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you could quite easily fudge that working (just write out the expression for -2P/Q, then multiply each part of P by -2, on the previous line, maybe even simplify that on the line before it..)

it just seems like you're doing to much in the one step for a show question, but I'm not sure how marking works, so maybe it would be acceptable I suppose if you showed the derivatives of the separate fractions, and then went from there to the 2nd line it would be a bit more clear.. but I...

I think you would need more working than that for the differentiation

b) 1-cos45/1+cos45 = (1-1/rt2)/(1+1/rt2) * rt2/rt2 * (rt2 - 1)/(rt2 - 1) = (rt2 - 1)2/(2-1) tan222.5 = (rt2 - 1)2 hence tan22.5 = rt2 - 1

it's a revolution about the y-axis, so you would need pi[f(y)2 - g(y)2] to be able to inegrate one function, but the shaded area is between functions of x, not y, so you have to inegrate the y functions individually where they are the closest to the y axis. edit: and because the volume you're...

sin-1x + cos-1x = pi/2, so the third graph of b will just be the line y = pi/2 and a) is just 1/sqrt2 if you didn't know that by inspection, your working could be as follows; let x = sin@ = cos(pi/2 - @) sin-1(sin@) = cos-1(cos(pi/2 - @)) @=pi/2-@ @=pi/4 sin pi/4 = 1/sqrt2

I don't get what you mean. They didn't assume N was on the parabole - they solved for the point of intersection of the tangent to the parabola, and the normal of this tangent which passes through the focus, which gives N=(ap, 0), so the locus of N is y=0. Why is this incorrect, and how exactly...

yeah that's right. F = ma is the net force; F = m(a+g) is the thrust force.

If you want an idea of the material, the past papers are here: http://www.aso.edu.au/www/index.cfm?itemid=31 A mark of over 50% is apparently considered to be very good (high distinction worthy?), so don't worry if you can't do a lot of the questions.

you sure that's the right integral? calc101 couldn't do it, and wolfram used an 'exponential integral Ei'

There are 8 places to put vowels. The ways of ordering the vowels equals the number of combinations of 4 from 8, as the order is given. 8C4 = 8!/4!4!

is the vertex (0, 4a), or am I way off? x=-2ap, p = -x/2a y = ap2 + 4a = x2/4a + 4a dy/dx = x/2a = 0 when x = 0 (vertex) y = 0 + 4a so coordinates are (0,4a)

7!/2!2! (arrange the consonanats) 8!/4!4! (arrange the vowels among them) total = 88200 do you know if that's right?

1/4 ?

b) is just 4/25*3/7 = 12/175. I have no idea where they pulled those numbers from (same answer btw when simplified)

d) there are 6 possible groups of 3. There are 8C3 = 56 combinations of 3 boys, so the chance of a particular combination is 6/56 = 3/28

a) could be boy first or girl first. 4! ways to order the boys, 4! to order the girls, so the answer is 4!*4!*2 b) first treat the 2 boys as one person. There is now 7 people that can be ordered in 7! ways - and 2! ways of ordering the 2 boys c) treat all the guys as one person. Now you have 5...

1-cos2x = 2sin2x lim x->0 sinx/x = 1 so lim x->0 2sin2x/x2 = 2

a) force to acceleration would just have a gradient of the moon's mass b) not too sure exactly what this would be. I suppose you would just have increasing acceleration against decreasing mass. c) r3 is proportional to T2 (Kepler's third law) where r = distance from centre of the central body...

ii) dI/dx > 0 if P<0 both bracketed terms of P are > 0. First one obviously is since all terms>0 for the second bracket x2 + 64 + 225 - 16.sqrt[289] > 0 x2 + 17*17 - 16*17 > 0 x2 + 17 > 0 is true therefore it is always greater than 0 so P<0 for 2x<0 and P>0 for 2x>0 so brightness is increasing...