I'm attempting to make a formula that will toggle a hyperlink on or off depending on the argument. Basically, I want to make it argue the following:  If there is text in (Tab A, Cell A), then create a hyperlink with the title (Text of Tab A, Cell A) that links to (Tab B, Cell A). If no text is present in (Tab A, Cell A), then just put the text "Not in Use" with no hyperlink.  I was attempting this formula, but it's not quite working. As long as there is text in SETUP!D39, it displays the text and activates the hyperlink, but the text is not blue or underlined. If there is no text in SETUP!D39, then it displays the text (Not In Use) properly, but the hyperlink is still there and gives the error "Cannot open the specified file". I'm pretty sure it's just a syntax error in my formula, but I can't figure it out:  =IF(ISTEXT(SETUP!D39),HYPERLINK("#'DEDICATED FUNDS (10)'!A1", (SETUP!D39)), "(NOT IN USE)")   So my questions are:   Any idea why the hyperlink is not turning blue and underlining? Any suggestions on why the hyperlink still exists in the second argument?   Any help is greatly appreciated.
ColA                  Col B              ColC Project A            A.pdf               A.pdf Project B                                No Link   I have a list of project records. One column (e.g., B) may have a reference to a pdf file. If I create a formula   =if(isblank(B2),"No Link",hyperlink(folderURL + B2),B2)   What I want is to place a hyperlink on only those rows where column B contains a pdf reference, and insert text "No Link". I get the proper text values, but in all rows have the hyperlnk default appearance and behavior -- Excel seems to be creating "phantom links" out of the "No Link" cells.     Clicking on the "No Link" value in Col C activates an attempt to open the file listed in Col B, which returns a No file found error.
Concatenate all arguments and wrap them with double quotes UK visit visa: What documents prove genuine relationship with the visitor? In the UK immigration rules for visitor visa  it states if;      Funds, maintenance and accommodation provided by a third party      V 4.3 A visitor’s travel, maintenance and accommodation may be   provided by a third party where the decision maker is satisfied that   they:      (a) have a genuine professional or personal relationship with the   visitor;and      (b) are not, or will not be, in breach of UK immigration laws at the   time of decision or the visitor’s entry to the UK; and      (c) can and will provide support to the visitor for the intended   duration of their stay.   I'm a UK  citizen. I'm planning to invite my brother-in-law and another close friend to the UK. In my invitation letter, I have clearly mentioned that I'm going to provide them accommodation only. They have prove sufficient funds to support themselves. I have sent them an invitation letter along with a copy of my passport and tenant agreement.   What other documents do I need to provide to indicate that it is a genuine professional or personal relationship with the visitor? iPhone stuck on Connect to iTunes My iPhone is stuck on Connect to iTunes Screen (support.apple.com/iphone/restore iTunes logo and up arrow with lightning cable) I have Tried :-   Connecting it to iTunes and Updating, It updates Successfully.But on the same screen again   Restoring, It restores successfully.But on the same screen again  Tried Recovery Mode Screen Method and DFU method Tried Closing and Restarting iTunes Tried Deleting 'iPhone Software Update' folder in iTunes and Re-Downloding Updates    Nothing Works... Any Help will be appreciated Chemical potential of particles with zero mass Why massless particles have zero chemical potential? Share menu hangs iOS app The Share menu in iOS 13 got a redesign, and the Stack Exchange iOS app is no longer compatible with it. I can still use it to copy the link, but when I try close the menu, it remains stuck. Opening another app and then returning automatically dismisses the menu, but then I can't use the back button anymore to return to the question list. (I'm still able to refresh the page, so it's a navigation issue.)    Ironically, I discovered this bug while trying to mark  as a duplicate; in the app, the quickest way to do that is to use the Share menu to copy the link to the original question to the clipboard...   App Version: 1.6.6.2 Device: iPhone 7 Plus OS Version: Version 13.1 (Build 17A844) generic function with a "has property X" constraint? A union of two subspaces not equal to the vector space. Find the exact value of the infinite sum $\sum_{n=1}^\infty \big\{\mathrm{e}-\big(1+\frac1n\big)^{n}\big\}$ I am wondering if my proof is correct? Thank you for whoever willing to take a look at it for me.     Proof $\liminf E_k \subset \limsup E_k $   If $\{E_k\}_{k=1}^\infty$ is a sequence of sets, we define \begin{align*} \limsup E_k &amp; = \bigcap_{j=1}^\infty\left(\bigcup_{k=j}^\infty E_k\right)\\ &amp;= \bigcap_{j=1}^\infty(E_j \cup E_{j+1} \cup \cdots)\\ &amp;= (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots. \end{align*}  Therefore, $\limsup E_k$ consists of those points in $\mathbb{R}^n$ which belong to infinitely many $E_k$. \begin{align*} \liminf E_k &amp; = \bigcup_{j=1}^\infty\left(\bigcap_{k=j}^\infty E_k\right)\\ &amp;= \bigcup_{j=1}^\infty(E_j \cap E_{j+1} \cap \cdots)\\ &amp;= (E_1 \cap E_2 \cap \cdots) \cup (E_2 \cap E_3 \cap \cdots) \cup \cdots. \end{align*} Therefore, $\liminf E_k$ consists of those points in $\mathbb{R}^n$ which belong to all $E_k$ for $k \geq k_0$.  Now consider $x \in \liminf E_k$, then $x \in E_1 \cap E_2 \cap \cdots$, or $x \in E_2 \cap E_3 \cap \cdots$ and so on, that is to say, $x \in E_{k_0} \cup E_{k_0+1} \cup \dots$ for some $k_0$. More specifically, $x \in E_k \forall k \geq k_0$. Therefore, $x \in (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots$, so we showed $x \in \limsup E_k$. Duplicate questions in feed, and questions incorrectly marked as unanswered I see duplicate questions in the feed (Why is the book still wanted if Aida has memorized it?) and the question is marked as "unanswered" even though it has an upvoted answer.  I would expect unique questions in the feed, as well as questions with answers to not be marked as unanswered.   App Version: 1.6.3.4 Device: iPhone 6 OS Version: Version 10.2 (Build 14C92) yum install error I can't run a yum command on my server without getting this:   CRITICAL:yum.cli:Config Error: Error accessing file for config file:///etc/yum.conf   This is what my yum.conf looks like:  # This is the fedorafaq.org yum.conf for Fedora Core 5. # # Note that you also need the rest of the configuration # as described at http://www.fedorafaq.org/#yumconf  [main] cachedir=/var/cache/yum keepcache=0 debuglevel=2 logfile=/var/log/yum.log exactarch=1 obsoletes=1 gpgcheck=1 plugins=1 metadata_expire=1800  # PUT YOUR REPOS HERE OR IN separate files named file.repo # in /etc/yum.repos.d   This is what my yum.repos.d looks like:  fedora.repo fedora-updates.repo fedora-updates-testing.repo   The permissions on those files are 0666 and it won't allow me to make them root:root I have  today and one member was voted as duplicate and private feedback was there, I choose no and submitted the form. When I revisit the question or refresh the page same notice reappears with yes/no options.  In-network tab I got 200 OK status and response as "success":true  Environment: OS: Windows 10 Browser: Chrome 79.0.3945.117 (Official Build) (64-bit)  From the comments, the same issue is happing with   (might be with whole StackExchange network)      From network tab What is the difference of 3 object declarations of javascript How do I do a clean install of Yosemite? Can jQuery get all CSS styles associated with an element? Prove $\operatorname{rank}(AB) = \operatorname{rank}(B)$ I am given the info that A is an invertible  m by m matrix and B is a m by n matrix. By solving for the null space $ABx=0$, I get $A^{-1}ABx=A^{-1}0$, therefore $Bx=0$. I concluded that the null spaces of A and AB must be equivalent, and therefore the nullity(AB) = nullity(B) = some number p. Therefore rank(AB) = n-p = rank(B)   (because both AB and B are m by n matrices).   Is this proof valid, specifically the null space step? I have had a question since childhood.  Why do we always get circular waves (ripples) in water even when we throw irregularly shaped object in it? There is already a similar post, that has an answer to my question, but it is quite short and I don't get it... So I am trying to show that  \begin{align} \{e_K |~ K \subset G~ conjugacy~ class \} \end{align} is a basis for $Z(\mathbb{Z}[G])$. Serre even writes "one immediately checks that the $e_K$ form a basis", but even reading the other post  I don't see that...  My approach to this is:  Be $x \in Z(\mathbb{Z}[G])$, which is equivalent to: $x = gxg^{-1}$ for all $g \in G$. Writing  $x = \sum\limits_{h \in G} \lambda_h h$, we get $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$. I don't really know how to go on at this point. I mean, I know that when $a,b \in K$ (same conjugacy class) you get $\sum\limits_{s \in K} \mu_s s = \mu\sum\limits_{s \in K} s$, since \begin{align} a = hbh^{-1} \Rightarrow \mu_a = \mu_b ~\forall a,b\in K \end{align} So all elements in the same conjugacy class have the same scalar.  But from here I don't really know how I can generate the $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$ with the $e_K$ ...  I would be really thankful, if someone could give me quite a basic proof that even I understand... Getting image dimensions without reading the entire file How can I remove the white pixels around the edges when exporting a transparent GIF?