Why can't I log into the old Minecraft: Java Edition launcher? Throws InvalidCredentialsExeption
Username and Password not recognized when logging into Minecraft
Lightning component quick action - send parameters to the lightning component I have a lightning component that is available for use as lightning component quick action. I need to use this component from various object actions, for example, I will add a lightning component quick action to account object and to contact object. Now on the lightning component itself, I want to know if it was invoked from the account object or from the contact object, but I cannot find a way to do that. I have found a way to get the record Id by adding &lt;aura:attribute name=&quot;recordId&quot; type=&quot;String&quot; /&gt; (see code snippet below) It is working but not good enough. What I really need is to send a parameter from the lightning component quick action invocation:  And get it in my component: &lt;aura:component implements=&quot;force:lightningQuickAction,force:hasRecordId,flexipage:availableForAllPageTypes&quot;&gt;     &lt;aura:attribute name=&quot;recordId&quot; type=&quot;String&quot; /&gt;     &lt;aura:attribute name=&quot;param&quot; type=&quot;String&quot;/&gt; &lt;/aura:component&gt;  Is this possible? And if not - any workaround for achieving that? Is there an overall view of your cities? Why does Ubuntu force users to create a password upon installation? Stringification of a macro value derivative bounded by a constant multiple of the function $f$ is differentiable on $[a,b]$, $f&#39;(x) \leq A|f(x)|$ where $A$ is a non-negative constant.  If $f(a)=0$ show $f(x)=0, \forall x\in [a,b]$  I imagine the proof uses the Mean Value Theorem but I have not been able to get it to work.  I know $|f(x)|=|f&#39;(c)|(x-a)$ where $c \in [a,x]$, so $|f(x)| \leq A\  |f(c)|(x-a)$ where $c\leq x$ And I guess I could sort of iterate this to keep getting a smaller $c$ but I don't see why it must go all the way to zero. /usr/sbin/grub-mkconfig: 1: /etc/default/grub: If: not found Labeling (one-way) streets in QGIS? What plot points are missing in the Harry Potter movies that are in the novels? General Information about Eigenvalues for an 3x3 symmetric matrix Is it possible to scale up an image to increase the the pixel density? In Texas Hold'em poker, there are some cases where a particular card, often called kicker, acts as a tiebreaker between players to determine who wins the pot, or if the pot has to be shared.  For some combinations such as three or four of a kind, there is no doubt: the kicker is the 5th card of the chosen combination, and determines the winner.  Example     Alice has 7 and 10   Bob has 7 and K   On the board are 7, 7, 2, 3, 4   In that case, both have a three of a kind. But because Bob has a king and Alice only a 10, he wins.  Most of poker websites explaining the rules clearly mention the role of the kicker applying to three of a kind and double pairs. But much fewer say something about colors, straights and full houses, and I found many contradicting answers. I also asked the question to some used online poker players and they also gave me contradicting answers.  I have tried to search for official tournament rules, but most of them only explain what happens with bad behaviors, bad deals, incorrect or confusing betting, showing cards when you shouldn't, acting when it's not your turn, etc. without mentioning subtleties about combinations at all.  To simplify my question, I will take three examples; I think it's better to start with examples before getting to the general answer if one exists. So, what's happening in the following 3 examples? Could you give a more generalized answer?  Example 1 - Flushes     Alice has 2&clubs; and 3&clubs;   Bob has 4&clubs; and 5&clubs;   On the board are 6&clubs;, 8&clubs;, 10&clubs;, 2&hearts;, 5&diams;   Contradicting answer 1: it's a tie, because the highest card included in the flush is the 10, which everybody chooses to include in their 5 showdown cards. Contradicting answer 2: Bob wins, because he has the greatest private card that is part of the flush  Example 2 - Straight     Alice has 6 and K   Bob has 6 and J   On the board are 4, 5, 7, 8, 10   Contradicting answer 1: it's a tie, because the greatest card in the straight is the 8 for both Alice and Bob Contradicting answer 2: Alice wins, because she owns an extra king, compared to the jack of Bob  Example 3 - Full house     Alice has 3 and 7   Bob has 3 and 6   On the board are 3, 3, 2, 2, 5   Contradicting answer 1: it's a tie, because one is supposed to choose only five cards to make a combination, and a full house is already five cards. There couldn't be any kicker, and thus their showdowns are strictly equals. Contradicting answer 2: Alice wins because of her extra 7, compared to the 6 of Bob  Thank you very much for your answers.  Note: I'm unable to post next to you, so I edit my own post; strange not be able to answer to an answer.  Ok, So if I summarize what you are saying :   In the flush case, Bob wins because at some point, their hand differs. Technically, I can continue comparing the highest private card to decide who wins. I had it correct. In the straight case, if the highest card of the straight is public, then it's technically always a tie, no matter what the players had as second private card (asuming that only one of the two cards was part of the straight). In a full house if both players have the same triplet and the same pair, it's always a tie, no matter what the players had as second private card (assuming again that only one of the two was part of the full house).   Please confirm me that what I summarized is correct.  Thank you for your answer. Enable Careers and Jobs in app "Authoring" versus "Writing" In my area of work the word author as a verb has become quite common. However, it seems to have a subtly different meaning than plain vanilla write. For example, one does not write software unit tests; one authors them.  Can anyone offer a clear distinction between the two words? Has author as a verb become a common usage outside of the field of computer software? What's the best way to find the inverse of datetime.isocalendar()? What happens if you call erase() on a map element while iterating from begin to end? In the following code I loop through a map and test if an element needs to be erased.  Is it safe to erase the element and keep iterating or do I need to collect the keys in another container and do a second loop to call the erase()?  map&lt;string, SerialdMsg::SerialFunction_t&gt;::iterator pm_it; for (pm_it = port_map.begin(); pm_it != port_map.end(); pm_it++) {     if (pm_it-&gt;second == delete_this_id) {         port_map.erase(pm_it-&gt;first);     } }   UPDATE: Of course, I then  which I didn't think would be related but answers my question. High-Swing Cascode Current Mirror I am currently dealing with current mirrors and came across the circuit shown below, a High-Swing Cascode Current Mirror. I read that this implementation, as the name suggests, has the advantage of a high voltage headroom, since the minimum potential at the drain terminal of M4 is: $$ V_{D4\_min}=V_{DS4}+V_{DS2}=(V_{GS4}-V_{TH})+(V_{GS2}-V_{TH}) $$ So the drain potential at M4 only needs to be twice the overdrive voltage, in case of same transistor dimensions, in order to hold M2 and M4 in saturation. But cannot this be said for many other cascode current mirrors, since that behaviour simply occurs by "stacking" together two transistors (M2 and M4)? Additionally, why is there a connection from the gate of M1 to the drain of M3? I have a hard time understanding whats going on in this circuit, help is greatly appreciated.    EDIT: A bias voltage source is setting the potential at the gate of M3/4. Sorry its not shown in the circuit.  EDIT 2: I just read that the mentioned bias voltage is implemented by a MOSFET in diode configuration, . For further discussion lets call it M5. This is a story that was read on BBC radio about thirty years ago, and this is about as right as I can remember it: The human pilot of a spacecraft is doing some reconnaissance on an alien planet. He crashes but survives. For some reason, no one will know where to look for him. He sees something in the distance that looks like a building of some sort. When he gets there, it seems to be a home. Very much like a human home. There's furniture, but it's strange and uncomfortable . There's food, but it's inedible. There's a faucet but what comes out of it seems more like acid. Everything looks as though it should be right, but it's not. He leaves the house, wanders and after a few days is desperate for water. But there isn't any water anywhere. He's close to death, crawling on the ground and then something happens. There are cracks in the soil and there appears to be water bubbling up from perhaps some underground stream. The water revives him. He goes back to the house and tries the food again. It's delicious and he eats enough to gain back his strength. Then he decides to try the water again. It's pure and wonderful. He takes a shower and the last line of the story is how refreshing the water is on his skin and something about him lifting his tail. Let $T,S$ be linear transformations, $T:\mathbb R^4 \rightarrow \mathbb R^4$, such that $T^3+3T^2=4I, S=T^4+3T^3-4I$. Then S is:   one-one but not onto onto but not one one invertible non-invertible   (One or more correct options)  My attempt: $S=T^4+3T^3-4I=T(T^3+3T^2)-4I=T(4I)-4I=4T-4I$  How do I go about proving or disproving my options?   My thoughts on options # 1. and 2.: I'm guessing $S$ is one-one if $Ker(S)=\theta$ is one way to go, but how do I obtain the kernel when I don't know what the transformation is? $Ker(S)=\theta$ if $T=I$, which does satisfy $T^3+3T^2=4I$. But so can other $T$s.   My thoughts on options # 3. and 4.: $S$ is invertible if $det(S)\neq0$, which is possible $det(T-I)\neq0$, i.e. $T\neq I$. How do I show that?  Please help! Preprocess categorical variables with many values How do I make my screenshot ask me what to do next every time? I take screenshots a lot. I usually take screenshot of an area, I then recheck if I have the desired area and then copy it to clipboard.  I want to change this behaviour so that immediately upon  pressing Shift+PrntScr the screen capture application allows me to select an area, and immediately afterwards it provides me with the option to save to a particular location or clipboard.  I want Shift+PrntScr to bring up a pop-up screen like this: