💡To determine the number of floors in the first wing of Linda's hotel, let's break down the problem step by step.1. **Define Variables:** - Let ( F ) represent the number of floors in the first wing.2. **Calculate Rooms in the First Wing:** - Each floor in the first wing has 6 halls. - Each hall in the first wing contains 32 rooms. - Therefore, the total number of rooms in the first wing is: [ F times 6 times 32 ]3. **Calculate Rooms in the Second Wing:** - The second wing has 7 floors. - Each floor in the second wing has 9 halls. - Each hall in the second wing contains 40 rooms. - Therefore, the total number of rooms in the second wing is: [ 7 times 9 times 40 = 2520 ]4. **Set Up the Equation for Total Rooms:** - The total number of rooms in the hotel is 4248. - Thus, the equation becomes: [ F times 6 times 32 + 2520 = 4248 ]5. **Solve for ( F ):** - Subtract 2520 from both sides: [ F times 192 = 1728 ] - Divide both sides by 192: [ F = frac{1728}{192} = 9 ]**Conclusion:** The first wing of the hotel has 9 floors.

💡Okay, so I need to figure out how many roses Tom sent his wife in total. Let's see, he sends 2 dozen roses every day for a week. Hmm, a dozen is 12, right? So 2 dozen would be 2 times 12, which is 24 roses each day. Now, a week has 7 days, so if he sends 24 roses every day for 7 days, I guess I need to multiply 24 by 7 to get the total number of roses. Let me do that: 24 times 7. Okay, 24 times 7 is... let's see, 20 times 7 is 140, and 4 times 7 is 28, so adding those together, 140 plus 28 equals 168. So, Tom sent his wife a total of 168 roses over the week. That seems right. I don't think I missed anything here. It's just a simple multiplication problem, but it's good to break it down step by step to make sure I didn't make any mistakes.

💡Okay, so I have this function f(x) = x³ - a x - 1, and I need to do two things. First, when a is 8, find the equation of the tangent line at x = 0. Second, discuss the monotonicity of the function for general a. Hmm, let's start with the first part.Alright, for part 1, when a = 8, the function becomes f(x) = x³ - 8x - 1. I remember that to find the tangent line at a specific point, I need two things: the slope of the tangent line at that point and a point through which the tangent line passes. The slope is given by the derivative of the function at that point, and the point is (x, f(x)).So, first, I need to find the derivative of f(x). The derivative of x³ is 3x², the derivative of -8x is -8, and the derivative of -1 is 0. So, f'(x) = 3x² - 8. Now, I need to evaluate this derivative at x = 0 to get the slope of the tangent line at that point.Plugging in x = 0, f'(0) = 3*(0)² - 8 = -8. So, the slope of the tangent line is -8.Next, I need a point on the tangent line. Since we're looking at x = 0, I plug x = 0 into the original function f(x). So, f(0) = (0)³ - 8*(0) - 1 = -1. Therefore, the point is (0, -1).Now, with the slope and a point, I can write the equation of the tangent line using the point-slope form: y - y₁ = m(x - x₁). Plugging in the values, we get y - (-1) = -8(x - 0), which simplifies to y + 1 = -8x. Subtracting 1 from both sides gives y = -8x - 1.Wait, the problem asks for the equation of the tangent line. I think that's it, but maybe they want it in a different form. If I rearrange it, it would be 8x + y + 1 = 0. That seems like a standard linear equation form. I think either form is acceptable, but maybe the second one is preferred because it's written in terms of x and y without fractions or decimals.Okay, that seems solid for part 1. Now, moving on to part 2: discussing the monotonicity of f(x) = x³ - a x - 1. Monotonicity refers to where the function is increasing or decreasing. To determine this, I need to look at the derivative of the function, which we already found earlier: f'(x) = 3x² - a.The sign of the derivative tells us about the function's behavior. If f'(x) > 0, the function is increasing at that point; if f'(x) < 0, it's decreasing. So, let's analyze f'(x) = 3x² - a.First, let's consider different cases for the value of a.Case 1: a ≤ 0. If a is zero or negative, then -a is non-negative. So, f'(x) = 3x² - a becomes 3x² + |a| (since a is negative, -a is positive). Since 3x² is always non-negative and we're adding a non-negative number, f'(x) is always non-negative. Therefore, the function is always increasing when a ≤ 0.Case 2: a > 0. Now, this is more interesting because the derivative becomes 3x² - a, which can be positive or negative depending on the value of x.Let's set f'(x) = 0 to find critical points: 3x² - a = 0 ⇒ x² = a/3 ⇒ x = ±√(a/3). These are the points where the derivative is zero, so the function could be changing its monotonicity here.To determine where the derivative is positive or negative, let's analyze intervals around these critical points.For x < -√(a/3): Let's pick a test point, say x = -√(a/3) - 1. Plugging into f'(x), we get 3*(-√(a/3) - 1)² - a. Expanding this, (-√(a/3) - 1)² = (a/3) + 2√(a/3) + 1. Multiplying by 3 gives a + 6√(a/3) + 3. Subtracting a, we get 6√(a/3) + 3, which is positive. So, f'(x) > 0 in this interval.For -√(a/3) < x < √(a/3): Let's pick x = 0 as a test point. f'(0) = 0 - a = -a, which is negative since a > 0. So, f'(x) < 0 in this interval.For x > √(a/3): Let's pick x = √(a/3) + 1. Plugging into f'(x), we get 3*(√(a/3) + 1)² - a. Expanding, (√(a/3) + 1)² = (a/3) + 2√(a/3) + 1. Multiplying by 3 gives a + 6√(a/3) + 3. Subtracting a, we get 6√(a/3) + 3, which is positive. So, f'(x) > 0 in this interval.Putting this all together, when a > 0, the function f(x) is increasing on the intervals (-∞, -√(a/3)) and (√(a/3), ∞), and decreasing on the interval (-√(a/3), √(a/3)).Wait, let me double-check that. So, when a is positive, the derivative is positive outside the interval between -√(a/3) and √(a/3), meaning the function is increasing there, and negative inside that interval, meaning decreasing. Yeah, that makes sense because the function is a cubic, which typically has one local maximum and one local minimum, leading to this kind of increasing-decreasing-increasing behavior.So, summarizing:- If a ≤ 0, f(x) is always increasing.- If a > 0, f(x) is increasing on (-∞, -√(a/3)) and (√(a/3), ∞), and decreasing on (-√(a/3), √(a/3)).I think that's a thorough analysis. I don't see any mistakes in my reasoning, but let me just recap to make sure.For part 1, I found the derivative at x=0 when a=8, got the slope, found the point, and wrote the tangent line equation. For part 2, I considered the derivative and analyzed its sign based on the value of a, breaking it into cases where a is non-positive or positive, and then finding critical points and testing intervals. Everything seems consistent.**Final Answer**1. The equation of the tangent line is boxed{8x + y + 1 = 0}.2. The function is increasing on boxed{(-infty, -sqrt{frac{a}{3}}) cup (sqrt{frac{a}{3}}, infty)} and decreasing on boxed{left(-sqrt{frac{a}{3}}, sqrt{frac{a}{3}}right)} when ( a > 0 ). When ( a leq 0 ), the function is increasing on boxed{mathbb{R}}.

💡Okay, so I have this problem: I need to find all values of ( a ) for which the sum of the roots of the equation ( x^2 - 2a(x - 1) - 1 = 0 ) is equal to the sum of the squares of the roots. Hmm, let me try to figure this out step by step.First, I remember that for a quadratic equation of the form ( x^2 + bx + c = 0 ), the sum of the roots is ( -b ) and the product of the roots is ( c ). This is known as Vieta's formulas. So, maybe I can rewrite the given equation in the standard quadratic form and then apply Vieta's formulas.Let me start by expanding and simplifying the given equation:( x^2 - 2a(x - 1) - 1 = 0 )Expanding the middle term:( x^2 - 2a cdot x + 2a - 1 = 0 )So, now it's in the form ( x^2 + (-2a)x + (2a - 1) = 0 ). Using Vieta's formulas, the sum of the roots ( x_1 + x_2 ) is equal to ( 2a ) (since the coefficient of ( x ) is ( -2a ), and Vieta's formula says sum is ( -(-2a) = 2a )).The product of the roots ( x_1 x_2 ) is equal to ( 2a - 1 ).Now, the problem states that the sum of the roots is equal to the sum of the squares of the roots. So, I need to express the sum of the squares of the roots in terms of the sum and product of the roots.I recall that ( x_1^2 + x_2^2 ) can be written as ( (x_1 + x_2)^2 - 2x_1 x_2 ). Let me verify that:( (x_1 + x_2)^2 = x_1^2 + 2x_1 x_2 + x_2^2 )So, subtracting ( 2x_1 x_2 ) from both sides gives:( (x_1 + x_2)^2 - 2x_1 x_2 = x_1^2 + x_2^2 )Yes, that's correct. So, the sum of the squares of the roots is ( (x_1 + x_2)^2 - 2x_1 x_2 ).Given that the sum of the roots is equal to the sum of the squares of the roots, we have:( x_1 + x_2 = x_1^2 + x_2^2 )Substituting the expressions we have from Vieta's formulas:( 2a = (2a)^2 - 2(2a - 1) )Let me compute the right-hand side step by step:First, ( (2a)^2 = 4a^2 )Then, ( 2(2a - 1) = 4a - 2 )So, substituting back:( 2a = 4a^2 - (4a - 2) )Wait, no, hold on. The expression is ( (2a)^2 - 2(2a - 1) ), which is ( 4a^2 - 2(2a - 1) ). So, let me compute that:( 4a^2 - 2(2a - 1) = 4a^2 - 4a + 2 )So, the equation becomes:( 2a = 4a^2 - 4a + 2 )Now, let's bring all terms to one side to form a quadratic equation:( 4a^2 - 4a + 2 - 2a = 0 )Simplify the terms:( 4a^2 - 6a + 2 = 0 )Hmm, this is a quadratic equation in terms of ( a ). Let me write it as:( 4a^2 - 6a + 2 = 0 )I can try to simplify this equation by dividing all terms by 2 to make the numbers smaller:( 2a^2 - 3a + 1 = 0 )Now, this is a simpler quadratic equation. I can solve for ( a ) using the quadratic formula. The quadratic formula is:( a = frac{-b pm sqrt{b^2 - 4ac}}{2a} )Wait, hold on, that might be confusing because the variable is also ( a ). Let me rewrite the equation as:( 2a^2 - 3a + 1 = 0 )So, here, the coefficients are:- ( A = 2 )- ( B = -3 )- ( C = 1 )So, applying the quadratic formula:( a = frac{-B pm sqrt{B^2 - 4AC}}{2A} )Plugging in the values:( a = frac{-(-3) pm sqrt{(-3)^2 - 4 cdot 2 cdot 1}}{2 cdot 2} )Simplify step by step:First, compute the numerator:- ( -(-3) = 3 )- Inside the square root: ( (-3)^2 = 9 ), and ( 4 cdot 2 cdot 1 = 8 )- So, ( 9 - 8 = 1 )So, the expression becomes:( a = frac{3 pm sqrt{1}}{4} )Since ( sqrt{1} = 1 ), this simplifies to:( a = frac{3 pm 1}{4} )Therefore, there are two solutions:1. ( a = frac{3 + 1}{4} = frac{4}{4} = 1 )2. ( a = frac{3 - 1}{4} = frac{2}{4} = frac{1}{2} )So, the values of ( a ) that satisfy the condition are ( a = 1 ) and ( a = frac{1}{2} ).Let me just double-check my steps to make sure I didn't make any mistakes.1. Expanded the original equation correctly: ( x^2 - 2a(x - 1) - 1 = 0 ) becomes ( x^2 - 2ax + 2a - 1 = 0 ). That seems right.2. Applied Vieta's formulas correctly: sum of roots ( x_1 + x_2 = 2a ), product ( x_1 x_2 = 2a - 1 ). That seems correct.3. Expressed sum of squares as ( (x_1 + x_2)^2 - 2x_1 x_2 ). That identity is correct.4. Substituted the values into the equation ( 2a = (2a)^2 - 2(2a - 1) ). That seems right.5. Expanded and simplified to get ( 4a^2 - 6a + 2 = 0 ), then divided by 2 to get ( 2a^2 - 3a + 1 = 0 ). That's correct.6. Applied quadratic formula correctly, got solutions ( a = 1 ) and ( a = frac{1}{2} ). That seems correct.I think all the steps are correct. So, the values of ( a ) are 1 and 1/2.**Final Answer**The values of ( a ) are boxed{1} and boxed{dfrac{1}{2}}.

💡First, I need to calculate the number of logs Jerry gets from each type of tree.For the pine trees, Jerry cuts up 8 trees, and each pine tree produces 80 logs. So, the total number of logs from pine trees is:8 pine trees × 80 logs per pine tree = 640 logsNext, for the maple trees, Jerry cuts up 3 trees, and each maple tree produces 60 logs. Therefore, the total number of logs from maple trees is:3 maple trees × 60 logs per maple tree = 180 logsThen, for the walnut trees, Jerry cuts up 4 trees, and each walnut tree produces 100 logs. Thus, the total number of logs from walnut trees is:4 walnut trees × 100 logs per walnut tree = 400 logsFinally, to find the total number of logs Jerry has, I add up the logs from all three types of trees:640 logs (from pine) + 180 logs (from maple) + 400 logs (from walnut) = 1220 logsTherefore, Jerry gets a total of 1220 logs.

💡First, I need to determine the value of ( a ) in the equation ( ax - 8 = 20 + a ) given that ( x = 5 ) is a solution.I'll start by substituting ( x = 5 ) into the equation:[5a - 8 = 20 + a]Next, I'll move all terms containing ( a ) to one side of the equation and the constant terms to the other side. I'll subtract ( a ) from both sides:[5a - a - 8 = 20]This simplifies to:[4a - 8 = 20]Now, I'll isolate the term with ( a ) by adding 8 to both sides:[4a = 28]Finally, I'll solve for ( a ) by dividing both sides by 4:[a = 7]Therefore, the value of ( a ) is 7.