W00t! I did question 10... (1 Viewer)

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Q10 was a simple simpsons formula and lots of algebra, but yea Q8-9 sucked
 

Xayma

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Lots of algebra, not really.

The last bit of question ten just involved a bit manipulation. 8 and 9 were easier still.
 

The Bograt

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Yeah well I messed up 8 big time, and it took me like 45 minutes, 9 wasn't great either. I got some things, and I'll get marks for working, just couldn't finish them off like all the others...
 

cheers_big_ears

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question ten was fairly good
the only part i couldnt get was b)i show xy=1/2bc

even withou t that the rest was fairly straight forward
cosine rule here and there and tricky differentiation.

part a was simple


it was a 'no dramas' paper for me.
 

disco_dave

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congrates... i had nothing. stared blankly for a while and gave it a go, vut it didnt get me far... im feeling quite depressed right now! grr
 

acmilan

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nicko04 said:
question ten was fairly good
the only part i couldnt get was b)i show xy=1/2bc

even withou t that the rest was fairly straight forward
cosine rule here and there and tricky differentiation.

part a was simple


it was a 'no dramas' paper for me.
Since Area(AST)=Area(BCST) then

Area(ABC)=2Area(AST)

1/2(bcsinA) = 2.1/2(xysinA)

1/2(bcsinA) = xysinA

hence xy = 1/2(bc)
 

ShaunSmith

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w00t your a fagget die, thanks.
quesion 10 B) was the worst thing since the day before sliced bread.. or something!
 

The Bograt

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nicko04 said:
question ten was fairly good
the only part i couldnt get was b)i show xy=1/2bc

even withou t that the rest was fairly straight forward
cosine rule here and there and tricky differentiation.

part a was simple


it was a 'no dramas' paper for me.
I think you used
A1 = 1/2 xy sin A
A2 = 1/2 bc sin A x 2 (since the area is split into 2, that triagle is both their areas)
cancel the sin A
A1: A2 = 1:2
=1/2
 

superbird

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10B iii was good. all u had to do was differentiate it twice and then sub in the x value
 

cheers_big_ears

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acmilan1987 said:
Since Area(AST)=Area(BCST) then

Area(ABC)=2Area(AST)

1/2(bcsinA) = 2.1/2(xysinA)

1/2(bcsinA) = xysinA

hence xy = 1/2(bc)
thanks but i really dont care what the answer was anymore (cant change it even if i did)

besides it was only worth one mark
 

xeriphic

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nicko04 said:
question ten was fairly good
the only part i couldnt get was b)i show xy=1/2bc

even withou t that the rest was fairly straight forward
cosine rule here and there and tricky differentiation.

part a was simple


it was a 'no dramas' paper for me.

for that one just equate the area together using 1/2absinC

2 x of the small area = total area then solve

I couldn't do that last two ahhhhhhhhh of ten, the rest was okai
 

Komaticom

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So how'd you do the last part? I'm guessing I had to use all the formulae proven beforehand? I tried working backwards, to no avail. Bah, help.
 

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