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range = 2v^2sin@cos@/g = v^2sin2@/g This is max when sin2@ = 1, because thats the maximum this sine function can be. so max range = 80 = v^2/g so now you know v. your assumption abt q10 is correct.

adv, ext1, ext2 94, 47, 48

wow, wish i'd gotten those sorta marks for internals last year. my adv eng marks were 80%. i also got 65% in my ext1 half yearlies. ha.

I think there is too much comment, which is psuedo-english as you said. I mean what, 20,30 marks of unseen? That's far too little. But it does make the latin course very easy. I never did extension, but I would have liked to.

1 unit = 10 cm 1 unit^3 = 1000 cm^3

really? i just get 1/sqrt(1+u) edit: yep acmilan edited. he's method works fine

no acmilan was right both substitutions work i think wait...mayb not, its too messy no it works fine lol

for 2 i get 116/135

Just to add a note of explanation to that method: What we're arguing here is that as the product is 1, if one factor approaches +/-infinity, the other factor approaches 1/+-infinity = 0. So consider x. As it approaches +-infinity, y-x, the other factor, must approach 0. That is, y-->x. So...

For A) http://www.boredofstudies.org/community/showthread.php?t=73284

Where did the W come from?

i though you only needed 5 minutes?

there was a conics in last years paper?

My material on this isn't as good as the others, but here's a few points: The Invention of the Transistor: Research Questions The transistor replaces an unreliable technology. Gather, process and present secondary information to: 1. Discuss how shortcomings in available...

1) int. (2x-3)dx= from (a+2) and (a) = (a+2)^2 - 3(a+3) - a^2 + 3a = 2a - 5 = 10 so a = 15/2

Theres 1/2 of getting either box For box A, its 12/21.11/20 for RR, and 9/21.8/20 for GG For box B, its 8/15.7/14 for RR, and 7/15.6/14 for GG So prob = 1/2 (12/21.11/20 + 9/21.8/20) + 1/2 (8/15.7/14 + 7/15.6/14) = 10/21

For 4), f'(x) = 12x^2 + 36x + 27 f''(x) = 24x + 36 Putting them equal to one another, and simplifying, we get 4x^2 + 4x - 9 = 0 (2x - 1)(2x + 3) = 0 x = 1/2, -3/2

3) As AE = 2, AO = 1 Also BO = 1 (isos triangle) Now angle AOB = 360/8 (revolution divided by 8) = 45 So are of AOB = 1/2 ab sinC = 1/2.1.1.1/sqrt2 = 1/2sqrt2 So are of octagon = 8/2sqrt2...

Here's some stuff on number 2.

Here's some material: