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This question is pretty tricky, if you draw the time line, it would be clearer. a) So, it goes as like this: At time 1 (which is one year from now), the pension pays $15000 straight away. At time 2, the pension increases according to the inflation index, which is going to be $15000(1.035) ...

yep, that's right.

I can understand how you feel, if the hsc question is vague, it's hard to answer what they want if you don't understand the question. But keep this in mind, there are a board of people whose job is to design the paper so that no matter how you read it, there is only one meaning out of it...

I'm not saying the student is breaking the law.... it's the teacher that's not suppose to teach it....... even if you did finish the prelim course. But again, since for maths there are no clear boundary between prelim and year 12 (unless about 4u), they can't really catch you for it.

Erm, isn't it obvious it should be at the beginning? I mean, what is the point of investing at the end........ you get no interest for 1 period and start to only accumulate interest next period? It's like superannuation funds, if you have the money now, why do you keep it til the end to save it...

Legally speaking though, you can't start year 12 topics in preliminary..... So your teacher is breaking the law there...... You can only start 4U in term 4. But of coz, you can always start early in tutoring centre as they are not "required" to follow the schedule and you can do whateva u want...

I totally agree to what you said. That's why when I tutor my students (and lots of them agree with it, even though they don't memorise the proves), I always emphasis on to understand where the formulas and concepts derived from, not to exactly memorise them, but to use them to understand the...

Hi, this question I agree it's hard to know where to start, but the best way to start is to draw up the diagram, and it shouldn't be too hard after that. So for a), I assume you know it already? Since the bee started together with bulldozer A on one end with velocity 2V m/s, it will be...

Well, I think the major difference between 2 unit and 3 unit is if you are good with algebra. In 3 unit, you manipulate algebra at a more sophisticated level than 2 unit kids. Also with 3 unit, you learn more probability and something called counting techniques. So ya, if you are thinking...

haha, thx funnytomato, and thx for the tip, ya, I forgot to copy the credentials across, but this was suppose to be just a small thing, my advertisement is on the other thread.

Well, first of all, this question does not give you the formula of the Q(t), so you need to derive it from knowing only the rate of change of mass is proportional to the mass. Here goes: Let Q be the mass, \frac{dQ}{dt}=kQ \int \frac{dQ}{Q}=\int kdt \ln \left | Q \right |=kt+c As t=0, Q=60...

If you use substitution of u=2x, then you will get the book's answer

HSC is near now, Trial is even closer~! Ever feel like you learnt everything, but do not know how to apply them, feeling lost when you see a question and do not know where to start? Come to our workshop at Kogarah then!! We offer free first trial class~! Basically how this will work is...

Hi, You should note that not every probability is binomial, and you need to learn how to recognise it. Characteristics of Binomial probability are: 1. There are n independent trials (in here, 5 winter days means 5 trials) 2. Constant probability for every trial (in here, it is, because...

Erm, you can't say there's a boundary to questions' difficulty, but all I can say is, the difference between the difficulty in questions you see in 3u and questions in 4u is that in 3u, it's about calculation and derivation of formulas. But in 4u, you need to be good at algebra and manipulation.

That's kool:), I was just trying to clarify too.

Hope you understand and find the solutions useful :)

For 11b, you do nearly the same thing, so again first get: x=\frac{V\sin 2\alpha }{g}, Then substitute the correct conditions (and letting R be the correct position), you get: R+a=\frac{V\sin 2\alpha }{g}...........[1] R-b=\frac{V\sin 2\beta }{g}...........[2] R=\frac{V\sin 2\theta...

For 11a, First get the equation: x=\frac{V\sin 2\alpha}{g} I assume you can get that ya? Then let the correct position of the target be R metres from the origin, therefore: At \; \alpha=45^{\circ},x=R+30, R+30=\frac{V\sin \left ( 2\times 45^{\circ} \right ) }{g} R+30=\frac{V}{g}...