The bluetooth adapter is visible and not blocked, but does not detect any devices at all while attempting to pair.   rfkill list  0: dell-wifi: Wireless LAN     Soft blocked: no     Hard blocked: no 1: dell-bluetooth: Bluetooth     Soft blocked: no     Hard blocked: no 2: phy0: Wireless LAN     Soft blocked: no     Hard blocked: no 3: brcmwl-0: Wireless LAN     Soft blocked: no     Hard blocked: no 6: hci0: Bluetooth     Soft blocked: no     Hard blocked: no   lsusb :  Bus 001 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub Bus 002 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub Bus 003 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub Bus 004 Device 001: ID 1d6b:0003 Linux Foundation 3.0 root hub Bus 001 Device 002: ID 8087:0024 Intel Corp. Integrated Rate Matching Hub Bus 002 Device 002: ID 8087:0024 Intel Corp. Integrated Rate Matching Hub Bus 003 Device 002: ID 04ca:0061 Lite-On Technology Corp.  Bus 001 Device 003: ID 0a5c:21d7 Broadcom Corp.  Bus 001 Device 004: ID 0bda:0129 Realtek Semiconductor Corp.  Bus 001 Device 005: ID 0bda:58c2 Realtek Semiconductor Corp.   I searched and found that issue lies with Broadcom BT. Is there no way around it?
Bluetooth is on but can't find or be found by any other devices.  Output of sudo service bluetooth status:  ● bluetooth.service - Bluetooth service    Loaded: loaded (/lib/systemd/system/bluetooth.service; enabled; vendor preset: enabled)    Active: active (running) since ხუთ 2015-06-04 22:33:18 GET; 13min ago  Main PID: 26678 (bluetoothd)    CGroup: /system.slice/bluetooth.service            └─26678 /usr/sbin/bluetoothd -n  ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPAG ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPHS ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSource ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSink ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPAG ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPHS ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSource ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSink ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Adapter /org/bluez/26678/hci0 has been enabled ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Adapter /org/bluez/26678/hci0 has been enabled   Output of lsusb:  Bus 003 Device 005: ID 0a5c:21d7 Broadcom Corp. BCM43142 Bluetooth 4.0   How can I fix it?
When I measure a 9V DC wall adapter power supply with a multimeter it shows 11.8V. When connected to my board this voltage lowers to 10V. Is this normal? Should I be afraid of toasting my  supplied board when using this adapter?  Edit: This regulator accepts up to 12V inputs. So that will be no problem. A second "9V" adapter yields 15V to the multimeter. Will this one burn my circuit? How to copy file path to clipboard? Is there a way to copy the currently selected filename, including the path, to the clipboard? Ideally this would be through the right-click menu.  Some degree of flexibility would be useful, like the ability to copy just the filename, path+filename, filename minus extension, etc. This isn't essential, though, as long as it can copy the path+filename.  The solution must be compatible with Windows XP, Windows Vista and Windows 7. Differences among Doctoral Degrees/titles Flag indicator present on Meta, but no flags visible Combinatorial argument for the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$ I am looking for the combinatorial argument for the identity:  \begin{equation} k\binom{n}{k} = n\binom{n-1}{k-1} \end{equation}  This is easy to show algebraically as:  \begin{equation} \binom{n}{k} = \dfrac{n(n-1)(n-2)(n-k+1)}{k(k-1)!} \end{equation}   What is the combinatorial argument? What are some general ideas to get started?   Here is a clarification of 2. From what I have seen so far, proving (combinatorially) an identity with an addition sign usually implies that we need to partition a set (this makes sense because of the addition rule and provides a nice visual). On the contrary, the previous observation leads me to believe that multiplication in identities can be resolved with the multiplication principle, but what is the "visual/interpretation" for this? Could someone provide such an interpretation for the example identity given above? Give moderators the option to select a deletion reason when removing posts How do I compile a C# script at runtime and attach it as a component to a game object The idea here is to let users program the behaviour of an agent in a way that reminds of the "Screeps" game, where you get to develop code which later runs and dictates the behaviour of your units.  My idea is to let users develop the behaviour or their units with C# code, compile that code at runtime and attach it as a MonoBehavior to an existing game object.  How can I achieve this? Map composer issue I'm planning to use ZFS on my system (HP ML370 G5, Smart Array P400, 8 SAS disk). I want ZFS to manage all disks individually, so it can utilize better scheduling (i.e. I want to use software RAID feature in ZFS).  The problem is, I can't find a way to disable RAID feature on the RAID controller. Right now, the controller aggregates all of the disks into one big RAID-5 volume. So ZFS can't see individual disk.  Is there any way to acomplish this setup? Geometry with Trigonometry if command in find -exec I was just trying to list all directories and files under current directory and also write if they are file or directory with the following command:  find -exec echo `echo {} : ;if [ -f {} ]; then echo file; else echo directory;fi` \;   I know it is a silly command, I can use other things like -type f or -type d, but I want to learn why that piece of code did not work as I expected. It just prints directory to all of them. For example while output of find is:  . ./dir ./dir/file   output of my code is :  . : directory ./dir : directory ./dir/file : directory   And output of   echo `echo dir/file : ;if [ -f dir/file ]; then echo file; else echo directory;fi`   is  dir/file : file   I am working on Ubuntu 14.10 and using find (GNU findutils) 4.4.2 Turn off text message (SMS) sync to Exchange email account I recently started using a Samsung Galaxy S4 and one of the features in a synchronization between text messages and email accounts.  When a text message comes in it throws a copy into my Exchange email accounts that are one the phone. I want to turn this feature off but cannot find a setting for it anywhere.  Any ideas? Can I use grep matches to execute command? I need to upgrade php from 5.3 to 5.4. I've installed it with rpm. I want clean install: delete all 5.3 packages, and install all necessary 5.4. Here is the command to see all installed packages:  rpm -qa |grep php   It prints out a list of packages, and I want to remove all them with:  rpm -e *pakage.name*   Can I iterate over all matches in grep results and use each match in:   rpm -e *grep.match* Given the following problem     Let $U$ and $V$ be finite dimensional vector spaces, such that $\dim(U) = \dim(V)$. Prove that $U$ and $V$ are isomorphic.   I was just wondering if somebody could critique and validate my proof.  Proof  Two vector spaces $U$ and $V$ are isomorphic if and only if there exists a bijection $\phi: U \to V$. That is, an invertible linear map between the two. Let $\dim(U) = \dim(V) = n$ and let $\left\{u_1, \dots, u_n\right\}$ and $\left\{v_1, \dots, v_n\right\}$ be bases for $U$ and $V$, respectively. It follows that any vector $u \in U$ can be written as  $$ u = c_1u_1 + \cdots + c_nu_n $$  We define $\phi: U \to V$ as  $$ \phi(c_1u_1 + \cdots + c_nu_n) = c_1v_1 + \cdots + c_nv_n $$  Observe that $\phi$ is invertible. That is, given a vector  $$ v = d_1v_1 + \cdots + d_nv_n $$  in $V$, we define $\phi^{-1}: V \to U$ as  $$ \phi^{-1}(d_1v_1 + \cdots + d_nv_n) = d_1u_1 + \cdots + d_nu_n $$  Since there exists a bijection between $U$ and $V$, they are isomorphic. Identification card for foreigners in trains in Germany I want to go to Berlin to Gottingen by train. I found that to avail myself of the &ldquo;savings fare&rdquo;, I have to book a ticket at least one day in advance. And I saw a requirement on the Bahn website to have a identification card for online tickets.  I am a foreigner in Germany, so does my passport count as valid identification in German trains? How to install the Boot-Repair tool in an Ubuntu live disc? When the bootloader (GRUB) is broken, I cannot access neither Ubuntu nor Windows.  I heard that an easy solution to fix this is to use the Boot-Repair tool from an Ubuntu liveCD (or liveUSB).  What are the steps to do this? Would you ever stop rolling the die? Why does replication require primers while transcription does not? From the man page on git-merge, there are a number of merge strategies you can use.     resolve -  This can only resolve two heads (i.e. the current branch and another branch you pulled from) using 3-way merge algorithm. It tries to carefully detect criss-cross merge ambiguities and is considered generally safe and fast. recursive -  This can only resolve two heads using 3-way merge algorithm. When there are more than one common ancestors that can be used for 3-way merge, it creates a merged tree of the common ancestors and uses that as the reference tree for the 3-way merge. This has been reported to result in fewer merge conflicts without causing mis-merges by tests done on actual merge commits taken from Linux 2.6 kernel development history. Additionally this can detect and handle merges involving renames. This is the default merge strategy when pulling or merging one branch. octopus -  This resolves more than two-head case, but refuses to do complex merge that needs manual resolution. It is primarily meant to be used for bundling topic branch heads together. This is the default merge strategy when pulling or merging more than one branches. ours -  This resolves any number of heads, but the result of the merge is always the current branch head. It is meant to be used to supersede old development history of side branches. subtree -  This is a modified recursive strategy. When merging trees A and B, if B corresponds to a subtree of A, B is first adjusted to match the tree structure of A, instead of reading the trees at the same level. This adjustment is also done to the common ancestor tree.   When should I specify something different than the default?  What scenarios are each best for? Quantum teleportation and no-communication theorem According to the Wikipedia article for the :     In very rough terms, the theorem describes a situation that is analogous to two people, each with a radio receiver, listening to a common radio station: it is impossible for one of the listeners to use their radio receiver to send messages to the other listener. This analogy is imprecise, because quantum entanglement suggests that perhaps a message could have been conveyed; the theorem replies 'no, this is not possible'.   According to the Jet Propulsion's recent article on quantum teleportation:   they can effect an entangled photon (B) with another photon (A) to change the state of the other entangled photon.  Doesn't this contradict what the no-communication theorem states ?