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Unless the question specifically asks for it, it's not necessary to use e here. Let the initial cost C_0 be 100\%, or 1 . We don't actually need to know what the initial cost is, we're only interested in its percentage change. Then, the cost C_t after t months is given, as a...

We may assume that if \frac{dA}{dt} = kA then A = A_0 e^{kt} for some constant A_0 . _______________________________________________ Applying this to your question: \frac{dS}{dt} = -kS \implies S = S_0e^{-kt} and we know the initial amount of salt to be 20 grams, that...

There is one situation where it's useful to memorise, and that's if you're running out of time on an exam. You will lose marks for assuming the result but you might be able to get more marks on the other question(s).

That's correct. I'm not sure what your mistake is because if you did that then you should have arrived at the right answer. The restriction is, essentially, that A and B cannot both be on the committee. You can have either one or the other but not both. A or B but not AB. If you were to split...

Case 1: B is not included. Then, you have to select the whole committee from nine others (without restriction). There are \binom{9}{6} choices. Case 2: B is included. Now, we have to select five more people from nine. But A is one of the nine, and we cannot select A. Therefore we must...

Re: HSC 2018 MX2 Marathon For simplicity, let P(x) =x^3-x^2-24x . First, sketch the graph of y = P(x) (the quickest way to do this is to find the turning points of the graph). The graph of y = P(x)+k can be obtained by shifting the graph of y = P(x) by k units up or down. Finally, a...

For me the most difficult thing about probability/counting is that you have almost no idea whether you're right or wrong, and there's so many sources of error, such as double counting and leaving out some cases. Experience is ultimately the best way to overcome those issues. I find it helpful...

Re: Interesting parametrics question! i) Using the general equation of the parabola x^2=-4ay and the given point A(-6,-9), we can deduce a=1. ii) Let B be given by B\left(u, -\frac{u^2}{4}\right) The cartesian equation of the normal to the parabola is y-y_1 =...

\int^{x=1}_{x=-3} \frac{1}{\sqrt{x+3}} \, dx Let x = u^2-3, then dx = 2u\, du The integral becomes \int^{u=2}_{u=0} \frac{2u}{\sqrt{u^2}} \, du (you can choose either u=2 or u=-2, it doesn't change the answer) = \int^{u=2}_{u=0} \frac{2u}{|u|} \, du = \int^{u=2}_{u=0} \frac{2u}{u}...

Re: HSC 2018 MX2 Integration Marathon We can get the volume of the solid by taking the volume of x = y^2-2 from 0 to 1 (rotated about the y-axis) and then subtracting the volume of the cylinder inside, which has radius and height 1. \begin{aligned} V &= \pi \int_0^1 (y^2-2)^2\, dy -...

$LHS$ = \frac{2-\sqrt 2}{4} $RHS$ = 2-\sqrt {2+\sqrt 2} Subtract 2: $LHS$ = \frac{-6-\sqrt 2}{4} $RHS$ = -\sqrt {2+\sqrt 2} Multiply by -1 (which flips the inequality sign) $LHS$ = \frac{6+\sqrt 2}{4} $RHS$ = \sqrt {2+\sqrt 2} Which is the result in b)

depends on how hard your school's assessment tasks are.

Don't square it - you want to get it to the result in part b). You'll be able to see the answer by doing just one subtraction.

Given \sqrt{2+\sqrt2}<\frac{6+\sqrt 2}{4} R.T.P. 4\sqrt{2-\sqrt2} < 8\sqrt{2-\sqrt{2+\sqrt 2}} Square both sides $LHS$ = 16(2-\sqrt 2) $RHS$ = 64(2-\sqrt {2+\sqrt 2}) Divide both sides by 64: $LHS$ = \frac{2-\sqrt 2}{4} $RHS$ = 2-\sqrt {2+\sqrt 2} This is starting...

(a +b)^2 = a^2+2ab+b^2 With a = 6, b = \sqrt 2: (6 + \sqrt 2)^2 = 6^2 + 6\sqrt2 + 6\sqrt 2 + 2^2 = 38 + 12\sqrt 2

They're pretty much equivalent, as long as the angles subtended by the arc are on the circumference.

b) R.T.P. \sqrt{2+\sqrt2} < \frac{6+\sqrt 2}{4} Squaring both sides, \mathrm{LHS} = 2 +\sqrt 2 \mathrm{RHS} = \frac{38 + 12\sqrt2}{16} Multiplying both sides by 16, \mathrm{LHS} = 32 + 16\sqrt 2 \mathrm{RHS} = 38 + 12\sqrt2 Can you finish it from here? c) seems to be missing...

i) This means that she hits a bullseye with one and misses it with her other two. There are three ways this can happen (she can hit it with either her first, second or third shot) and the probability of each is 3/5 \cdot (2/5)^2. ii) Method 2: Hitting at least two bullseyes is the...

Yes.

With these sorts of questions it may help to remember your definitions and think things through slowly and methodically. f'(x) is the gradient function. That means that the value of f'(x) is the gradient of f(x) at x . A gradient of 0 means that there is a stationary point. As the...