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its better, you can also reference your piece in reflection by name

It really depends on the quote / excerpt, you probably don't have to write anything verbatim just close enough so that the marker knows your referencing the quote.

Nah, I think that's a waste. But its better than not addressing the quote / excerpt you should have part of it in there, at least the general idea of the excerpt.

same

1 over anything has a horizontal asymptote, but yeah you have to visualise the graph

discriminant < 0 means no x intercepts so that when you do 1 over, there is no vertical asymptotes but there is still one horizontal asymptote. does that clear the confusion?

this is basically picking the answer that fits the best, which is why you might be confused. b^2-16a^2<0 b^2<16a^2 -4a<b<4a since a>1 then we can say that -1 \leq b \leq 4 satisfies the condition

Try solving it without considering the limits at all. You get the same answer, maybe its wrong since doing u=x^3 changes the limits so that you can't cancel out the integrals but I don't know how you properly account for that. Since wolfram alpha can't get it the antiderivative is probably...

so the answer to the indefinite integral is \frac{\ln(x+1) \ln(x^3+1)}{3} ? Can you differentiate to check?

How do you do this? \int_0^1 \frac{x^2 \ln(1+x+x^9+x^{10})}{1+x^3}dx Wolfram alpha can't get a closed form answer for this but I think it's \frac{\ln^2{2}}{3}

they got backlash from making it hard last year so they'll definitely make it easier. Not that much easier though.

Its fine, I spent about 3-4 hours when I did on each exercise when I first did them (will probs get high E4). Some longer, some shorter so 5hrs is ok. I'd advise not spending more than like 20 minutes on a single question unless you really want to, just to save time. Take your time to soak in...

dy/dx= f'(x) - g'(x) = 0

lets say y=f(x)-g(x) we have dy/dx=0 that means at every point the function doesn't change and therefore is a constant function. You can check this by integrating dy/dx=0. The only problem with this is that if the function is discontinuous, the value can jump without changing, think |x|/x...

its monic therefore it should be going upwards on each side therefore either c or d. since its divisible by x^2+x+1 you can think of it as p(x)=(x^2+x+1)(x-a)^2 so only the double root exists so it cant be D therefore C

this is how I did it. Maybe there's a faster way to prove the roots have magnitude 1 before finding them. I don't think there is but if that's given you could say z+1/z =2costheta and not have to solve the second quadratic.

ill do it one sec

that would be so slow, just do tywebb's method first to find complex roots then use that to find two quadratics

smart

How about terminal velocity? like capital V and lowercase v.