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Are two (or N) resistors in series more precise than one big resistor? Let's say I have one 2 k&ohm; resistor with 5% tolerance. If I replace it with two 1 k&ohm; resistors with 5% tolerance, will resulting tolerance go up, down, or remain unchanged?  I'm bad with probabilities, and I'm not sure what exactly tolerance says about resistance and its distribution.  I am aware that in the 'worst case' it will be the same; I'm more interested in what will happen on average. Will the chance of getting a more precise value increase if I use a series of resistors (because deviations will cancel each other out)?  On 'intuitive level' I think that it will, but I have no idea how to do the math with probabilities and find out if I'm actually right. Anyone ever seen a "wok adapter" for induction cooktops? In theory a cheap pressed steel wok would work on an , but the "contact patch" for a traditional round bottomed wok would be so small it would be ineffective.  Has anyone seen a steel or cast-iron "wok adapter" with a flat bottom and a concave "bowl" on top to sit the wok in, serving dual purposes of supporting the wok on the stovetop and transferring the induction heat to the wok?  Could anyone speculate if this would this even work? Or wouldn't you be able to get the wok hot enough? Adding coordinate system to layout in QGIS composer? Where are the key wardens and where are my Infernal Machines to create a Hellfire Ring/Hellfire Amulet? In order to obtain a Hellfire Ring or Amulet, one has to work their way through the Uber Boss system of harvesting machines and organs to produce a legendary ring or amulet ().  To create the infernal machine required to open the portals to the Ubers, the keys are drops from mobs nick named Key Wardens.  Where are the various key wardens?  Are there any requirements for making the machines drop? What's the drop rate on different difficulty settings? Easiest way to solve $y''+y=\frac{1}{\cos x}$ Examples:        Why, I'd love to.   Why, of course!      I get the concept of starting a sentence with a word not completely related to the overall response, but this one seems to be a particularly odd choice&nbsp;— is there any semantic meaning I'm missing? 3/4 circle with plus in it? After I recently updated my phone about a day ago, this strange sign showed up on the top left that I have never seen before. Does anyone know what it means? Or how to get rid of it? From Wikipedia: Fermi was known for his ability to make good approximate calculations with little or no actual data, hence the name. One example is his estimate of the strength of the atomic bomb detonated at the Trinity test, based on the distance travelled by pieces of paper dropped from his hand during the blast. Fermi's estimate of 10 kilotons of TNT was remarkably close to the now-accepted value of around 20 kilotons, a difference of less than one order of magnitude.  I have not been able to find any references explaining how he made this calculation. Please provide a reference or an example calculation.  See also: What is the difference between \emph{...} and {\em ...}?  When to use each of them? Variable doesn't get returned from AJAX function A few years ago Umber Ferrule wrote that the toolbars in the taskbar are defined in   HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\Explorer\Streams.   This looks very plausible but cannot be the entire story.  Deleting this key will in no way affect the taskbar and on rebooting the system the key magically reappears.  Why do I want to know?  Simple.  I would like to export the key to a reg. file as a convenient backup. Do I need a transit visa to collect and re-check my luggage at Istanbul Ataturk airport? I'm Egyptian and flying to Cairo from Tbilisi, via Istanbul. The first flight Tbilisi-Istanbul is on Atlasjet, then from Istanbul to Cairo I am flying on Egyptair. For that, I have to recheck in at Istanbul to my Egyptair flight  Will I need to pass passport control at Istanbul airport to collect my bags and check in on Egyptair? Will I need a transit visa for that? Or it can be all done through neutral zone? Event handler not working on dynamic content PHP __PHP_Incomplete_Class Object with my $_SESSION data I thought that the quotation mark (") was simply a type of grouping marker but I'm debugging some NHibernate code and notice that while  SELECT * FROM site WHERE site_id = 3;   Works fine  SELECT * FROM "site" WHERE site_id = 3;   fails with a table or view does not exist error.  What gives? How do I handle collision detection so fast objects are not allowed to pass through walls? I'm creating a 2d sidescrolling shooter, and I'm having a little trouble with collision detection for the bullets. Everything including the bullets are objects with their own polygons / update methods.  The trouble is that bullets go fast, and at 60 frames per second (what the game runs at) a bullet will often skip right through a wall -- since it moves more than the width of the wall during in the update interval -- and continue happily on its way since the polygons will never actually overlap.  What can I do about this? The only thing I've been able to come up with is to draw a line from the old position to the new position and do collision detection on that, but line drawing for collision detection is recommended against by the slick2d documentation. How can I solve this? Math Induction Proof: $(1+\frac1n)^n < n$ Find the Values of $p$ and $q$ such that $\int_0^{\infty} x^p\ln(1+x)^q dx$ Converges or Diverges. The question is to find the value values of $p$ and $q$ such that the improper integral converges or diverges.  My friend indeed found some values of $p$ and $q$ such that it converges, but after many trials, I am still getting divergent for all values.  Via L'Hospital, we have the following:  $$\lim_{x \to 0} x^c\ln(1+x)^q = \infty \space \text{when} \space  c &lt; 0$$ $$\lim_{x \to 0} x^c\ln(1+x)^q = 0 \space \text{when} \space  c &gt; 0$$ $$\lim_{x \to \infty} x^c\ln(1+x)^q = \infty \space \text{when} \space  c &gt; 0$$ $$\lim_{x \to \infty} x^c\ln(1+x)^q = 0 \space \text{when} \space  c &lt; 0$$  Thus, $$\int_1^{\infty} x^p\ln(1+x)^q dx = \int_1^{\infty} x^{p-c}x^c\ln(1+x)^q dx &gt; \int_1^{\infty} x^{p-c} dx$$ which diverges when $p &gt; -1$.  Here, we set $c = \frac{1}{2}(p+1) &gt; 0$.  On the other hand, $$\int_0^1 x^p\ln(1+x)^q dx = \int_0^1 x^{p-c}x^c\ln(1+x)^q dx &gt; \int_0^1 x^{p-c} dx$$ which diverges when $p &lt; -1$.  Here, we set $c = \frac{1}{2}(p+1) &lt; 0$.  Thus, the integral diverges whenever $p \ne –1$.  Then, let $p = -1$.  $$\int_0^{\infty} \frac{1}{x}\ln(1+x)^q dx = \int_0^1 \frac{1}{x}\ln(1+x)^q dx + \int_1^{\infty} \frac{1}{x}\ln(1+x)^q dx $$ $$ \ge \int_0^1 \frac{1}{x+1}\ln(1+x)^q dx + \int_1^{\infty} \frac{1}{x+1}\ln(1+x)^q dx $$  The first summand diverges when $q &lt; -1$ and the second diverges when $q &gt; -1$.  Finally, let $ p = -1$ and $q = -1$.  The integral $\int_0^{\infty} \frac{1}{x\ln(1+x)} dx ≥ \int_0^{\infty} \frac{1}{(x+1)\ln(1+x)} dx$ diverges on both ends.  Did I make a mistake? Derivation of $E=pc$ for a massless particle? In classical mechanics, massless particles don't exist because for $m=0$, $p=0$.  The relativistic relation between energy, mass and spatial momentum is: $E^2= (pc)^2 + (mc^2)^2$ . So it is said that setting $m=0$ in the first equation you get $E=pc$.  How could setting $m=0$ in that equation give you $E=pc$ whilst $p$ appears in the equation and we know $p=γmu$? If you set $m=0$ you will have indeterminacy due to "$γm$". It seems to me like we are doing a "trick" in order to get the $E=pc$. Perhaps there is another proof for this relation? Create a .desktop file that opens and execute a command in a terminal