answer:First, we need to convert the speed of the train from kilometers per hour (kmph) to meters per second (m/s) because the lengths of the train and the bridge are given in meters. To convert kmph to m/s, we use the conversion factor: 1 kmph = 1000 meters / 3600 seconds = 5/18 m/s So, the speed of the train in m/s is: 72 kmph * (5/18) m/s/kmph = 20 m/s Next, we need to find the total distance the train needs to travel to completely cross the bridge. This is the length of the train plus the length of the bridge: 250 meters (train length) + 300 meters (bridge length) = 550 meters (total distance) Now, we can calculate the time it will take for the train to cross the bridge by dividing the total distance by the speed of the train: Time = Total Distance / Speed Time = 550 meters / 20 m/s = 27.5 seconds Therefore, it will take the train boxed{27.5} seconds to cross the bridge.

answer:If Kayden's business delivered an equal number of the 400 cartons of milk to four customers, then each customer received: 400 cartons / 4 customers = 100 cartons per customer If each of the four customers returned 60 cartons damaged, then the number of cartons accepted by each customer is: 100 cartons - 60 damaged cartons = 40 cartons accepted per customer Since there are four customers, the total number of cartons accepted by all customers is: 40 cartons/customer * 4 customers = boxed{160} cartons accepted in total

answer:Given overrightarrow{AB}=({2,3}) and overrightarrow{BC}=({1,-4}), to find overrightarrow{AC}, we use the fact that overrightarrow{AC} = overrightarrow{AB} + overrightarrow{BC}. Starting with the given vectors: - overrightarrow{AB} = (2,3) - overrightarrow{BC} = (1,-4) We add these vectors component-wise: - For the x-components: 2 + 1 = 3 - For the y-components: 3 + (-4) = -1 Therefore, overrightarrow{AC} = overrightarrow{AB} + overrightarrow{BC} = (3, -1). So, the correct answer is boxed{A}.

answer:1. Given that ( P ) is an internal point of ( triangle ABC ) such that ( angle PAB = angle PBC = angle PCA ), we want to prove that: [ a^{2} + b^{2} + c^{2} leqslant 3left(PA^{2} + PB^{2} + PC^{2}right). ] 2. Consider the expressions for ( a^2, b^2, ) and ( c^2 ) based on the given angle conditions. Using the Law of Cosines, we get: [ begin{aligned} a^2 &= PB^2 + PC^2 - 2 cdot PB cdot PC cdot cos(pi - C) &= PB^2 + PC^2 + 2 cdot PB cdot PC cdot cos(C), b^2 &= PC^2 + PA^2 + 2 cdot PC cdot PA cdot cos(A), c^2 &= PA^2 + PB^2 + 2 cdot PA cdot PB cdot cos(B). end{aligned} ] 3. Add these three expressions together: [ begin{aligned} a^2 + b^2 + c^2 &= (PB^2 + PC^2 + 2 cdot PB cdot PC cdot cos(C)) + (PC^2 + PA^2 + 2 cdot PC cdot PA cdot cos(A)) + (PA^2 + PB^2 + 2 cdot PA cdot PB cdot cos(B)) &= (PB^2 + PB^2) + (PC^2 + PC^2) + (PA^2 + PA^2) + 2 cdot PB cdot PC cdot cos(C) + 2 cdot PC cdot PA cdot cos(A) + 2 cdot PA cdot PB cdot cos(B) &= 2 (PB^2 + PC^2 + PA^2) + 2 cdot (PA cdot PB cdot cos(B) + PB cdot PC cdot cos(C) + PC cdot PA cdot cos(A)). end{aligned} ] 4. Using the assumptions about the angles, we have the following inequity due to optimization conditions for the cosine terms: [ PA cdot PB cdot cos(B) + PB cdot PC cdot cos(C) + PC cdot PA cdot cos(A) leq PB^2 + PC^2 + PA^2. ] 5. Substituting this inequality into our previous equation, we get: [ 2 (PB^2 + PC^2 + PA^2) + 2 (PA cdot PB cdot cos(B) + PB cdot PC cdot cos(C) + PC cdot PA cdot cos(A)) leq 3 left(PA^2 + PB^2 + PC^2right). ] 6. Hence, we obtain the desired inequality: [ a^2 + b^2 + c^2 leq 3 left(PA^2 + PB^2 + PC^2right). ] # Conclusion: [ boxed{a^2 + b^2 + c^2 leq 3 left(PA^2 + PB^2 + PC^2right)} ]