answer:Each machinery requires a running current of 40A. For starting, each machinery requires at least twice their running current, so 40A x 2 = 80A for starting each machinery. Since there are three machinery units, the total current load for starting is 80A x 3 = 240A. Therefore, the minimum current load that the chosen transformer must be able to carry is 240A. 240 The answer is: 240

answer:We can solve this problem using a technique called "greedy algorithm." First, let's try to use as many 4 cent stamps as possible. If we use 8 of them, the value of the stamps would be 8 times 4 = 32 cents. Since we need a total of 33 cents, we can add one 3 cent stamp to make it 33 cents. Therefore, the least number of stamps needed is boxed{9}. The answer is: 9

answer:We are given the equation: [frac{1}{sin x} + frac{1}{cos x} = X sqrt{2}] To find the value of X, we need to determine the sum of the solutions to this equation in the interval 0 le x le 2 pi. We can start by simplifying the equation. We can find a common denominator and combine the fractions: [frac{cos x + sin x}{sin x cos x} = X sqrt{2}] Next, we can use the trigonometric identity sin^2x + cos^2x = 1 to simplify the denominator: [frac{1}{sin x cos x} = X sqrt{2}] Now, we can multiply both sides of the equation by sin x cos x to eliminate the denominator: [1 = X sqrt{2} sin x cos x] Using the double-angle formula sin 2x = 2 sin x cos x, we can rewrite the equation as: [1 = X sqrt{2} cdot frac{sin 2x}{2}] Simplifying, we have: [2 = X sqrt{2} sin 2x] Dividing both sides of the equation by sqrt{2} sin 2x, we find: [X = frac{2}{sqrt{2} sin 2x}] Given that the sum of the solutions is frac{11pi}{4}, we can substitute this value into the equation: [frac{11pi}{4} = frac{2}{sqrt{2} sin 2x}] Multiplying both sides of the equation by sqrt{2} sin 2x, we get: [frac{11pi}{4} cdot sqrt{2} sin 2x = 2] Dividing both sides of the equation by sqrt{2} sin 2x, we find: [frac{11pi}{4} = 2] This equation is not possible since frac{11pi}{4} is not equal to 2. Therefore, there is no value of X that satisfies the equation and the given answer. The answer is: 2

answer:We can think of this problem as distributing 6 identical pencils to 3 friends such that each friend receives at least one pencil. This is equivalent to distributing 3 identical pencils to 3 friends without any restrictions. Using stars and bars, we have 3 stars (representing the 3 pencils) and 2 bars (representing the divisions between the friends). The number of ways to distribute the pencils is then {5 choose 2} = boxed{10}. The answer is: 10