mklink /d link_folder target_folder    How to add link_folder to Windows 10 Explorer quick access list? Now if you drag it to quick access, the added folder point to target_folder .
Scenario:   Create SymLink: C:\Data\workspaces --&gt; C:\Users\****\Documents\workspaces Navigate to C:\Data\Workspaces in explorer Click and drag the the path from the address bar onto "Quick Access" in the navigation pane.   Expected Result:   Clicking on the shortcut takes you to C:\Data\workspaces   Actual Result:   Clicking on the shortcut takes you to C:\Users\****\Documents\workspaces   How do you make it so that the shortcut points to the correct path. (the SymLink)  Command Used to create the SymLink:  mklink /D C:\Data\workspaces C:\Users\****\Documents\workspaces   Updated with pictures:  Please note the workspaces symlink inside C:\Data:    When I double-click on the symlink it works correctly. (note the address bar.    When i pin it to the Quick access bar, however, it uses the address that the symlink points to and NOT the symlink address.
How to batch resize an image and create multiple sizes (with Photoshop)? I want to batch resize an image in multiple sizes with Photoshop.  The sizes are most of the time the same so I would like to have profiles for a set of fixed sizes. Ideally I just need to open my source image and click on "Batch Profile #1" and then Photoshop is creating/saving X different sizes of that image as an output in a specific folder.  This is what i want: How can I make a macOS Mojave bootable USB installer in Windows 10? No field properly between $\mathbb Q$ and $E$ iff $G(K/\mathbb Q) \cong A_4$ or $S_4$ Can't preview smoke simulation on Blender 2.82 Whenever I add a Quick Smoke effect, I can't preview it on the viewport unless I bake it.   And Yes, I added the Smoke Domain on the object.   I've seen tutorials like this  and this  that clearly shows the possibility of previewing the smoke before baking it.   I'm using Blender on a Mac with AMD GPU. I would guess that this is something related to OpenGL, as thoroughly explained here -  but that's for Windows.   Does anyone have any idea how to fix this? jquery selector not working with [ ] Closed form for $\int_0^\infty\frac{1}{(1+x^2)^s}\,dx$ when $s\in (0.5,\infty)\setminus\mathbb{N}$ I'm preparing for my exam, but I have difficulties in perceiving why there is a $\frac{1}{2}$ in the distance formula $d=\frac{1}{2}at^2$? How do you compute negative numbers to fractional powers? My teachers have gone over rules for dealing with fractional exponents. I was just wondering how someone would compute say: $$(-5)^{2/3}$$ I have tried a couple ways to simplify this and I am not sure if the number stays negative or turns into a positive. I know that if a negative number is raised to an odd power it is negative, but fractional powers are neither odd or even. Is there a general rule for dealing with these types of problems? Suppose $Y_1, \ldots, Y_n$ is a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$. Let $S^2$ be the sample variance, which is unbiased for $\sigma^2$.  GOAL: Show that $S = \sqrt{S^2}$ is a biased estimator of $\sigma$.  I realize that $(n-1)S^2/ \sigma ^2$ is chi-square with $(n-1)$ degrees of freedom.  I am given that $$E(S) = E\left[\frac{\sigma}{\sqrt{n-1}} \left[\frac{(n-1)S^2}{\sigma^2}\right]^{1/2}\right] = \frac{\sigma}{\sqrt{n-1}} \int_0^\infty v^{1/2} \frac{1}{\Gamma[(n-1)/2] 2^{(n-1)/2}} v^{(n-1)/2} e^{-v/2} \, dv$$  Can anyone help me understand why they have $v^{1/2}$ and $v^{(n-1)/2}$ (combining them $v^{n/2}$ as I think it should be $v^{1/2}$ and $v^{\frac{(n-1)}{2}-1}$ (combining both terms give $v^{(n-2)/2}$ since the pdf for chi-square distribution is $$ f(v) = \frac{v^{(x/2)-1} e^{-v/2}}{2^{x/2}\Gamma(x/2)},$$ where $x$ is degree of freedom?  Could anyone please tell me which one is the correct answer with explanation?  Thank you How do I get rid of my bounty? I accidentally killed a noble, now every guard wants to kill me. Is there any way I can get rid of the bounty? Can I backup my entire Ubuntu installation to ISO DVD? Can anyone come up with an explicit example of two functions $f$ and $g$ such that: $f\circ g$ is bijective, but neither $f$ nor $g$ is bijective?  I tried the following: $$f:\mathbb{R}\rightarrow \mathbb{R^{+}} $$ $$f(x)=x^{2}$$  and  $$g:\mathbb{R^{+}}\rightarrow \mathbb{R}$$ $$g(x)=\sqrt{x}$$  $f$ is not injective, and $g$ is not surjective, but $f\circ g$ is bijective  Any other examples? What do you call a person who does/thinks the opposite of the crowd for just that reason? Such a person does not hold in high regard the decisions taken by a group (not a team) of people. She believes that groups tend to be anarchic and don't serve their purpose.  An example would be - for outings with friends, this person will not like planning it with the group since they will mess it up.  Therefore, this person has convinced herself that when more people are involved with something, it will be worse. She distrusts crowds instinctively basically. Reverse Cauchy Schwarz for integrals Let $f,g$ be two continuous positive functions over $[a,b]$ Let $m_1$ and $M_1$ be the minimum and maximum of $f$ Let $m_2$ and $M_2$ be the minimum and maximum of $g$ Prove that $$\sqrt{\int_a^bf^2 \int_a^b g^2}\leq \frac{1}{2}\left(\sqrt{\frac{M_1M_2}{m_1m_2}}+\sqrt{\frac{m_1m_2}{M_1M_2}}\right)\int_a^bfg$$  I can't make any significant progress on this one... Thanks for any hint. Is it possible to scale up an image to increase the the pixel density? I have taken a picture whose resolution is less. I can resize the image easily (i.e I can scale to higher resolution) but the image quality goes for a toss, which is true because the image might not have enough pixels to scale properly.   Is resizing a means of increasing the resolution of the image ? Is it possible to increase the pixel density of the image and then resize, so that quality of the image is not affected much?   I mainly use GIMP and Picasa. The assumptions are:    Alice and Bob have perfectly synchronized clocks Alice and Bob have successfully exchanged a pair of entangled photons   The idea is simply to have Alice and Bob perform the Quantum Eraser Experiment (doesn't need to be the delayed choice).   Alice and Bob agree on a specific time when Bob's photon will be between the "path marker" (which is usually just after the slits) and the detector.  If Alice acts collpasing the wave-function on her photon, the interference pattern will disappear. If not it won't.  Alice and Bob can be spatially separated...  What am I misunderstanding?   The only meaningful difference from this spatially separated quantum eraser experiment to one done on tabletop is that you won't be able to use a coincidence detector, but that is not impeditive to identifying the interference pattern, just will make errors more probable. Which we should be able to deal with a appropriate protocol...  There is a experimental paper with a small amount of citations pointing out to the breaking of complementarity in a very similar setup: I found many related questions on this site but I am still not sure about the rules. How can I decide if I need a visa to transit? Schengen members as of May 2021 are as follows:  Austria Belgium Czech Republic Denmark (excluding Greenland and the Faroe Islands - but an open border with the Schengen Area is maintained) Estonia Finland France (excluding overseas departments and collectivities) Germany Greece Hungary Iceland Italy Latvia Liechtenstein Lithuania Luxembourg Malta Netherlands (excluding Aruba, Curaçao, Sint Maarten and the Caribbean Netherlands) Norway (excluding Svalbard) Poland Portugal Slovakia Slovenia Spain (except Ceuta and Melilla) Sweden Switzerland Solve $f(x)f(2x^2) = f(2x^3+x)$ Why does factoring eliminate a hole in the limit? $$\lim _{x\rightarrow 5}\frac{x^2-25}{x-5}  = \lim_{x\rightarrow 5} (x+5)$$  I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and that the original limit is equal to the new and simplified limit. I understand how to do this procedurally, but I'd like to know why this works. I've only been told the methodology of expanding the $x^2-25$ into $(x-5)(x+5)$, but I don't just want to understand the methodology which my teacher tells me to "just memorize", I really want to know what's going on. I've read about factoring in abstract algebra, and about irreducible polynomials (just an example...), and I'd like to get a bigger picture of the abstract algebra in order to see why we factor the limit and why the simplified is equal to the original if it's missing the $(x-5)$, which has been cancelled. I don't want to just memorize things, I would really like to understand, but I've been told that this is "just how we do it" and that I should "practice to just memorize the procedure." I really want to understand this in abstract algebra terms, please elaborate. Thank you very much. Basically this suggestion for the Android app:   Many times I'll see that a post has been edited on the Android app. I'll want to see exactly what was edited, and the Android app doesn't show this. Can the revision history be added to the Android app?