Search results

Re: HSC 2015 3U Marathon What do you mean by ends of a function? Also dv/dt is the rate of change of velocity, aka acceleration, so dv/dt=0 would give you where acceleration is zero, not where it is greatest. You do have the right answer though.

Standard cell potential calculations.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Let $\angle{DPA}=\alpha$. $\angle{BPE}=\alpha$, as $\angle{DPA}$ and $\angle{BPE}$ are vertically opposite angles. $\angle{PQA}=\angle{DPA}=\alpha$, and $\angle{BQP}=\angle{BPE}=\alpha$, as per the alternate segment theorem...

If James has to search all three levels before finding his car, then it has to be on the last level he searches. At first, there are three levels he can choose from to search. Two of the levels don't have his car, so there is a 2/3 chance that James will search a level that does not have his...

Re: HSC 2015 2U Marathon Take the cube root of both sides, since \sin^3{x}=\left (\sin{x} \right )^3.

Must be something wrong with that graphing calculator. Here's what WolframAlpha shows: http://i.imgur.com/SSPcBxH.png

HCl + H2O -> Cl- + H3O+. HCl is a strong acid, meaning it ionises completely. Because of this, the reaction can't go the opposite direction. NH3 + H2O <--> NH4+ + OH-. NH3 is a weak bsae, meaning it ionises partially. Because of this, the reaction can go either direction, allowing for the...

$Managed to find the equations of the two circles in the first quadrant. Since the x-axis and y-axis are both tangent to the circle, the x-coordinate of the centre is equivalent to the x-intercept of the circle, and the y-coordinate of the centre is equivalent to the y-intercept of the circle...

Re: HSC 2015 3U Marathon Okay I think I've figured out your problem. Your calculator isn't in degrees, it is in gradians. EDIT: Nevermind didn't see your post.

Re: HSC 2015 3U Marathon Well when I use my calculator to find \theta, I get 26^{\circ}34'=27^{\circ}. Are you sure you entered it into your calculator correctly?

Re: HSC 2015 3U Marathon Did you end up with \tan{\theta}=\frac{1}{2} though?

From the reaction, you can see that CO and H2 react in a 1:2 ratio. This means that for every 1 mole of CO, 2 moles of H2 is required to produce CH3OH. This also means that their molar concentrations are also in the ratio 1:2. Let's number the y-axis from 0M to 10M, with increments every 1M...

As CH3OH is a product, its initial concentration would be zero as the reactants haven't reacted to form it yet. This means the bottom dashed line represents CH3OH. Over time as the CH3OH is being made, the middle dashed line decreases so that it has a value half that of the solid line. This...

Re: HSC 2015 3U Marathon Which expansions are you comparing the coefficients of?

Re: HSC 2015 3U Marathon Pretty much what I did. You have skipped steps, but I'll assume you did it correctly. Next question: $Using the binomial theorem, prove that $ r\binom{n}{r}=n\binom{n-1}{r-1}$.

Re: HSC 2015 3U Marathon $Show that $(\cos{t})^{2n+1}=\cos{t}(1-\sin^2{t})^n$. Hence, using the binomial theorem to expand $(1-\sin^2{t})^n$ where $n$ is a positive integer, show that $\int^{\frac{\pi}{2}}_0 (\cos{t})^{2n+1} \mathrm{d}t=\sum_{r=0}^{n}(-1)^r \cdot \frac{1}{2r+1} \cdot \binom{n}{r}$.

$Since B lies on the line $y=1-3x$, the coordinates of B can be written as $(x,1-3x)$. Using this, the gradient of OB is $m_1=\frac{1-3x}{x}$, and the gradient of AB is $m_2=\frac{1-3x}{x-5}$. Since these two lines are perpendicular, $m_1\cdot m_2=-1$. Solving this...

\begin{align*}6\sin{x}\cos{x}+3\sin{x}&=2\cos{x}+1 \\3\sin{x}(2\cos{x}+1)-(2\cos{x}+1)&=0 \\(3\sin{x}-1)(2\cos{x}+1)&=0 \\\sin{x}&=\frac{1}{3} \to x=19^{\circ}28',160^{\circ}32' \\\cos{x}&=-\frac{1}{2} \to x=120^{\circ},240^{\circ} \\\end{align*}

\begin{align*}\cos{x}\tan{x}+\tan{x}&=\cos{x}+1 \\\tan{x}(\cos{x}+1)-(\cos{x}+1)&=0 \\(\tan{x}-1)(\cos{x}+1)&=0 \\\tan{x}&=1 \to x=45^{\circ},225^{\circ} \\\cos{x}&=-1 \to x=180^{\circ} \\\end{align*}

$Thanks! So just to confirm: if I use $T_k=\binom{n}{k}x^{n-k}$, where $k=0,1,2...$ for $0\leq k \leq n$, and $T_{k-1}=\binom{n}{k-1}x^{n-(k-1)}$, where $k=1,2,3...$ for $0\leq k-1 \leq n$, the greatest coefficient can be determined using $\frac{T_k}{T_{k-1}}$? And this gives me a value of $k$...