I never got the blueprint for a hyper drive parts and now I am stuck in my first star system. I have tried deleting local game content (with and without steam cloud sync) and going into the programs and deleting everything related to NMS. Please help.
Is there a way to start over?  No spoilers here, but after a certain point in the game I wish I had done something different, so I'd like to perhaps start over and make some different decisions based on my current game play experience.
What makes a field not groupable I'm trying to use an aggregate query, but one of the fields I want to group by is not groupable (saw it in the soql error and then verified groupable=false in the metadata).  It's just a number field that holds an integer (number 2,0 in salesforce field spec-speak).  Is there a trick to make it groupable?  Any way around this problem? I don't want to write hundreds of lines of code just to average a list by this field. How to run Visual Studio post-build events for debug build only How can I limit my post-build events to running only for one type of build?  I'm using the events to copy DLL files to a local IIS virtual directory, but I don't want this happening on the build server in release mode. Evaluating the limit of a sequence given by recurrence relation $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Is my solution correct? Problem  The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:   $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$.   Evaluate the limit $\lim_{n\to\infty} a_n$.  Solution   Show that the sequence $(a_n)_{n=1}^\infty$ is monotonic. The statement $$V(n): a_n &lt; a_{n+1}$$ holds for $n = 1$, that is $\sqrt2 &lt; \sqrt{2+\sqrt2}$. Let us assume the statement holds for $n$ and show that $V(n) \implies V(n+1)$. We have that $$a_n &lt; a_{n+1}.$$ Adding 2 to both sides and taking square roots, we have that $$\sqrt{2+a_n} &lt; \sqrt{2+a_{n+1}},$$ that is $a_{n+1} &lt; a_{n+2}$ by definition. Find bounds for $a_n$. The statement $$W(n): 0 &lt; a_n &lt; 2$$ holds for $n=1$, that is $0 &lt; \sqrt2 &lt; 2$. Let us assume the statement holds for $n$ and show that $W(n) \implies W(n+1)$. We have that $$0 &lt; a_n &lt; 2.$$ Adding two and taking square roots, we have that $$0 &lt; \sqrt2 &lt; \sqrt{2+a_n} &lt; \sqrt4 = 2.$$ The limit $\lim_{n\to\infty} a_n$ exists, because $(a_n)_{n=1}^\infty$ is a bounded monotonic sequence. Let $A = \lim_{n\to\infty} a_n$. Therefore the limit $\lim_{n \to\infty} a_{n+1}$ exists as well and $\lim_{n \to\infty} a_{n+1} = A$. (For $(n_k)_{k=1}^\infty = (2,3,4, \dots)$, we have that $(a_{n_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, from which the statement follows.) We have that $a_{n+1} = f(a_n)$. That means that $A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}$. Solving the equation $A = \sqrt{2 + A}$, we get $A = -1 \lor A = 2$. Putting it all together, we get that $A = 2$, because the terms of the sequence are increasing and $a_1 &gt; 0$.   Is my solution correct? How to set QGIS language other then OS I am currently in the process of trying to repair a power supply that has had one of its DC rails fail. A resistor in the PSU has exploded and I am trying to find what value it is.    Band 1: White Band 2: Brown Band 3: Silver (Possibly Grey?) Band 4: Gold Band 5: Green    The weird thing is that the middle band appears to be silver, which should be invalid as far as I can tell. If you assume the silver is grey (although this is unlikely, as there still needs to be a grey colour as well and the band is metallic), it produces a value of resistor which is incredibly uncommon to the point it does not exist (91.8 Ohms 0.5%).  Apologies for the noobie question, I'm just stumped after hours of trying to understand why this is and coming up short :). How to set a random value for multiple shape keys? My question is related to my previous asked one:   I have about 300 shapekeys and want to set the random value for all of them . How to do it using scripting? The current is maximum through those segments of a circuit that offer the least resistance. But how do electrons know beforehand that which path will resist their drift the least? Confidence interval for Bernoulli sampling I have a random sample of Bernoulli random variables $X_1 ... X_N$, where $X_i$ are i.i.d. r.v. and $P(X_i = 1) = p$, and $p$ is an unknown parameter.  Obviously, one can find an estimate for $p$: $\hat{p}:=(X_1+\dots+X_N)/N$.  My question is how can I build a confidence interval for $p$? Behavior of $\|f\|_p$ when $p$ is near the endpoint of $I$ Given measure space $(X, \mu)$ and measurable function $f$. If $p \in (0, +\infty]$, below are the facts I know about $\Vert f \Vert_p$:  (log convex in $\frac 1 p$) $\forall p_1, p_2\in (0, +\infty] \text{ such that } p_1 &lt; p_2. \forall t \in(0, 1), \text{ let } \frac 1 p = \frac t {p_1} + \frac {1-t} {p_2}$, then $\Vert f \Vert_p \le \Vert f \Vert_{p_1}^t\Vert f \Vert_{p_2}^{1-t}$ ($0\cdot(+\infty)$ is defined to be $0$). (lower semi-continuity) $\forall p_0 \in (0, +\infty], \liminf_{p\to p_0}\Vert f\Vert_p \ge \Vert f \Vert_{p_0}$. If $0&lt;\mu(X) &lt; +\infty$, then $ \Vert f \Vert_p/\mu(X)^{\frac 1 p}$ is monotone increasing in $p$, combine it with (2) we get $\lim_{p\to p_0-}\Vert f\Vert_p = \Vert f \Vert_{p_0}$.  From (1) we know the set of $p$ such that $\Vert f \Vert_p &lt; +\infty$ is an interval $I$ (possibly empty or a singleton).  If $I \not\in a = \inf I$ and $a \neq 0$, from (2) we know $\lim_{p\to a+} \Vert f \Vert_p = +\infty$. If $I \not\in b = \sup I$, then $\lim_{p\to b-} \Vert f \Vert_p = +\infty$.  From (1) and the dominated convergence theorem we know that if $I \ni p_0 \neq \inf I$ and $p_0 &lt; +\infty$, then $\lim_{p\to p_0+} \Vert f \Vert_p = \Vert f \Vert_{p_0}$. If $I \ni p_0 \neq \sup I$, then $\lim_{p\to p_0-} \Vert f \Vert_p = \Vert f \Vert_{p_0}$.  If $I$ is nondegenerate and $+\infty \in I$, then $\lim_{p\to +\infty}\|f\|_p = \|f\|_{\infty}$.  If $\inf I = 0$, from (1) we know $\lim_{p\to 0+}\|f\|_p$ exists.   Question: How $\Vert f \Vert_p$ behaves when $p$ is the endpoint of $I$? To be more precise, assume $I$ is nonempty. Let $p_0 = \sup I$ (only right endpoint is considered for simplicity):  $p_0\in I$ or $p_0 \not\in I$? Can $I$ be a singleton? What if $p_0 = +\infty$? What if $0&lt;\mu(X) &lt; +\infty$?  Please also notify me if there is any mistake above. Access is denied because the NTFS volume is already exclusively opened print IP address with interface name I want to show the IP address like below  lo : 127.0.0.1  eth0 : 192.168.5.123 eth1 : 192.172.0.212 wlan0 : 10.1.0.124   I'm able to print all IP address by ifconfig | awk '/inet addr/{print substr($2,6)}'. But it is only printing the IPs. Every system have their own interface names and addresses. So my script has to show their interfaces with respect to IP addresses. Suppose $n$ is an even positive integer and $H$ is a subgroup of $\mathbb Z/n \mathbb Z$. Prove that either every element of $H$ is... Prove that $\sup \{-x \mid x \in A\} = -\inf\{x\mid x \in A\}$ I need to prove that $\sup \{-x \mid x \in A\} = -\inf\{x \mid x \in A\}$ and am having trouble moving the $-x$ out of the $\sup$ to $\inf$. Another thing is that I don't quite know how to prove $b = \inf$ (see below). Any help would be appreciated! Here my attempt:  Let $a = \sup\{-x \mid x \in A\}$  Then by definition $a \geq -x$ for $x \in A \implies -a\leq x$ for $x \in A$  Thus $-a$ is a lower bound of $A$. Assume $b$ is a lower bound of $A$  Then $b \leq -a \implies  a \geq -b$ and $a = \sup \{−x \mid x\in A\} = -\inf(A)$ Do I need a transit visa at Istanbul Airport to enter Istanbul for city sightseeing or hotel accommodation?. All services provided free by the Turkish Airline. I am Nigerian citizen and booked Turkish Airline flight to Nigeria with 12hours layover at Istanbul Airport. I need help with this:  If a is an element of $\mathbb Z_n$ and $\gcd(a,n) &gt; 1$ , then a is not invertible.   First you show that if $a$ is an element of $\mathbb Z_n$ and $\gcd(a,n)&gt;1$, then there  is an element $b$ of $\mathbb Z_n$ and ($b$ is not equal to the zero element) from which $ab = 0$.   The second part I cant get. i.e. show if $b$ is not the zero element  and $ab=0$ then $a$ is not invertible. How to prove $B-A \succeq 0 \Leftrightarrow$ ellipsoid $x^TBx \leq 1$ contains $x^TAx \leq 1$? Assume $A \in \textbf{S}^n_{++}$, an ellipsoid centered at the origin given by $$\mathcal{E}_A = \{x\mid x^TA^{-1}x \leq 1\}$$ Then we have $\mathcal{E}_A \subseteq \mathcal{E}_B $ if and only if $B-A \succeq 0$. This is the proposition in the Boyd &amp; Vandenberghe's  (pages 45-46). The authors gave only the result, without proof. How to prove it? Travel to Australia with new passport (with visa in old passport) I was granted a visa to Australia few months ago. I’d like to travel to Australia in few days. However, I’ve just renewed my passport and I don’t have enough time to inform them about my passport renewal. Can I travel to Australia by carrying both my old and my new passports? When is $\mathbb{F}_p[x]/(x^2-2)\simeq\mathbb{F}_p[x]/(x^2-3)$ for small primes? I think everybody with a history of participation can't help but notice the explosive growth of repeated duplicate questions. The work of looking for duplicates and then closing questions has overtaken the work of actually answering questions.  It's become virtually impossible to ask something which hasn't been yet asked. Lately I feel boredom each time I see a new question and start answering it only to realize after the first written sentence that I have answered similar questions five times over or seen them answered some ten times or more. I know where it's going to be going and what answers with what votes it is going to receive before any answer is even written.  It's obviously not the administration's fault, it is simply so. The questions is, what to do about it? Engaging more moderators to close and close duplicates won't work because it's already starting to repel users who get mad at their questions closed and become afraid of asking more or perhaps just go away.  The original idea of Jeff stated the the site is for questions that can be answered. Meaning no open-ended discussions. When I first read this I had a casual thought - well, and what's gonna happen when they all are asked and answered? Will it become a museum, a monument of wisdom and knowledge or what?  I'm not sure what to do about it. Perhaps deprecation/versioning of older questions so that their new versions can sparkle the tiresome evening...  POSTSCRIPTUM.  I recall Star Trek Voyager episode  when another Q asked Captain Janeway for asylum. He was tired of immortality and wanted to die. They opened a hearing and went on to visit the continuum to evaluate the living conditions there which caused the suffering and the wish to escape it in death. A lifeless picture presented itself to the visitors... a house lone in an open space, two people going about their business, nobody talking. When asked why they were silent, the following was said. It's been so long, all has already been said, all stories told, all jokes are made... there is nothing left to do...  Are we there as well?  UPDATE: Just saw this answer in another question:       Also, some of the core questions got asked early on in the life of the site. Great questions still come around from time to time, but many have already been asked.   Doesn't that confirm my point? Code in beamer presentation Good day! So, I need to present my disertation code to my thesis advisors, but I'm quite unsure how to actually do it in a proper way. I found  showing some ideas.  Is there a way of making the slides 48-67 in beamer without using tikz? Any ideas to implement the whole ideas the author propose?  Thanks in advance.