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A raw pass will easily align to an E3 AT LEAST. So if you can confidently get to that threshold (that's something you should honestly ask yourself), then your ATAR would be scaled pretty high.

Well I hope you can see that A_{60} = 500[(1.003)^{60} + (1.003)^{59}(1.01) + \dots + (1.01)^{59}(1.003)] You can take a factor of 1.003^{60} to get 500(1.003)^{60}[1+ \frac{1.01}{1.003} + \frac{1.01^2}{1.003^2} + \dots + \frac{1.01^{59}}{1.003^{59}}] From there it's a sum of a...

Well the answer is 3 since ln(1) = 0 \\ |x^2-1| = 1 \Rightarrow x^2-1 = 1, \ \ \ x^2 -1 = -1 First equation has 2 solutions, second equation has one.

1. You have to find P(2), assuming you know your remainder theorem. P(x) = Q(x)(x+1)(x-2) + 18x+17 Sub x = 2 to get your answer 2. You have to see that f'(x) >0 and therefore you use the discriminate on f'(x) and make that negative. \Delta <0 3. That's not even an equation

Mostly the latter for me really, there's a lot of pattern recognition going on when I approach them.

i) you expand x(1+x)^n , differentiate both sides then sub in x = 1 ii) you expand (1+x)^n , integrate both sides twice (make sure you can find the constants) then sub x = -1 and do some re arrangement.

For 2U and Ext 1 Mathematics, I don't think so. Pretty much all of topics you'll learn in HSC builds up or uses the concepts taught in Preliminary, except for maybe probability.

If you compared the answers in i) and ii) The last n terms on the left side of the equation are gone, the right side looks something pretty close to something being divided by 2 since 4^n/2 = 2^{2n-1} The second term looks pretty close to \binom{2n}{n} and then a bit of intuition comes...

There can never be an even number of terms, only an odd amount. You're expanding (1+x)^{even} So expanding (1+x)^2 generates 3 terms. expanding (1+x)^4 generates 5 terms etc... You would get a pretty different scenario if you do get an even number of terms.

NAH Jks, this is a pretty cool problem. You start by applying the symmetrical co-efficient property \binom{n}{r} = \binom{n}{n-r} with the LAST n-1 terms so you have \binom{2n}{0} = \binom{2n}{2n} \\ \binom{2n}{1} = \binom{2n}{2n-1} \\ \binom{2n}{2} = \binom{2n}{2n-2} \\ \binom{2n}{3}...

Yeah and even if you don't make it, it's not the end of the world, not even close. I failed to make it to a selective school back in year 6, and also failed to transfer in subsequent years (although some schools had me on the "reserved" list),mainly because I'm just terrible at english and...

So you first consider the case where the eldest women is chosen, that means there's 2 other women to choose from 6 and 4 girls from 5 since one of them is out. \binom{6}{2}\binom{5}{4} = 75 Then you consider the other case where the eldest women isn't chosen, that means there's 3 women from...

It really depends what equation you're given, if you're given a displacement equation (the example form the khan academy link) it involves subbing points in as you can see. However, if you're given a velocity equation, you have to find the area under the velocity curve using integration e.g...

First part is just algebra bashing, I'm pretty sure you'll be fine with that The second part can be re-written as \sum_{r=1}^n \frac{r * \binom{n}{r}}{\binom{n}{r-1}} \\ \\ = \sum_{r=1}^n (n-r+1) = n(n) - (1+2+\dots +n) + n(1) = n(n+1) - \frac{n}{2}(n+1) = \\ \\ \frac{n}{2}(n+1) The...

I have no idea what the heck is \theta in your first part I know that you should to consider \frac{d \theta}{dt} = \frac{dh}{dt} \frac{dx}{dh} \frac{d \theta}{dx} But I can't seem to get an equation of theta in terms of x.... But for your 2nd screenshot, you basically sub x = 12 and y =...

Use the equation in part iii) Sub in x = 40 and y = 0 (instead of y = 20 like you did in part iv) ) Solve for theta and you'll get 15 (your lower limit) and 75 (the higher limit)

Let a,b,c,d be the roots of P(z) Consider sum of each of the square of the roots i.e a^2 + b^2+ c^2 + d^2 = (a+b+c+d)^2 - 2($sum of roots 2 at a time$) You'll see that this is negative, that proves that there is at least one complex root, since P(z) has real coefficients, the 2nd...

It's the mark for the whole course, and very likely that it will not change upon getting your final marks

You'd definitely need to improve those marks that's for sure, though I'm curious as to how you got a much higher mark in Ext 1 math than 2U math.

Brah I recall seeing a Shakespeare question worth 50k in the millionaire hot seat, it might be worth it you never know ;)