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Simpsons rule question (1 Viewer)

sPideS

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I can do a simpsons rule question when there are 2 distances between the others (i.e 'h' of the formula) but I cant do it when there are 3 distances.

Please help
 

bazookie

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have you got an example?

If there are more than 2 distances, you must use two applications of simpsons rule. Post an example and i will show you if you don't understand.
 

bazookie

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lol wtf... i've never seen a question like that. K lemme go try work it out.

edit: k yea i dunno how to do it. Are you sure the question is right? i don't think you need to worry about it too much. I've never come across a question like that. They usually just ask for the normal one, or one where you can apply the rule 2 times and add the two values together.
 

sPideS

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haha... im sure a question like that was in my trial..... too bad i didn't get my paper back :(
 

xJennax

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Oh I've had ones like these before, you just work out one of them, and then the 2nd one and add them together, I'll try now and scan the answer if it's right.

edit, actually I've only seen ones with 3 lengths or 5...I'm lookin in my textbook now.
 

sPideS

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I can just see a question like that in the paper tomorrow
 

mz`ds

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i doubt a question like that will be in the hsc. i remember i asked my teacher before and he told me that they will never ask u to work out the simpsons rule with the 3 distances. they might ask with 2 or 4 but not 3.
 

Smithyo

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working out:

Area 1: 12/3(8+4x10+19)
= 268

Area 2: 12/3(19+4x40+0)
= 716

Total Area = 268 + 716
= 984
 

UNZPME

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Smithyo said:
working out:

Area 1: 12/3(8+4x10+19)
= 268

Area 2: 12/3(19+4x40+0)
= 716

Total Area = 268 + 716
= 984
r u sure......wer did u get the 0 from its doesnt work out lik dat :S
 

Smithyo

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Yeap, if nothing is there - it becomes a 0 :)
 

michael1990

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sPideS said:
Oh, so you leave one of the measurements out?
Its not that your leaving a measurement out.
I will post up the answer.

If it doesn't make sense, ask away.
 

tomjones

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Smithyo said:
working out:

Area 1: 12/3(8+4x10+19)
= 268

Area 2: 12/3(19+4x40+0)
= 716

Total Area = 268 + 716
= 984
I am pretty sure that Simpsons rule does not work with odd number of areas and I'm am 100% sure they will only ask us a questions about it when there are even number of areas. Smithyo putting zero in there does not work as it will still calculate two areas but with the end length being zero if that makes sense lol.
 
Last edited:

savio23q

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Smithyo said:
working out:

Area 1: 12/3(8+4x10+19)
= 268

Area 2: 12/3(19+4x40+0)
= 716

Total Area = 268 + 716
= 984
Are you sure? I'm getting...

Area 1: 12/3(8+4x10+19)
= 268

Area 2: 12/3(10+4x19+40)
= 504

Total Area = 268 + 496
= 772m^2

I think the OP left a vertical line out between 19 and 8. =\
 
Last edited:

NEVAGIVEUP

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I'm highly dubious that you can calculate using the Simpsons rule for such a question;

And who ever said to use zero; you can't because there are four lengths given; which is not correct
 

Smithyo

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hmm yeah i think there is a length missing.
 

sPideS

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Nope, no length is missing. Perhaps you cant do solve a question like that with Simpsons rule
 

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