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it appeared in marking reports for 2001, so it's allowed I guess

0 is a constant :p

the '3 unit method' is easier, since the '2 unit method' requires you too see it in 1 step

here's another relation: 1) find the first derivative f'(x) for the function f(x) from first principles 2) find the second derivative by the definition of the derivative 3) find the third derivate, using the definition of the derivative etc. the nth derivative should turn out to be...

the way teh 2 unit teacher in our school teaches: notice d/dx [ e^(x^2)] = 2x(e^(x^2)) (1/2)d/dx [ e^(x^2)] = x(e^(x^2)) (1/2)(e^(x^2)) is the answer. the 'proper' way let u=x^2 du/dx = 2x (1/2) du = x dx (integrate) xe^x^2 dx = (integrate) (1/2) e^u du = (1/2)e^u + C =...

going in without a compass would be courageous

those schools should be grateful that their students are so enthusiastic. With me encouraging all my friends last year, only 2 joined extension 2

lol

binomials is a good bet. trig related rates those 3 in 4u books simpson's rule/trapezoidal rule?

almost it's (x-h)^2 = 4a(y-k) not (x-h)^2 =4a(y-k)^2

that's erh... 'obvious', it's something you should remember some 'reasons': 1.) cotangent means complementary tangent, since (pi/2 - t) and t are complimentary (adds up to 90), you have that identity 2.) tan (Pi/2 - t) = sin(pi/2 - t)/cos(pi/2 -t) = cos(t)/sin(t) = cot(t)

for P(x,y) to be on the locus, insert the conditions the gave you in equation form solve the equations, eliminating the parameters so that you get an equation with x,y being the only variables

because I know with that choice, the circle will be centered at origin and I can save myself doing algebra I am just choosing the orgin so that it divides the line AB externally in ratio of 1 to 4

I live near parramatta too

1) the charge (q) held by a capacitor proportional to the capacitance (C) of a capacitor and the voltage across it (V). (q=CV) a charged capacitor is connected in series to a resistor of resistance R ohms. a) explain why dq/dt = V/R b) show that dq/dt = q/RC c) initially the...

they are 2 unit level, I tried soemthing along those lines with a friend doing general, she had no problems.

no.. it's (1-0.3)^n <= 0.01 0.7^n <= 0.01 n log(0.7) <= log(0.01) n >= log 0.01 / log 0.7 n>= 12.9... n = 13 ======================= for my questions: a) the probability is 1 - (364/365)^100 so the odds are about 76 to 24 against. b) 19, yes not 20, 19! think why :P...

the affinity trial school certificate 2003 question 0 40% chance a) affinity's birthday party is on december 17th. and suppose affinity has 100 acquaintances. what are the odds of having two birthday parties clashing on that day? Assume of of affinity's acquaintances are party animals...

hmm might be next exam, might take for ever you left something out from the question you approach this by finding the probability for the minimum number of Times to spin such that the probability of getting NO E is less than 0.01 which is equivalent to the question (4/5)^T <= 0.01...

hmm, better if you choosed wisely so that the origin becomes the center let A be the point (4n,0) and B be the point (n,0) for some positive real n. let P(x,y) satisfies: |PA| = 2|PB| |PA|^2 = 4|PB|^2 (x-4n)^2 + y^2 = 4[(x-n)^2 + y^2] x^2 -8nx + 16n^2 + y^2 = 4x^2 - 8nx + 4n^2...