nicholash747
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- Aug 7, 2012
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- HSC
- 2012
Worked solutions someone?
I left my answer in degrees... do you think this will affect anything?i got pi/6 for the min question . was a bit hard and i left some parts out.
i did that. but didnt remove r. fmllllllllllllll.I completely BSed it until it happened to give me the answer but I had the equation from (i) and another equation and I simultaneously solved them, then made x^2 > 0 which happened to give me a half. No idea what the other equation was, though :\
Yay, I got pi/6 tooi got pi/6 for the min question . was a bit hard and i left some parts out.
Haha this.BOS trolled us. Gave most people the feeling they'll ace the exam and a "state rank" mentality. NEK MINUT Q16..
nice work . did you do all other questions? (potential state rank)c)ii)
4c = 1+4r^2
c=1/4 +r^2
c>0
Therefore, 1/4 +r^2 >0
rearrange and r>1/2.... because r >0
But since c>r then c>1/2 !!
If you move the 1/4 across you get a negative 1/4 and you can't root negatives :Sc)ii)
4c = 1+4r^2
c=1/4 +r^2
c>0
Therefore, 1/4 +r^2 >0
rearrange and r>1/2.... because r >0
But since c>r then c>1/2 !!
did this tooc)ii)
4c = 1+4r^2
c=1/4 +r^2
c>0
Therefore, 1/4 +r^2 >0
rearrange and r>1/2.... because r >0
But since c>r then c>1/2 !!
Beast Haha, yep think I'll definitely get zero for C.Hi! so i can do worked solutions if people want but basically here was how to do 16c i. took me a solid 30min to work out how to do it. Anyways.
equate the two graphs. so you get
y + (y-c)^2 = R^2.
y + y^2 - 2yc + c^2 - R^2 = 0
y^2 + y(1-2c) + (c^2 - R^2) = 0
now there are 2 equal and identical solutions to the above equation (because they intersect twice at the same location, just on oposite sides of the y axis)
hence the discriminate = 0
(1-2c)^2 - 4(c^2 - R^2) = 0 will then simplify down into the required answer, 4c = 1-4C^2
No idea how to do part ii though...