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It's like if you want to be a good pianist, you have to do a lot of scales and a lot of practice, and a lot of that is kind of boring, it's work. But you need to do that before you can really be very expressive and really play beautiful music. You have to go through that phase of practice...

1 + \left(\frac{y}{x}\right)^3 = \frac{8}{x^3} \therefore \frac{y}{x} = \sqrt[3]{\frac{8}{x^3} - 1} |x| \rightarrow \infty\, , \,\, \sqrt[3]{\frac{8}{x^3} - 1} \rightarrow \sqrt[3]{0 - 1} = -1 \therefore \frac{y}{x} \rightarrow -1

I completely disagree with the above post. Every preliminary course builds up to the HSC course. If you want to do well in the HSC the most important thing is understanding. And to understand the HSC course you must understand the preliminary course. For example if you didnt pay attention in...

There are 4 general cases, you should be able to adapt it to present a solution. \frac{d}{dx}(sinx) = cosx \frac{d^2}{dx^2}(sinx) = \frac{d}{dx}(cosx) = -sinx \frac{d^3}{dx^3}(sinx) = \frac{d}{dx}(-sinx) = -cosx \frac{d^4}{dx^4}(sinx) = \frac{d}{dx}(-cosx) = sinx As it is cyclic...

(a) y^2(x^2 + 1) = x^2 y^2 = \frac{x^2}{x^2+1} We must show x \geq \pm \sqrt{\frac{x^2}{x^2+1}} ie x^2 \geq \frac{x^2}{x^2+1} x^2(x^2+1) \geq x^2 x^2(x^2 + 1 - 1) \geq 0 x^4 \geq 0 which is true (b) Note that because y is plus and minus then the graph is the graph of...

A voltage will register if there is a current. There is a current when there is a full path. H2O does not in fact conduct electricity, this is because it is a molecule and taking away electrons would be difficult. But when it combines with other elements we get a solution which is conductive...

Our school has done (in order) complex graphs polynomials conics

You should use more than 1 text book. That way if there is something you dont understand in one it may be better explained in the other. Also you have twice (and with cambridge three) times as many problems of difficulties ranging from super easy (MIF) to extra hard (cambridge).

Its a tossup between princeton and harvard.

If you have a lot of time: Cambridge If you want to get help with difficult problems preparing you for the level of difficulty of HSC: Fitzpatrick If you want to do really simple problems and not really understand mathematics: Maths in focus

Cant find it, could you provide a link?

I was wondering what mark out of 100 in the HSC Physics exam would you need to get to achieve a state rank?

He found the modulus from the complex number. That is to say, given a complex number x+iy, whose co-ordinates are (x,y) on the Argand plane, the modulus |x+iy| is equal to the distance from (0,0) and (x,y) ie \sqrt{x^2+y^2} and the same for |a+ib| *sorry nrlwinner for butting in ;S*

a^2b + ax^2 + abx = 2bx^2 + a^2x (ax^2 - 2bx^2) + (abx -a^2x) + a^2b = 0 (a-2b)x^2 + (ab - a^2)x +a^2b = 0 x = \frac{-(ab - a^2) \pm \sqrt{(ab - a^2)^2 - 4(a-2b)(a^2b)}}{2(a-2b)} = \frac{a(a-b) \pm \sqrt{a^2(b-a)^2 - 4a^2(a-2b)(b)}}{2(a-2b)} = \frac{a(a-b) \pm a \sqrt{(b-a)^2 -...

I hated Bio (droped like a bag of sh!t)!

Allow me to explain ;P lol 1 + w + w^2 = 0 \implies 1 + w^2 = -w \therefore (1 + 3w + w^2)^2 = (-w + 3w)^2 = 4w^2 ;)

You have to know these two rules 1 + w + w^2 = 0 w^3 = 1 ie w^3 \cdot w = 1 \cdot w These are explained in both Excel MX2 and Cambridge (i think) So (1 + 3w + w^2)^2 = (1 + w + w^2 + 2w)^2 = (0 + 2w)^2 = 4w^2 and (1 + w + 3w^2)^2 = (1 + w + w^2 + 2w^2)^2 = (0 + 2w^2) = 4w^4 = 4w EDIT...

Recall that the initial quantity is "A", ie V = A So When its three quarters full, V = 3/4 V. .: 3/4 V = V e^-5k .: 3/4 = e^-5k Then just solve via logs for k

Could i do a B Science with majors in maths and psych and minor physics?

Hi. I am interested in doing an BSc(adv. maths) with honours. Is it possible to 'minor' (?) in physics? Also, could i get a BPsy(Hons) at the same time as my BSc(adv. maths)? Thanks =)