Computer Basic Answers Please?
117726
Jimmyr's answer is partially correct! He is right that the image or file was not retrieved and the Red X means "not available".\nHowever,very often the file is "not Available" because it was blocked by a firewall, security program (such as Norton, McAfee etc.) a SpyBot program, Ad-Aware etc.\nIf I know that it is an image or file from a friend or trusted source that I definitely want I can try one of two things:\n1. I will turn off my security and firewalls temporarily (this is a pain and can be dangerous) and then retry. More often than not, however, I will use the next option.\n2. You will note that when you see the Red X on the screen you will also see the complete URL in the address box above. I copy that URL into my download manager and download the file, image etc. from that program. Note: I only use this option when I really want the file and I know it's safe!\nIf you don't have a powerful download manager get one for free at:\nhttp://www.freedownloadmanager.org/support.htm\nNote the above URL extension (.org), this is a non-profit site, not a commercial one. Highly reliable and professional.\nLast, of course, the file may have been truly removed from the server and really might not be available. If this is the case "FreeDownload will tell you.
336716
Try this tutorial below...
310681
Basic Input Output System.
522800
The value stopped changing at 3.141384 but\nI killed it at cnt= 148337\nOf course, to run it I converted it to GW Basic \nsince I don't have the version of basic this is for.  here it is with the minor changes.\n5 CLS\n6 KEY OFF\n10 A=0\n20 CNT=0\n30 C=1\n40 A=A+(1/C-1/(2+C))\n50 C=C+4\n60 CNT=CNT+1\n70 LOCATE 2,1\n75 PRINT 4*A,"cnt="CNT\n80 IF INKEY$="" GOTO 40
165476
How about (a+c)/(b+d)  ? more interesting!!\n\nbtw this only work when a/b != c/d
383056
what kind of program are you doing?? all i know is visual basic stuff
1104745
Since we're in the computer section, when we say BASIC grammar, i think we are talking about the language syntax for the programming language, "basic".  I think there is just a little communication.\n\nHowever there is a check spelling button we should all become a little more familiar with.
658966
I searched online and it looks like a program called VCDGear could do it. Check out the source link below and hopefully this will solve it for you.
598093
There really isn't anything that I know of that u can do in java, but try not having the scripts embedded. It sounds like the computer can't handle so much info.
418804
I'm a bit rusty with BASIC but both of them look ok (assuming the syntax is good).\n\nThe second one is more efficient because of the FOR loop.\n\nIf you always need to calculate the average for 6 students then you don't need to prompt for the number of students, although it does make the program more useful.
162512
http://www.intelligentedu.com/newly_researched_free_training/Visual_Basic.html
66876
Here's a pretty good one I wrote:\n\nhttp://www.planetsourcecode.com/vb/scripts/ShowCode.asp?txtCodeId=58476&lngWId=1\n\nYou'll need Visual Basic to compile it or you can e-mail me and I'll send it to you...
816046
Here you have the answer:\nhttp://askcomputerexpert.ws43.com/answers/
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
15588
This may sound like a foolish question, but: are you sure you're entering a program executable file?  Just be sure you're using the right filename.\n\nAlso, depending on the program, you may need to type the entire file path to it.
232131
#include<iostream.h>\nint main() \n{ \n   int stop=false;    \nchar pl='X';    \nchar p; //is used for checking when a game is won  \n  \nint v; //value    \nchar list[]="123456789&#92;0"; //create our array that holds all the values \n\n   do { //start our while loop        //INPUT \n       cin >> v;        //MARKING \n       if (list[v-1]!='X' and list[v-1]!='O')\n {       \n    list[v-1]=pl;       \n }        \nelse\n {  \n     cout << "That place is already taken.&#92;n";   \n      goto skip; //if its already taken skip the turn switch             \n}  \n    if (pl=='X')\n {  \n          pl='O';  \n   }\n   else\n { \n \n          pl='X';  \n      }        skip:;        //win checks   \n     for (int w=0; w<2; w++)\n{    \n        if (w==0)\n{       \n        p='X';   \n         } \n           else{ \n                p='O'; \n           }\n     for (int z=0; z<9; z=z+3)\n {\n     if (list[z]==p and list[z+1]==p and list[z+2]==p) {                 \n  stop=true; \n        goto GAME_END;    \n            } \n           }\n            for (int z=0; z<3; z++) { \n        if (list[z]==p and list[z+3]==p and list[z+6]==p) {\n                   stop=true;            \n       goto GAME_END;        \n        }  \n          }        \n    if (list[0]==p and list[4]==p and list[8]==p or list[2]==p and list[4]==p and list[6]==p)\n{     \n          goto GAME_END;   \n         }   \n     } \n       //PRINTING OF THE VALUES       \n for (int x=0; x<9; x++) {  \n           cout << list[x];   \n          if (x==2 or x==5 or x==8) {  \n              cout << "&#92;n";    \n         }    \n    }   \n     //END PRINTING OF THE VALUES \n       //system("PAUSE");   \n }\n while (stop==false); \n//End the while loop    \nGAME_END:;      \n  cout << p << " wins!&#92;n";      \n  system("PAUSE");\n}
226603
http://www.learnthat.com/computers/learn.asp?id=425&index=0