x is the time symbol. explain
125381
r is probably 27 and t is 8. Here's my reasoning:\n\n18 * 18 * 18 = r * r * t\n5832/r = r * t\nLets factorize 5832\n5832 = 27 * 27 * 8\nAnd Since 27 > 8, it can be accepted as the solution.\n\nI don't know if this is the correct way of doing it. Anyway, in my opinion worth a shot.
144529
#include<iostream.h>\n#define isleapyear(year) ((!(year % 4) && (year % 100)) || (!(year % 400) && (year % 1000)))\n\nclass Date\n{\nprivate:\nint  month, day, year;\n \npublic:\nDate(int month,int day,int year):month(month),day(day),year(year){};\nint isdatevalid();\nint weekday();\n\n};\n\n//\n//   return 1 if date is valid, 0 otherwise.\n//\nint Date::isdatevalid()\n{\n  if (day <= 0) return 0 ;\n  switch( month )\n{\n  case 1  :\n  case 3  :\n  case 5  :\n  case 7  :\n  case 8  :\n   case 10 :\n  case 12 : if (day > 31) return 0 ; else return 1 ;\n  case 4  :\n  case 6  :\n  case 9  :\n  case 11 : if (day > 30) return 0 ; else return 1 ;\n  case 2  : \n    if ( day > 29 ) return 0 ;\n      if ( day < 29 ) return 1 ;\n      if (isleapyear(year)) return 1 ;   // leap year\n    else return 0 ;\n}\n  return 0 ;\n}\n\n//\n// given month, day, year, returns day of week, eg. Monday = 0 etc.\n// tested for 1901 to 2099 (seems to work from 1800 on too)\n// \nint Date::weekday()\n{\n  int ix, tx, vx;\n \n  switch (month) {\n    case 2  :\ncase 6  : vx = 0; break;\ncase 8  : vx = 4; break;\ncase 10 : vx = 8; break;\ncase 9  :\ncase 12 : vx = 12; break;\ncase 3  :\ncase 11 : vx = 16; break;\ncase 1  :\ncase 5  : vx = 20; break;\ncase 4  :\ncase 7  : vx = 24; break;\n  }\n  if (year > 1900)  // 1900 was not a leap year\n    year -= 1900;\n  ix = ((year - 21) % 28) + vx + (month > 2);  // take care of February \n  tx = (ix + (ix / 4)) % 7 + day;              // take care of leap year\n  return (tx % 7);\n}\n \n \nint main()\n{\n  char week[7][10] = {\n  "Monday","Tuesday","Wednesday","Thursday",\n"Friday","Saturday","Sunday"\n};\n  Date *d;\n  int  month, day, year;\n  cout<<"Return the day of the week given the date."<<endl;\n  cout<<"Enter date in the form mm/dd/yyyy : ";\n  scanf("%d/%d/%d",&month,&day,&year);\n  d = new Date(month,day,year);\n  if (d->isdatevalid())\n  {\n    cout<<"The day of the week for this date is "<<week[d->weekday()]<<endl;\n  }\n  else\n    printf("%d/%d/%d not a valid date!",  month,day,year);\n  \n  getchar();   // wait \n  getchar();   // 2nd wait needed\n  return 0;\n}
449551
Insert this for the Date:\n<h4>It is now  \n<script type="text/javascript">\n<!--\nvar currentTime = new Date()\nvar month = currentTime.getMonth() + 1\nvar day = currentTime.getDate()\nvar year = currentTime.getFullYear()\ndocument.write(month + "/" + day + "/" + year)\n//-->\n</script>\n</h4>\n\nInsert this for Time:\n<h4>It is now  \n<script type="text/javascript">\n<!--\nvar currentTime = new Date()\nvar hours = currentTime.getHours()\nvar minutes = currentTime.getMinutes()\nif (minutes < 10)\nminutes = "0" + minutes\ndocument.write(hours + ":" + minutes + " ")\nif(hours > 11){\ndocument.write("PM")\n} else {\ndocument.write("AM")\n}\n//-->\n</script>\n</h4>
328533
given\nt=# of hours past midday(meaning 12:00pm I assume)\n =15:00-12:00=3\n\nD=3.5\n\nso\n\nthe simplied equation becomes\n1.5*A -.5*A*k=-2\nthere are alot of solutions to this problem...\nfor example \nif k=1, A=-2\nif k=2, A=-4... \nmore information is needed to narrow the solution to only two numbers.\nDoes it tell you the depth at midday(t=0, D=?)
363019
I found X.  It is right in front of the ?\n\nI think that what you believe to be the value of X is also what I think about the imagination of the people who posted the following.  Thanks for adding to my list.\n\nhttp://answers.yahoo.com/question/;_ylt=Apmq_.hERAjJ7oLEmCZLdsEjzKIX?qid=1006040601312\n\nhttp://answers.yahoo.com/question/;_ylt=ApehPmyXrBDiVbvXoaAUoBkjzKIX?qid=1006012405311\n\nhttp://answers.yahoo.com/question/;_ylt=AuYvVkm2FuuEnkXAxhv7H6ojzKIX?qid=1006042414796\n\nhttp://answers.yahoo.com/question/;_ylt=AlTkJo2lkPejzTI0bDnkMqIjzKIX?qid=20060909154312AATEBpz\n\nhttp://answers.yahoo.com/question/;_ylt=ApN9sEMM28MHkA_vH6XK7EsjzKIX?qid=1006013010415\n\nhttp://answers.yahoo.com/question/;_ylt=AvL4qWl22bVT.6SvRkNL20MjzKIX?qid=20060611231741AAT6Umg\n\nhttp://answers.yahoo.com/question/;_ylt=ArIrpvEbLZGe10UGUcyX9q8jzKIX?qid=1006040513133\n\nhttp://answers.yahoo.com/question/;_ylt=Ai_Yp3YVHrIv_sps8mrdsLkjzKIX?qid=20060619124924AA9y7ww\n\nhttp://answers.yahoo.com/question/;_ylt=AqY3p57fbV4YRC3z699c0hUjzKIX?qid=1006031514517\n\nhttp://answers.yahoo.com/question/;_ylt=Antx58lvGy6USSq04b69FeEjzKIX?qid=20060803070419AASqp83\n\nhttp://answers.yahoo.com/question/;_ylt=AlG2R1zJEQCH0g5yYcd5PrQjzKIX?qid=1006020706760\n\nhttp://answers.yahoo.com/question/;_ylt=AjIAz7nYxFNt.XaEnu5tWlsjzKIX?qid=1005122402394\n\nhttp://answers.yahoo.com/question/;_ylt=AlfQ2T5Xx9CHgdTXkSAn1ZQjzKIX?qid=1006032008690\n\nhttp://answers.yahoo.com/question/;_ylt=AuGVJPrZ6F29eC5Lnw9x3KIjzKIX?qid=1006052329295\n\nhttp://answers.yahoo.com/question/;_ylt=AjINH6a1jmravo6q54cc2awjzKIX?qid=20060710114620AA81Ct6\n\nhttp://answers.yahoo.com/question/;_ylt=AnYtW.lJ5J9QXOzeKl45kVgjzKIX?qid=1006043024244\n\nhttp://answers.yahoo.com/question/;_ylt=Am8k.X_YPy6SVRVlDrJW6hgjzKIX?qid=1006021106437\n\nhttp://answers.yahoo.com/question/;_ylt=AsxKyvTJCJ2Mri0WyG78AhsjzKIX?qid=1005123102716\n\nhttp://answers.yahoo.com/question/;_ylt=As_4eOB6Pa1Swe1QKRzbZfEjzKIX?qid=1006040309767\n\nhttp://answers.yahoo.com/question/;_ylt=AqNb7svYM7Da9CiRp9XqlocjzKIX?qid=1006031501175\n\nhttp://answers.yahoo.com/question/index;_ylt=AoPVIBF8CgzpqJddQ2pukzEjzKIX?qid=20061016171444AARISR0\n\nhttp://answers.yahoo.com/question/index;_ylt=AsDfFnkfbBEhvwdO2UhNiksjzKIX?qid=20061020201152AANV3K4\n\nhttp://answers.yahoo.com/question/index;_ylt=AvUMyPB.Mvhb6iz0xUxTT3IjzKIX?qid=20061019071741AA6ATaj\n\nhttp://answers.yahoo.com/question/index;_ylt=AoR_pWjmV7AR17DYIJubGJgjzKIX?qid=20061029115414AAgQRAd\n\nhttp://answers.yahoo.com/question/index;_ylt=Ap_fN_LiwUl8IlZzaZXa_8QjzKIX?qid=20061101174924AAsjnZf\n\nhttp://answers.yahoo.com/question/index;_ylt=AjtBGiC9otkPClr4WKAGjhMjzKIX?qid=20061030085926AA47rRt\n\nhttp://answers.yahoo.com/question/index;_ylt=AtDzRP73zG4NaEJcxpePkVsjzKIX?qid=20061031225359AAesSen\n\nhttp://answers.yahoo.com/question/index;_ylt=AkTtldS_TFLncSM9gkKrIIwjzKIX?qid=20061103122304AAyFLKJ\n\nhttp://answers.yahoo.com/question/index;_ylt=AhiSBmKy.HqSbNOspH.w9kYjzKIX?qid=20061107013950AAKInc4\n\nhttp://answers.yahoo.com/question/index;_ylt=AuGYkAWzTWZN_N3lws_fXugjzKIX?qid=20061106133704AAp1Q4V
129107
x can be any value.\n\ntan x = sin x / cos x\n\ncot x = cos x / sin x\n\nso tan x * cot x = sin x/cos x * cos x/sin x\n\nsin x's cross out each other\n\ncos x's cross out each other\n\nso you're left with 1.
158132
emmmmmm\n\nd. its not a fuction of x in terms of y explicitly. It needs to be rearranged
474273
50:4000=\n1:80 - -answer
523330
36\n\nFirst time= x-(x/2+2) = x/2-2\nSecond time=x/2-2-(1/2(x/2-2)+2)=x/4-3\nThird time=x/4-3-(1/2(x/4-3)+2)=x/8-3.5\nx/8-3.5=1
483499
The 12-hour clock is a timekeeping convention in which the 24 hours of the day are divided into two periods called ante meridiem (a.m., from Latin, literally "before midday") and post meridiem (p.m., "after midday").\n\nTaken from wikipedia at:\n\nhttp://en.wikipedia.org/wiki/12-hour_clock
514903
Hello,\n\nThe symbols in relationship to the numbers change each time. \nThe symbol you choose is the correct one  for all answers. No matter the number you choose there is one correct symbol for all answers.
498452
a.x=[-8+/-rt(64-24)]/2\n\nb.x=[-5+/-rt(25-56)]/4\n\nc.x=[-7+/-rt(49-64)]/4\n\nd.x=[-5+/-rt(25-72)]/4
324089
Let x = 0.999...\n\n10x = 9.999....\n\n10x - x = 9.999... - 0.999...\n\n9x = 9\n\nx = 1.\n\nThis is the 3rd time in two weeks I've encountered this question here. So next time this is posted, do me the favor of answering it.
35697
int main()\n{\nint minutes,hours;\nprintf("Enter the number of minutes /n");\nscanf("%d",&minutes);\n\nhours=minutes%60;\nminutes=minutes-(60*hours);\n\nprintf("Time is %d hours and %d minutes",hours,minutes);\n}
260676
Santosh is correct, both work...\n\nI want to know how you use the ∫ symbol!! I just copied & pasted from yours..what's the keystroke for it??
470859
I don't see any x  in the equations but I think the problem is that xmax< = xmin or Y max<=Y min  as a result the calulator could not calulate it that way try making Xmin < Xmax and Ymin < Ymax  this should solve the problem.
338771
S = southbound speed\nN = northbound speed\nd = distance\nx = time from 8am\nS(x) + N(x-1) = d\n\nS*2 + N*1 = 500  => N = 500 - 2S\nS*3 + N*2 = 850 = 3S + (1000 -4S) = 1000 - S\nS = 150, N = 200