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What do you want to do in IT? USYD is better. Not because the course is better, or that it has a higher ranking (predominantly determined by citations to their postgraduate research papers), but because you'll be surrounding yourself with more brilliant, competitive and motivated people (who...

Let me preface this by saying that I did a software engineering degree at UNSW so account for bias as you see fit. But UTS sucks. Their computing graduates have no skills, no programming ability and no motivation. Probability suggests that there are exceptions to this rule, but I have yet to...

Out of the three you will only need to do physics in computer engineering. You are not considered an engineer if you do computer science. There is a lot of maths to be done in first year regardless, but you'll find that most of what you learn in first year math is not as important as they make...

you also need to test for the zero vector, which also proves that the set is non-empty because the zero vector can exist without closure under addition or scalar multiplication: eg let S = {x in P3(R) | x13 + x22 + x3 = 0} If the zero vector (0,0,0) does not exist then clearly it's not a...

i got @=n(pi), @=n(pi)+(-1)^n(pi/2), @=n(pi)+(-1)^n(-pi/2) the answer is @=npi/2 if you take a look closely they're both the same answer... but in any case you're both wrong the answer should be x = n.pi +- pi/4 and sin3x=sin2x, i thought it would be quite simple, like...

there are questions which are designed to separate the better students to the best students who usually just happen to be 3 unit students... but in any case there will never be a question in the 2 unit paper which is out of its syllabus i.e. any extension-only topics, if that's what you were...

hint: sin(2x) = 2sin(x)cos(x)

given: 1. DCF is right angled. 2. D and F are circle centres. (normal of tangent to a circle passes circle centre) rtp: DEF is right angled. let E be the 2nd intersection pt between the circles. construct DE & FE. 1. CF = EF (equal radii) 2. CD = ED (similarly) 3. DF is common...

same concept just draw the diagram and it should become clear what to do

since acceleration is the rate of change of velocity, the values of the acceleration curve correspond to the gradient of the velocity curve at a given t so when the velocity curve has a turning point, the acceleration curve crosses the x-axis(time axis). it's better to understand why in your...

basically all you need to know is that v = dx/dt, a = dv/dt the velocity-time graph is the derivative of the displacement-time graph so every value in your velocity curve corresponds to the gradient of the displacement curve for any given t. as a result, when the displacement curve has...

that can't be right... it's the area bounded by x-axis, parabola and tangent rotated around the x-axis, so the parabola does affect it. draw up a picture if you can't visualise it you need to find the volume of the area under the parabola, and subtract the volume under the line (which...

cost = (150 + v^2/80) * time velocity = displacement/time c = (150 + 500^2/(80t^2))t c = 150t + 3125/t dc/dt = 150 - 3125/t^2 stat points occur when dc/dt = 0 i.e. when 3125/t^2 = 150 -> t^2 = 3125/150 = 125/6 t = ~4.56 (t > 0) d/dt(dc/dt) = 3125/3t^3 > 0 (t > 0) so this is a...

f'(x) = e^x + e^-x -> f(x) = e^x - e^-x + C if the function has a y-intercept at 3, then f(x) = 3 when x = 0, 3 = e^0 - e^0 + C -> C = 3, f(x) = e^x - e^-x + 3

also, f(x) = sin(1/x) domain: all real x, x != 0 range: -1 <= y <= 1

1. f(x) = Sqrt(9 - x^2), -3 <= x <= 0 x = Sqrt(9 - y^2) x^2 = 9 - y^2 y = +-Sqrt(9 - x^2) f^-1 = -Sqrt(9 - x^2) (Dom(f): -3 <= x <= 0 -> Range(f^-1): -3 <= y <= 0) so range: -3 <= y <= 0 at y = 0, x = +-3 at y = -3, x = 0 domain: -3 <= x <= 3 2. domain and range of a function which...

the range is the set of all possible y values for given domain y = 2^(-x) = 1/2^x the domain of the function is x > 0 since the function is monotonic (decreasing), we can expect the extremes of the range to correspond to the extremes of the domain (0 to infinity) by checking the limiting...

1. y = 2^(-x), x > 0 at x = 0, y = 1 as x -> infinity, y -> 0 so the range is 0 < y < 1 f^-1(x) = -ln(x)/ln(2) from the original function, domain: 0 < x < 1 range: y > 0 2. y = x^2 + 2x, x > 0 x = y^2 + 2y x = (y + 1)^2 - 1 y = +-Sqrt(x + 1) - 1 f^-1 = Sqrt(x + 1) - 1...

<font face = "courier new"> 3*Sqrt((x + 6)^2 + (y - 5)^2) = 2*Sqrt((x - 3)^2 + (y + 1)^2) 9(x^2 + 12x + 36 + y^2 - 10y + 25) = 4(x^2 - 6x + 9 + y^2 + 2y + 1) 9x^2 + 108x + 324 + 9y^2 - 90y + 225 = 4x^2 - 24x + 36 + 4y^2 + 8y + 4 5x^2 + 132x + 5y^2 - 98y + 509 = 0 </font> the...

the arc length would be given by 2 * pi * r * angle/2pi = r * angle then area = pi * r^2 * angle/2pi = 1/2 * angle * r^2 equate and solve simultaneously.. let angle = @ r@ = pi/5 ..(1) (@r^2)/2 = 3pi/10 ..(2) (2)-> @r^2 = 3pi/5 @ = 3pi/(5r^2) ->(1) 3pi/(5r) = pi/5, r = 3cm ->(1) @...