answer：Alright, so I've got this problem here where I need to find the minimum value of the expression ( dfrac{(x+1)^2 + 3}{x+2} + dfrac{y^2}{y+1} ) given that ( x geq 0 ), ( y geq 0 ), and ( x + y = 2 ). Hmm, okay, let's break this down step by step.First, I notice that both terms in the expression are fractions, and they involve ( x ) and ( y ) respectively. Since ( x + y = 2 ), maybe I can express one variable in terms of the other to reduce the problem to a single variable. That might make it easier to handle.Let me try expressing ( y ) in terms of ( x ). Since ( x + y = 2 ), that means ( y = 2 - x ). Okay, so I can substitute ( y ) with ( 2 - x ) in the expression. Let's rewrite the expression with this substitution:( dfrac{(x+1)^2 + 3}{x+2} + dfrac{(2 - x)^2}{(2 - x) + 1} )Simplify the second term's denominator: ( (2 - x) + 1 = 3 - x ). So now the expression becomes:( dfrac{(x+1)^2 + 3}{x+2} + dfrac{(2 - x)^2}{3 - x} )Hmm, that looks a bit complicated, but maybe I can simplify each term individually.Let's look at the first term: ( dfrac{(x+1)^2 + 3}{x+2} ). Let me expand the numerator:( (x+1)^2 = x^2 + 2x + 1 ), so adding 3 gives ( x^2 + 2x + 4 ). Therefore, the first term becomes:( dfrac{x^2 + 2x + 4}{x + 2} )I wonder if I can perform polynomial division here to simplify this fraction. Let me try dividing ( x^2 + 2x + 4 ) by ( x + 2 ).Dividing ( x^2 ) by ( x ) gives ( x ). Multiply ( x + 2 ) by ( x ) to get ( x^2 + 2x ). Subtracting this from the numerator:( (x^2 + 2x + 4) - (x^2 + 2x) = 4 )So, the division gives ( x ) with a remainder of 4. Therefore, the first term simplifies to:( x + dfrac{4}{x + 2} )Nice, that's simpler. Now, let's look at the second term: ( dfrac{(2 - x)^2}{3 - x} ). Let me expand the numerator:( (2 - x)^2 = 4 - 4x + x^2 ). So, the second term becomes:( dfrac{x^2 - 4x + 4}{3 - x} )Again, maybe I can perform polynomial division here. Let me rewrite the denominator as ( -x + 3 ) to make it easier.Divide ( x^2 - 4x + 4 ) by ( -x + 3 ). Hmm, dividing ( x^2 ) by ( -x ) gives ( -x ). Multiply ( -x + 3 ) by ( -x ) to get ( x^2 - 3x ). Subtracting this from the numerator:( (x^2 - 4x + 4) - (x^2 - 3x) = (-4x + 4) - (-3x) = -x + 4 )Now, divide ( -x + 4 ) by ( -x + 3 ). Dividing ( -x ) by ( -x ) gives 1. Multiply ( -x + 3 ) by 1 to get ( -x + 3 ). Subtracting this:( (-x + 4) - (-x + 3) = 1 )So, the division gives ( -x + 1 ) with a remainder of 1. Therefore, the second term simplifies to:( -x + 1 + dfrac{1}{3 - x} )But wait, ( 3 - x = (2 - x) + 1 ), which is the same as ( y + 1 ) since ( y = 2 - x ). So, the second term can be written as:( -x + 1 + dfrac{1}{y + 1} )Putting it all together, the entire expression becomes:( left( x + dfrac{4}{x + 2} right) + left( -x + 1 + dfrac{1}{y + 1} right) )Simplify this by combining like terms:( x - x + dfrac{4}{x + 2} + 1 + dfrac{1}{y + 1} )Which simplifies to:( 1 + dfrac{4}{x + 2} + dfrac{1}{y + 1} )Okay, so now the expression is simplified to ( 1 + dfrac{4}{x + 2} + dfrac{1}{y + 1} ). Since ( x + y = 2 ), maybe I can express ( y + 1 ) in terms of ( x ). Let's see:( y = 2 - x ), so ( y + 1 = 3 - x ). Therefore, the expression becomes:( 1 + dfrac{4}{x + 2} + dfrac{1}{3 - x} )Now, I have an expression in terms of a single variable ( x ). To find the minimum value, I can take the derivative with respect to ( x ) and set it equal to zero to find critical points.Let me denote the expression as ( f(x) = 1 + dfrac{4}{x + 2} + dfrac{1}{3 - x} ).First, find the derivative ( f'(x) ):( f'(x) = 0 + dfrac{-4}{(x + 2)^2} + dfrac{1}{(3 - x)^2} )Set ( f'(x) = 0 ):( dfrac{-4}{(x + 2)^2} + dfrac{1}{(3 - x)^2} = 0 )Move one term to the other side:( dfrac{1}{(3 - x)^2} = dfrac{4}{(x + 2)^2} )Take square roots of both sides (since both denominators are positive because ( x geq 0 ) and ( y geq 0 ), so ( x + 2 geq 2 ) and ( 3 - x geq 1 )):( dfrac{1}{3 - x} = dfrac{2}{x + 2} )Cross-multiplying:( (x + 2) = 2(3 - x) )Expand the right side:( x + 2 = 6 - 2x )Bring all terms to one side:( x + 2x = 6 - 2 )( 3x = 4 )( x = dfrac{4}{3} )Okay, so ( x = dfrac{4}{3} ). Then ( y = 2 - x = 2 - dfrac{4}{3} = dfrac{2}{3} ).Now, let's plug ( x = dfrac{4}{3} ) back into the expression ( f(x) ):( fleft( dfrac{4}{3} right) = 1 + dfrac{4}{dfrac{4}{3} + 2} + dfrac{1}{3 - dfrac{4}{3}} )Simplify each term:First, ( dfrac{4}{dfrac{4}{3} + 2} ). Convert 2 to thirds: ( 2 = dfrac{6}{3} ). So:( dfrac{4}{dfrac{4}{3} + dfrac{6}{3}} = dfrac{4}{dfrac{10}{3}} = 4 times dfrac{3}{10} = dfrac{12}{10} = dfrac{6}{5} )Next, ( dfrac{1}{3 - dfrac{4}{3}} ). Convert 3 to thirds: ( 3 = dfrac{9}{3} ). So:( dfrac{1}{dfrac{9}{3} - dfrac{4}{3}} = dfrac{1}{dfrac{5}{3}} = dfrac{3}{5} )Now, add all terms together:( 1 + dfrac{6}{5} + dfrac{3}{5} = 1 + dfrac{9}{5} = dfrac{5}{5} + dfrac{9}{5} = dfrac{14}{5} )So, the minimum value is ( dfrac{14}{5} ).Wait, but just to make sure, I should check the endpoints as well because sometimes the minimum can occur at the boundaries of the domain.The domain for ( x ) is from 0 to 2 since ( x geq 0 ) and ( y = 2 - x geq 0 ).Let's evaluate the expression at ( x = 0 ):( f(0) = 1 + dfrac{4}{0 + 2} + dfrac{1}{3 - 0} = 1 + 2 + dfrac{1}{3} = 3 + dfrac{1}{3} = dfrac{10}{3} approx 3.333 )And at ( x = 2 ):( f(2) = 1 + dfrac{4}{2 + 2} + dfrac{1}{3 - 2} = 1 + 1 + 1 = 3 )Comparing these with the critical point value ( dfrac{14}{5} = 2.8 ), which is indeed smaller than both endpoint values. So, the minimum occurs at ( x = dfrac{4}{3} ) and ( y = dfrac{2}{3} ), giving the minimum value of ( dfrac{14}{5} ).I think that's thorough enough. I considered simplifying the expression, used calculus to find the critical point, and checked the endpoints to ensure it's the minimum. Looks solid!

answer：Okay, so I have this problem here that I need to solve. It has two parts. The first part is about logarithms and exponents, and the second part is a logarithmic equation that I need to solve for x. Let me take them one at a time.Starting with the first part: If ( a = log_{4}3 ), then find ( 2^{a} + 2^{-a} ). Hmm, okay. So, I know that ( log_{4}3 ) means the exponent to which we raise 4 to get 3. So, ( 4^{a} = 3 ). That seems straightforward.But the question is asking for ( 2^{a} + 2^{-a} ). Hmm, since 4 is ( 2^2 ), maybe I can express ( 4^{a} ) in terms of base 2. Let me try that.So, ( 4^{a} = (2^2)^{a} = 2^{2a} ). And we know that ( 4^{a} = 3 ), so ( 2^{2a} = 3 ). That means ( 2^{a} ) is the square root of 3, right? Because if ( (2^{a})^2 = 3 ), then ( 2^{a} = sqrt{3} ). Okay, that makes sense.So, ( 2^{a} = sqrt{3} ). Then, ( 2^{-a} ) would be the reciprocal of that, which is ( 1/sqrt{3} ). So, ( 2^{-a} = 1/sqrt{3} ).Now, adding them together: ( 2^{a} + 2^{-a} = sqrt{3} + 1/sqrt{3} ). Hmm, that's a sum of a number and its reciprocal. Maybe I can combine these terms over a common denominator.Let me compute that: ( sqrt{3} + 1/sqrt{3} = (sqrt{3} times sqrt{3})/sqrt{3} + 1/sqrt{3} = (3 + 1)/sqrt{3} = 4/sqrt{3} ). Wait, that doesn't seem right because ( sqrt{3} + 1/sqrt{3} ) isn't equal to ( 4/sqrt{3} ). Let me check my steps again.Wait, actually, if I have ( sqrt{3} + 1/sqrt{3} ), to combine them, I can write ( sqrt{3} ) as ( 3/sqrt{3} ). So, ( 3/sqrt{3} + 1/sqrt{3} = (3 + 1)/sqrt{3} = 4/sqrt{3} ). Oh, that's correct. So, ( 4/sqrt{3} ) is the sum.But usually, we rationalize the denominator. So, multiplying numerator and denominator by ( sqrt{3} ), we get ( (4sqrt{3}) / 3 ). So, ( 2^{a} + 2^{-a} = frac{4sqrt{3}}{3} ). That seems like the answer for the first part.Alright, moving on to the second part: Solve the equation ( log_{2}(9^{x-1} - 5) = log_{2}(3^{x-1} - 2) + 2 ).Hmm, okay. So, this is a logarithmic equation. I remember that if two logarithms with the same base are equal, then their arguments must be equal. But here, on the right side, there's a "+2" which complicates things a bit. I need to figure out how to handle that.Let me recall that ( log_{2}(A) + B = log_{2}(A) + log_{2}(2^{B}) ) because ( B = log_{2}(2^{B}) ). So, maybe I can rewrite the right side as a single logarithm.So, the right side is ( log_{2}(3^{x-1} - 2) + 2 ). Let me express 2 as ( log_{2}(4) ) because ( 2 = log_{2}(4) ). So, the right side becomes ( log_{2}(3^{x-1} - 2) + log_{2}(4) ).Now, using the logarithm property that ( log_{b}(A) + log_{b}(C) = log_{b}(A times C) ), I can combine these two logs into one. So, the right side becomes ( log_{2}(4 times (3^{x-1} - 2)) ).So, now the equation is ( log_{2}(9^{x-1} - 5) = log_{2}(4 times (3^{x-1} - 2)) ).Since the logarithms are equal and they have the same base, their arguments must be equal. So, I can set the arguments equal to each other:( 9^{x - 1} - 5 = 4 times (3^{x - 1} - 2) ).Let me simplify this equation. First, expand the right side:( 9^{x - 1} - 5 = 4 times 3^{x - 1} - 8 ).Now, let's bring all terms to one side to see if we can form a quadratic equation or something similar. Let me add 8 to both sides:( 9^{x - 1} - 5 + 8 = 4 times 3^{x - 1} ).Simplify the left side:( 9^{x - 1} + 3 = 4 times 3^{x - 1} ).Hmm, okay. I notice that ( 9^{x - 1} ) is ( (3^2)^{x - 1} = 3^{2(x - 1)} ). So, ( 9^{x - 1} = (3^{x - 1})^2 ). Let me let ( y = 3^{x - 1} ). Then, ( 9^{x - 1} = y^2 ).Substituting into the equation, we have:( y^2 + 3 = 4y ).Now, this is a quadratic equation in terms of y:( y^2 - 4y + 3 = 0 ).Let me solve this quadratic equation. The quadratic is ( y^2 - 4y + 3 = 0 ). I can factor this:Looking for two numbers that multiply to 3 and add up to -4. Hmm, -1 and -3. So, ( (y - 1)(y - 3) = 0 ).Therefore, the solutions are ( y = 1 ) and ( y = 3 ).But remember, ( y = 3^{x - 1} ). So, let's substitute back:Case 1: ( 3^{x - 1} = 1 ).Since ( 3^{0} = 1 ), this implies ( x - 1 = 0 ), so ( x = 1 ).Case 2: ( 3^{x - 1} = 3 ).Since ( 3^{1} = 3 ), this implies ( x - 1 = 1 ), so ( x = 2 ).So, the solutions are ( x = 1 ) and ( x = 2 ). But wait, I need to check if these solutions are valid in the original equation because sometimes when dealing with logarithms, we might get extraneous solutions.Let me check ( x = 1 ):Original equation: ( log_{2}(9^{x-1} - 5) = log_{2}(3^{x-1} - 2) + 2 ).Substitute ( x = 1 ):Left side: ( log_{2}(9^{0} - 5) = log_{2}(1 - 5) = log_{2}(-4) ).Wait, logarithm of a negative number is undefined. So, ( x = 1 ) is not a valid solution.Now, check ( x = 2 ):Left side: ( log_{2}(9^{1} - 5) = log_{2}(9 - 5) = log_{2}(4) = 2 ).Right side: ( log_{2}(3^{1} - 2) + 2 = log_{2}(3 - 2) + 2 = log_{2}(1) + 2 = 0 + 2 = 2 ).Both sides are equal, so ( x = 2 ) is a valid solution.Therefore, the solution to the equation is ( x = 2 ).So, summarizing:1. ( 2^{a} + 2^{-a} = frac{4sqrt{3}}{3} ).2. The solution to the equation is ( x = 2 ).**Final Answer**The value of (2^{a} + 2^{-a}) is boxed{dfrac{4sqrt{3}}{3}} and the solution to the equation is boxed{2}.

answer：First, I need to calculate the value of ( P ) by dividing 4020 by 4.Next, I'll find ( Q ) by multiplying ( P ) by 2.Finally, I'll determine ( Y ) by subtracting ( Q ) from ( P ).

answer：First, I need to determine the total area that requires waterproofing. This includes the exterior sidewall and the two triangular ends of the fort.For the rectangular sidewall:- Width = 9 feet- Height = 7 feet- Area = Width × Height = 9 ft × 7 ft = 63 square feetFor each triangular roof face:- Base = 9 feet- Height = 6 feet- Area of one triangle = (1/2) × Base × Height = (1/2) × 9 ft × 6 ft = 27 square feet- Since there are two identical triangles, the total area for both = 27 sq ft × 2 = 54 square feetNow, summing up the areas:- Total area to be covered = Area of sidewall + Total area of triangular ends = 63 sq ft + 54 sq ft = 117 square feetNext, I need to determine how many 10-feet by 15-feet panels are required to cover this area.- Area of one panel = 10 ft × 15 ft = 150 square feetSince the total area to be covered (117 sq ft) is less than the area of one panel (150 sq ft), only one panel is needed.Finally, calculating the total cost:- Cost per panel = 32- Total cost = 1 panel × 32 = 32

answer：First, I need to determine the total number of shirts Jack sews. Since he has 3 kids and sews 3 shirts for each kid, the total number of shirts is 3 multiplied by 3, which equals 9 shirts.Next, I know that Jack uses a total of 63 buttons for all the shirts. To find out how many buttons are sewn into each shirt, I divide the total number of buttons by the total number of shirts. So, 63 buttons divided by 9 shirts equals 7 buttons per shirt.

answer：Okay, so I have this problem about vectors in a plane. Let me try to understand it step by step. We have three vectors: (overrightarrow{a}), (overrightarrow{b}), and (overrightarrow{c}). The magnitudes of (overrightarrow{a}) and (overrightarrow{b}) are both (sqrt{5}), and the magnitude of (overrightarrow{c}) is 1. The condition given is that ((overrightarrow{a} - overrightarrow{c}) cdot (overrightarrow{b} - overrightarrow{c}) = 0). I need to find the range of (|overrightarrow{a} - overrightarrow{b}|). First, let me recall what the dot product means. If two vectors are perpendicular, their dot product is zero. So, the condition ((overrightarrow{a} - overrightarrow{c}) cdot (overrightarrow{b} - overrightarrow{c}) = 0) implies that the vectors (overrightarrow{a} - overrightarrow{c}) and (overrightarrow{b} - overrightarrow{c}) are perpendicular to each other. Let me write down the given condition:[(overrightarrow{a} - overrightarrow{c}) cdot (overrightarrow{b} - overrightarrow{c}) = 0]Expanding this dot product, I get:[overrightarrow{a} cdot overrightarrow{b} - overrightarrow{a} cdot overrightarrow{c} - overrightarrow{b} cdot overrightarrow{c} + overrightarrow{c} cdot overrightarrow{c} = 0]Since (overrightarrow{c} cdot overrightarrow{c}) is just (|overrightarrow{c}|^2), which is 1, the equation simplifies to:[overrightarrow{a} cdot overrightarrow{b} - overrightarrow{a} cdot overrightarrow{c} - overrightarrow{b} cdot overrightarrow{c} + 1 = 0]Let me rearrange this:[overrightarrow{a} cdot overrightarrow{b} + 1 = overrightarrow{a} cdot overrightarrow{c} + overrightarrow{b} cdot overrightarrow{c}]Hmm, I can factor the right-hand side:[overrightarrow{a} cdot overrightarrow{b} + 1 = (overrightarrow{a} + overrightarrow{b}) cdot overrightarrow{c}]Okay, so now I have:[(overrightarrow{a} + overrightarrow{b}) cdot overrightarrow{c} = overrightarrow{a} cdot overrightarrow{b} + 1]I remember that the dot product of two vectors is also equal to the product of their magnitudes times the cosine of the angle between them. So, maybe I can use that to find some bounds.Let me denote (overrightarrow{d} = overrightarrow{a} + overrightarrow{b}). Then, the equation becomes:[overrightarrow{d} cdot overrightarrow{c} = overrightarrow{a} cdot overrightarrow{b} + 1]The left-hand side is the dot product of (overrightarrow{d}) and (overrightarrow{c}), which can be written as:[|overrightarrow{d}| |overrightarrow{c}| cos theta = overrightarrow{a} cdot overrightarrow{b} + 1]Where (theta) is the angle between (overrightarrow{d}) and (overrightarrow{c}). Since (|overrightarrow{c}| = 1), this simplifies to:[|overrightarrow{d}| cos theta = overrightarrow{a} cdot overrightarrow{b} + 1]Now, the maximum value of (cos theta) is 1, and the minimum is -1. So, the absolute value of the left-hand side is less than or equal to (|overrightarrow{d}|). Therefore:[|overrightarrow{a} cdot overrightarrow{b} + 1| leq |overrightarrow{d}| = |overrightarrow{a} + overrightarrow{b}|]So, we have:[|overrightarrow{a} cdot overrightarrow{b} + 1| leq |overrightarrow{a} + overrightarrow{b}|]Let me square both sides to eliminate the absolute value and the square root:[(overrightarrow{a} cdot overrightarrow{b} + 1)^2 leq |overrightarrow{a} + overrightarrow{b}|^2]Expanding the left-hand side:[(overrightarrow{a} cdot overrightarrow{b})^2 + 2 overrightarrow{a} cdot overrightarrow{b} + 1 leq |overrightarrow{a} + overrightarrow{b}|^2]Now, let's compute (|overrightarrow{a} + overrightarrow{b}|^2):[|overrightarrow{a} + overrightarrow{b}|^2 = |overrightarrow{a}|^2 + 2 overrightarrow{a} cdot overrightarrow{b} + |overrightarrow{b}|^2]Given that (|overrightarrow{a}| = |overrightarrow{b}| = sqrt{5}), this becomes:[|overrightarrow{a} + overrightarrow{b}|^2 = 5 + 2 overrightarrow{a} cdot overrightarrow{b} + 5 = 10 + 2 overrightarrow{a} cdot overrightarrow{b}]So, plugging this back into the inequality:[(overrightarrow{a} cdot overrightarrow{b})^2 + 2 overrightarrow{a} cdot overrightarrow{b} + 1 leq 10 + 2 overrightarrow{a} cdot overrightarrow{b}]Subtracting (2 overrightarrow{a} cdot overrightarrow{b}) from both sides:[(overrightarrow{a} cdot overrightarrow{b})^2 + 1 leq 10]Subtracting 10 from both sides:[(overrightarrow{a} cdot overrightarrow{b})^2 - 9 leq 0]Which factors as:[(overrightarrow{a} cdot overrightarrow{b} - 3)(overrightarrow{a} cdot overrightarrow{b} + 3) leq 0]This inequality holds when (overrightarrow{a} cdot overrightarrow{b}) is between -3 and 3. So:[-3 leq overrightarrow{a} cdot overrightarrow{b} leq 3]Now, I need to find the range of (|overrightarrow{a} - overrightarrow{b}|). Let's compute the square of this magnitude:[|overrightarrow{a} - overrightarrow{b}|^2 = |overrightarrow{a}|^2 + |overrightarrow{b}|^2 - 2 overrightarrow{a} cdot overrightarrow{b}]Substituting the known magnitudes:[|overrightarrow{a} - overrightarrow{b}|^2 = 5 + 5 - 2 overrightarrow{a} cdot overrightarrow{b} = 10 - 2 overrightarrow{a} cdot overrightarrow{b}]Since (overrightarrow{a} cdot overrightarrow{b}) is between -3 and 3, let's substitute these bounds:1. When (overrightarrow{a} cdot overrightarrow{b} = 3):[|overrightarrow{a} - overrightarrow{b}|^2 = 10 - 2(3) = 10 - 6 = 4]So, (|overrightarrow{a} - overrightarrow{b}| = 2).2. When (overrightarrow{a} cdot overrightarrow{b} = -3):[|overrightarrow{a} - overrightarrow{b}|^2 = 10 - 2(-3) = 10 + 6 = 16]So, (|overrightarrow{a} - overrightarrow{b}| = 4).Therefore, the range of (|overrightarrow{a} - overrightarrow{b}|) is from 2 to 4. Looking back at the options, option B is [2,4], which matches our result. So, the correct answer should be B.**Final Answer**boxed{B}