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$\noindent The simultaneous equations are \\ \begin{align*}\quad 2M + 2N + G &= 10 \\ M + N + G &= 6 \\ 2G &= 4\end{align*} \\ I don't think there is a way to solve for $M$ and $N$ but it is possible to know which possible values they can take.

Exactly six of these 'unusually shaped portions' fit into the regular octahedron by symmetry (one at each vertex) so the volume of one portion is one-sixth the volume of the octahedron, ie. 1/6 * 120 = 20

It depends on the question. For example if they say a point P(x,y) moves so that it is equidistant from the line y = -a and the point S(0, a) then we can say that PM = PS to find the locus of P, where PM is the perpendicular distance from P to the line. From here we can say that (PM)^2 = (PS)^2...

In the distance formula the whole expression is under a square root. So in locus problems where two distances are equal or proportional then we often square both sides at the beginning to immediately eliminate the square root.

From where? (Don't have access to the AMC problems)

$\noindent \textbf{(i)} Since $h$ increases at a constant rate, then $h = at$, for some constant $a$. Also, $v = \frac{dx}{dt}$ so $\frac{dx}{dt} = \frac{A}{h} = \frac{A}{at}$. If we let $k = \frac{A}{a}$ where $k$, $A$ and $a$ are constants then $\frac{dx}{dt} = \frac{k}{t} $\noindent...

$\noindent The area of the gasket is $A = y^2 - 9x^2$ \\ The perimeter of the gasket is $P = 4(y-3x) + 24x = 4y + 12x$. So, \\ \begin{align*} y^2 - 9x^2 &= 4y + 12x \\ (y-3x)(y+3x) &= 4(y+3x) \\ y - 3x &= 4 \text{ since } y+3x \neq 0 \end{align*} \\ If $x = 2$, $y = 10$ which is not prime but if...

Is the answer 9 cm? (I did it by finding the area of the annulus which represents the length of the paper, so I halved this area since half the paper was used and then found the new diameter which gives such area)

$\noindent Is the question written correctly? Because why wasn't it just written $f(x-1)$? \\\\ If this is the case then $\int_0^4 f(x)dx = \int_3^7 f(x-3)dx$ as the graph along with the bounds have all been shifted 3 units to the right. You can then split up the integral and work out...

$\noindent Okay, so assuming 50 is correct, then this is how I did it: The upper left quadrant of the figure can be translated and rotated $90\degree$ anticlockwise onto the upper right part, and similarly the lower left part of the figure can be translated and rotated $90\degree$ clockwise onto...

$\noindent \textbf{Q3.} Are you sure the answer is $25\pi - 25$? I got 50 square centimetres.

$\noindent $x = \frac{1}{2}(2x-1) + \frac{1}{2}$ so $\int_1^2\frac{x}{(2x-1)^2}dx = \int_1^2\left(\frac{1}{2(2x-1)}+\frac{1}{2(2x-1)^2}\right)dx $\noindent Or just write $x = \frac{1}{2}(u+1)

P belongs to the slanted edge OB but it's not the midpoint. AP is just perpendicular to OB. If the triangle was equilateral or if AB = OA then it would be a perpendicular bisector but it isn't.

$\noindent If $AC$ is the diagonal of the square base $ABCD$ and $O$ is the vertex of the pyramid and $AP$ is drawn such that $AP \perp OB$ then the required angle is $\angle APC$.

$\noindent I think you are after the size of $\angle CAF$. In $\triangle AEF$, $\cos{35\degree} = \frac{AE}{AF}$. While in $\triangle BCF$, $\sin{20\degree} = \frac{CF}{BF} = \frac{CF}{AE}$ since $BF = AE$. \\\\ Solve these two equations for $AE$ and conclude $AF\cos{35\degree} =...

Thanks I did this but I let V = 108 in V = 6/5 * (t-5) when I should've just found the area under the horizontal part. Why does substituting V = 108 in V = 6/5 * (t-5) not work? Edit: (btw, I got the question out of a HSC Trial)

On a factory production line a tap opens and closes to fill containers with liquid. As the tap opens, the rate of flow increases for the first 10 seconds according to the relation R = 6t/50, where R is measured in L/sec. The rate of flow then remains constant until the tap begins to close. As...

$\noindent $\frac{dy}{dx} = \frac{x}{2}$. If $P\left(2p, p^2\right)$ then the normal at $P$ has gradient $-\frac{1}{p}$. So the equation of the normal at $P$ is $y - p^2 = -\frac{1}{p}\left(x-2p\right)$. ie. $y = -\frac{1}{p}x + 2 + p^2$. To find $Q$ solve this simultaneously with $x^2 = 4y$ to...

Double-check the LHS is written correctly ('sins', and there may be an exponent because there are parentheses)

Q1. Aren't they equal? The magnitude of velocity is speed Q2. There has been a change in direction and therefore a change in velocity. I think a change in velocity indicates an acceleration