what does the "T" mean in launches..((."T-30 minutes" for eg.)?
337986
T = time of launch...which means that its Time of launch: in 30 minutes
333065
Start -> Run: shut down -s -t followed by the period (in seconds)\n;)
864099
Date\n\n(Company Name) has decided that (Vendor Company Name)'s service is no longer required. Please discontinue your services immediately. \n\nIn the event that your firm's services are required in the future, we will contact you.\n\nBest Regards,\n\nxxxxxxxxxxxx
45472
"Seven minutes past" is short for "Seven minutes past the hour" or seven minutes after the last "o'clock" e.g. it's 10:30, someone asks, "At what time did we come in?"  you say, "About seven minutes past."  means you came in about 10:07.
979156
It means that the startup built in test detected a problem. Could be memory.\n\n**********************************************************\n\nDid you get this solved??
232596
as far as i think its just that after using shift key, to put ! (exclamation marks) and when done, they release the shift first.. so the 1 remains ...\n\ni dont think there is a meaning to it\n\n!!!!!!!!!!!!!!!!!!1
1005235
It means the person has not been on the computer.\nFor instance if you step away from your comp, your screensaver goes on.\nWell, your idle sign goes up if you do that on AIM
190943
Trial period 30days. usually in the FINE PRINT
29302
The system clock is set thru BIOS.  Therefore, you'll have to run an external command to set the clock.  \n\nString cmd="time 11:30:00";\nProcess proc = Runtime.getRuntime().exec(cmd);\n\nYou will have to make sure the 'command' is correct; my example is just a guess.
424294
Depends on your internet connection.\n\nAnywhere from 45 mins on a 56k modem to seconds on a T1 connection.
418748
reaction and operation time of a task < allowed delay\n\nnot only how fast can it be done, but can it be done as specific times and can these time be accurately reported.\n\n\nis buffering or queuing of data possible.\nmulti-tasking/priority (can not do everything at same time)
723034
It means:\n"You have 3 days to accept this mission, or this message will self-destruct!"
1177657
It's an abbreviation of an abstraction of an abbreviation\n\n'til later.\nTilly Tah.\nT' Tah. or Ta Ta
367619
it might be at www.launch.com
300905
The "T" in the company's name stands for "Telekom." All subsidiaries of Deutsche Telekom have names beginning with "T-".\n\nT-Online (DSL in Germany)\nT-Mobile (Mobile Phone)\nT-Systems (Database etc)\nT-Com (Landline Phone)\nT-Internet (Dial-Up Internet)
72876
He means does the angle of launch (theta) affect the range (R), when you launch an object at some constant speed (u)?\nThis is a basic question in applied mathematics (mechanics).\n\nYes it does.\nWrite the equations of motion for the object.\n(Neglect wind resistance, which is secondary.)\nAssuming the launch surface is flat and you did not launch at an elevation, thus y(0)=0,then the components of velocity in x-,y-directions are:\n u_x = u*cos(theta)\n u_y = u*sin(theta)\n\nNoting the acceleration: a_x(t) = 0, a_y(t) = -g (always downwards), then the velocity components are:\n v_x(t) = u*cos(theta)\n v_y(t) = u*sin(theta) - g*t\n\nHence distance traveled is:\n s_x(t) = u*t*cos(theta)\n s_y(t) = u*t*sin(theta) - 1/2 g*t^2\n\nThe object lands at some time t=t' when s_y(t') = 0 again (not the t=0 solution, obviously). Thus:\n 2*u*t'*sin(theta) = g*t'^2\n 2*u*sin(theta)/g  = t'\n...\nand the range R is the s_x traveled at this time t=t' : \n R = s_x(t')\n   = u*t'*cos(theta)\n   = 2*u^2*sin(theta)*cos(theta)/g\n\nNow u,g are constant here, hence range R=R(theta) is purely a function of theta:\n R = (u^2/g) * 2*sin(theta)*cos(theta)\n R = (u^2/g) * sin(2*theta)          ; [a trigonometric equality that sin(2*theta) = 2*sin(theta)*cos(theta)]\n\nIn this case you can show (without even needing differentiation) that R achieves a maximum when:\n sin(2*theta) = 1\n i.e. theta = 45 deg\n which is what you might have suspected.\n\n============================================================\n\n[For the general solution of launching at an elevation, y(0) = y_0 , then you need to rewrite s_y(t), solve as above, using differentiation to show R(theta) achieves a max when R'(theta)=0 for some theta which does not have a neat solution].\n\n[If you want to be a smartass, add the second-order air resistance terms into s_x(t), s_y(t), so you get acceleration due to a viscous force:\n a_x(t) =      -nu * v_x(t)^2\n a_y(t) = -g + -nu * sgn(v_y)*v_y(t)^2\n   [Note we have to use sgn(y_t) to ensure the viscous forces acts against direction of motion. Basically: down when the projectile is rising, and up when it's falling.\nBut you can usually neglect the air resistance component of a_y, since |v_x|>>|v_y| for most if not all of the flight, unless u is small and launch angle theta is large.]\n\nIf you neglect the air resistance component of a_y(t), then you can still get an analytic solution, otherwise I think now you get two difference equations and you have to solve approximately e.g. by Runge-Kutta approximation.\n\nIf you want to be an utter smartass, add in the effect of wind with a vector (w_x,w_y), such that the air resistance force depends on the relative velocity of projectile to wind:\n( v_x(t)-w_x , v_y(t)-w_y ) \n\nAlso, we neglected terrain (mountains etc.), these degenerate into piecewise tests to make sure you always maintain positive altitude s_y(t).\n\nIf you want to get silly, do the ballistic projectile case, where g decreases as a function of altitude, as does nu, and over long distances you have to factor in the earth's curvature, and probably you want to write it in polar, rather than Cartesian, coordinates.\n g(s_y) = g_0  - k_g(s_y(t))   (linearized approximation)\n nu     = nu_0 * exp (-k_nu* (s_y(t)+Earth's radius)) (atmospheric density)\n [but you can linearize this too, for the lower atmosphere anyway: http://www.aerospaceweb.org/design/scripts/atmosphere/ ]\n\nAnd remember this was just for a projectile, where mass=constant. For a rocket, mass is decreasing (non-linearly) over time, and the dm/dt change in mass is proportional to the acceleration force. The direction of the acceleration is of course along the current velocity vector.\nYou may have design limitations imposed on the maximum dm/dt (e.g. the fuel pumps have a maximum throughput). And you could have two- or three-stage rockets.\n\nIsn't it fun how complicated you can make simple things?\n\nHere's a
753296
|t - 78.3| ≤ 38.3\n\nSo either t -78.3 ≤ 38.3 whence t ≤ 116.6\nOR - (t - 78.3) ≤ 38.3\nwhence - t ≤ - 40 ie 40 ≤ t\n\nSo 40 ≤ t ≤ 116.6