For some reason, there appears to be a cut-off point on my stage beyond which nothing renders. I run into it when I'm rotating the camera around my objects (I keep the "camera fixed to view" option on so I don't get lost so easily).    It's as if there was a curtain or something there, impeding the view, but I can find no obvious reason for it.  Any ideas?
I have created a simple model here   When rendering it the end of it is cut off   How can I get the camera to render a longer distance?
How does Ubuntu partition the disk when we choose "Erase disk and install Ubuntu" during installation? How do I install KDE? I want to have both KDE and Gnome shells on Ubuntu. Ubuntu only has Gnome installed by default. How do I install KDE? Return FileName Only when using OpenFileDialog I'm trying to use Linq to return a list of ids given a list of objects where the id is a property. I'd like to be able to do this without looping through each object and pulling out the unique ids that I find.  I have a list of objects of type MyClass and one of the properties of this class is an ID.  public class MyClass {   public int ID { get; set; } }   What I want to do is write a Linq query to return me a list of those Ids  How do I do that given an IList&lt;MyClass&gt; such that it returns an IEnumerable&lt;int&gt; of the ids?  I'm sure it must be possible to do it in one or two lines using Linq rather than looping through each item in the MyClass list and adding the unique values into a list.  Any help in creating an elegant solution would be much appreciated! Finding $\int_0^\infty \frac{\cos(ax)-\cos(bx)}{x}dx$ I am practicing calculus section of GRE Math Subject test, and I can't figure out a way to do the following integral: $$\int_0^\infty \frac{\cos (ax) - \cos (bx)}{x}dx.$$  I absolutely have no clue how to do this. Can someone show me the solution explicitly? Thank you. A positive alternative to "smelling" to describe something with a pleasant odor When one hears that something smells, one would generally assume that it smells bad.   Isn't there a word which wouldn't bring to mind the idea of a bad odor? For example, how would you describe pot-pourri (assuming you like the smell of pot-pourri)? If possible, I would like an adjective, or a word that could takes the place of an adjective. Lower bound for $\phi(n)$: Is $n/5  1$? Is it true that :  $\frac {n}{5} &lt; \phi (n) &lt; n$ for all $n &gt; 1$ where $\phi (n)$ is  .  Since $\phi(n)$ has maximum value when $n$ is a prime it follows that maximum value of $\phi(n)$ in term of $n$ is $n-1$ , therefore $\phi(n)&lt; n$ for all $n$. What is the best lower bound for $\phi(n)$? How can an 18 month old be taught that bullying is not ok? He seems to like causing the trouble it brings upon him, and bullying, mean acts also seem to please him. He likes to sit on his younger cousin (my son), push him down, and steal his toys. His parents seem to be at a loss about it, and I don't really have any good suggestions to give. Thanks. Convergence to a Uniform Distribution $\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor} $  Show that if $P(X_n = i/n)=1/n$ for every $i = 1,...,n$, then $X_n$ converges in distribution to a uniformly distributed random variable $X$.  Convergence in distribution is defined as   $$P(X_n \leq x) \rightarrow P(X \leq x)$$  and I have that a uniform distribution with parameters $a$ and $b$ implies that  $$P(X \leq x) = \frac{x-a}{b-a}$$  I cannot go any further however. I feel like I don't know anything about $X_n$ - I don't even know if it is continuous or discrete. What I can just say is that   $$P(X_n \leq x) = \sum_{i=1}^\floor{x} \frac{1}{n}$$  where $\floor{x}$  means rounding $x$ to the closest lower integer. Proof verification: column rank = row rank $\newcommand{\span}{\operatorname{span}}\newcommand{\rank}{\operatorname{rank}}$I would like to prove that for every martix, column rank = row rank Let $A\in M_{m\times n}$ be some matrix. fix $R_A$ = vector space of the rows of $A$ , $C_A$ = vector space of the columns of $A$  $$\rank (R_A) = dim \ (\span \{R_1, R_2, \ldots, R_m\})$$ $$\rank (C_A) = dim \ (\span \{C_1, C_2, \ldots, C_n\})$$  let $x$ be some vector. $$x\in \operatorname{Null}(A) \Leftrightarrow \forall i  \ \ (1 \leq i \leq m) : \langle x,R_i\rangle = 0$$ (to be clear - I'm referring to the inner product of $x$ with each row of $A$) Using the rank nullity theorem, $\dim \operatorname{Null}(A) =  n - \rank(R_A)$ as $n$ =  number of columns. what I would like to do is to say: $$\dim \operatorname{Null}(A) =  n - \rank(R_A)$$ $$\dim \operatorname{Null}(A) = n - \rank(C_A)$$ therefore,  $\dim \operatorname{Null}(A) = n - \rank(R_A)= n - \rank(C_A)  \Longrightarrow \rank(R_A) = \rank(C_A) $ it that false? is using the rank- nullity  theorem this way is cheating? or just doesn't prove what needs to be proven formally? There is a bar of chocolate of size $N$ x $M$. Little square $(x, y)$ is poisoned. There are 2 players taking turns. In each move they can cut the bar along any of the horizontal or vertical lines, then the part which does not contain $(x, y)$ is being eaten. The player who is left only with square $(x, y)$ loses (beacause it is poisoned). Which player has the winning strategy? I guess that this game is equivalent to game of Nim due to Sprauge-Grundy theorem, but I don't know how. Using "nor" in a list without "neither" CSS two divs next to each other To show that $2$ | $x^{2n} - x^n$, $n \in \mathbb{Z^+}$ is the following adequate? Suppose that G is a group and N &lt; G a normal subgroup. Assume that there is no normal subgroup M &lt; G with N &lt; M. Prove that G/N is simple.  This problem is really throwing me for a loop. The concept of quotient groups and how they relate to simple subgroups is confusing me, and so I am not really sure how to begin. If a simple subgroup's only normal subgroup is itself (and the trivial group), then G/N being simple would be proving that when multiplying the entire subgroup N by an element g for all g in G, that the collection of those sets cannot be broken into smaller normal subgroups? Boolean and Texts I have used Hexagon for years but find that it sometimes crashes when trying to do Boolean operations with texts.  I have been trying to make the two tables of the Ten Commandments and have been trying to use Blender.  I did it about two years ago in Blender and lost everything on my computer so I am trying again with the latest version of Blender and am having difficulty.  I start with a cube and then get it close to the dimensions I would think the Ten Commandments might look like. I then bevel the top using the bevel modifier. But in doing it this way I don't have any cross sub divisions on the face of the object. I have tried sub dividing first but then I find that I can't bevel the top arches as far as I would like. I still don't know how to add divisions after I bevel the top. Yes I am still a novice at this program.  In any case I did figure out how to make text and turn them into an object. When I make the texts I do all four of the commandments for one side and then the other six for the other side. So this makes allot of texts to cut out.  When doing the Boolean operation in Difference sometimes nothing happens. I am thinking I need to have some sub - divisions on the face of the sized cube after it is beveled but am not really sure if this is the problem.  This whole thing about normal's is still Greek to me. I have watched some videos on how to cut letters into a cube but for some reason it does not work this easy after I bevel the top of the cube. Or it could be that Blender does not like a bunch of texts all at once.  Could someone show me a video or give me some simple steps to make my project work? Taylor's Remainder $x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) <x-\frac{x^2}{2}+\frac{x^3}{3}$ Prove that   $\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}&lt;\log(1+x) &lt;x-\frac{x^2}{2}+\frac{x^3}{3}$  My attempt: I can prove this by taking one side at a time and assuming $\displaystyle f(x) = \log (1+x)-x+\frac{x^2}{2}-\frac{x^3}{3}$   and then proving that it is a decreasing function and then the same for the other side.  But I am looking for a better solution using Lagrange's Mean Value Theorem or using Taylor's remainder. I'm used to always hearing or seeing a  before certain nouns.  Yet on certain occasions the article is totally omitted, and it bothers me.  I'm wondering what the justification for omitting the article is, especially when the definite article would seem to be required, or if it is just bad English either on the part of the individual in question or more likely by tradition.  The most frequent place I hear this is at the doctor or dentist's office. In every one I have ever visited, the receptionist or assistant always says things like "Doctor will see you shortly" instead of "The doctor will see you shortly."  The wikipedia article linked above says this: "A definite article indicates that its noun is a particular one (or ones) identifiable to the listener".  In the case of a clinical setting, I know who the doctor is, especially if this is not my first visit.  The people in the office know the doctor even better than I, so it would seem to be inappropriate to omit the article. Even the indefinite article would be incorrect. It further says, under "Zero article":  "In languages having a definite article, the lack of an article specifically indicates that the noun is indefinite."  Since the noun is not indefinite, why are they omitting the article?  Another place I see the article omitted is with the word "bar".  Bartenders apparently don't say "I tend the bar" they say "I tend bar".  Unless a bartender works in multiple bars, it would seem like an article would be required.  Furthermore, "the bar" can mean any bar a person goes to, so even in a case like , it seems like the word "the" should be in the sentence.  The only thing I can think of in this case is maybe the speaker doesn't want to confuse the listener into thinking they work in the legal field.  Is this just a bad habit, a historical corruption or tradition, or is there really a valid reason (or perhaps more than one) for these omissions? Asserting that Atheism and Theism are both faith based positions I am recently testing an assertion that I have concluded namely that atheism is a faith based position just like theism is a faith based position.  The reason I arrived at this conclusion is that theists have no proof that God actually exists. It's purely a matter of faith. Yes, there maybe strong implicit proof that God exists in their opinion (scriptures, history, etc) but there is no concrete proof of God's existence. In my view, atheists also suffer from the same problem of providing a proof of their position. Specifically that they cannot prove that God is non-existent.  The typical rebuttal I get is that the burden of proof is on the theists. But I view this as a cop out and they hide behind the wall of burden of proof which is just a bias in the debate.  Given the above, I claim that both atheism and theism are positions based on faith. Would it be incorrect to claim that? linebreak in description title