💡Okay, so I have this problem about a bag with balls of different colors, and I need to figure out which two events are mutually exclusive but not complementary. Let me try to break this down step by step.First, let me understand the setup. The bag contains 3 red balls, 2 white balls, and 1 black ball. So, in total, there are 3 + 2 + 1 = 6 balls. We are drawing two balls from this bag. The question is asking about four different pairs of events (A, B, C, D) and wants to know which pair is mutually exclusive but not complementary.I remember that two events are mutually exclusive if they cannot happen at the same time. That means their intersection is empty; there's no overlap between the two events. On the other hand, complementary events are a pair where one event happens if and only if the other does not. So, if two events are complementary, they cover all possibilities, and one must occur while the other cannot.So, for the events to be mutually exclusive but not complementary, they must not overlap (they can't both happen), but they also shouldn't cover all possibilities. That is, there should be some outcomes where neither event happens.Let me look at each option one by one.**Option A: At least one white ball; At least one red ball**Hmm, so the first event is drawing at least one white ball, and the second is drawing at least one red ball. Are these mutually exclusive? Well, can I draw two balls where one is white and one is red? Yes, for example, one white and one red. So, these two events can happen together. That means they are not mutually exclusive. So, option A is out.**Option B: At least one white ball; One red ball and one black ball**Alright, the first event is at least one white ball, and the second is specifically one red and one black ball. Let me see if these can happen at the same time. If I draw one red and one black ball, that means I have no white balls. So, the first event requires at least one white ball, but the second event has none. Therefore, these two events cannot happen together. So, they are mutually exclusive.Now, are they complementary? Let's think. If I draw two balls, the possible outcomes are:1. Both white2. One white and one red3. One white and one black4. Two reds5. One red and one black6. Two blacks (but there's only one black ball, so this isn't possible)So, the second event is specifically one red and one black. The first event is at least one white. Are there any outcomes that are neither? Yes, for example, drawing two red balls. That doesn't fall into either event. So, these two events are not complementary because there are outcomes where neither occurs. Therefore, option B is mutually exclusive but not complementary. That seems like a possible answer.**Option C: Exactly one white ball; One white ball and one black ball**The first event is exactly one white ball, which means one white and one non-white. The second event is specifically one white and one black. So, can these happen at the same time? Yes, if I draw one white and one black, that satisfies both events. So, they are not mutually exclusive. Therefore, option C is incorrect.**Option D: At least one white ball; Both are white balls**The first event is at least one white ball, and the second is both being white. Are these mutually exclusive? Well, if both are white, that certainly means at least one is white. So, these two events can happen together. Therefore, they are not mutually exclusive. So, option D is out.So, after going through each option, it seems like option B is the correct answer because it's the only pair that is mutually exclusive but not complementary.Just to double-check, let me think about option B again. If I have at least one white ball, that includes scenarios like one white and one red, one white and one black, or two white balls. The other event is specifically one red and one black. So, these two events don't overlap because one requires at least one white, and the other requires no white. Also, as I thought earlier, there are outcomes where neither happens, like drawing two red balls. So, yes, they are mutually exclusive and not complementary.I think that makes sense. So, I feel confident that option B is the right answer.**Final Answer**boxed{B}

💡First, I need to identify the tenths and hundredths places in the number 4.65. The tenths place is the first digit after the decimal point, which is 6, and the hundredths place is the second digit, which is 5.Next, I'll apply the rounding rule. If the digit in the hundredths place is 5 or greater, I round up the tenths place by one. In this case, since the hundredths digit is 5, I will increase the tenths place from 6 to 7.Finally, after rounding, the number becomes 4.7.

💡Alright, so I've got this problem here with two propositions, p and q, and I need to figure out which of the options A, B, C, or D is correct. Let me try to break this down step by step.First, let's look at proposition p: "There exists an x₀ in the real numbers such that sin(x₀) equals sqrt(5)/2." Hmm, okay. I know that the sine function has a range between -1 and 1. So, sin(x) can never be greater than 1 or less than -1. Now, sqrt(5)/2 is approximately 1.118, which is definitely greater than 1. That means there's no real number x₀ where sin(x₀) equals sqrt(5)/2 because it's outside the possible range of the sine function. So, proposition p is false.Alright, moving on to proposition q: "For all x in the interval (0, π/2), x is greater than sin(x)." Okay, so this is saying that if I pick any number x between 0 and π/2 (which is about 1.5708), then x is bigger than its sine. I remember that for small angles, sin(x) is approximately equal to x, but as x increases, sin(x) grows slower than x. Let me test this with a couple of values to see if it holds.Let's take x = π/6, which is 30 degrees. sin(π/6) is 0.5, and π/6 is approximately 0.5236. So, 0.5236 > 0.5. That works. How about x = π/4, which is 45 degrees. sin(π/4) is sqrt(2)/2, approximately 0.7071, and π/4 is about 0.7854. So, 0.7854 > 0.7071. Still holds. Let's try x = π/3, which is 60 degrees. sin(π/3) is sqrt(3)/2, approximately 0.8660, and π/3 is about 1.0472. So, 1.0472 > 0.8660. Yep, that's true too.Maybe I should think about why this is always true. I recall that the sine function is concave on the interval (0, π/2). That means the graph of sin(x) lies below the line connecting (0,0) and (π/2,1). Since the line connecting those two points has a slope of 2/π, which is approximately 0.6366, and the derivative of sin(x) at 0 is 1, which is steeper. So, initially, sin(x) grows faster than x, but since it's concave, it eventually slows down and stays below the line y = x.Wait, but actually, at x = 0, both sin(x) and x are 0. The derivative of sin(x) at 0 is 1, same as the derivative of x. So, they start off increasing at the same rate, but since sin(x) is concave, it starts curving downward, meaning its growth rate decreases. Therefore, for x > 0 in (0, π/2), x will always be greater than sin(x). That makes sense.So, proposition q is true because for every x in (0, π/2), x is indeed greater than sin(x). Therefore, the negation of q, which would be "There exists an x in (0, π/2) such that x ≤ sin(x)," is false. So, ¬q is false.Now, looking back at the options:A: p is true. Well, we already determined p is false because sin(x) can't be sqrt(5)/2. So, A is incorrect.B: ¬q is false. Since q is true, its negation ¬q is false. So, B is correct.C: p ∧ q is true. Since p is false and q is true, p ∧ q would be false ∧ true, which is false. So, C is incorrect.D: p ∨ q is false. Since q is true, p ∨ q is true ∨ false, which is true. So, D is incorrect.Therefore, the correct judgment is option B.**Final Answer**boxed{B}

💡First, I need to determine the speed of the stream given the boat's speed in still water and the distances covered downstream and upstream in the same time.Let's denote the speed of the stream as ( v ) kmph.When the boat is moving downstream, its effective speed is the sum of its speed in still water and the speed of the stream, which is ( 36 + v ) kmph. Conversely, when moving upstream, the effective speed is the difference between the boat's speed in still water and the speed of the stream, which is ( 36 - v ) kmph.The time taken to travel a certain distance is equal to the distance divided by the speed. Therefore, the time taken to cover 80 km downstream is ( frac{80}{36 + v} ) hours, and the time taken to cover 40 km upstream is ( frac{40}{36 - v} ) hours.According to the problem, these times are equal. This gives us the equation:[frac{80}{36 + v} = frac{40}{36 - v}]To solve for ( v ), I'll cross-multiply:[80(36 - v) = 40(36 + v)]Expanding both sides:[2880 - 80v = 1440 + 40v]Combining like terms:[2880 - 1440 = 80v + 40v][1440 = 120v]Finally, solving for ( v ):[v = frac{1440}{120} = 12]So, the speed of the stream is 12 kmph.

💡First, I need to evaluate each term in the expression separately. For (3^2 - 3), I calculate (3^2 = 9) and then subtract 3 to get 6.Next, for (4^2 - 4), I calculate (4^2 = 16) and then subtract 4 to get 12.Then, for (5^2 - 5), I calculate (5^2 = 25) and then subtract 5 to get 20.Finally, for (6^2 - 6), I calculate (6^2 = 36) and then subtract 6 to get 30.Now, I substitute these values back into the original expression:[6 + 12 - 20 + 30]I perform the addition and subtraction step by step:- (6 + 12 = 18)- (18 - 20 = -2)- (-2 + 30 = 28)So, the value of the expression is 28.

💡Okay, so I have this geometry problem here, and I need to find the length of BD in a triangle setup. Let me try to visualize and understand the problem step by step.First, there's a triangle ABC with vertices A, B, and C. Point D is on side AC. The given angles are:- Angle ADB is 90 degrees, so triangle ADB is a right-angled triangle at D.- Angle ABD is 30 degrees, which is one of the angles in triangle ABD.- Angle BAC is 45 degrees, which is the angle at vertex A in triangle ABC.Also, the length of AD is given as 6 units. I need to find the length of BD.Alright, let me try to sketch this mentally. Triangle ABC has point A, and from A, there's a point D on AC such that when connected to B, it forms a right angle at D. So, triangle ABD is right-angled at D, with angle ABD being 30 degrees. That means the other angle at A in triangle ABD must be 60 degrees because the angles in a triangle add up to 180 degrees.Wait, angle BAC is 45 degrees. So, in triangle ABC, angle at A is 45 degrees. But in triangle ABD, angle at A is 60 degrees. Hmm, that seems a bit conflicting. Maybe I need to clarify that.Wait, no. In triangle ABC, angle BAC is 45 degrees, which is the angle between sides AB and AC. But in triangle ABD, angle at A is part of triangle ABD, which is a different triangle. So, maybe angle BAD is 60 degrees, and angle BAC is 45 degrees. That would mean that angle CAD is 45 - 60 = negative 15 degrees, which doesn't make sense. Hmm, I must be making a mistake here.Wait, perhaps angle BAC is 45 degrees, which is the angle between BA and CA. So, in triangle ABC, angle at A is 45 degrees. Then, in triangle ABD, angle at A is 60 degrees because angle ABD is 30 degrees and angle ADB is 90 degrees. So, angle BAD is 60 degrees.Therefore, angle BAC is 45 degrees, which is the angle between BA and CA. But angle BAD is 60 degrees, which is the angle between BA and DA. So, point D is on AC such that angle BAD is 60 degrees, and angle BAC is 45 degrees. That would mean that angle between DA and CA is 45 - 60 = negative 15 degrees, which again doesn't make sense.Wait, maybe I'm misinterpreting the angles. Let me clarify.In triangle ABD, angle ABD is 30 degrees, angle ADB is 90 degrees, so angle BAD must be 60 degrees because 180 - 90 - 30 = 60. So, angle BAD is 60 degrees.But in triangle ABC, angle BAC is 45 degrees. So, angle BAC is the angle at A between BA and CA, which is 45 degrees. But angle BAD is 60 degrees, which is the angle between BA and DA. Therefore, point D must be located such that angle BAD is 60 degrees, but angle BAC is 45 degrees. That seems impossible because if angle BAD is 60 degrees, then angle BAC cannot be 45 degrees unless point D is beyond point C, which contradicts the given that D is on AC.Wait, maybe I'm misunderstanding the configuration. Let me try to draw it out step by step.1. Draw triangle ABC with point A at the bottom left, B at the bottom right, and C at the top, forming a triangle with angle BAC = 45 degrees.2. Point D is on AC. So, from A, going towards C, we have point D somewhere along that line.3. Connect point D to point B, forming triangle ABD.4. In triangle ABD, angle at D is 90 degrees, angle at B is 30 degrees, so angle at A must be 60 degrees.But in triangle ABC, angle at A is 45 degrees. So, angle BAD is 60 degrees, but angle BAC is 45 degrees. That would mean that point D is located such that angle BAD is 60 degrees, but since angle BAC is only 45 degrees, point D must be beyond point C, which contradicts the given that D is on AC.Hmm, this is confusing. Maybe I need to reconsider the configuration.Alternatively, perhaps angle BAC is 45 degrees, and angle BAD is 60 degrees, meaning that D is beyond C on the extension of AC. But the problem states that D is on AC, so that can't be.Wait, maybe I'm misinterpreting the angles. Let me check the problem again."Point D is on AC such that angle ADB = 90 degrees, angle ABD = 30 degrees, and angle BAC = 45 degrees."So, angle ABD is 30 degrees, which is the angle at B in triangle ABD. Angle ADB is 90 degrees, so triangle ABD is a right-angled triangle at D with angle at B being 30 degrees. Therefore, the sides of triangle ABD are in the ratio 1 : √3 : 2.Given that AD is 6 units, which is the side opposite the 30-degree angle, so AD = 6. Therefore, BD, which is opposite the 60-degree angle, should be 6√3. Wait, but that might not be correct because I need to confirm which side is which.Wait, in triangle ABD, angle at D is 90 degrees, angle at B is 30 degrees, so angle at A is 60 degrees. Therefore, side opposite 30 degrees is AD, which is 6 units. Therefore, the hypotenuse AB would be twice that, so AB = 12 units. Then, the other side BD, opposite the 60-degree angle, would be 6√3 units.But wait, in triangle ABC, angle BAC is 45 degrees. So, triangle ABC has angle at A of 45 degrees, and side AB is 12 units. If I can find the length of AC, then I can find the length of AD and DC.But wait, AD is given as 6 units, so AC = AD + DC = 6 + DC. But I don't know DC yet.Alternatively, maybe I can use the Law of Sines or Cosines in triangle ABC.Wait, let me try to find the coordinates of the points to make it clearer.Let me place point A at the origin (0,0). Since angle BAC is 45 degrees, and assuming AC is along the x-axis, then point C would be at (c,0), and point B would be somewhere in the plane such that angle BAC is 45 degrees.But since D is on AC, let's say point D is at (d,0), where d is between 0 and c.But wait, in triangle ABD, angle at D is 90 degrees, so point D cannot be on AC if AC is along the x-axis because then angle at D would be between AD and BD, but if D is on AC, which is along the x-axis, then BD would have to be vertical to make angle ADB 90 degrees. So, point D would be at (d,0), and point B would be at (d, b), making BD vertical.But then angle ABD is 30 degrees. Let me try to model this.Let me set point A at (0,0), point D at (d,0), and point B at (d, b). Then, triangle ABD is right-angled at D, with AD = d units, BD = b units, and AB = sqrt(d² + b²).Given that angle ABD is 30 degrees, which is the angle at B between AB and BD. Since BD is vertical, the angle between AB and BD is 30 degrees. So, the slope of AB would correspond to a 30-degree angle from the vertical.Wait, if BD is vertical, then the line AB makes a 30-degree angle with the vertical line BD. Therefore, the slope of AB would be tan(60 degrees) because it's the complement of 30 degrees from the vertical.Wait, tan(theta) = opposite/adjacent. If the angle between AB and BD is 30 degrees, and BD is vertical, then the horizontal component from B to A would be adjacent, and the vertical component is BD.Wait, maybe it's better to use trigonometry in triangle ABD.In triangle ABD, right-angled at D, angle at B is 30 degrees. Therefore, side opposite 30 degrees is AD, which is 6 units. Therefore, AD = 6 = opposite side, BD = adjacent side, and AB = hypotenuse.In a 30-60-90 triangle, the sides are in the ratio 1 : √3 : 2. So, opposite 30 degrees is the shortest side, which is AD = 6. Therefore, the hypotenuse AB would be twice that, so AB = 12 units. Then, the other side BD, opposite 60 degrees, would be 6√3 units.So, BD = 6√3.But wait, in triangle ABC, angle BAC is 45 degrees. So, triangle ABC has angle at A of 45 degrees, side AB = 12 units, and side AC = AD + DC = 6 + DC.But I don't know DC yet. Maybe I can find AC using the Law of Sines in triangle ABC.Wait, but I don't know any other angles or sides in triangle ABC except angle BAC = 45 degrees and side AB = 12 units.Alternatively, maybe triangle ABC is a 45-45-90 triangle, but that would require angles at B and C to be 45 degrees as well, which isn't necessarily the case here.Wait, no. Triangle ABC has angle at A of 45 degrees, but we don't know the other angles. So, perhaps I need more information.Wait, but point D is on AC, and we've already used the information about triangle ABD to find BD = 6√3. So, maybe that's all that's needed, and triangle ABC isn't necessary for finding BD.Wait, but the problem mentions triangle ABC, so perhaps I need to ensure that the configuration is consistent.Wait, if AB = 12 units, and angle BAC = 45 degrees, then in triangle ABC, using the Law of Sines, we can relate sides AB, AC, and BC.Law of Sines states that AB / sin(angle ACB) = AC / sin(angle ABC) = BC / sin(angle BAC).But we don't know angles at B and C, so that might not help directly.Alternatively, maybe using coordinates would help.Let me place point A at (0,0), point D at (6,0) since AD = 6 units. Then, since triangle ABD is right-angled at D, point B must be at (6, y) for some y. Then, angle ABD is 30 degrees.So, point B is at (6, y). Then, angle ABD is the angle at B between points A, B, and D.Wait, point A is at (0,0), point B is at (6, y), and point D is at (6,0). So, angle ABD is the angle at B between points A, B, and D.So, vectors BA and BD would form a 30-degree angle.Vector BA is from B to A: (0 - 6, 0 - y) = (-6, -y).Vector BD is from B to D: (6 - 6, 0 - y) = (0, -y).The angle between vectors BA and BD is 30 degrees.The formula for the angle between two vectors u and v is:cos(theta) = (u . v) / (|u| |v|)So, cos(30°) = [(-6)(0) + (-y)(-y)] / [sqrt((-6)^2 + (-y)^2) * sqrt(0^2 + (-y)^2)]Simplify:cos(30°) = (0 + y²) / [sqrt(36 + y²) * sqrt(y²)]cos(30°) = y² / [sqrt(36 + y²) * y]Simplify denominator:sqrt(36 + y²) * y = y * sqrt(36 + y²)So,cos(30°) = y² / (y * sqrt(36 + y²)) = y / sqrt(36 + y²)We know that cos(30°) = √3 / 2, so:√3 / 2 = y / sqrt(36 + y²)Cross-multiplying:√3 * sqrt(36 + y²) = 2ySquare both sides:3 * (36 + y²) = 4y²108 + 3y² = 4y²108 = y²So, y = sqrt(108) = 6√3Therefore, point B is at (6, 6√3). Therefore, BD is the distance from B(6,6√3) to D(6,0), which is 6√3 units.So, BD = 6√3.Wait, but earlier I thought BD was 6√3, but in the initial thought process, I thought it was 3√3. So, which one is correct?Wait, in the initial thought process, I considered AD as the hypotenuse, but actually, in triangle ABD, AD is opposite the 30-degree angle, so it's the shorter leg, making BD the longer leg, which is AD * √3 = 6√3.Yes, that makes sense. So, BD = 6√3.But wait, in the Asymptote code provided earlier, the coordinates were different. Let me check that.In the Asymptote code, point A is at (0,0), point B is at (-1,0), point C is at (sqrt(6),0), and point D is at (0,sqrt(6)). So, in that diagram, BD is from (-1,0) to (0,sqrt(6)), which would have length sqrt(1 + 6) = sqrt(7), but that's not matching our calculation.Wait, perhaps the Asymptote code is just a generic diagram and not to scale. So, maybe I shouldn't rely on it for exact measurements.Therefore, based on the calculations, BD = 6√3 units.