What is the equivalent of cin.ignore() in Java?
How do I compare strings in Java?
Credit score just tanked ~50 points and I'm not sure why I check my credit score monthly using a credit tracking app provided by my credit card company. I use a reminder to check on the same day each month, and have been tracking my credit score for almost two years.  In the last year, the app switched from a TransUnion model to VantageScore 3.0. I have TransUnion data for the entire duration, but only about a year's worth of VantageScore data.  I have good credit, and my TransUnion score has increased steadily from ~740 to ~780 in the last two years.  My VantageScore started around ~760 and increased to ~800 last month. This morning when I checked my credit score, it had dropped just over 50 points to ~750.  My TransUnion score has stayed solid at ~780.  My questions are:  Should I be concerned about this sudden change in score?  Is there any way I can determine what changed to account for the sudden change in credit score? Assume $x \in \mathbb{R}$ and $(x + \frac{1}{x}) \in \mathbb{Z}$.   Use Induction to prove that for all $n \in \mathbb{N},~ (x^n + \frac{1}{x^n}) \in \mathbb{Z}$.  I'm not sure how to use the information given to prove the induction step. What I have so far is.  $P(n) : (x^n + \frac{1}{x^n}) \in \mathbb{Z}$  Base Case Then $(x^0 + \frac{1}{x^0}) = 2 \in \mathbb{Z}$  Induction Hypothesis: $x^k + \frac{1}{x^k}\in \mathbb{Z}$  Want to prove: $x^{k+1} + \frac{1}{x^{k+1}} \in \mathbb{Z}$  My first step was  $x^{k+1} + \frac{1}{x^{k+1}} = x \cdot x^k + \frac{1}{x} \cdot \frac{1}{x^k}$ Troubleshoot High CPU usage by the "System" process If a blocking creature leaves the battlefield, is the attacker still blocked, or will it damage the opponent? Suppose Player 1 has a Dragon on the table and  in hand, while Player 2 has two Spiders on the table.  Player 1 chooses to attack with the Dragon and Player 2 chooses to block with one of the Spiders.  Before combat damage is to be dealt, Player 1 uses Unsummon on the blocking Spider.    Does Player 2 take any life damage from the attacking Dragon? Sum of the series $\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\dots$ I am recently struck upon this question that asks to find the sum until infinite terms $$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+.....∞$$ I tried my best to get something telescoping or something useful, but I failed. I even made a recurrence as $t_n=t_{n-1}\frac{2n-1}{2n+2}$, but this question was expected to be done with simple logic of series (Also that the recurrence on solving gives a higher order charasteristic polynomial, which might be difficult to solve without calculator). So, thus I ended up being confused with this question. So, can anyone prode a small solution to this (might be easy) problem  . I am using k-means clustering algorithm to cluster a given dataset into two. I took two different initial points and run k-means clustering algorithm for each initial points. With the first initial points I got “within cluster sum of squared distances” (error) =23874.341. With the second initial points I got “within cluster sum of squared distances” (error) = 42913.684. 23874.341 is less than 42913.684. So, can we say the first one is the better clustering quality? If yes, is it always hold true? Or can we say less within cluster sum of squared distances is always the better clustering quality? Prove the equality: $\det\left[\begin{smallmatrix} -2a &a+b &a+c \\ b+a&-2b &b+c \\ c+a&c+b &-2c \end{smallmatrix}\right] = 4(a+b)(b+c)(c+a)$ i have to prove:   $$\begin{vmatrix} -2a &amp;a+b  &amp;a+c \\   b+a&amp;-2b  &amp;b+c \\   c+a&amp;c+b  &amp;-2c  \end{vmatrix} = 4(a+b)(b+c)(c+a)$$   I have tried many calculations between the rows and columns of the determinant to get to the answer i want to solve the exercise however none of them gave me something right. Can someone help? Restrict certain mods to certain worlds Render scene with background jpg in python Fairly new to python and blender, so apologies if this is super basic. I've searched around and haven't found an answer.  I'd like to render some procedurally generated objects with a background image. In case it makes a difference, I'm using:    I don't need any interaction between the generated object and the background image so I've left a skeleton script I'm using to render below  import os DIR ="/home/path/to/folder" bg_file = os.path.join(DIR, "myimage.jpg") bg_img = bpy.data.images.load(bg_file) bpy.data.scenes['Scene'].render.filepath = os.path.join(DIR, "image") bpy.context.scene.camera = bpy.data.objects['Camera'] bpy.ops.render.render(write_still=True)   I managed to find some scripts that seem to be trying to do what I want online, but none that I could get to work. e.g.  import bpy filepath = "/home/path/to/folder" img = bpy.data.images.load(filepath)  for area in bpy.context.screen.areas:     if area.type == 'VIEW_3D':         space_data = area.spaces.active         bg = space_data.background_images.new()         bg.image = img         break  cam = bpy.data.cameras.new("Camera") cam_ob = bpy.data.objects.new("Camera", cam) bpy.context.scene.camera = cam_ob bpy.context.scene.objects.link(cam_ob) bpy.context.scene.objects.active = bpy.context.scene.objects["Camera"]   Any thoughts/tips would be greatly appreciated! Can I use the "Scan" button with SANE? I have an HP Deskjet with integrated scanner. It has a "Scan" button on its top, but this button does not interact with the xsane program on my Debian machine: To get a page scanned, I have to initiate the scanning process from xsane.  Is there a setting (or a scan program) that allows me to press the scan button and cause the program to scan a page? How can I view and examine the Android log? Working in Cycles.  Say I have a "disco ball" with a glossy shader. I'm pointing a spot light to it, and I'm trying to have the reflections of that light illuminate the diffuse material on the walls with the patterns from the ball.    I get this:    but I'd like to get something like:    For the illustration I placed a point light inside the ball an removed faces. But what I'd like to do is move the light around the ball so that the light reflected from it changes like it would do in real life.   Is that possible at this point?  EDIT:  Just for the record here's the image rendered in LuxRender (yet another program to learn...) How to forbid a specific package to be installed? Getting an error in W3C Markup Validation when trying to get a whole  element as a link in a page solve $T(n)=T(n-1)+T(\frac{n}{2})+n$ Using the recursion tree i tried solving this: $T(n)=T(n-1)+T(\frac{n}{2})+n$; the tree has two parts (branches) one that of $T(n-1)$ and other branch is of $T(\frac{n}{2})$. But as the term T(n-1) reach higher depth, I solved that part of the tree. Finally it resulted in O$(2^n)$ is it correct or is it $O(n^2)$? How can we solve it by substitution method? How to toggle background image on button click? Is phonetic typing for Tamil available in Ubuntu 12.04?  How can I install it? Too big addon list, what should I do? How expand names and keep it all? Note: this post is different to , which just cites part from somewhere else, without any further analysis and thinking. Furthermore, none of the answers to that post answers my question - the situations where &quot;will (going to) be doing&quot; is more appropriate.   I will be participating in IELTS test next Tuesday afternoon.  In an online class, the teacher says the following at the beginning of the lesson.  I will be using NotePod (pointed out by red rectangle) to write relevant points we talk about today.   I guess &quot;will be using&quot; is more appropriate than &quot;will use&quot; there, right?  A  explains  This &quot;tense&quot; is used for reporting a scheduled activity with special focus on its timing— as if one is standing in the future and experiencing the flow of the time. The focus is on the temporary, ongoing, or repetitive nature of the activity.   I understand the grammar rules of those verb tenses and the difference between this group  I am going to drive home tomorrow.   I am going to be driving home tomorrow.  The former is just convey my activity in the near future while the latter probably makes an implication that &quot;don't disturb me as I am driving in that time&quot; However, in the following situations, I don't see the difference between 2 verb tenses.  Welcome back to this class. Today's lesson is for the IELTS students who are going to be taking the IELTS test. it's for everybody to be honest with you. this lesson is useful for anybody who needs to do some writing.   who are going to take the IELTS test ... Evaluating $\int_{-\infty}^{\infty}\frac{1}{(x^2+b^2)^2}dx$