answer：Okay, so I have this limit to compute: lim _{n rightarrow infty} frac{(3-n)^{2}+(3+n)^{2}}{(3-n)^{2}-(3+n)^{2}}Hmm, let's see. I remember that when dealing with limits as n approaches infinity, especially with polynomials, the highest degree terms dominate. So maybe I should try to simplify the numerator and the denominator separately first.Starting with the numerator: (3 - n)^2 + (3 + n)^2. Let me expand both squares.First, (3 - n)^2 is 9 - 6n + n^2, and (3 + n)^2 is 9 + 6n + n^2. So adding them together:9 - 6n + n^2 + 9 + 6n + n^2Let me combine like terms. The constants: 9 + 9 = 18. The linear terms: -6n + 6n = 0. The quadratic terms: n^2 + n^2 = 2n^2. So the numerator simplifies to 18 + 2n^2.Now the denominator: (3 - n)^2 - (3 + n)^2. Again, I'll expand both squares.We already have (3 - n)^2 = 9 - 6n + n^2 and (3 + n)^2 = 9 + 6n + n^2. Subtracting the second from the first:(9 - 6n + n^2) - (9 + 6n + n^2)Let me distribute the negative sign: 9 - 6n + n^2 - 9 - 6n - n^2Combine like terms: 9 - 9 = 0. -6n -6n = -12n. n^2 - n^2 = 0. So the denominator simplifies to -12n.So now the expression becomes:frac{18 + 2n^2}{-12n}Hmm, okay. So as n approaches infinity, the dominant terms in the numerator and denominator will be the highest degree terms. In the numerator, that's 2n^2, and in the denominator, it's -12n.So if I factor out the highest degree terms, I can write the numerator as 2n^2(1 + 9/n^2) and the denominator as -12n(1). Wait, actually, let me see:Wait, numerator is 2n^2 + 18, so factor out 2n^2:2n^2(1 + 9/n^2)Denominator is -12n, so factor out -12n:-12nSo the expression becomes:frac{2n^2(1 + 9/n^2)}{-12n}Simplify the constants: 2 / (-12) = -1/6.And n^2 / n = n.So now it's:-frac{1}{6} cdot n cdot (1 + frac{9}{n^2})As n approaches infinity, the term 9/n^2 approaches 0, so the expression inside the parentheses approaches 1. Therefore, the whole expression behaves like -frac{1}{6} cdot n as n becomes very large.But wait, if the expression behaves like -frac{1}{6}n, then as n approaches infinity, the expression will go to negative infinity. So is the limit negative infinity?Wait, but let me double-check my steps because sometimes when simplifying, especially with signs, it's easy to make a mistake.So starting again:Numerator: (3 - n)^2 + (3 + n)^2 = 2n^2 + 18Denominator: (3 - n)^2 - (3 + n)^2 = -12nSo the expression is:frac{2n^2 + 18}{-12n}Divide numerator and denominator by n:frac{2n + 18/n}{-12}As n approaches infinity, 18/n approaches 0, so the expression simplifies to:frac{2n}{-12} = -frac{n}{6}Which indeed goes to negative infinity as n approaches infinity.But wait, let me think again. Maybe I can factor n^2 in the numerator and n in the denominator.So numerator: 2n^2 + 18 = 2n^2(1 + 9/n^2)Denominator: -12n = -12n(1)So the expression is:frac{2n^2(1 + 9/n^2)}{-12n} = frac{2n^2}{-12n} cdot (1 + 9/n^2) = -frac{n}{6} cdot (1 + 9/n^2)As n approaches infinity, 9/n^2 approaches 0, so the term in the parentheses approaches 1. Therefore, the expression behaves like -n/6, which tends to negative infinity.So, yeah, the limit is negative infinity.But wait, let me check if I did the denominator correctly.Original denominator: (3 - n)^2 - (3 + n)^2Let me compute this again:(3 - n)^2 = 9 - 6n + n^2(3 + n)^2 = 9 + 6n + n^2Subtracting the second from the first:(9 - 6n + n^2) - (9 + 6n + n^2) = 9 - 6n + n^2 - 9 - 6n - n^2Simplify: 9 -9 = 0; -6n -6n = -12n; n^2 - n^2 = 0. So yes, it's -12n. Correct.So denominator is -12n.Numerator: 2n^2 + 18.So expression is (2n^2 + 18)/(-12n) = (-2n^2 - 18)/(12n) = (-2n^2)/(12n) - 18/(12n) = (-n/6) - (3)/(2n)As n approaches infinity, the first term goes to negative infinity, the second term goes to 0. So overall, the limit is negative infinity.Therefore, the limit is negative infinity.Wait, but sometimes when you have a negative sign, you have to be careful with the direction.Alternatively, maybe I can divide numerator and denominator by n^2 to see.So numerator: (2n^2 + 18)/n^2 = 2 + 18/n^2Denominator: (-12n)/n^2 = -12/nSo expression becomes:(2 + 18/n^2)/(-12/n) = (2 + 0)/0 as n approaches infinity.Wait, that's 2 divided by 0, but the denominator is approaching 0 from the negative side because -12/n approaches 0 from below.So 2 divided by a very small negative number is negative infinity.Therefore, the limit is negative infinity.So yeah, that's consistent with the previous result.So, to recap:1. Expand numerator and denominator.2. Simplify both.3. Factor out dominant terms.4. Observe behavior as n approaches infinity.5. Confirm with alternative method of dividing numerator and denominator by n^2.Both methods lead to the conclusion that the limit is negative infinity.Therefore, the answer is negative infinity.**Final Answer**The limit is boxed{-infty}.

answer：First, I need to determine how much profit Javier makes per dozen donuts. He buys a dozen for 2.40 and sells each donut for 1, which means he sells a dozen for 12. Next, I'll calculate the profit by subtracting the cost from the selling price: 12 minus 2.40 equals 9.60 profit per dozen.Javier wants to raise 96, so I'll divide his goal by the profit per dozen: 96 divided by 9.60 equals 10.Therefore, Javier needs to buy and sell 10 dozen donuts to reach his goal.

answer：First, I need to determine how many nickels Ray has initially. Since each nickel is worth 5 cents and Ray has 175 cents, I divide 175 by 5 to find the total number of nickels.Next, Ray gives 30 cents to Peter. To find out how many nickels this is, I divide 30 by 5, which equals 6 nickels.Then, Ray gives twice as many cents to Randi as he gave to Peter. Since he gave 30 cents to Peter, he gives 60 cents to Randi. Dividing 60 by 5 gives me 12 nickels for Randi.Finally, to find out how many more nickels Randi has than Peter, I subtract the number of nickels Peter received from the number Randi received: 12 minus 6 equals 6 more nickels.

answer：Okay, so I have this problem here: Given that x is greater than 0, y is greater than 0, and x plus 2y equals 1, I need to find the minimum value of 1 over x plus 1 over (2y). Hmm, okay. Let me think about how to approach this.First, let me write down what I know. The constraints are x > 0, y > 0, and x + 2y = 1. The function I need to minimize is f(x, y) = 1/x + 1/(2y). So, I have a function of two variables with a linear constraint. This seems like a problem that can be solved using methods from calculus, maybe Lagrange multipliers, or perhaps even using the AM-HM inequality since it's about minimizing a sum of reciprocals.Let me try both approaches and see which one works better.Starting with the calculus approach. Since I have a constraint, I can express one variable in terms of the other and substitute it into the function. Let's solve the constraint equation for x: x = 1 - 2y. Since x must be greater than 0, 1 - 2y > 0, which implies y < 1/2. Similarly, since y must be greater than 0, y is in the interval (0, 1/2).Now, substitute x = 1 - 2y into the function f(x, y):f(y) = 1/(1 - 2y) + 1/(2y).So now, f is a function of a single variable y, and I can find its minimum by taking the derivative and setting it equal to zero.Let me compute the derivative f'(y). First, f(y) = (1 - 2y)^{-1} + (2y)^{-1}.The derivative of (1 - 2y)^{-1} with respect to y is (-1)*(1 - 2y)^{-2}*(-2) = 2/(1 - 2y)^2.Similarly, the derivative of (2y)^{-1} is (-1)*(2y)^{-2}*2 = -1/(2y^2).So, f'(y) = 2/(1 - 2y)^2 - 1/(2y^2).To find critical points, set f'(y) = 0:2/(1 - 2y)^2 - 1/(2y^2) = 0.Let me move one term to the other side:2/(1 - 2y)^2 = 1/(2y^2).Multiply both sides by 2y^2(1 - 2y)^2 to eliminate denominators:2 * 2y^2 = (1 - 2y)^2.Simplify left side: 4y^2.Right side: (1 - 2y)^2 = 1 - 4y + 4y^2.So, 4y^2 = 1 - 4y + 4y^2.Subtract 4y^2 from both sides:0 = 1 - 4y.So, 1 - 4y = 0 => y = 1/4.Okay, so y = 1/4 is a critical point. Let's check if this is a minimum.First, let's compute the second derivative or test the intervals around y = 1/4.Alternatively, since it's the only critical point in the interval (0, 1/2), and the function tends to infinity as y approaches 0 or 1/2, it must be a minimum.So, y = 1/4. Then, x = 1 - 2y = 1 - 2*(1/4) = 1 - 1/2 = 1/2.So, x = 1/2, y = 1/4.Now, compute f(x, y) at this point:f(1/2, 1/4) = 1/(1/2) + 1/(2*(1/4)) = 2 + 1/(1/2) = 2 + 2 = 4.So, the minimum value is 4.Wait, let me check if that's correct. Alternatively, maybe I can use the AM-HM inequality or Cauchy-Schwarz.Let me try the AM-HM inequality approach.We have two positive numbers, x and 2y, since x > 0 and y > 0, so 2y > 0. Their sum is x + 2y = 1.We need to minimize 1/x + 1/(2y). Let me denote a = x and b = 2y, so a + b = 1, and we need to minimize 1/a + 1/b.This is a classic problem where the harmonic mean is involved. The expression 1/a + 1/b is related to the harmonic mean.Recall that for positive numbers a and b, the harmonic mean H is 2ab/(a + b), and the arithmetic mean A is (a + b)/2. We know that H ≤ A, but that might not directly help here.Alternatively, using the Cauchy-Schwarz inequality:(1/a + 1/b)(a + b) ≥ (1 + 1)^2 = 4.Since a + b = 1, this gives:(1/a + 1/b)*1 ≥ 4 => 1/a + 1/b ≥ 4.So, the minimum value of 1/a + 1/b is 4, achieved when a = b.Since a = x and b = 2y, a = b implies x = 2y.But from the constraint, x + 2y = 1, substituting x = 2y:2y + 2y = 1 => 4y = 1 => y = 1/4, so x = 2*(1/4) = 1/2.So, same result as before. Therefore, the minimum value is indeed 4.So, whether I use calculus or the Cauchy-Schwarz inequality, I get the same answer. That gives me more confidence that it's correct.Alternatively, I could also use substitution and then apply the method of completing the square or another algebraic method, but in this case, calculus and Cauchy-Schwarz both worked smoothly.Let me just recap the steps:1. Recognize that the problem involves minimizing a function subject to a linear constraint.2. Use substitution to express the function in terms of a single variable.3. Take the derivative, set it to zero, solve for the critical point.4. Verify that the critical point is indeed a minimum, either by second derivative test or by considering the behavior of the function.5. Alternatively, use an inequality like Cauchy-Schwarz to find the minimum directly.Both methods led me to the same conclusion, so I think it's safe to say that the minimum value is 4.**Final Answer**The minimum value is boxed{4}.

answer：First, I'll define variables for each team's recruitment:- Let ( t ) represent the number of people recruited by the track team.- The choir recruits twice as many as the track team, so choir recruits ( 2t ).- The dance team recruits 10 more than the choir, so dance team recruits ( 2t + 10 ).Next, I'll set up the equation based on the total recruitment:[t + 2t + (2t + 10) = 100]Combining like terms:[5t + 10 = 100]Subtracting 10 from both sides:[5t = 90]Dividing both sides by 5:[t = 18]Now, calculating the dance team's recruitment:[2t + 10 = 2 times 18 + 10 = 46]Therefore, the dance team recruits 46 people.

answer：Alright, so I have this problem here: given the function ( f(x) = frac{1}{2x + b} ), I need to find the value of ( b ) such that the inverse function ( f^{-1}(x) ) is equal to ( frac{1 - 2x}{2x} ). Hmm, okay. Let me try to figure this out step by step.First, I remember that to find the inverse of a function, you typically switch the roles of ( x ) and ( y ) and then solve for ( y ). So, let me start by writing the function as an equation:( y = frac{1}{2x + b} )To find the inverse, I'll swap ( x ) and ( y ):( x = frac{1}{2y + b} )Now, I need to solve this equation for ( y ). Let's do that step by step. Starting with:( x = frac{1}{2y + b} )I can take the reciprocal of both sides to get rid of the fraction:( frac{1}{x} = 2y + b )Now, subtract ( b ) from both sides:( frac{1}{x} - b = 2y )Then, divide both sides by 2:( y = frac{frac{1}{x} - b}{2} )Simplify that expression:( y = frac{1 - b x}{2x} )So, the inverse function ( f^{-1}(x) ) that I found is ( frac{1 - b x}{2x} ). But the problem states that ( f^{-1}(x) = frac{1 - 2x}{2x} ). Hmm, so I need to set the expression I found equal to the given inverse function and solve for ( b ).Let me write that out:( frac{1 - b x}{2x} = frac{1 - 2x}{2x} )Since the denominators are the same and non-zero (as long as ( x neq 0 )), I can set the numerators equal to each other:( 1 - b x = 1 - 2x )Now, subtract 1 from both sides:( -b x = -2x )Multiply both sides by -1 to make it simpler:( b x = 2x )Now, assuming ( x neq 0 ), we can divide both sides by ( x ):( b = 2 )So, it looks like ( b ) must be 2. Let me double-check my steps to make sure I didn't make a mistake.Starting with ( f(x) = frac{1}{2x + b} ), I found the inverse by swapping ( x ) and ( y ), which gave me ( x = frac{1}{2y + b} ). Then, solving for ( y ), I correctly took reciprocals, subtracted ( b ), divided by 2, and simplified to get ( y = frac{1 - b x}{2x} ). Setting this equal to the given inverse function ( frac{1 - 2x}{2x} ), I equated the numerators and solved for ( b ), which gave me ( b = 2 ). Just to be thorough, let me plug ( b = 2 ) back into the original function and see if its inverse is indeed ( frac{1 - 2x}{2x} ).Original function with ( b = 2 ):( f(x) = frac{1}{2x + 2} )Let's find its inverse. Start with:( y = frac{1}{2x + 2} )Swap ( x ) and ( y ):( x = frac{1}{2y + 2} )Take reciprocals:( frac{1}{x} = 2y + 2 )Subtract 2:( frac{1}{x} - 2 = 2y )Divide by 2:( y = frac{frac{1}{x} - 2}{2} )Simplify:( y = frac{1 - 2x}{2x} )Which matches the given inverse function. So, yes, ( b = 2 ) is correct.I think that's solid. I don't see any errors in my reasoning, and plugging it back in confirms the result. So, I'm confident that ( b = 2 ) is the right answer.**Final Answer**The value of ( b ) is boxed{2}.