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p and q are parameters. If you don't have a good understanding of what parameters are then you should definitely take some time to go over the textbook again. Start with what you know and what is given to you. If you can't think of anything, try writing down an equation that is true of the...

As much time as necessary to learn the content. There is no one figure that will work for everyone. For example, the only revision I ever did for 3 unit was in the weeks before an exam block and it didn't have a negative effect on my marks. Past papers are usually supposed to refine your...

Yes.

No - your two equations are \begin{cases} x = -2(6)t \\y = -(6)t^2 \end{cases} \begin{cases} x = 2(6)t \\y = (6)t^2 \end{cases} (given a = 6) These are two different parabolas. If you wanted the same answer, then yes, you would have to use y = -at^2 .

Re: HSC 2018 MX2 Integration Marathon I = \int_{-1}^{1}\frac{x^{2018}\sqrt{1-\cos^2(\frac{\pi}{2}x^{2019})} \log_2(\sec(\frac{\pi}{4}x^{2019}))}{(3+\cos(\pi x^{2019}))(1+2018^x)}\, dx =\int_{-1}^{1}\frac{x^{2018}}{1+2018^x} \cdot \frac{|\sin(\frac{\pi}{2}x^{2019})| }{ (2+2...

But the substitution is not there to force you to use it, right? It's there because in 3U integration one is not expected to be able to find substitutions, only apply them.

The ratio of this GP is given by r= \frac{\sqrt 5 -\sqrt3}{\sqrt 5 + \sqrt 3} = \frac{8-2 \sqrt{15}}{2} The sum of a geometric series, if it converges, is given by S_{\infty} =\frac{a}{1-r} And of course, a = \sqrt 5 + \sqrt 3.

Re: HSC 2018 MX2 Integration Marathon I've reduced the integral to \frac{1}{4038 \pi \log 2} \int_{0}^1 \frac{\tan^{-1}x}{1+x}\,dx if someone else wants to finish it from this, but it seems very difficult. Maybe a different approach might be necessary?

Use the definition of a factorial: n! = n \times (n-1) \times (n-2) \times ... \times 1 In particular, n! = n \times (n-1)! This means we can write k! + (k+1)! = k(k+1)! + 1(k+1)!

Re: HSC 2018 MX2 Integration Marathon a very nice integral.

1) (k+1)! - k! = k!(k+1) - k!(1) Now take out a common factor... 2) Break up 100! into factors. The number of factors of 7 is the amount of times it can be divided by 7 . 100! = 1 \times 2 \times ... \times 100 How many numbers from 1 to 100 are divisible by 7? 7...

This question should go in the Maths subforum. x = -p \pm \sqrt{p^2-q} Because p^2 - q < 0, we can use the identity \sqrt{ab} = \sqrt a \sqrt b with a = -1 and b = q - p^2 . (Note that this identity is not valid if both a and b are negative) Hence, x = -p \pm i\sqrt{q-p^2}...

Let T_n denote the amount of water in the pond at the start of the n^{\rm th} week, expressed as a percentage of the original amount of water. For geometric series we have T_n = ar^{n-1} If the pond loses 7\% of its water each week, then that means that at the end of week there...

97.40, very happy because i was expecting 95-96.

already?!

Chemistry 86/100 88/100 87 5 English (Advanced) 79/100 81/100 80 5 Mathematics Extension 1 99/100 96/100 98 E4 Mathematics Extension 2 95/100 95/100 95 E4 Software Design and Development 94/100 95/100 95 6 I wish my 4 unit mark was a bit better. It tied with my SDD mark despite the...

Recall that z-w represents the vector from w to z . \arg (z-w) is the angle that this makes with the positive real axis. Draw an Argand diagram and mark on it the points -2 and i . Now imagine a variable point z moving around the diagram. We want to find the values of z for...

Cambridge Enhanced.

We have \frac 16 k(k+1)(2k+1) = \frac 1 6(2k^3+3k^2+k) and we want to obtain \frac 16 (k+1)(k+2)(2k+3) = \frac 1 6(2k^3+9k^2+13k+6) Now, \left[\frac 16 k(k+1)(2k+1) \right]+ (k+1)^2 = \frac 1 6(2k^3+3k^2+k) + \frac 1 6(6k^2+12k+6) ...

Re: HSC 2018 MX2 Integration Marathon I put my solution as an attachment so that it won't spoil the answer for other people attempting to solve this.