💡Okay, so I have this problem with two parts. Let me start by understanding what's given and what I need to find.First, there are two vectors, m and n. Vector m is (sin x, -1) and vector n is (√3 cos x, -1/2). The function f(x) is defined as the dot product of (m + n) and m. So, I need to find f(x) and then determine its interval of monotonic increase. That's part (1).Then, in part (2), I'm given a triangle ABC where a, b, c are the sides opposite angles A, B, C respectively. A is an acute angle, a is 2√3, c is 4, and f(A) is the maximum value of f(x) in the interval [0, π/2]. I need to find angle A, side b, and the area S of triangle ABC.Alright, starting with part (1). Let me write down the vectors:m = (sin x, -1)n = (√3 cos x, -1/2)So, m + n would be the sum of the corresponding components. That is:m + n = (sin x + √3 cos x, -1 + (-1/2)) = (sin x + √3 cos x, -3/2)Now, f(x) is the dot product of (m + n) and m. The dot product is calculated as:(m + n) · m = (sin x + √3 cos x) * sin x + (-3/2) * (-1)Let me compute each part:First component: (sin x + √3 cos x) * sin x = sin²x + √3 sin x cos xSecond component: (-3/2) * (-1) = 3/2So, f(x) = sin²x + √3 sin x cos x + 3/2Hmm, that's a trigonometric expression. Maybe I can simplify it using some identities. Let me recall that sin²x can be written as (1 - cos 2x)/2, and sin x cos x is (sin 2x)/2.So, substituting these in:f(x) = (1 - cos 2x)/2 + √3*(sin 2x)/2 + 3/2Let me combine the constants:(1/2) + (3/2) = 2So, f(x) = - (cos 2x)/2 + (√3 sin 2x)/2 + 2Hmm, this looks like a single sine function with a phase shift. The expression a sin θ + b cos θ can be written as R sin(θ + φ), where R = √(a² + b²) and tan φ = b/a.Wait, actually, in this case, it's (√3/2) sin 2x - (1/2) cos 2x. So, let me write it as:(√3/2) sin 2x - (1/2) cos 2x = sin(2x - π/6)Because sin(A - B) = sin A cos B - cos A sin B. Let me verify:sin(2x - π/6) = sin 2x cos π/6 - cos 2x sin π/6cos π/6 is √3/2, and sin π/6 is 1/2. So,sin(2x - π/6) = sin 2x*(√3/2) - cos 2x*(1/2)Which matches exactly with our expression. So, the expression simplifies to sin(2x - π/6).Therefore, f(x) = sin(2x - π/6) + 2.Great, that's the analytical expression for f(x).Now, I need to find the interval where f(x) is monotonically increasing. Since f(x) is a sine function with amplitude 1, shifted up by 2, its derivative will give me the rate of change.Let me compute the derivative f’(x):f(x) = sin(2x - π/6) + 2f’(x) = 2 cos(2x - π/6)To find where f(x) is increasing, I need f’(x) > 0.So, 2 cos(2x - π/6) > 0Which implies cos(2x - π/6) > 0The cosine function is positive in the intervals (-π/2 + 2kπ, π/2 + 2kπ) for any integer k.So, let's set up the inequality:-π/2 + 2kπ < 2x - π/6 < π/2 + 2kπLet me solve for x.First, add π/6 to all parts:-π/2 + π/6 + 2kπ < 2x < π/2 + π/6 + 2kπCompute -π/2 + π/6:-π/2 is -3π/6, so -3π/6 + π/6 = -2π/6 = -π/3Similarly, π/2 + π/6 is 3π/6 + π/6 = 4π/6 = 2π/3So, the inequality becomes:-π/3 + 2kπ < 2x < 2π/3 + 2kπDivide all parts by 2:-π/6 + kπ < x < π/3 + kπSo, the function f(x) is increasing on the intervals (-π/6 + kπ, π/3 + kπ) for any integer k.But since x is a real number, we can consider these intervals for all k. However, if the problem is expecting a specific interval, maybe within a certain range, but since it's not specified, I think it's safe to present the general solution.So, for part (1), f(x) = sin(2x - π/6) + 2, and it's increasing on intervals (-π/6 + kπ, π/3 + kπ) for any integer k.Moving on to part (2). We have triangle ABC with sides a, b, c opposite angles A, B, C respectively. A is acute, a = 2√3, c = 4, and f(A) is the maximum value of f(x) in [0, π/2].First, I need to find the maximum value of f(x) on [0, π/2]. Since f(x) = sin(2x - π/6) + 2, the maximum occurs when sin(2x - π/6) is maximum, which is 1.So, the maximum value of f(x) is 1 + 2 = 3.Therefore, f(A) = 3. So, sin(2A - π/6) + 2 = 3 => sin(2A - π/6) = 1.So, 2A - π/6 = π/2 + 2kπ, but since A is in [0, π/2], let's solve for A.2A - π/6 = π/2So, 2A = π/2 + π/6 = (3π/6 + π/6) = 4π/6 = 2π/3Therefore, A = (2π/3)/2 = π/3.So, angle A is π/3 radians, which is 60 degrees.Now, in triangle ABC, we have sides a = 2√3, c = 4, and angle A = π/3. We need to find side b and the area S.I can use the Law of Sines here, which states that a/sin A = b/sin B = c/sin C.Given a = 2√3, c = 4, angle A = π/3.First, let's find angle C.From the Law of Sines:a / sin A = c / sin CSo,2√3 / sin(π/3) = 4 / sin CCompute sin(π/3) = √3/2.So,2√3 / (√3/2) = 4 / sin CSimplify 2√3 divided by (√3/2):2√3 * (2/√3) = 4So, 4 = 4 / sin CTherefore, 4 = 4 / sin C => sin C = 1So, sin C = 1 implies that angle C is π/2, which is 90 degrees.Therefore, triangle ABC is a right-angled triangle at C.Now, since the sum of angles in a triangle is π, angle B = π - A - C = π - π/3 - π/2 = π - 5π/6 = π/6.So, angle B is π/6.Now, to find side b, we can use the Law of Sines again:a / sin A = b / sin BSo,2√3 / sin(π/3) = b / sin(π/6)Compute sin(π/3) = √3/2, sin(π/6) = 1/2.So,2√3 / (√3/2) = b / (1/2)Simplify left side:2√3 * (2/√3) = 4So, 4 = b / (1/2) => b = 4 * (1/2) = 2So, side b is 2.Alternatively, since it's a right-angled triangle at C, we can use Pythagoras theorem:a² + b² = c²But wait, in this case, c is the side opposite angle C, which is π/2, so c is the hypotenuse.Given a = 2√3, c = 4, so:(2√3)² + b² = 4²Compute:(4*3) + b² = 16 => 12 + b² = 16 => b² = 4 => b = 2Same result.Now, the area S of triangle ABC can be calculated as (1/2)*base*height. Since it's a right-angled triangle at C, sides a and b are the legs.So, area S = (1/2)*a*b = (1/2)*(2√3)*2 = (1/2)*4√3 = 2√3.Alternatively, using formula (1/2)*ab*sin C, but since sin C = 1, it's the same as above.So, summarizing:Angle A = π/3, side b = 2, area S = 2√3.Wait, but in the problem statement, it says "find A, b, and the area S". So, I think that's all.But just to make sure, let me recap:Given f(A) is maximum, which is 3, so A is π/3. Then, using Law of Sines, found angle C is π/2, angle B is π/6, side b is 2, and area is 2√3.Everything seems consistent.**Final Answer**(1) The analytical expression for ( f(x) ) is ( boxed{sinleft(2x - frac{pi}{6}right) + 2} ) and its interval of monotonic increase is ( left[-frac{pi}{6} + kpi, frac{pi}{3} + kpiright] ) for any integer ( k ).(2) The angle ( A ) is ( boxed{frac{pi}{3}} ), the side ( b ) is ( boxed{2} ), and the area ( S ) of ( triangle ABC ) is ( boxed{2sqrt{3}} ).

💡Okay, so I have this problem: I need to find the maximum value of the function ( y = cos^3 x + sin^2 x - cos x ) where ( x ) is a real number. The options given are (A) ( frac{4}{27} ), (B) ( frac{8}{27} ), (C) ( frac{16}{27} ), and (D) ( frac{32}{27} ). Hmm, let me think about how to approach this.First, I remember that to find the maximum value of a function, especially a trigonometric one, I might need to use calculus. That is, take the derivative of the function with respect to ( x ), set it equal to zero, and solve for ( x ) to find critical points. Then, evaluate the function at those points to see which gives the maximum value.But before jumping into calculus, maybe I can simplify the function a bit using some trigonometric identities. Let me write down the function again:( y = cos^3 x + sin^2 x - cos x )I notice that there are both ( cos^3 x ) and ( sin^2 x ) terms. Maybe I can express everything in terms of ( cos x ) or ( sin x ) to make it easier. Let me try to express ( sin^2 x ) in terms of ( cos x ). I remember that ( sin^2 x = 1 - cos^2 x ). So, substituting that in:( y = cos^3 x + (1 - cos^2 x) - cos x )Simplify that:( y = cos^3 x + 1 - cos^2 x - cos x )Combine like terms:( y = cos^3 x - cos^2 x - cos x + 1 )Hmm, that seems a bit simpler. Maybe I can factor this expression. Let me factor out ( cos x ) from the first three terms:( y = cos x (cos^2 x - cos x - 1) + 1 )Wait, that doesn't seem particularly helpful. Maybe I can factor the entire expression differently. Let me see:( y = cos^3 x - cos^2 x - cos x + 1 )Let me group the terms:( y = (cos^3 x - cos^2 x) + (-cos x + 1) )Factor out ( cos^2 x ) from the first group and factor out -1 from the second group:( y = cos^2 x (cos x - 1) - 1 (cos x - 1) )Now, notice that both terms have a common factor of ( (cos x - 1) ). So, factor that out:( y = (cos x - 1)(cos^2 x - 1) )Wait, ( cos^2 x - 1 ) is equal to ( -sin^2 x ). So, substituting that in:( y = (cos x - 1)(-sin^2 x) )Which simplifies to:( y = (1 - cos x)sin^2 x )Oh, that's a nice simplification! So now, the function is:( y = (1 - cos x)sin^2 x )Hmm, maybe I can express this in terms of a single trigonometric function. Let me think. I know that ( sin^2 x = 1 - cos^2 x ), but that might not help directly. Alternatively, I can express ( 1 - cos x ) in terms of ( sin ) or ( cos ) of half-angle identities. Let me recall that:( 1 - cos x = 2sin^2 left( frac{x}{2} right) )And ( sin x = 2sin left( frac{x}{2} right)cos left( frac{x}{2} right) ), so ( sin^2 x = 4sin^2 left( frac{x}{2} right)cos^2 left( frac{x}{2} right) ). Let me substitute these into the expression for ( y ):( y = 2sin^2 left( frac{x}{2} right) times 4sin^2 left( frac{x}{2} right)cos^2 left( frac{x}{2} right) )Multiply those constants:( y = 8sin^4 left( frac{x}{2} right)cos^2 left( frac{x}{2} right) )Hmm, so now ( y ) is expressed in terms of ( sin ) and ( cos ) of ( frac{x}{2} ). Maybe I can set ( t = sin left( frac{x}{2} right) ) or ( t = cos left( frac{x}{2} right) ) to make it a single variable function.Let me set ( t = sin left( frac{x}{2} right) ). Then, ( cos left( frac{x}{2} right) = sqrt{1 - t^2} ). So, substituting into ( y ):( y = 8t^4 (1 - t^2) )Simplify that:( y = 8t^4 - 8t^6 )So now, the function is ( y = 8t^4 - 8t^6 ), where ( t ) is in the range ( [-1, 1] ) because ( t = sin left( frac{x}{2} right) ) and sine functions have outputs between -1 and 1. However, since ( t^4 ) and ( t^6 ) are even functions, the function ( y ) will be the same for ( t ) and ( -t ). So, I can consider ( t ) in the range ( [0, 1] ) to simplify the problem.So, now I have a function ( y(t) = 8t^4 - 8t^6 ) for ( t in [0, 1] ). To find its maximum, I can take the derivative with respect to ( t ), set it equal to zero, and solve for ( t ).Let's compute the derivative:( y'(t) = frac{d}{dt}(8t^4 - 8t^6) = 32t^3 - 48t^5 )Set ( y'(t) = 0 ):( 32t^3 - 48t^5 = 0 )Factor out common terms:( 16t^3(2 - 3t^2) = 0 )So, the critical points are when ( 16t^3 = 0 ) or ( 2 - 3t^2 = 0 ).Solving ( 16t^3 = 0 ) gives ( t = 0 ).Solving ( 2 - 3t^2 = 0 ):( 3t^2 = 2 )( t^2 = frac{2}{3} )( t = sqrt{frac{2}{3}} ) or ( t = -sqrt{frac{2}{3}} )But since we're considering ( t in [0, 1] ), we take ( t = sqrt{frac{2}{3}} ).So, the critical points are at ( t = 0 ) and ( t = sqrt{frac{2}{3}} ). We should also check the endpoints of the interval, which are ( t = 0 ) and ( t = 1 ).Now, let's evaluate ( y(t) ) at these critical points and endpoints.1. At ( t = 0 ):( y(0) = 8(0)^4 - 8(0)^6 = 0 )2. At ( t = sqrt{frac{2}{3}} ):First, compute ( t^4 ) and ( t^6 ):( t = sqrt{frac{2}{3}} )( t^2 = frac{2}{3} )( t^4 = left( frac{2}{3} right)^2 = frac{4}{9} )( t^6 = left( frac{2}{3} right)^3 = frac{8}{27} )So,( yleft( sqrt{frac{2}{3}} right) = 8 times frac{4}{9} - 8 times frac{8}{27} )Compute each term:( 8 times frac{4}{9} = frac{32}{9} )( 8 times frac{8}{27} = frac{64}{27} )So,( y = frac{32}{9} - frac{64}{27} )To subtract these, convert to a common denominator:( frac{32}{9} = frac{96}{27} )So,( y = frac{96}{27} - frac{64}{27} = frac{32}{27} )3. At ( t = 1 ):( y(1) = 8(1)^4 - 8(1)^6 = 8 - 8 = 0 )So, evaluating ( y(t) ) at the critical points and endpoints, we get:- ( y(0) = 0 )- ( yleft( sqrt{frac{2}{3}} right) = frac{32}{27} )- ( y(1) = 0 )Therefore, the maximum value of ( y(t) ) is ( frac{32}{27} ), which occurs at ( t = sqrt{frac{2}{3}} ).But wait, let me double-check my steps to make sure I didn't make any mistakes. So, I started by expressing the original function in terms of ( cos x ) and ( sin x ), then used trigonometric identities to rewrite it in terms of half-angle functions. Then, I set ( t = sin left( frac{x}{2} right) ) and transformed the function into a polynomial in ( t ). I found the critical points by taking the derivative and solving for ( t ), then evaluated the function at those points and the endpoints.Everything seems to check out. The maximum value occurs at ( t = sqrt{frac{2}{3}} ), and the value is ( frac{32}{27} ), which is option (D).Alternatively, just to be thorough, maybe I can approach this problem without substituting ( t ). Let me try another method.Starting again with the original function:( y = cos^3 x + sin^2 x - cos x )I can express this in terms of ( cos x ) only. Let me let ( u = cos x ). Then, since ( sin^2 x = 1 - u^2 ), the function becomes:( y = u^3 + (1 - u^2) - u )Simplify:( y = u^3 - u^2 - u + 1 )So, ( y(u) = u^3 - u^2 - u + 1 ), where ( u in [-1, 1] ) because ( cos x ) ranges between -1 and 1.To find the maximum of this function, take the derivative with respect to ( u ):( y'(u) = 3u^2 - 2u - 1 )Set the derivative equal to zero to find critical points:( 3u^2 - 2u - 1 = 0 )Solve this quadratic equation for ( u ):Using the quadratic formula:( u = frac{2 pm sqrt{(-2)^2 - 4 times 3 times (-1)}}{2 times 3} )Compute the discriminant:( (-2)^2 - 4 times 3 times (-1) = 4 + 12 = 16 )So,( u = frac{2 pm sqrt{16}}{6} = frac{2 pm 4}{6} )Thus, two solutions:1. ( u = frac{2 + 4}{6} = frac{6}{6} = 1 )2. ( u = frac{2 - 4}{6} = frac{-2}{6} = -frac{1}{3} )So, the critical points are at ( u = 1 ) and ( u = -frac{1}{3} ). We should also check the endpoints of the interval ( u = -1 ) and ( u = 1 ).Now, evaluate ( y(u) ) at these points:1. At ( u = -1 ):( y(-1) = (-1)^3 - (-1)^2 - (-1) + 1 = -1 - 1 + 1 + 1 = 0 )2. At ( u = -frac{1}{3} ):Compute each term:( u^3 = left( -frac{1}{3} right)^3 = -frac{1}{27} )( -u^2 = -left( frac{1}{3} right)^2 = -frac{1}{9} )( -u = -left( -frac{1}{3} right) = frac{1}{3} )So,( yleft( -frac{1}{3} right) = -frac{1}{27} - frac{1}{9} + frac{1}{3} + 1 )Convert all terms to 27 denominators:( -frac{1}{27} - frac{3}{27} + frac{9}{27} + frac{27}{27} )Add them up:( (-1 - 3 + 9 + 27) / 27 = (32)/27 )So, ( yleft( -frac{1}{3} right) = frac{32}{27} )3. At ( u = 1 ):( y(1) = 1^3 - 1^2 - 1 + 1 = 1 - 1 - 1 + 1 = 0 )So, evaluating ( y(u) ) at the critical points and endpoints, we get:- ( y(-1) = 0 )- ( yleft( -frac{1}{3} right) = frac{32}{27} )- ( y(1) = 0 )Therefore, the maximum value is ( frac{32}{27} ), which is option (D).Wait, so both methods lead me to the same conclusion. That gives me more confidence that the answer is indeed ( frac{32}{27} ). Just to make sure, let me think if there's another way to approach this problem.Another approach might be to use the method of Lagrange multipliers or some inequality, but since this is a single-variable function, calculus seems appropriate. Alternatively, maybe using the AM-GM inequality? Let me see.Looking back at the expression ( y = (1 - cos x)sin^2 x ), perhaps I can apply AM-GM. Let me consider the terms ( 1 - cos x ) and ( sin^2 x ). But I'm not sure how to directly apply AM-GM here because they are multiplied together, and AM-GM is more about sums. Maybe if I can express this product as a sum, but that might complicate things.Alternatively, since we have ( y = 8t^4(1 - t^2) ) where ( t = sin left( frac{x}{2} right) ), perhaps I can write this as ( y = 8t^4 - 8t^6 ) and then consider it as a function of ( t ). But that's essentially what I did earlier.Alternatively, maybe using substitution with ( t = cos x ), but that's similar to the second method.Wait, another thought: since ( y = (1 - cos x)sin^2 x ), maybe I can express this in terms of ( sin x ) only. Let me try:( y = (1 - cos x)sin^2 x )Express ( 1 - cos x ) as ( 2sin^2 left( frac{x}{2} right) ) and ( sin^2 x = 4sin^2 left( frac{x}{2} right)cos^2 left( frac{x}{2} right) ). So, substituting:( y = 2sin^2 left( frac{x}{2} right) times 4sin^2 left( frac{x}{2} right)cos^2 left( frac{x}{2} right) )Which simplifies to:( y = 8sin^4 left( frac{x}{2} right)cos^2 left( frac{x}{2} right) )Now, let me set ( t = sin left( frac{x}{2} right) ), so ( cos left( frac{x}{2} right) = sqrt{1 - t^2} ). Then,( y = 8t^4(1 - t^2) )Which is the same expression as before. So, again, taking the derivative and finding the maximum gives ( frac{32}{27} ).Alternatively, maybe I can use substitution ( u = t^2 ), so ( y = 8u^2(1 - u) ). Then, ( y = 8u^2 - 8u^3 ). Taking derivative with respect to ( u ):( dy/du = 16u - 24u^2 )Set to zero:( 16u - 24u^2 = 0 )Factor:( 8u(2 - 3u) = 0 )Solutions: ( u = 0 ) or ( u = frac{2}{3} )So, ( u = 0 ) gives ( y = 0 ), and ( u = frac{2}{3} ) gives:( y = 8left( frac{2}{3} right)^2 left( 1 - frac{2}{3} right) = 8 times frac{4}{9} times frac{1}{3} = 8 times frac{4}{27} = frac{32}{27} )Same result.So, through multiple methods, I consistently get that the maximum value is ( frac{32}{27} ), which is option (D). Therefore, I'm confident that the answer is (D).**Final Answer**boxed{D}

💡Okay, so I have this problem where there are 2021 non-constant real polynomials, P₁ through P₂₀₂₁, and they satisfy the condition that the composition of P₁ with P₂ is equal to the composition of P₂ with P₃, and so on, all the way around to the composition of P₂₀₂₁ with P₁. The goal is to show that all these polynomials have the same degree.Hmm, let me try to unpack this. First, I know that when you compose two polynomials, the degree of the resulting polynomial is the product of the degrees of the two original polynomials. So, if P has degree m and Q has degree n, then P∘Q has degree m*n. That seems important here.Given that all these compositions are equal, their degrees must be equal as well. So, if I take the degree of P₁∘P₂, that should be equal to the degree of P₂∘P₃, and so on. Let me write that down:deg(P₁∘P₂) = deg(P₂∘P₃) = ... = deg(P₂₀₂₁∘P₁)But since the degree of a composition is the product of the degrees, this translates to:deg(P₁) * deg(P₂) = deg(P₂) * deg(P₃) = ... = deg(P₂₀₂₁) * deg(P₁)So, each of these products must be equal. Let me denote deg(Pᵢ) as dᵢ for simplicity. Then, the equation becomes:d₁*d₂ = d₂*d₃ = ... = d₂₀₂₁*d₁Now, since all these products are equal, I can set them equal to some constant value, say k. So, d₁*d₂ = k, d₂*d₃ = k, and so on, up to d₂₀₂₁*d₁ = k.This seems like a system of equations. Each adjacent pair of d's multiplies to k. So, d₁*d₂ = d₂*d₃ implies that d₁ = d₃, right? Because if d₂ is non-zero (which it is, since the polynomials are non-constant), we can divide both sides by d₂, giving d₁ = d₃.Similarly, d₂*d₃ = d₃*d₄ implies d₂ = d₄, and so on. Continuing this way, we can see that all the degrees must be equal. For example, d₁ = d₃ = d₅ = ... and d₂ = d₄ = d₆ = ..., but since the sequence is cyclic (it goes all the way around to d₂₀₂₁*d₁), the even and odd indexed degrees must also be equal. Therefore, all degrees are equal.Wait, let me make sure I'm not missing something. If the degrees are equal in pairs, but because of the cyclic nature, the even and odd ones must also be equal. So, suppose d₁ = d₃ = d₅ = ... and d₂ = d₄ = d₆ = ..., but since d₂₀₂₁ is connected back to d₁, which is an odd index, and d₂₀₂₁ is even (since 2021 is odd, the last term is d₂₀₂₁*d₁, which would be d₂₀₂₁*d₁ = k, and d₂₀₂₁ is connected to d₂₀₂₀, which is even). Hmm, maybe I need to think about the parity.Wait, 2021 is an odd number, so the indices go from 1 to 2021, which is odd. So, the last composition is P₂₀₂₁∘P₁, which would be d₂₀₂₁*d₁ = k. But since d₂₀₂₁ is connected to d₂₀₂₀, which is even, and d₁ is odd. So, if d₂₀₂₀ = d₂, and d₂₀₂₁ = d₁, then d₂₀₂₁*d₁ = d₁*d₁ = k. But earlier, we had d₁*d₂ = k. So, d₁*d₂ = d₁², which implies that d₂ = d₁.Therefore, all degrees must be equal. So, d₁ = d₂ = ... = d₂₀₂₁.I think that makes sense. The key is that because the number of polynomials is odd, the cyclic condition forces the even and odd indexed degrees to be equal, hence all degrees are equal.Another way to think about it is to consider the ratios. From d₁*d₂ = d₂*d₃, we get d₁ = d₃. Similarly, d₂ = d₄, and so on. But since the number of polynomials is odd, when we get back to the beginning, we have d₂₀₂₁ = d₁, which connects back to d₂, forcing d₂ = d₁. Therefore, all degrees must be equal.Yeah, I think that's solid. So, the conclusion is that all the degrees are equal.

💡First, I need to determine the distance traveled by the first horse. The distance traveled in one revolution is the circumference of the circle it moves along, which can be calculated using the formula ( C = 2pi r ), where ( r ) is the radius.For the first horse, the radius is 30 feet. So, the circumference ( C_1 ) is:[C_1 = 2pi times 30 = 60pi text{ feet}]Since the first horse makes 40 revolutions, the total distance traveled ( D ) is:[D = 40 times 60pi = 2400pi text{ feet}]Now, for the second horse, which is 10 feet from the center, the radius is 10 feet. The circumference ( C_2 ) for the second horse is:[C_2 = 2pi times 10 = 20pi text{ feet}]To find out how many revolutions ( N ) the second horse needs to make to cover the same distance ( D ), I set up the equation:[N times 20pi = 2400pi]Solving for ( N ):[N = frac{2400pi}{20pi} = 120]Therefore, the second horse needs to make 120 revolutions to travel the same distance as the first horse.

💡Okay, so I have this problem here where I need to find the number of ordered quadruples (x₁, x₂, x₃, x₄) of positive odd integers such that their sum is 84. Then, I have to divide that number by 100 and provide the result. Hmm, let me think about how to approach this.First, I remember that when dealing with problems involving sums of integers with certain constraints, like being odd or even, it's often helpful to use substitution to simplify the problem. Since all the x_i are positive odd integers, I can express each x_i in terms of another variable that represents a natural number. I recall that any positive odd integer can be written as 2k - 1, where k is a positive integer. So, if I let x_i = 2y_i - 1 for each i from 1 to 4, then each y_i will be a positive integer. This substitution should help because it transforms the problem into one involving sums of natural numbers, which is a more familiar territory for me.Let me write that down:x₁ = 2y₁ - 1 x₂ = 2y₂ - 1 x₃ = 2y₃ - 1 x₄ = 2y₄ - 1 Now, substituting these into the equation x₁ + x₂ + x₃ + x₄ = 84, we get:(2y₁ - 1) + (2y₂ - 1) + (2y₃ - 1) + (2y₄ - 1) = 84Let me simplify this equation step by step. First, I'll expand the left-hand side:2y₁ + 2y₂ + 2y₃ + 2y₄ - 1 - 1 - 1 - 1 = 84Combining like terms, the constants add up to -4:2(y₁ + y₂ + y₃ + y₄) - 4 = 84Now, I'll isolate the term with y_i:2(y₁ + y₂ + y₃ + y₄) = 84 + 4 2(y₁ + y₂ + y₃ + y₄) = 88Divide both sides by 2:y₁ + y₂ + y₃ + y₄ = 44Okay, so now the problem is transformed into finding the number of ordered quadruples (y₁, y₂, y₃, y₄) of positive integers such that their sum is 44. This is a classic stars and bars problem in combinatorics.The stars and bars theorem states that the number of ways to distribute n identical items into k distinct bins is given by the combination formula C(n-1, k-1). In this case, n is 44 (the total sum) and k is 4 (the number of variables y₁ to y₄). So, the number of solutions should be C(44 - 1, 4 - 1) = C(43, 3).Let me compute that:C(43, 3) = 43! / (3! * (43 - 3)!) = (43 * 42 * 41) / (3 * 2 * 1) = (43 * 42 * 41) / 6Calculating the numerator:43 * 42 = 1806 1806 * 41 = Let's compute that step by step: 1806 * 40 = 72,240 1806 * 1 = 1,806 Adding them together: 72,240 + 1,806 = 74,046Now, divide by 6:74,046 / 6 = 12,341So, the number of ordered quadruples (y₁, y₂, y₃, y₄) is 12,341. Since each y_i corresponds to an x_i, the number of ordered quadruples (x₁, x₂, x₃, x₄) is also 12,341. Therefore, m = 12,341.The problem asks for m divided by 100, so I need to compute 12,341 / 100. That's straightforward:12,341 / 100 = 123.41So, the final answer is 123.41. Let me just double-check my steps to make sure I didn't make a mistake.1. Substituted each x_i with 2y_i - 1, which is correct because it transforms odd integers into natural numbers.2. Expanded the equation correctly and simplified to y₁ + y₂ + y₃ + y₄ = 44.3. Applied the stars and bars formula correctly, computing C(43, 3).4. Calculated the combination step by step, ensuring that the arithmetic was correct.5. Divided the result by 100 as per the problem's requirement.Everything seems to check out. I think I'm confident with this answer.**Final Answer**The value of (frac{m}{100}) is boxed{123.41}).

💡Okay, so I have this problem here about a convex polygon where the sum of all its interior angles except one is 3150 degrees. I need to figure out how many sides the polygon has, and the options are from 18 to 22. Hmm, let me think about how to approach this.First, I remember that the sum of the interior angles of a polygon with n sides is given by the formula (n-2)*180 degrees. That's a standard formula, right? So, if I can find the total sum of the interior angles, I can then figure out how many sides the polygon has.But wait, the problem says the sum of all angles except one is 3150 degrees. That means one angle is missing from the total sum. Since it's a convex polygon, all interior angles are less than 180 degrees. So, the missing angle must be less than 180 degrees.Let me denote the total sum of the interior angles as S. Then, according to the problem, S - one angle = 3150 degrees. Since the missing angle is less than 180 degrees, S must be slightly more than 3150 degrees. So, I can write:S = 3150 + x, where x is the missing angle, and x < 180 degrees.But I also know that S = (n-2)*180. So, putting it together:(n-2)*180 = 3150 + xSince x is less than 180, I can say that:(n-2)*180 > 3150Let me solve for n:(n-2) > 3150 / 180Calculating 3150 divided by 180. Let me do that:3150 ÷ 180. Well, 180*17 = 3060, and 3150 - 3060 = 90. So, 3150 / 180 = 17.5So, (n-2) > 17.5Therefore, n > 19.5Since n must be an integer (number of sides can't be a fraction), the smallest possible value for n is 20.Wait, but let me verify that. If n is 20, then the total sum S would be (20-2)*180 = 18*180 = 3240 degrees.Given that the sum of all angles except one is 3150 degrees, then the missing angle would be 3240 - 3150 = 90 degrees.Since 90 degrees is less than 180 degrees, it fits the condition of a convex polygon. So, that works.But just to be thorough, let me check the other options to make sure.If n were 19, then S would be (19-2)*180 = 17*180 = 3060 degrees.Then, the missing angle would be 3060 - 3150 = -90 degrees. Wait, that doesn't make sense. You can't have a negative angle. So, n can't be 19.If n were 21, S would be (21-2)*180 = 19*180 = 3420 degrees.Then, the missing angle would be 3420 - 3150 = 270 degrees. But that's more than 180 degrees, which would mean the polygon is concave, not convex. So, that's not possible either.Similarly, n=22 would give an even larger missing angle, which is also not possible.n=18: S=(18-2)*180=16*180=2880. Missing angle=2880-3150=-270. Negative again, so that's not possible.So, the only feasible answer is n=20.Therefore, the number of sides must be 20.