A lake lies between A and C and lines AB and DE run north-south.  If AB =8.3 mi De=2.4 mi and DC =2.2 mi?
813707
We do not have enough information to solve the problem.\nWhat connects A or B to  C D or E.
458976
The answers for these questions:\n\nAnswer 1: (c) 27 meters\nAnswer 2: (d) .5 km\n\nAlthough the answer to 2nd question should be 0.4 kms, but I reckon it's just approximate. \n\nYou can email me for further help by clicking Contact Nitro in my answers profile.
1028343
a little unclear but ...\n\nthe 'usual' equation is :\ns=1/2   * a  * t^2\n= .5 * 9.8 * 5^2\n= 4.9 * 25\n= 122.5 meters\n\noops .. in feet\ns=.5 * 32 * 5^2 = 16 * 25 = 400 feet (which "should" equal 122.5 meters)
745771
first find the midpoint of AB which is C(9,-2).\nNow obviously the one-fourth distance will be the midpoint of AC which is ...well I let you do this. This is simple knowing the formula for the midpoint .\nMathguy did well too but you are supposed to apply, I think, midpoint formula.
791122
When you see"of" it means multiply.  So you multiply 4/7 x 39/1 and get 156/7 which reduces to 22-2/7 miles.  Then you multiply 2/3 x 39/1 and get 78/3 which reduces to 26 miles.  Then you subtract (because it is asking for the "difference.")  26 - 22-2/7 is 3-5/7 miles.
160821
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s
958697
10.4 m/sec\n\nV(t) = (30+a*t) m/sec\n\nwhere a = - 9.8m/sec^2\n\nV(2) = (30 - 09.8*2) m/sec\n\nV(2) = 30-19.6 = 10.4 m/sec
716347
Of three points, A,B,C\n Check the slope of A-B,  B-C, A-C.  If any of the slopes are the same, you have parallel lines and thus no triangle.  So, to do (2), do (4) first.\n  If you know the distance between A-B, B-C, A-C, you can calculate the area with the formula, so (5) also depends on (4).\n\n You write your own code.  yA isn't a site to do your homework, since if you don't do it yourself, you don't pass the class!
1169918
1 ->25000\n2cm ->2*25000cm=50000cm=500m\n\nans: b:500m
827600
Let the time of the appointment be t in hours, with now being time zero.  Then in the first case, she travels a distance of 66t kilometers.  In the second case, she travels 90(t - 2/3) kilometers, because 40 minutes is 2/3 hours.  Just solve the algebraic equations 66t = 90(t - 2/3) for t, and then evaluate either of the distance expressions.
566525
let V1= 4 mph, V2= 5 mph\n\nV1*t + V2*t = 3\nt= 20 min\n\n19. a 20 min\n\n\n14.16/3 = 4.72\n\n4.72*14 = 66.08\n\n20 c 66.08
792509
At 1 p.m the 1st bus starts moving west with a velocity of 44 mph  that means at 2 pm the bus will be 44 miles away from the station because  D = V x t = 44 mph x 1 h = 44 miles .\nAt 2 pm the 2nd bus starts moving east i.e in the opposite direction with a velocity of 48 mph \n              \nB1  ____________B2      ( B1 B2 = 44 miles )\n                      \n             \n\nAfter 2 pm the distance between the 2 buses at any time is equal \nto : \nD = 44 +( V1 x t ) + (V2 x t ) = 44 + (44 x t ) + (48 x t ) = 44 + 92 t \nwhere D >> miles , t >> hrs \n                                         \n                                          B1    _________S___________ B2\nWhere B1 S = 44 miles + 44 t \n            B2 S = 48 t \n                                             \nPut D = 274 miles to get  t \n274 = 44 + 92 t     >>  t = 2.5 h\nTherefore the two buses will be a distance of 274 miles apart at : \n2 pm + 2.5 h =  4 : 30 pm
910005
.8 *10= 8\n\nso the question is at a speed of 18 miles per hour how long would it take to run 8 miles. \n\n18 miles / 60 minutes = .3 miles per minute\n8 miles / .3 miles per minute = 26.667 minutes
337691
if the question is: If it takes 3 hours to travel 24 miles downstream and 6 hours to travel UPstream, how fast is the current? then,\nlets call the travelling speed Vt and the current speed Vc\ndownstream we have Vt+Vc=24miles/3hours=8mph\nupstream we have Vt-Vc=24/6=4mph\nto solve the system of two equations we can add the equations and get 2Vt=12, Vt=6mph\nVc=8-6=6-4=2mph\nThe current is 2mph fast.
48038
distance = rate x time, \n\nLet x be the distance she can go out on the lake, and t the time it takes to get there.\nGoing out she paddles a 3km/h so d = 3 x t\nComing back, the distance is the same, the rate is 2 km/h\nand the time she has is whatever is left of the 2.5 hrs.\nShe use t amount of time going out so (2.5 -t)\n\nGoing out   d = 3 x t\nComing back d = 2 x ( 2.5 - t)\nThe distance is the same both ways, so \n3 x t = 2 x (2.5 - t), now solve for t and then figure out d.
550744
If I'm reading your description right, I think your first boat is heading up and to the right ("direction N.68 degrees E") making an angle of 68 degrees with north, or 22 degrees with east.  And I think your second boat is headed directly east ("S.90 degrees E").  So I'm assuming that they form a triangle, with one angle being 22 degrees and another one being a 90 degree angle, since we're looking for the point where one boat will be directly south of the other.\n\nThe first boat traveled 12t miles so far.  The second boat left an hour later and traveled 16(t-1) miles.\n\nThe first boat's path turns out to be the hypotenuse of the triangle, and the second boat's path will be one of the adjacent legs to the 22 degree angle.  So we can set up an equation with trig identities:\n\ncos 22 = 16(t-1)/12t\n0.9272 = 4(t-1)/3t\n2.7816t = 4t - 4\n4 = 1.2184t\nt = 3.3 hours\n\nSo after 3.3 hours, one boat will be directly south of the other.\n\nFinding that distance can be done one of several ways: using the Pythagorean theorem, sine relationships, law of cosines, etc.  I'll use sine, since we're already doing trig.  After 3.3 hours, the northernmost boat will have traveled 39.6 miles.  The distance between the boats becomes our side opposite the 22 degree angle.  So,\n\nsin 22 = distance between/39.6\ndistance = 39.6sin 22\n= 14.83 miles\n\nNow if I read the angle wrong, then simply go back and change all the "22" angles to the correct one and recalculate.
325281
OK, this is a simple unit conversion problem, buddy. First, get your measurements into similar units: in this case, change your minutes into hours (1 hr/60 min). You'll end up with.\n\n(40 km/h * .5 h) + (60 km/h * .25 h)\n\nThe hours will cancel out, and you'll be left with the kilometers.\n\n20 km + 15 km = 35 km\n\nIn this instance, the direction of travel doesn't matter: the question asks the distance traveled, not how far away from the point of origin you are.