|a|e - |a| = |a|(e-1) = 1
Therefore |a| = 1
a= 1 or a = -1
Oh and also you can do the other vertex:
|a|e + |a| = |a|(1+e) = 1
Therefore |a| = 1/3
a = 1/3 or a = -1/3
So I've gotten bored in the last couple of months and have learned a lot of kanji which aren't in the syllabus. I can use these right? They're not very complicated ones, I've just learned basically all the kanji that are taught in grades 1 and 2 (going of Jisho.com) that are not in the regular...
a)
5V between the electodes, Energy = Vq = 5x1.602x10^(-19) = 8.01 x 10^(-19).
Hence, (mv^2)/2 = 8.01 x 10^(-19). v = 1,326,159 m/s = 1.33 x 10^6 m/s. This is the minimum. So the maximum speed will be this plus the speed of the electrons which are emitted by the highest frequency of light...
I like math, and I want to study math. Hence, I'll probably do the Dalyell Scholar's Advanced Science course at USYD and major in math, though I'll probably end up double majoring in physics as well. I also have space engo as my second preference, because that's awesome, but I think I'll go with...
I go to a school that was ranked in the 400's last year, and in my 4U class, I got a 98 as well as another kid, the behind us was a 91, 3 people in the mid 80's, and a person in the mid 70's. So, even in a bad ranking school, you can get a pretty good class in one year.
Thank you!! I'll look at the tutorial questions. So, is it not worth my time actually pracitising past papers?
Oh I definitely understand this haha, which is why I'm resorting to asking here. It was obviously extremely hard to find papers online to practice.
Hi, does anyone know where I can find past papers for the courses MATH1901 and MATH1903 at USYD or any other similar courses at other universities? It's basically just calculus. Keep in mind I am NOT a university student, I've just been doing some learning by myself and wanted to do some...
Re: HSC 2017 MX2 Marathon ADVANCED
For Q1 instead of (n-1) at the top it should just be n-i.
$Anyway for that one in the prove for $k+1$ part (the induction step is it called?) get $(fg)^{(k+1)}$ which will be the derivative of $(fg)^{(k)}$, and this can value can be derived from the...
i) L = 2B, so dL/dt = 2dB/dt = 1 cm/min
ii) A = LB = 2B^2, so dA/dt = 4B*dB/dt = 2B cm/min
iii) P = 2L + 2B = 6B, dP/dt = 6*dB/dt = 3 cm/min, which is independent of B, and so when B=8 the perimeter changes at 3 cm/min.
EDIT: Nevermind look above
$Ok, so this is saying that $ ax^4 + bx^3 + x^2 + 4x + 2 = (2x^2 + 2x - 1)Q(x) + (2x + 3)$ where $Q(x)$ is some polynomial. \\Now, substitute in the zeroes of $(2x^2 + 2x - 1)$, which are $x=\frac{-1+\sqrt(3)}{2}$ and $x=\frac{-1-\sqrt(3)}{2}$. As $(2x^2 + 2x - 1)Q(x)=0$ for these values you can...
Coincidentally I've been playing that Atheist album so much lately, it's really good. And yeah that's some terrible cover art, so here's one with an amazing art: