I know this is probably a super easy answer, but I can't find it anywhere. How do I turn the shading off on objects when I got them in texture mode? I want to fully see the texture without the crazy dark shadows; preferably no shadows at all.  Thanks.
I'm trying to display models that have shading drawn or baked directly on the textures, but I can't figure out how to disable Blender's default scene lighting. If there are no lights in the scene, everything is black, but I want to see the textured objects fully bright. How do I disable the lighting?  NOTE: Just before posting this, I found the "Shadeless" material property, which does more or less what I want. However, is there a way to globally toggle the default lighting?
TypeError: 'str' does not support the buffer interface Should I use the first or last bits from a SHA-256 hash? I have the following code:  Using cmd As SqlCommand = Connection.CreateCommand     cmd.CommandText = "UPDATE someTable SET Value = @Value"     cmd.CommandText &amp;= " WHERE Id = @Id"     cmd.Parameters.AddWithValue("@Id", 1234)     cmd.Parameters.AddWithValue("@Value", "myValue")     cmd.ExecuteNonQuery End Using   I wonder if there is any way to get the final SQL statment as a String, which should look like this:  UPDATE someTable SET Value = "myValue" WHERE Id = 1234   If anyone wonders why I would do this:   for logging (failed) statements for having the possibility to copy &amp; paste it to the Enterprise Manager for testing purposes Is it common to review papers assigned to your supervisor? Is it common to supervisors to email their students papers to review (which were apparently assigned to the supervisor, and not to the student). The field is Computer Science, but I  would also like to hear from the other fields. Find the Unit Vector of a Three Dimensional Vector Creating label in QGIS? I am using QGIS 2.18.16 on Windows 10.    What I want is a label to appear within the boundaries of a polygon.    I select a layer, then select command Layer>Labelling, and I get:    From here, I can change "No Labels" to "Show labels for this layer".  But I can't actually create label text.  I find a clue in a prior question:    “Labelling polygon using QGIS?”    The answer was:  The layer you are trying to label must have at least one attribute field populated with data. You can then turn on labelling in Layer Properties and select which attribute field to use for your labels.  I can enter command Layer>Open Attribute Table, and what I get is:    If I need something in the attribute table in order to apply a label, how do I accomplish that? In linear regression, when is it appropriate to use the log of an independent variable instead of the actual values? Am I looking for a better behaved distribution for the independent variable in question, or to reduce the effect of outliers, or something else? The meaning of Bit-wise NOT in Python IVP Differential Equation I have stumbled across a very old exam question from my linear algebra course and the solutions are not available. I was wondering if my working/logic is correct and if any improvements can be made.     Let  $\ B=\begin{pmatrix}     1 &amp; -1 &amp; 4 \\     6 &amp; -7 &amp; 2 \\     -3 &amp; 1 &amp; -6       \end{pmatrix}\ $ and $\ \vec{v_1}=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.      Part (i): Calculate $(B+5I)\vec{v_1}$ and $(B+5I)^2\vec{v_1}$.   Using simply matrix multiplication, I calculated $(B+5I)\vec{v_1}=\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and $(B+5I)^2\vec{v_1}=\vec{0}$.     Part (ii): Find all eigenvalues and eigenvectors of B.   From our calculations in part (i), we can see that $\ \lambda=-5,-5$ are two of the eigenvalues, with corresponding eigenvector $\ \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and generalised eigenvector $\ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. From the trace of B, the remaining eigenvalue is $\lambda=-2$ with corresponding eigenvector $\begin{pmatrix} -2 \\ -2 \\ 1 \end{pmatrix}$.     Part (iii): Giving reasons, write down a basis for the generalised eigenspace $GE_{-5}$ of B.   I am unsure of this part of the question. If I had to guess, I would say a basis for $GE_{-5}=\left\{\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix},\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\right\}$. Is this correct?     Part (iv): Solve the initial value problem $\vec{y'}=B\vec{y}$ where $\ B=\begin{pmatrix}     1 &amp; -1 &amp; 4 \\     6 &amp; -7 &amp; 2 \\     -3 &amp; 1 &amp; -6       \end{pmatrix}\ $ and $\vec{y}(0)=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.   Well, the solution to this IVP will take the form $\vec{y}=e^{tB}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. I noticed that $\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ was a generalised eigenvector and hence we can write $\vec{y}=e^{-5t}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}+e^{-5t}\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}t$. Thus $\vec{y}=e^{-5t}\begin{pmatrix} 1+t \\ 1+2t \\ -1-t \end{pmatrix}$ Countable additivity implies countable subadditivity? Suppose $\mu$ is a measure in a ring $R$ of a set $X$. So for any mutually disjoint subsets $A_1,A_2,\ldots$ such that $\sum_{i=1}^\infty A_i\in R$, we have that $\mu(\sum_{i=1}^\infty A_i) = \mu(A_1)+\mu(A_2)+\cdots$. (i.e. countable additivity)  I'm wondering: Does that imply countable subadditivity? I.e. for any subsets $A_1,A_2,\ldots$ such that $\sum_{i=1}^\infty A_i\in R$, is it always true that $\mu(\sum_{i=1}^\infty A_i)\le \mu(A_1)+\mu(A_2)+\cdots$?  I believe it should be true, because you can take the mutually disjoint subsets $A_1,A_2-A_1,A_3-(A_1\cup A_2),A_4-(A_1\cup A_2\cup A_3),\cdots$, each of which is in $R$ (because unions and differences of sets in $R$ are again in $R$), then apply countable additivity, then use monotonicity on the fact that $A_n-(A_1\cup\cdots\cup A_{n-1})\subseteq A_n$.  Is that right? I just want to make sure I didn't make any mistake in my logic. Calculate: $$\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)\mathrm{d}t$$ My try: $\displaystyle\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)dt\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\cos(\mathfrak{I\textrm{z}}+3(-i\log z)dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}}^{(\mathfrak{I\textrm{z}}+3(-i\log z))i}dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}}z^{3}dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}R}z^{3}dz$ and there is nothing here that is not holomorphic, therefore according to Cauchy theorem it must be exactly $0$. Show by a counterexample that for a family $(X_i)_{i\in I}$ of random variables the independence of all pairs $(X_i,X_j)$ with $i,j\in I, i\neq j$ does not imply the independence of the family (It is enough to consider $\mbox{card}(I)=3$.).      Hello, I am searching for a simple example. The professor gave the hint that one might consider random variables which only can take two different values, but I do not know exactly what is meant.  By the way:  We call the family $(X_i)_{i\in I}$ of random variables independent, if $(\sigma(X_i))_{i\in I}$ is independent, whereat we call a family $(\mathcal{E}_i)_{i\in I}$ with $\mathcal{E}_i\subset\mathcal{A}$ independent, if for any finite subset $I_0\subset I$ and any choice $E_i\in\mathcal{E}_i, i\in I_0$, it is $$ \mathbb{P}(\bigcap_{i\in I_0}E_i)=\prod_{i\in I_0}\mathbb{P}(E_i). $$ I'm out of close votes. Can I use moderator flags instead? Is it correct to say "I slump/slumped on/onto the stool/barstool?" I wrote this sentence: "I buy a bottle of beer and slump on/onto the last stool."  A native English speaker told me that "slump" doesn't work here. That you can't slump on a bar/barstool.  Is this correct? If so, why? For $f$ in dual space, there exists $x$ with norm 1 and $f(x)=\|f\|$ if space is reflexive (and nontrivial) Let $X\ne\{0\}$ be a reflexive space and let $f\in X^*$, where $X^*$ is the dual of $X$. I want to know: in general, does there exist an $x\in X$ with $\|x\|=1$, and $f(x)=\|f\|$, where $\|f\|$ is defined as $\sup\{|f(x)|:x\in X,\|x\|=1\}$?  I know this is true for $\mathbb{R}^n$ with the norm from the standard inner product, but I'm wondering if it is true in general. I always thought of current as the time derivative of charge, $\frac{dq}{dt}$. However, I found out recently that it is the ampere that is the base unit and not the coulomb. Why is this? It seems to me that charge can exist without current, but current cannot exist without charge. So the logical choice for a base unit would be the coulomb. Right? Number of solutions of $x_1+2x_2+\cdots+kx_k=n$? "who" vs. "whom" and when to omit the relative pronoun This is the boy who was asked by the teacher to stay back after school. This is the boy whom the teacher asked to stay back after school. This is the boy the teacher asked to stay back after school.   Are the sentences above correct ? Already awarded bounty refunded on question migration? In March 2011 I offered a 500 bounty on one of my StackOverflow questions () and awarded it to Filip de Vos.   This question was later migrated to Database Administrators and has the following messages on it.     migrated to dba.stackexchange.com by Will♦ Dec 28 '11 at 12:25      This   question belongs on our site for database professionals who wish to   improve their database skills and learn from others in the community.   .     deleted Dec 31 '11 at 1:01   I appear to have been   on whereas Filip .  Is this a bug? Definitely doesn't seem fair or correct to me. The question still shows up in both our bounty tabs despite the migration. I would have hoped that the previously awarded bounty reputation points could still be honoured in this situation. Click to hide in the Hero content is not aligned correctly for the non joined communities The "Click to hide" icon in the hero content (welcoming message) for the non joined communities are not aligned correctly for most of the private beta and some other communities.  A few examples are:        Can the icon be aligned to the right hand side with correct padding?