I'm trying to create a form in SharePoint Designer 2013, I want to be able to hide a specific field in the form unless the answer to another field has been changed to a specific option. Here is my code:  &lt;tr&gt;     &lt;td width="25%" class="ms-vb"&gt;         &lt;b&gt;External Sales?:&lt;/b&gt;     &lt;/td&gt;     &lt;td width="75%" class="ms-vb"&gt;         &lt;SharePoint:FormField runat="server" id="ff438747{$Pos}" ControlMode="New" FieldName="External_x0020_Sales_x003f_" __designer:bind="{ddwrt:DataBind('i',concat('ff438747',$Pos),'Value','ValueChanged','ID',ddwrt:EscapeDelims(string(@ID)),'@External_x0020_Sales_x003f_')}" /&gt;         &lt;SharePoint:FieldDescription runat="server" id="ff219374description219373{$Pos}" FieldName="External_x0020_Sales_x003f_" ControlMode="Edit" /&gt;     &lt;/td&gt; &lt;/tr&gt; &lt;xsl:if test="@External_x0020_Sales_x003f_ = 'Yes'"&gt; &lt;tr&gt;     &lt;td width="25%" class="ms-vb"&gt;         &lt;b&gt;PPE:&lt;/b&gt;     &lt;/td&gt;     &lt;td width="75%" class="ms-vb"&gt;         &lt;SharePoint:FormField runat="server" id="ff438748{$Pos}" ControlMode="New" FieldName="PPE" __designer:bind="{ddwrt:DataBind('i',concat('ff438748',$Pos),'Value','ValueChanged','ID',ddwrt:EscapeDelims(string(@ID)),'@PPE')}" /&gt;         &lt;SharePoint:FieldDescription runat="server" id="ff219375description219373{$Pos}" FieldName="PPE" ControlMode="Edit" /&gt;     &lt;/td&gt; &lt;/tr&gt; &lt;/xsl:if&gt;   As you can see, I want to be able to make it so when "External Sales?" = Yes, then it shows the field "PPE".  Any help on this matter would be greatly appreciated.
I need to hide field 'B' when the value in field 'A' is NOT equal to 'Other'.  This is a custom list form.  Is jquery the way to go?
A star's collapse can be halted by the  due to the Pauli exclusion principle. In extreme relativistic conditions, a star will continue to collapse regardless of the degeneracy pressure to form a black hole. Does this violate the Pauli exclusion principle? If so, are theorists ok with that? And if it doesn't violate the Pauli exclusion principle, why not? Swapping two variable value without using third variable How to plot a graph from its adjacency matrix and coordinates of vertices? A similar question is . If we give all the data of the vertices's coordinates and the graph's adjacency matrix, how to plot it with tikz, pstrick or other tool in tex?   here are the data of adjacency matrix and coordinates  Coordinates:{{0.809,0.588},{0.309,0.951},{-0.309,0.951},{-0.809,0.588},{-1.,0.},{-0.809,-0.588},{-0.309,-0.951},{0.309,-0.951},{0.809,-0.588},{1.,0.}}  Adjacency matrix: {{0,1,0,0,1,0,1,0,0,1},{1,0,1,0,0,1,0,1,0,0},{0,1,0,1,0,0,1,0,1,0},{0,0,1,0,1,0,0,1,0,1},{1,0,0,1,0,1,0,0,1,0},{0,1,0,0,1,0,1,0,0,1},{1,0,1,0,0,1,0,1,0,0},{0,1,0,1,0,0,1,0,1,0},{0,0,1,0,1,0,0,1,0,1},{1,0,0,1,0,1,0,0,1,0}}   Sometimes the number of data is too large to put the data in the main tex file. It would be better to import the data from external files. . $$f(x,y) = \left\{\begin{array}{cc} \frac{xy}{x^2+y^2} &amp; (x,y)\neq(0,0) \\ f(x,y) = 0 &amp; (x,y)=(0,0) \end{array}\right.$$  In order to verify if this function is differentiable, I tried to prove it by the theorem that says that if $\frac{∂f}{∂x}$ and $\frac{∂f}{∂y}$ exist and are continuous at the point $(x_0,y_0)$, then the function is differentiable at this point. So I did:  $$\frac{\partial f}{\partial x}(0,0) = \lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h} = 0$$ $$\frac{\partial f}{\partial y}(0,0) = \lim_{h\to 0}\frac{f(0,0+h)-f(0,0)}{h} = 0$$  so we have that the partial derivatives at point $(0,0)$ is $0$. Now, if we take the derivative at $(x,y)\neq (0,0)$ and then take the limit of it as $(x,y)\to(0,0)$, we can see if the derivatives are continuous or not. So here it is:  $$\frac{\partial f}{\partial x}(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)}$$  but   $$\lim_{(x,y)\to(0,0)} \frac{y(y^2-x^2)}{(x^2+y^2)} $$ does not exist (by wolfram alpha... but can anybody tell me an easy way to prove this limit does not exist? easier than taking the limit in different directions?), therefore the derivative is not continuous at $(0,0)$, so we can't say $f$ is differentiable at $(0,0)$, but for $(x,y)\neq (0,0)$ the function is continuous, as it is a quotient of continuous functions. So $f$ is at least differentiable at $(x,y)\neq (0,0)$.   Now, to verify differentiability at $(0,0)$ I think we must use the limit definition of differentiablity: A function is differentiable at $(0,0)$ iff: $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = 0$$  Let's calculate this limit:  $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = \\ \lim_{(h,k)\to (0,0)} \frac{\frac{hk}{h^2+k^2}}{\sqrt{h^2+k^2}} = \\ \lim_{(h,k)\to (0,0)} \frac{hk}{(h^2+k^2)\sqrt{h^2+k^2}}$$  which I think, it's a limit that does not exist, therefore the function isn't differentiable at $(0,0)$ Why did Snape need to appear well-informed? "You will have to give Voldemort the correct date of Harry's departure from his aunt and uncle's," said Dumbledore. "Not to do so will raise suspicion, when Voldemort believes you so well informed."   Snape had just killed Dumbledore, why would it raise suspicion about the movements of Harry's departure? it would seem likely that once Dumbledore died at the hands of Snape the Order would surely have changed any plans they would have had in place for Harry's exit from Privet Drive? What is a "callable"? How can I stop Lollipop ignoring my wifi network? Android 5.0 has a new "feature": If a wifi network has no connection to the internet, it will remain connected to it but route traffic over mobile data instead. A small exclamation mark shows in the status bar over the wifi symbol.  Unfortunately, for some reason Android has decided that my home wifi does not have a good internet connection - which is not true, as I'm using it to access Stackexchange at the moment - and is, instead, trying to send everything over GPRS.  With mobile data disabled, things work over wifi just fine, despite the exclamation mark - so it isn't a problem with the wifi not working, per se.  Is there any way to override this "smart" feature? The only way I have found to make things work at the moment is to disable mobile data while I am in my house - which is not a practical long-term solution.   EDITING to add that I have now solved this problem by buying a new router! I was using something very old... But, I'll leave the question here for the benefit of anybody else with the same difficulty. When do exactly we choose to use have or take? They look sort of having similar context     I have a bath      I take a bath   And by the way, I saw that      I take a chance   Can't be transformed to     I have a chance   Can it?   *Moreover, If there's a verb used like the bath, it means the bath is no more a verb but a noun right?  Also very important is that , can they be applied randomly to *a verb or a noun?  Like :  *  - I take a choose   Or    I take an option ( this one makes sense but I mean in the context of delexical verb)   ?  *EDIT: My mistake , bath is actually not "a verb transformed to a noun, but its absolutely a noun."  So my concerns would be if they can be used interchangeably? And be applied to random noun? How can I install a package without installing some dependencies? I'm trying to install the package LaTeXila, and the output looks like this:  $ sudo apt-get install latexila --no-install-recommends Reading package lists... Done Building dependency tree        Reading state information... Done The following extra packages will be installed:   latexila-data latexmk luatex tex-common texlive-base texlive-binaries   texlive-common texlive-doc-base texlive-latex-base Suggested packages:   rubber texlive-latex-extra debhelper Recommended packages:   texlive texlive-latex-recommended texlive-luatex lmodern   texlive-latex-base-doc The following NEW packages will be installed:   latexila latexila-data latexmk luatex tex-common texlive-base   texlive-binaries texlive-common texlive-doc-base texlive-latex-base 0 upgraded, 10 newly installed, 0 to remove and 0 not upgraded. Need to get 29.3 MB of archives. After this operation, 74.5 MB of additional disk space will be used. Do you want to continue [Y/n]?   I don't want to install the texlive packages. I've installed texlive manually from . Any suggestions? Is the Doctor Who episode 'Blink' the first use of the phrase "timey-wimey"? An Olympiad problem in prime numbers :can we proof that "n" is prime in this case " Let x,n ∈ $\mathbb{N}$  such that :  1+$x^1$+$x^2$+$x^3$+....+$x^{n-1}$ is prime .  Proof that n is prime .  I actually suppose that n isn't prime absurdly so n=p×k such as p is prime ,I also use the $a^n$-$b^n$ formula  and find that $x^k$-1/$x^1$-1 so that means that k=1 and then  n=p which is absurd. Does Anyone has another answer ? Finding center and radius of circumscribed triangle How can you handle an IN sub-query with LINQ to SQL? Let $Z_i,Z_2,\ldots$ be IID Random Variables with density $f$. Suppose that $P(Z_i&gt;0)=1$ and that $\lambda=\lim_{x \to 0+} f(x)&gt;0$. How can I show that $X_n=n \times \min\{Z_i\}$ has a limiting exponential distribution with mean$1/\lambda$?  I know that the CDF of the first order Statistic is for $t&gt;0$  $$F_{min \{ Z_i \}} (t)= 1- \left[ 1-F(t) \right]^{n}$$  Convergence in distribution requires that $F_n (t) \to F(t)$ for some distribution function $F$ but I cannot quite take the limit of the above, with $t$ replaced by $x/n$, as the distribution is not known. I guess I have to insert the second piece of information somewhere but I do not recognise where.  Could you please give me a hint or two?  Thank you. Is it possible to instantiate an object at runtime if I only have the DLL name and the class name, without adding a reference to the assembly in the project? The class implements a interface, so once I instantiate the class, I will then cast it to the interface.  Assembly name:     library.dll    Type name:      Company.Project.Classname     EDIT: I dont have the absolute path of the DLL, so Assembly.LoadFile won't work. The DLL might be in the application root, system32, or even loaded in the GAC. Where can I leave luggage in Frankfurt, either airport or train station? I need to find a place to leave my luggage in lockers for 2 days in Frankfurt, either the airport (Terminal 1) or at the train station. I have to go by train to Stuttgart for 2 days without luggage. The luggage is 5 bags, 28 inches. Flagging summary improvement: filters Here is my flagging summary on SU:   Your flagging history   32 posts flagged as requiring moderator attention    2 are waiting for review   3 were marked valid   1 were marked invalid   (older flags were not recorded valid/invalid)    2 posts marked as spam    It is too complicated to look for invalid flags. Would it be possible to make filters on the top of the page (like on favorites for example) with:  By date, By type (off-topic, spam, dupes), By status (accepted, pending, rejected)  It would make it easier for reviewers to check their flagging and eventually know what to improve on.  Or is there another easy way to audit this? How to calculate 95% CI for cumulative incidence rate using R? When I gain control of a creature, do I also gain control of any auras or equipment attached to it? Is "can not" unambiguous? I always try to remember to use "cannot" when applicable as in "I cannot take an umbrella" (because I do not have one).  I also thought that the problem with "I can not take an umbrella" was that it was ambiguous: it could also mean that I could decide not to take an umbrella. Trying to write unambiguously most of the time, I would therefore never use it.  Now I see it claimed that it is in fact not ambiguous, and only carries the second meaning. Usually I would accept such advice at face value and start using "can not" when the second meaning is unambiguously meant, but this comes from a style guide I have other reasons not to respect very much.  Does "can not" have a single unambiguous meaning?  (I have read  but my question is more about current usage. The length of the answer to that question does confirm my prejudice a bit, though)