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lol no need to rlly prep for the test unless ur aiming for high level maths olympiad. There should be some past papers on the amt website though

u can -2x + 2x on the numerator instead, then u get (x-1)/x + 2x/(x^2-x) which is a very simple integral. for c tho u can literally split it up then evaluate integral of 1/(x^3+x) by dividing both sides by x^3 and doing reverse chain rule. +-x^2 on the numerator also works pre nicely. d, you can...

also there are qs like a in which u should be able to avoid partial fracs pre easily but the decomposition is so free u might as well just do it

I disagree there are certain qs that definitely require partial fractions it just so happens that when u have degree of numerator = denominator u can usually avoid it

hmm interestingly b and c don't require partial fractions at all

q11a it's intended for u to use the fact that a = dv/dt = dv/dx x dx/dt = vdv/dx (or alternate form which is d/dx(1/2v^2) since suvat equations are non-assumable in 4u

well hopefully not 😊

they set hard tests 🤷‍♂️

I think my school is just too good🤷‍♂️

who even are you and how can you be so sure that is a lie?

I assume you'd be fine with part a, and most of part b except perhaps the finding s part. For s, one way to approach it would be to let the vector = [x, y, z]. Then since it's perpendicular to the face PQR, it is perpendicular to both PQ and PR. We find both of these, then dot [x, y, z] with...

looks like it ye

typing this on word is a bit messy so lemme know if i made a mistake!

u.v/|u||v| = v.w/|w||v| = u.w/|u||w| = ½ w = (a, 0, c) v = (0, b, c) u = (a, b, 0) using these and the first line, we get: b^2/sqrt(a^2+b^2)sqrt(b^2+c^2) = ½ b^2 = ½sqrt(a^2+b^2)sqrt(b^2+c^2)=1/2(sqrt4b^4) c^2 = 1/2sqrt(c^2+a^2)sqrt(c^2+b^2) = ½(sqrt4c^4) a^2 =...

hi I would like to get free mx1 as well as free tutoring for english advanced, maths extension 2, chemistry and physics. I just recently got back my half yearlies, and honestly, I'm devastated. I came 3rd last for physics, failed chem and mx1, got below average in english, and got 20% below...

GRUUUUUUUU

Me and quite a few people ik started 4u rlly early (like in yr 10) and the start can be very confusing at first, but if u just stick through it and grind out a bunch of qs, you should be fine. Although Complex I is prolly one of the easier topics in 4u, it could be difficult especially if u...

you don't actually need to construct a triangle although that's probably the most obvious method. You can also note tan2theta = +-sqrt(sec^22theta-1). since tan 2theta is in this case negative since 2arcsin3/4 is in the 2nd quad, tan2theta = -sqrt(sec^2(2theta)-1) = -sqrt(1/cos^2(2theta) -1 )) =...

yep looks legit

change sin2x into (sinx+cosx)^2-1