Will my following laptop run Farming Simulator 15 ?   LENOVO IdeaPad 110 15.6" Laptop   Intel® Celeron® N3060 Processor Dual-core 1.6 GHz / 2.48 GHz (Burst) 2 MB cache   Memory (RAM) : 4 GB DDR3 (8 GB maximum installable RAM)   Storage : 1 TB HDD, 5400 rpm     Thanks in advance
Is there a service that I can use to check if my machine will be able to run a certain game?
Creating the Nabla operator (also known as Del operator) as an operator In classical logic, why is $(p\Rightarrow q)$ True if both $p$ and $q$ are False? I am studying entailment in classical first-order logic.  The Truth Table we have been presented with for the statement $(p \Rightarrow q)\;$ (a.k.a. '$p$ implies $q$') is: $$\begin{array}{|c|c|c|} \hline p&amp;q&amp;p\Rightarrow q\\ \hline T&amp;T&amp;T\\ T&amp;F&amp;F\\ F&amp;T&amp;T\\ F&amp;F&amp;T\\\hline \end{array}$$  I 'get' lines 1, 2, and 3, but I do not understand line 4.      Why is the statement $(p \Rightarrow q)$ True if both p and q are False?  We have also been told that $(p \Rightarrow q)$ is logically equivalent to $(~p || q)$ (that is $\lnot p \lor q$).  Stemming from my lack of understanding of line 4 of the Truth Table, I do not understand why this equivalence is accurate.       Administrative note. You may experience being directed here even though your question was actually about line 3 of the truth table instead. In that case, see the companion question  And even if your original worry was about line 4, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines. I'm new to Ubuntu. I installed it today, but the first thing I encounter is that I can only connect to the Internet through wifi.  Can someone help me?   iwconfig result :  eth0      no wireless extensions.  lo        no wireless extensions.   lspci:  00:00.0 Host bridge: Intel Corporation 3rd Gen Core processor DRAM     Controller (rev 09) 00:01.0 PCI bridge: Intel Corporation Xeon E3-1200 v2/3rd Gen Core processor PCI Express Root Port (rev 09) 00:02.0 VGA compatible controller: Intel Corporation 3rd Gen Core processor Graphics Controller (rev 09) 00:14.0 USB controller: Intel Corporation 7 Series/C210 Series Chipset Family USB xHCI Host Controller (rev 04) 00:16.0 Communication controller: Intel Corporation 7 Series/C210 Series Chipset Family MEI Controller #1 (rev 04) 00:1a.0 USB controller: Intel Corporation 7 Series/C210 Series Chipset Family USB Enhanced Host Controller #2 (rev 04) 00:1b.0 Audio device: Intel Corporation 7 Series/C210 Series Chipset Family High Definition Audio Controller (rev 04) 00:1c.0 PCI bridge: Intel Corporation 7 Series/C210 Series Chipset Family PCI Express Root Port 1 (rev c4) 00:1c.1 PCI bridge: Intel Corporation 7 Series/C210 Series Chipset Family PCI Express Root Port 2 (rev c4) 00:1c.3 PCI bridge: Intel Corporation 7 Series/C210 Series Chipset Family PCI Express Root Port 4 (rev c4) 00:1d.0 USB controller: Intel Corporation 7 Series/C210 Series Chipset Family USB Enhanced Host Controller #1 (rev 04) 00:1f.0 ISA bridge: Intel Corporation HM76 Express Chipset LPC Controller (rev 04) 00:1f.2 SATA controller: Intel Corporation 7 Series Chipset Family 6-port SATA Controller [AHCI mode] (rev 04) 00:1f.3 SMBus: Intel Corporation 7 Series/C210 Series Chipset Family SMBus Controller (rev 04) 01:00.0 3D controller: NVIDIA Corporation GF117M [GeForce 610M/710M/820M / GT 620M/625M/630M/720M] (rev a1) 03:00.0 Network controller: MEDIATEK Corp. MT7630e 802.11bgn Wireless Network Adapter 04:00.0 Unassigned class [ff00]: Realtek Semiconductor Co., Ltd. Device 5289 (rev 01) 04:00.2 Ethernet controller: Realtek Semiconductor Co., Ltd. RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller (rev 0a)   lspci -nnk | grep 0280 -A2:  03:00.0 Network controller [0280]: MEDIATEK Corp. MT7630e 802.11bgn Wireless Network Adapter [14c3:7630]     Subsystem: Foxconn International, Inc. Device [105b:e074] 04:00.0 Unassigned class [ff00]: Realtek Semiconductor Co., Ltd. Device [10ec:5289] (rev 01) 80's Animated movie: An old man joins mission to save a boy stranded on a hostile world sequence $x_{n+1}=x_n+e^{-2018x_n}$ Let's assume a very simplified situation. There is only one seller asking for 10$ and there is only one buyer bidding 20$. At what price would the transaction go?   Some answers say it won't, because the ask price has to match the bid price. () Other answers say, that the time ordering is important. () Which answer is correct? What is the final price in the presented situation? I installed Dev C++ using Wine on Ubuntu 18.04 and the program is compiling but it not showing any output screen. What exactly is "base ISO" and how do I find what is base ISO on my camera? If I understood correctly, base ISO is in full stop steps from the lowest possible ISO setting on my camera. For example if the lowest setting on my camera is ISO 100, than the following table would be base ISO:    100, 200, 400, 800, 1600, 3200, 6400   While steps in between like   125, 160, 250, 320, 500, 640, 1000, 1300, 2000, 2500, 4000, 5000   would be third-stop "pulled" values, which result more noise than perhaps higher base ISO values (for example 2500 is worse than 3200).   Did I get this right or missed by a mile, what did I miss? Do non-base ISO values generate more noise than next higher base-value ISO stop? How do I find what are my base-value ISO stops (is it the lowest ISO value multiplied by 2n)? How are the AES S-Boxes calculated? Safari Extension to Manage Tabs? How to arrive at $\int_{0}^{1} f^3(x) \, dx < (\int_{0}^{1} f(x) \, dx)^2$? Define $f : [0,1] \to \mathbb{R} $ differentiable satisfying $f(0)=0$ and $0&lt;f'(x)&lt;1$. Prove:  $\displaystyle\int_{0}^{1} f^3(x) \, dx &lt; \left(\displaystyle\int_{0}^{1} f(x) \, dx\right)^2$ I found the following answer,  Define $F(x)=\displaystyle\int_{0}^{x} f^3(t) \, dt - \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)^2$ We have $$F'(x)=f^3(x)-2 \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)f(x)=f(x)\left[f^2(x)-2 \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)\right]$$ and $$\left[f^2(x)-2 \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)\right]'=2f(x)\left[f'(x)-1\right]$$ from which we can arrive at the conclusion.  But I felt a little confused. In the past I usually proved the inequality by classical inequalities and something like Taylor expansion to use the condition about derivative. But this answer is little out of my expectation. Do we have other proofs or ideas to arrive at $\displaystyle\int_{0}^{1} f^3(x) \, dx &lt; \left(\displaystyle\int_{0}^{1} f(x) \, dx\right)^2$? And other thoughts about solving these types of issues? I think when we talk about $\left(\displaystyle\int_{0}^{1} f(x) \, dx\right)^2 &gt;\ldots \,$ it's a little difficult to deal with. I don't quite know what tools to use. A 2D game called Deer Hunter has projectiles like this:    Are the projectiles 2D or 3D models? How could I implement something similar? Failed to save undefined: No COMPONENT named markup://svg found : I got  question in review. First sentence is:     I know there is a  question, but no answers.   And indeed the two questions are virtually identical. I started to flag as duplicate - but the original question had no answer at the time, so it wouldn't be correct and apparently . (I didn't go so far as to try to submit the flag).  What is the proper action here? How do we want such a question to be handled? Should both (or more) questions stay open indefinitely until either is answered and somebody gets the good idea to do the duplicate linking?   As the new question is essentially a bump to the front page (perhaps with re-tagging), especially in the case where the OP knows that a previous question exist, perhaps something should be done to enable bumping re-emphasizing without writing the new question.  As I wrote this,  actually got an answer. So the old one should now be closed as a duplicate of the new - but my question remains for the case where all known/found duplicate questions have no answer?  Ironically, just before asking this, the "Questions that may already have your answer" gave me ... (Although the suggestion has no answer). So it has been asked before - the comment there even suggest the ability to re-emphasize. Using while loop to ssh to multiple servers I have a file servers.txt, with list of servers:  server1.mydomain.com server2.mydomain.com server3.mydomain.com   when I read the file line by line with while and echo each line, all works as expected. All lines are printed.  $ while read HOST ; do echo $HOST ; done &lt; servers.txt server1.mydomain.com server2.mydomain.com server3.mydomain.com   However, when I want to ssh to all servers and execute a command, suddenly my while loop stops working:  $ while read HOST ; do ssh $HOST "uname -a" ; done &lt; servers.txt Linux server1 2.6.30.4-1 #1 SMP Wed Aug 12 19:55:12 EDT 2009 i686 GNU/Linux   This only connects to the first server in the list, not to all of them. I don't understand what is happening here. Can somebody please explain?  This is even stranger, since using for loop works fine:  $ for HOST in $(cat servers.txt ) ; do ssh $HOST "uname -a" ; done Linux server1 2.6.30.4-1 #1 SMP Wed Aug 12 19:55:12 EDT 2009 i686 GNU/Linux Linux server2 2.6.30.4-1 #1 SMP Wed Aug 12 19:55:12 EDT 2009 i686 GNU/Linux Linux server3 2.6.30.4-1 #1 SMP Wed Aug 12 19:55:12 EDT 2009 i686 GNU/Linux   It must be something specific to ssh, because other commands work fine, such as ping:  $ while read HOST ; do ping -c 1 $HOST ; done &lt; servers.txt Cardioid in coffee mug? Isomorphism between dual space and bilinear forms Studying from Roman's Advanced Linear Algebra, I want to prove that $$U^* \otimes V^* \cong (U \otimes V)^* \cong \hom(U,V;\mathbb{F})$$ The author proves that $U^* \otimes V^* \cong (U \otimes V)^*$ by showing that there is an unique linear transformation   $$\theta:U^* \otimes V^* \to (U \otimes V)^*$$  defined by $\theta(f \otimes g)=f \odot g$ where $(f \odot g)(u \otimes v)=f(u)g(v)$  I want to prove that $U^* \otimes V^* \cong \hom(U,V;\mathbb{F})$. I tried to use the universal property by fixing $f \in U^*, g\in V^*$ and defining $S: U^*\times V^* \to \hom(U,V;\mathbb{F})$ as $S(f,g)=F_{f,g}$, where $F_{f,g}(u,v)=f(u)g(v)$, but I got stuck. Can't download any app from AppStore! I'm using iPhone 6s 16 GB. My storage isn't full but still I can't download anything. Every time I wanted to start a download it keeps telling:  app name* could not be download at this time Is it normal to get hundreds of break-in attempts per day? I just checked my server's /var/log/auth.log and found that I'm getting over 500 failed password/break-in attempt notifications per day! My site is small, and its URL is obscure. Is this normal? Should I be taking any measures? ID a book about a mirror world and 4th dimension I read a book about 10-15 years ago about a boy and a girl who find a mirror. What made the mirror special was, they were able to enter the reflection and be a part of the reverse image world. They end up going to school and when asked to write something on the board, it ends up being backwards, but to them looks normal.  Later, the two end up in a world with 4 dimensions. With the added dimension they can't see as they normally could. If one moved a certain way the other would see their insides. Eventually a native of that dimension made some type of device that lets them see regularly.  Other than that my memory is blank. If I remember anything else I will be sure to edit it in.