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About this!This is a known problem with Windows executables written in the Go languauge which is documented at the Go website. the Go web site:Why does my virus-scanning software think my Go distribution or compiled binary is infected?This is a common occurrence, especially on Windows machines, and is almost always a false positive.Commercial virus scanning programs are often confused by the structure of Go binaries, which they don't see asoften as those compiled from other languages.If you've just installed the Go distribution and the system reports it is infected, that's certainly a mistake.To be really thorough, you can verify the download by comparing the checksum with those on the downloads page.In any case, if you believe the report is in error, please report a bug to the supplier of your virus scanner.Maybe in time virus scanners can learn to understand Go programs.5.2 AMD CPUsA patched macOS AMD kernel must be used to run on older AMD systems, but there is a workaround if you have a modernAMD Ryzen CPU. The unlocker cannot patch this but we can recommend settings for the VMX file that allows macOS torun on recent AMD CPUs. The tests are being recorded in this issue, and it would be useful if more can reportsuccess or failures in that issue.DrDonk#33You must have Hyper-V disabled on Windows or VMware falls back to a mode called ULM. CPUID masking is not available in ULM mode as it is pushed to Windows for the low level VMX/SVM operations. You can see if VMware is in ULM mode on Hyper_V by searching the guest's vmware.log file for these 2 lines:vmx IOPL_Init: Hyper-V detected by CPUIDMonitor Mode: ULMHere is a link to a Microsoft artcile on disabling Hyper-V. the VMX fileRead this KB article to learn how to edit a guest's VMX file safely the following lines to the VMX file:cpuid.0.eax = "0000:0000:0000:0000:0000:0000:0000:1011"cpuid.0.ebx = "0111:0101:0110:1110:0110:0101:0100:0111"cpuid.0.ecx = "0110:1100:0110:0101:0111:0100:0110:1110"cpuid.0.edx = "0100:1001:0110:0101:0110:1110:0110:1001"cpuid.1.eax = "0000:0000:0000:0001:0000:0110:0111:0001"cpuid.1.ebx = "0000:0010:0000:0001:0000:1000:0000:0000"cpuid.1.ecx = "1000:0010:1001:1000:0010:0010:0000:0011"cpuid.1.edx = "0000:0111:1000:1011:1111:1011:1111:1111"vhv.enable = "FALSE"vpmc.enable = "FALSE"Make sure there are no duplicate lines in the VMX file or the guest will not start and a dictionary error willbe displayed by VMware.You can now install and run macOS as a gu6. VMware DownloadsThese URLs will link to the latest versions of VMware's hosted products:VMware Fusion Workstation for Windows Workstation for Linux Player for Windows Player for Linux VMware ToolsThe Unlocker provides the VMware tools ISO images. There can be Related searches » nanosoft smart inv.net 7.0 » spread studio for net 7.0 v4 » spread studio for net 7.0 v4 version » secret net 7.0 скачать » rsnetworx for device net 7.0 » net 7.0 desktop runtime » dot net 7.0 » net 7.0 runtime » aspose.cells for net 2003 v4.7.0 » net extender 7.0 More Microsoft .NET Framework 6.0.11 Editor's Review: Microsoft .NET FrameworkThe Microsoft .NET Framework is a software development platform developed by Microsoft that offers a comprehensive and consistent programming model for building applications. more info... 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Length 52, bad cksum 0 (->376d)!) 192.168.1.105.57010 > 192.168.1.105.http: Flags [.], cksum 0x8449 (incorrect -> 0x7a07), ack 1, win 65328, options [nop,nop,TS val 317572454 ecr 317572454], length 0 0x0000: 4500 0034 7f34 4000 4006 0000 c0a8 0169 E..4.4@[email protected] 0x0010: c0a8 0169 deb2 0050 6a91 b7b5 5fc3 67ad ...i...Pj..._.g. 0x0020: 8010 ff30 8449 0000 0101 080a 12ed c566 ...0.I.........f 0x0030: 12ed c566 ...f10:56:06.872906 IP (tos 0x0, ttl 64, id 1603, offset 0, flags [DF], proto TCP (6), length 52, bad cksum 0 (->b05e)!) 192.168.1.105.57011 > 192.168.1.105.http: Flags [.], cksum 0x8449 (incorrect -> 0xe6a1), ack 1, win 65328, options [nop,nop,TS val 317572454 ecr 317572454], length 0 0x0000: 4500 0034 0643 4000 4006 0000 c0a8 0169 E..4.C@[email protected] 0x0010: c0a8 0169 deb3 0050 e79e 5105 5b14 e963 ...i...P..Q.[..c 0x0020: 8010 ff30 8449 0000 0101 080a 12ed c566 ...0.I.........f 0x0030: 12ed c566 ...f10:56:06.877862 IP (tos 0x0, ttl 64, id 42608, offset 0, flags [DF], proto TCP (6), length 64, bad cksum 0 (->1025)!) 192.168.1.105.57012 > 192.168.1.105.http: Flags [S], cksum 0x8455 (incorrect -> 0x4db9), seq 1017984763, win 1, options [mss 16344,nop,wscale 0,nop,nop,TS val 317572454 ecr 0,sackOK,eol], length 0 0x0000: 4500 0040 a670 4000 4006 0000 c0a8 0169 [email protected]@[email protected] 0x0010: c0a8 0169 deb4 0050 3cad 36fb 0000 0000 ...i...PFor further visual evidence, here is a screen shot of the Server-Status page right when the attack is stopped:Notice that all of the threads are in a "W" Sending Reply state.Mitigating Slow-Read DoS Attacks with ModSecurityModSecurity v2.6 introduced a new directive called SecWriteStateLimit, which will place a limit on the concurrent number of threads (per IP address) in a SERVER_BUSY_WRITE state. This was originally created to help mitigate the Slow Request Body DoS attacks as Apache moves these threads into this state when it is reading request body payloads. Well, as it turns out, when Apache is sending response. Parameter: Value: Comments : 5010: 0 : 5012: 0 : 5011: 1111 0101 = Baud Rate 2400 . 5013: 1111 0101 : See below for other baud rates : 1111 0011 = 600 1111 0100 = 1200

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Thus the RCM constraint can be satisfied and modulated by resolving the joint angle variable \(\theta \).4 Simplified RNN for Redundancy Resolution with RCM ConstraintsIn this section, the modeling and theoretical analysis of the proposed SIMPLIFIED RNN for redundancy resolution with RCM constraints are addressed.4.1 Optimization ParadigmIn order to deal with the RCM constrained redundancy resolution issue, in this work, Lagrange function which is simultaneously associated with the kinematic objective function and the RCM constraint is defined as follows$$\begin{aligned} \begin{aligned} L=&\dot{\theta }^T\dot{\theta }/2+c_2\dot{k}^2/2+\lambda _1^T[\dot{k}(r_A-r_B)+kJ_1\dot{\theta }+(1-k)J_2\dot{\theta }\\&+c_1(r_P-r_P(0))]+\lambda _2^T[J_1{\dot{\theta }}+c_3(r_A-r_d)-\dot{r}_d] \end{aligned} \end{aligned}$$ (10) where \(\lambda _1\) and \(\lambda _2\) denote the Lagrange multiplier vectors. The partial derivatives of the Lagrange function with respect to the unknown variables \(({\dot{\theta }},\dot{k},\lambda _1,\lambda _2)\) are$$\begin{aligned} \left\{ \begin{array}{ll} \frac{\partial L}{\partial {\dot{\theta }}}={\dot{\theta }}+[kJ_1+(1-k)J_2]^T\lambda _{1}+J_1^T\lambda _2\\ \frac{\partial L}{\partial \dot{k}}=c_2\dot{k}+ \lambda _{1}^T(r_A-r_B)\\ \frac{\partial L}{\partial \lambda _{1}}=(kJ_1+(1-k)J_2)\dot{\theta }+\dot{k}(r_A-r_B)\\ ~~~~~~~+c_1(r_P-r_P(0))\\ \frac{\partial L}{\partial \lambda _{2}}=J_1{\dot{\theta }}+c_3(r_A-r_d)-\dot{r}_d \end{array} \right. \end{aligned}$$ (11) According to the Karush-Kuhn-Tucker (KKT) conditions [25] , the optimization above can be solved by force the following equations$$\begin{aligned} \left\{ \begin{array}{ll} \frac{\partial L}{\partial {\dot{\theta }}}=0\\ \frac{\partial L}{\partial \dot{k}}=0\\ \frac{\partial L}{\partial \lambda _{1}}=0\\ \frac{\partial L}{\partial \lambda _{2}}=0 \end{array} \right. \end{aligned}$$ (12) to be zero and obtain the desired solution \(({\dot{\theta }}^*,\dot{k}^*,\lambda _1^*,\lambda _2^*)\) for satisfying the optimization objective and constraints.4.2 Original ZD-based RNN MethodIn order to solve such optimization paradigm which reflects the redundancy resolution with RCM constraints, according to the general design principle of the ZD-based RNN model to make the aforementioned partial derivatives of the Lagrange function being 0, we thus need to construct the following error-monitoring function$$\begin{aligned} \begin{aligned} \Xi&=\begin{bmatrix} {\dot{\theta }}+[kJ_1+(1-k)J_2]^T\lambda _{1}+J_1^T\lambda _2\\ c_2\dot{k}+ \lambda _{1}^T(r_A-r_B)\\ (kJ_1+(1-k)J_2)\dot{\theta }+\dot{k}(r_A-r_B)+c_1(r_P-r_P(0))\\ J_1{\dot{\theta }}+c_3(r_A-r_d)-\dot{r}_d \end{bmatrix}\\&=AZ+B \end{aligned} \end{aligned}$$ (13) where$$\begin{aligned} A= & {} \begin{bmatrix} I_{n\times n} &{} 0_{n\times 1} &{} kJ_1^T+(1-k)J_2^T &{} J_1^T\\ 0 &{} c_2 &{} r_A^T-r_B^T &{} 0_{1\times m} \\ kJ_1+(1-k)J_2 &{} r_A-r_B &{} 0 &{} 0 \\ J_1 &{} 0_{m\times 1} &{} 0 &{} 0 \end{bmatrix} \end{aligned}$$ (14) $$\begin{aligned} Z= & {} \begin{bmatrix} {\dot{\theta }}\\ \dot{k}\\ \lambda _1\\ \lambda _2 \end{bmatrix}~~\text {and}~~B=\begin{bmatrix} 0_{n\times 1}\\ 0\\ 0_{m\times 1}\\ c_3(r_A-r_d)-\dot{r}_d \end{bmatrix} \end{aligned}$$ (15) Our goal of applying ZD-based method is to force \(\varXi =0\) eventually and thus the resultant optimal solution of Z can be got.Based on the design discipline of the ZD method, we have the following error-processing formula for redundancy resolution of manipulator with RCM constraints$$\begin{aligned} \dot{\varXi }=-\gamma \varPsi (\varXi ) \end{aligned}$$ (16) where the convergence scaling parameters can be configured as \(\gamma >0\) for unity, \(\varPsi (\cdot ):R^{(n+7)\times (n+7)}\rightarrow R^{(n+7)\times (n+7)}\) denotes the general nonlinear activation function array with its entries being monotonously-increasing odd functions. Therefore, we would have the following ZD-based neural network model for redundancy resolution of manipulator with RCM constraints$$\begin{aligned} \dot{A}Z+A\dot{Z}+\dot{B}=-\gamma \varPsi (AZ+B) \end{aligned}$$ (17) Due to existence of time derivative of coefficient matrix A in the system equation above, such ZD model needs to obtain the time-derivatives of Jacobian matrices analytically during its solution to update the model states,i.e., the coefficient matrix \(\dot{A}\) that is consisted of Jacobian matrices \(\dot{J}_1\) and \(\dot{J}_2\). However, the analytical time-derivative As we depend on some tiff features that are not implemented, I’m using ImageMagick via Magick.NET.NET 5 Generate Captcha imageAnd you think what your images will be x10 stronger?) Btw, you can use cross-platform Magick.NET, but don't do that use any of nuget packages captchas, most of then have image difficulty settings.Resizing an image without System.DrawingIf you're happy with using image magick there's always Magick.net Posts with mentions or reviews of SkiaSharp. We have used some of these posts to build our list of alternatives and similar projects. The last one was on 2023-06-18.Jellyfin: The Free Software Media SystemIt looks like the port was compiled with a binary SkiaSharp [0] since that requires Google tooling to build. Interesting to see the committer allowing this. Apparently it is not the first time.0. just dosen't workThe library that I've had the best luck with is SkiaSharp with .net 3.1. That said, I agree with others that I'd update the version of .net if possible.Open-source cross-platform C# library that encode .BMP files?Found this issue that basically say it's not supported in skia directly, and it's right, from this page it seems only supported on iOS/Mac. This is pretty frustrating considering the documentation of SkiaSharp Encode() doesn't mention it.Easy-to-use 2D graphics librariesCreating 2d Interactive Building Map with Blazor.Yes, I'm not aware of any way to use either the canvas api or webgl directly. This project has bindings for the canvas API, but it hasn't been updated in two years: SkiaSharp seems a nice option too: It supports graphics across a whole load of platforms, including Blazor on the web.What library should I use to make basic 2D graphics which is simple to use and that is able to quickly draw a bunch of particles (circles) on the screen with updating positions every frame? (Not the best description ever. I explain it better in the post's text)Compress/resize imagesIf you want pure .NET, cross-platform, and future compatibility, SkiaSharp is the new way to do it.ImageSharp leaving the .NET Foundation due to licensing changeProblem using Plugin.Firebase on IOSThis seems to be a clash with skiasharp. The other

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Select Topic AreaGeneralBodyI have had declining downloads speeds from all github project over the past few months. My ISP is Optimum with the account located in Texas. I have a 300Mbs up/25Mbs down connection with fixe IPv4 address. The download speeds I get from github a basically dial-up speeds of less than 100Kbps.For example, the last download of mongodb 4.4.29 source took nearly 10 minutes to download 50MB, e.g. Downloading r4.4.29.tar.gz... % Total % Received % Xferd Average Speed Time Time Time Current Dload Upload Total Spent Left Speed 0 0 0 0 0 0 0 0 --:--:-- --:--:-- --:--:-- 0100 49.2M 0 49.2M 0 0 90525 0 --:--:-- 0:09:30 --:--:-- 149k"> -> Downloading r4.4.29.tar.gz... % Total % Received % Xferd Average Speed Time Time Time Current Dload Upload Total Spent Left Speed 0 0 0 0 0 0 0 0 --:--:-- --:--:-- --:--:-- 0100 49.2M 0 49.2M 0 0 90525 0 --:--:-- 0:09:30 --:--:-- 149kThe individual hop-times seem okay in traceroute between my computer and github.com, but there are 30 hops listed, e.g.# traceroute github.comtraceroute to github.com (140.82.114.4), 30 hops max, 60 byte packets 1 _gateway (192.168.6.11) 0.751 ms 1.021 ms 0.700 ms 2 * * * 3 173.219.152.4 (173.219.152.4) 13.881 ms 13.866 ms 14.342 ms 4 173.219.197.86 (173.219.197.86) 17.576 ms 17.693 ms 17.549 ms 5 dls-b23-link.ip.twelve99.net (213.248.88.228) 16.846 ms 16.832 ms 19.765 ms 6 dls-bb1-link.ip.twelve99.net (62.115.136.118) 20.769 ms * * 7 * * * 8 * atl-bb1-link.ip.twelve99.net (62.115.137.54) 34.364 ms 35.094 ms 9 rest-bb1-link.ip.twelve99.net (62.115.138.70) 53.207 ms 54.298 ms 53.931 ms10 rest-b2-link.ip.twelve99.net (62.115.123.41) 53.472 ms 52.949 ms 50.715 ms11 github-ic-368832.ip.twelve99-cust.net (213.248.67.47) 64.777 ms 50.298 ms 53.864 ms12 * * *13 * * *14 * * *15 * * *16 * * *17 * * *18 * * *19 * * *20 * * *21 * * *22 *

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More information on my problem;# lsmodCode: Select allModule Size Used byparport_pc 32068 0 lp 12580 0 parport 35656 2 parport_pc,lpsnd_pcm_oss 40992 0 snd_seq_dummy 3844 0 snd_seq_oss 31360 0 snd_seq_midi_event 7168 1 snd_seq_osssnd_seq 47952 5 snd_seq_dummy,snd_seq_oss,snd_seq_midi_eventsnd_seq_device 8204 3 snd_seq_dummy,snd_seq_oss,snd_seqsnd_mixer_oss 16640 1 snd_pcm_ossusb_storage 85952 1 serio_raw 6788 0 squashfs 46856 0 unionfs 73572 0 fuse 48284 0 nls_iso8859_1 4096 1 ohci1394 31792 0 nls_cp437 5760 1 pcspkr 2944 0 ieee1394 89272 1 ohci1394rt2860sta 522456 1 snd_hda_intel 333204 0 snd_pcm 73736 2 snd_pcm_oss,snd_hda_intelforcedeth 48268 0 snd_timer 22276 2 snd_seq,snd_pcmsnd 52772 8 snd_pcm_oss,snd_seq_oss,snd_seq,snd_seq_device,snd_mixer_oss,snd_hda_intel,snd_pcm,snd_timersoundcore 7520 1 sndsnd_page_alloc 10376 2 snd_hda_intel,snd_pcmehci_hcd 33932 0 ohci_hcd 25860 0 usbcore 136560 4 usb_storage,ehci_hcd,ohci_hcdssb 34436 1 ohci_hcdwmi 7720 0 fan 5508 0 evdev 11776 0 thermal 18716 0 button 8080 0 processor 32816 1 thermal# lsmod in Slax (sound working)Code: Select allModule Size Used bysnd_seq_dummy 6788 0 snd_seq_oss 32896 0 snd_seq_midi_event 10112 1 snd_seq_osssnd_seq 50416 5 snd_seq_dummy,snd_seq_oss,snd_seq_midi_eventsnd_seq_device 10380 3 snd_seq_dummy,snd_seq_oss,snd_seqsnd_pcm_oss 40608 0 snd_mixer_oss 18048 2 snd_pcm_osspcmcia 36012 0 pcmcia_core 35092 1 pcmciaagpgart 32200 0 lp 13444 0 ppdev 11396 0 parport_pc 27940 0 parport 34668 3 lp,ppdev,parport_pcpcspkr 6400 0 snd_hda_intel 363276 1 snd_pcm 69892 2 snd_pcm_oss,snd_hda_intelsnd_timer 22792 2 snd_seq,snd_pcmpsmouse 40208 0 snd_page_alloc 11272 2 snd_hda_intel,snd_pcmsnd_hwdep 10756 1 snd_hda_intelsnd 50596 9 snd_seq_oss,snd_seq,snd_seq_device,snd_pcm_oss,snd_mixer_oss,snd_hda_intel,snd_pcm,snd_timer,snd_hwdepserio_raw 9220 0 sg 30260 0 soundcore 9952 2 sndshpchp 32916 0 forcedeth 55184 0 evdev 13216 3 fuse 54300 7 aufs 137224 1 squashfs 47748 12 sqlzma 6788 1 squashfsunlzma 8320 1 sqlzma#lspcmcia#lspcmcia in Slax#lspciCode: Select all00:00.0 Class 0500: 10de:0754 (rev a2)00:01.0 Class 0601: 10de:075c (rev a2)00:01.1 Class 0c05: 10de:0752 (rev a1)00:01.2 Class 0500: 10de:0751 (rev a1)00:01.3 Class 0b40: 10de:0753 (rev a2)00:01.4 Class 0500: 10de:0568 (rev a1)00:02.0 Class 0c03: 10de:077b (rev a1)00:02.1 Class 0c03: 10de:077c (rev a1)00:04.0 Class 0c03: 10de:077d (rev a1)00:04.1 Class 0c03: 10de:077e (rev a1)00:06.0 Class 0101: 10de:0759 (rev a1)00:07.0 Class 0403: 10de:0774 (rev a1)00:08.0 Class 0604:. Parameter: Value: Comments : 5010: 0 : 5012: 0 : 5011: 1111 0101 = Baud Rate 2400 . 5013: 1111 0101 : See below for other baud rates : 1111 0011 = 600 1111 0100 = 1200 Download Magick Desktop for Windows for free. Good bunch of beautiful wallpapers for your desktop. Magick Desktop is a full-featured application that Google Chrome; Ommoo. 0. 0 reviews . 17.8 k

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× 10-1 Concept:32-bit floating-point representation of a binary number in IEEE- 754 is Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) Calculation:Given binary number is00111110011011010000000000000000Here, sign bit is 0. So, number is positive. 0 01111100 11011010000000000000000 Exponent bits = E = 01111100 = 124 (in decimal)Mantissa bits M = 11011010000000000000000In IEEE-754 format, 32-bit (single precision) (-1)s × 1.M × 2E – 127 = (-1)0 × 1.1101101 × 2124 – 127= 1.1101101 × 2-3= (1 + 2-1 + 2-2 + 2-4 + 2-5 + 2-7) × 2-3= 0.231 = 2.31 × 10-1 ≈ 2.27 × 10-1 In IEEE floating point representation, the hexadecimal number 0xC0000000 corresponds to –3.0–1.0–4.0–2.0Answer (Detailed Solution Below) Option 4 : –2.0 Concept:32-bit floating-point representation of a binary number in IEEE- 754 is Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) Calculation:Binary number is0xC0000000 = (11000000000000000000000000000000)2Here, the sign bit is 1. So, the number is negative. 1 10000000 00000000000000000000000 Exponent bits = E = 10000000 = 128 (in decimal)Mantissa bits M = 00000000000000000000000In IEEE-754 format, 32-bit (single precision)(-1)s × 1.M × 2E – 127= (-1)1 × 1. 0 × 2128 – 127= -1 × 1.0 × 2= -2In IEEE floating-point representation, the hexadecimal number 0xC0000000 corresponds to -2. Let R1 and R2 be two 4-bit registers that store numbers in 2’s complement form. For the operation R1 + R2, which one of the following values of R1 and R2 gives an arithmetic overflow? R1 = 1011 and R2 = 1110R1 = 1100 and R2 = 1010R1 = 0011 and R2 = 0100R1 = 1001 and R2 = 1111Answer (Detailed Solution Below) Option 2 : R1 = 1100 and R2 = 1010 The correct answer is option 2.Concept:Stored numbers in registers R1 and R2 are in 2's complement form. Register size is 4 bits. The range of numbers in 2's complement form is -8 to +7. If R1 + R2, the result is out of the above range, then it is overflow.The given data,Given two four-bit registers R1 and R2.Option 1: R1 = 1011 and R2 = 1110False, R1 = 1 0 1 1 = -(0101)= -5+ R2 = 1 1 1 0 = -(0010)= -2----------------------------------------------- 1 0 0 1 = = -7 Here No overflow occurred, because sign bit is same for (R1 + R2 ).Option 2: R1 = 1100 and R2 = 1010True,R1 = 1 1 0 0 = -(0100)= -4+ R2 = 1 0 1 0 = -(0110)= -6 -------------------------------------------- 0 1 1 0 = = -10 Here Overflow occurred because the sign bit is different for (R1 + R2 ).Option 3: R1 = 0011 and R2 = 0100False,R1 = 0 0 1 1 = +(0011)= +3+ R2 = 0 1 0