answer:1. **Identify the Coordinates of Points**: - A = (0,0). - overline{BC} is parallel to the x-axis, and vertices lie on y = 2x^2. Let x-coordinates of B and C be -b and b, respectively. Thus, B = (-b, 2b^2) and C = (b, 2b^2). 2. **Calculate the Length of overline{BC}**: - Length of overline{BC} is b - (-b) = 2b. 3. **Set Up the Area Formula for triangle ABC**: - Area = frac{1}{2} times text{base} times text{height}, here base is 2b and height is 2b^2. - Thus, area = frac{1}{2} times 2b times 2b^2 = 2b^3. 4. **Solve for b Given the Area**: - Given area 128, set up: [ 2b^3 = 128 implies b^3 = 64 ] - Solve for b: [ b = sqrt[3]{64} = 4 ] 5. **Find the Length of overline{BC}**: - b = 4, thus length of overline{BC} = 2b = 2 times 4 = 8. Conclusion: - The length of overline{BC} is 8. The final answer is boxed{textbf{(C)} 8}

answer:The speed of the train excluding stoppages is 50 kmph, and including stoppages, it is 30 kmph. This means that due to stoppages, the train's effective speed is reduced by 20 kmph (50 kmph - 30 kmph). To find out how many minutes the train stops per hour, we need to calculate the time lost due to stoppages. First, we'll find out how far the train would travel in one hour without any stoppages, which is 50 km. Next, we'll find out how far the train actually travels in one hour with stoppages, which is 30 km. The difference in distance traveled is 20 km (50 km - 30 km), which is the distance the train would have covered if it hadn't stopped. Now, we need to convert the speed lost due to stoppages into time. Since the train could have traveled 20 km in one hour without stoppages, we'll calculate the time it would take to travel this distance at the original speed of 50 kmph. Time = Distance / Speed Time = 20 km / 50 kmph To convert kmph into km per minute, we divide by 60 (since there are 60 minutes in an hour): 50 kmph = 50/60 km per minute Now, we can calculate the time: Time = 20 km / (50/60 km per minute) Time = 20 km / (5/6 km per minute) Time = 20 km * (6/5 minutes per km) Time = (20 * 6) / 5 minutes Time = 120 / 5 minutes Time = 24 minutes Therefore, the train stops for boxed{24} minutes per hour.

answer:1. **Initial Evaluation at Specific Points:** We start by evaluating cos(sin x) and sin(cos x) at three key points x=0, x=frac{pi}{2}, and x=pi. - When x=0: [ cos(sin 0) = cos(0) = 1 ] [ sin(cos 0) = sin(1) approx 0.8415 ] Clearly, cos(sin 0) > sin(cos 0). - When x =frac{pi}{2}: [ cos(sin frac{pi}{2}) = cos(1) approx 0.5403 ] [ sin(cos frac{pi}{2}) = sin(0) = 0 ] Clearly, cos(sin frac{pi}{2}) > sin(cos frac{pi}{2}). - When x = pi: [ cos(sin pi) = cos(0) = 1 ] [ sin(cos pi) = sin(-1) = -0.8415 ] Clearly, cos(sin pi) > sin(cos pi). From these evaluations, we have cos(sin x) > sin(cos x) at the points x=0, frac{pi}{2}, pi. 2. **Checking the Interval left(frac{pi}{2}, piright):** For x in the interval left(frac{pi}{2}, piright): - We know 0 < sin x < 1 - Also, -1 < cos x < 0 Since sin x is between 0 and 1, cos(sin x) will be positive because the cosine function of a value between 0 and 1 lies in (0, 1]. As for sin(cos x), cos x will be between -1 and 0, and hence sin(cos x) will be non-positive as the sine of any angle between -1 and 0 is non-positive. Therefore, for any x in left(frac{pi}{2}, piright), we have cos(sin x) > 0 > sin(cos x). 3. **Checking the Interval left(0, frac{pi}{2}right):** For x in the interval left(0, frac{pi}{2}right): - We know 0 < x < frac{pi}{2}, which implies 0 < sin x < 1 and 0 < cos x < 1. We need to prove that: [ cos(sin x) > sin(cos x) ] Considering that both sin x and cos x lie in (0, 1), let's transform the inequality: [ cos(sin x) = sinleft(frac{pi}{2} - sin xright) ] So we must show: [ sinleft(frac{pi}{2} - sin xright) > sin(cos x) ] Since 0 < frac{pi}{2} - sin x < frac{pi}{2}, it follows that: [ sinleft(frac{pi}{2} - sin xright) = cos(sin x) ] We now need to show: [ frac{pi}{2} - sin x > cos x ] This is equivalent to proving: [ sin x + cos x < frac{pi}{2} ] We know that sin x + cos x leq sqrt{2} considering the maximum value addition formula in trigonometry. Since sqrt{2} approx 1.414 < frac{pi}{2} approx 1.571, this shows: [ sin x + cos x < frac{pi}{2} ] 4. **Conclusion:** From the above evaluations and inequalities, we have shown that for all x in [0, pi]: [ cos(sin x) > sin(cos x) ] boxed{cos(sin x) > sin(cos x)}

answer:**Analysis** This question examines the judgment of sufficient and necessary conditions and the issue of constant truth. First, identify the necessary and sufficient condition for the proposition to be true, which is {a|a geqslant 4}. A sufficient but not necessary condition would be a proper subset of {a|a geqslant 4}, which allows for the solution. **Answer** Given the proposition "For all x in [1,2], x^2-a leqslant 0" to be true, it can be transformed into "For all x in [1,2], a geqslant x^2" always holds, thus a geqslant (x^2)_{max} = 4, meaning the necessary and sufficient condition for "For all x in [1,2], x^2-a leqslant 0" to be true is a geqslant 4, and the sufficient but not necessary condition we are looking for is a proper subset of {a|a geqslant 4}, from the options, it is clear that C meets the criteria, therefore, the correct choice is boxed{C}.