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well it makes it easier to post your solution on here if you do write those lines out also, in an exam, would you just write Int cotx.dx = log(sinx) + C [by inspection] ? or would you have to show the steps like Aerath did? (can a teacher mark you down for not including working in this question)

cos(A-B-2@) =cos(A-B)cos2@ + sin(A-B)sin2@ = (cos2@ - sin2@)(cosAcosB + sinAsinB) + 2sin@cos@(sinAcosB - sinBcosA) I wouldn't imagine there'd be a simpler way to do that one than regular expanding

wouldn't doing this by inspection still use the same method that Aerath used, just in your head? "cotx = cosx/sinx = u'/u. where u = sinx, so u' = cosx, so the answer must be log(sinx) + C"

how do you do it? (the show AFDC is cyclic bit)

4. total of 10 bolts P = (1/10)3(9/10)7*10!/7!3! 5. 4/10 * 3/9 * 2/8 = 1/30 b) combiantions: 4 combinations of 3 red 6*6 = 36 combinations of 2 red, 1 black 4*15 = 60 combinations of 1 red, 2 black 6*5*4/6 = 20 combinations of 3 black so greatest probability of 1 red 6. P4 = 1/4 = 3/12, P2 =...

kk, thought it was. btw, an easy way to solve this type of locus would be to notice that the locus will have the same gradient but the y intercept will be K times as much as the old intercept, where K is the ratio of the line intersected on the origin-side eg. for "2x-3y+6=0 and the point Q...

you've never used pythag in 3-d? say you have 3 lines; x, y and z and you want to find the diagonal xyz then for the xy diagonal (xy) xy2 = x2 + y2 and for the the xyz diagonal (xyz) xyz2 = xy2 + z2 = x2 + y2 + z2 (hopefully the way I explained the diagonals makes sense) + what doesn't make...

1. let the x coordinate be x1 then 2x1 = x y1 = 0.5(8x1 + 3) 2. 5/3x1 = x y1 = 2/5(5/3 x1+3) 3. just use pythagoras

yeah.. or you could just

I suppose if there's nothing else you can do, you could first prove that the altitudes are concurrent, as shown here - http://jwilson.coe.uga.edu/EMT668/EMT668.Student.Folders/BrombacherAarnout/write-up%203.aarnout/orthocenterproof.html

for the second one, would we be allowed to use the rule that the altitudes of a triangle intersect at a common point?

isn't similar |||? I reckon if a certain abbreviation isn't going to come up much just write the whole word, and if it is then putting a key might be acceptable, like "|| means parallel"

we most likely don't, but no harm in learning it anyway

I think it's measured by placing the substance between the plates of a capacitor and then measuring the change in the electric field of the capacitor (caused by the lining up of the molecules so that +ve end faces -ve capacitor plate, -ve end faces +ve capacitor plate) then do some calculations...

q is energy so you would expect the units to be joules - and the expression backs this up: mass*energy per gram per degree*number of degrees

yeah, great band. too bad the vocals are a bit.. ehhh

Re: Stupid chem assignment! intramolecular = within the molecule (eg C to O) intermolecular = between molecules (eg one glucose molecule to another)

in 1. you should get pq = -2, and y0 = apq = -2a, so the midpoint of 2a and -2a is the x axis

lol why do you have a inside and outside the square root? if you square your formula you get v2 = 2ad which is the same as v2=u2 + 2as when the intial velocity is 0

focus is at (0,a) AB is (ap, 0) , (aq, 0) it's midpoint is (a(p+q)/2 , 0) gradient of line through focus and midAB is -a/[a(p+q)/2] = -2/(p+q) y-a = -2x/(p+q) 2x + y(p+q) - a(p+q) = 0 chord of contact is the line joining P(2ap,ap^2) and Q(2aq,aq^2). it's gradient is a(p+q)(p-q)/2a(p-q) =...