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lol yeah. I mean, physics as elite? isn't it designed so someone doing general maths can still go well in it? mostly rote learning and plug-and-chug equations. MX1 and chemistry have been really easy so far too. I don't really know that there are any year 11 courses that you can really describe...

Re: maths preliminary we've done absolute value/inequalities, trig, co-ordinate/plane/circle geometry, functions

you want to show that tanxtan2x = -(2) + [cotx][tan2x], yeah? = (2tan^2 x + 2 - 2)/(1 - tan^2 x) = -2 + 2cotx tanx/(1 - tan^2 x) = -2 + cotx tan2x

yeah d2v/dt2 is the 'acceleration' (rate of change of rate of change in volume) - and this is what is used to find the greatest 'velocity' (change in volume) to find the greatest (or least/stationary) value of something you find when it's change is zero - so you will be looking at acceleration...

first one c[c^2 - sc + t]/c[c^2 + sc - t] = [c^3 - s[1 - s^2] + s]/[c^3 + s[1 - s^2] - s] = [c^3 + s^3]/[c^3 - s^3] = [1 + tan3]/[1 - tan3] second [s/s + 1/s + c/s]/[s/s + 1/s - c/s] cancel out s and multiply top and bottom by [s + 1 - c] = [s^2 + c^2 +1 + 2(sc + s c)]/[s^2 + 2s + 1 - (1-s^2)]...

net force is the unbalanced for acting on the object that causes it to accelerate - the sum of the vector forces. An object in orbit does have a net force since it is constantly accelerating

make a list of the mistakes you usually make, and check each question according to the common mistakes. if you keep trying that you should get quite efficient at it, and hopefully decrease in the number of mistakes you end up making

The distance from a point (Q) to a point that it is rotated from (F) will remain constant. A 180* turn makes a straight line..

eh? ln x = a, therefore e^a = x. is true by a definition. I was just saying that e^ln x = x is also true basically by definition, so you don't really need to prove it

q1,3 what Trebla said ^ q2. you don't really need to do it like that. It's just using substitution and evaluating dx/du so it will be in a form you can inegrate q4. the book is right. I forgot to combine logs.

e to the power [the power e must be put to to make it equal to x] = x or you could say ln x = a e^a = x e^ln x = e^a = x but the definition is sufficient imo

it won't help when differentiating a linear function because there is no max/min rate of change - so instead you find whether the least or greatest possible value as implied by the question has the greater gradient (t = initial or final) if you're finding thg greatest rate of change in a...

y' = 6/(5-2x) dx = -3(-2/(5-2x)) y = -3log(5-2x) +C 2. y' = 6(1-2x)^-2 dx let u = (1-2x) y' = 6u^-2.du.(dx/du) dx/du = -1/2 y = -1/2*-6u^-1 +C y = 3(1-2x)^-1 +C 3. y' = (4-3x)^-1 dx = -1/3 -3/(4-3x).dx y = -1/3 log(4-3x) +C 4. dy/dx= 1/(2x) and y=2 when x=e^2 y' = 1/2*2/2x => y = 1/2 log...

the fastest rate according to your second derivative is NOT 2 seconds. It implies that there is never a fastest rate. This is mathematically true because if time were negative, then the rate of volume decrease would be greater than at t=0. In this case, you find the time of fastest decrease not...

you're saying that dA/dt = -9.8 ms^-2 shouldn't it be -9.8 ms^-3 like how dv/dt = ms^-2, whereas v = ms^-1 would you agree that it is a very poorly written question?

why does it have the same units as acceleration?

wtf how do we even learn enough stuff to get anywhere near that much per week

I mean the + bit, which you left out (just checking that I solved it correctly)

weren't we just given A = -9.8, not dA/dt? (obviously you were right in what you did, but did the question state that?)

is it supposed to be [ax.x1 + by.y1]?