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2002 winter diet - HSN forum

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2002 winter diet


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#1 Larry

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Posted 03 May 2005 - 08:20 PM

does any website have the 2002 winter diet fully worked solutions?? cheers xxxx

#2 Dave

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Posted 03 May 2005 - 09:01 PM

i doubt it i personally have never seen one

whats wrong with the SQA answers

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#3 Larry

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Posted 03 May 2005 - 10:32 PM

it just takes ages to work out the answer if you get it wrong, and worked solutions are so handy! i haven´t done the paper yet so i´m not stuck with anything, looks like i´ll have to post if i can´t do them though.

#4 AppleCore

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Posted 15 May 2005 - 03:05 PM

what was the "winter diet" paper about anyways???... how come they dont have it every year???

#5 tupacshakur

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Posted 15 May 2005 - 03:47 PM

QUOTE(itsmereally @ May 15 2005, 03:05 PM)
what was the "winter diet" paper about anyways???... how come they dont have it every year???

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they wanted re-sit candidates to resit the exam earlier i think.
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#6 Dave

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Posted 15 May 2005 - 03:48 PM

there is another thread on the history of the winter diet somewhere

anyway you can forget about it because it doesnt exist anymore and was for people to sit there final exam around December time. The idea was for people to resit quickly and for colleges but it didnt work out plus can you imagine producing 2 papers for each subject each year...nightmare

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#7 Larry

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Posted 18 May 2005 - 03:40 PM

where could i get the answers to the winter diet papers??? i've lost my answer sheet!!

#8 Vixus

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Posted 19 May 2005 - 09:16 AM

I'm curious about Q11 on 2002WD. How am I supposed to answer it? I have just stated explicitly the answer is 0 because a.(b+c) is perpendicular.

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Regular hexagon of sides 2 units. Value of a.(b+c)?

#9 dfx

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Posted 19 May 2005 - 10:45 AM

a.(b+c) = a.b + a.c

= |a||b|Cos Theta + |a||c|Cos Theta

For Theta, you need the value of the exterior angle when a is extended so that they both project outwards from the same origin (Q) . The exterior angle is 360/n =360/6 = 60

= 2 x 2 x Cos60 + 2 x 2 x Cos60

= 4 x (1/2) + 4 x (1/2)

= 4?????

HUH ?

#10 ermd

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Posted 19 May 2005 - 11:20 AM

Angles in a hexagon don't add up to 360 degree.gif . Problem solved biggrin.gif

#11 dfx

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Posted 19 May 2005 - 11:40 AM

The exterior angle of any regular polygon is 360/n, where n is the number of sides in the polygon. And we do need the exterior angle, cause the interior angles aren't between, say a and b, for a.b, or b and c for b.c, so we extend the vector, hence the exterior angle. Problem back again sad.gif

#12 ermd

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Posted 19 May 2005 - 12:50 PM

Here is my solution.

#13 dfx

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Posted 19 May 2005 - 12:58 PM

Oh RIGHT, a.c is -1/2 and its 120, not 60. Cheers! biggrin.gif

#14 The Wedge Effect

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Posted 19 May 2005 - 02:03 PM

QUOTE(dfx @ May 19 2005, 11:45 AM)
a.(b+c) = a.b + a.c

= |a||b|Cos Theta + |a||c|Cos Theta

For Theta, you need the value of the exterior angle when a is extended so that they both project outwards from the same origin (Q) . The exterior angle is 360/n =360/6 = 60

= 2 x 2 x Cos60 + 2 x 2 x Cos60

= 4 x (1/2)  + 4  x  (1/2)

= 4?????

HUH ?

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It's 120 degrees, not 60! tongue.gif

#15 gary

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Posted 19 May 2005 - 02:13 PM

Someone please tell me there will be no paper like this hell tomorrow.

#16 tupacshakur

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Posted 19 May 2005 - 02:36 PM

yeah thats what i was worrying about in the "how hard is maths gonna be this year" thread i made lol

someone please re assure us!!!!!!!!!!! sad.gif sad.gif sad.gif sad.gif sad.gif
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#17 gary

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Posted 19 May 2005 - 02:41 PM

I was thinking the SQA just made that winter diet exam hard because all the people who sat it woud have been revising non stop from August-January so they should have been prepared for something so hard. I don't think they could give us something that difficult and get away with it seeing as there was suppost to be an outrage in 2000.

#18 tupacshakur

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Posted 19 May 2005 - 02:43 PM

QUOTE(gary @ May 19 2005, 02:41 PM)
it seeing as there was suppost to be an outrage in 2000.

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did they have angry protests outside SQA headquarters or something? biggrin.gif
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