🧠:<think>Okay, so I have this question here about identifying which of the given events is a random event. The options are A, B, C, and D. Let me try to understand what a random event means first. From what I remember, a random event is something that happens without a predictable pattern or outcome. It's not certain and can vary each time it occurs.Alright, let's look at each option one by one.Option A: A coin thrown from the ground will fall down.Hmm, when you throw a coin up, it always comes back down because of gravity. That seems pretty certain. I don't think there's any randomness here because the outcome is always the same. So, this doesn't sound like a random event.Option B: A shooter hits the target with 10 points in one shot.Okay, this one is about a shooter hitting a target. I know that in shooting sports, hitting the bullseye or scoring 10 points is not always guaranteed. It depends on the shooter's skill, the conditions, maybe even luck. So, this outcome isn't certain every time. There's an element of unpredictability here, which makes me think this could be a random event.Option C: The sun rises from the east.The sun rising from the east is something that happens every day, right? It's a natural phenomenon due to the Earth's rotation. I don't see any randomness here because it's consistent and predictable. So, this isn't a random event either.Option D: A horse runs at a speed of 70 meters per second.Wait, 70 meters per second is really fast. I know that horses can run pretty quickly, but 70 meters per second seems extremely high. Maybe I'm misremembering, but I think the fastest horses run around 20-30 meters per second. So, this event seems impossible or at least highly unlikely. Since it's not something that can happen, it doesn't fit the definition of a random event either.So, after going through all the options, it seems like Option B is the only one that fits the definition of a random event because it involves an unpredictable outcome based on various factors.</think>

🧠:<think>First, I need to find the area of a sector with a central angle of 45 degrees in a circle with a radius of 4 units.I recall that the formula for the area of a sector is:S_{text{sector}} = frac{theta}{360^circ} times pi r^2where ( theta ) is the central angle in degrees and ( r ) is the radius of the circle.Substituting the given values into the formula:S_{text{sector}} = frac{45^circ}{360^circ} times pi times 4^2Simplifying the fraction:frac{45^circ}{360^circ} = frac{1}{8}Calculating the squared radius:4^2 = 16Now, multiply the values together:S_{text{sector}} = frac{1}{8} times pi times 16Simplifying further:S_{text{sector}} = 2piTherefore, the area of the sector is ( 2pi ) square units.</think>

🧠:<think>Okay, so I have this problem about four gears, A, B, C, and D, that are meshed together in sequence. Each gear has a certain number of teeth: x for A, y for B, z for C, and w for D. I need to figure out the proportion of their angular speeds, ω_A, ω_B, ω_C, and ω_D. The options are given as A, B, C, D, each with different proportions.First, I remember that when two gears mesh together, their angular velocities are inversely proportional to their number of teeth. That is, if gear A meshes with gear B, then ω_A / ω_B = y / x. This is because the linear speed at the point of contact must be the same for both gears, so the gear with more teeth will rotate slower.Since the gears are meshed sequentially, this relationship should hold for each pair. So, ω_A / ω_B = y / x, ω_B / ω_C = z / y, and ω_C / ω_D = w / z.I think I can express each angular speed in terms of the next one. Starting from ω_A, I can write ω_B as (x / y) * ω_A. Then, ω_C would be (y / z) * ω_B, which is (y / z) * (x / y) * ω_A = (x / z) * ω_A. Similarly, ω_D would be (z / w) * ω_C = (z / w) * (x / z) * ω_A = (x / w) * ω_A.So, if I express all angular speeds in terms of ω_A, I get:- ω_A = ω_A- ω_B = (x / y) * ω_A- ω_C = (x / z) * ω_A- ω_D = (x / w) * ω_ATo find the proportion, I can write them as:ω_A : ω_B : ω_C : ω_D = 1 : (x / y) : (x / z) : (x / w)But I need to express this without fractions. To do that, I can multiply each term by yzw to eliminate the denominators:- 1 * yzw = yzw- (x / y) * yzw = xzw- (x / z) * yzw = xyw- (x / w) * yzw = xyzSo, the proportion becomes yzw : xzw : xyw : xyz.Looking back at the options, option D is yzw : xzw : xyw : xyz, which matches what I derived.I think that makes sense because each step of the proportion accounts for the inverse relationship between the number of teeth and the angular speed, and by multiplying through by yzw, I cleared the denominators to get a clean proportion.I don't see any mistakes in my reasoning, so I feel confident that option D is correct.</think>

🧠:<think>Alright, so I have this problem here: there's a function f(x) whose derivative is given by f'(x) = a(x + 1)(x - a), and I know that a is less than zero. The function f(x) reaches a maximum at x = a, and I need to figure out the range of values for a. Hmm, okay, let's break this down step by step.First, I remember that if a function has a maximum at a certain point, the derivative changes from positive to negative as it passes through that point. So, at x = a, the derivative f'(x) should be zero, and to the left of a, the derivative should be positive, and to the right of a, it should be negative. That makes sense because if the function is increasing before a and decreasing after a, it must have a peak at a.Given that f'(x) = a(x + 1)(x - a), and a is negative, I need to analyze the sign of the derivative around x = a. Let's consider the critical points where the derivative is zero. Setting f'(x) = 0, we get:a(x + 1)(x - a) = 0Since a is not zero (because a < 0), the critical points are at x = -1 and x = a. So, we have two critical points: one at x = -1 and another at x = a.Now, I need to determine the intervals where the derivative is positive or negative. Since a is negative, let's consider different cases for a to see how the sign of the derivative behaves.Case 1: a is between -1 and 0, so -1 < a < 0.In this case, the critical points are at x = -1 and x = a, with a being closer to zero than -1. Let's test the sign of f'(x) in the intervals:1. x < -12. -1 < x < a3. x > aFor x < -1: Let's pick a number less than -1, say x = -2. Plugging into f'(x):f'(-2) = a(-2 + 1)(-2 - a) = a(-1)(-2 - a)Since a is negative, (-2 - a) is negative because a is less than zero, so -2 - a = -2 + |a|, which is still negative because |a| < 2 (since a > -1). So, (-1)(-2 - a) is positive, and multiplying by a (negative) gives negative. So, f'(-2) is negative.For -1 < x < a: Let's pick x = 0 (assuming a is between -1 and 0, so 0 is greater than a). Wait, actually, if a is between -1 and 0, then a is less than 0, so x = 0 is greater than a. Maybe pick x = (a - 1)/2, which is between -1 and a. Let's say a = -0.5, then x = (-0.5 - 1)/2 = -0.75.f'(-0.75) = a(-0.75 + 1)(-0.75 - a) = a(0.25)(-0.75 - a)Since a is negative, (-0.75 - a) is negative because a is negative, so -0.75 - a = -0.75 + |a|, which is still negative if |a| < 0.75. Wait, but a is between -1 and 0, so |a| is between 0 and 1. So, if |a| is less than 0.75, then -0.75 - a is negative, otherwise, it's positive. Hmm, this might complicate things.Maybe instead of plugging in specific numbers, I can analyze the sign algebraically.f'(x) = a(x + 1)(x - a)Since a < 0, the sign of f'(x) depends on the product (x + 1)(x - a).Let's analyze the intervals:1. x < -1: Both (x + 1) and (x - a) are negative. Negative times negative is positive. But since a is negative, f'(x) = a*(positive) = negative.2. -1 < x < a: (x + 1) is positive, and (x - a) is negative (since x < a). Positive times negative is negative. So, f'(x) = a*(negative) = positive (because a is negative).3. x > a: Both (x + 1) and (x - a) are positive. Positive times positive is positive. So, f'(x) = a*(positive) = negative.So, in this case, when -1 < a < 0, the derivative changes from negative to positive as x increases through -1, and then from positive to negative as x increases through a. Therefore, at x = a, the derivative changes from positive to negative, indicating a local maximum. That's exactly what we want!Case 2: a = -1.In this case, the derivative becomes f'(x) = -1(x + 1)(x - (-1)) = -1(x + 1)(x + 1) = -1(x + 1)^2.So, f'(x) = - (x + 1)^2.Since (x + 1)^2 is always non-negative, f'(x) is always non-positive. It's zero at x = -1, but negative everywhere else. So, at x = a = -1, the derivative doesn't change sign; it's zero there but negative on both sides. Therefore, x = -1 is not a maximum; it's actually a point of inflection or a saddle point. So, this doesn't satisfy the condition of having a maximum at x = a.Case 3: a < -1.Here, a is less than -1, so the critical points are at x = -1 and x = a, with a being to the left of -1. Let's analyze the sign of f'(x) in the intervals:1. x < a: Both (x + 1) and (x - a) are negative. Negative times negative is positive. So, f'(x) = a*(positive) = negative (since a < 0).2. a < x < -1: (x + 1) is negative (since x < -1), and (x - a) is positive (since x > a). Negative times positive is negative. So, f'(x) = a*(negative) = positive (since a < 0).3. x > -1: Both (x + 1) and (x - a) are positive. Positive times positive is positive. So, f'(x) = a*(positive) = negative.So, in this case, the derivative changes from negative to positive as x increases through a, and then from positive to negative as x increases through -1. Therefore, at x = a, the derivative changes from negative to positive, indicating a local minimum, not a maximum. That's the opposite of what we want.Therefore, the only case where x = a is a local maximum is when -1 < a < 0.Wait, let me double-check. If a is between -1 and 0, then at x = a, the derivative changes from positive to negative, which is a maximum. If a is less than -1, it's a minimum. If a is exactly -1, it's not a maximum. So, yeah, the range should be -1 < a < 0.I think that makes sense. So, the range of values for a is all real numbers between -1 and 0, not including -1 and 0 themselves.</think>