For some reason my GTA V still keeps crashing even after the update which should "fix" the problem with 2 graphics cards.  Also when I try to run only with my Intel/AMD graphics card it ends up giving an error that DirectX 9 wasn't found on my device? Which I get for my Intel but not my AMD. It has to be installed on at least one of them cause it doesn't give this error with both enabled.  Specs: i7 4700MQ AMD HD8970m 8GB of ram Windows 7 64 bit  My GTA is installed on a hdd and there is over 400gb of space left. I also tried repairing my files with the uninstaller you can also repair your game there instead of uninstalling.
I recently ordered the GTA V PC from Steam and when I finished installing and launch the game. It gave me "Grand Theft Auto V has exited unexpectedly". I tried "safe mod" and every option available, but none of them work. I checked some forums and found a lot of people having the same problems. What could be the possible cause? I'm using a laptop with these specs:  CPU: i7 4700m GPU: AMD HD8970M RAM: 8GB  I'm pretty sure my laptop is way beyond the minimum specs. Does someone know how can I enter the game?
Where is the denoiser? connectedness of uniform topology How can I schedule timer to power on and off device? Is it necessary that an International cricket ground should has the boundary line in circle shape?Is it allowed by ICC to make it a square or ellipse or any other? What is the minimum and maximum area allowed for an international cricket ground? What is the maximum number of international cricket grounds that a country allowed to have? Does the Rogue's Assassinate feature grant criticals on multiple attacks? Does the Rogue's Assassinate feature (auto-crit on surprised targets) allow for multiple crits if you make multiple attacks that round, either through two-weapon fighting or extra attacks gained through multiclassing?  As an example: my 3rd level Human (Variant, TWF feat) Rogue-Assassin rolls high on his stealth roll, and gets up behind his target.  With two daggers, he gets 2 attacks per round if he uses his bonus action as well.  Do both of these attacks qualify for Assasinate's auto-crit feature?  That is: would I do 2 criticals for       4d4+4d6+2*DEX damage   or a single critical and a non-critical for       3d4+4d6+2*DEX damage?   Additionally, if I gained the Extra Attack feature through multiclassing, would I get crits on all 3 of my attacks, or just the first one? Test if a package (or package option) is loaded [french]{babel} (aka frenchb) is incompatible with some other packages, such as tikz, since it tries to insert \thinspace before semi-colons.  In order to make these two play nice, you have to call \shorthandoff{;} in the document.  Now I have a package that uses tikz and I would like to locally disable the semi-colon for babel, instead of disabling it for the whole document. I could hack the .sty of the package to do that, knowing that I'm using frenchb, but I would rather make something clean and portable.  Is there a way in my package that I can test if frenchb is loaded and disable semi-colon management with babel locally, so the package doesn't have to be hacked?  Edit:  Martin's solution gives me something like this in general:  \RequirePackage{ifthen} \newcommand{\@babel@loaded}{0} \AtBeginDocument{\@ifpackageloaded{babel}                                                    {\renewcommand{\@babel@loaded}{1}}                                                         {\renewcommand{\@babel@loaded}{0}}                                                       } \newcommand{\somecommandusingsemicolon}{   \ifthenelse{\@babel@loaded=1}{\shorthandoff{;}}{}     % code with semi-colon here }   and in the specific case of tikz:  \AtBeginDocument{\@ifpackageloaded{babel}   {%     \g@addto@macro\tikz@installcommands{%       \shorthandoff{;}%     }     \g@addto@macro\tikz@uninstallcommands{%       \shorthandon{;}%     }   }   {\relax} } Symfony plugin sfDoctrineActAsTaggablePlugin not working $A$ and $B$ are $n\times n$ matrices and $A+B=AB$. I have an interesting proof that this implies $A$ commutes with $B$, but the proof only works when $||B|| \lt 1$. I'm looking for a way to salvage the proof so that it works for all $B$.  $$A=AB-B$$ $$A=(A-I)B$$ by repeatedly substituting $A$ with $(A-I)B$ on the RHS, $$A= (A-I)B^N - \sum_{i=1} ^{N-1} B^i$$  Since the space of $n\times n$ matrices is Banach (hence complete) under the standard operator norm, the limit $\sum_{i=1}^{\infty}B^i$ is well defined given $||B|| \lt 1$. Hence, taking the limit as $N \rightarrow \infty$, $$A =  -\sum_{i=1}^{\infty}B^i$$  Now the LHS commutes with B, hence so does the RHS.    Is there some trick to extend the method to work with any $B$ where $||B|| \geq 1$? I know that the result is true due to another simple algebraic proof, but I'd like to see if this method can be salvaged. Existence of a simple closed curve which is not null-homotopic Problem. Assume that $U$ is an open and connected subset of $\mathbb R^2$, and $\gamma :[0,1]\to U$ is a closed curve, which is not null-homotopic in $U$ and not necessarily simple closed. Show that there exists a simple closed curve, which is not null-homotopic in $U$.  I am sure that this is a standard problem, and I am looking for an elegant, if possible, solution or a reference. I have in mind the following sketch of proof:  a. Construct a polygonal closed curve $\tilde\gamma$ which is homotopic with $\gamma$ in $U$.  b. Construct $\tilde\gamma$ so that it intersects itself in finitely many points, and not whole segments.  c. Use induction on the number of self–intersections of $\tilde\gamma$, in order to show that $\tilde\gamma$ is written as a product (in the fundamental group's sense) of simple closed polygonal curves. Clipping raster with mask in QGIS gives AAIGrid error? If, one fine day, someone found a contradiction in ZFC (or even ZF), what implications would such an event have for mathematicians? Is there currently any backup axiomatic system on par with ZFC that would enable mathematicians to continue pursuing mathematics, should the unthinkable occur? formal proof from calulus Given $f:R \to R$, $f$ is differentiable on $R$ and $\lim_{x \to \infty}(f(x)-f(-x))=0$. I need to show that there is $x_0 \in R$ such that $f'(x_0)=0$  I am trying to prove it by contradiction .... so i assume there is no $x_0 \in R$ such that  $f'(x_0)=0$ this means this function is strictly monotonic because the derivative "respects" the mean value theorem so it cannot be negative and positive without passing a zero value ....  (from what i understand the derivative function of a continuous/differentiable functions isn't necessarily continuous but it still respect the mean value theorem i think it's called darboux's theorem )  now from $\lim_{x \to \infty}(f(x)-f(-x))=0$ ,we get $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)$  now the last step is the contradiction its very intuitive and even maybe obvious that we can't get $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)$ for strictly monotonic function but i cant think of a formal way to prove it . Here is the problem:  Suppose that $*$ is an associative binary operation on a set $S$. Let  $$H:= \{a \in  S\mid a * x = x * a \mbox{ for all }x\in S\}.$$  In other words, $H$ is consisting of all the elements of $S$ that commute with every element in $S$.  Show that $H$ is closed under $*$. For context, see .  In essence:     "This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation (not including the ) on this site."   Suggested addition to existing message is in bold. product distribution of two uniform distribution, what about 3 or more If I take a loss when I sell my car, can I claim a capital loss deduction on my income tax return? How are long-term capital gains taxed if the gain pushes income into a new tax bracket? ubuntu 18.04 how do I install drivers for my NVIDIA GeForce GTX 1050? How does entropy depend on location and scale? Convergence of a Harmonic Continued Fraction