How long would it take by car? \n(I'm not from the U.S.)
132556
Charleston to DC is 536 miles - 8.5 hours.\n\nColumbia to DC is 483 miles - 7.5 hours\n\nGreenville to DC is 496 miles - 8.0 hours\n\nFlorence to DC is 407 miles - 6.5 hours
629557
depends with the country
113789
Here is a solution.  Let x be the distance the motorist traveled at 20km/h on city road, and let y be the distance the motorist traveled at 50km/h on the highway.  We are looking for y.\n\nFirst, we know that the total distance is 155km\n\nx+y=155     (eq1)\n\nSecond we know that if you travel x km at 20km/h, it will take you x/20 hours.  Since the total hours is 4, then:\n\nx/20 + y/50 = 4      (eq2)\n\nThat is two equations, two unknowns, so we can solve the problem.  We'll rearrange eq1 to be:\n\ny=155-x     (eq3)\n\nThen substituting this into eq4 yields\n\nx/20 + (155-x)/50 = 4 \n\nor\n\n5x/100 + 2(155-x)/100 = 4\n5x+2(155)-2x=400\n3x=90\n\nx=30\nThat gives us x, and we can solve for y from (eq 3)\n\ny=155-30\n\ny=125\n\nAnswer: 125 km
422530
hi, it is 3:30pm (3 mins since your post- it tells me)and I'm in the south west of Western Australia enjoying a lazy week-end after a hard week of work...hope you enjoyed the carshow.
479020
as long as it takes for you to make certain that there are no other cars coming your way.\nusually about 3-5 seconds. \n10 seconds is way too much, is my opinion\nthe cars behind you will start honking...
673313
Tim is right.  i just want to explain it further so you will understand.\n\nso the problem is we are looking for the time spent in both trips. lets call it "X". one is 1 hour longer than the other so we can call it "X + 1". \n\nif trip1 with a distance of 180 miles is taken in X with an additional 1 hour we can call this equation as 180 per X+1 or in mathematical form its 180/(X+1)\n\nif trip 2 with a distance of 135 miles is taken in X we can call it 135 per X or 135/X\n\nwe have now 2 equations 180/X+1 and 135/X\n\nsince the rates are the same we can say that the rate for 180/(X+1) is equal to rate for 135/X.  or in equational form it is\n180/(X+1)=135/X.\n\nWe now have a working equation.  If you notice, we are dealing with fractions and to equate it we need to transpose it using cross multiplication.  Cross multiplication is done by pairing the numerator of one side to the denominator of the other side and then proceed to multiplication.\n\nlooking at the working equation, we will pair 180 and X or 180X and another pair which is 135 and (X+1) or 135(X+1)\n\nnow we have a new working equation 135(X+1) = 180X\n\nto simplify 135(X+1) we multiply 135 with X and with 1. 135 and X is equal to 135X and 135 and 1 is equal to 135\n\nnow we have simplified the equation to 135X + 135 = 180X\n\nto get the value of X, we combine similar terms.  they are 135X and 180X.  We combine this by transposition.  Remember that in order not to change the value of the equation we must change the sign of what is moved to another side.  In this case i will move 180X to pair it with 135X and move 135 to the other side. so now we get 135X - 180X = -135\n\nwe simplify the like terms and we get -45X.\n\nnow we have -45X = -135\n\nbut we are only looking for X so we must eliminate -45 from the X.  identify the numerical coefficient of X which is -45 and devide it from the original. what you do to one side, you should also do it to the other.\n\n-45X/-45 = -135/-45 with this equation we can simplify it into\nX=3\n\nnow we have the value for X\n\nso one trip will be 3 and since another is 1 hour longer then it must 3 + 1 = 4\n\nheres the solution\n\n180/(X+1)    =      135/X\n135(X+1)     =      180X\n135X + 135   =    180X\n135X - 180X   =    -135  \n            -45X   =    -135\n         -45X/-45=    -135/-45\n                   X =  3\n          \n             X + 1 =  trip 2\n             3+1 = 4
526515
In this equation, you have a right triangle.  After 3 hours, the first car has traveled 150 km south and the 2nd car ahs traveled 210 km west.  You can find the between these two cars at 3 hours by using the Pythagorean Theorem since you have a right triangle.\n\nDistance = sqrt(150^2+210^2) = 258.07 km\n\nThe general form of the Pythagorean Theorem is:\n\na^2 + b^2 = c^2, where:\n\na = distance of car 1\nb = distance of car 2\nc = distance between cars.\n\nTake the derivative to find how the rates are related to one another.  Take the derivative with respect to time:\n\n2a(da/dt) + 2b(db/dt) = 2c(dc/dt)\n\nIn this problem, we are looking for dc/dt.  These are our known values:\n\na = 150 km\nb = 210 km\nc = 258.07 km\nda/dt = 50 km/hr\ndb/dt = 70km/hr\n\nPlugging these values for the equation above, solve for dc/dt:\n\ndc/dt = [2a(da/dt) + 2b(db/dt)]/[2c] ------->\n\n[(2*150*50) + (2*210*70)]/(2*258.07)\n\nTherefore, dc/dt = 86.02 km/hr.  Now change to km/sec:\n\n(86.02 km / 1 hr)(1 hr / 60 min)(1 min / 60 sec) = 0.024 km/sec\n\n-----------\n\nHope this helps
574908
it shouldn't take that long.  Try calling them...depending on what service they are using they should have had one the next day or possibly up to two weeks, but no longer than that i would say!
523844
Total distance to be covered:690miles\nJulia+Maria:60+55\n=115miles\nTotal time taken:690/115\n=6hours
270648
An alternative method:\nThe times taken by A, B and C to complete one lap around the track would be 10/7, 10/8 and 10/9 mins, respectively. Now we just need the LCM of these three to find out when they will all catch up with each other. \n\nThis turns out to be 10 mins.
580104
yes you can for a limited time like one-three months depends on where in USA. just search for the state you are going to and check for how long you can drive
459745
I don't know, but how long does it take to eat a possum?\nit depend on how much traffic there is!
259698
As you can tell, there's not really enough info within the problem itself to solve it.  You'll need to get a map, atlas, or some internet source to determine how much of the 481 miles will be driven in Oregon, and how much in California.  I won't look that up for you, but let's pretend that it's 121 miles in Oregon, and the other 481-121=360 miles in Cali.\n\nAssuming you constantly maintain the speeds indicated, you can calculate the time it takes you for each leg of the trip by using the formula:\n\n   distance = speed x time\n\nOr, equivalently,\n\n   time = distance / speed\n\nSo, for the 121 miles in Oregon, over which you will be traveling 65 mph,\n\n   time = 121 miles / 65 mph = (approximately) 1.86 hours\n\nand for the 360 miles in California, over which you will be traveling 70 mph,\n\n   time = 360 miles / 70 mph = (approx.) 5.14 hours\n\nAltogether, then, you'd be traveling 1.86 + 5.14 = (approx.) 7.00 hours.\n\nIf you want to write that out as one long problem, you could write it this way:\n\nTotal Time = Time1 + Time 2\n           = Dist1/Speed1 + Dist2/Speed2\n           = 121 mi/65 mph + 360 mi/70 mph\n           ~ 1.86 hr + 5.14 hr\n           ~ 7.00 hr\n\nNow, don't forget, I picked the distance values, 121 and 360 miles, out of thin air.  You'll need to look them up and do the problem yourself.  But hopefully that should give you an idea of how.  Good luck!
366261
SRY!! Matt is going to a basketball game out of town. If he could travel 75mph, He would arrive one hour early. If he could travel 45mph he arrives one hour too Late. How far is Matt's trip\n\ndis=75*(time-1)\ndis=45*(time+1)\n\nthen\ndis= 225mile,    time=4hour
540172
d=r*t\nhighway= 60*highwaytime\ncity = 30*citytime\ncity = highway -  100\ncitytime =highway time - 5/6  50 min = 5/6 hour.\nsub in\n\nhighway - 100 = 30 (highwaytime- 5/6)\n60 * highwaytime - 100 = 30 * highwaytime - 30*5/6\nsimplify\n30* highwaytime = 75\nhighwaytime = 2 1/2 hours\nh  =2.5*60 = 150 miles\ncity =30* 1 2/3 = 50 miles\ntotal 200 miles
51513
It will take you 10 hours. \n\nIf you ride only 2 km ever hour, then after one hour you will be 2 km closer to your destination. Take 20 and divide it by 2, and that gives you 10 hours. \n\nIt doesn't matter for your question, but that's a pretty slow bike! A person could walk faster than that.
41125
You need to know the speed of each car to answer the question. Assuming car 1 has a speed of s1 miles/hour and car 2 has a speed of s2 miles/hour. The time they will take to meet each other will be 8.5/(s1+s2). You can now substitute the values to arrive at the answer!