I did not found many results when searching in Google and the items that I found did not help me very much.  How can I add a label "New" to posts that are not older than 24 hours?  Is there a function I can add in my functions.php I can use to manage this?
Every time i publish a new post i want to display an icon like a badge (NEW) and after a specific time it will disappear.  Lets say i publish a new post and i want to display that it is a new post. So i want to insert an icon that for example a star, in each new post in the blog list view into the metadata of the posts. Also i want to be removed after a specific time will pass. Lets say in one day or 24 hours the icon should be removed automatically.  How can i do this?
What happens if Superman is near a white dwarf star? With the Superman lore, we know that light from a yellow star gives him his powers, light from blue stars makes him stronger, and light from red stars weakens him.  However, what happens if he is near a white dwarf, or a neutron star?  Has there been any evidence of Superman visiting a star system that wasn't red/yellow/blue? On the meta home page I noticed the  is listed as doing something, but cannot figure out what it is.  There is nothing on the  or its , but the recent activity is listed with the same timestamp. Gluing Lemma for Closed Sets: Infinite Cover Counter-Example Is there a way to unlock PDF documents? I realize most PDF documents are locked for a reason, but I spend several hours a week editing datasheets provided directly from my vendors to be sent to customers. Obviously it's very annoying and somewhat of a show-stopper when I download a datasheet in PDF format, then I am not able to draw boxes, add comments, or draw arrows to the supplied product.  One of our vendors told us the way around that was to print a physical copy, then scan it in as a PDF - not the most technological or aesthetically pleasing solution I have found.  I at least would like to be able to print the PDF to another PDF so that the new copy is unlocked for editing. Is there any way around this? using trial and error in math problems Suppose one bacterium is in a jar at 12AM, and then suppose there are two bacteria at 12:30AM,  and then there are four bacteria at 1AM, etc.  (a)  How many bacteria will be in the jar at 12PM that day?  (b)  Use trial and error to estimate at what time there will be 1 billion (1,000,000,000) bacteria in the jar.  for a) is it going to be 44 bacterias ? or am i wrong? Please help with b), i have no idea how to figure it out in short term Thanks parsing NSJSONReadingAllowFragments Determine the Set of a Sum of Numbers I want to determine the set of natural numbers that can be expressed as the sum of some non-negative number of 3s and 5s.  $$S=\{3k+5j∣k,j∈\mathbb{N}∪\{0\}\}$$  I want to check whether that would be: 0,3, 5, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and so on.  Meaning that it would include 0, 3, 5, 8. Then from 9 and on, every Natural Number. But how would I explain it as a set? or prove that these are the numbers in the set? Should I always use a comma after "e.g." or "i.e."? Proving that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain We're proving that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain, using the norm function $$\nu (a + b\sqrt{2} ) = |a^2 - 2b^2|$$ and the first part says that since $\nu (a + b\sqrt{2} ) = |(a + b\sqrt{2})(a - b\sqrt{2})|$ it's clear that $\nu (xy) = \nu(x) \nu(y)$? ... Can someone please explain to me how this is clear? Even my main monitor is shown as unknown. I have tried setting up dual monitors using Nvidia settings with no joy however it does recognize that I have 2 monitors.  This was not an issue in 11.04. My problem with the Nvidia settings appears to be that my 2 monitors require different resolutions but even when I set them to a resolution that they both can handle, it is then only appearing in cloned mode. I really hope someone can help or I will have to roll back to 11.04.  Why won't Ubuntu 11.10 recognize that I have 2 Monitors?   Here is the contents of my xorg.conf:  # nvidia-settings: X configuration file generated by nvidia-settings # nvidia-settings:  version 280.13  (buildd@rothera)  Thu Aug 11 17:28:49 UTC 2011  Section "ServerLayout"     Identifier     "Layout0"     Screen      0  "Screen0" 0 0     InputDevice    "Keyboard0" "CoreKeyboard"     InputDevice    "Mouse0" "CorePointer"     Option         "Xinerama" "0" EndSection  Section "Files" EndSection  Section "InputDevice"     # generated from default     Identifier     "Mouse0"     Driver         "mouse"     Option         "Protocol" "auto"     Option         "Device" "/dev/psaux"     Option         "Emulate3Buttons" "no"     Option         "ZAxisMapping" "4 5" EndSection  Section "InputDevice"     # generated from default     Identifier     "Keyboard0"     Driver         "kbd" EndSection  Section "Monitor"     # HorizSync source: xconfig, VertRefresh source: xconfig     Identifier     "Monitor0"     VendorName     "Unknown"     ModelName      "CRT-0"     HorizSync       30.0 - 83.0     VertRefresh     50.0 - 76.0     Option         "DPMS" EndSection  Section "Device"     Identifier     "Device0"     Driver         "nvidia"     VendorName     "NVIDIA Corporation"     BoardName      "GeForce 7025 / nForce 630a" EndSection  Section "Screen"     Identifier     "Screen0"     Device         "Device0"     Monitor        "Monitor0"     DefaultDepth    24     Option         "TwinView" "1"     Option         "TwinViewXineramaInfoOrder" "CRT-0"     Option         "metamodes" "CRT: 1280x1024 +1024+0, DFP: 1024x768 +0+0"     SubSection     "Display"         Depth       24     EndSubSection EndSection"   ok I have tried disper -e. Problem is that Nvidia settings appear to be treating both monitors as the same one. so changing the settings to one has no effect, but change the other and it affects both screens identically.  Results of disper -l as follows:      carter@carter-desktop:~$ disper -l display DFP-0: HP vs17  resolutions: 320x240, 400x300, 416x312, 512x384, 576x432, 680x384, 640x480, 720x450, 640x512, 800x512, 840x525, 800x600, 960x540, 832x624, 960x600, 896x672, 1024x768, 1152x864, 1280x960, 1280x1024 display CRT-0: CRT-0  resolutions: 320x240, 400x300, 416x312, 512x384, 576x432, 680x384, 640x480, 720x450, 640x512, 800x512, 840x525, 800x600, 960x540, 832x624, 960x600, 896x672, 1024x768, 1152x864, 1360x768, 1280x960, 1440x900, 1280x1024, 1400x1050, 1600x1024, 1680x1050, 1600x1200, 1920x1080, 1920x1200, 1792x1344 carter@carter-desktop:~$ How to prove that the center of the dihedral group $D_{2n}$ is $\{1,r^{n}\}$ and the center of $D_{2n-1}$ is $\{1\}$?  I don't know how to prove it in this general case. Show that $\left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right|$. I'm trying to prove that  $$ \left| \frac{1}{x}-\frac{1}{y} \right| $$  is a metric on $X = \mathbb R^+$.  The properties of non-negativity and symmetry are obvious (do I actually even need to prove them?), but the tricky one is, as usual the inequality property, i.e.  $$ d(x,z) \le d(x,y) + d(y,z) \text{ for any } z \in X. $$  So, ok, I have that  $$ d(x,z) \le d(x,y) + d(y,z) \\ \iff \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| \\ \iff \left| \frac{1}{x} \right|- \left| \frac{1}{z} \right| \le \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| $$  ...and that's about as far as I've gotten–which is to say, not very far. I also tried multiplying both sides by $ \left| \frac{1}{x}-\frac{1}{z} \right| $, but that wasn't very fruitful. If someone could point me in the right direction, it'd be greatly appreciated. Thanks. Does it look good to edit the question you're answering? I have a question about editing practices:  Does it look good to edit a question you have just answered?  For example: say I had just answered a question, then saw that the question had some minor formatting issues. Should I edit it? Does it turn up voters off when they see that I am answering a question I changed, possibly to better suit my answer? Show "Insufficient Privileges" directly when clicking 'Edit' (override), not on 'Save' If a user clicks 'Edit' on an opportunity which he isn't allowed to edit, the "Insufficient Privileges" page directly appears. This is good. If a user clicks 'Edit' on a custom object (which has an edit override that points to a visualforce page), he first sees the form without problems, and the "Insufficient Privileges" page only appears when he tries to save it. This behaviour is misleading for the users, because they think they can manipulate the data and put effort into it, and only afterwards they realize that this wasn't possible after all.  An idea was to check at page load if the user has edit permissions and otherwise redirect him to the "Insufficient Privileges" page.  Might there be another way? Thanks! Do quest items weigh anything? Someone said that quest items don't count against your carrying capacity in Skyrim, even if they have a weight listed. Is this false? Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Apparently $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$.  How? What's the proof? Or maybe it is self apparent just looking at the above?   PS: This problem is known as "The sum of the first $n$ positive integers". Accessing an array out of bounds gives no error, why? Can I turn off macOS Catalina's internal display without close the lid while using an external monitor? Could not list layers for this store error when creating new ImageMosaic Proving that a list of perfect square numbers is complete Well, I have a number  $n$ that is given by:  $$n=1+12x^2\left(1+x\right)\tag1$$  I want to find $x\in\mathbb{Z}$ such that $n$ is a perfect square.  I found the following solutions:  $$\left(x,n\right)=\left\{\left(-1,1^2\right),\left(0,1^2\right),\left(1,5^2\right),\left(4,31^2\right),\left(6,55^2\right)\right\}\tag2$$     Is there a way to prove that this a complete set of solutions? So I mean that the solutions given in formula $(2)$ are the only ones?     My work:   We know that: $$ 1 + 12x^2 \left(1+x \right) \ge 0   \space \Longleftrightarrow \space   x \ge -\frac{1+2^{-2/3}+2^{2/3}}{3}   \approx -1.07245 \tag3 $$ So we know that for $x&lt;-1$ there are definitely no solutions.