402 + 442 + 252)0.5 / 4) = 778 ± 17 [a,] = {x ∈ ℝ ∣ a ≤ x ≤ b} [a,] = {x ∈ ℝ ∣ a ≤ x ≤ b}![]() ![]() ![]() ℝ* = ℝ \ {0} = 123456.7 = 1.234567 * 10^5 101.7654 = 1.017654 * 10^2 = 0.001017654 * 10^5 Hence: 123456.7 + 101.7654 = (1.234567 * 10^5) + (1.017654 * 10^2) = (1.234567 * 10^5) + (0.001017654 * 10^5) = (1.234567 + 0.001017654) * 10^5 = 1.235584654 * 10^5 |
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Jul 15 2009, 12:59 AM
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Junior Member
33 posts Joined: Jan 2009 |
QUOTE(rapeace @ Jul 15 2009, 12:56 AM) u into astronomy???QUOTE 2.54 ± 0.06 Mly (778 ± 17 kpc) ~2.6 × 1010 L☉ ⟨a1, a2, …, an⟩ (787 ± 18, 770 ± 40, 772 ± 44, 783 ± 25) = ((787 + 770 + 772 + 783) / 4) ± ((182 + 402 + 442 + 252)0.5 / 4) = 778 ± 17 this formula related to calculation andromeda galaxy trajectory. 2 billion years before it crashes to milky way. |
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Jul 15 2009, 01:00 AM
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Junior Member
33 posts Joined: Jan 2009 |
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Jul 15 2009, 01:00 AM
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All Stars
18,672 posts Joined: Jan 2003 From: Penang |
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Jul 15 2009, 01:00 AM
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Senior Member
1,542 posts Joined: Jul 2005 From: cheeseland |
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Jul 15 2009, 01:01 AM
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Junior Member
152 posts Joined: Jun 2008 From: Maria Ozawa House |
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