What's the difference between "Files" and "Notes &amp; Attachments" related Lists?  They seem to be a carbon copy of each other whenever I upload a file to one or the other.
We have been using Notes and Attachment to store files related to the record. I also read about Files in salesforce.   How are Files different from Attachment? Are there any limitations or benefits of using Files instead of Attachments?  Are Files available for all editions of Salesforce?  Is it possible to link Files to particular records so they are visible as related list? I tried looking into the content version  but cant seem to find anything to relate to a record
self referential struct definition? Is there any way to make a hotkey turn the screen off? How can I make a script (or make some changes) to allow me to turn the screen off in Ubuntu (12.04) whenever I want, the way thevscreen can be turned off in Cellphones?  And how can I make a combination of keys (a shortcut) to execute that script?  I want to conserve power.  I've seen this question    but there is no hotkey for it.  Moreover, in the answer, it is said that there are many ways to accomplish this so I'd like to know more. For any natural number $n\ge 1$, given pairs $(a_1,b_1),(a_2,b_2),...,(a_n,b_n)$ of integer numbers, there exist integer number $c$ and $d$ such that  $$\prod_{i=1}^{n}(a_i^2+b_i^2) = c^2+d^2$$    My initial approach is  Base Case: $(a_1^2+b_1^2) = a_1^2+b_1^2$ which is true. (Although it is trivial)  Prove the statement is true when $n=2$: We have $$(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$$ (Thanks André Nicolas for pointing it out)   So if $a,b,c,d$ are integers, $ac,bd,ad,bc$ are all integers and integers are closed under addition and subtraction. Hence $(ac-bd),(ad+bc)$ are integers.   Inductive Hypothesis: $\prod_{i=1}^{n}(a_i^2+b_i^2) = c^2+d^2$ is true  Inductive Step: $$\prod_{i=1}^{n+1}(a_i^2+b_i^2) = \prod_{i=1}^{n}(a_i^2+b_i^2)\cdot (a_{n+1}^2+b_{n+1}^2) = (c^2+d^2)\cdot (a_{n+1}^2+b_{n+1}^2)$$  Where $c$ and $d$ are integers.   But when we apply $n=2$, we have  $(c^2+d^2)\cdot (a_{n+1}^2+b_{n+1}^2) = (e^2 + f^2)$ where $e$ and $f$ are integers.  Hence, by the principle of induction, the statement we needed to prove is true. Google's CDN is nice, but it's not 100% reliable for all users. It occasionally fails to load for me. Other users , and some .  It's one more possible point of failure. It would be preferable if Stack Exchange didn't depend on it.  If you don't want to stop using it, consider adding a fallback if it fails to load. Something like this:  &lt;script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"&gt;&lt;/script&gt; &lt;script&gt;   if (!window.jQuery) {     document.write('&lt;script src="https://cdn.sstatic.net/stackoverflow/content/js/jquery-1.7.1.min.js"&gt;&lt;\/script&gt;');   } &lt;/script&gt; Prove that there exists at least one $x_0\in\mathbb{R}$, such that $f(x_0)+f''(x_0)=0$ Let $f:\mathbb{R}\to\mathbb{R}$ be a function, two times   differentiable with $\left|f(x)\right|\leq1,\forall x\in\mathbb{R}$   and $f^2(0)+\left(f'(0)\right)^2=4$. Prove that there exists at least   one $x_0\in\mathbb{R}$, such that $f(x_0)+f''(x_0)=0$.   I have tried the usual Rolle method by $e^x$, but didn't go far... Any help available? Is it possible to be admitted to a masters in Germany with a 3-year bachelors from another country? Are there any expressions that describe going from a bad to a worse situation? Are there idioms or expressions in English that describe going from one bad situation to one that's even worse?  I heard "between a rock and hard place" but this describes a dilemma not really a transition. I am looking for an expression that describes someone trying to get out of a bad situation but after much effort, the situation only got worse. My Android phone is completely reset, whats the best way to recover the deleted data? using apt-cache search I am learning how to use apt. When I do an apt-cache search git to see all packages matching git, I see a bunch of packages whose descriptions have nothing to do with git. Why is this so and how can I fix it? (Ubuntu 12.04.2 LTS).  Also, why is the search functionality found in a command that has "cache" in its name? What does search functionality have to do with caches?  Thanks. How much energy does a photon need to form a black hole? I was wondering if it is possible for a single photon to form a black hole if it has a small enough wavelength.If so, what would this wavelength be? I came across this question because I am reading about loop quantum gravity, and have heard that it is 'untestable'. I know that the loops making up spacetime in loop quantum gravity are thought to have a size in the order of $10^{-35}$m, and I was wondering whether making a photon with a wavelength small enough to investigate this would form a black hole.  I am aware of the two posts here, but neither seem to really answer my question because they refer to several photons:  and .  Thank you. Procedure cache (plan cache) size Bounded sequence with divergent Cesaro means Is there a bounded real-valued sequence with divergent Cesaro means (i.e. not Cesaro summable)?  More specifically, is there a bounded sequence $\{w_k\}\in l^\infty$ such that $$\lim_{M\rightarrow\infty} \frac{\sum_{k=1}^M w_k}{M}$$ does not exist?  I encountered this problem while studying the limit of average payoffs criterion for ranking payoff sequences in infinitely repeated games. How can I recover from a "AMD-Action:downloadProduct:SP" during installation of macOS? Just wiped my hard drive on my MacBook Pro from 2013. Went to install a new copy of MacOS Sierra. I get the following error:  AMD-Action:downloadProduct:SP   I looked online and tried change the date using the Utility Terminal and the following command:  date mmddHHMMyy   Then did   csrutil enable csrutil clear   Then I tried clearing the NVRAM using command + option + P + R on startup.  Still no luck. Still getting the same error message. Apple support told me to go to a "computer repair store".  Any advice? Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic I'm reading a book about abstract algebra, but I'm having trouble solving this excercise: "Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic"  Where $(\mathbb{Q}^*,\cdot)$ is the group of nonzero rational numbers under multiplication and $(\mathbb{R}^*,\cdot)$ is the group of nonzero real numbers under multiplication.    Here is my attempt for the first.  Suppose $(\mathbb{Q}^*,\cdot)$ is cyclic, then $\mathbb{Q}^*=\langle\frac{p}{q}\rangle=\{(\frac{p}{q})^n,n\in\mathbb{Z}\}$, where $p$ and $q$ are coprime.  $\frac{2p}{q}$ is also in $\mathbb{Q}^*$ so it must be equal to $(\frac{p}{q})^n$ for some $n\in\mathbb{Z}$.  To solve $\frac{2p}{q}=(\frac{p}{q})^n$, I take a logarithm of both sides and end up with $1+\log_\frac{p}{q}(2)=n$, since $n$ is an integer $\log_\frac{p}{q}(2)$ must be an integer too, but it is possible only when $\frac{p}{q}=2^{\frac{1}{k}}, k\in\mathbb{N}$, (i.e. $\frac{p}{q}$ is a k-th root of $2$), but $k$ must be $1$ for $2^\frac{1}{k}$ to be rational so $\frac{p}{q}=2$ contradicting the hypothesis of $p$ and $q$ being coprime.    However I don't know whether this is a proper proof and the same reasoning cannot be applied to $\mathbb{R}^*$, I'd like you to just give me an hint towards a proof, without telling me the whole proof, if possible. Where is the Layers panel in Blender 2.8? I am starting working on Blender 2.8, but I can't find the layer panel. Where is it? Number of $n^2\times n^2$ permutation matrices with a 1 in each $n\times n$ subgrid I found the following question in a paper I was trying to solve:  The following figure shows a $3^2 \times 3^2$ grid divided into $3^2$ subgrids of size $3 \times 3$. This grid has $81$ cells, $9$ in each subgrid.    Now consider an $n^2 \times n^2$ grid divided into $n^2$ subgrids of size $n \times n$. Find the number of ways in which you can select $n^2$ cells from this grid such that there is exactly one cell coming from each subgrid, one from each row and one from each column.  My try:  Since we have $n^2$ rows, $n^2$ columns and $n^2$ subgrids in total, we have to choose one and only one cell from each of them. Let's choose them one at a time. We can choose the first cell in $n^4$ many ways. Then, we'll have to avoid that subgrid, that column and that row that we've chosen the first one from when choosing the second cell. So, we have $n^4-n^2-2n(n-1)$ choices. We can continue this to get the total number of possible ways. But, I think there's a hole. Say, we've chosen the first cell from the subgrid of the up-left corner and the second from the subgrid just right to it so that it doesn't violate any rules. Then, when finding the number of ways we can choose the third cell, we would have substracted some of the cells twice. I think you get it. Please, if anyone can help me solving this problem, it'd be greatly appreciated. Why do we get the same differential equations from both principles? Surely there is a fundamental connection between them? When written out, the two seem to have nothing in common.   $$\sum _i ( \mathbf F _i - \dot{\mathbf p}_i) \cdot \delta \mathbf r _i = 0$$  $$S[q(t)] = \int ^{t_2} _{t_1} \mathcal L (q,\dot{q},t)dt$$  After playing with d'Alembert's principle  we find that we can rewrite the whole thing as $$\sum _i \left[ \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_i} \right) - \frac{\partial T}{\partial q_i}-Q_i\right] \delta q_i$$  This can further be rewritten under certain conditions so we get the exact form of the E-L equation.   It seems to me that both ways of arriving at the result are fundamentally different.   A function must obey the E-L equations in order to minimize the action over a path, but when we look at the virtual work, it appears that they come from the fact that (quoting Goldstein) "particles in the system will be in equilibrium under a force equal to the actual force plus a 'reversed effective force'."  I think I understand the principle of stationary action, I can see how it leads to the E-L equations, but d'Alembert's Principle seems so arbitrary, I can't see any motivation for it. How can I split a polygon by network line in QGIS? I have a road network as line layer and a land cover as polygon layer, I want to find number of subpolygon for each polygon that divided by line layer in QGIS?  I used union function in Geoprocessing Tools in Vector menu of QGIS and then polygonize function to convert union result to polygon but in table of new polygon there is NULL data. "You're too clever a man" You're too clever a man to imagine this.   The above sentence was , a man of excellent rhetorical skills.   Since he said it, I doubt it's wrong, grammatically. But, I wonder if there is an explanation for this. Because adjectives always come after the article not before it. e.g. You're a clever man.   How could this be, grammatically? How to make a film in black and white and add color to some objects? How is a movie made to show selective colors. For instance, in  a sister and brother are transported to the universe of an old black and white TV show, and the movie switches to being B&amp;W. Then, as the protagonists cause the inhabitants to have 'real' emotions, like fright, anger, or love, things (people, places, etc.) start having color. In the middle of the movie, many objects are still B&amp;W, yet many more are now colorized.   also used this effect (the whole film was initially shot in full color, and was converted to black-and-white. Colorization is used on certain subjects in a scene), for example.  How is this effect achieved? Is it post-production editing? What techniques might a director employ to do this?