Search results

Whole question was for when breadth is 8 cm. (I.e. Do (i), (ii) and (iii) at the moment when breadth is 8 cm.)

(Dots refer to time derivatives.)

$\noindent Let the length be $2x$ and the breadth $x$, where $x=x(t)$ is a function of time $t$ with $\dot{x}(t) \equiv 0.5$.$ $\noindent i) Just double the breadth's constant rate of growth, i.e. 1 cm/min.$ $\noindent ii) We have area being $A = 2x\cdot x = 2x^{2}$, so $\dot{A} = 4x\times...

Hint for part (ii): if dy/dt = dx/dt on the curve, we are looking for where dy/dx = 1. (In principle, a point where dx/dt and dy/dt are both zero would do, but we'd need to be given x(t) explicitly for that. (And note dy/dt = (dy/dx)*(dx/dt) = (8 – 2x(t))*(dx/dt), so dy/dt is automatically zero...

Yeah, e.g. reverse the above steps. :p

$\noindent One method is as follows. Note that $\cos^{2}x = \frac{1}{2}\left(1+c\right)$ and $\sin^{2}x = \frac{1}{2}\left(1-c\right)$, where $c \equiv \cos 2x$. Thus$ $$\begin{align*}\cos^{6} x + \sin^{6}x &= \left(\cos^{2} x\right)^{3} + \left(\sin^{2} x\right)^{3}\\ &=...

I think essentially the only ranks that really matter in that sense are first and last. Other than those, I think your marks are based on the distribution of internal marks and relative gaps etc. (from memory).

You probably would only need to write it in a step where you did something like cancellation. The value of a limit is unaffected by the function's value at the point we are taking the limit, and only affected by the behaviour of the function 'near' that point, which is the reason why we can do...

In the context of congruent triangles, corresponding angles refers to 'matching angles' in the congruent triangles (i.e. if you moved one of the congruent triangles so that it lies on top of the other (which is possible since they are congruent), corresponding angles are the ones that would...

Re: International Baccalaureate Marathon $\noindent If the cost per guest is $G(n)$, then the total cost for $n$ guests is $nG(n)$. So the answer that did $45\times G(45)$ is correct.$

Analysis of pendulum motion (e.g. relationship between the period and the length of pendulum, and whether the mass of the bob influences the period, or whether the amplitude does).

$\noindent a) $1 < m \leq \frac{\pi}{2}$$

Techniques are an absolute must.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent In that case, the distance of $P(x,y)$ to the origin is $\sqrt{x^{2} + y^{2}}$ and the distance to the $x$-axis is $y$ (since $P$ is to lie above the $x$-axis). We want \emph{the sum} of these to be $2$, so...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread What is the question (don't have a copy of the textbook nearby)?

Another method. $\noindent Write$ $$ax^4 + bx^3 + x^2 + 4x + 2 = \left(Cx^2 + Dx + E\right)\left(2x^{2} + 2x -1\right) + (2x+3)\quad (\star)$$ $\noindent and solve first for the constants $C,D,E$. Equating constant terms yields $-E+3 = 2\Rightarrow \boxed{E = 1}$. Equating $x$ coefficients...

Re: HSC 2017 Maths (Advanced) Marathon Yes, that's right (it's common practice to drop the '1' and write it like they've written it).

In other words, inspection. Should be fine as long as it's clear what you did. $\noindent (If using the method of equating real and imaginary parts of $a^2 -b^2 + 2abi = 8+6i$, you are allowed to just find $a$ and $b$ by inspection.)$

$\noindent The lines intersect at $x = -3$, so we are after the area between the curve $y=1-x$ and $y=4$ from $x=-3$ to $0$. The upper curve is always $y=4$ here and the lower curve $y= 1-x$, and so the area could be computed as $\int_{-3}^{0} \left(4-(1-x)\right)\,\mathrm{d}x$.$

$\noindent If we sketch the curve, if is clear we want the area under the curve between $x=0$ and $4$. So the area will be given by $A = \int_{0}^{4}2x^{2}(4-x)\,\mathrm{d}x$. Expanding, we have $A = \int_{0}^{4}\left(8x^{2} -2x^{3}\right)\, \mathrm{d}x$. You should be able to evaluate the...