How do I show that my solution to the congruence is the only solution? I'm asked to solve, $\mu,\lambda$ that satisfy $$89 \lambda+55 \mu=1$$ Using Euclid's algorithm I found $\lambda=-21,\mu=34$. Then I'm asked to find the solution to $$89 x \equiv 7\pmod{55}$$ Using $\lambda 89 \equiv1 \pmod{55}$ I found that $7\lambda89 \equiv7 \pmod{55} \implies x=7\lambda +55k$ for $k\in \mathbb{Z}$. But how do I show, that this is the only solution to the congruence?
I want to ask a question about modular arithmetic. I know, that modular multiplicative inverse exists only if modulo and integer are relatively prime. I want to know, are there any ways of division in modular arithmetic, if modulo and integer aren't relatively prime? I tried to find info about that, but failed.
How old is SpongeBob? How old is SpongeBob? I don't mean how old is the series, I mean what is SpongeBob's age if he were a real person. I ask because his age seems to be ambiguous in the show.   He has a full-time job He has his own house He is always referred to as a kid (especially in the movie) He acts childish and immature In one episode, he was pretending to be an adult/grown-up The bar at the very bottom of the screen has gone away? A skydiver jumps out of a plane from a height of $2200$m. The skydiver's height, $h$ meters above the ground, can be modeled by the function $$s = 2200 - 4.9t^2$$ How fast is the skydiver falling at $t = 4$s?  If the skydiver is approaching a height of $2200$m, I reasoned that I could use the following definition to find the solution $${f(x+h)-f(x)\over h}$$ I entered the function into this definition and got stuck when I got to this$$f'(x) = \lim_{x\to 0} {2200-4.9t^2-9.8th-4.9h^2\over h}$$ Where can I go from here? Can I continue with this method or did I make a mistake a few steps back? New Solution  How does this look? This is the proof, which I mostly understand except for one bit:  You have $h_1 \in H_1$ and $h_2 \in H_2$.   We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$. Similarly, we have $(h_1^{-1}h_2^{-1}h_1)h_2 \in H_2$. Therefore  $$ h_1^{-1}h_2^{-1}h_1h_2 \in H_1 \cap H_2 = \{1_G\} $$  and so $h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$. Let's first multiply everything on the left by $h_1$  $$h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$$ $$ h_1 h_1^{-1}h_2^{-1}h_1h_2 = h_1 \{1_G\}$$ $$ h_2^{-1}h_1h_2 = h_1 \{1_G\}$$   Multiply both sides on the left by $h_2$ giving us  $$ h_1h_2 = h_2 h_1 $$  The bit I don't get is right at the beginning. Why is this correct: "We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$." What's the difference between a projection and a datum? Google Drive storage mismatch I have a paid 100GB Google Drive plan. It is reporting that I am at 99% capacity.  According to the the storage both within the desktop and web application, I have used 90.53GB within Drive.    My Drive is mapped to my OS X filesystem, and when fully synched it reports only 61.28GB used.    So what is going on here? I need to do a cleanup but cannot account for the 30GB difference. Let $f$ be defined on $\mathbb{R}$ by $f(x) = e^{-1/x^2}$ for $x$ not equal to $0$. and $f(0)= 0$. Prove that $f^{(n)}(0)=0$ for all $n = 1, 2,3$ ...  Do I need to use Taylor expansion from calculus class?  Any hint would be appreciated. \@seccntformat only for section I would like to add a dot between the number of the section and section name. I used   \makeatletter \renewcommand*{\@seccntformat}[1]{% \csname the#1\endcsname .} \makeatother   The problem is that, obviously, it put a dot also among subsection number and subsection name which I do not want. Is it possible to add the dot only between section number and section using low level instructions i.e. without any additional package? jQuery ajax load() java script not executing? Understanding downside of triple-leveraged ETF What happens to the price of the triple-leveraged ETF if the S&amp;P 500 loses more than one-third of its value (i.e., the return is below -33 1/3%)?  Does it mean the value is equal to zero? I tried to use this for the last part of my table, but I can't get a vertical line at the end. Does anyone know?   \begin{tabular}{|CC|C|C|C|C|C|C|C|C|C|C|}\hline \Tab{Method Based on \\\hline       Radiography\\Magnetic Particle (A.C.)\\ \\Eddy Current\\Dye Penetrants\\ Ultrasonics} &amp; \Tab{ \\\hline \\ Wet \\ Dry \\ \\ \\ \\} &amp; \Tab{\rotatebox{90}{~\parbox{5cm}{Minute Surface Cracks }~}  \\\hline       N\\ G\\F\\F\\F\\P}  &amp; \Tab{\rotatebox{90}{~\parbox{5cm}{Deeper Surface Cracks      }~}  \\\hline            F\\G\\G\\G\\G\\G} &amp; \Tab{\rotatebox{90}{~\parbox{5cm}{Internal Cracks      }~}  \\\hline            F\\N\\N\\N\\N\\G} &amp; \Tab{\rotatebox{90}{~\parbox{5cm}{Fatigue Cracks      }~}  \\\hline            P\\G\\G\\N\\G\\G} &amp; \Tab{\rotatebox{90}{~\parbox{5cm}{Internal Voids      }~}  \\\hline            G\\N\\N\\N\\N\\G}           &amp; \Tab{\rotatebox{90}{~\parbox{5cm}{Porosity and Slag in Welds      }~}  \\\hline            G\\N\\N\\P\\N\\F}           &amp; \Tab{\rotatebox{90}{~\parbox{5cm}{Thickness      }~}  \\\hline            F\\N\\N\\P\\N\\G} &amp; \Tab{\rotatebox{90}{~\parbox{5cm}{Stress Corrosion      }~}  \\\hline            F\\G\\F\\N\\G\\F} &amp; \Tab{\rotatebox{90}{~\parbox{5cm}{Blistering      }~}  \\\hline            P\\N\\N\\N\\N\\F} &amp; \Tab{\rotatebox{90}{~\parbox{5cm}{Corrosion Pits      }~}  \\\hline            G\\N\\P\\N\\F\\P} \\\hline &amp; \multicolumn{11}{l}{G=Good;F=Fair;P=Poor;N=Not suitable.} \\ \hline \end{tabular} \label{tab:NDETechniques} \end{table}   It comes out like this: How do you find the Lie algebra of a Lie group (in practice)? Given a Lie group, how are you meant to find its Lie algebra?  The Lie algebra of a Lie group is the set of all the left invariant vector fields, but how would you determine them?  My group is the set of all affine maps $x \rightarrow A.x+v$ from $R^n$ to $R^n$ under function composition. When $z\neq 0$, $\arg z$ is defined to be the set $\{\theta \in \mathbb{R} : z=|z|e^{i\theta}\}$.  What if $z=0$? Usually does one leave the argument of $0$ undefined? Or is $\arg 0 = \mathbb{R}$? Variance of a linear combination of vectors Since Chrome version 38, there's a pretty ugly button in the title bar of the Chrome window that allows you to switch profiles quickly.  Since I'm the only one that uses my computer, this is not really useful to me, and I'd rather reclaim the space for more open tabs.  I've looked in the settings but don't see a way to disable it.  How can I get rid of the button? We know that if you (using Polyjuice), take the form of someone who had been injured, you assume that injured form (e.g. Barty/Moody). If you are injured, presumably you would be uninjured while under Polyjuice. My question is, is this in real-time? That is, if you take a Polyjuice, and while you're still under its effects, the person being transformed into gets injured (or even dies), will that injury be transferred onto you? And conversely, if you get injured while in someone else's body, will that transfer over to your real body after the Polyjuice wears off, will it disappear, or will it go to the person you turned into? Proof verification of a number theory problem involving sequences. Has a movie from a movie series ever been removed/retconned due to fan/general outcry? I recently upgraded to 12.04 from 11.10 and since upgrading have been unable to log in to my user profile. The upgrade went okay and I can login to a guest session fine but whenever I try to login to my profile, after entering my password, I just return to the login screen.  I've changed my password in Root (passwd &lt;username&gt;) and can log in to tty1 with no issues, it's just in GUI I'm having problems.  I'm using a HP dv7 laptop, 32-bit Ubuntu install, Intel® Core™2 Duo CPU P7350 @ 2.00GHz × 2, Nvidia graphics.  Any help would be appreciated. What is that component from the ADP5091 Schematic? I am looking for an explanation what that component does inside of that circuit.  It can be found on page 3 of the Datasheet of ADP5091:   It would help me out a lot. Thank you in advance!