Question about drives (Partition does not start on physical sector boundary)
"Misaligned partition" - Should I do repartition (how?)
Is this mold? Black? White? Guys I need to find out what this is- it's also on my wooden bed racks Why does the Cauchy distribution have no mean? I would like to evaluate the sum  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$  Here is my attempt:    Letting    $$f(z)=\frac{1}{az^2+bz+c}$$  The poles of $f(z)$ are located at   $$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$  and  $$z_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}$$  Then  $$ b_0=\operatorname*{Res}_{z=z_0}\,\pi \cot (\pi z)f(z)= \lim_{z \to z_0} \frac{(z-z_0)\pi\cot (\pi z)}{az^2+bz+c}= \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b} $$  Using L'Hopital's rule. Continuing, we have the limit is  $$ \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}= \frac{\pi\cot (\pi z_0)}{2az_0+b} $$  For $z_0 \ne 0$  Similarly, we find  $$b_1=\operatorname*{Res}_{z=z_1}\,\pi \cot (\pi z)f(z)=\frac{\pi\cot (\pi z_1)}{2az_1+b}$$  Then   $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c} = -(b_0+b_1)=\\ -\pi\left( \frac{\cot (\pi z_0)}{2az_0+b} + \frac{\cot (\pi z_1)}{2az_1+b}\right)=  -\pi\left( \frac{\cot (\pi z_0)}{\sqrt{b^2-4ac}} + \frac{\cot (\pi z_1)}{-\sqrt{b^2-4ac}}\right)= \frac{-\pi(\cot (\pi z_0)-\cot (\pi z_1))}{\sqrt{b^2-4ac}}= \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{\sqrt{b^2-4ac}} $$  Then we have  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c} = \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{2\sqrt{b^2-4ac}}$$    Is this correct?  I feel like I made a mistake somewhere.  Could someone correct me?  Is there an easier way to evaluate this sum? Find all continuous functions $f$ satisfying   $$\int_0^xf=(f(x))^2+C$$ for some constant $C \neq 0$, assuming that $f$ has at most one $0$.  I have a question about the solution to this problem. It says that   Clearly $f^2$ is differentiable everywhere, its derivative at $x$ is $f(x)$ (I'm assuming that this is because the left hand side is differentiable, the right hand side must be differentiable as well). So $f$ is differentiable at $x$ whenever $f(x) \neq 0$, and  $$f(x)=2f(x)f'(x),$$ so $f'(x) = \frac{1}{2}$ at such points. Thus, $f= \frac{1}{2}x+b$ for some $b$ on any interval where it is non-zero; if $f$ has a zero, this gives two possible solutions, with two possible values of $b$, but since $f$ is assumed continuous, they must be the same. So we need $$\int_0^x(\frac{1}{2}t+b)dt=(\frac{1}{2}x+b)^2+C$$  so we must have $b=\sqrt{-C}$ or $-\sqrt{-C}$, leading to two solutions for $C \lt 0$.  I don't understand the bolded statements. First, how does the differentiability of $f^2$ guarantee that $f$ is differentiable, and why only at $x$ where $f(x) \neq 0$?  Next, why are there two possible solutions if $f$ has a zero, but since $f$ is continuous they must be the same?  Finally, does the last statement mean that if $C \gt 0$, then there is no solution, since we're restricted to real numbers here?  I would appreciate it if anyone answers my questions. Get index of parent div from child divs Help with multirow table I am trying to wrap text in the multirow column (1st one) and only have the vertical lines inside the table if that makes sense.   Can anyone help me? `  \begin{table*}[!htbp] \centering \renewcommand{\arraystretch}{1.4} \begin{tabular}{|c| p{2cm}|c|c|c|c|} \cline{3-6} &amp;&amp;\multicolumn{2}{c|}{\textbf{Gravité du Dommage G1}} &amp; \multicolumn{2}{c|}{\textbf{Gravité du Dommage G2}}\\  \cline{3-6} &amp;&amp;\textbf{Fréquence F1} &amp; \textbf{Fréquence F2} &amp; \textbf{Fréquence F1} &amp; \textbf{Fréquence F2}\\ \hline \multirow{ 2}{*}{\textbf{Probabilité Occurence O1}}&amp;\textbf{Possibilité Evitement P1}&amp;&amp;&amp;&amp;\\ \cline{2-6} &amp;\textbf{Possibilité Evitement P2}&amp;&amp;&amp;&amp;\\ \hline \multirow{ 2}{*}{\textbf{Probabilité Occurence O2}}&amp;\textbf{Possibilité Evitement P1}&amp;&amp;&amp;&amp;\\ \cline{2-6} &amp;\textbf{Possibilité Evitement P2}&amp;&amp;&amp;&amp;\\ \hline \multirow{ 2}{*}{\textbf{Probabilité Occurence O3}}&amp;\textbf{Possibilité Evitement P1}&amp;&amp;&amp;&amp;\\ \cline{2-6} &amp;\textbf{Possibilité Evitement P2}&amp;&amp;&amp;&amp;\\ \hline \end{tabular} \end{table*} How to make a file (e.g. a .sh script) executable, so it can be run from a terminal Problem understanding the logistic regression link function Prove that if $a&gt;0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $.   Using the fact that $a \gt 0$, multiply by $a$ on both sides and get everything to one side we have; $a^{11}-a^{10}-a+1 \geq 0$. By factoring $(a^{10}-1)(a-1) \geq 0 $.  I am not sure how to proceed any further. Is there a typo in the name of the "Joga Bounito" hat? We have currently a hat called "Joga Bounito":    Bounito? Which language is that?  It is clearly a latin-derived language, so not English, German, Arabic, Chinese...   In Portuguese it should be "Joga Bonito" (which means "Plays beautifully"). I'm not sure about Spanish, but I guess that it should be then "Juega Bonito". I'm not sure about Italian either, but I guess that it should be "Gioca Bel". In French (not sure) it should be "Joue Beau". In Romen (really unsure) it should be "Joacă Frumos".   So, the best guess is that it is in Portuguese. Further this hat is basically a soccer ball, and in this year we had the FIFA World Cup in Portuguese-speaking Brazil, which is known for being a country where soccer is the main sport. So "Joga Bounito" would really be Portuguese.  But there is a small problem. It has a typo: It should not be "Bounito", it should be "Bonito". Are there any rules on the positioning adverbs should take in a sentence? For example:     Ever wish you could share information broadly   Could it be rewritten to:     Ever wish you could broadly share information   Are there any rules for the position of the adverbs. Irreducibility of a polynomial if it has no root (Capelli) Let $F$ be a field of arbitrary characteristic, $a\in F$, and $p$ a prime number. Show that $$f(X)=X^p-a$$ is irreducible in $F[X]$ if it has no root in $F$.   This  to a related question mentions the result is due to Capelli.  I can prove the result if $F$ has characteristic $p$ as follows.  Suppose $f$ is reducible: $f(X)=g(X)h(X)$ with $g(X)$ an irreducible factor of degree $m$, $1\le m&lt;p$. Then if $\alpha$ is a root of $g$ in some extension field $K$ of $F$, we have $$f(X)=X^p-\alpha^p=(X-\alpha)^p$$ so its divisor $g(X)$ must be of the form $(X-\alpha)^m$. Since the coefficient of $X^{m-1}$ in $g$ is in $F$, we have $m\alpha\in F$.  So $\alpha\in F$ because $m$ is invertible modulo $p$.  How would you show the result in other characteristics? I have been running  over some C# code, and it keeps reporting that my using directives should be inside the namespace.  Is there a technical reason for putting the using directives inside instead of outside the namespace? I'm trying to fill out the skills section on my first character, but when I look at the character sheets in the Starter Set as examples of how to make a character, the skills that are circled have higher modifiers than their Dexterity or Strength or whatever should make them.  I've filled out all skills but those. What do I add to the ability modifier to get those numbers?   My abilities are Strength 13, Dexterity 14, Constitution 11, Intelligence 15, Wisdom 8, and Charisma 8. Proficiencies are Arcana, Athletics, History, Investigation and Stealth. My saving throws are Strength and Dexterity.   I just started D&amp;D and have the Player's Handbook, DM's Guide, Monster Manual, and the Starter Set. Cannot select vertices from different object while in edit mode Whenever I enter edit mode, with 2 separate meshes (cubes), I can only select vertices/edges/faces of 1 mesh.  Example (the screen shot did not include the mouse cursor):    Also, when I enter edit mode with both meshes selected, one of the meshes has a red border and cannot be selected (it also has this border in Object mode): Prove by induction. How do I prove $\sum\limits_{i=0}^n2^i=2^{n+1}-1$ with induction? Use of inequality $1 - \cos (x) \leq x^2 /2$ In  answer the value of $1 - \cos(x)$ has to be evaluated in order to find its upper limit, if it exists.  In particular, $x = 2 \pi / n$. The answer is related to the length of a side of a regular $n$-gon inscribed into a unit-radius circumference; because the perimeter of the $n$-gon is always less than $2 \pi$, the single side must always be less than $2 \pi / n$.  The inequality  $$1 - \cos (x) \leq \displaystyle \frac{x^2}{2}$$ (1)  is used and the proof is completed with   $$2(1 - \cos(x)) \leq (2 \pi / n)^2$$  $$\sqrt{2(1 - \cos(x))} \leq 2 \pi / n$$  But it is well known that the cosine is a function $f(x) \in [-1;1]$, so $1 - \cos (x) \in [0,2]$. By using this information, we would obtain  $$1 - \cos (x) \leq 2$$ (2)  The proof would provide  $$2(1 - \cos(x)) \leq 4$$  $$\sqrt{2(1 - \cos(x))} \leq 2$$  which is a completely different result.   Why in that case it is preferable to use (1) instead of (2)? How to choose when it is convenient to use (1) and when to use (2) in a proof? Let $p &gt; 3$ be a prime such that $p \equiv 2 \pmod 3$. Define the elliptic curve $E$ over $\mathbb{F}_p$ by $y^2 = x^3 + 1$. Prove that $E(\mathbb{F}_p)$ consists of $p+1$ points.  Using Fermat's little theorem you can prove that $x^3 + 1$ is a bijection on $\mathbb{F}_p$. Hence, $$\#E(\mathbb{F}_p) = \#\{(x,y) \in \mathbb{F}_p^2 : y^2 = x^3 + 1\} + \#\{\infty\} = \#\{(x,y) \in \mathbb{F}_p^2 : y^2 = x\} + 1.$$ But then I am stuck trying to prove that $y^2 = x$ has $p$ solutions $(x,y) \in \mathbb{F}_p^2$. How can I count unique selections from multiple selections in a drop down list (Excel) I would like to count the number of unique selections from a column that contains multiple selections from a drop down list.  For example, column B3 contains,     Monday, Tuesday, Wednesday   The count function returns a value of 1 instead of 3 - is there any way to count the three days distinctly? Thank you!  Multiple selections from a drop down list was made possible using the VBA code from: How to remove ubuntu partitions and use them to install windows