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Oh yeah. I should've subbed those roots into y. That seems long, is there another method?

Yep the sign of k changes. Thanks :punch: Can someone show me how to do Q8 (a) (i)?

http://4unitmaths.com/sbhs2002.pdf Q6 (a) (iii) There's no solutions and couldn't find my mistake. I got a result that contradicted theirs (I got k<0 for 3 real roots) Can someone help spot my mistake? -- (ignore the first 5 lines)

Re: HSC 2015 2U Marathon x = 2t - 3 \ln(t+1) v = 2 - \frac{3}{t+1} a = \frac{3}{(t+1)^2 } $ Initial conditions are: $ x = 0, t = 0 $ At $ t= 0, v = 2-3 = -1 < 0 $ This means the particle will move into the negative region. $ $ Note that acceleration is always positive for all...

Hm, I think it would still be inclusive. It's continuous

Yeah, just relax a bit beforehand. Past papers could induce stress My friends want to do a hakk session an hour beforehand to relieve the stress

How are you sure that the distance AE is in fact the perpendicular distance from A to line BE? Gradient of line AE = -5/2 (if I were to construct a line) Gradient of BE = 5 since -5/2 * 5 =/= -1, that means line AE and line BE cannot be perpendicular. Hence, the distance A to E cannot be...

(i) When it asks for range, it's asking for "what values of y exist for this graph?". Looking at the graph, the y values can go all the way down to -infinity (shown by the arrow on the curve on the right hand side) and it can go up to y = 3 (stationary point. It doesn't go to +infinity because...

re: HSC Chemistry Marathon Archive If it's concentrated, there would be little amount of ionisable [H+] so wouldn't adding water cause the 'burning' sensation from acids to start occuring, (giving the acid opportunity to ionise) until it's diluted enough?

re: HSC Chemistry Marathon Archive In dotpoints: - From t = 0 to t = 4, the system is at equilibrium (constant molarities) - At t = 4, the reaction was heated. By Le Chatelier's Principle, the system will shift towards the endothermic reaction, that is, the forward reaction. This causes the...

re: HSC Chemistry Marathon Archive Ozone and O2 have similar physical properties due to their similar molecular weight. Thus, they have similar intermolecular (dispersion) forces and causes them to require similar amounts of energy input in order to break these forces. Thus, ozone and O2 have...

For OP, this is because in the quadratic formula: x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ If the the thing under the Squar3root $ (b^2 - 4ac) $ is less than zero, then it gives unreal roots since you cannot square root negatives in the real field. $

Re: HSC 2015 4U Marathon $ let $ z_{1} = x_{1} + iy_{1} , z_{2} = x_{2} + iy_{2} $ The conditions given are equivalent to $ (x-2)^2 + (y-2)^2 = 23 \longrightarrow x^2 - 4x + y^2 - 4y = 15 (x-8)^2 + (y-5)^2 = 38 \longrightarrow x^2 - 16x + y^2 - 10y = -51 $ Subtracting bottom...

Re: HSC 2015 4U Marathon $ Ah cool. I used some relationship that didn't even know existed. Just drew it out and inferred. $ arg( \frac{-1}{z}) = \frac{-2 \pi}{3} arg(-z) = \frac{2 \pi}{3} arg(z) + \pi = arg(-z) \therefore arg(z) = \frac{- \pi}{3} \therefore $ z must have...