I am currently using Ubuntu 20.04. My Laptop uses Intel Wireless-AC 9560 chipset.     I am getting the above comment in the Additional Drivers menu of "Software &amp; Updates". As you can see I am able to connect to my WiFi perfectly without any problems still I see the "This device is not working" message in the Additional Drivers section.  Is this a bug of some kind? How do I solve it such that I am able to see something similar to what I see right now for my Nvidia Graphics.   I can also confirm that I see the required firmware installed in my Laptop (double checked from the Intel website).   Note : I have my exact question . I thought I'll edit it because the image was not in English and didn't state any information except that it wasn't working. But some people (not all :) ) just rejected my review without even giving any proper explanation. Actually, there was the automated reason that I should have written it as a comment or an answer (wow, do people just review edit suggestions for the sake of it?). I hope that at least you'll look at my profile and understand that I don't have commenting powers as of yet (ouch). I request you to not mark this question as duplicate as I need some solution (enter an innocent puppy face).
Recently I've installed Kubuntu in a newly bought laptop with an Intel WiFi 6 AX200 network card, I can connect to the wifi without apparent problems but when I open the driver manager in the settings I face this weird situation:  Intel Corporation: Wi-Fi 6 AX200 This device is not working.  [ ] Using iwlwifi driver backport in DKMS format from backport-iwlwifi-dkms (open source)  [ ] Continue using a manually installed driver  [x] Do not use the device   Note that "Continue using a manually installed driver" is the only voice selectable in the list.  Here is some additional information:  $ lspci -nnk | grep 0280 -A3 02:00.0 Network controller [0280]: Intel Corporation Wi-Fi 6 AX200 [8086:2723] (rev 1a)         Subsystem: Intel Corporation Wi-Fi 6 AX200 [8086:0084]         Kernel driver in use: iwlwifi         Kernel modules: iwlwifi  $ rfkill list all 0: hci0: Bluetooth         Soft blocked: yes         Hard blocked: no 1: phy0: Wireless LAN         Soft blocked: no         Hard blocked: no  $ dmesg | grep iwl [    2.888227] iwlwifi 0000:02:00.0: enabling device (0000 -&gt; 0002) [    2.945179] iwlwifi 0000:02:00.0: Direct firmware load for iwlwifi-cc-a0-50.ucode failed with error -2 [    2.945217] iwlwifi 0000:02:00.0: Direct firmware load for iwlwifi-cc-a0-49.ucode failed with error -2 [    2.947244] iwlwifi 0000:02:00.0: TLV_FW_FSEQ_VERSION: FSEQ Version: 43.2.23.17 [    2.947250] iwlwifi 0000:02:00.0: Found debug destination: EXTERNAL_DRAM [    2.947252] iwlwifi 0000:02:00.0: Found debug configuration: 0 [    2.947697] iwlwifi 0000:02:00.0: loaded firmware version 48.4fa0041f.0 op_mode iwlmvm [    3.145585] iwlwifi 0000:02:00.0: Detected Intel(R) Wi-Fi 6 AX200 160MHz, REV=0x340 [    3.159226] iwlwifi 0000:02:00.0: Applying debug destination EXTERNAL_DRAM [    3.160060] iwlwifi 0000:02:00.0: Allocated 0x00400000 bytes for firmware monitor. [    3.336456] iwlwifi 0000:02:00.0: base HW address: 50:eb:71:d7:0b:40 [    3.353553] iwlwifi 0000:02:00.0 wlp2s0: renamed from wlan0 [    4.457920] iwlwifi 0000:02:00.0: Applying debug destination EXTERNAL_DRAM [    4.621843] iwlwifi 0000:02:00.0: FW already configured (0) - re-configuring [    4.629999] iwlwifi 0000:02:00.0: BIOS contains WGDS but no WRDS   Someone have idea why this happened?
How do Idle games progress while not active? What happens to the energy when waves perfectly cancel each other? What happens to the energy when waves perfectly cancel each other (destructive interference)? It appears that the energy "disappear" but the law of conservation of energy states that it can't be destructed. My guess is that the kinetic energy is transformed into potential energy. Or maybe it depends on the context of the waves where do the energy goes? Can someone elaborate on that or correct me if I'm wrong? When I apply a glossy shader to the material of my text, the text becomes just black. Why?    I have lamp and everything, but the problem persists. The text is red before I add the glossy shader.  I'm using Cycles render.   Thanks! I'm considering finite groups only. Cayley's theorem says the a group $G$ is isomorphic to a subgroup of $S_{|G|}$. I think it's interesting to ask for smaller values of $n$ for which $G$ is a subgroup of $S_n$. Obviously, it's not always possible to do better than Cayley's theorem. But sometimes it is possible (for example, $\mathbb{Z}_6$ as a subgroup of $S_5$).  So I'm asking:        Given a finite group $G$, is there an algorithmic way to find or approximate the minimal $n$ for which $G$ is isomorphic to a subgroup of $S_n$?   If the answer to $(1)$ is not known, is it known for specific classes of groups?   In particular, for finite abelian groups, is it true that for a prime $p$, the minimal $n$ for $\mathbb{Z}_{p^{t_1}} \times \mathbb{Z}_{p^{t_2}}$ is $p^{t_1}+p^{t_2}$ (I can prove that is is true for different primes $p_1$ and $p_2$, but have problems when it's the same prime in both factors).      Thanks! Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$ I need to show that  $$ \int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3} $$  I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from.  Using a change of variable $ x \mapsto 2x $ :  $$ \int_0^\infty \frac{\sin^2(2x)}{x^2}dx = \pi $$  Now using the identity $\sin^2(2x) = 4\sin^2x - 4\sin^4x $, we obtain  $$ \int_0^\infty \frac{\sin^2x - \sin^4x}{x^2}dx = \frac{\pi}{4} $$ $$ \frac{\pi}{2} - \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4} $$ $$ \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4} $$  But I am now at a loss as to how to make $x^4$ appear at the denominator.  Any ideas appreciated.  Important: I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and use the change of variable and identity mentioned above If the lines intersect the vertices of the square. The area of the square is $1$ and the lines also intersect the midpoints of the square lines. How to find the area of shaded region? Yesterday, Asaf Karagila's  sparked an extensive discussion on ways of proving that all countable ordinals are embeddable in $\Bbb Q$, and whether particular solutions to this use the Axiom of Choice.  Since the arguments used to prove this can be shady on their use of Choice when formulated tersely, fully fleshed out answers are appreciated.  To be concrete, this question asks for proofs of the following fact:     All countable ordinals $\alpha &lt; \omega_1$ are order-embeddable in the rationals $\Bbb Q$.   Especially answers avoiding the Axiom of Choice are appreciated.    Of relevance and related interest is  (but it's not a duplicate, since that question asks for explicit embeddings). Is Object constructor called when creating an array in Java? In Java, an array IS AN Object. My question is... is an Object constructor called when new arrays is being created? We would like to use this fact to instrument Object constructor with some extra bytecode which checks length of array being constructed. Would that work? How do I find the Arch-mage at the end of the quest "Under Sarhaal"? Different voltage characteristics of CMOS NAND gate for different connections Recently, I've played with a simple 2-input 1-output NAND gate realised in CMOS technology as shown in Fig. 1.     Fig. 1. CMOS NAND scheme.  I took some measurements of volteges U(output) vs. U(input). Using two independent voltage sources (constant high level Udd and Uin varying in 0-to-Udd range), I considered three different connection possibilities so to investigate all four input configurations. Those were:  a)  Input terminal A connected to Udd, input terminal B connected to Uin.  b)  Input terminal A connected to Uin, input terminal B connected to Udd.  c)  Both input terminals connected to Uin.  I've obtained following characteristics:    Fig. 2. CMOS NAND voltage characteristics with different connection methods.  As long as a difference between case c) and other cases I could have expected, I was surprised to see a) and b) differ as well. Could someone please explain to me why would changing voltages connection between A and B significantly changes threshold voltage of the gate? Is it a widely observed effect or have I just picked a defective NAND? The proof I have starts of with $\;xy=0\;$ in a field.   Then $x^{-1}$ exists because it is a field.  Then $x^{-1} xy=x^{-1} 0$.  Therefore $y=0$.  But surely if an integral domain can not have any zero divisors, how can we end the proof by saying $y=0$? Surely then y is a zero divisor and hence the field is not an integral domain? If $[G' : G'']\leq p^2$, then $G'$ is abelian. Problem : Let $G$ be a p-group and $G'$ denote the commutator subgroup of $G$. If $[G' : G'']\leq p^2$, then $G'$ is abelian.  It is easy to prove it for the case of $[G' : G'']=1$ since G is solvable. But it is too hard to prove the other cases. Please give me any advice or solutions. Thanks in advance. Display Tag Excerpt tooltips when editing a question What is an "Emergency Call," and can I  set number for it? I know it is not recommended, but is it at all possible to pass the user's password to scp?  I'd like to copy a file via scp as part of a batch job and the receiving server does, of course, need a password and, no, I cannot easily change that to key-based authentication. How do I detect what .NET Framework versions and service packs are installed? A similar question was asked , but it was specific to .NET 3.5. Specifically, I'm looking for the following:   What is the correct way to determine which .NET Framework versions and service packs are installed? Is there a list of registry keys that can be used? Are there any dependencies between Framework versions? jsp:useBean scope Explain how (or if) a box full of photons would weigh more due to massless photons I understand that mass-energy equivalence is often misinterpreted as saying that mass can be converted into energy and vice versa.  The reality is that energy is always manifested as mass in some form, but I struggle with some cases:  Understood Nuclear Decay Example  In the case of a simple nuclear reaction, for instance, the total system mass remains the same since the mass deficit (in rest masses) is accounted for in the greater relativistic masses of the products per $E=\Delta m c^2$.  When a neutron decays and you are left with a fast proton and a relativistic electron.  If you could weigh those two without slowing them down, you would find it weighed as much as the original neutron.  Light in a Box  This becomes more difficult for me when moving to massless particles like photons.  Photons can transmit energy from one heavy particle to another.  When a photon is absorbed the relativistic mass (not the rest mass) of the (previously stationary) particle that absorbs it increases.  But if my understanding is correct, the energy must still be manifested as mass somehow while the photon is in-flight, in spite of the fact that the photon does not have mass.  So let's consider a box with the interior entirely lined with perfect mirrors.  I have the tare weight of the box with no photons in it.  When photons are present the box has an additional quantifiable amount of energy (quantified below) due to the in-flight photons.  Say there are $N$ photons... obviously assume $N$ is large.  $$\Delta m = \frac{ E }{ c^2 } = \frac{ N h }{ \lambda c}$$  Interactions are limited to reflections with the wall, which manifest as a constant pressure on the walls.  If I hold this box in a constant gravitational field (like the surface of Earth) then there will be a gradient in the pressure that pushes down slightly.  Is this correct?  Wouldn't there still technically be mass as the photons are in-flight, which would cause its own gravitational field just as all matter does?  How is this all consistent with the assertion that photons are massless?  Is it really correct to say that photons don't have mass?  It seems to be a big stretch.  Please offer a more complete and physically accurate account of this mirror-box. Is it possible to play FIFA 2015, 4v4 on 2 Consoles? I've the following situation: My friends and I, we're 8 in total, want to play Fifa 15 4v4. I know that the Playstation 4 just supports 4 controllers and thats why we think about a 2 console + 2 TV's setup, so 4 players per console.   So my question is: Is it possible or is there a game mode available were we could play such a setup?  Update: After I got the first answer I've to add that I know that the only way to play on 2 consoles would be through the psn online features. So we did not plan to play on Lan. We just want to sit in the same room with 2 consoles and play against each other over psn. I have being using Skype for a long time on my Ubuntu 14.04 LTS. From this morning on, Skype simple cannot connect to my account anymore.  That is my Skype version:     Skype 4.3.0.37   © 2014 Skype and/or Microsoft   My Ubuntu system:  LSB Version:    core-2.0-amd64:core-2.0-noarch:core-3.0-amd64:core-3.0-noarch:core-3.1-amd64:core-3.1-noarch:core-3.2-amd64:core-3.2-noarch:core-4.0-amd64:core-4.0-noarch:core-4.1-amd64:core-4.1-noarch:cxx-3.0-amd64:cxx-3.0-noarch:cxx-3.1-amd64:cxx-3.1-noarch:cxx-3.2-amd64:cxx-3.2-noarch:cxx-4.0-amd64:cxx-4.0-noarch:cxx-4.1-amd64:cxx-4.1-noarch:desktop-3.1-amd64:desktop-3.1-noarch:desktop-3.2-amd64:desktop-3.2-noarch:desktop-4.0-amd64:desktop-4.0-noarch:desktop-4.1-amd64:desktop-4.1-noarch:graphics-2.0-amd64:graphics-2.0-noarch:graphics-3.0-amd64:graphics-3.0-noarch:graphics-3.1-amd64:graphics-3.1-noarch:graphics-3.2-amd64:graphics-3.2-noarch:graphics-4.0-amd64:graphics-4.0-noarch:graphics-4.1-amd64:graphics-4.1-noarch:languages-3.2-amd64:languages-3.2-noarch:languages-4.0-amd64:languages-4.0-noarch:languages-4.1-amd64:languages-4.1-noarch:multimedia-3.2-amd64:multimedia-3.2-noarch:multimedia-4.0-amd64:multimedia-4.0-noarch:multimedia-4.1-amd64:multimedia-4.1-noarch:printing-3.2-amd64:printing-3.2-noarch:printing-4.0-amd64:printing-4.0-noarch:printing-4.1-amd64:printing-4.1-noarch:qt4-3.1-amd64:qt4-3.1-noarch:security-4.0-amd64:security-4.0-noarch:security-4.1-amd64:security-4.1-noarch Distributor ID: Ubuntu Description:    Ubuntu 14.04.3 LTS Release:    14.04 Codename:   trusty   Any suggestion why or how to fix it?  Thanks in advance.