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ex-seftoner here as well... damn, i feel so old.

Degrees are the same. do something before you use partial fractions. long division is one way to go.

There are ways to find out which ports are open; but is there a way to force open a port? say for example, i want to open port 19 (Character Generator), is it possible? If not, how do you go about finding out how to open certain ports? thanks.

by subsitution, the idea is to get rid of all the old variables (x) and make everything the new variable (u), and this applies also to the dx's and du's... so if you have something like du=3dx, you want to find an expression for dx, like dx=1/3 du and then go back to the integral and...

a couple on the resources page, but not as many as Dr. Buchanan's page which goes back as far as 1982.

Just wondering if there is another website apart from Dr. Buchanan's 4u page that has a list of the CSSA papers. I don't like the re-typed setting, preferring the original spacing of the exams, even though the re-typed setting is much more neater and all... thanks.

use the apparent/absolute magnitude formula: m-M = 5log(d) - 5 m = apparent, M= absolute, d = distance in parsecs. To use this, though, you need to estimate the absolute magnitude. One way to do this is: 1. Look at the star, get a spectrum of the energy/light emitted from the star...

you get ur student card straight away, from what i remember.

just like u said... seeing that 3ax^2 + 2bx = 0, then either x=0 or the other nonzero one (too lazy to type it out). clearly, x=0 is not the double root, since P(0)=d (ok...assume d to be nonzero) then the double root must be the other one. so, P(other one)=0 and this gives you what you...

I suppose that explains it! thanks...

I don't know how long it's been, but no-one's jumped onto them yet...so here goes: 1. ON.OT = a^2 is what you want. Results falls out once coordinates of N and T are found. Try using P(acos theta, bsin theta) and then working out N and T. 2. if ellipse is x^2/a^2 + y^2/b^2 = 1, then...

it's done in Principia mathematica by Alfred North Whitehead and Bertrand Russell, though i don't remember which volume it is in.

congratulations! is it possible for a breakdown of the marks you got?

Nothing wrong with geometry (in general). And with money, just ignore all $ signs and you'll be fine.

how much time was given?

it's a consequence of the existence of the (additive) zero, and the distributivity of multiplication: 0+x=x x(y+z) = xy+xz i think. it's been a while since i looked at the field axioms.

1. convert \sqrt 3 + i into mod arg form first the answer is 2 cis (pi / 6) 2. Recall De Moivre's theorem which is (r cis(theta))^n =r^n cis (n.theta) 3. Apply to answer in part 1 with n = 1/2 you get \sqrt 2 cis {(pi/6 + 2kpi)/2} putting k=0 and k=2 gives you the answers you want...

hmmm...ok. if w is a cube root of unity, then you know that w^3=1. then for (a+bw+cw^2) / (b+cw+aw^2), multiply top and bottom by w...you get w(a+bw+cw^2) / w(b+cw+aw^2) the denominator expands to give a+bw+cw^2 which cancels out with the top, leaving you with just w.

multiply top and bottom by w, you get w as the answer.

it's been quite a while for me, but i remember when doing studies for trials and HSC, i would aim at getting through the paper in two hours for 4U...of course, this was done at home or wherever, without the added pressure of it actually counting... when you do sit the exam though, there are...