💡Okay, so I have this problem here about Al, Bob, and Carl's ages. Let me try to figure it out step by step. First, the problem says, "Al's age is 20 more than the sum of Bob's age and Carl's age." So, if I let A be Al's age, B be Bob's age, and C be Carl's age, then I can write that as:A = 20 + B + CGot that down. Now, the second part says, "the square of Al's age is 1800 more than the square of the sum of Bob's and Carl's age." Hmm, okay. So, the square of Al's age is A², and the square of the sum of Bob's and Carl's age is (B + C)². So, the equation would be:A² = 1800 + (B + C)²Alright, so now I have two equations:1. A = 20 + B + C2. A² = 1800 + (B + C)²I need to find the sum of their ages, which would be A + B + C. Let me see how I can use these equations to find that.From the first equation, I can express A in terms of B and C. Maybe I can substitute that into the second equation. Let me try that.So, from equation 1, A = 20 + B + C. Let me plug that into equation 2:(20 + B + C)² = 1800 + (B + C)²Alright, let's expand the left side:(20 + B + C)² = 20² + 2*20*(B + C) + (B + C)²= 400 + 40(B + C) + (B + C)²So, substituting back into the equation:400 + 40(B + C) + (B + C)² = 1800 + (B + C)²Hmm, I notice that (B + C)² appears on both sides, so maybe I can subtract that from both sides to simplify:400 + 40(B + C) = 1800Now, let's solve for (B + C):40(B + C) = 1800 - 40040(B + C) = 1400Divide both sides by 40:B + C = 1400 / 40B + C = 35Okay, so the sum of Bob's and Carl's ages is 35. Now, going back to equation 1, A = 20 + B + C, which is:A = 20 + 35A = 55So, Al is 55 years old. Now, the question asks for the sum of the ages of Al, Bob, and Carl, which is A + B + C.We already know A is 55, and B + C is 35, so:A + B + C = 55 + 35 = 90So, the total sum of their ages is 90. Let me just double-check my steps to make sure I didn't make any mistakes.1. I defined A, B, C correctly.2. I set up the equations based on the problem statement.3. Substituted A from the first equation into the second.4. Expanded and simplified correctly.5. Solved for B + C, then found A.6. Added them all together.Everything seems to check out. I think 90 is the right answer.**Final Answer**The sum of the ages of Al, Bob, and Carl is boxed{C}.

💡Okay, so I have this problem here with two parts. Let me try to work through it step by step. First, the function is given as ( f(x) = x^{-k^2 + k + 2} ) where ( k ) is an integer. And we're told that ( f(2) < f(3) ). We need to find the value of ( k ). Alright, so let's start with part (1). **Part (1): Finding the value of ( k )**Given ( f(2) < f(3) ), let's write out what that means. ( f(2) = 2^{-k^2 + k + 2} )( f(3) = 3^{-k^2 + k + 2} )So, the inequality is:( 2^{-k^2 + k + 2} < 3^{-k^2 + k + 2} )Hmm, okay. So both sides have the same exponent, which is ( -k^2 + k + 2 ). Let me denote that exponent as ( n ) for simplicity. So, ( n = -k^2 + k + 2 ).So, the inequality becomes:( 2^n < 3^n )Now, when is ( 2^n < 3^n )? Well, since 2 and 3 are both positive numbers greater than 1, the function ( a^n ) is increasing if ( a > 1 ). So, if ( n > 0 ), then ( 2^n < 3^n ) because 3 is larger than 2. If ( n = 0 ), both sides are 1, so they are equal. If ( n < 0 ), then ( 2^n > 3^n ) because taking reciprocals reverses the inequality.So, for ( 2^n < 3^n ), we need ( n > 0 ). Therefore:( -k^2 + k + 2 > 0 )Let me rewrite that:( -k^2 + k + 2 > 0 )Multiply both sides by -1 (remembering to reverse the inequality):( k^2 - k - 2 < 0 )Okay, so we have a quadratic inequality: ( k^2 - k - 2 < 0 ). Let's solve this.First, find the roots of the quadratic equation ( k^2 - k - 2 = 0 ).Using the quadratic formula:( k = frac{1 pm sqrt{1 + 8}}{2} = frac{1 pm 3}{2} )So, the roots are:( k = frac{1 + 3}{2} = 2 )and( k = frac{1 - 3}{2} = -1 )So, the quadratic ( k^2 - k - 2 ) factors as ( (k - 2)(k + 1) ).Now, the inequality ( (k - 2)(k + 1) < 0 ) is satisfied when ( k ) is between -1 and 2 because the product of two numbers is negative when one is positive and the other is negative.So, the solution is ( -1 < k < 2 ).But wait, ( k ) is an integer. So, the integers between -1 and 2 are 0 and 1.Therefore, ( k = 0 ) or ( k = 1 ).Okay, that seems straightforward. So, part (1) is done, and ( k ) can be 0 or 1.**Part (2): Determining if there exists a positive number ( p ) such that ( g(x) ) has a specific range**Now, moving on to part (2). We need to determine if there exists a positive number ( p ) such that the function ( g(x) = 1 - p f(x) + (2p - 1)x ) has its range on the interval ([-1, 2]) as ([-4, frac{17}{8}]). If it exists, find ( p ); otherwise, explain why not.First, let's note that from part (1), ( k ) is either 0 or 1. Let's see what ( f(x) ) becomes in each case.If ( k = 0 ):( f(x) = x^{-0^2 + 0 + 2} = x^{2} )If ( k = 1 ):( f(x) = x^{-1^2 + 1 + 2} = x^{-1 + 1 + 2} = x^{2} )So, in both cases, ( f(x) = x^2 ). That's helpful.Therefore, ( g(x) = 1 - p x^2 + (2p - 1)x ).Let me write that out:( g(x) = -p x^2 + (2p - 1)x + 1 )So, ( g(x) ) is a quadratic function in terms of ( x ). Since the coefficient of ( x^2 ) is ( -p ), which is negative because ( p ) is positive, the parabola opens downward. Therefore, it has a maximum point at its vertex.But we are told that the range of ( g(x) ) on the interval ([-1, 2]) is ([-4, frac{17}{8}]). So, the minimum value is -4 and the maximum is ( frac{17}{8} ).Since it's a downward-opening parabola, the maximum occurs at the vertex, and the minimum occurs at one of the endpoints of the interval.So, let's first find the vertex of this quadratic function.The general form is ( g(x) = ax^2 + bx + c ), so here ( a = -p ), ( b = 2p - 1 ), and ( c = 1 ).The vertex occurs at ( x = -frac{b}{2a} ).Plugging in the values:( x = -frac{2p - 1}{2(-p)} = frac{2p - 1}{2p} )So, the vertex is at ( x = frac{2p - 1}{2p} ).Now, we need to check whether this vertex lies within the interval ([-1, 2]). Because if it does, then the maximum value of ( g(x) ) on ([-1, 2]) is at the vertex, and the minimum will be at one of the endpoints. If the vertex is outside the interval, then the maximum will be at one of the endpoints.So, let's consider different cases based on where the vertex is.**Case 1: Vertex inside the interval ([-1, 2])**So, ( frac{2p - 1}{2p} in [-1, 2] )Let me solve this inequality:First, ( frac{2p - 1}{2p} geq -1 )Multiply both sides by ( 2p ). But since ( p > 0 ), the inequality sign remains the same.( 2p - 1 geq -2p )Bring all terms to one side:( 2p - 1 + 2p geq 0 )( 4p - 1 geq 0 )( 4p geq 1 )( p geq frac{1}{4} )Second inequality: ( frac{2p - 1}{2p} leq 2 )Again, multiply both sides by ( 2p ):( 2p - 1 leq 4p )Bring terms to one side:( 2p - 1 - 4p leq 0 )( -2p - 1 leq 0 )Multiply both sides by -1 (reverse inequality):( 2p + 1 geq 0 )Since ( p > 0 ), this is always true. So, the only constraint is ( p geq frac{1}{4} ).Therefore, when ( p geq frac{1}{4} ), the vertex is within the interval ([-1, 2]). So, the maximum is at the vertex, and the minimum is at one of the endpoints.**Case 2: Vertex outside the interval on the right (i.e., ( frac{2p - 1}{2p} > 2 ))**Let's see when this happens.( frac{2p - 1}{2p} > 2 )Multiply both sides by ( 2p ):( 2p - 1 > 4p )Bring terms to one side:( 2p - 1 - 4p > 0 )( -2p - 1 > 0 )( -2p > 1 )( p < -frac{1}{2} )But ( p ) is positive, so this case is impossible.**Case 3: Vertex outside the interval on the left (i.e., ( frac{2p - 1}{2p} < -1 ))**Let's solve this:( frac{2p - 1}{2p} < -1 )Multiply both sides by ( 2p ):( 2p - 1 < -2p )Bring terms to one side:( 2p - 1 + 2p < 0 )( 4p - 1 < 0 )( 4p < 1 )( p < frac{1}{4} )So, when ( p < frac{1}{4} ), the vertex is to the left of the interval ([-1, 2]). Therefore, the maximum of ( g(x) ) on ([-1, 2]) is at the left endpoint ( x = -1 ), and the minimum is at the right endpoint ( x = 2 ).Wait, but since the parabola opens downward, if the vertex is to the left of the interval, then the function is increasing on the interval. So, the maximum would be at the right endpoint, and the minimum at the left endpoint? Wait, no, hold on.Wait, if the vertex is to the left of the interval, the function is increasing on the interval. So, as ( x ) increases, ( g(x) ) increases. Therefore, the minimum is at ( x = -1 ), and the maximum is at ( x = 2 ).Wait, but the parabola opens downward, so it's increasing to the left of the vertex and decreasing to the right. So, if the vertex is to the left of the interval, then on the interval ([-1, 2]), the function is decreasing. So, the maximum is at the left endpoint ( x = -1 ), and the minimum is at the right endpoint ( x = 2 ).Wait, I think I confused myself. Let me think again.Since the parabola opens downward, it has a maximum at the vertex. If the vertex is to the left of the interval, then on the interval, the function is decreasing. So, as ( x ) increases from -1 to 2, ( g(x) ) decreases. Therefore, the maximum is at ( x = -1 ), and the minimum is at ( x = 2 ).Similarly, if the vertex is to the right of the interval, the function is increasing on the interval, so the maximum is at ( x = 2 ), and the minimum is at ( x = -1 ). But in our case, the vertex can't be to the right because when ( p geq frac{1}{4} ), the vertex is inside the interval, and when ( p < frac{1}{4} ), it's to the left.Wait, no, when ( p < frac{1}{4} ), the vertex is to the left of the interval, so the function is decreasing on the interval, so maximum at ( x = -1 ), minimum at ( x = 2 ). When ( p geq frac{1}{4} ), the vertex is inside the interval, so maximum at the vertex, and minimum at one of the endpoints.So, let's summarize:- If ( p geq frac{1}{4} ): maximum at vertex, minimum at either ( x = -1 ) or ( x = 2 ).- If ( p < frac{1}{4} ): maximum at ( x = -1 ), minimum at ( x = 2 ).But in our problem, the range is given as ([-4, frac{17}{8}]). So, the minimum is -4, and the maximum is ( frac{17}{8} ).So, we need to find ( p ) such that:Either:1. When ( p geq frac{1}{4} ), the maximum is at the vertex, which is ( frac{17}{8} ), and the minimum is either ( g(-1) = -4 ) or ( g(2) = -4 ).Or:2. When ( p < frac{1}{4} ), the maximum is ( g(-1) = frac{17}{8} ), and the minimum is ( g(2) = -4 ).But let's see if both cases are possible.Let me first handle Case 1: ( p geq frac{1}{4} ).**Case 1: ( p geq frac{1}{4} )**In this case, the maximum is at the vertex, which is ( gleft( frac{2p - 1}{2p} right) = frac{17}{8} ), and the minimum is either ( g(-1) ) or ( g(2) ), which should be -4.So, let's compute ( g ) at the vertex.First, let's find the value of ( g ) at the vertex.The vertex is at ( x = frac{2p - 1}{2p} ).So, ( gleft( frac{2p - 1}{2p} right) = -p left( frac{2p - 1}{2p} right)^2 + (2p - 1) left( frac{2p - 1}{2p} right) + 1 ).Let me compute this step by step.First, compute ( left( frac{2p - 1}{2p} right)^2 ):( left( frac{2p - 1}{2p} right)^2 = frac{(2p - 1)^2}{(2p)^2} = frac{4p^2 - 4p + 1}{4p^2} )So, ( -p times frac{4p^2 - 4p + 1}{4p^2} = -p times frac{4p^2 - 4p + 1}{4p^2} = -frac{4p^2 - 4p + 1}{4p} )Simplify:( -frac{4p^2 - 4p + 1}{4p} = -frac{4p^2}{4p} + frac{4p}{4p} - frac{1}{4p} = -p + 1 - frac{1}{4p} )Next, compute ( (2p - 1) times frac{2p - 1}{2p} ):( (2p - 1) times frac{2p - 1}{2p} = frac{(2p - 1)^2}{2p} = frac{4p^2 - 4p + 1}{2p} )So, putting it all together:( gleft( frac{2p - 1}{2p} right) = left( -p + 1 - frac{1}{4p} right) + left( frac{4p^2 - 4p + 1}{2p} right) + 1 )Let me compute each term:First term: ( -p + 1 - frac{1}{4p} )Second term: ( frac{4p^2 - 4p + 1}{2p} = frac{4p^2}{2p} - frac{4p}{2p} + frac{1}{2p} = 2p - 2 + frac{1}{2p} )Third term: 1So, adding them all together:( (-p + 1 - frac{1}{4p}) + (2p - 2 + frac{1}{2p}) + 1 )Combine like terms:- ( -p + 2p = p )- ( 1 - 2 + 1 = 0 )- ( -frac{1}{4p} + frac{1}{2p} = frac{1}{4p} )So, overall:( p + frac{1}{4p} )Therefore, ( g ) at the vertex is ( p + frac{1}{4p} ), and we know this equals ( frac{17}{8} ).So, set up the equation:( p + frac{1}{4p} = frac{17}{8} )Multiply both sides by ( 4p ) to eliminate the denominator:( 4p^2 + 1 = frac{17}{8} times 4p )Simplify the right side:( frac{17}{8} times 4p = frac{17}{2} p )So, equation becomes:( 4p^2 + 1 = frac{17}{2} p )Multiply both sides by 2 to eliminate the fraction:( 8p^2 + 2 = 17p )Bring all terms to one side:( 8p^2 - 17p + 2 = 0 )Now, solve this quadratic equation for ( p ).Using the quadratic formula:( p = frac{17 pm sqrt{(-17)^2 - 4 times 8 times 2}}{2 times 8} )Compute discriminant:( D = 289 - 64 = 225 )So,( p = frac{17 pm 15}{16} )Thus,First solution:( p = frac{17 + 15}{16} = frac{32}{16} = 2 )Second solution:( p = frac{17 - 15}{16} = frac{2}{16} = frac{1}{8} )So, ( p = 2 ) or ( p = frac{1}{8} ).But remember, in Case 1, we assumed ( p geq frac{1}{4} ). So, ( p = 2 ) is valid, but ( p = frac{1}{8} ) is less than ( frac{1}{4} ), so it doesn't belong to this case. So, only ( p = 2 ) is a candidate here.Now, we need to check whether with ( p = 2 ), the minimum of ( g(x) ) on ([-1, 2]) is -4.So, let's compute ( g(-1) ) and ( g(2) ) when ( p = 2 ).First, ( g(x) = -2x^2 + (4 - 1)x + 1 = -2x^2 + 3x + 1 ).Compute ( g(-1) ):( g(-1) = -2(-1)^2 + 3(-1) + 1 = -2(1) - 3 + 1 = -2 - 3 + 1 = -4 )Compute ( g(2) ):( g(2) = -2(2)^2 + 3(2) + 1 = -8 + 6 + 1 = -1 )So, with ( p = 2 ), ( g(-1) = -4 ) and ( g(2) = -1 ). The vertex gives the maximum ( frac{17}{8} ).Therefore, the range is from -4 to ( frac{17}{8} ), which matches the given range. So, ( p = 2 ) is a valid solution.Now, let's check Case 2.**Case 2: ( p < frac{1}{4} )**In this case, the vertex is to the left of the interval, so the function is decreasing on ([-1, 2]). Therefore, the maximum is at ( x = -1 ), and the minimum is at ( x = 2 ).Given that the range is ([-4, frac{17}{8}]), we have:- ( g(-1) = frac{17}{8} )- ( g(2) = -4 )So, let's compute ( g(-1) ) and ( g(2) ) in terms of ( p ) and set up the equations.First, ( g(-1) = -p(-1)^2 + (2p - 1)(-1) + 1 = -p(1) + (-2p + 1) + 1 = -p - 2p + 1 + 1 = -3p + 2 )Set this equal to ( frac{17}{8} ):( -3p + 2 = frac{17}{8} )Solve for ( p ):( -3p = frac{17}{8} - 2 = frac{17}{8} - frac{16}{8} = frac{1}{8} )So,( p = -frac{1}{24} )But ( p ) is supposed to be positive, so this is invalid. Therefore, there is no solution in this case.Therefore, the only valid solution is ( p = 2 ).**Conclusion**So, summarizing:1. The possible values of ( k ) are 0 and 1.2. There exists a positive number ( p = 2 ) that satisfies the given condition.**Final Answer**(1) The value of ( k ) is boxed{0} or boxed{1}.(2) The value of ( p ) is boxed{2}.

💡First, I need to understand the operation defined as ( a textdollar b = (a - b)^2 ). This means that for any two expressions ( a ) and ( b ), the operation ( textdollar ) involves subtracting ( b ) from ( a ) and then squaring the result.Next, I'll apply this operation to the given expressions ( (2x + 3y) ) and ( (3x - 4y) ). So, I'll set up the expression as follows:[(2x + 3y) textdollar (3x - 4y) = ((2x + 3y) - (3x - 4y))^2]Now, I'll simplify the expression inside the square:[(2x + 3y) - (3x - 4y) = 2x + 3y - 3x + 4y = -x + 7y]With the simplified expression, I'll now square it:[(-x + 7y)^2 = x^2 - 14xy + 49y^2]So, the result of ( (2x + 3y) textdollar (3x - 4y) ) is ( x^2 - 14xy + 49y^2 ).

💡First, I need to determine the total earnings from both Wednesday's and Saturday's games. The combined earnings are 7,250.Next, I'll define variables for the earnings from each day:- Let ( W ) represent the earnings from Wednesday's game.- Let ( S ) represent the earnings from Saturday's game.From the problem, I know that:[ W + S = 7250 ]The problem also provides the percentage distributions for each revenue source on both days:- **Wednesday:** - Ticket Sales: 60% - Merchandise Sales: 30% - Food Concessions: 10% - **Saturday:** - Ticket Sales: 50% - Merchandise Sales: 35% - Food Concessions: 15%Additionally, it's given that Wednesday's ticket sales are 142.50 less than Saturday's ticket sales. This can be expressed as:[ 0.60W = 0.50S - 142.50 ]Now, I have a system of two equations:1. ( W + S = 7250 )2. ( 0.60W = 0.50S - 142.50 )I'll solve this system to find the value of ( S ), which represents Saturday's total earnings.First, solve equation 2 for ( W ):[ 0.60W = 0.50S - 142.50 ][ W = frac{0.50S - 142.50}{0.60} ]Substitute this expression for ( W ) into equation 1:[ frac{0.50S - 142.50}{0.60} + S = 7250 ]To eliminate the fraction, multiply the entire equation by 0.60:[ 0.50S - 142.50 + 0.60S = 4350 ]Combine like terms:[ 1.10S - 142.50 = 4350 ]Add 142.50 to both sides:[ 1.10S = 4492.50 ]Finally, divide both sides by 1.10 to solve for ( S ):[ S = frac{4492.50}{1.10} ][ S = 4084.09 ]Therefore, the total money taken in at Saturday's game from all revenue sources is approximately 4,084.09.

💡Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem gives me an equation: b cos C + (2a + c) cos B = 0. I need to find the measure of angle B. Hmm, let's see.First, I remember that in a triangle, the sum of the angles is π radians, so A + B + C = π. That might come in handy later. Also, I recall the Law of Sines, which says that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. Maybe I can use that to express b and c in terms of sine functions.So, let me rewrite the given equation using the Law of Sines. Since b = 2R sin B and c = 2R sin C, and a = 2R sin A, I can substitute these into the equation.Starting with the original equation:b cos C + (2a + c) cos B = 0Substituting the Law of Sines expressions:(2R sin B) cos C + (2*(2R sin A) + 2R sin C) cos B = 0Simplify each term:2R sin B cos C + (4R sin A + 2R sin C) cos B = 0I can factor out 2R from each term:2R [sin B cos C + (2 sin A + sin C) cos B] = 0Since 2R is not zero (because R is the radius of the circumscribed circle, which can't be zero in a valid triangle), the expression inside the brackets must be zero:sin B cos C + (2 sin A + sin C) cos B = 0Let me rearrange this:sin B cos C + sin C cos B + 2 sin A cos B = 0Wait, I notice that sin B cos C + sin C cos B is equal to sin(B + C) because of the sine addition formula. So, that simplifies to:sin(B + C) + 2 sin A cos B = 0But in a triangle, B + C = π - A, so sin(B + C) = sin(π - A) = sin A. Therefore, the equation becomes:sin A + 2 sin A cos B = 0Factor out sin A:sin A (1 + 2 cos B) = 0Now, sin A can't be zero because A is an angle in a triangle, so 0 < A < π, meaning sin A > 0. Therefore, the other factor must be zero:1 + 2 cos B = 0Solving for cos B:2 cos B = -1cos B = -1/2So, what angle B has a cosine of -1/2? I remember that cos(2π/3) = -1/2 because 2π/3 is 120 degrees, which is in the second quadrant where cosine is negative. Since angles in a triangle are between 0 and π radians, 2π/3 is a valid solution.Therefore, angle B is 2π/3 radians.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the given equation, I substituted using the Law of Sines correctly. Then, I used the sine addition formula appropriately, and since B + C = π - A, that substitution was correct. Factoring out sin A was the right move, and since sin A isn't zero, setting 1 + 2 cos B to zero was correct. Solving for cos B gave me -1/2, and 2π/3 is indeed the angle with that cosine value. So, I think my solution is correct.I don't see any errors in my reasoning, so I'm confident that angle B is 2π/3 radians.**Final Answer**The measure of angle ( B ) is boxed{dfrac{2pi}{3}}.

💡First, I need to determine the total amount of work done by both the men and the women. For the men:- There are 15 men working for 21 days, with each day consisting of H hours.- The total work done by the men is calculated as 15 men * 21 days * H hours/day.For the women:- There are 21 women working for 90 days, with each day consisting of 2 hours.- The total work done by the women is calculated as 21 women * 90 days * 2 hours/day.Next, I need to account for the work equivalence between men and women. The problem states that 3 women do as much work as 2 men. This means that the work done by 3 women in one hour is equal to the work done by 2 men in one hour. Therefore, I can establish a ratio where 3 women are equivalent to 2 men in terms of work capacity.Using this ratio, I can convert the number of women into the equivalent number of men to ensure both groups are compared on the same basis. Substituting the equivalent number of men into the equation for the women's work:- 21 women * (2 men / 3 women) = 14 men.Now, I can set up the equation equating the total work done by men to the total work done by women:15 men * 21 days * H hours/day = 14 men * 90 days * 2 hours/day.Since the number of men is the same on both sides of the equation, I can cancel it out:21 days * H hours/day = 90 days * 2 hours/day.Finally, I solve for H:H = (90 days * 2 hours/day) / (21 days) = 180 hours / 21 days ≈ 8.57 hours/day.Therefore, the men work approximately 8.57 hours per day.