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I guess so. If you're in doubt, just use the equals sign, since the HSC doesn't really care if you just use equals (whereas some markers may find it strange if you use triple bars in a place where it'd be better to use equals). (Of course for geometry, use triple bars, not equals, for...

I just did that because I was saying it was defined to be r^2 – z^2 (sometimes the triple bar is used when you're defining something equal). You don't really "need" to do it though, you can just use the equals sign if you want, and you certainly wouldn't lose marks if you did.

It should be equals. I don't think you lose marks in the HSC if you just always use the equals sign for identities or equations.

$\noindent Yes, it returns to the origin. Let $f(x)\equiv 16+6x-2x^3$ (the function for $v^2$). Since $f(2) = -2\times \left(-3\times 2\right) = 12>0$ and $f(3) = -2 \times \left(27 -9-8\right) < 0$, $f$ (being a continuous function) attains the value $0$ for some $x^{*}\in \left(2,3\right)$. At...

Re: MATH1231/1241/1251 SOS Thread Row-reduce that augmented matrix (get it into row-echelon form) and you should find a zero row at the bottom, with some linear expression involving x,y and z in this row in the right-hand augmented part. There will be solutions, i.e. we will have v be in S, if...

The above by Fizzy_Cyst is what I was doing too, but it is an answer well smaller than the one leehuan said is apparently the right answer. It would be: mgh – 1/2 mv^2 = m(g – 1/2 v^2) (as h = 1) = 0.1 * (g – 1/2 v^2). Since v = 3.8, v^2 = (4 – 0.2)^2 = 16 – 1.6 + 0.04 = 14.44 (didn't have...

What do you mean by letting one side be positive? The "up" direction was treated as positive, if that's what you were referring to.

$\noindent a) The equation for the vertical displacement as a function of time is $y = ut -\frac{1}{2}gt^2$, using $u=100,g=9.8$ and SI units for all units. The time of flight is found by solving $y(t)=0, t\neq 0$. This happens when $u-\frac{1}{2}gt =0$, i.e. $t= \frac{2u}{g}$. Sub. in $u=100...

Coming second wouldn't affect it too much, especially if your cohort is relatively big/good and if your internal mark was relatively close to that of first place.

Try using the formula E = V/d (check your textbook/notes etc. if you haven't seen this before). Edit: done in above working by jathu123.

That emboldened part was the " – f(x)".

$\noindent Let $f(x):= x^3$, then $f(x+h) = (x+h)^3 = x^3 + 3x^2 h + 3xh^2 + h^3$, so $f(x+h) -f(x) = x^3 + 3x^2 h + 3xh^2 + h^3 - x^3 = 3x^2 h + 3xh^2 + h^3$. Hence $\frac{f(x+h)-f(x)}{h} = \frac{3x^2 h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2$, for any $h\neq 0$. Taking the limit as $h\to 0$, we...

Use the fact that the resistance is inversely proportional to the cross-sectional area (along with the formula for the area of a circle, assuming the wire's cross section is circular). $\noindent (The formula is $R=\frac{\rho L}{A}$, as explained here in the section ``Resistivity and...

Re: MATH1231/1241/1251 SOS Thread $\noindent Here's the other method for doing these types of Q's that'll work even if the surface isn't something like a sphere where we can easily see a normal. We don't need to try and solve for $z$. We just need to make use of the following fact. If a...

Re: MATH1231/1241/1251 SOS Thread $\noindent There's an easier way to do this particular one. This surface is a sphere centred at the origin, so the normal to the surface at any point $\left(x,y,z\right)$ on the surface can geometrically be seen to just be that same vector...

Re: MATH1231/1241/1251 SOS Thread Compute the partial derivatives (which they've done) and sub. in the given (x,y) point. $\noindent E.g. They found $F_x \left(x,y\right) = \pi y^2 \cos \left(\pi xy^2\right)$. So at $(x,y) = (2,-1)$, $F_x (2,-1) = \pi \cdot (-1)^2 \cos \left(\pi \cdot 2\cdot...

Re: MATH1231/1241/1251 SOS Thread The solution they gave is essentially a more 'general' sort of solution than (x – c)^3, which is a shifted cubic. The more 'general' solution given is essentially taking this cubic but 'widening out' the horizontal point of the cubic to any arbitrary length...

Re: MATH1231/1241/1251 SOS Thread When you divided by y in separating variables, you had to assume y =/= 0, making you lose that solution of y = 0. And what did you want clarification with for the solution from the back of the book? You can check that it is differentiable everywhere and...

If the calculator gives you the graph, then those graphing questions wouldn't be testing anything. It's like allowing primary school students to use a calculator on an arithmetic test.

Well they ask Q's about sketching the graphs of given functions, so they probably wouldn't want students having graphing calculators in the exam if there's Q's like that.