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| 16 Mar 2012 10:26 PM |
Title
inb4 im not doing your homework inb4 op cant inb4 |
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EXcellent
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Storm9802
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| 16 Mar 2012 10:27 PM |
17 I think.
This is stupid - Andrew Hussie |
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| 16 Mar 2012 10:27 PM |
5, 17, blah blah.
This has been done before. |
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Regi8401
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| 16 Mar 2012 10:27 PM |
16/2[8-3(4-2)]+1 8[8-3(4-2)]+1 8[5(2)]+1 8[10]+1 80+1 81
~Cannot The Kingdom Of Salvation Take Me Home?~ |
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Regi8401
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| 16 Mar 2012 10:28 PM |
of course there are a few other possible answers though
~Cannot The Kingdom Of Salvation Take Me Home?~ |
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Storm9802
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| 16 Mar 2012 10:29 PM |
@Regi
WRONG
There are no 'possible answers'
You do it in this order:
Parentheses Exponents Muliplication/Divition Addition/Subtraction
This is stupid - Andrew Hussie |
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cafeolay2
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Regi8401
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| 16 Mar 2012 10:30 PM |
@storm well i still had the correct answer before so it's not incorrect i was thinking of some other problem that has two possible answers
~Cannot The Kingdom Of Salvation Take Me Home?~ |
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Storm9802
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| 16 Mar 2012 10:31 PM |
@not
No, you frogot to do the 3(4-2)
4-2 is 2 as it is in the parentheses, then you multiply it by 3 which is 6. 8-6 = 2, so it would now be
16/2 (2) + 1
Then it would be
8(2) + 1
That would be 16, plus 1,
17.
This is stupid - Andrew Hussie |
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Inks13
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| 16 Mar 2012 10:32 PM |
if the quantum of a physics beyond that of the means of a brain nueron cell being shot forward in exess of 20.568, Removing the matter of mass from the Quantum and making it whole, Squared by 57, and then divided by pi, Multiplied by The atomic Sub-pulse of a Nuetronic Vein. This therefore in causing the Malfunction of time iteself.
So, in the end, If you were to move the 20.568 to 318.368, and then square the matter of mass from the whole quantum, and then squaring it by 556 instead of 57, What in the end is the Atomic Sub-pulse Amount in Nuetronic Scale? |
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cafeolay2
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Storm9802
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| 16 Mar 2012 10:33 PM |
@Inks
42
This is stupid - Andrew Hussie |
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PsychoBob
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| 16 Mar 2012 10:37 PM |
16/2[8-3(4-2)]+1 16/2[8-3(2)]+1 16/2[8-6]+1 16/2[2]+1 8[2]+1
17
(2^4)/(2^1)[(2^3)-(1+2^1)((2^2)-(2^1))]+(2^0) |
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