I'm in India where AC adapter of Y510p(Original) is not available. Ac adapter of Y510p has rating of 19.5v and 6.15a, I found  AC adapter having same rating of Hp, is it secure to use on lenovo laptop?
I have seen a "Universal Laptop Adapter" sold in a computer shop before. I am thinking of getting one for a laptop. Is it safe to use? What differences are there between one of these and a typical laptop adapter?
WARNING DO NOT ATTEMPT TO RUN THIS ON A PRODUCTION MACHINE   In reading the  I generally follow what's going on with the following code:  :(){ :|:&amp; };:   excerpt of description     The following fork bomb was presented as art in 2002;   its exact origin is unknown, but it existed on Usenet prior to 2002.   The bomb is executed by pasting the following 13 characters into a    shell such as  or . It operates by defining   a function called ':', which calls itself twice, once in the   foreground and once in the background.   However the last bit isn't entirely clear to me. I see the function definition:  :(){ ... }   But what else is going on?  Also do other shells such as ksh, csh, and tcsh also suffer the same fate of being able to construct something similar? Showing diffeomorphism between $S^1 \subset \mathbb{R}^2$ and $\mathbb{RP}^1$ I am trying to construct a diffeomorphism between $S^1 = \{x^2 + y^2 = 1; x,y \in \mathbb{R}\}$ with subspace topology and $\mathbb{R P}^1 = \{[x,y]: x,y \in \mathbb{R}; x \vee y \not = 0  \}$ with quotient topology and I am a little stuck.  I have shown that both are smooth manifolds, and I used stereographic projection for $S^1$, but now I am runing into trouble when I give the homeomorphism between $S^1$ and $\mathbb{RP}^1$ as the map that takes a line in $\mathbb{RP}^1$ to the point in $S^1$ that you get when letting the parallel line go through the respective pole used in the stereographic projection.   If I use the south and north poles I get a potential homeomorphism, but I cannot capture the horizontal line in my image, but when I pick say north and east then my map is not well defined as I get different lines for the same point in $S^1$. Can somebody give me a hint how to make this construction work, or is it better to move to a different representation of $S^1$ ? Can't Use /Summon Command Monitor cell value and copy it to another cell Case expressions in Access I am going to change my machine: from a laptop with Ubuntu to another laptop where I am going to install Ubuntu...  I am quite happy with the current system, what I am afraid of is having to rebuild everything from 0.  So what do people often do to clone an Ubuntu system? That may include the data, packages, shortcut settings, etc... Does equality of characteristic polynomials guarantee equivalence of matrices? I have a qualifying exam coming up in a couple days and I am just trying to understand some pathological examples I have in my notes. I will list a similar problem which I know the solution to and then the question.  True or False:   let $A,B \in \mathbb{Q}^{n \times n }$  Suppose $xI-A$ and $xI-B$ are equivalent then $\det(xI-A) = \det(xI-B)$.     I wrote in my notes this is true because any two matrices are equivalent in $\mathbb{Q[x]}^{n \times n}$ if and only if they have the same invariant factors.  Since the characteristic polynomial is a product of the invariant factors it follows that $\det(xI-A) = \det(xI-B)$.        True or False:   let $A,B \in \mathbb{Q}^{n \times n }$. If $\det(xI-A) = \det(xI-B)$ (in $\mathbb{Q}[x]$) then $xI-A$ and $xI-B$ are equivalent  in $\mathbb{Q}[x]^{n \times n}$   I think this is false because of the fact that equality of determinants is not strong enough to guarantee all invariant factors are equal but I do not have an example. Find the angle ADE of the given triangle. Given :  Angle ABE = 50, Angle DAC = 20, Angle BAD = 60, Angle ACB = 20 and Side CD = Side DA.Unit of angle values given is degree.  Find : Angle ADE.    I'm stuck and I need help here. Childrens fantasy, blind monk leading boy who can control the elements What is Snell's law's main formula? My book states it as $(\sin i)/(\sin r)=n$. However, University Physics has a different say.  It says, $n_{a} \sin a =n_{b} \sin b$. Which of the following is correct? The prepositions “with”, “at” and “for” are also used to associate a business title with a company's name. It seems that they are interchangeable, with no (significant) difference in meaning. The following examples taken from The New York Times website seem to confirm that:      Until now, said Justin Nielson, a senior analyst for the media   research firm SNL Kagan, “there really hasn't been any direct   competitor . . .      “The consumer is still deleveraging,” said Jason Goldberg, a senior   analyst with Barclays. “It's not a lack of supply; it's a lack of   demand.”      “There’s the party of fear and the party of despair,” Nikos Xydakis, a   political analyst and an editor at Kathimerini daily, said.   Are they really the same as for meaning? If it isn't so, what is the difference between them? If "I'm thinking" is correct, why isn't "I'm knowing" correct in formal English?  think+ing = thinking is correct know+ing  = knowing isn't correct   Why? Find a tangent to two quadratic bezier curves I have two quadratic bezier curves and I want to find whether they have a common tangent, and if so where it is.  If the first bezier is defined by points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, then the points on the curve have coordinates:  $x = (x_1 - 2x_2 + x_3)t^2 + (2x_2 - 2x_1)t + x_1 \\ y = (y_1 - 2y_2 + y_3)t^2 + (2y_2 - 2y_1)t + y_1$  Or to simplify things:  $x = A_xt^2 + B_xt + C_x \\ y = A_yt^2 + B_yt + C_y$  Likewise, I can get an equation for the second bezier:  $x = D_xs^2 + E_xs + F_x \\ y = D_ys^2 + E_ys + F_y$  The gradient of the tangent, $m$, to the first curve at $t$ is $\dfrac{2A_yt + B_y}{2A_xt + B_x}$,  which must be equal to the tangent to the second curve at $s$, $\dfrac{2D_ys + E_y}{2D_xs + E_x}$.  So $m = \dfrac{2A_yt + B_y}{2A_xt + B_x} = \dfrac{2D_ys + E_y}{2D_xs + E_x}$  Which I rearranged to get:  $t = \dfrac{2(B_yD_x - D_yB_x)s + B_yE_x - B_xE_y}{4(A_xD_y - A_yD_x)s + 2(A_xE_y - A_yE_x)}$  The gradient of the tangent is also equal to the difference in the y-coordinates over the difference in x-coordinates:  $m = \dfrac{D_ys^2 + E_ys + F_y - (A_yt^2 + B_yt + C_y)}{D_xs^2 + E_xs + F_x - (A_xt^2 + B_xt + C_x)}$  But that's as far as I got. The algebra looks too messy from here. Can anyone solve this, or is there a better approach? How to set default style for arrow tips in TikZ? Cardinal of quotient rings of gaussian integers. It is known that $\mathbb{Z}[i]$ is a PID and that $\mathbb{Z}[i]/(a+bi)\mathbb{Z}[i]$ is finite for all $(a,b) \in \mathbb{Z}^2\backslash \{(0,0)\}$.  My question :      Is there any result on the cardinal of  $\mathbb{Z}[i]/(a+bi)\mathbb{Z}[i]$ ? Every finitely generated algebra over a field is a Jacobson ring Knowing that the polynomial ring in $n$ variables over a field $k$ is a Jacobson ring, how can we prove from it that every finitely generated $k$-algebra is a Jacobson ring?   EDIT: We define a ring to be Jacobson if every prime ideal is an intersection of maximal ideals. Proving $a^2$ is even $\implies$ $a$ is even by contradiction If I have an adjacency matrix for a graph, can I do a series of matrix operations on the adjacency matrix to find the connected components of the graph? I am trying to list all the files from dir1, dir2, dir3 and dir4 which might be anywhere in as a sub directory of my cwd using the find command. I tried the following with no success:  find . -type f -regextype posix-egrep -regex 'dir1/.+|dir2/.+|dir3/.+|dir4/.+'   I tried posix-extended as well. How can I list these files? If we remove the square in the upper right corner of a $8\times 8$ chessboard.  The question is: Is it possible to cover the entire remaining area, with $1\times 3$ chocolate bars? (they can be laid on the chessboard vertically and horizontally aswell)  For the answer: We know that there will be $63$ squares left on the chessboard, of which $32$ are white and $31$ are black, since the one in the upper right corner was black. $63$ is divisible by $3$, so it could be possible to cover the area, but that doesn't prove that an arrangement exists to do it. I'm not sure how to continue and am seeking help. Thanks.