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Q16 (2 Viewers)

medicore

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wow, you guys are boss. I couldn't do 16 B and C LOL, A was good.
 

mytstephen

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radians/degrees?=/

Hey
For question 16)b)iv)... about finding theta that gives the minimum area of the trapezium... I wrote 30 degrees instead of in radian form (pi/6)...

Do you think I'll loose marks for this?
I surprised myself by even getting that far... it'd suck if I don't get the mark for it=/

THANKS
 

Kimyia

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Re: radians/degrees?=/

I'd say it depends on the marking critera, but it did say 0 < theta < pi/2 so I think they would have preferred radians :\
But then again, its still the same thing so I'm not sure.
 

cookeemonstah

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Re: radians/degrees?=/

iono cant remember if question asked for radians. If it did then ... gg

cant remember if they gave out half marks or not.
 

freeeeee

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dam was it pi/6, haha i got like 47degrees
 
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usmas

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C(I) that was just probability: P=(1/n^k-1){k(n-1)/(n-k)x(n-2)/(n-k-1)x...(n-k)/(...)} then simplify
C(II) substitition of k+1 in I, then solve simultaneous
(III) look for discrimant
(iv)??
 

Timske

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C(I) that was just probability: P=(1/n^k-1){k(n-1)/(n-k)x(n-2)/(n-k-1)x...(n-k)/(...)} then simplify
C(II) substitition of k+1 in I, then solve simultaneous
(III) look for discrimant
(iv)??
16C had 2 parts, wrong thread
 

megaman64

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Re: radians/degrees?=/

did you have to prove this is a minimum by doing 2nd derivative because i just finished the question by writing pi/6
 

eat_well

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For 16.c)ii)

Rearrange 4c=1+4r^2
c=1/4+r^2
r^2=c-1/4
r=(c-1/4)^2

If you take the square root of 1/4, you get 1/2
Since r>0 then c>1/2

I made silly mistakes everywhere else in the paper though and i didn't get 16.c)i)
 

mytstephen

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Re: radians/degrees?=/

did you have to prove this is a minimum by doing 2nd derivative because i just finished the question by writing pi/6
Yeah I proved it was a minimum and everything... it's just that it's in degrees...
 

DefiningTheta

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Re: radians/degrees?=/

The domain was between 0 < theta < pi/2
 

gr_111

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This is for 16b)

gradient OT = tan theta, therefore gradient tp = -1/tan theta...

point T is (cos theta, sin theta) and use equation of a line formula
 
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nirukk

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This is for 16b)

gradient OT = tan theta, therefore gradient tp = -1/tan theta...

point T is (sin theta, cos theta) and use equation of a line formula
Point T is (cos theta, sin theta) or the other way around ?
 

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