💡Okay, so I have this problem where I need to calculate the sum of a function evaluated at a bunch of points. The function is a cubic function, and I remember that cubic functions have something called an inflection point, which is also their center of symmetry. The problem mentions this, so I think that's going to be important.First, let me write down the function again to make sure I have it right: ( f(x) = frac{1}{3}x^3 - frac{1}{2}x^2 + 3x - frac{5}{12} ). I need to find the sum ( fleft(frac{1}{2017}right) + fleft(frac{2}{2017}right) + ldots + fleft(frac{2016}{2017}right) ). That's a lot of terms, 2016 in total. Since the function is cubic, I recall that it has a center of symmetry at its inflection point. To find this inflection point, I need to compute the second derivative of the function and set it equal to zero. Let me try that.First, find the first derivative ( f'(x) ). The derivative of ( frac{1}{3}x^3 ) is ( x^2 ), the derivative of ( -frac{1}{2}x^2 ) is ( -x ), the derivative of ( 3x ) is 3, and the derivative of a constant like ( -frac{5}{12} ) is zero. So, putting that together, ( f'(x) = x^2 - x + 3 ).Next, find the second derivative ( f''(x) ). The derivative of ( x^2 ) is ( 2x ), the derivative of ( -x ) is ( -1 ), and the derivative of 3 is zero. So, ( f''(x) = 2x - 1 ).To find the inflection point, set ( f''(x) = 0 ): ( 2x - 1 = 0 ). Solving for x, I get ( x = frac{1}{2} ). So, the inflection point is at ( x = frac{1}{2} ). Now, since the function is symmetric about this point, that means for any x, the function satisfies ( fleft(frac{1}{2} + aright) + fleft(frac{1}{2} - aright) = 2fleft(frac{1}{2}right) ). Let me compute ( fleft(frac{1}{2}right) ) to see what that value is.Plugging ( x = frac{1}{2} ) into the function: ( fleft(frac{1}{2}right) = frac{1}{3}left(frac{1}{2}right)^3 - frac{1}{2}left(frac{1}{2}right)^2 + 3left(frac{1}{2}right) - frac{5}{12} ).Calculating each term:1. ( frac{1}{3} times frac{1}{8} = frac{1}{24} )2. ( -frac{1}{2} times frac{1}{4} = -frac{1}{8} )3. ( 3 times frac{1}{2} = frac{3}{2} )4. ( -frac{5}{12} )Adding them all together:( frac{1}{24} - frac{1}{8} + frac{3}{2} - frac{5}{12} ).To add these fractions, I need a common denominator. 24 seems like a good choice.Convert each term:1. ( frac{1}{24} ) stays the same.2. ( -frac{1}{8} = -frac{3}{24} )3. ( frac{3}{2} = frac{36}{24} )4. ( -frac{5}{12} = -frac{10}{24} )Now, add them:( frac{1}{24} - frac{3}{24} + frac{36}{24} - frac{10}{24} = frac{1 - 3 + 36 - 10}{24} = frac{24}{24} = 1 ).So, ( fleft(frac{1}{2}right) = 1 ). Therefore, the symmetry condition is ( fleft(frac{1}{2} + aright) + fleft(frac{1}{2} - aright) = 2 times 1 = 2 ).This is useful because in the sum I need to compute, each term is ( fleft(frac{k}{2017}right) ) where k goes from 1 to 2016. I can pair terms such that their arguments add up to 1. For example, ( frac{1}{2017} ) and ( frac{2016}{2017} ) add up to 1. Similarly, ( frac{2}{2017} ) and ( frac{2015}{2017} ) add up to 1, and so on.Since there are 2016 terms, pairing them like this will give me 1008 pairs. Each pair sums to 2 because of the symmetry. Therefore, the total sum should be ( 1008 times 2 = 2016 ).Wait, let me double-check that. If I have 2016 terms, pairing them would mean 2016 / 2 = 1008 pairs. Each pair sums to 2, so 1008 * 2 = 2016. That seems right.But just to make sure, let me think about a specific example. Let's say k = 1 and k = 2016. Then, ( fleft(frac{1}{2017}right) + fleft(frac{2016}{2017}right) = 2 ). Similarly, for k = 2 and k = 2015, it's also 2. This pattern continues all the way to the middle, which would be k = 1008 and k = 1009, since 1008 + 1009 = 2017. So, each of these pairs adds up to 2, and there are 1008 such pairs.Therefore, the total sum is indeed 2016. I think that's the answer. It makes sense because the function is symmetric around the inflection point, and pairing the terms appropriately allows me to use that symmetry to simplify the sum.**Final Answer**boxed{2016}

💡Okay, so I have this integral to solve: the definite integral from 0 to 1 of (2x + e^x) dx. Hmm, let's see. I remember that to solve definite integrals, I need to find the antiderivative of the function and then evaluate it at the upper and lower limits, subtracting the lower limit result from the upper one. First, let me break down the integral into two separate parts because I know that integrals are linear, so I can integrate each term individually. That means I can write this as the integral from 0 to 1 of 2x dx plus the integral from 0 to 1 of e^x dx. Starting with the first part: the integral of 2x dx. I think the antiderivative of x is (1/2)x^2, right? So if I have 2x, multiplying by 2 should just give me x^2. Let me check: the derivative of x^2 is 2x, which matches. So, the integral of 2x dx is x^2. Now, moving on to the second part: the integral of e^x dx. I remember that the antiderivative of e^x is e^x itself because the derivative of e^x is also e^x. So, that part is straightforward. Putting it all together, the antiderivative of (2x + e^x) is x^2 + e^x. Next, I need to evaluate this from 0 to 1. That means I plug in 1 into the antiderivative and then subtract the value of the antiderivative at 0. Calculating at the upper limit, x = 1: (1)^2 + e^(1) = 1 + e. Now, calculating at the lower limit, x = 0: (0)^2 + e^(0) = 0 + 1, since e^0 is 1. Subtracting the lower limit result from the upper limit result: (1 + e) - (0 + 1) = 1 + e - 1 = e. So, the value of the definite integral is e. Looking back at the options, that's option C. Wait, let me double-check to make sure I didn't make any mistakes. The antiderivative of 2x is definitely x^2, and the antiderivative of e^x is e^x. Evaluating at 1 gives 1 + e, and at 0 gives 0 + 1. Subtracting 1 from 1 + e leaves e. Yeah, that seems right. I think I got it. The answer should be e, which is option C.

💡Okay, so I have this problem where I need to find two real numbers, p and q, such that when I take the cross product of the vectors [3, p, -8] and [6, 5, q], I get the zero vector. Hmm, cross product equals zero... I remember that if the cross product of two vectors is the zero vector, then those two vectors must be parallel or scalar multiples of each other. So, that means one vector is just a multiple of the other.Let me write down the vectors:First vector: [3, p, -8]Second vector: [6, 5, q]Since they are parallel, there must be some scalar, let's call it k, such that each component of the second vector is k times the corresponding component of the first vector. So, that would mean:6 = k * 35 = k * pq = k * (-8)Alright, so I can solve for k first using the first component. If 6 = k * 3, then k = 6 / 3 = 2. Okay, so k is 2. Now I can use this value of k to find p and q.Starting with the second component: 5 = k * p. Since k is 2, that becomes 5 = 2 * p. To solve for p, I divide both sides by 2: p = 5 / 2. So, p is 2.5 or 5/2.Now, moving on to the third component: q = k * (-8). Again, k is 2, so q = 2 * (-8) = -16. So, q is -16.Let me double-check to make sure this makes sense. If I substitute p = 5/2 and q = -16 back into the second vector, it should be twice the first vector.First vector: [3, 5/2, -8]Second vector: [6, 5, -16]Is [6, 5, -16] equal to 2 times [3, 5/2, -8]? Let's see:2 * 3 = 6, which matches.2 * (5/2) = 5, which matches.2 * (-8) = -16, which matches.Yes, that seems correct. So, the cross product being zero implies that the vectors are scalar multiples, and we found the scalar k = 2. Using that, we determined p and q correctly.I think that's it. I don't see any mistakes in my reasoning. The key was recognizing that a zero cross product means the vectors are parallel, so one is a scalar multiple of the other. Then, by setting up the proportions and solving for the scalar, I could find the unknowns p and q.**Final Answer**The ordered pair is boxed{left( dfrac{5}{2}, -16 right)}.

💡First, I need to determine the number of messages sent on each day from Monday to Wednesday.On Monday, there were 300 messages.On Tuesday, there were 200 messages.On Wednesday, there were 300 more messages than on Tuesday, which means 200 + 300 = 500 messages.Next, I'll calculate the total number of messages sent from Monday to Wednesday:300 (Monday) + 200 (Tuesday) + 500 (Wednesday) = 1000 messages.The total number of messages after Thursday was 2000. To find out how many messages were sent on Thursday, I'll subtract the total messages from Monday to Wednesday from the overall total:2000 - 1000 = 1000 messages on Thursday.Finally, to find the ratio of messages sent on Thursday to those sent on Wednesday:Thursday messages : Wednesday messages = 1000 : 500.Simplifying this ratio by dividing both numbers by 500 gives:2 : 1.

💡Okay, so I have this problem: for any integer n that's at least 2, I need to show that the expression n⁴ + n² + 1 isn't a prime number. Hmm, primes are numbers greater than 1 that have no positive divisors other than 1 and themselves. So, if I can show that n⁴ + n² + 1 can be factored into smaller integers greater than 1, then it's not prime.Let me start by looking at the expression: n⁴ + n² + 1. It looks a bit like a quadratic in terms of n². Maybe I can rewrite it to see if it factors nicely. Let me think about how to factor polynomials. Sometimes, adding and subtracting terms can help factor expressions.Wait, I remember that n⁴ + n² + 1 can be related to the formula for factoring a sum of cubes or something similar. Let me recall the formula for a³ + b³ = (a + b)(a² - ab + b²). But here, I have n⁴, which is (n²)². Maybe I can manipulate it to fit a known factoring formula.Let me try to write n⁴ + n² + 1 as a product of two quadratics. Suppose it factors as (n² + an + b)(n² + cn + d). If I expand this, I get n⁴ + (a + c)n³ + (ac + b + d)n² + (ad + bc)n + bd. Comparing this to n⁴ + n² + 1, I can set up equations for the coefficients:1. a + c = 0 (since there's no n³ term)2. ac + b + d = 1 (coefficient of n²)3. ad + bc = 0 (since there's no n term)4. bd = 1 (constant term)From equation 1, c = -a. Plugging this into equation 3: ad + b(-a) = 0 ⇒ a(d - b) = 0. So either a = 0 or d = b.If a = 0, then c = 0. Then equation 2 becomes 0 + b + d = 1, and equation 4 is bd = 1. So, we have b + d = 1 and bd = 1. The solutions to this would be b and d such that they multiply to 1 and add to 1. The only possibilities are b = 1 and d = 1, but 1 + 1 = 2 ≠ 1. So, that doesn't work. Therefore, a cannot be 0, so d = b.Now, with d = b, equation 4 becomes b² = 1, so b = 1 or b = -1.Let's try b = 1. Then d = 1. From equation 2: ac + 1 + 1 = 1 ⇒ ac = -1. Since c = -a, we have a(-a) = -1 ⇒ -a² = -1 ⇒ a² = 1 ⇒ a = 1 or a = -1.If a = 1, then c = -1. So, the factors would be (n² + n + 1)(n² - n + 1). Let me check if this works:(n² + n + 1)(n² - n + 1) = n⁴ - n³ + n² + n³ - n² + n + n² - n + 1 = n⁴ + n² + 1. Perfect, that's the original expression.If a = -1, then c = 1. The factors would be (n² - n + 1)(n² + n + 1), which is the same as above, just reversed. So, either way, the factorization works.Therefore, n⁴ + n² + 1 factors into (n² + n + 1)(n² - n + 1). Now, I need to check if both of these factors are greater than 1 for n ≥ 2.Let's evaluate n² + n + 1 for n ≥ 2. For n = 2: 4 + 2 + 1 = 7, which is greater than 1. For larger n, n² grows faster, so it's definitely greater than 1.Similarly, n² - n + 1 for n = 2: 4 - 2 + 1 = 3, which is greater than 1. For n = 3: 9 - 3 + 1 = 7, and so on. Since n² - n is always positive for n ≥ 2 (because n² > n), adding 1 keeps it greater than 1.Therefore, both factors are integers greater than 1, which means n⁴ + n² + 1 is composite, not prime, for any n ≥ 2.

💡First, I need to determine between which two consecutive integers the square root of 30 falls. To do this, I'll consider the perfect squares that are closest to 30.I know that 5 squared is 25 and 6 squared is 36. Since 30 is between 25 and 36, it means that the square root of 30 must be between 5 and 6.Therefore, the square root of 30 lies between the integers 5 and 6.