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what's the answer if it's not 19?

the OP was confused (asking why ln2 isn't differentiated) jetblack was just pointing out that ln2 is a number/constant, not a function, so it is treated the same as any other constant when differentiating

dm isn't rock :(

'at least two different scores' - what do you mean by that? the question doesn't sound complete

"So for all values of t, you have all possible values for an argument." θ1 =/= ±pi/2, so 2θ =/= pi, so you don't have -all- possible values (the point (-1,0) is missing from the unit circle). you can see from your algebraic solution that x=/=-1

for the second one z = x + iy z* = conjugate of z = x - iy zz* = |z|2 = x2 + y2 (|z|- iz) / (|z|+ iz) * (|z| - iz*)/(|z| - iz)* = (|z|2 - zz* - i|z|(z+z*)/(|z|2 + zz* + i(z-z*)) = -i(2x)/(2|z| + i2iy) = -i(x)/(|z| - y) * (|z| + y)/(|z| + y) = -i(x|z| + xy)/(x2 + y2 - y2) = -i(|z|/x + y/x) =...

for it to be a function, there can only be one y value for every x value so you can't have the ±. I'm not too sure why it's - rather than + though edit: oh right.. didn't notice the x<=1 part

well there's one intersection at m=ke, so if m > ke there would be 2, if 0 =< m < ke there would be 0. drawing a graph may help in understanding it: y=e^kx lies above y=kex for all values except one, so if the line increases in gradient it will lie above the exponential for a range of values...

yeah it's m/(m+n). they must have taken m+n to be 1 for some reason (if m=2, n=3 for example, obviously XY is not twice as big as BC: it is 2/5 the size)

if you have an inverse function, the x and y change places, since y=f-1(x), x = f(y) dy/dx at x=-1 in f(x) is 3-12 = -9. In the inverse function, this is dx/dy, so dy/dx is 1/(-9) = -1/9

to order the vowels, rather than treating them as different letters I would just straight away consider them as identitcal (since they can't be ordered), so to order them it is 8C4 rather than 8P4/2 (ie. it is the number of ways of choosing 4 of the 8 spots to have a vowel in them). this...

you would integrate both sides, so you get 1/2 v2 = F(x) + C

wtf was the argument even about :confused:

when i have dv/dt = f(x) I just multiply out by .dx to make it possible to integrate the RHS, and go from there (dv/dt .dx = dv .dx/dt = v.dv) do we really need to write it as d(v²/2)/dx / is there really any difference?

difference of 2 squares, then difference and sum of 2 cubes.. always do the difference of 2 squares before the 2 cubes, or it will be much harder

i was just writing out the answer and thread got deleted :( d(arctan 3/x)/dx = (-3/x2)/(9/x2 + 1) just using derivative of inverse tan ..and you can do the rest

your answer sounds correct to me. they give you the order of days so you don't need to bother with the combinations (5C3)

you could do it like cosec2θ = sec2[pi/2-θ] | sec2[pi/2-θ] .dθ = -tan[pi/2-θ] = -cotθ

from there you would just go = | cot2θ.dθ = | (csc2θ-1).dθ = -cotθ - θ + C = -[√16-x2]/x - sin-1[x/4] + C

incantation - onward to golgotha