Buy Cars and Trucks in West Point, Mississippi
 |
| Ford : F-250 f-250 2004 FORD F250 EXTENDED CAB SERVICE TRUCK KNAPHEIDE BODY
$4,550.00 $12,500.00 
Time Left: 3d 19h 8m |
|
| Ford : F-250 f-250 2004 FORD F250 EXTENDED CAB SERVICE TRUCK KNAPHEIDE BODY
$4,550.00 $12,500.00
Time Left: 3d 19h 8m |
|
two cars start from the same point. Car A travels north at a speed of 40 mph and car b travelsdue west....? Question: two cars start from the same point. Car A travels north at a speed of 40 mph and car b travels due west at a speed of 30 mph. how many hours will the cars have to travel before they are 500 miles apart? d= r X t Answer: Since they are traveling at a right angle to each other, this is just a problem of solving the pythagorean theorem for the value C for a right triangle: A^2 + B^2 = C^2, where A & B are the distances traveled by the cars and C is the distance between them that youre solving for (the hypotenuse of the right triangle) which is 500 miles. (note that the actual compass headings are irrelevant, the only part that matters is the angle between their two directions) Let X = hours traveling, then car A has traveled 40mph * X, similarly car B has traveled 30mph * X. Now we are working with only distances, so we can use the pythagorean theorem: Substituting we get: [40*X]^2 + [30*X]^2 = 500^2 Simplifying, we can reduce this to: (1,600)*(X^2) + (900)*(X^2) = 250,000 (2,500)*(X^2) = 250,000 X^2 = 100 X = 10 So after 10 hours, the cars will be 500 miles apart. (you can verify this by plugging 10 hours back into pythagorean theorem ) Hope it helps. |
|
Two cars start moving from the same point. One travels south at 24 mi/h and the other travels west at 18 mi/h.? Question: Two cars start moving from the same point. One travels south at 24 mi/h and the other travels west at 18 mi/h. At what rate is the distance between the cars increasing two hours later? Answer: 60 miles apart, as Pythagoras might have said had he knew modern English. |
|
Two cars start moving from the same point. One travels south at 25 mi/hr and the other travels west at 53 mi/? Question: Two cars start moving from the same point. One travels south at 25 mi/hr and the other travels west at 53 mi/ hr . At what rate is the diatance between the cars increasing 4 hours later? ( Use calculator to approximate square-root of a number) Please show work so I can understand it Answer: the diatance between the cars after 4 hrs is 234.4013 m. the diatance increasing @ 58.6003 m/hr |
|
How relaxed is west point on its rules? Question: I know that some of the academy's are really hard but others are easier on rules. Like are you allowed to get a car junior year at west point? Or what privileges do you receive and how early are these privileges given to you? Answer: West Point has very, very strict rules and trying to break any of their rules will get you into the very deepest trouble and would forever be on your record. West Point is THE worst place to break rules. |
|
Two cars start moving from the same point. One travels south at 27 mph and the other travels west at 50 mph.? Question: At what rate is the distance between the cars increasing 3 hours later? Round your answer to the nearest hundredth. Answer: coordinate system has origin at the location of the start of the two cars. Positive x is west, positive y is south. Let t be the time in hours after the cars being moving. Car 1: x = 0 y = 27*t Car 2: x = 50*t y= 0 d = sqrt( (x1-x2)^2 + (y1-y2)^2) d = sqrt( (50t)^2 + (27t)^2) d = sqrt(3229*t^2) take the derivative d' = .5*(3229*t^2)^-.5 * (2*3229*t) let t=3 d' = 56.8 mph |
|
two cars start from the same point. Car A travels north at a speed of 40 mph and car b travels at a speed of..? Question: two cars start from the same point. Car A travels north at a speed of 40 mph and car b travels at a speed of 30 mph due west . how many hours will the cars have to travel before they are 500 miles apart? d= r X t Answer: Set up your variables, always explain them if not given. So... D = Distance x = Time passed ax = Distance traveled by Car A bx = Distance traveled by Car B Set up the equation... D = ax + bx Plug in the values... 500 = 40x + 30x 500 = 70x Divide both sides by 70 7.14 is about equal to x |
|
Two cars start moving from the same point. One travels south at 60mih and the other travels west at 25mih. How? Answer: The driver starts the engine, puts it in drive and presses the gas pedal. |
|
Two cars start moving from the same point. One travels south at 100km/hr, the other west at 50km/hr.? Question: How far apart are they two hours later? Is the distance formulal used here? Answer: Use the Pythagorean theorem: for a right triangle h^2 = a^2 + b^2, where h is the length of the hypotenuse, and a and b are the lengths of the sides adjacent to the right angle. The cars' paths form a right angle with lengths of 200 and 100 km. The distance between them (by air) is the hypotenuse. d^2 = 200^2 + 100^2 d = sqrt (50000) = 100 * sqrt(5) = ca. 223.6 km. |
|
A car is driven 120 km west and then 95 km southwest. What is the displacement of the car from the start point? Question: physics question: A car is driven 120 km west and then 95 km southwest. What is the displacement of the car from the point of origin? Answer: If you don't do your own homework you won't learn things you need to learn. |
|
Car is traveling along a parabola with vertex at origin. Car starts at point 100m west and 100m north? Question: of the origin and travels in a easterly direction. There is a statue at 100m east and 50 m north of the origin. (statue is not on the highway) At what point on the highway will the car's headlights illuminate the statue. Express answer as ordered pair. I know I need to find the derivative at which the line intersects the statue. Answer: The start point is NW of the vertex; I'm assuming the parabola opens upwards because then the car continues east throughout. The general equation for an up-opening parabola with vertex at the origin is y = ax² (-100, 100) is a point on the parabola, so a=0.01 and the equation becomes y = 0.01x² The car's headlights are tangent to the parabola. The slope of the tangent is y' = 0.02x The statue is at (100, 50). Let (x₁, 0.01x₁²) be the coordinates of the car when its headlights hit the statue. The slope of the parabola at this point is 0.02x₁. The slope of the line joining the car and statue is (50-0.01x₁²)/(100-x₁). (50-0.01x₁²)/(100-x₁) = 0.02x₁ 0.01x₁² -2x₁+50 = 0 Using the quadratic formula, x₁ = 29.3 or 170.7 170.7 is invalid because then the statue is behind the car, so x₁ = 29.3 y₁ = (0.01)29.3² = 8.6 The headlights hit the statue when the car is at (29.3, 8.6). http://www.flickr.com/photos/dwread/2864627978/ |
