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Is it Maths In Focus?

You can use them to get the given angle down to a number that you known the exact trig. value of.

Note sin(-225 deg) = -sin(225 deg) (sin is odd) = -sin((180-225) deg) (recall the identity sin(180 deg – X) = sin X) = -sin(-45 deg) = sin(45 deg) = 1/(sqrt(2)).

$\noindent Here's some explanations for the main steps. The reason that$ $$A_{n} = \left(1 + \frac{0.06}{12}\right)^{6}A_{n-1} + 2000$$ $\noindent is true is that the value just after the $n$-th deposit is the value just after the previous ($(n-1)$-st) deposit (which is $A_{n-1}$), multiplied...

$\noindent Let $A_{n}$ be the value of the money (in dollars) in fund just after the $n$-th deposit, for $n \in \mathbb{Z}^{+}$. Note that $A_{1} = 2000$ (initial amount deposited). Also, $A_{n} = \left(1 + \frac{0.06}{12}\right)^{6}A_{n-1} + 2000$. Let $a = \left(1 + \frac{0.06}{12}\right)^{6}$...

$\noindent The answer is $n^{2}$. One method is to use the formula for the sum of an AP. Another method is to use induction.$ $\noindent You can also prove it pictorially.$

This is not the case.

$\noindent If you draw (or visualise) a diagram, you'll see it's $180^\circ + 65^\circ$, i.e. $245^\circ$. This follows from the given data, definition of bearing, and the fact that alternate angles in parallel lines are equal.$

$\noindent He means make sure the right-hand inequality wasn't $<$, and really was $\leq$. If it was $<$, we'd need to exclude $360^{\circ}$ from our solution set.$

$\noindent Also, if you think $360^{\circ}$ is a solution, you can check whether it is by just subbing it in to the equation and seeing if it holds.$

So for your example (42 total types of Pokémon and 5 minute mean arrival time), E[M] comes out to be about 908.6, so about 15 hours, 8 minutes and 36 seconds: http://www.wolframalpha.com/input/?x=0&y=0&i=42*5*(sum+from+k+%3D+1+to+42+of+1%2Fk) .

$\noindent Alternatively, if you only care about the expected time to catch all types of Pok\'emon, you can note that $\mathbb{E}[M] = \frac{n}{\lambda}\sum_{k = 1}^{n}\frac{1}{k}$. One way to show this is to consider the time to arrival of a new type of Pok\'emon at each stage that you just...

$\noindent Note that if $X_{1},\ldots , X_{n}$ are iid random variables with CDF $F_{X}$, then you can show quite easily that the CDF of their maximum is $\left(F_{X}(x)\right)^{n}$. This was used near the end above.$

$\noindent We can find the distribution of the desired time in the following way. Let $n$ be the total types of Pok\'emon (in your case, $n = 42$). I will assume Pok\'emon arrive according to a Poisson process with mean interarrival time 5 minutes and conditional on an arrival of a Pok\'emon...

$\noindent Another method would be to note that $w = z^{-1} = u + iv$, where $u = \frac{x}{x^{2} + y^{2}}$ and $v = \frac{-y}{x^{2} + y^{2}}$. On the line $x+y = 4$, we have $y = 4-x$. We can substitute this into the equations for $u$ and $v$ to obtain a pair of parametric equations for $u$ and...

$\noindent The map $z\mapsto z^{-1}$ maps lines not through the origin to circles through the origin. You could use this fact to find the desired image.$

The method is as follows. We want to get rid of the terms with odd r in the sum in part b). To do this, sub. x = -1 into the function you used to obtain the part b) result (where you probably subbed x = +1 before), which will give us an alternating version of the sum in part b), with the odd r...

Must do n = 4 as base case and go from there (after having verified n = 1, 2, 3).

$\noindent You should do (remark on) terms $n = 1, 2, 3$ separately and use $ n = 4$ as the base case, because the 1-3 terms are just generally arbitrarily given (initial conditions) and don't form part of the sequence that gets forced by the recurrence relation, so it could have been that one...

I will make the assumptions that all subject marks are whole numbers and all subjects are equally weighted. (If these assumptions don't both hold, you would have to provide us with what assumptions to use I suppose.) The fractional part of the new Average is 0.462 (at least to three decimal...