answer：Alright, so I've got this geometry problem here, and I need to prove that four points form a trapezoid. Let me try to unpack the problem step by step.First, let's parse the given information:- There's a circle ( k ) with center ( M ).- ( t ) is a tangent line to ( k ) at point ( T ).- Point ( P ) is on this tangent line ( t ).- There's another line ( g ) (not equal to ( t )) passing through ( P ) and intersecting the circle ( k ) at points ( U ) and ( V ).- ( S ) is the point on ( k ) that bisects the arc ( UV ) not containing ( T ).- ( Q ) is the reflection of ( P ) over the line ( ST ).We need to prove that ( Q ), ( T ), ( U ), and ( V ) form a trapezoid. Remember, a trapezoid is a quadrilateral with at least one pair of parallel sides.Okay, so let's try to visualize this. I imagine a circle with center ( M ). The tangent ( t ) touches the circle at ( T ). Point ( P ) is somewhere on this tangent. Then, from ( P ), we draw another line ( g ) that cuts the circle at ( U ) and ( V ). The point ( S ) is the midpoint of the arc ( UV ) that doesn't include ( T ). Then, reflecting ( P ) over ( ST ) gives us ( Q ).I think it might help to draw a diagram. Since I can't draw here, I'll try to imagine it and describe it.First, the circle ( k ), center ( M ). Tangent ( t ) at ( T ). Point ( P ) on ( t ). Line ( g ) through ( P ) intersects ( k ) at ( U ) and ( V ). So, ( g ) is a secant. Then, ( S ) is the midpoint of arc ( UV ) not containing ( T ). So, ( S ) is opposite to ( T ) in some sense.Reflecting ( P ) over ( ST ) gives ( Q ). So, ( Q ) is such that ( ST ) is the perpendicular bisector of segment ( PQ ). That means ( Q ) is as far from ( ST ) as ( P ) is, but on the opposite side.Now, to show that ( Q ), ( T ), ( U ), and ( V ) form a trapezoid. So, we need to show that either ( QT parallel UV ) or ( QU parallel TV ), or maybe ( QV parallel TU ). But since ( U ) and ( V ) are on the circle, and ( T ) is a point of tangency, perhaps there's some symmetry or parallelism we can exploit.Let me recall some properties:1. The tangent at ( T ) is perpendicular to the radius ( MT ).2. The reflection over ( ST ) will preserve distances and angles, so ( Q ) is a mirror image of ( P ) across ( ST ).3. Since ( S ) is the midpoint of arc ( UV ), the line ( ST ) might have some symmetrical properties with respect to ( U ) and ( V ).Maybe I can use power of a point. The power of point ( P ) with respect to circle ( k ) is ( PT^2 = PU cdot PV ). Since ( P ) is on the tangent, this is true.Also, since ( S ) is the midpoint of arc ( UV ), line ( ST ) is the angle bisector of angle ( UTV ). Hmm, not sure if that's directly useful.Wait, maybe looking at angles. Since ( S ) is the midpoint of arc ( UV ), then ( SU = SV ) in terms of arc length, so ( angle USV = 2angle UTV ). But I'm not sure.Alternatively, since ( Q ) is the reflection of ( P ) over ( ST ), then ( ST ) is the perpendicular bisector of ( PQ ). So, ( SP = SQ ) and ( TP = TQ ). Wait, no, reflection over a line doesn't necessarily mean that ( SP = SQ ), but rather that ( ST ) is the axis of reflection.Wait, actually, reflecting ( P ) over ( ST ) gives ( Q ), so ( ST ) is the perpendicular bisector of segment ( PQ ). Therefore, ( SP = SQ ) and ( TP = TQ ). Hmm, not necessarily, because reflection over a line doesn't imply equal distances from the center unless the line is a diameter.Wait, maybe I need to think about triangles. Let's consider triangle ( PST ) and its reflection over ( ST ) to get triangle ( QST ). So, ( Q ) is such that ( ST ) is the axis, so ( Q ) is the mirror image of ( P ).Therefore, angles formed by ( P ) and ( Q ) with respect to ( ST ) are equal. So, ( angle PST = angle QST ).Hmm, maybe we can use some cyclic quadrilateral properties here.Alternatively, since ( S ) is the midpoint of arc ( UV ), the line ( ST ) might be related to the angle bisector or something.Wait, another idea: maybe we can show that ( QT ) is parallel to ( UV ). If that's the case, then ( Q ), ( T ), ( U ), ( V ) would form a trapezoid.To show ( QT parallel UV ), we can show that the angles formed by a transversal are equal. For example, if we can show that ( angle QTV = angle UVT ), then by the converse of the corresponding angles theorem, ( QT parallel UV ).Alternatively, maybe using power of a point or similar triangles.Wait, let's consider the reflection. Since ( Q ) is the reflection of ( P ) over ( ST ), then ( PQ ) is perpendicular to ( ST ), and ( ST ) bisects ( PQ ). So, ( PQ perp ST ), and the midpoint of ( PQ ) lies on ( ST ).Given that, perhaps we can find some similar triangles involving ( P ), ( Q ), ( U ), ( V ), ( T ), ( S ).Wait, another approach: maybe using inversion. But that might be too advanced for this problem.Alternatively, let's consider the polar of point ( P ) with respect to circle ( k ). Since ( P ) lies on the tangent ( t ), its polar line is the tangent itself. But I'm not sure if that helps.Wait, maybe harmonic division or projective geometry concepts, but again, perhaps overcomplicating.Let me try to think about angles. Since ( S ) is the midpoint of arc ( UV ), then ( SU = SV ) as arcs, so ( angle SUT = angle SVT ). Also, since ( S ) is the midpoint, ( ST ) is the angle bisector of ( angle UTV ).Wait, since ( S ) is the midpoint of arc ( UV ), then ( ST ) is the bisector of angle ( UTV ). So, ( angle UTS = angle VTS ).Hmm, maybe we can use this to find some congruent angles.Also, since ( Q ) is the reflection of ( P ) over ( ST ), then ( angle QST = angle PST ). So, perhaps triangles ( QST ) and ( PST ) are congruent.Wait, but ( ST ) is the axis of reflection, so reflecting ( P ) over ( ST ) gives ( Q ). Therefore, ( SP = SQ ) and ( TP = TQ ). Wait, is that true? If ( ST ) is the axis, then the distances from ( P ) and ( Q ) to ( ST ) are equal, but not necessarily the distances from ( S ) or ( T ).Wait, no. Reflection over a line preserves distances. So, the distance from ( P ) to ( S ) is the same as from ( Q ) to ( S ), and similarly for ( T ). So, ( SP = SQ ) and ( TP = TQ ). That's correct.So, ( SP = SQ ) and ( TP = TQ ). Therefore, triangles ( SPT ) and ( SQT ) are congruent by SSS, since ( SP = SQ ), ( TP = TQ ), and ( ST ) is common.Therefore, ( angle SPT = angle SQT ).Hmm, interesting. So, the angles at ( P ) and ( Q ) in triangles ( SPT ) and ( SQT ) are equal.Now, let's consider the line ( g ) passing through ( P ) and intersecting the circle at ( U ) and ( V ). So, ( PU ) and ( PV ) are segments of this secant.Since ( S ) is the midpoint of arc ( UV ), perhaps we can use the fact that ( SU ) and ( SV ) are equal in length, or that angles subtended by ( SU ) and ( SV ) are equal.Wait, another idea: maybe we can use the fact that ( ST ) is the angle bisector of ( angle UTV ), and since ( Q ) is the reflection of ( P ), maybe ( QT ) is parallel to ( UV ).Wait, let's try to see if ( QT ) is parallel to ( UV ). For that, we need to show that the angles formed by a transversal are equal.Consider line ( TV ) as a transversal cutting ( QT ) and ( UV ). If we can show that ( angle QTV = angle UVT ), then ( QT parallel UV ).Alternatively, consider line ( TU ) as a transversal cutting ( QT ) and ( UV ). If ( angle QTU = angle UVU ), then ( QT parallel UV ).Wait, maybe using the alternate interior angles theorem.Alternatively, since ( S ) is the midpoint of arc ( UV ), the line ( ST ) is the angle bisector of ( angle UTV ). So, ( angle UTS = angle VTS ).Given that, and knowing that ( Q ) is the reflection of ( P ) over ( ST ), perhaps we can relate angles at ( Q ) and ( P ).Wait, let's consider triangle ( PTV ). Since ( P ) is on tangent ( t ), and ( TV ) is a chord, then ( angle PTV = angle TVU ) because of the tangent-chord angle theorem.Wait, actually, the angle between tangent and chord is equal to the angle in the alternate segment. So, ( angle PTV = angle TVU ).Similarly, since ( Q ) is the reflection of ( P ) over ( ST ), perhaps ( angle QTV = angle PTV ), which would then equal ( angle TVU ), implying ( QT parallel UV ).Wait, let's see:Since ( Q ) is the reflection of ( P ) over ( ST ), then ( angle QTS = angle PTS ). Because reflection preserves angles.But ( angle PTS = angle PTV ) because ( S ) lies on ( ST ).Wait, no, ( S ) is on the circle, so ( ST ) is a radius? Wait, no, ( S ) is a point on the circle, so ( MS ) is a radius, but ( ST ) is not necessarily a radius unless ( S ) and ( T ) are endpoints of a diameter, which they aren't because ( S ) is the midpoint of arc ( UV ) not containing ( T ).Wait, but ( S ) is the midpoint of arc ( UV ), so ( ST ) is the bisector of angle ( UTV ). So, ( angle UTS = angle VTS ).Given that, and since ( Q ) is the reflection of ( P ) over ( ST ), then ( angle QTS = angle PTS ).But ( angle PTS ) is equal to ( angle PTV ) because ( S ) lies on ( ST ), which is the angle bisector.Wait, maybe I'm getting confused here.Let me try to write down the angles step by step.1. Since ( S ) is the midpoint of arc ( UV ), ( ST ) bisects ( angle UTV ). So, ( angle UTS = angle VTS ).2. ( Q ) is the reflection of ( P ) over ( ST ). Therefore, ( ST ) is the perpendicular bisector of ( PQ ). So, ( PQ perp ST ), and the midpoint of ( PQ ) lies on ( ST ).3. Therefore, ( angle QST = angle PST ) because reflection preserves angles.4. Also, ( SP = SQ ) and ( TP = TQ ) because reflection preserves distances.Now, let's consider triangle ( PTV ). The angle ( angle PTV ) is equal to the angle in the alternate segment, which is ( angle TVU ). So, ( angle PTV = angle TVU ).Similarly, since ( Q ) is the reflection of ( P ) over ( ST ), the angle ( angle QTV ) should be equal to ( angle PTV ) because reflection preserves angles. Therefore, ( angle QTV = angle PTV = angle TVU ).Therefore, ( angle QTV = angle TVU ). This implies that ( QT parallel UV ) because the corresponding angles are equal when cut by transversal ( TV ).Hence, ( QT parallel UV ), which means that quadrilateral ( QTVU ) is a trapezoid.Wait, but the problem states that ( Q ), ( T ), ( U ), and ( V ) are the vertices. So, the trapezoid is ( Q ), ( T ), ( U ), ( V ), with ( QT parallel UV ).Therefore, we've shown that ( QT ) is parallel to ( UV ), so ( Q ), ( T ), ( U ), ( V ) form a trapezoid.I think that makes sense. Let me just recap:- ( S ) is the midpoint of arc ( UV ), so ( ST ) bisects ( angle UTV ).- Reflecting ( P ) over ( ST ) gives ( Q ), preserving angles and distances.- Using the tangent-chord angle theorem, ( angle PTV = angle TVU ).- Since reflection preserves angles, ( angle QTV = angle PTV ).- Therefore, ( angle QTV = angle TVU ), implying ( QT parallel UV ).- Hence, ( Q ), ( T ), ( U ), ( V ) form a trapezoid.Yeah, that seems to hold up. I think I've got it!

answer：Okay, so I have this problem about arithmetic sequences and another sequence defined based on it. Let me try to understand what's given and what I need to find.First, there's an arithmetic sequence {a_n}. I know that in an arithmetic sequence, each term increases by a constant difference. So, if I denote the first term as a_1 and the common difference as d, then the nth term can be written as a_n = a_1 + (n-1)d.Then, there's another sequence {b_n} defined in a piecewise manner. For odd n, b_n is a_n minus 6, and for even n, it's twice a_n. So, b_n = { a_n - 6, if n is odd; 2a_n, if n is even }.They also mention S_n and T_n, which are the sums of the first n terms of {a_n} and {b_n}, respectively. We are given that S_4 = 32 and T_3 = 16. The problem has two parts: first, find the general formula for {a_n}, and second, prove that for n > 5, T_n > S_n.Starting with part (1): Finding the general formula for {a_n}.Since {a_n} is an arithmetic sequence, as I mentioned, a_n = a_1 + (n-1)d. So, I need to find a_1 and d.Given S_4 = 32. The sum of the first n terms of an arithmetic sequence is given by S_n = n/2 * (2a_1 + (n-1)d). So, plugging n=4:S_4 = 4/2 * (2a_1 + 3d) = 2*(2a_1 + 3d) = 4a_1 + 6d = 32.So, equation (1): 4a_1 + 6d = 32.Simplify this equation: divide both sides by 2: 2a_1 + 3d = 16.Now, the other condition is T_3 = 16. T_n is the sum of the first n terms of {b_n}. Let's compute T_3.Since n=3, which is odd, so the terms are:b_1 = a_1 - 6,b_2 = 2a_2,b_3 = a_3 - 6.So, T_3 = b_1 + b_2 + b_3 = (a_1 - 6) + 2a_2 + (a_3 - 6).Let me write this as:T_3 = a_1 - 6 + 2a_2 + a_3 - 6 = (a_1 + 2a_2 + a_3) - 12.Now, since {a_n} is arithmetic, a_2 = a_1 + d, and a_3 = a_1 + 2d.So, substitute these into the expression:T_3 = (a_1 + 2(a_1 + d) + (a_1 + 2d)) - 12.Simplify inside the parentheses:a_1 + 2a_1 + 2d + a_1 + 2d = (1 + 2 + 1)a_1 + (2d + 2d) = 4a_1 + 4d.So, T_3 = 4a_1 + 4d - 12 = 16.Therefore, equation (2): 4a_1 + 4d - 12 = 16.Simplify equation (2): 4a_1 + 4d = 28. Divide both sides by 4: a_1 + d = 7.So, now we have two equations:1) 2a_1 + 3d = 16,2) a_1 + d = 7.Let me solve this system of equations. From equation (2), a_1 = 7 - d.Substitute this into equation (1):2*(7 - d) + 3d = 16,14 - 2d + 3d = 16,14 + d = 16,d = 2.Then, from equation (2), a_1 + 2 = 7, so a_1 = 5.So, the first term is 5, and the common difference is 2. Therefore, the general formula for {a_n} is:a_n = a_1 + (n - 1)d = 5 + (n - 1)*2 = 5 + 2n - 2 = 2n + 3.So, a_n = 2n + 3.Let me double-check this. If a_n = 2n + 3, then:a_1 = 2*1 + 3 = 5,a_2 = 2*2 + 3 = 7,a_3 = 2*3 + 3 = 9,a_4 = 2*4 + 3 = 11.Compute S_4: 5 + 7 + 9 + 11 = 32. Correct.Compute T_3: b_1 = a_1 - 6 = 5 - 6 = -1,b_2 = 2a_2 = 2*7 = 14,b_3 = a_3 - 6 = 9 - 6 = 3.So, T_3 = -1 + 14 + 3 = 16. Correct.Great, so part (1) is solved.Now, part (2): Prove that for n > 5, T_n > S_n.Hmm. So, I need to compare the sum of the first n terms of {b_n} and {a_n} for n > 5.First, let's recall that S_n is the sum of the first n terms of {a_n}, which is an arithmetic sequence. So, S_n = n/2*(2a_1 + (n - 1)d). Since a_1 = 5 and d = 2, S_n = n/2*(10 + 2(n - 1)) = n/2*(10 + 2n - 2) = n/2*(2n + 8) = n(n + 4).So, S_n = n(n + 4).Now, let's compute T_n, the sum of the first n terms of {b_n}. Since {b_n} is defined differently for odd and even n, we need to consider whether n is odd or even.But since n > 5, n can be either even or odd. So, perhaps we can express T_n in terms of n, considering the number of odd and even terms.Alternatively, maybe we can find a general formula for T_n.Let me think.Each term b_k is defined as:- If k is odd: b_k = a_k - 6,- If k is even: b_k = 2a_k.So, for each term, depending on whether k is odd or even, we have a different expression.Therefore, to compute T_n, we can separate the sum into odd and even terms.Let me denote m as the number of odd terms in the first n terms. If n is even, then m = n/2. If n is odd, m = (n + 1)/2.Similarly, the number of even terms is n - m.So, T_n can be written as:T_n = sum_{odd k} (a_k - 6) + sum_{even k} (2a_k).Which is equal to:sum_{odd k} a_k - 6*m + sum_{even k} 2a_k.So, T_n = (sum_{odd k} a_k + sum_{even k} 2a_k) - 6m.Now, let's compute sum_{odd k} a_k and sum_{even k} a_k.Since {a_n} is an arithmetic sequence, the odd-indexed terms form another arithmetic sequence, as do the even-indexed terms.Specifically, the odd terms are a_1, a_3, a_5, ..., and the even terms are a_2, a_4, a_6, ...Each of these is also an arithmetic sequence. Let's find their common differences.The original sequence {a_n} has a common difference d = 2.The odd terms: a_1, a_3, a_5, ... The difference between consecutive terms is a_3 - a_1 = (5 + 2*2) - 5 = 4, which is 2d. So, the common difference for odd terms is 4.Similarly, the even terms: a_2, a_4, a_6, ... The difference is a_4 - a_2 = (5 + 2*3) - (5 + 2*1) = 11 - 7 = 4, which is also 2d. So, the common difference for even terms is 4.Therefore, both the odd and even subsequences have a common difference of 4.Now, let's compute sum_{odd k} a_k and sum_{even k} a_k.First, for the odd terms:If n is even, say n = 2p, then the number of odd terms is p, and the last odd term is a_{2p - 1}.Similarly, if n is odd, say n = 2p + 1, then the number of odd terms is p + 1, and the last odd term is a_{2p + 1}.Similarly for the even terms.But perhaps it's easier to express in terms of m, the number of odd terms.Wait, maybe I can find a general formula for sum_{odd k} a_k and sum_{even k} a_k in terms of n.Alternatively, perhaps I can express T_n in terms of S_n.Wait, let's see.Since T_n = sum_{odd k} (a_k - 6) + sum_{even k} (2a_k).Which is equal to sum_{odd k} a_k + sum_{even k} 2a_k - 6m.But sum_{odd k} a_k + sum_{even k} a_k = S_n.So, T_n = S_n + sum_{even k} a_k - 6m.Because sum_{even k} 2a_k = sum_{even k} a_k + sum_{even k} a_k.So, T_n = S_n + sum_{even k} a_k - 6m.Alternatively, T_n = S_n + sum_{even k} a_k - 6m.Hmm, maybe that's a helpful expression.Let me denote E_n as the sum of even-indexed terms in {a_n} up to n terms.Similarly, O_n as the sum of odd-indexed terms.So, S_n = O_n + E_n.Then, T_n = (O_n - 6m) + 2E_n = O_n + 2E_n - 6m.But since S_n = O_n + E_n, then T_n = S_n + E_n - 6m.So, T_n = S_n + E_n - 6m.Therefore, T_n - S_n = E_n - 6m.So, to prove that T_n > S_n for n > 5, it suffices to show that E_n - 6m > 0 for n > 5.So, if I can show that E_n > 6m, then T_n > S_n.So, let's compute E_n and m.First, m is the number of odd terms in the first n terms.If n is even, m = n/2.If n is odd, m = (n + 1)/2.Similarly, the number of even terms is n - m.So, let's compute E_n, the sum of even-indexed terms.The even-indexed terms form an arithmetic sequence with first term a_2 = 7, common difference 4, and number of terms equal to the number of even terms in the first n terms.Similarly, the sum of an arithmetic sequence is given by (number of terms)/2 * (2*first term + (number of terms - 1)*common difference).So, for E_n:If n is even, number of even terms is n/2.So, E_n = (n/2)/2 * [2*7 + (n/2 - 1)*4] = (n/4)*(14 + 2n - 4) = (n/4)*(2n + 10) = (n/4)*2(n + 5) = (n/2)(n + 5).If n is odd, number of even terms is (n - 1)/2.So, E_n = [(n - 1)/2]/2 * [2*7 + ((n - 1)/2 - 1)*4] = [(n - 1)/4]*(14 + (n - 3)*2) = [(n - 1)/4]*(14 + 2n - 6) = [(n - 1)/4]*(2n + 8) = [(n - 1)/4]*2(n + 4) = [(n - 1)/2](n + 4).So, summarizing:If n is even, E_n = (n/2)(n + 5).If n is odd, E_n = [(n - 1)/2](n + 4).Similarly, m is:If n is even, m = n/2.If n is odd, m = (n + 1)/2.So, let's compute E_n - 6m for both cases.Case 1: n is even.Let n = 2p, where p is an integer >=1.Then, E_n = (2p/2)(2p + 5) = p*(2p + 5).m = 2p/2 = p.So, E_n - 6m = p*(2p + 5) - 6p = 2p^2 + 5p - 6p = 2p^2 - p.We need to show that 2p^2 - p > 0 for p >=3 (since n >5, so p >2.5, so p >=3).2p^2 - p = p(2p -1). For p >=1, this is positive because both p and 2p -1 are positive.But specifically, for p >=3, 2p^2 - p >= 2*9 -3 = 18 -3 =15 >0.So, for even n >5, E_n -6m >0.Case 2: n is odd.Let n = 2p +1, where p is an integer >=3 (since n >5, so 2p +1 >5 => p >=3).Then, E_n = [(2p +1 -1)/2]*(2p +1 +4) = [2p/2]*(2p +5) = p*(2p +5).m = (2p +1 +1)/2 = (2p +2)/2 = p +1.So, E_n -6m = p*(2p +5) -6*(p +1) = 2p^2 +5p -6p -6 = 2p^2 -p -6.We need to show that 2p^2 -p -6 >0 for p >=3.Let's compute for p=3: 2*9 -3 -6=18 -3 -6=9>0.For p=4: 2*16 -4 -6=32 -4 -6=22>0.In general, for p >=3, 2p^2 -p -6 is increasing since the quadratic coefficient is positive, and at p=3, it's already positive.Therefore, for odd n >5, E_n -6m >0.Hence, in both cases, whether n is even or odd, for n >5, E_n -6m >0, which implies T_n - S_n >0, so T_n > S_n.Therefore, we have proven that for n >5, T_n > S_n.I think this makes sense. Let me just check with n=6 and n=7 to be sure.For n=6 (even):a_n =2n +3, so a_1=5, a_2=7, a_3=9, a_4=11, a_5=13, a_6=15.Compute S_6 = sum of a_1 to a_6 = (5 +15)*6/2=20*3=60.Compute T_6: b_1=5-6=-1, b_2=2*7=14, b_3=9-6=3, b_4=2*11=22, b_5=13-6=7, b_6=2*15=30.Sum: -1 +14 +3 +22 +7 +30 = (-1 +14)=13, (13 +3)=16, (16 +22)=38, (38 +7)=45, (45 +30)=75.So, T_6=75, S_6=60. 75>60, correct.For n=7 (odd):a_7=2*7 +3=17.S_7 = sum from a_1 to a_7 = (5 +17)*7/2=22*3.5=77.Compute T_7: b_1=-1, b_2=14, b_3=3, b_4=22, b_5=7, b_6=30, b_7=17-6=11.Sum: -1 +14=13, +3=16, +22=38, +7=45, +30=75, +11=86.So, T_7=86, S_7=77. 86>77, correct.So, the proof holds for n=6 and n=7, which are the first n>5. Therefore, it seems correct.

answer：Okay, so I have this problem where a cone is formed from a 270-degree sector of a circle with a radius of 24 by aligning the two straight sides. I need to calculate the volume of the cone divided by π. Hmm, let me try to figure this out step by step.First, I remember that when you form a cone from a sector of a circle, the radius of the sector becomes the slant height of the cone. So, in this case, the slant height (let's call it 'l') of the cone is 24 units. Got that.Next, the arc length of the sector becomes the circumference of the base of the cone. The original circle has a radius of 24, so its full circumference would be 2π times 24, which is 48π. But we're only using a 270-degree sector, which is three-quarters of the full circle. So, the arc length of the sector is (270/360) times 48π. Let me calculate that:270 divided by 360 is 0.75, so 0.75 times 48π is 36π. Okay, so the arc length is 36π. This arc length becomes the circumference of the base of the cone.Now, the circumference of the base of the cone is 2πr, where 'r' is the radius of the base of the cone. So, I can set up the equation 2πr = 36π. If I divide both sides by π, I get 2r = 36, so r = 18. Got it, the radius of the base of the cone is 18 units.Now, I need to find the height of the cone to calculate the volume. I know the slant height (l) is 24, and the radius (r) is 18. I remember from the Pythagorean theorem that in a right triangle, the square of the hypotenuse (which is the slant height here) is equal to the sum of the squares of the other two sides (the radius and the height). So, l² = r² + h².Plugging in the values I have: 24² = 18² + h². Let's compute that:24 squared is 576, and 18 squared is 324. So, 576 = 324 + h². Subtracting 324 from both sides gives h² = 252. Taking the square root of both sides, h = sqrt(252). Hmm, sqrt(252) can be simplified. Let me see: 252 is 36 times 7, so sqrt(36*7) is 6*sqrt(7). So, the height of the cone is 6√7 units.Now, I can find the volume of the cone. The formula for the volume of a cone is (1/3)πr²h. Plugging in the values I have: (1/3)π*(18)²*(6√7). Let me compute that step by step.First, 18 squared is 324. Then, 324 multiplied by 6 is 1944. So, now I have (1/3)π*1944√7. Dividing 1944 by 3 gives 648. So, the volume is 648π√7.But the problem asks for the volume divided by π. So, I just need to take 648π√7 and divide it by π. The π cancels out, leaving me with 648√7.Wait, let me double-check my steps to make sure I didn't make a mistake. The arc length was 36π, which became the circumference of the base, giving me a radius of 18. Then, using the Pythagorean theorem, I found the height to be 6√7. Plugging into the volume formula, I got 648π√7, and dividing by π gives 648√7. Yeah, that seems right.I think I've got it. The volume divided by π is 648√7.**Final Answer**The result when the volume of the cone is divided by π is boxed{648sqrt{7}}.

answer：First, I need to determine the radius of circle ( C ) using the given circumference. The formula for the circumference of a circle is ( 2pi r ). Given that the circumference is ( 18pi ) inches, I can set up the equation:[2pi r = 18pi]Solving for ( r ):[r = frac{18pi}{2pi} = 9 text{ inches}]Now that I know the radius is 9 inches, I can identify the lengths of segments ( AC ) and ( AB ). Since both are radii of the circle, they are each 9 inches long.Next, I observe that triangle ( ACB ) has two sides of equal length (( AC = AB = 9 ) inches) and the included angle ( angle ACB = 30^circ ). This makes triangle ( ACB ) an isosceles triangle with sides ( AC ) and ( AB ) equal.To find the length of segment ( CB ), I can use the Law of Cosines, which relates the lengths of the sides of a triangle to the cosine of one of its angles. The formula for the Law of Cosines is:[CB^2 = AC^2 + AB^2 - 2 cdot AC cdot AB cdot cos(angle ACB)]Plugging in the known values:[CB^2 = 9^2 + 9^2 - 2 cdot 9 cdot 9 cdot cos(30^circ)]Calculating each term:[CB^2 = 81 + 81 - 162 cdot frac{sqrt{3}}{2}]Simplifying further:[CB^2 = 162 - 81sqrt{3}]To find ( CB ), I take the square root of both sides:[CB = sqrt{162 - 81sqrt{3}}]This can be simplified by factoring out 81:[CB = sqrt{81(2 - sqrt{3})} = 9sqrt{2 - sqrt{3}}]Therefore, the length of segment ( CB ) is ( 9sqrt{2 - sqrt{3}} ) inches.

answer：To determine how many positive four-digit integers of the form (__25) are divisible by 25, we can analyze the structure of such numbers.1. **Understanding the Form**: - A four-digit integer of the form (__25) can be represented as (AB25), where (A) and (B) are digits. - This means the number is (1000A + 100B + 25).2. **Divisibility by 25**: - A number is divisible by 25 if its last two digits are either 00, 25, 50, or 75. - In this case, the last two digits are fixed as 25, which satisfies the divisibility condition.3. **Range of Four-Digit Numbers**: - The smallest four-digit number is 1000, and the largest is 9999. - For the number (AB25) to be a four-digit number, (A) must be between 1 and 9 (since (A) cannot be 0), and (B) can be between 0 and 9.4. **Counting the Possibilities**: - There are 9 possible choices for (A) (1 through 9). - There are 10 possible choices for (B) (0 through 9).5. **Total Number of Valid Numbers**: - Multiply the number of choices for (A) and (B): (9 times 10 = 90).Therefore, there are 90 positive four-digit integers of the form (__25) that are divisible by 25.

answer：Okay, so I have this problem here: For all real values of ( t ), the expression ( sqrt{t^4 + t^2} ) is equal to one of the given options. The options are ( t^3 ), ( t^2 + t ), ( |t^2 + t| ), ( tsqrt{t^2 + 1} ), and ( |t|sqrt{1 + t^2} ). I need to figure out which one is correct.First, I should probably try to simplify the expression ( sqrt{t^4 + t^2} ). Let me see if I can factor something out from inside the square root. Both terms have a ( t^2 ) in them, so maybe I can factor that out.So, factoring ( t^2 ) from ( t^4 + t^2 ) gives me ( t^2(t^2 + 1) ). That means the expression becomes ( sqrt{t^2(t^2 + 1)} ).Now, I remember that the square root of a product is the product of the square roots, so I can split this into ( sqrt{t^2} times sqrt{t^2 + 1} ). The square root of ( t^2 ) is ( |t| ) because square roots are always non-negative. So, that part becomes ( |t| ). Putting it all together, the expression simplifies to ( |t|sqrt{t^2 + 1} ). Looking back at the options, that's exactly option E: ( |t|sqrt{1 + t^2} ). Wait, let me double-check to make sure I didn't make a mistake. The original expression is ( sqrt{t^4 + t^2} ). Factoring ( t^2 ) gives ( t^2(t^2 + 1) ). Taking the square root, we get ( |t|sqrt{t^2 + 1} ). Yeah, that seems right. I also want to make sure that none of the other options could be equivalent. Let's see:Option A: ( t^3 ). If I plug in ( t = 1 ), ( sqrt{1 + 1} = sqrt{2} ), but ( t^3 = 1 ). Not equal. So A is out.Option B: ( t^2 + t ). Let's test ( t = 2 ). The original expression is ( sqrt{16 + 4} = sqrt{20} approx 4.47 ). Option B gives ( 4 + 2 = 6 ). Not equal. So B is out.Option C: ( |t^2 + t| ). Let's test ( t = -1 ). Original expression is ( sqrt{1 + 1} = sqrt{2} approx 1.414 ). Option C gives ( |1 - 1| = 0 ). Not equal. So C is out.Option D: ( tsqrt{t^2 + 1} ). Let's test ( t = -1 ). Original expression is ( sqrt{1 + 1} = sqrt{2} approx 1.414 ). Option D gives ( -1 times sqrt{1 + 1} = -sqrt{2} approx -1.414 ). Not equal because square roots are non-negative. So D is out.Therefore, E is the correct answer.**Final Answer**boxed{E}