💡To determine how long it takes Anne to clean the house on her own, we'll analyze the given information step by step.First, let's denote:- ( B ) as Bruce's cleaning rate (houses per hour)- ( A ) as Anne's cleaning rate (houses per hour)**Step 1: Combined Cleaning Rate**When Bruce and Anne work together, they can clean the house in 4 hours. This gives us the equation:[(B + A) times 4 = 1 quad Rightarrow quad 4B + 4A = 1]**Step 2: Anne's Doubled Speed**If Anne's speed is doubled, her new rate becomes ( 2A ). With this increased rate, they can clean the house in 3 hours:[(B + 2A) times 3 = 1 quad Rightarrow quad 3B + 6A = 1]**Step 3: Solving the System of Equations**We have the system:[begin{cases}4B + 4A = 1 3B + 6A = 1end{cases}]To eliminate ( B ), multiply the first equation by 3 and the second equation by 4:[begin{cases}12B + 12A = 3 12B + 24A = 4end{cases}]Subtract the first equation from the second:[12A = 1 quad Rightarrow quad A = frac{1}{12}]**Step 4: Time for Anne to Clean Alone**Anne's rate is ( frac{1}{12} ) houses per hour. Therefore, the time she takes to clean one house alone is:[text{Time} = frac{1}{A} = 12 text{ hours}]

💡First, I need to identify the components of the arithmetic series. The first term ( a ) is (-45), and the common difference ( d ) is ( 3 ).Next, I'll determine the number of terms ( n ) in the series. Since the last term is ( 0 ), I can set up the equation for the ( n )-th term of an arithmetic sequence:[a + (n-1)d = 0]Substituting the known values:[-45 + (n-1) cdot 3 = 0]Solving for ( n ):[(n-1) cdot 3 = 45 n-1 = 15 n = 16]There are 16 terms in the series.Now, I'll use the sum formula for an arithmetic series:[S = frac{n}{2} [2a + (n-1)d]]Plugging in the values:[S = frac{16}{2} [2(-45) + (16-1) cdot 3] S = 8 [-90 + 45] S = 8 cdot (-45) S = -360]

💡Alright, so I've got this logarithmic expression to evaluate: frac {lg8 + lg125 - lg2 - lg5}{lg sqrt {10} cdot lg0.1}Hmm, okay. Let me try to break this down step by step. First, I know that "lg" usually stands for logarithm base 10, so all these are log base 10 functions. That helps because I can use properties of logarithms to simplify this expression.Starting with the numerator: lg8 + lg125 - lg2 - lg5. I remember that logarithms of products can be broken down into sums, and logarithms of quotients can be broken down into differences. So maybe I can combine these terms somehow.Looking at lg8 and lg125, I notice that 8 is 2^3 and 125 is 5^3. So, using the power rule of logarithms, which says that lg(a^b) = b cdot lga, I can rewrite these terms:lg8 = lg(2^3) = 3 cdot lg2lg125 = lg(5^3) = 3 cdot lg5So substituting back into the numerator:3 cdot lg2 + 3 cdot lg5 - lg2 - lg5Now, let's combine like terms:For the lg2 terms: 3 cdot lg2 - lg2 = 2 cdot lg2For the lg5 terms: 3 cdot lg5 - lg5 = 2 cdot lg5So the numerator simplifies to:2 cdot lg2 + 2 cdot lg5I can factor out the 2:2(lg2 + lg5)Now, I recall that lg2 + lg5 is the same as lg(2 cdot 5) because the sum of logs is the log of the product. So:lg2 + lg5 = lg(2 cdot 5) = lg10And since lg10 is log base 10 of 10, which is 1, because any log base a of a is 1. So:lg2 + lg5 = 1Therefore, the numerator becomes:2 cdot 1 = 2Okay, so the numerator simplifies nicely to 2. Now, let's tackle the denominator: lg sqrt{10} cdot lg0.1First, let's simplify each logarithm separately.Starting with lg sqrt{10}. I know that the square root of 10 is the same as 10^{1/2}. Using the power rule again:lg sqrt{10} = lg(10^{1/2}) = frac{1}{2} cdot lg10And since lg10 = 1, this simplifies to:frac{1}{2} cdot 1 = frac{1}{2}Next, lg0.1. I know that 0.1 is the same as frac{1}{10}, which is 10^{-1}. So:lg0.1 = lg(10^{-1}) = -1 cdot lg10 = -1 cdot 1 = -1So now, the denominator is:frac{1}{2} cdot (-1) = -frac{1}{2}Putting it all together, the entire expression becomes:frac{2}{-frac{1}{2}}Dividing by a fraction is the same as multiplying by its reciprocal, so:2 div -frac{1}{2} = 2 times (-2) = -4So, the value of the expression is -4.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Simplified the numerator correctly using logarithm properties.2. Correctly identified that lg2 + lg5 = lg10 = 1.3. Simplified the denominator by expressing square root and 0.1 as powers of 10.4. Calculated each logarithm correctly and multiplied them.5. Divided the simplified numerator by the simplified denominator correctly.Everything seems to check out. I think -4 is the right answer.**Final Answer**The value of the expression is boxed{-4}.

💡First, identify the two possible positions for athlete A: either the first or the fourth runner.**Case 1: A is the first runner**If A is the first runner, the remaining three positions (second, third, and fourth) need to be filled by athletes B, C, and D. The number of ways to arrange three athletes in three positions is calculated using the factorial of 3:3! = 3 × 2 × 1 = 6So, there are 6 possible sequences when A is the first runner.**Case 2: A is the fourth runner**If A is the fourth runner, the first three positions need to be filled by athletes B, C, and D. Again, the number of ways to arrange three athletes in three positions is:3! = 3 × 2 × 1 = 6So, there are 6 possible sequences when A is the fourth runner.**Total Number of Sequences**To find the total number of different sequences, add the number of sequences from both cases:6 (when A is first) + 6 (when A is fourth) = 12Therefore, there are 12 different sequences of runners.

💡Okay, so I need to find the remainder when 2,368,297 is divided by 3. Hmm, I remember there's a trick with the sum of the digits for checking divisibility by 3. Let me try to recall how that works.First, I think I should add up all the digits of the number. The number is 2,368,297. Let me write down the digits separately: 2, 3, 6, 8, 2, 9, 7. Now, I'll add them one by one.Starting with 2, then adding 3 gives me 5. Next, adding 6 makes it 11. Then, adding 8 brings the total to 19. Adding another 2 makes it 21. Now, adding 9 gives me 30, and finally, adding 7 brings the sum up to 37. So, the sum of the digits is 37.Now, I need to see what the remainder is when 37 is divided by 3. Let me do that division. 3 goes into 37 how many times? Well, 3 times 12 is 36, which is just one less than 37. So, 37 divided by 3 is 12 with a remainder of 1. That means the remainder when 2,368,297 is divided by 3 should also be 1.Wait, let me double-check my addition to make sure I didn't make a mistake. 2 + 3 is 5, plus 6 is 11, plus 8 is 19, plus 2 is 21, plus 9 is 30, plus 7 is 37. Yep, that seems right. And 37 divided by 3 is definitely 12 with a remainder of 1. So, I think I'm confident that the remainder is 1.

💡Okay, so I have this function f(x) defined piecewise. For x less than 0, it's a quadratic function x² + 2x + a, and for x greater than 0, it's the natural logarithm ln x. There are three parts to the problem, and I need to tackle them one by one.Starting with part (I): Identify the intervals of monotonicity for f(x). Monotonicity refers to where the function is increasing or decreasing. Since f(x) is piecewise, I can analyze each piece separately.First, for x < 0, f(x) = x² + 2x + a. To find where this quadratic is increasing or decreasing, I should find its derivative. The derivative f’(x) for x < 0 is 2x + 2. To find critical points, set f’(x) = 0: 2x + 2 = 0 → x = -1. So, the function changes its monotonicity at x = -1.Now, let's test intervals around x = -1. For x < -1, say x = -2, f’(-2) = 2*(-2) + 2 = -4 + 2 = -2, which is negative, so the function is decreasing on (-∞, -1). For x between -1 and 0, say x = -0.5, f’(-0.5) = 2*(-0.5) + 2 = -1 + 2 = 1, which is positive, so the function is increasing on (-1, 0).Next, for x > 0, f(x) = ln x. The derivative here is f’(x) = 1/x, which is always positive for x > 0. So, the function is increasing on (0, ∞).Putting it all together, f(x) is decreasing on (-∞, -1), increasing on [-1, 0), and increasing on (0, ∞). So, that's part (I) done.Moving on to part (II): If the tangents to the graph of f(x) at points A and B are perpendicular, and x₂ < 0, find the minimum value of x₂ - x₁.Given that A and B are points on the graph with x₁ < x₂, and both x₁ and x₂ are less than 0 since x₂ < 0. So both points are on the quadratic part of the function.The tangent lines at A and B are perpendicular, which means the product of their slopes is -1. The slope of the tangent at any point x on the quadratic is f’(x) = 2x + 2.So, the slopes at A and B are 2x₁ + 2 and 2x₂ + 2 respectively. Therefore, (2x₁ + 2)(2x₂ + 2) = -1.Let me denote m₁ = 2x₁ + 2 and m₂ = 2x₂ + 2. So, m₁ * m₂ = -1.Since x₁ < x₂ < 0, let's see what m₁ and m₂ are. For x < 0, 2x + 2 is a linear function. At x = -1, it's 0. For x < -1, 2x + 2 is negative, and for x between -1 and 0, it's positive.Given x₁ < x₂ < 0, and since x₁ < x₂, x₁ could be less than -1, and x₂ could be between -1 and 0, or both could be on the same side. But since the product of the slopes is -1, which is negative, one slope must be positive and the other negative. So, one of m₁ or m₂ is positive, and the other is negative.Given that x₁ < x₂, and x₂ < 0, it's possible that x₁ < -1 and x₂ > -1, so m₁ is negative and m₂ is positive. That would make their product negative, which is what we need.So, let's assume x₁ < -1 and x₂ > -1, so m₁ = 2x₁ + 2 < 0 and m₂ = 2x₂ + 2 > 0.We have m₁ * m₂ = -1. Let me write that as:(2x₁ + 2)(2x₂ + 2) = -1.Let me expand this:4x₁x₂ + 4x₁ + 4x₂ + 4 = -1.Simplify:4x₁x₂ + 4x₁ + 4x₂ + 4 + 1 = 0 → 4x₁x₂ + 4x₁ + 4x₂ + 5 = 0.Hmm, not sure if this helps directly. Maybe another approach.Since m₁ * m₂ = -1, and m₁ = 2x₁ + 2, m₂ = 2x₂ + 2.Let me denote u = 2x₁ + 2 and v = 2x₂ + 2. Then u * v = -1.Also, since x₁ < x₂, and x₁ < -1, x₂ > -1, so u < 0 and v > 0.We need to find the minimum value of x₂ - x₁.Express x₁ and x₂ in terms of u and v:x₁ = (u - 2)/2,x₂ = (v - 2)/2.So, x₂ - x₁ = [(v - 2)/2] - [(u - 2)/2] = (v - u)/2.We need to minimize (v - u)/2, given that u * v = -1, u < 0, v > 0.So, we can consider u = -k, where k > 0, and v = l, where l > 0. Then, (-k) * l = -1 → k * l = 1.So, we have k * l = 1, and we need to minimize (l - (-k))/2 = (l + k)/2.So, minimize (k + l)/2, given that k * l = 1, k > 0, l > 0.This is a standard optimization problem. The minimum of k + l given k * l = 1 occurs when k = l = 1, by AM ≥ GM.So, the minimum of (k + l)/2 is (1 + 1)/2 = 1.Therefore, the minimum value of x₂ - x₁ is 1.Wait, let me verify. If k = l = 1, then u = -1, v = 1.So, x₁ = (u - 2)/2 = (-1 - 2)/2 = -3/2,x₂ = (v - 2)/2 = (1 - 2)/2 = -1/2.So, x₂ - x₁ = (-1/2) - (-3/2) = (-1/2 + 3/2) = 1. Yep, that checks out.So, the minimum value is 1.Now, part (III): If the tangents to the graph of f(x) at points A and B coincide, determine the range of values for a.So, the tangents at A and B are the same line. That means both the slopes and the y-intercepts must be equal.Given that A and B are points on the graph, and x₁ < x₂. Since f(x) is defined differently for x < 0 and x > 0, we need to consider the cases where A and B are both on the quadratic part, both on the logarithmic part, or one on each.But, if both are on the quadratic part (x < 0), then their derivatives would be 2x + 2. For the tangents to coincide, their slopes must be equal, so 2x₁ + 2 = 2x₂ + 2 → x₁ = x₂, which contradicts x₁ < x₂. Similarly, if both are on the logarithmic part (x > 0), their derivatives are 1/x, so 1/x₁ = 1/x₂ → x₁ = x₂, again a contradiction. Therefore, A and B must be on different pieces: A on x < 0 and B on x > 0.So, A is (x₁, f(x₁)) with x₁ < 0, and B is (x₂, f(x₂)) with x₂ > 0.The tangent at A: f(x) = x² + 2x + a, so f’(x) = 2x + 2. The slope at A is m = 2x₁ + 2.The equation of the tangent line at A is:y = f(x₁) + m(x - x₁) = (x₁² + 2x₁ + a) + (2x₁ + 2)(x - x₁).Simplify:y = x₁² + 2x₁ + a + (2x₁ + 2)x - (2x₁ + 2)x₁= x₁² + 2x₁ + a + (2x₁ + 2)x - 2x₁² - 2x₁= -x₁² + a + (2x₁ + 2)xSo, the tangent line is y = (2x₁ + 2)x + (-x₁² + a).Similarly, the tangent at B: f(x) = ln x, so f’(x) = 1/x. The slope at B is m = 1/x₂.The equation of the tangent line at B is:y = f(x₂) + m(x - x₂) = ln x₂ + (1/x₂)(x - x₂)Simplify:y = ln x₂ + (x/x₂) - 1= (1/x₂)x + (ln x₂ - 1)So, the tangent line is y = (1/x₂)x + (ln x₂ - 1).For these two tangent lines to coincide, their slopes and y-intercepts must be equal.So, set the slopes equal:2x₁ + 2 = 1/x₂. (Equation 1)And set the y-intercepts equal:-x₁² + a = ln x₂ - 1. (Equation 2)From Equation 1: 2x₁ + 2 = 1/x₂. Let me solve for x₂:x₂ = 1/(2x₁ + 2). Since x₂ > 0, the denominator must be positive. So, 2x₁ + 2 > 0 → x₁ > -1.But x₁ < 0, so x₁ is in (-1, 0).From Equation 1, x₂ = 1/(2x₁ + 2). Let me denote t = x₁, so t ∈ (-1, 0).Then, x₂ = 1/(2t + 2).Now, substitute x₂ into Equation 2:- t² + a = ln(1/(2t + 2)) - 1.Simplify ln(1/(2t + 2)) = -ln(2t + 2).So, Equation 2 becomes:- t² + a = -ln(2t + 2) - 1.Solving for a:a = t² - ln(2t + 2) - 1.So, a is expressed in terms of t, where t ∈ (-1, 0).We need to find the range of a as t varies from -1 to 0.Let me analyze the function a(t) = t² - ln(2t + 2) - 1.First, let's consider the domain of t: t ∈ (-1, 0).As t approaches -1 from the right, 2t + 2 approaches 0 from the positive side, so ln(2t + 2) approaches -∞, so -ln(2t + 2) approaches +∞. Therefore, a(t) approaches +∞.As t approaches 0 from the left, 2t + 2 approaches 2, so ln(2t + 2) approaches ln 2. Therefore, a(t) approaches 0² - ln 2 - 1 = -1 - ln 2.Now, let's check if a(t) is increasing or decreasing in (-1, 0). Compute the derivative a’(t):a’(t) = 2t - [1/(2t + 2)]*2 - 0 = 2t - 2/(2t + 2).Simplify:a’(t) = 2t - 1/(t + 1).Let me analyze the sign of a’(t):For t ∈ (-1, 0), t + 1 ∈ (0, 1). So, 1/(t + 1) > 1.And 2t is negative since t < 0.So, 2t - 1/(t + 1) is negative minus positive, which is negative. Therefore, a’(t) < 0 for all t ∈ (-1, 0).This means a(t) is decreasing on (-1, 0). Therefore, as t increases from -1 to 0, a(t) decreases from +∞ to -1 - ln 2.Hence, the range of a is (-1 - ln 2, +∞).Wait, but let me double-check. As t approaches -1+, a(t) approaches +∞, and as t approaches 0-, a(t) approaches -1 - ln 2. Since a(t) is decreasing, the range is all real numbers greater than -1 - ln 2.So, the range of a is (-1 - ln 2, ∞).I think that's correct.**Final Answer**(I) The function is decreasing on boxed{(-infty, -1)} and increasing on boxed{[-1, 0)} and boxed{(0, infty)}.(II) The minimum value of x_2 - x_1 is boxed{1}.(III) The range of values for a is boxed{(-1 - ln 2, +infty)}.