Let $\alpha(t)$ be a regular curve. Suppose there is a point a in R^3 such that $\alpha(t)-a$ is orthogonal to T(t) for all t. Prove that $\alpha(t)$ lies on a sphere.  So we know   $&lt;\alpha(t)$ - a, T(t) > $= 0$  and so it follows that   $(\alpha(t)$ - a) ($\alpha'(t)) =0$  But where do I proceed?
Let $\alpha(t)$ be a regular curve. Suppose there is a point $a$ in $R^3$ such that $\alpha(t)-a$ is orthogonal to $T(t)$ for all t. Prove that $\alpha(t)$ lies on a sphere.  So, I let $\alpha(t)=(x(t), y(t),z(t))$ and $a=(a,b,c)$.  Since, $\alpha(t)-a$ is orthogonal to $T(t)$,  $&lt;\alpha(t)-a,T(t)&gt;=0$.  But I don't know how to solve this..
What happened to Finn's blaster at Maz Kanata's? When entering, we see that Finn has the blaster that was given to him earlier:    However, when the fighting begins, Finn tells Maz that he needs a weapon and no longer has his blaster with him.  It seems unlikely to have been lost in the first shots, given how it is secured across his body.  Also, Han and Chewie didn't lose their weapons and they are arguably less secure.  Now, when the team gets to the Resistance base, Finn again has his blaster:    So where was this blaster during the battle? Title does not specify "On Hold" Testing the app, I noticed the title does not specify if a question is "On Hold" ... I haven't checked specifically, but would assume it doesn't specify as "Duplicate" either ... Or am I missing it somewhere? Is it not in the title but located/displayed some other way? If it is, this could be very confusing to users. If it isn't there, it's definitely confusing to users :-)  Using: Version 1.0.1.125 on iPad3 with iOS8 JavaScript open in a new window, not tab Why are neural networks becoming deeper, but not wider? In recent years, convolutional neural networks (or perhaps deep neural networks in general) have become deeper and deeper, with state-of-the-art networks going from 7 layers () to 1000 layers ( in the space of 4 years. The reason behind the boost in performance from a deeper network, is that a more complex, non-linear function can be learned. Given sufficient training data, this enables the networks to more easily discriminate between different classes.  However, the trend seems to not have followed with the number of parameters in each layer. For example, the number of feature maps in the convolutional layers, or the number of nodes in the fully-connected layers, has remained roughly the same and is still relatively small in magnitude, despite the large increase in the number of layers. From my intuition though, it would seem that increasing the number of parameters per layer would give each layer a richer source of data from which to learn its non-linear function; but this idea seems to have been overlooked in favour of simply adding more layers, each with a small number of parameters.  So whilst networks have become "deeper", they have not become "wider". Why is this? QtCreator and Command Line Arguments Declare a global variable in a Blender session Can anyone tell me how Google decides which image to use when it shows an image with the search results (as is the case with mobile search, sometimes)?  If you look at these mobile search results …    … and take the AliExpress.com result as an example, by inspecting  you can see that  used by Google seems to have no special declaration, etc. It is just the first JPG that appears in that page's HTML - so I was thinking maybe that's why it's the image that is used in the Google search results.  However, when you look at these mobile search results …    … and consider the Backpack Billboards result,  reveals that  being used is actually the seventh image declared in the HTML.  So can anyone shed any light on how Google determines which image to be displayed? I would like to know so I can then have control over which image is displayed for my site.  Update / Solution  Thanks to the answers, below, this is the code I have now added to my head HTML in order to get the thumbnail shown in Google when searching for backpack billboards:  &lt;script type="application/ld+json"&gt; {   "@context": "http://schema.org/",   "@type": "Product",   "name": "Regular Backpack Billboard",   "image": [     "https://backpackbillboards.com/wp-content/uploads/2017/12/backpack-billboards-1x1.jpg",     "https://backpackbillboards.com/wp-content/uploads/2017/12/backpack-billboards-4x3.jpg",     "https://backpackbillboards.com/wp-content/uploads/2017/12/backpack-billboards-16x9.jpg"    ],   "description": "Backpack Billboards feature a rechargeable battery which powers the internal LED lights providing a bright illuminated display for up to 8 hours.",   "sku": "BPBB8",   "brand": {     "@type": "Thing",     "name": "Backpack Billboards"   },   "offers": {     "@type": "Offer",     "priceCurrency": "USD",     "price": "299.00",     "priceValidUntil": "2020-11-11",     "itemCondition": "http://schema.org/NewCondition",     "availability": "http://schema.org/InStock",     "seller": {       "@type": "Organization",       "name": "Backpack Billboards"     }   } } &lt;/script&gt; This is an example from a question sheet (non-assessed) of a university class. If M is a non-orientable, closed, connected 3 manifold, prove $H_1(M;\mathbb{Z})$ is an infinite group. I know that since $M$ is non-orientable it follows that $H_3(M;\mathbb{Z}) = 0$ and since $M$ is connected $H_0(M;\mathbb{Z}) = \mathbb{Z}$. I want to apply a combination of Poincare duality and Universal coefficient theorem to try and find a contradiction if $H_1(M;\mathbb{Z})$ has no free part. However poincare duality doesn't apply here so I'm stuck.  I would appreciate any hints as to how to proceed. Prove: $$ (a+b)^\frac{1}{n} \le a^\frac{1}{n} + b^\frac{1}{n}, \qquad \forall n \in \mathbb{N} $$ I have have tried using the triangle inequlity $ |a + b| \le |a| + |b| $,  without any success. 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? Product of two cyclic groups is cyclic iff their orders are co-prime Say you have two groups $G = \langle g \rangle$ with order $n$ and $H = \langle h \rangle$ with order $m$. Then the product $G \times H$ is a cyclic group if and only if $\gcd(n,m)=1$.  I can't seem to figure out how to start proving this. I have tried with some examples, where I pick $(g,h)$ as a candidate generator of $G \times H$. I see that what we want is for the cycles of $g$ and $h$, as we take powers of $(g,h)$, to interleave such that we do not get $(1,1)$ until the $(mn)$-th power. However, I am having a hard time formalizing this and relating it to the greatest common divisor.  Any hints are much appreciated! How do I set a monitor resolution that is not available in the Display Settings dialog? Decomposition of semisimple Lie algebra via its roots Exercise 14.33 in Fulton and Harris's Representation Theory claims that  If the roots of a semisimple Lie algebra lie in a collection of mutually orthogonal subsets, one sees that the Lie algebra decomposes into a direct sum accordingly.  Although intuitively I can see why this is true, could anybody direct me to a proof of this assertion, or outline how I should go about doing it? Show that if $f:X\to Y$ is a continuous function if and only if the graph of $f$ is a closed subset of $X\times Y$ When I ordered this TP-Link AC1300 Wireless Dual Band PCI-Express Adapter (Archer T6E), I looked into customers reviews and they said it worked fine with no drivers on Ubuntu. I didn't have a CD/DVD player on my desktop so I (on a laptop with a DVD drive) copied and pasted the DVD support disk content on to a USB drive and used that. But when I tried to run autorun.exe it puts up archive manager and says An error occurred while loading the archive Ubuntu verison- Ubuntu 16.04 LTS.  Output from: lspci -vnn  02:00.0 Ethernet controller [0200]: Realtek Semiconductor Co., Ltd. RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller [10ec:8168] (rev 11) Subsystem: ASRock Incorporation Motherboard (one of many) [1849:8168] Kernel driver in use: r8169  Kernel modules: r8169   I am fairly new to Ubuntu (1 month) and have learned alot (sudo commands) and now know that windows is overpriced and that ubuntu is better in most fields.  Thanks for the help. How does SO's (or other SE sites) versioning work of questions and answers? Do they use some standard piece of code or is it something homebrew?  With versioning I meant if for example a question is edited you'll see all the changes (a diff). I have a directory on my Windows 7 machine that has hundreds if not thousands of sub-directories.  Some of them have files, some do not.  I want to delete all the empty directories.  Looking at the del and rmdir DOS command, it does not look like you can recursively do this without deleting all the files.  Is there a way to do this from the command line? Introduction to Statistical Learning defines the F-statistic as follows: $$\frac{(TSS - RSS) / p}{RSS / (n - p - 1)}$$ I am trying to interpret this formula intuitively - the numerator looks like the $ESS$ per regressor, and the denominator looks like the $RSS$ per observation. This does not seem like an apples to apples comparison; can anyone explain why it makes sense?  They also say that the expectation of the denominator is equal to the variance of the irreducible error if the linear model assumptions hold (I get this, as the denominator is really the Residual Standard Error, which is an unbiased estimator). They also say the numerator is equal to the variance of the irreducible error if the null hypothesis is true. Therefore the F-statistic will be close to 1 if the null hypothesis is true.  But if you let p = 1, that is, apply the F-statistic to single linear regression, it becomes:  $$\frac{TSS - RSS}{RSS / (n-2)}$$  According to the book, if the regressor has no explanatory power, the F-statistic should be close to 1. But if you imagine a data set where the coefficient on X is 0 (i.e. no explanatory power), $TSS$ will be equal to $RSS$, so the numerator and hence the F-statistic should be 0, not 1 as they claim. What is going on?  Furthermore, if you accept that the F-statistic is 1, as they claim, then $(TSS - RSS) = RSS / (n-2)$, which means $ESS = RSS / (n-2)$. If you think of the F-statistics as comparing explained vs. unexplained variation, this does not seem like a fair comparison since this decomposition makes it the ratio of TOTAL explained variation summed across ALL observations vs. the unexplained variation PER observation. Again, what am I missing?  I am just trying to make sense of it in a layman way, apologies if I am missing something obvious. Noun for "receive"? Receipt?  Receiving?  I'm trying to come up with a name for "the acceptance of a delivery."  I'd like to be able to say      Ten deliveries were made but there are only five [fill_in_the_blank_here] on record.   Any suggestions? javax.persistence.PersistenceException : [PersistenceUnit: vodPersistenceUnit] class or package not found