How was it ? CSSA Trial (1 Viewer)

[tigerboi]

hahah . you beat me? i let you beat me.
Yeah . since its the third year of the new hsc, its going to be the hardest

 

[Huy]

Originally posted by tigerboi
since its the third year of the new hsc, its going to be the hardest

What about the first year, since the introduction of the New HSC?
They'd have to learn outcomes, revised topics, dot points, etc.

Didn't think about that one, did you?
You'd think that with 2 years under the BOS's belt, they'd know how to revise syllabi and 'dumb it down' for us kids.

(Generally speaking, as the years go on, the HSC will become easier with further adjustments/modifications)

 

[Ragerunner]

the easier the HSC the higher the UAI requirements it seems to be

 

[SoCal]

Originally posted by nerdd
isn't it kinda common sense....?
that if y=cos^-1 (blah)
then cos y = blah??
surely you would've been taught that!

Not that I can ever remember.

 

[hereSIR]

Look at it in the reverse sense
If COS (A) = 2/3
than
. A = Cos^-1 (2/3)
It's exactly the same thing right!
??? I think????

Can you smell what the ROCK is cooking! Ke ke ke ke

 

[wogboy]

Look at it in the reverse sense
If COS (A) = 2/3
than
. A = Cos^-1 (2/3)
It's exactly the same thing right!

Wrong. If cosA = 2/3, then A = 2pi*n +- arccos(2/3) (where n can be any integer).

On the other hand it's correct to say:
if A = arccos(2/3), then cosA = 2/3.

Try to think why this is true (hint: a graph of f(x)=cosx and f(x) = arccos x will help).

(sorry I hate writing cos^-1 due to ambiguity so I write arccos instead)

 

[nerdd]

wtf is arc?!

 

[wogboy]

wtf is arc?!

I repeat:

(sorry I hate writing cos^-1 due to ambiguity so I write arccos instead)

arcsin & arccos & arctan are the inverse functions of sin, cos, tan respectively. I can't stand writing cos^-1(x) since it should really mean 1/cosx, which is not the inverse function of cosx.

cos^2(x) is equivalent to (cosx)^2, so going by that cos^-1(x) should really be equivalent to (cosx)^-1. In most (if not all?)computer programs, whenever calculating the inverse sine of something, you're required to enter arcsin(x) rather than sin^-1(x). The person who invented the notation sin^-1(x) for inverse sine should be shot hehe

 

[TimTheTutor]

Originally posted by Huy
2.5

just put in large values of n
lim as n -> infinity

5(10^n) + 3
----------------
2(10^n) - 1

say n = 10
5(10^10) + 3
-----------------
2(10^10) - 1

= 2.5
that's what i should have done, but instead i wrote something like "3(10^n - 1)"

Your answer is correct but your method leaves a bit to be desired.
Just as with limits with polynomials divide numerator/denominator through by the term with the highest power and then apply the basic rule that lim (x->infinity) [1/x] = 0

[5(10^n)+3]/(10^n)
-------------------------
[2(10^n)+1]/(10^n)

5 + 3/(10^n)
= ----------------
2 + 1/(10^n)

5 + 0
= -------
2 + 0

= 5/2