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She was just trying to explain what an asymptote was with the graph y = 1/x as an example.

This is a graph with an oblique asymptote. An asymptote is a line that the curve approaches but never reaches. y = x + 1/x lim x —> ±∞ [x + 1/x] = x so there's an asymptote at y = x then just differentiate for the stationary points as usual and sub in a few points near 0 to see how...

There are two values for the gradient at the corner points B and C. Subbing a value into a gradient-function always gives 1 gradient so this curve cannot be differentiated at those points..

I don't understand what you're getting at... It was a coincidence that the way you interpreted the question worked with n = 0 However it doesn't work with n > 0 so you're trying to prove an incorrect statement. cosθ * cos2θ * cos4θ......cos(2nθ) = sin[(2n+1)θ]/(2n+1)sinθ when n = 0...

here is your answer:

That induction question you guys are trying to prove does not work with n > 0. This is why brackets are essential: cosθ.cos2θ.cos4θ......cos2^n θ = sin (2^(n+1)θ)/2^(n+1)sinθ try again: cosθ * cos2θ * cos4θ * ... * cos2nθ = sin(2n+1θ)/2n+1sinθ

The 4-aces is a 3u probability question and aren't asked in the mathematics exam.. You can always work towards the binomial probability taught in 3u from mathematics level anyway: A = ace, O = other you can either get AAAAO, AAAOA, AAOAA, AOAAA and OAAAA... so with 5 combinations...

u² - 3u + 2 = 0 isn't an inequality but anyway: From rama's 2 lines: -20 - 20tan2x + 60tanx - 20 = 0 20tan2x - 60tanx + 40 = 0 He multiplied by -1 here so he knows he should reverse the inequality later on... ... but there should have been inequality signs here essentially: -20 -...

Could somebody help me with these questions: 1. Prove that (1 + sin2x)/(1 - sin2x) = tan²(x + pi/4) 2. File attached - (2unit?) Maxmin question I can get x = ac/(a + b) after some messy algebra but can't show it's the shortest path... also please do show any better methods to approach...

consider vertical motion of projectile: a = -10 v = -10t + 30sinθ y = -5t² + 30t*sinθ --[1] horizontal: a = 0 v = 30cosθ x = 30t*cosθ so t = x/(30cosθ) --[2] sub [1] --> [2] y = -x²sec²θ/180 + xtanθ at x = 60 you want y > 20 so -20sec²θ + 60tanθ > 20

In an arithmetic progression, whose first term and common difference are both non-zero, Un denotes the nth term and Sn denotes the sum of n terms. If U6, U4, U10 for a geometric progression: i) show that S10=0 ii) show that S6 + S12 = 0 iii) deduce that U7 + U8 + U9 + U10 = U11 +U12 i)...

Subbing it does give the answer ... pi / 4sqrt[2]

x = 80°2' and 325°11'