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mojako

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FinalFantasy did it.. he's getting better and better everyday

Ive finished my hsc.. with aligned marks of 97 (96.5) for ext1 and 92 (91.5) for ext2
marks of around 92 for ext2 are the most common marks which the 2004 candidature got :(
 

Trev

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mojako said:
FinalFantasy did it.. he's getting better and better everyday

Ive finished my hsc.. with aligned marks of 97 (96.5) for ext1 and 92 (91.5) for ext2
marks of around 92 for ext2 are the most common marks which the 2004 candidature got :(
hes smarter than us :) lol, probly just me tho haha
congrads., 92 is a good mark; im doin it in 264 days according to bos (or wateva the count is up to now)
 

FinalFantasy

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lol im not smart, i can only do dis cuz I levelled up trig. equations heaps long time ago.
 

2rko

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ok heres my problem ..
when justin borrows $8500 at 12% p.a. over 5 years he doesn't have to make any repayments for the first 2 months. What is the amount of each monthly repayment?
i keep getting $189.08 ... the answers $197.75 :eek:
 

Trev

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A1 = 8500(1.01) - M
A2 = [A1](1.01) - M
= 1.01[8500(1.01) - M] - M
= 8500(1.01^2) - M(1.01^1) - M
continued for (5 times 12)-2 = 58
A58 = 8500(1.01^58) - (1.01^57)M - ... - (1.01)M - M

A58 = 0
therefore:
8500(1.01^58) = (1.01^57)M - ... - (1.01)M - M
8500(1.01^58) = M[(1.01^57) - ... - (1.01) - 1]
M = [8500(1.01^58)]/[(1.01^57) - ... - (1.01) - 1]

since [(1.01^57) - ... - (1.01) - 1] is a geometric series, a=1; r = 1.01; n=58
Sn = a(r^n - 1)/(r-1)
S58 = 1(1.01^58 - 1)/(1.01-1)

therefore:
M = [8500(1.01^58)]/S58
= [8500(1.01^58)]/[1(1.01^58 - 1)/(1.01-1)]
= 193.85
hmm *cries*

i swear thats right, tho i havnt done this for like 2 mnths...
 
Last edited:
D

ddm

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Oh gawd, its soooooo hard!!!!!!! lol

Nah,,,,,,,

"A parabolic satelitte dish has a diameter of 4m at a depth of 0.4 m. Find the depth at which its diameter is 3.5 m, correct to 1 dp."

Locus is annoying me......
 

mojako

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Try the following.

A1 = 8500(1.01)
A2 = [A1](1.01)
A3 = [A2](1.01) - M
A4 = [A3](1.01) - M

BTW I don't really understand that satellite question.
 
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Templar

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ddm said:
"A parabolic satelitte dish has a diameter of 4m at a depth of 0.4 m. Find the depth at which its diameter is 3.5 m, correct to 1 dp."

Locus is annoying me......
The parabola would just have roots at (0,0), which would have the equation y=kx^2. When x=2 (half of the diameter) y=0.4. From there you can work k out and use it to solve for x=1.75
 

goan_crazy

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whats the best way to go about locus and parabola
its a topic im not good at...
thats my prob
 

Trev

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try and draw it accurately, remember the basic equations so you dont have to derive them every time coz that shits me (ie. the equation of tangent and perpendicular of P(2ap,ap^2))...
i duno, i find locus to be a 'rote' topic, u just follow the same procedure every time, theres not really any tricks like in permbs/combs n wateva else
 

Trev

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joe_m_2000 said:
theres like x^2=4ay
and x-h=4a(y-k) and stuff eh
well you cant memorise tangents/perpendicular equations for equations that do not have a vertex as the origin [well you can, but it would be very stoopd and there r an infinite amount of equations.... you'll just have to derive it everytime with the other parabolas)
its just a rote process and doesnt really change for questions (plus the questions they ask you to prove are basically the same), for our harder 3u section in the 4u exam there were 5 questions, 1 of them was locus and the parabola and it was the only question i got full marks for even tho i hadnt even looked at it since preliminary coz u dont need to remember much.....
 

FinalFantasy

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joe_m_2000 said:
theres like x^2=4ay
and x-h=4a(y-k) and stuff eh
it's (x-h)²=4a(y-k) btw=P
just know ur x²=4ay and use shifting to get the (x-h)²=4a(y-k)
dun even need to remember dat:p, juz remember vertex is (h,k)
those 2u locus q. often use that equation though
 

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