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if the answers say that 1. gives a negative result, then the answers are wrong for 2. you got a negative (unless you're saying the answers give a positive?)

\frac{d }{d x}x arcsin[x] = acrsin[x] + \frac{x}{\sqrt{1-x^2}}\\ xarcsin[x] = \int arcsin[x]dx + \int \frac {x}{\sqrt{1-x^2}}dx\\ \int arcsin[x]dx = xarcsin[x] - \int \frac {x}{\sqrt{1-x^2}}dx\\ \int arcsin[x]dx = xarcsin[x] + \sqrt{1-x^2} + C

you limit the domain first off with the inverse, the range must be the same as the domain you chose (the range and domain swap around, when you interchange the x and y), so choose the root that satisfies this eg. if you started by limiting the domain to x<0 and for the inverse you got to y =...

y' = xa y = xa+1/(a+1) +C just do this for them all

kujah is correct (his post under this) and as his method shows, it's usually best to set these up as algebra problems, letting the unknown quantity be x (for example)

the only topic that will probably be at all difficult will be harder 3u

I mostly think of it as ordering (multiply by n!) and unordering (divide by n!) rather than P and C +also, perm/comb = best 3u topic

yeah, it's integration

missingno was craaaaazy lvl 255 pokemon ftw yo

you have: y = (x+5)² what would be a better way to get x on its own than expanding?

I lost like 12 lvl 100's one time :( (I don't remember if I saved over the file, or if the game reset itself)

^getting to sleep maybe? (though could just be a case of needing to do that)

\int_{0}^{a} f(2a-x)dx\\ let\ u = 2a-x\\ du = -dx\\ when\ x=0,\ u=2a,\ when\ x=a,\ u=a\\ \int_{2a}^{a}f(u)[-du]\\ \int_{a}^{2a}f(u)du\ =\int_{a}^{2a}f(x)dx\\ \int_{0}^{a}f(x)dx+\int_{a}^{2a}f(x)dx\\ =\int_{0}^{2a}f(x)dx you should start with the LHS side though

the first part of the question is equivalent to saying "how many ways can you choose 3 marbles from 7 marbles to be made blue?" which is 7C3 = 7!/3!4! alternatively, arrange the marbles in a row (7!) then divide out by 3! to unorder the blue marbles, and divide out by 4! to unorder the yellow...

@kurt: do you know induction (for divisibility)? (you let the expression for k=n equal, say 9m, then show that the expression is also true for k=n+1 if it is true for k=n) so yeah, you get the k+1 expression in the form of 9M, but in order to do this you will need to do a separate induction...

yeah, the alkanes dissolve due to the non-polar end of the ethanol, which water doesn't have there would still be dispersion forces due to the water however

do it the normal way, but I think at some point you need to do an induction proof within the induction proof

say the projectory of the stone is represented by y=-Cx2, so that the origin is the maximum point/height if d is the distance the stone travels to get from a height of 2p (max) to p, then p = Cd2. The distance that the stone travels in going from 2p to the ground would then be rt2 d (same...

^ the point is, in 2 unit you apparently don't do substitution at all