I'm not sure why, but I can't get my iPhone 4 to talk to my new MacBook Air 13".  I turn bluetooth on for both devices, open the Bluetooth Preferences and pair the devices.  This works - the code is displayed on the Mac and the phone, I click OK and it looks busy.  After that I get a message on the iPhone saying "[mac name] is not supported".  I have also tried clicking 'Connect to Network' on the Mac and that results in the error "The Bluetooth network is unavailable: There was an error connecting to your device. Make sure the device is turned on and in range".  I'm running 10.7.2 and iOS 5.0.  Well the device (iPhone) is on and in range.  Can anyone help?  I want to be able to control Keynote on my Mac using the Keynote Remote app on my phone.
I can successfully tether my Mac (10.7.3) to my iPhone4 (5.0.1) via wifi, no problem, been doing it for months.  However, when I try to tether via bluetooth instead, I can't seem to see the internet from my mac.  Network Settings reports a green light for Bluetooth PAN, but my browser can't connect to anything at all.  My Mac reports very similar IP addresses for both, starting with 172.20.10 and differing only in the last octet.  Any idea why I can't successfully tether via bluetooth?
I want that my monitor turns off when I am inactive for a while, but I do not want that I am prompted for a password when I revoke my pc. In Ubuntu 16.04 it was possible to disable the lock but still enable the screensaver as explained .  In Ubuntu 18.04 I entered the following in terminal:  dconf-editor   and disabled ubuntu-lock-on-suspend:    However, since I made this change, my monitor never suspends anymore. I tried restart but it didn't help.  Any ideas how I can disable lock but still enable suspend mode? Proving that $\mathbb{Z}_m\oplus \mathbb{Z}_n \cong \mathbb{Z}_d\oplus \mathbb{Z}_l $ as groups, where $l=\mathrm{lcm}(m,n)$ and $d=\gcd(m,n)$ How would one go about proving that $\mathbb{Z}_m\oplus \mathbb{Z}_n \cong \mathbb{Z}_d\oplus \mathbb{Z}_l $ as groups, where  $l=\mathrm{lcm}(m,n)$ and $d=\gcd(m,n)$?    I am attempting to use the fundamental theorem of finitely generated abelian groups but am struggling. In the interest of honesty, this is a past exam question that I am attempting for which the solutions are not available. EDIT: As the question was asked in the paper, the Chinese Remainder Theorem would not be permitted since this is only proved in the follow-up course. Proof that every finite dimensional normed vector space is complete Can you read my proof and tell me if it's correct? Thanks.  Let $V$ be a vector space over a complete topological field say $\mathbb R$ (or $\mathbb C$) with $\dim(V) = n$, base $e_i$ and norm $\|\cdot\|$. Let $v_k$ be a Cauchy sequence w.r.t. $\|\cdot\|$. Since any two norms on a finite dimensional space are equivalent, $\|\cdot\|$ is equivalent to the $l^1$-norm $\|\cdot\|_1$ which means that for some constant $C$, $\varepsilon &gt; 0$, $k,j$ large enough, $$ \varepsilon &gt; \|v_j - v_k\| \geq C \|v_j - v_k\|_1 e_i= C \sum_{i=1}^n |v_{ji} - v_{ki}| \geq |v_{ji} - v_{ki}|$$ for each $1 \leq i \leq n$. Hence $v_{ki}$ is a Cauchy sequence in $\mathbb R$ (or $\mathbb C$) for each $i$. $\mathbb R$ (or $\mathbb C$) is complete hence $v_i = \lim_{k \to \infty} v_{ki} $ is in $\mathbb R$ (or $\Bbb C$) for each $i$. Let $v = (v_1, \dots , v_n) = \sum_i v_i e_i$. Then $v$ is in $V$ and $\|v_k - v\| \to 0$:  Let $\varepsilon &gt; 0$. Then $$ \|v_k - v\| \leq C \|v_k - v\|_1 = C \sum_{i=1}^n |v_{ki} - v_i| \leq C^{'}n \varepsilon$$  for $k$ large enough. Does a conducting wire give off measurable radiation? A question about long exact sequence of finite dim vector spaces Should we avoid negative words when writing error messages? Isomorphism of a product $C_n \times C_m$ of cyclic groups with the cyclic group $C_{mn}$ How do you clean a seasoned cast iron skillet? I have a seasoned cast iron skillet, and I don't want to mess up the seasoning when I'm cleaning it.  What do I use and what don't I use to get it back to clean?  I've heard not to use soap and to make sure it is dry, but nothing beyond that. Can ubuntu be installed in intel kabylake 7th gen processor? I am using a 7th generation i5 processor(7200,2.50 ghz).The processor also have intel hd graphics 620 and nvidia geforce 940mx 2gb graphics card. I read previously that only windows 10 can be installed to this specific processor.  Is it true no version of ubuntu can be used with this processor?  If it's possible to install ubuntu using this processor, what additional steps other than the normal installation process should I undertake to get Ubuntu working? How do you choose Spells Known when creating a new character above 1st level? It's not clear to me how to choose Spells Known when creating a new character at a higher level.  Let's take the Bard as a working example first:  Bard level 2:   Spells Known: 5 Slots per Spell level: 1st &times; 3   Bard level 3:   Spells Known: 6 Slots per Spell level: 1st &times; 4, 2nd &times; 2   I'm interpreting this to mean that, if an existing 2nd-level Bard gains a new level (level 3), he's gaining 1 (one) new spell known. He can then choose the new spell known to be either 1st or 2nd, correct?  What I don't understand then, is the case of creating a new character starting at a higher level; say, creating a new Bard at level 3.  In that case you're forced to choose 6 spells, which must be from the list of 1st- and 2nd-level spells, correct?  When making this 3rd-level Bard from scratch, though, can I select 2nd level spells for all 6 spells known?  As another example, say you're creating a very high level Bard from scratch, starting at level 15. The book says your Spells Known is 19, and your spell slots are 1st × 4, 2nd × 3, 3rd × 3, 4th × 3, 5th × 2, 6th × 1, 7th × 1, and 8th × 1.  So, for those 19 Spells Known, you can choose any spells from 1st through 8th spell level. Am I free then to select as many spells as I can at 8th spell level (5 in number), working backwards, and in the end ignoring the 1st, 2nd, 3rd level spells entirely?  As another example, the Ranger doesn't know any spells at 1st level, nor have any spell slots, but the book says “You know two 1st-level spells of your choice from the ranger spell list.” Does it mean at 2nd level and above?  Long story short: When choosing Spells Known for a spellcasting character created above 1st level, am I allowed to choose my Spells Known at whatever spell level I like, just so long as I have at least one spell slot of that level?  I'm working on a D&amp;D 5e character generator and need to clarify this. In my online brokerage account, I want to buy a particular stock and I see the following:  Bid:  13.20 x200           Ask:  13.27 x1,000   the current stock price is 13.22.   Can someone explain what the bid and ask prices mean relative to the current price?  If I buy 1000 shares, why would I pay more? Do the beacon lighters really live on top of the mountains? Following this image I found online:    Do the beacon lighters really live on top of the mountains?  It seems really cold on top of some of these mountains and there are no villages or whatsoever. how to calculate $$\int_0^\infty \frac{x^{3}}{e^{x}-1} \; dx$$   Be $q:= e^{z}-1 , p:= z^{3}$ , then $e^{z} = 1 $ if $z= 2\pi n i $, so the residue at 0 is : $$\frac{p(z_{0})}{q&#39;(z_{0})} = 2\pi i n ^{3}$$  problem is that this is not symmetric, so how does one find the definite integral? Female main character in an ice world earning gems/runes Here's what I remember:   The main character's female (99% sure) There is one part where she has to cross a big tundra and she has to kill a big ice creature The kids in the story go to a school-like place where they train for something. I'm pretty sure they earn gems or runes or something like that There's one part where the main character has to play a piece of music perfectly to earn her gem/rune thing It's definitely a series; young-adult (I read them in middle school)   Thanks! "...his parents' dream of *him* achieving a Cambridge degree." What is the function of "him" here? I have a problem analysing this sentence from the point of finite/nonfinite clauses, clause elements and their functions:     He does not want to destroy his parents' dream of him achieving a Cambridge degree.   I am especially interested in the: dream of him achieving a Cambridge degree.  I know that 'achieving a Cambridge degree' is a non-finite -ing participle clause. However what is its function? And what is the function of 'of him'? Is it a postmodification? Let $F$ be a field and $F[x]$ a polynomial ring. Let $p(x)$ be an irreducible polynomial.  Show that $\gcd(p(x),q(x)) = 1\Longrightarrow \exists r(x),s(x)$ such that $r(x)p(x)+s(x)q(x) = 1$.  I know the proof for Bezout's identity for , but this proof uses the notion of absolute value, which cannot be applied to a polynomial ring.   I am thinking about a way to prove the statement: Use the Euclidean algorithm to compute $\gcd(p(x),q(x))$ and show that the result is $1$, and somehow go backwards to obtain $r(x)p(x)+s(x)q(x) = 1$, but it is a bit messy, and I'm not yet able to see it completely.  Please offer some help. If you have some other methods, that would be great, too. Is it possible to do something like this  #ifdef SOMETHING #define foo // #else #define foo MyFunction #endif   The idea is that if SOMETHING is defined, then calls to foo(...) become comments (or something that doesn't get evaluated or compiled), otherwise it becomes a call to MyFunction.  I've seen __noop used, but I don't believe I can use that.  EDIT(s):  I don't think I can really use a macro here, because MyFunction takes a variable number of arguments.  Also, I'd like to make it so the arguments are NOT evaluated!  (So doing something like commenting out the body of MyFunction doesn't really give me what I need, as the arguments will still be evaluated) How to make an international emergency call? How to make Portable QGIS 3 Is it possible to make QGIS 3 portable with changed location for Qgis setup parameters folder.  For QGIS 2 it looked like this in qgis.bat file:     qgis --optionspath "X:\qgis" --configpath "X:\qgisini"   For QGIS 3 it doesn't work. Setup parameters are default placed in:      C:\Users\xxx\AppData\Roaming\QGIS Show that the series $\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$ converges