I can't take over barclay mills I've done all the objectives in Barclay Mills, but it only took the damage scale down to $2550.  I've scoured the map for leftover objectives but no dice.  People tell me to find some enemies in Barclay Mills to kill, but I can't find them.
Mafia 3 no more damage to be dealt barclay mills I did most of the barclay mills missions and brought the damage down to 2550 but now I can find anything else to damage or steal, so the meter is stuck at 2550 and the objective says "find and destroy garbage barges" but when I click it, no waypoint.  Please help.
Sum of two uniform random variables MTMFS consuming an entire CPU core and fans running full bore with Lion You may be aware that Lion introduced MTMFS, the Mobile Time Machine File System.  I've had a couple of instances now where MTMFS has been chewing up an entire core of my dual-core (mid 2009) MacBook Pro for hours on end with no evidence that it's actually doing anything constructive.  As a consequence, CPU temperature rises and the cooling fans run fast and noisy.  Restarting the machine quietened things down again, but I'm expecting it to start again.  Does anyone understand MTMFS well enough to suggest why it's keeping itself busy (or is it most likely, simply a bug), or if there's currently any solution? I fear I may have gone deaf or been scalded to death before Apple release an update... Variable variable class extensions in PHP--is it possible? Is something like the following possible in PHP?  $blah = 'foo1';  class foo2 extends $blah {     //... }  class foo1 {     //... }   This gives an error.  I want to dynamically set $blah so I can extend whatever class I want.  Edit: The reason for wanting to do this because I wanted to use a function out of another class in a related class. In the end it would have been something like:  Final extends foo1 extends foo2 extends foo3 extends foo4 extends parent { ... }   In the end I decided to instantiate the other class within the class and use it. Not the best options because they both you 2 of the same classes, but this won't be used that often, so it will work for now. Android: View.setID(int id) programmatically - how to avoid ID conflicts? How can gravity affect light? What is the difference between linear regression and logistic regression? What is the difference between linear regression and logistic regression?  When would you use each? How to change kernel at boot I installed the latest kernel (4.1) and want to boot into it instead of 3.9. How can i do this on ubuntu 15.04? What is the resolution of tapping a blocking creature? Let's say I have declared a blocker during the correct phase and during the activate abilities step I choose to tap the blocker to use its ability.  All the abilities resolve and now we move to the declare damage step.  I believe the resolution is that the attacking creature remains blocked and that it deals damage to my blocker; however, the blocker does not deal damage to the attacker.  How would this change if the blocker had first strike? How would this change if the blocker was sacrificed? How can I press "Insert" on a keyboard without that key? Get number of frames in scene from the command line? 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For example, looking over a few other questions on MSE:      From : evaluate $$ \int x(x^2+2)^4\,\mathrm{d}x. $$    Via the substitution $u = x^2+2$, this becomes $$ \int x(x^2+2)^4\,\mathrm{d}x       = \frac{1}{10}x^{10} + x^8 + 4x^6 + 8x^4 + 8x^2 + \frac{32}{5} + C.     $$ However, multiplying out the polynomial and integrating using the power rule gives $$ \int x(x^2+2)^4\,\mathrm{d}x       = \frac{1}{10}x^{10} + x^8 + 4x^6 + 8x^4 + 8x^2 + C    $$     From :  evaluate   $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x. $$    Simplifying the integrand using partial fractions then integrating gives $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x       = \frac{-2}{(x+1)} - \ln|x+1| + C.    $$ Via integration by parts, we get $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x       = \frac{x-1}{(x+1)} - \ln|x+1| + C.    $$     From : evaluate   $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x. $$    Using the substitution $u = \sec(\pi x)$, this becomes $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x       = \frac {\sec^2(\pi x)}{4\pi} + C.    $$ Using the substitution $u = \tan(\pi x)$, this becomes $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x       = \frac {\tan^2(\pi x)}{4\pi} + C.    $$ Why do we require radians in calculus? I think this is just something I've grown used to but can't remember any proof.  When differentiating and integrating with trigonometric functions, we require angles to be taken in radians. Why does it work then and only then? I flagged a post and (choose one)   my flag was marked as helpful, yet nothing was done my flag was marked as declined, yet the post was deleted/edited/closed   Why? How do you access the matched groups in a JavaScript regular expression? How can I make a plane's edges taper inwards? How can I deform one edge or multiple edges of a plane or cube so they are tapered in and rounded equally? Can the term "etymology" be applied to a phrase or only individual words? I have always heard the term used in referring to a single word. When browsing questions on this site, I've seen it used applied to entire phrases, and have suppressed the compulsion to edit them and replace the term with origin. Prove there are no other solutions of functional equation $f(x+y) = \frac{f(x)+f(y)}{1-f(x)f(y)}$ I have the following functional equation. Find all continuous functions $f:(-1,1) \to \mathbb R$ such that $$ f(x+y)=\frac{f(x) + f(y)}{1 - f(x)f(y)} $$ The first obvious solution is $f(x) \equiv 0$. Another one I guessed, it is $f(x) = \pm \tan x$. I suspect, that there are no more solutions. The problem is that I don't know how to prove that.  Since we have a rational equation, I have no idea how to make any substitutions in order to get expression for $\tan x$ (as it is not a rational expression).  P.S. I can also show that it is true that (a) $f(0) = 0$ (take $x=y=0$), and (b) $f(-x) = -f(x)$ (take $y=-x$).  Update: there is a solution of this problem also here: Polynomials with integer coefficients such that $p(a)=b,p(b)=c,p(c)=a$ While studying polynomials a question struck in my mind that  To have a polynomial $p(x)$ with integer coefficients and $a,b,c$ are three distinct integers, then is it possible to have $$p(a)=b$$$$p(b)=c$$$$p(c)=a$$ Assistive technology not found AWTError $ java -jar aprof-plot.jar Exception in thread "main" java.awt.AWTError: Assistive Technology not found: org.GNOME.Accessibility.AtkWrapper     at java.awt.Toolkit.loadAssistiveTechnologies(Toolkit.java:807)     at java.awt.Toolkit.getDefaultToolkit(Toolkit.java:886)     at java.awt.Toolkit.getEventQueue(Toolkit.java:1734)     at java.awt.EventQueue.invokeLater(EventQueue.java:1264)     at aprofplot.Main.newWindow(Main.java:33)     at aprofplot.Main.main(Main.java:359)   Possible explanations I saw here was to install Java-access-bridge. But I am unable to install libaccess-java-bridge. I am unable to find the bias of the sample variance estimator in this problem. Unfortunately I keep coming up with   Question: Suppose that the true linear model is   $y = X_1\beta_1 + X_2\beta_2 + e$,   where $e \sim MVN(0,\sigma^2 I)$, $X_1: N \times p_1, \beta_1: p_1 \times 1, X_2: N \times p_2, \beta_2: p_2 \times 1$. But we wrongly fit the model  $y = X_1\beta_1+e$ and calculate the estimators of $\beta_1$ and $\sigma^2$ as:  $\hat{\beta}_1 = (X_1'X_1)^{-1}X_1'y$,  $s^2 = \large \frac{y'[I-X_1(X_1'X_1)^{-1}X_1']y}{N-p_1}$.  (a) Show that, in general, $\hat{\beta}_1$ is a biased estimator with  Bias$(\hat{\beta}_1) = (X_1'X_1)^{-1}X_1'X_2\hat{\beta}_2$  Under what condition on $X_1$ and $X_2$ is $\hat{\beta}_1$ unbiased?  (b) Show that, in general, $s^2$ is a biased estimator with   Bias$(s^2) = \frac{1}{N-p_1}\beta_2'X_2'[I-X_1(X_1'X_1)^{-1}X_1']X_2\beta_2$  Is the bias positive or negative? Under what condition on $\beta_2$ is $s^2$ unbiased?  Attempt:  For part (a):  Bias$(\hat{\beta}_1) = E((X_1'X_1)^{-1}X_1'y - \beta_1)$  $\hspace{15mm}= (X_1'X_1)^{-1}X_1'E(y) - \beta_1$  $\hspace{15mm}= (X_1'X_1)^{-1}X_1'(X_1\beta_1 + X_2\beta_2) - \beta_1$  $\hspace{15mm}= (X_1'X_1)^{-1}X_1'X_1\beta_1 + (X_1'X_1)^{-1}X_1'X_2\beta_2) - \beta_1$  $\hspace{15mm}= I\beta_1 + (X_1'X_1)^{-1}X_1'X_2\beta_2 - \beta_1$  $\hspace{15mm}= \beta_1 + (X_1'X_1)^{-1}X_1'X_2\beta_2 - \beta_1$  $\hspace{15mm}= (X_1'X_1)^{-1}X_1'X_2\beta_2$  If $X_1$ and $X_2$ are orthogonal, then $\hat{\beta}_1$ is unbiased.  For part (b), I am unable to figure out how to derive:  Bias$(s^2) = \frac{1}{N-p_1}\beta_2'X_2'[I-X_1(X_1'X_1)^{-1}X_1']X_2\beta_2$  I can get   $\frac{1}{N-p_1}\beta_1'X_1'+\beta_2'X_2'[I-X_1(X_1'X_1)^{-1}X_1']X_1\beta_1+X_2\beta_2$  after taking the expectation of y on both sides.   Any help would be much appreciated.