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4 dni temu · Solved examples to find the perpendicular distance of a given point from a given straight line: 1. Find the perpendicular distance between the line 4x - y = 5 and the point (2, - 1). Solution: The equation of the given straight line is 4x - y = 5 or, 4x - y - 5 = 0
5 dni temu · Let the horizontal distance between the helicopter and the island be \( d \), and the width of the island \( w \). Then \( \tan 24^\circ = \frac{1000}{d+w} \) and \( \tan 31^\circ = \frac{1000}{d} \), implying \( d = \frac{1000}{\tan 31^\circ} \).
4 dni temu · Problems to calculate the motion or speed of train: 1. Find the time taken by a train 150 m long, running at a speed of 90 km/hr in crossing the pole. Solution: Length of the train = 150m. Speed of the train = 90 km/hr = 90 × 5/18 m/sec
5 dni temu · The formula to calculate distance from rate and time is straightforward: \ [ D = R \times T \] Where: \ (D\) is the distance traveled, \ (R\) is the rate or speed of travel, and. \ (T\) is the time spent traveling. Example Calculation.
3 dni temu · Single-source shortest path algorithms operate under the following principle: Given a graph \ (G\), with vertices \ (V\), edges \ (E\) with weight function \ (w (u, v) = w_ {u, v}\), and a single source vertex, \ (s\), return the shortest paths from \ (s\) to all other vertices in \ (V\).
4 dni temu · Solution: Average Speed = \(\frac{\textrm{Total Distance Covered}}{\textrm{Total Time Taken}}\) = \(\frac{325 km}{5 hours}\) = 65 km/hr 2. A bus covers a distance of 420 kms in 6 hours. Find its average speed.
23 godz. temu · Given a horizontal distance \(D = 70\) and an angle \(a = 30^\circ\), the height \(H\) can be calculated as: \[ H = 70 \times \tan(30^\circ) \approx 40.4508 \] This example demonstrates how to determine the height from a known distance and angle, using the tangent function.
