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$\noindent From your line$ $$1 - d = -1.75,$$ $\noindent the next line should be$$ $$-d = -2.75.$$ $\noindent This is because we are subtracting $1$ from both sides, and $-1.75 -1 = -2.75$.$

$\noindent Just in general, suppose $V$ is a vector space over a field $\mathbb{F}$, and let $\phi$ be a function and $W$ a set such that $\phi : V \to W$ is a bijection (so of course the inverse $\phi^{-1} : W \to V$ is also a bijection). Then we can make $W$ be a vector space over $\mathbb{F}$...

Yes. (I assume you meant the field to be R.)

$\noindent Of course, you can also do the SHM one as follows (this is also how to do it if it weren't a multiple choice question). Recall that if the particle's motion satisfies $v^2 = \omega^2 \left(A^2 - x^2\right)$ (where $A, \omega$ are positive constants), then the particle is undergoing...

$\noindent For the binomial sum, recall the following general series fact:$ $$\sum_{r = 0}^{n}a_{r} = \sum_{r = 0}^{n} a_{n-r}$.$ $\noindent (To see why this is true, you can write out both sums in long notation and you'll see that one is just written in reverse order of the other, with...

For the SHM one, the answer must be (D) — this is the only option with the correct units for a period (seconds). Note from the equation v^2 = a^2 - b^2 x^2, the LHS has units m^2 s^(-2). So the RHS does do. Thus a must have units m/s. Also, b^2 x^2 must have units m^2 s^(-2), so bx has units...

You can see a definition here: https://www.mathsisfun.com/definitions/locus.html .

$\noindent If you use the $u$-substitution method, you don't need to do the above. To answer your original question, the $x$ originally in the numerator can be removed by using $x = \frac{1}{2}(u+1)$ (i.e. solve for $x$ in the substitution you made).$

$\noindent This (finding constants $a$ and $b$ such that $x \equiv \underbrace{a(2x-1) + b}_{= 2ax + (b-a)}$) can be done by inspection or by equating coefficients in the identity $x \equiv 2a x + (b-a)$.$

$\noindent Welcome!$ $\noindent We should not equate $V$ to $108$ in that formula, but rather to $114$. This is because while the tap closes off in the ``final stage'', it imparts a final 6 L of liquid, and denoting $T \geq 10$ the time at which the final stage (tap closing off) begins, $V(T) =...

"Fully open" means its flow rate is 6/5 L/sec, so we want the time for which the tap is operating at 6/5 L/sec.

It's due to how you interpreted the question. Have a read of the discussion here (same question got asked and people appeared to misinterpret it): http://community.boredofstudies.org/13/mathematics-extension-1/350318/rates-change-question-help.html .

I don't think they'll put calculus into HSC Physics.

The sum of the first n positive odd numbers is n^2, so let k = 2n-1 (so there are n terms in the sum), so n^2 = 1000000, so n = 1000. So k = 2n-1 = 1999.

$\noindent So for example, $x = 390^{\circ}$ is a solution to the equation $\cos x = \frac{\sqrt{3}}{2}$, but since $390^\circ$ isn't between $0^\circ$ and $360^\circ$, we are not going to include it as a solution to the given equation with the given domain. We are only looking for the solutions...

$\noindent It means we are looking for the solutions to the equation $\cos x = \frac{\sqrt{3}}{2}$ that satisfy $0^{\circ}\leq x \leq 360^{\circ}$. If we didn't make this restriction on $x$, there would generally be other solutions to the equation (outside the given domain).$

Just use the Pender and/or Fitzpatrick textbooks (if you have access to them).

$\noindent The reason that $\omega$ still has dimensions of inverse time is that angles are dimensionless (being the ratio of two lengths). And you could do this question using usual methods (resolving forces to get simultaneous equations, remembering that the centripetal force is...

$\noindent Since $5^2 + 12^2 = 13^2$, we know from the auxiliary angle method that $x = 13\sin (4t +\phi)$, for some constant $\phi$. So differentiating this gives $v = 52\cos(4t + \phi)$, which we know has maximum value $52$.$

Quick answer: C is the only option with the right units (or dimensions), so it's the answer. :) (As we know, the dimensions of angular velocity are inverse time (units of which is s-1); units of option C is ((m s-2)/m)1/2 = s-1. It is quick to check that none of the other options have the right...