answer：Alright, so I have this math problem here about functions and maximum values. Let me try to wrap my head around it step by step. Okay, the problem says:"Given that the function ( f(x) ) has a domain of ( I ) and ( M ) is a constant, then 'for all ( x ) in ( I ), ( f(x) leqslant M )' is ( ) ' ( M ) is the maximum value of ( f(x) ) on ( I )'."And then there are four options: A, B, C, D, which correspond to different types of conditions—sufficient but not necessary, necessary but not sufficient, necessary and sufficient, or neither necessary nor sufficient.Hmm, okay. So, I need to figure out what kind of condition the statement 'for all ( x ) in ( I ), ( f(x) leqslant M )' is for the statement ' ( M ) is the maximum value of ( f(x) ) on ( I )'.Let me recall some definitions to make sure I'm on the right track. A condition is **sufficient** if it guarantees the truth of another statement. It's **necessary** if the other statement can only be true if this condition is met. If a condition is both necessary and sufficient, then they imply each other.So, in this case, the first statement is 'for all ( x ) in ( I ), ( f(x) leqslant M )'. The second statement is ' ( M ) is the maximum value of ( f(x) ) on ( I )'.Let me think about what it means for ( M ) to be the maximum value of ( f(x) ) on ( I ). It means two things: first, that ( f(x) leqslant M ) for all ( x ) in ( I ), and second, that there exists at least one ( x ) in ( I ) such that ( f(x) = M ). So, the maximum value is not just an upper bound; it's actually attained by the function somewhere in the domain.Now, looking back at the first statement, 'for all ( x ) in ( I ), ( f(x) leqslant M )', this only tells me that ( M ) is an upper bound for the function ( f(x) ) on the domain ( I ). It doesn't necessarily mean that ( M ) is the maximum because ( f(x) ) might never actually reach ( M ). For example, consider the function ( f(x) = 1 - x^2 ) on the domain ( mathbb{R} ). Here, ( f(x) leqslant 1 ) for all ( x ), but the maximum value is actually 1, which is attained at ( x = 0 ). However, if I take another function like ( f(x) = 1 - e^{-x} ) on ( mathbb{R} ), this function approaches 1 as ( x ) approaches infinity but never actually reaches 1. So, in this case, ( f(x) leqslant 1 ) for all ( x ), but 1 is not the maximum value because it's never attained.Wait, so does that mean that 'for all ( x ) in ( I ), ( f(x) leqslant M )' is not sufficient to conclude that ( M ) is the maximum? Because in the second example, ( M = 1 ) is an upper bound but not the maximum. So, the condition is not sufficient on its own.But is it necessary? Well, if ( M ) is the maximum value of ( f(x) ) on ( I ), then by definition, ( f(x) leqslant M ) for all ( x ) in ( I ). So, the condition 'for all ( x ) in ( I ), ( f(x) leqslant M )' must hold if ( M ) is the maximum. Therefore, it's a necessary condition.So, putting it all together, 'for all ( x ) in ( I ), ( f(x) leqslant M )' is a necessary condition for ( M ) being the maximum value of ( f(x) ) on ( I ), but it's not sufficient because it doesn't guarantee that ( M ) is actually attained by the function.Let me double-check this reasoning with another example. Suppose ( f(x) = sin(x) ) on the domain ( mathbb{R} ). The maximum value of ( sin(x) ) is 1, which is attained at ( x = pi/2 + 2pi k ) for any integer ( k ). So, in this case, ( f(x) leqslant 1 ) for all ( x ), and 1 is indeed the maximum. Now, if I take ( M = 2 ), then ( f(x) leqslant 2 ) for all ( x ), but 2 is not the maximum value of ( f(x) ). This reinforces the idea that ( f(x) leqslant M ) is necessary for ( M ) being the maximum but not sufficient.Another angle to consider: if ( M ) were not an upper bound, then it couldn't be the maximum. So, the condition is necessary. But even if it is an upper bound, unless the function actually reaches ( M ), it's not the maximum. Hence, the condition is necessary but not sufficient.I think I'm confident with this conclusion now.

answer：Alright, so I've got this problem here about a boat crossing a river. Let me try to understand it step by step. The boat's maximum speed is 4 km/h, and the river's current is 3 km/h. The shortest time it takes for the boat to reach the opposite bank is 2 hours. I need to find the width of the river.First, I need to visualize the scenario. There's a river flowing with a current speed of 3 km/h. The boat is trying to cross this river, and its maximum speed is 4 km/h. I assume that the boat's speed is relative to the water, not the ground. So, if the boat is moving directly across the river, the current will carry it downstream.But wait, the problem mentions the shortest time to reach the opposite bank. That makes me think that the boat isn't necessarily trying to go straight across; instead, it might be heading upstream at an angle to counteract the current. That way, it can reach the point directly opposite without being carried downstream. Hmm, but if it's going upstream, wouldn't that take longer? Or maybe not, because it's trying to minimize the time, not the distance.Let me think about how to approach this. I remember from physics that when dealing with vectors, like the boat's velocity and the river's current, I can use vector addition to find the resultant velocity relative to the ground. The boat's velocity relative to the water is 4 km/h, and the river's current is 3 km/h downstream.If the boat wants to reach the opposite bank in the shortest time, it should aim upstream at an angle such that the component of its velocity upstream cancels out the river's current. That way, the boat's resultant velocity relative to the ground is directly across the river. This would mean that the boat's upstream component equals the river's downstream current.So, let's break down the boat's velocity into two components: one upstream (against the current) and one across the river. The upstream component needs to be equal to the river's current to cancel it out. The boat's speed is 4 km/h, so if we let θ be the angle upstream from the direction directly across the river, then the upstream component is 4 sin θ, and the across component is 4 cos θ.To cancel out the river's current of 3 km/h, we set 4 sin θ = 3. Solving for θ, we get sin θ = 3/4, so θ = arcsin(3/4). That gives us the angle the boat needs to head upstream.Now, the across component of the boat's velocity is 4 cos θ. Since sin θ = 3/4, we can find cos θ using the Pythagorean identity: cos θ = sqrt(1 - sin² θ) = sqrt(1 - 9/16) = sqrt(7/16) = sqrt7 / 4.So, the across component is 4 * (sqrt7 / 4) = sqrt7 km/h. That's the boat's effective speed across the river.Given that the shortest time to cross is 2 hours, we can find the width of the river by multiplying the across speed by time: width = sqrt7 * 2 ≈ 2 * 2.6458 ≈ 5.2915 km.Wait, but the problem mentions the shortest time, and I've calculated the width based on the effective across speed. Is there another way to interpret this? Maybe the boat isn't heading upstream at all and is just going straight across, letting the current carry it downstream, but that would take longer. But the problem specifies the shortest time, so I think my initial approach is correct.Alternatively, if the boat heads straight across without compensating for the current, its across speed would be 4 km/h, and the downstream drift would be 3 km/h. But in that case, the time to cross would be width / 4, and the downstream distance would be 3 * (width / 4). However, since the problem asks for the shortest time, the boat should aim upstream to cancel the current, resulting in a shorter crossing time.Wait, but if the boat aims upstream, it's actually reducing its across speed because part of its velocity is used to counteract the current. So, the across speed is sqrt7 km/h, which is approximately 2.6458 km/h, and the time is 2 hours, so the width is approximately 5.2915 km.But the problem states that the shortest time is 2 hours, so maybe I need to consider that the boat's maximum speed is 4 km/h relative to the water, and the river's current is 3 km/h. If the boat heads straight across, its speed relative to the ground is still 4 km/h across, because the current only affects the downstream drift, not the across speed. Wait, is that correct?Let me clarify. The boat's velocity relative to the water is 4 km/h. The water is moving downstream at 3 km/h. So, the boat's velocity relative to the ground is the vector sum of its velocity relative to the water and the water's velocity relative to the ground.If the boat heads straight across, its velocity relative to the ground would have two components: 4 km/h across and 3 km/h downstream. So, the resultant speed relative to the ground is sqrt(4² + 3²) = 5 km/h, but that's the speed relative to the ground, not the across speed.Wait, no. The across speed relative to the ground is still 4 km/h because the current doesn't affect the across component. The downstream component is 3 km/h, but that doesn't affect the time to cross the river, which depends only on the across speed.So, if the boat heads straight across, its across speed is 4 km/h, and the time to cross is width / 4. If the shortest time is 2 hours, then width = 4 * 2 = 8 km.But earlier, I thought that the boat should aim upstream to cancel the current, resulting in a shorter crossing time. But if the boat aims upstream, it's actually reducing its across speed because part of its velocity is used to counteract the current. So, the across speed becomes sqrt(4² - 3²) = sqrt(7) km/h, which is approximately 2.6458 km/h. Then, the time to cross would be width / sqrt7, and if that's 2 hours, then width = 2 * sqrt7 ≈ 5.2915 km.But which interpretation is correct? The problem says the shortest time to reach the opposite bank. If the boat aims upstream, it can reach the point directly opposite in 2 hours, but if it heads straight across, it would take longer because it's being carried downstream, but the time to cross is determined by the across speed.Wait, no. The time to cross is determined by the across speed relative to the ground. If the boat heads straight across, its across speed is 4 km/h, so time is width / 4. If it aims upstream, its across speed is sqrt7 km/h, so time is width / sqrt7. To minimize the time, the boat should aim upstream to increase the across speed? Wait, no, aiming upstream reduces the across speed because part of the velocity is used to counteract the current.Wait, I'm getting confused. Let me think again.If the boat aims upstream at an angle θ, its velocity relative to the water is 4 km/h at angle θ upstream. The water is moving downstream at 3 km/h. So, the boat's velocity relative to the ground is (4 cos θ, 4 sin θ - 3), where the first component is across and the second is downstream.To reach the opposite bank directly, the downstream component must be zero, so 4 sin θ - 3 = 0, which gives sin θ = 3/4, as I thought earlier. Then, the across component is 4 cos θ = 4 * sqrt(1 - (9/16)) = 4 * sqrt(7/16) = sqrt7 km/h.So, the across speed is sqrt7 km/h, and the time to cross is width / sqrt7 = 2 hours, so width = 2 * sqrt7 ≈ 5.2915 km.But earlier, I thought that if the boat heads straight across, its across speed is 4 km/h, and time is width / 4 = 2 hours, so width = 8 km.Which one is correct? The problem says the shortest time to reach the opposite bank. If the boat aims upstream, it can reach the point directly opposite in 2 hours, but if it heads straight across, it would take longer because it's being carried downstream, but the time to cross is determined by the across speed.Wait, no. The time to cross is determined by the across speed relative to the ground. If the boat aims upstream, it's reducing its across speed, so it would take longer to cross. But the problem says the shortest time is 2 hours, so the boat must be heading straight across, maximizing its across speed, which is 4 km/h, resulting in width = 4 * 2 = 8 km.But then, why does the current matter? If the boat is heading straight across, the current only affects the downstream drift, not the time to cross. So, the width is 8 km.Wait, but if the boat aims upstream, it can reach the opposite bank in less time because it's not being carried downstream, but in reality, aiming upstream reduces the across speed, so it takes longer. Therefore, to minimize the time, the boat should head straight across, maximizing the across speed, resulting in width = 8 km.So, I think the correct answer is 8 km.

answer：Alright, so I have this geometry problem here about triangle XYZ. The problem says that in triangle XYZ, the altitudes XK and YL intersect at point W. I'm given that angle XYZ is 30 degrees and angle XZY is 80 degrees. I need to find the measure of angle XWY. Hmm, okay, let me try to figure this out step by step.First, I remember that in any triangle, the sum of the interior angles is 180 degrees. So, if I know two angles, I can find the third one. Here, I know angle XYZ is 30 degrees and angle XZY is 80 degrees. Let me write that down:Angle XYZ = 30°Angle XZY = 80°So, to find angle YXZ, which is the third angle at vertex X, I can subtract the sum of the other two angles from 180 degrees.Angle YXZ = 180° - 30° - 80° = 70°Okay, so angle YXZ is 70 degrees. That's good to know.Now, the problem mentions altitudes XK and YL intersecting at point W. I remember that in a triangle, the point where all three altitudes intersect is called the orthocenter. So, W is the orthocenter of triangle XYZ.I need to find angle XWY. Let me visualize triangle XYZ with the orthocenter W inside it. Since W is the orthocenter, both XK and YL are altitudes, meaning they are perpendicular to the opposite sides.So, XK is perpendicular to YZ, and YL is perpendicular to XZ. That means angles XKZ and YLZ are both 90 degrees.Wait, but I need to find angle XWY. Let me think about how to approach this. Maybe I can use some properties of the orthocenter or cyclic quadrilaterals?I recall that in a triangle, the orthocenter creates several cyclic quadrilaterals. For example, the quadrilateral formed by the orthocenter and two vertices of the triangle is cyclic. Is that correct?Let me think. If I consider quadrilateral XWYL, since both XK and YL are altitudes, angles XKW and YLW are right angles (90 degrees). So, quadrilateral XWYL has two opposite right angles. Hmm, does that make it cyclic? Wait, a quadrilateral with two opposite right angles doesn't necessarily make it cyclic, but maybe there's another way.Alternatively, maybe I can consider triangle XWY and use some triangle properties. Let me try to find some angles in triangle XWY.I know that angle XWY is the angle at point W between points X and Y. Since W is the orthocenter, I can relate this angle to the angles of the original triangle.I remember that in the orthocenter configuration, the angles at the orthocenter are related to the original angles of the triangle. Specifically, the measure of angle XWY is equal to 180 degrees minus the measure of angle YXZ.Wait, is that correct? Let me verify. If I consider the orthocenter, the angles formed at the orthocenter are supplementary to the angles of the original triangle. So, angle XWY should be equal to 180° minus angle YXZ.Since angle YXZ is 70°, then angle XWY would be 180° - 70° = 110°. Is that right?Let me think again to make sure I'm not making a mistake. So, angle XWY is at the orthocenter between points X and Y. The original angle at X is 70°, and since the orthocenter creates supplementary angles, angle XWY should indeed be 180° - 70° = 110°.Alternatively, I can think about the quadrilateral formed by points X, W, Y, and L. Since both XK and YL are altitudes, angles XKW and YLW are 90°, as I mentioned earlier. So, in quadrilateral XWYL, two of the angles are 90°, and the angle at X is 70°. If I add up all the angles in a quadrilateral, they should sum to 360°.So, angle XWY + angle XKW + angle YLW + angle YXZ = 360°Plugging in the known values:Angle XWY + 90° + 90° + 70° = 360°Adding up the known angles: 90 + 90 + 70 = 250°So, angle XWY = 360° - 250° = 110°Yes, that confirms it. So, angle XWY is 110 degrees.I think that makes sense. The orthocenter creates angles that are supplementary to the original angles of the triangle. So, since angle YXZ was 70°, angle XWY is 110°. I can't think of any mistakes in this reasoning, so I'm pretty confident that the answer is 110 degrees.

answer：Let's determine the number of children in the Alvez family step by step.1. **Define Variables:** - Let ( m ) be the age of the mother. - Let ( x ) be the number of children. - Let ( y ) be the average age of the children. Therefore, the total age of the children is ( xy ).2. **Set Up Equations Based on Given Information:** - **Average Age of the Family:** [ frac{m + 50 + xy}{2 + x} = 22 ] Multiply through by ( 2 + x ): [ m + 50 + xy = 44 + 22x ] (Equation 1) - **Average Age of the Mother and Children:** [ frac{m + xy}{1 + x} = 15 ] Multiply through by ( 1 + x ): [ m + xy = 15 + 15x ] (Equation 2)3. **Solve the System of Equations:** - Subtract Equation 2 from Equation 1 to eliminate ( m + xy ): [ (m + 50 + xy) - (m + xy) = (44 + 22x) - (15 + 15x) ] Simplify: [ 50 = 29 + 7x ] [ 7x = 21 ] [ x = 3 ]Thus, there are three children in the Alvez family.

answer：Okay, so I have this problem where I need to find the distance from point D to the plane ABC. The coordinates given are A(0,0,2), B(0,2,1), C(2,1,0), and D(2,0,1). The options are A: 2√145/145, B: 2√5/25, C: 2√29/29, and D: 2/5. Hmm, let me think about how to approach this.First, I remember that the distance from a point to a plane can be found using the formula:Distance = |(vector from a point on the plane to the given point) · normal vector of the plane| / |normal vector|So, I need to find the normal vector of the plane ABC. To do that, I can use two vectors lying on the plane and take their cross product. Let me choose vectors AB and AC.Calculating vector AB: B - A = (0,2,1) - (0,0,2) = (0,2,-1)Calculating vector AC: C - A = (2,1,0) - (0,0,2) = (2,1,-2)Now, I need to find the cross product of AB and AC to get the normal vector n.The cross product formula is:n = AB × AC = |i   j   k|             0   2   -1             2   1   -2Calculating the determinant:i*(2*(-2) - (-1)*1) - j*(0*(-2) - (-1)*2) + k*(0*1 - 2*2)= i*(-4 + 1) - j*(0 + 2) + k*(0 - 4)= i*(-3) - j*(2) + k*(-4)So, n = (-3, -2, -4)Wait, but I think the direction of the normal vector doesn't matter for the distance formula, so I can also take (3, 2, 4) as the normal vector.Now, I need a point on the plane ABC to write the equation of the plane. Let's use point A(0,0,2). The general equation of the plane is:n · (r - A) = 0Where r = (x, y, z). Plugging in the normal vector and point A:(3, 2, 4) · (x - 0, y - 0, z - 2) = 03x + 2y + 4(z - 2) = 03x + 2y + 4z - 8 = 0So, the equation of the plane is 3x + 2y + 4z - 8 = 0.Now, to find the distance from point D(2,0,1) to this plane, I can use the distance formula:Distance = |3*2 + 2*0 + 4*1 - 8| / sqrt(3^2 + 2^2 + 4^2)= |6 + 0 + 4 - 8| / sqrt(9 + 4 + 16)= |2| / sqrt(29)= 2 / sqrt(29)To rationalize the denominator, multiply numerator and denominator by sqrt(29):= 2*sqrt(29) / 29Looking at the options, that's option C: 2√29/29.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, vectors AB and AC:AB = (0,2,-1), AC = (2,1,-2). Cross product:i*(2*(-2) - (-1)*1) = i*(-4 +1) = -3ij*(0*(-2) - (-1)*2) = j*(0 +2) = 2j, but since it's subtracted, it's -2jk*(0*1 - 2*2) = k*(0 -4) = -4kSo, n = (-3, -2, -4). Taking the absolute values, n = (3,2,4). That seems correct.Plane equation using A(0,0,2):3x + 2y + 4z + D = 0. Plugging in A: 0 + 0 + 8 + D = 0 => D = -8. So, 3x + 2y + 4z -8 =0. Correct.Distance from D(2,0,1):|3*2 + 2*0 +4*1 -8| / sqrt(9+4+16) = |6 +0 +4 -8| / sqrt(29) = |2| / sqrt(29) = 2/sqrt(29) = 2√29/29. That's option C.I think that's correct. Maybe I should check if there's another way to compute it, like using vectors from D to the plane.Alternatively, I could parametrize the line from D perpendicular to the plane and find where it intersects the plane, then compute the distance. But that might be more complicated.Alternatively, using the formula for the distance from a point to a plane, which I did. So, I think my answer is correct.**Final Answer**boxed{C}

answer：First, I need to determine the amount the user keeps as profit. They earn 50 in a day and want to keep 10% of this profit for themselves.Calculating 10% of 50:10% of 50 = 0.10 × 50 = 5Next, I subtract the amount kept as profit from the total profit to find out how much is left to distribute among the employees:50 - 5 = 45Finally, I divide the remaining amount equally among the 9 employees to find out how much each employee would receive:45 ÷ 9 employees = 5 per employeeTherefore, each employee would receive 5.