Does deleting a photo from iCloud on PC affect its copy on iPhone storage?
Does deleting iCloud photos affect them on devices?
Google chrome is reading characters backwards Google Chrome is my favorite browser, and I can't imagine going on without it. Suddenly (without closing and re-opening) the browser; chrome has been reading characters upside down, very weird:    Chrome version:    I tried restarting my browser and my MAC (running OS X Yosemite 10.10), but the error keeps on coming. If time standard clocks and any memories about the time standard are destroyed, can we recover the time standard again? Assume the time standard clocks and any memories about the time standard are destroyed. Can we recover the time standard again exactly?  Recovering the time standard again means we can determine the date  and time that are exactly the same as the destroyed time standard clocks will show (if they are not destroyed). Allowing only one download per Apple ID We have a Barbarian on our team that very much likes to switch up his weapons; he goes from a Dwarven Hammer, to a pair of Hand-Axes, to occasionally just going bare-fisted. No issues there, he's fairly adept with all 3.  However, I have only just recently picked up on the way that he rolls his attacks for the two Hand-Axes: He rolls the first attack, adds his modifiers etc. Then he rolls the second attack, this time without modifiers, which is all well and good. But I'm not sure if this action uses his bonus action or not, or is considered a separate action (though, being a level 4 Barbarian, he only has one action per turn), or is merely part of the same action.  If we have a look at the Two-Weapon Fighting feature from the Fighter's fighting styles:     Two-Weapon Fighting   When you engage in two-weapon fighting, you can add your ability modifier to the damage of the second attack. (PHB5e p.72)   Similarly, the Dual Wielder feat also uses the Two-Weapon Fighting feat:     Dual Wielder   You master fighting with two weapons, gaining the following benefits:      ...         You can use two-weapon fighting even when the one-handed melee weapons you are wielding aren’t light. (PHB5e p.164, emphasis mine)      So; yes, this gives a bonus to the second weapon attack, but having asked a  to clarify which action this secondary weapon attack belongs to, it was determined that this uses your bonus action;     [...] the second attack requires a bonus action   So after that long winded explanation, here is the question: Without any additional feats, can you wield and use two weapons at once, and if so, how many actions does it require to make a full attack with both weapons in a turn? What apps are people using to record voice meetings on the iPad? Blender is rendering an older version my project Email searches in Outlook 2013 work as expected. I am trying to refine it by excluding certain folders by default. I was hoping I could uncheck certain folders in Control Panel > Indexing Options > Outlook. Unfortunately, the checkbox is for Outlook itself. I can either search Outlook or not. There is no refinement beyond that. Is there another way to exclude searching of specific folders? How do I recover the default version of some configuration file? By accident, I messed up a configuration file. How do I recover the default version without reinstalling or affecting other things? I tried to reproduce the results from the  unfortunately the video link is broken. And I might have overlooked some related settings.  The spot light is barley visible on the ground.    Increasing the Density of the Volume Scatter Node darkens the sky texture but leads to some light cone from the car's headlight. (only 10 samples in preview).    Which settings do I need to change or should I try a different approach?  Integrator settings are default and spotlight settings were copied from the example found in the docs.  Related:    (pre 2.7)  (BI only) I would like to filter the markers on my (geojson) layer according to two features: year and type of event, ideally using both filters (year and eventType) as check-boxes without using layerGroup at all. I would like to be able to click on a checkbox and update dynamically the markers by filtering them according to feature.properties.eventType too, in addition to the year displayed, something like    &lt;input type="checkbox" id="all" name="all" value="all"&gt;   &lt;label for="all"&gt;All&lt;/label&gt;   &lt;input type="checkbox" id="funfair" name="funfair" value="newsletter"&gt;   &lt;label for="funfair"&gt;Funfair&lt;/label&gt;   &lt;input type="checkbox" id="trade" name="trade" value="trade"&gt;   &lt;label for="trade"&gt;Trade&lt;/label&gt;   &lt;input type="checkbox" id="exhibition" name="exhibition" value="exhibition"&gt;   &lt;label for="exhibition"&gt;Paid exhibition&lt;/label&gt;   etc... which would trigger a  filter: function(feature, layer) {   return feature.properties.eventType == value-of-checkbox; }   I have five types of eventType. So, for instance, if 1890 is checked, the map is showing all 1890 events, no matter what kind of event it is (because by default it would show all events). If the user checks the 'funfair' and the 'trade' checkbox, though, I want the map to show only the 1890 events of the funfair and trade kind, and not other kind of events.  This is my complete script, which so far is working fine with filtering by year.  var oldmap = L.tileLayer('address', {   maxZoom: 12 }),  var modernmap = L.tileLayer('https://api.tiles.mapbox.com/v4/{id}/{z}/{x}/{y}.png?access_token={accessToken}', {   maxZoom: 12,   id: 'mapbox.streets',   accessToken: 'mytoken' });  var map = L.map('map', {   center: [40.1857864,-4.5536861],   zoom: 7,   layers: [oldmap, modernmap] });  var baseMaps = {   "Old map": oldmap,   "Modern map": modernmap };  var jsontest = {"type": "FeatureCollection",   "features": [     {       "type": "Feature",       "geometry": {         "type": "Point",         "coordinates":  [ -16.5471268,28.4136726 ]       },       "properties": {         "Name":"Point1",         "year":'1892',         "eventType":"funfair",         "notes":""       }     },     {       "type": "Feature",       "geometry": {         "type": "Point",         "coordinates":  [ 2.1753895,41.3767925 ]       },       "properties": {         "Name":"Point2",         "year":'1890',         "eventType":"trade demonstration",         "notes":""       }     }   ]};  var mapLayerGroupsYear = [];  function onEachFeature(feature, layer) {   var year = mapLayerGroupsYear[feature.properties.year];   if (year === undefined) {     year = new L.layerGroup();      //add the layer to the map     year.addTo(map);      //store layer     mapLayerGroupsYear[feature.properties.year] = year;   }    //add the feature to the layer   year.addLayer(layer);    }   var myLayer = L.geoJSON(jsontest, {   onEachFeature: onEachFeature })  L.control.layers(baseMaps, mapLayerGroupsYear).addTo(map);   I am a beginner with leaflet so please forgive the dumb question, in case it is. tar files only, no directories How to get image size (height & width) using JavaScript? Textbook clarification: $\log = \ln$ Textbook reads:     All logarithms are natural logarithms: $\log = \ln$.   Does this mean $n\log(n) = n\ln(n)$? Model a simple 3D spiral shape? I need to model a simple spiral shape. I imagine the best way to do this is with curves, but I am unsure how to create a perfect spiral shape with curves. What's the best/easiest way to make a spiral tube? According to Hubble's Law, how can the expansion of the Universe be accelerating? Scientists today think the expansion of the universe is accelerating.   According to Hubble's law, objects further away are moving faster than objects closer to us. The further away an object is, the further back in time we are seeing, so in the past objects moved faster (is this sentence correct?).  So because objects moved faster before than they do now, surely that's deceleration not acceleration?  my textbook says "if Scientists today think the rate of expansion of the Universe were decreasing then distant objects should appear different to Hubble's Law predictions: universe is accelerating. More distant objects would seem to be receding faster (since expansion was faster in the past)" How to use paging in a WFS query? I am running a GetFeature request against a WFS server which does not support to download all data at once. Can I use ogc:PropertyIsGreaterThanOrEqualTo to split the dataset into chunks at download them step by step? I wonder how I can construct the actual URL to include the paging filter. Here is what I tried:  http://example.com/wfs.aspx?request=GetFeature&amp;service=WFS&amp;version=1.1.0 \ &amp;typeName=example:example&amp;maxFeatures=50000 \ &amp;FILTER=&lt;ogc:Filter xmlns:ogc="http://www.opengis.net/ogc"&gt; \ &lt;ogc:PropertyIsGreaterThanOrEqualTo&gt;&lt;ogc:PropertyName&gt;OBJECTID&lt;/ogc:PropertyName&gt; \ &lt;ogc:Literal&gt;50000&lt;/ogc:Literal&gt;&lt;/ogc:PropertyIsGreaterThanOrEqualTo&gt;&lt;/ogc:Filter&gt;   This is not a valid URI though... That's why uri-encoded the filter part:  http://example.com/wfs.aspx?request=GetFeature&amp;service=WFS&amp;version=1.1.0 \ &amp;typeName=example:example&amp;maxFeatures=50000 \     &amp;FILTER=%3Cogc%3AFilter%20xmlns%3Aogc%3D%22http%3A%2F%2Fwww.opengis.net \ %2Fogc%22%3E%3Cogc%3APropertyIsGreaterThanOrEqualTo%3E%3Cogc%3APropertyName%3EOBJECTID \ %3C%2Fogc%3APropertyName%3E%3Cogc%3ALiteral%3E50000%3C%2Fogc%3ALiteral \ %3E%3C%2Fogc%3APropertyIsGreaterThanOrEqualTo%3E%3C%2Fogc%3AFilter%3E   This actually works. Is there a "nicer" way to do this? What is the difference between the following class methods?  Is it that one is static and the other is not?  class Test(object):   def method_one(self):     print "Called method_one"    def method_two():     print "Called method_two"  a_test = Test() a_test.method_one() a_test.method_two() Isomorphism $f$ preserves the order of an element? Let's say $f$ is an isomorphism $f:G  \rightarrow G'$, where $G$ and $G'$ are multiplicative groups.  Then for $x\in G$, if $f(x) = x' \in G'$. Do we have always have $\text{ord}(x) = \text{ord}(x')$? Why? The  mentions the following matrix identity $$\det\begin{pmatrix}A&amp;B\\B&amp;A\end{pmatrix}=\det(A+B)\det(A-B),$$ valid for squared matrices $A$ and $B$ of the same size.  How is this result proved? If $a$ divides $bc$ and $\gcd(a,b) = d$ then $\frac a d$ divides c I'm trying to prove that if $a$ divides $bc$ and $\gcd(a,b) = d$ then $\frac a d$ divides c. I tried using Bezout identity but couldn't get anywhere.