Enlighten me about Killing vector (with basic results).?
349352
The Killing vectors of a given spacetime metric come from symmetries of the metric; that is, a Killing vector points along a direction in which the metric does not change.\n\nOne "basic result" is that the scalar product of a Killing vector with the 4-velocity of an object travelling a geodesic is conserved along that geodesic.
665854
yes in c++\nno side effects
987949
The issue is really the purpose behind the killing. Is it killing to get rid of injustice and oppression, or is it meaningless killing or killing based upon blind assumptions? \n\nIf its the first case, then killing is perfectly justified.
35459
You need to apply vectors.
1108682
U•V = |U||V|cosα  Where |U| is the norm of U and α is the angle between the vectors.\n\nIt also boils down to:\nU•V = x1y1+x2y2\nU•V= (-4)(1) + (6)(5)\nU•V = 26
767429
First each object has it's "own" part of memory so it doesn't matter what happens to one of them as long as we ar talking only related to the content of this array. The objects are not "interacting" with each-other so nothing will happen with the "name" of the object... As far as for the index this is another thing. If you only "destroy" the object without "removing" it from the array than the object will be NULL and no index will be changed. If you use a remove method than the other will be automaticaly "reindexed".
397803
/*\n\ni need to know an example of writting a program in c\nlanguage under linux using threads?\n\nwrite aprogram to calculate the result of the equation\nY=3sin(x)+6cos(x)+5x2 so each term accomplished by a\nthread. The main thread will calculate the result.\n\n*/\n\n\n#include <pthread.h>\n#include <math.h>\n#include <stdlib.h>\n#include <stdio.h>\n\n/* static global location -- all threads will read */\nstatic double x;\n\nstatic void*\nthree_sinx(void*p)\n{\n  double* result = (double*)p;\n  *result = 3 * sin(x);\n  return 0;\n}\n\nstatic void*\nsix_cosx(void*p)\n{\n  double* result = (double*)p;\n  *result = 6 * cos(x);\n  return 0;\n}\n\nstatic void*\nfive_x_squared(void*p)\n{\n  double* result = (double*)p;\n  *result = 5 * x * x;\n  return 0;\n}\n\n/*\n *\n * Y = 3sin(x) + 6cos(x) + 5x2\n *\n */\n\nint\nmain(int argc, char* argv[])\n{\n  double result[3]; /* one location for each thread */\n  pthread_t tid[3];\n  void *r;\n  int i;\n\n  /* Set "x" to double value passed in on command line */\n  x = argc>1 ? strtod(argv[1], 0) : 0;\n\n\n  /*\n   *  Since result areas are distinct, no mutex needed\n   */\n  if (pthread_create(&tid[0], NULL,\n    three_sinx, &result[0]) != 0) {\n    perror("pthread_create[0]"), exit(-1);\n  }\n\n  if (pthread_create(&tid[1], NULL,\n    six_cosx, &result[1]) != 0) {\n    perror("pthread_create[1]"), exit(-1);\n  }\n\n  if (pthread_create(&tid[2], NULL,\n    five_x_squared, &result[2]) != 0) {\n    perror("pthread_create[2]"), exit(-1);\n  }\n\n  /* Synchronize on the completing child threads */\n  for (i=0; i<3; i++) {\n    if (pthread_join(tid[i], &r) != 0) {\n      perror("pthread_join");\n      exit(-1);\n    }\n  }\n\n  /* main thread will now combine the result */\n  printf("For y=3sin(x)+6cos(x)+5x2, when x=%f, y=%f&#92;n",\n      x, result[0]+result[1]+result[2]);\n\n  return 0;\n}\n\n\n\nTo show that it works:\n\n$ cc pthr.c -lpthread -lm\n\n$ a.out 3.14159\nFor y=3sin(x)+6cos(x)+5x2, when x=3.141590, y=43.347947\n\n$ a.out 0\nFor y=3sin(x)+6cos(x)+5x2, when x=0.000000, y=6.000000
762129
Good Joke, Programmer must have died of Exhausion ! Ha Ha Ha...This reminds me, an old incidence, while persuing my Post Graduation, to submit my Dessertation thesis I had to work on "Fortran 20" Computer [Now it may be seen in Museums, if any of them ae left !] where we need to feed data in the form of Punch Cards. I submitted my data and was expecting the results within some minutes. To my surprise and also to the surprise of fellas working there, my results wouldn't come out foe more than 2 hours.......After the through labourious scanning and listing it was found that I had forgotten to add the end card to the programme which used to end the Programme ! And my calculations were getting repeated and repeated and repeated like our Programmer in the Joke !!!
1066466
unsigned int delaytimer = 0;\nvoid __stdcall delaytime(HWND, UINT, UINT, ULONG)\n{\n  if (delaytimer) {\n    KillTimer(0, delaytimer);\n    delaytimer=0;\n  }\n  AfxMessageBox("Hi!");\n}\n\nint main()\n{\n  delaytimer = SetTimer(0, 0, 10000 /* milliseconds */, delaytime);\n}
474388
Okay, first, since the vector given for B is a unit vector, we know that the vector from the origin to B must be given by (some constant)*(2,-2,1)/3.  So, what you want to do is find the value of that constant such that |A-B|=10.  Do this by setting A-B=(6-2x/3, -2+2x/3, -4-x/3).  Then just find the length of that vector in terms of x, set it to 10, and solve for x.
1095286
1.www.physics.uoguelph.ca/tutorials/vectors/vectors.html \n2.mathforum.org/~klotz/Vectors/vectors.html
469104
The algebraic sum of the currents entering a node is equal to zero. Where current entering is positive and exiting is negative (or vice versa). In other words, what comes in must come out.\nNote: this is only true for DC signals.
279743
in this case, the cross-product uses the sine of the angle between the two vectors to multiply the product of the magnitudes of the two vectors.   the angle between any vector and itself is zero.\n\nthe sine of zero is zero.\n\nthen, i x i = |i|*|i|*sin(0) = 1*1*0 = 0
796480
use v^2 =  GM/r\n\nv = root GM/2r\n\norbit = 2 pi r\n\nfor part 2 use suvat equations, s=vu+.5at^2 etc and it should fall out
849879
here you go...\n(it'll work with up to 255 different disks, if you need more, you'll have to replace chars with ints)\n\nm1(char *a,b) {memove(b+1,b,strlen(b)+1); *b=*a; memmove(a+1,a, strlen(a);}\nm(char *a,b,c) {if (*a)  {m(a+1,b,c);m1(a,c);m(b,c,a);}
573927
single replacement, double replacement, decomposition, synthesis
1020607
Microsoft doesn't do a good job at killing a zombie process sometimes.  Usually using the task manager to kill it will result in another process to try to kill it but it's not guaranteed that the killer process will succeed.  But eventually it will just keep trying.  If all else fails, reboot.