answer：Okay, so I have this problem where a point is moving along a straight line, and the distance it travels after t seconds is given by the equation s = (1/3)t³ - (1/2)t² - 2t + 1. I need to find the moment when the velocity is zero. The options are end of 1, 2, 3, or 4 seconds. Hmm, let me think about how to approach this.First, I remember that velocity is the derivative of the displacement with respect to time. So, if I can find the derivative of s with respect to t, that should give me the velocity function. Then, I can set that derivative equal to zero and solve for t to find when the velocity is zero.Alright, let's start by finding the derivative of s. The function s is given by:s = (1/3)t³ - (1/2)t² - 2t + 1To find the velocity, I need to differentiate this with respect to t. I think the derivative of t³ is 3t², so (1/3)t³ would differentiate to (1/3)*3t², which simplifies to t². Okay, that seems straightforward.Next, the term -(1/2)t². The derivative of t² is 2t, so multiplying by -(1/2) gives -(1/2)*2t, which simplifies to -t. Got that.Then, the term -2t. The derivative of t is 1, so multiplying by -2 gives -2*1, which is -2.Finally, the constant term +1. The derivative of a constant is zero, so that term disappears.Putting it all together, the derivative of s with respect to t, which is the velocity v(t), is:v(t) = t² - t - 2Okay, so now I have the velocity function: v(t) = t² - t - 2. I need to find when this velocity is zero, so I set up the equation:t² - t - 2 = 0Now, I need to solve this quadratic equation for t. Quadratic equations can be solved using factoring, completing the square, or the quadratic formula. Let me see if this one factors nicely.Looking at the equation t² - t - 2 = 0, I need two numbers that multiply to -2 and add up to -1. Let me think: factors of -2 are (1, -2) and (-1, 2). Let's try 1 and -2: 1 + (-2) = -1. Perfect! So, the equation factors as:(t + 1)(t - 2) = 0Setting each factor equal to zero gives the solutions:t + 1 = 0 => t = -1t - 2 = 0 => t = 2So, the solutions are t = -1 and t = 2. But since time cannot be negative in this context, t = -1 doesn't make sense here. Therefore, the only valid solution is t = 2 seconds.Let me double-check my work to make sure I didn't make any mistakes. Differentiating s(t) gives v(t) = t² - t - 2, which seems correct. Solving t² - t - 2 = 0 gives t = 2 and t = -1, which makes sense because the quadratic factors correctly. Since negative time isn't applicable, t = 2 is the answer.So, the moment when the velocity is zero is at the end of 2 seconds. Looking back at the options, that's option B.**Final Answer**The moment when the velocity is zero is at the end of boxed{B} seconds.

answer：Okay, so I have this problem here: Given the function f(x) = lg(x² - 4x - 5) is monotonically increasing on the interval (a, +∞), what is the range of values for a? The options are A: (2, +∞), B: [2, +∞), C: (5, +∞), and D: [5, +∞).Alright, let me try to figure this out step by step. First, I know that lg usually stands for logarithm base 10, so f(x) is a logarithmic function. Logarithmic functions are only defined when their arguments are positive. So, the expression inside the logarithm, which is x² - 4x - 5, must be greater than zero.Let me write that down: x² - 4x - 5 > 0.To solve this inequality, I can factor the quadratic. Let's see, x² - 4x - 5. I need two numbers that multiply to -5 and add up to -4. Hmm, that would be -5 and +1. So, the factored form is (x - 5)(x + 1) > 0.Now, to find where this product is positive, I can use a sign chart. The roots are x = 5 and x = -1. These divide the number line into three intervals: (-∞, -1), (-1, 5), and (5, +∞). I'll test each interval.For x < -1, say x = -2: (-2 - 5)(-2 + 1) = (-7)(-1) = 7, which is positive.For -1 < x < 5, say x = 0: (0 - 5)(0 + 1) = (-5)(1) = -5, which is negative.For x > 5, say x = 6: (6 - 5)(6 + 1) = (1)(7) = 7, which is positive.So, the inequality x² - 4x - 5 > 0 holds for x < -1 or x > 5. Therefore, the domain of f(x) is (-∞, -1) ∪ (5, +∞).Now, the problem states that f(x) is monotonically increasing on the interval (a, +∞). So, I need to find the values of a such that f(x) is increasing for all x > a.Since f(x) is a logarithmic function, its monotonicity depends on the monotonicity of its argument, which is the quadratic function g(x) = x² - 4x - 5. The logarithm function lg(t) is increasing when t is increasing, provided t > 0.So, f(x) will be increasing when g(x) is increasing, given that g(x) > 0.Let me find where g(x) is increasing. The quadratic function g(x) = x² - 4x - 5 is a parabola opening upwards because the coefficient of x² is positive. The vertex of this parabola is at x = -b/(2a) = 4/(2*1) = 2. So, the vertex is at x = 2.Since the parabola opens upwards, g(x) is decreasing on (-∞, 2) and increasing on (2, +∞).But wait, the domain of f(x) is (-∞, -1) ∪ (5, +∞). So, in the interval (-∞, -1), g(x) is decreasing, and in (5, +∞), g(x) is increasing.Therefore, f(x) = lg(g(x)) will be increasing where g(x) is increasing, which is in (5, +∞). So, f(x) is increasing on (5, +∞).Therefore, for f(x) to be monotonically increasing on (a, +∞), a must be at least 5. So, a can be 5 or any number greater than 5. Hence, the range of values for a is [5, +∞).Let me double-check. If a is 5, then the interval is (5, +∞), and we know f(x) is increasing there. If a is greater than 5, say 6, then (6, +∞) is still within (5, +∞), so f(x) is still increasing. If a were less than 5, say 4, then the interval (4, +∞) would include some points where g(x) is decreasing (between 4 and 5), so f(x) wouldn't be increasing throughout that interval. Therefore, a must be at least 5.So, the correct answer should be D: [5, +∞).

answer：First, I need to determine the original speed of the car when it covers 432 km in 6 hours. Speed is calculated by dividing the distance by the time.Next, with the new speed of 48 km/h, I need to find out how much time it would take to cover the same distance. Using the same formula for speed, I can rearrange it to solve for time.After finding the new time, I will calculate the fraction of the original time by dividing the new time by the original time. This will give me the fraction of the previous time required to cover the same distance at the reduced speed.

answer：First, I need to determine Joe's weight loss rate per month. He is projected to weigh 180 pounds in 3 months and currently weighs 198 pounds. The total weight he will lose in the next 3 months is 198 - 180 = 18 pounds. Dividing this by 3 months gives a monthly weight loss rate of 6 pounds per month.Next, I calculate the total weight Joe has lost over the 4 months he has been on the diet. Multiplying the monthly weight loss rate by 4 months results in a total weight loss of 24 pounds.Finally, to find Joe's initial weight before starting the diet, I add the total weight he has lost to his current weight. Adding 24 pounds to his current weight of 198 pounds gives an initial weight of 222 pounds.

answer：First, I need to determine how much ribbon John used on each box. He used a total of 5/12 of a roll of ribbon for five boxes, and the amount used on each box is the same.To find the amount used on one box, I will divide the total ribbon used by the number of boxes. This means dividing 5/12 by 5.Dividing by a whole number is the same as multiplying by its reciprocal. The reciprocal of 5 is 1/5.So, I will multiply 5/12 by 1/5:5/12 × 1/5 = (5 × 1) / (12 × 5) = 5/60.Finally, I will simplify 5/60 by dividing both the numerator and the denominator by 5, resulting in 1/12.Therefore, John used 1/12 of a roll of ribbon on each box.

answer：Okay, so I have this problem about a sequence {a_n} where S_n is the sum of the first n terms. The given equation is (2S_n)/n + n = 2a_n + 1. I need to prove that {a_n} is an arithmetic sequence, and then, given that a_4, a_7, and a_9 form a geometric sequence, find the minimum value of S_n.Alright, starting with part (1). I know that S_n is the sum of the first n terms, so S_n = a_1 + a_2 + ... + a_n. The given equation is (2S_n)/n + n = 2a_n + 1. Maybe I can manipulate this equation to find a relationship between S_n and a_n.Let me rewrite the equation:(2S_n)/n + n = 2a_n + 1Multiply both sides by n to eliminate the denominator:2S_n + n^2 = 2n a_n + nHmm, okay. So 2S_n + n^2 = 2n a_n + n. Let me write that as:2S_n = 2n a_n + n - n^2Simplify the right side:2S_n = 2n a_n + n(1 - n)Divide both sides by 2:S_n = n a_n + (n(1 - n))/2So, S_n = n a_n + (n - n^2)/2Hmm, that's interesting. I also know that S_n can be expressed in terms of a_1 and the common difference d if it's an arithmetic sequence. But wait, I don't know yet if it's arithmetic. Maybe I can find a recursive relation.Let me consider S_{n+1}. S_{n+1} = S_n + a_{n+1}Using the given equation for n+1:(2S_{n+1})/(n+1) + (n+1) = 2a_{n+1} + 1Multiply both sides by (n+1):2S_{n+1} + (n+1)^2 = 2(n+1)a_{n+1} + (n+1)So, 2S_{n+1} = 2(n+1)a_{n+1} + (n+1) - (n+1)^2Simplify the right side:2S_{n+1} = 2(n+1)a_{n+1} + (n+1)(1 - (n+1))Which is:2S_{n+1} = 2(n+1)a_{n+1} + (n+1)(-n)So, 2S_{n+1} = 2(n+1)a_{n+1} - n(n+1)Now, from the earlier equation, we have:2S_n = 2n a_n + n - n^2So, let me write both equations:1) 2S_n = 2n a_n + n - n^22) 2S_{n+1} = 2(n+1)a_{n+1} - n(n+1)But S_{n+1} = S_n + a_{n+1}, so 2S_{n+1} = 2S_n + 2a_{n+1}Substitute equation 1 into this:2S_n + 2a_{n+1} = 2(n+1)a_{n+1} - n(n+1)But 2S_n is equal to 2n a_n + n - n^2, so substitute that:2n a_n + n - n^2 + 2a_{n+1} = 2(n+1)a_{n+1} - n(n+1)Let me bring all terms to one side:2n a_n + n - n^2 + 2a_{n+1} - 2(n+1)a_{n+1} + n(n+1) = 0Simplify the terms:2n a_n + n - n^2 + [2a_{n+1} - 2(n+1)a_{n+1}] + n(n+1) = 0Factor out 2a_{n+1}:2n a_n + n - n^2 + 2a_{n+1}(1 - (n+1)) + n(n+1) = 0Simplify 1 - (n+1) = -n:2n a_n + n - n^2 - 2n a_{n+1} + n(n+1) = 0Now, let's distribute n(n+1):n(n+1) = n^2 + nSo, substituting back:2n a_n + n - n^2 - 2n a_{n+1} + n^2 + n = 0Combine like terms:2n a_n + n - n^2 - 2n a_{n+1} + n^2 + nThe -n^2 and +n^2 cancel out.2n a_n + n + n - 2n a_{n+1} = 0So, 2n a_n + 2n - 2n a_{n+1} = 0Factor out 2n:2n(a_n + 1 - a_{n+1}) = 0Since 2n is not zero for n >=1, we have:a_n + 1 - a_{n+1} = 0Which simplifies to:a_{n+1} = a_n + 1So, the difference between consecutive terms is 1, which is constant. Therefore, {a_n} is an arithmetic sequence with common difference d = 1.Alright, that proves part (1). Now, moving on to part (2). It says that a_4, a_7, and a_9 form a geometric sequence. I need to find the minimum value of S_n.First, since {a_n} is arithmetic with common difference d=1, let me write the general term.Let me denote the first term as a_1 = a. Then, the nth term is:a_n = a + (n - 1)d = a + (n - 1)*1 = a + n - 1So, a_n = a + n - 1Therefore, a_4 = a + 4 - 1 = a + 3a_7 = a + 7 - 1 = a + 6a_9 = a + 9 - 1 = a + 8Given that a_4, a_7, a_9 form a geometric sequence. So, the ratio between consecutive terms should be constant.In a geometric sequence, (a_7)^2 = a_4 * a_9So, (a + 6)^2 = (a + 3)(a + 8)Let me compute both sides:Left side: (a + 6)^2 = a^2 + 12a + 36Right side: (a + 3)(a + 8) = a^2 + 11a + 24Set them equal:a^2 + 12a + 36 = a^2 + 11a + 24Subtract a^2 from both sides:12a + 36 = 11a + 24Subtract 11a from both sides:a + 36 = 24Subtract 36:a = 24 - 36 = -12So, the first term a = -12.Therefore, the general term is:a_n = -12 + n - 1 = n - 13So, a_n = n - 13Now, I need to find the minimum value of S_n, which is the sum of the first n terms.Since the sequence is arithmetic, the sum S_n can be calculated by:S_n = n/2 [2a_1 + (n - 1)d]We know a_1 = -12, d = 1, so:S_n = n/2 [2*(-12) + (n - 1)*1] = n/2 [-24 + n - 1] = n/2 [n - 25]So, S_n = (n(n - 25))/2Alternatively, S_n = (n^2 - 25n)/2To find the minimum value of S_n, since it's a quadratic function in terms of n, opening upwards (coefficient of n^2 is positive), the minimum occurs at the vertex.The vertex of a quadratic function ax^2 + bx + c is at x = -b/(2a). Here, a = 1/2, b = -25/2.Wait, let me write S_n as:S_n = (1/2)n^2 - (25/2)nSo, a = 1/2, b = -25/2Thus, the vertex is at n = -b/(2a) = -(-25/2)/(2*(1/2)) = (25/2)/1 = 25/2 = 12.5But n must be an integer, so the minimum occurs at n = 12 or n = 13.Compute S_12 and S_13.Compute S_12:S_12 = (12^2 - 25*12)/2 = (144 - 300)/2 = (-156)/2 = -78Compute S_13:S_13 = (13^2 - 25*13)/2 = (169 - 325)/2 = (-156)/2 = -78So, both S_12 and S_13 are equal to -78. Therefore, the minimum value of S_n is -78.Wait, let me double-check. Since the terms of the sequence are a_n = n - 13, so the terms are:a_1 = -12, a_2 = -11, a_3 = -10, ..., a_12 = -1, a_13 = 0, a_14 = 1, etc.So, the terms are negative up to a_12, then zero at a_13, and positive beyond that. Therefore, the sum S_n will decrease until n=12 or 13, then start increasing.Since S_12 and S_13 are both -78, that's the minimum.So, the minimum value of S_n is -78.**Final Answer**The minimum value of ( S_n ) is boxed{-78}.