Math Help Plskthx

2009-05-13 14:56:37

#1

Noobpatty: ʎʇʇɐdqoou; +194|6527|West NY

There needs to be a math/homework help sticky/'Official' thread that gets buried once no one needs it anymore.
Anyway I'm having a total brain fart right now and need someone to point out how to do this again

The table below gives information about the radioactive decay of carbon-14. Part of the table has been deliberately left blank.

https://castlelearning.com/review/Courses%5CEarth%5Cq2639-1.gif

After how many years will 1/128 gram of the original carbon-14 remain?

1. 22,800 yr
2. 28,500 yr
3. 34,200 yr
4. 39,900 yr

Last edited by Noobpatty (2009-05-13 15:07:34)

2009-05-13 14:59:10

#2

Flaming_Maniac: prince of insufficient light; +2,490|6880|67.222.138.85

current mass = original mass (.5^t/5700)

where t is the number of years

2009-05-13 15:01:00

#3

KEN-JENNINGS: I am all that is MOD!; +2,977|6805|949

If that is the question verbatim, write down "not enough information"

2009-05-13 15:07:05

#4

blah: macaroni with cheeseeee; +111|5921|Croatia

shit isn't hard you just need right formulas
it's pretty much usage of logarythms and potencies

2009-05-13 15:08:01

#5

Noobpatty: ʎʇʇɐdqoou; +194|6527|West NY

Hahah sorry about all the confusion I missed the 1/128 part

2009-05-13 15:11:47

#6

bf2gammer: Member; +14|6394

39,990

2009-05-13 15:13:32

#7

DankmanHightimes: Member; +24|6673|atlanta GA

bf2gammer wrote:
39,990

2009-05-13 15:21:49

#8

Aries_37: arrivederci frog; +368|6749|London

blah wrote:
shit isn't hard you just need right formulas
it's pretty much usage of logarythms and potencies

u r gud at this one I can tell

2009-05-13 15:30:02

#9

Ty: Mass Media Casualty; +2,398|6948|Noizyland

Dude, you've go a 25% chance of getting it right - just guess.

[Blinking eyes thing]
Steam: http://steamcommunity.com/id/tzyon

2009-05-13 15:32:40

#10

Aries_37: arrivederci frog; +368|6749|London

Ty wrote:
Dude, you've go a 25% chance of getting it right - just guess.

or just put 39,900 and have 100% chance of getting it right lol

2009-05-13 15:41:42

#11

mcminty: Moderating your content for the Australian Govt.; +879|6895|Sydney, Australia

blah wrote:
shit isn't hard you just need right formulas
it's pretty much usage of logarythms and potencies

Not even.

The half life is 5,700 years. You can see that by simple subtraction..

So, when you have 1/(2ⁿ) th of the mass remaining, it will be n*5700 years. ie. 1/(2⁷) = 1/128 => 7*5,700 = 39,000

2009-05-13 15:45:51

#12

Gawwad: My way or Haddaway!; +212|6858|Espoo, Finland

As it's half times just figure out when (1/2)ⁿ = 1/128. With such small numbers you can just test it and the answer is 7.
Then you see the half time right on the table which is 5700a and just multiply that with your number of halvings recuired (7)

Flaming_Maniac wrote:
current mass = original mass (.5^t/5700)

where t is the number of years

That's a bit advanced for this problem, don't you think?

Last edited by Gawwad (2009-05-13 15:47:04)

2009-05-13 15:50:25

#13

VspyVspy: Sniper; +183|6846|A sunburnt country

bf2gammer wrote:
39,990

7 x 5700.

2009-05-13 17:31:08

#14

Ryan: Member; +1,230|7017|Alberta, Canada

Final = Initial(ratio)^(time/period)

Er somethin' like that. Then you can use logs to find t.

2009-05-13 17:36:58

#15

CammRobb: Banned; +1,510|6304|Carnoustie MASSIF

DankmanHightimes wrote:
bf2gammer wrote:
39,990

2009-05-13 17:38:06

#16

CammRobb: Banned; +1,510|6304|Carnoustie MASSIF

Ryan wrote:
Final = Initial(ratio)^(time/period)

Er somethin' like that. Then you can use logs to find t.

Why? It's just easy addition.

Goddamn people trying to look smart by writing big maths related words...

2009-05-13 17:55:29

#17

S.Lythberg: Mastermind; +429|6620|Chicago, IL

7*5700

its 7 half lives later

2009-05-13 20:07:00

#18

mcminty: Moderating your content for the Australian Govt.; +879|6895|Sydney, Australia

SirSchloppy wrote:
Ryan wrote:
Final = Initial(ratio)^(time/period)

Er somethin' like that. Then you can use logs to find t.
Why? It's just easy addition.

Goddamn people trying to look smart by writing big maths related words...

Welll.. radioactive decay should be modeled as exponential decay. It's just simple due to the nature of the question. If he was asked to find the amount left at a certain number of years (that is not a multiple of the half life), then it would be a little more involved...

2009-05-14 00:19:54

#19

kylef: Gone; +1,352|6667|N. Ireland

Where's Freeman when you need him.

2009-05-14 01:44:16

#20

bennisboy: Member; +829|6820|Poundland

Just continue the damn patterns. This question is barely maths, its basically logic/non-verbal reasoning

2009-05-14 01:52:11

#21

CammRobb: Banned; +1,510|6304|Carnoustie MASSIF

mcminty wrote:
SirSchloppy wrote:
Ryan wrote:
Final = Initial(ratio)^(time/period)

Er somethin' like that. Then you can use logs to find t.
Why? It's just easy addition.

Goddamn people trying to look smart by writing big maths related words...
Welll.. radioactive decay should be modeled as exponential decay. It's just simple due to the nature of the question. If he was asked to find the amount left at a certain number of years (that is not a multiple of the half life), then it would be a little more involved...

I know that if you had to find at a set year etcetcetc, but that's not what the question entails. It's just simple. Not trolling or anything, but it's rather obvious.

2009-05-14 04:48:58

#22

The A W S M F O X: I Won't Deny It; +172|5858|SQUID

58

BF2S Forums Math Help Plskthx

bf2gammer wrote:

blah wrote:

Ty wrote:

blah wrote:

Flaming_Maniac wrote:

bf2gammer wrote:

DankmanHightimes wrote:

bf2gammer wrote:

Ryan wrote:

SirSchloppy wrote:

Ryan wrote:

mcminty wrote:

SirSchloppy wrote:

Ryan wrote:

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