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Almost correct! (The y-intercept is not 9, but 3.) Does that help you with evaluating the integral (remembering that it is the "area under the curve")?

Hint for the first one: what shape is the graph of the function? (Alternative (slower) method: use a trigonometric substitution if you've learnt them.)

$\noindent Oh, I realised tazhossain99 answered the question above. And his working out is actually showing it for the new $u$, not new $u_y$.$

$\noindent If the angle of projection is $\theta$ and the speed of projection is $u$, then $ y = ( u\sin \theta)t - \frac{1}{2}gt^2$. The maximum height thus occurs when $t = \frac{u\sin \theta}{g}$.$ $\noindent Therefore, the maximum height is$ $$\begin{align*}H &= (u\sin \theta)\times...

Re: Questions that are driving me crazy cause I can't get the answers that match the For the latus rectum one, the point S should be (0, a), not (0, 1). Consequently, the integral's limits should be from 0 to a.

Re: Questions that are driving me crazy cause I can't get the answers that match the Possibly you have set up the correct integrals but are just making errors in evaluating them. What integrals did you set up?

$\noindent By the way, by either using a substitution or considering a translation of a graph, we have$ $$\int_{3}^{4}(y-3)\, dy = \int_{0}^{1}y \, dy,$$ $\noindent and you should find this latter integral is easier to calculate.$

I think you have linked the wrong picture for your question.

You have the correct integral and correct antiderivative. You have just made an error in evaluating $$\left[\frac{y^2}{2} -3y\right]_{3}^{4}.$$ So you can have another go at evaluating this (carefully). Make sure to get all signs right when evaluating it (you have a sign error in your...

What was the integral you obtained when trying to calculate the volume?

Critical points of a differentiable function should just be points where the first derivative is 0. You may want to consult your textbook to see what definition it gives for a critical point. Wikipedia: https://en.wikipedia.org/wiki/Critical_point_(mathematics)

This is effectively disclosing (at least some of) their names. Also, another game one can play is (for those who don't know the answer) to guess how many 99.95's were attained in James Ruse (or any other school of your choosing).

It's still very much possible to get 99.95 ATAR.

You should be able to use either formula, provided you understand why it's true. (You can even use it if you don't understand why it's true, but this is not recommended.)

I think it was there before. I remember answering it in my head but then forgetting to comment on it in my post. Haha

Well I did say I was sketching a method (i.e. not giving the full answer).

$\noindent A sketch of a method. Rewrite the quadratic form $ax^2 + 2bxy + cy^2 \equiv \mathbf{x}^{T}A\mathbf{x}$ $\Bigg{(}$where $\mathbf{x} = \begin{bmatrix}x \\ y \end{bmatrix}, A = \begin{bmatrix}a & b \\ b & c\end{bmatrix}\Bigg{)}$ as $\lambda_{1}u^{2} + \lambda_{2}v^{2}$, where...

$\noindent If you know in general how to sketch the graph of $f'(x)$ from $f(x)$, then you know how to sketch $f''(x)$ from $f'(x)$.$

I don't think this argument would explain why English needs to be made compulsory, but rather that some particular subject should be compulsory (for scaling purposes).

Probably best to use symmetry. Don't need to use conditional expectation. But for the particular question you asked (with blue and red balls), it's also easy to do it by conditioning on the first draw's colour, and noting the number of draws to get the remaining colour after that is just...