💡Okay, so I have this equation to solve:[frac{2}{x - 1} + frac{4}{x - 2} + frac{6}{x - 3} + dots + frac{200}{x - 100} = x.]Hmm, that's a lot of fractions added together, each with a different denominator. The numerators are increasing by 2 each time, starting from 2 up to 200. The denominators are linear terms, each one less than the next. So, the first term is 2 over (x - 1), the next is 4 over (x - 2), and so on until 200 over (x - 100).I need to find the number of real solutions to this equation. That is, how many real numbers x satisfy this equation.First, I should probably define a function f(x) as the left-hand side of the equation:[f(x) = frac{2}{x - 1} + frac{4}{x - 2} + frac{6}{x - 3} + dots + frac{200}{x - 100}.]So, the equation becomes f(x) = x. To find the number of real solutions, I can analyze the behavior of f(x) and see how it intersects with the line y = x.Let me think about the properties of f(x). Each term in f(x) is of the form (frac{2k}{x - k}) where k ranges from 1 to 100. Each of these terms has a vertical asymptote at x = k. So, f(x) will have vertical asymptotes at x = 1, 2, 3, ..., up to x = 100.That means the function f(x) is undefined at each integer from 1 to 100. Between each pair of consecutive integers, the function will have a certain behavior. Let me consider the intervals divided by these asymptotes:1. ( (-infty, 1) )2. ( (1, 2) )3. ( (2, 3) )4. ...5. ( (99, 100) )6. ( (100, infty) )In each of these intervals, f(x) is continuous because there are no asymptotes within the interval. So, I can analyze the behavior of f(x) in each interval separately.Now, let's think about the behavior near each asymptote. For each term (frac{2k}{x - k}), as x approaches k from the left (i.e., x approaches k⁻), the term approaches negative infinity. As x approaches k from the right (i.e., x approaches k⁺), the term approaches positive infinity.However, since f(x) is the sum of all these terms, the behavior near each asymptote is dominated by the term with the vertical asymptote at that point. For example, near x = 1, the term (frac{2}{x - 1}) will dominate, so as x approaches 1 from the left, f(x) approaches negative infinity, and as x approaches 1 from the right, f(x) approaches positive infinity.Similarly, near x = 2, the term (frac{4}{x - 2}) dominates, so as x approaches 2 from the left, f(x) approaches negative infinity, and as x approaches 2 from the right, f(x) approaches positive infinity. This pattern continues for each asymptote up to x = 100.Now, what about the behavior as x approaches positive or negative infinity? Each term (frac{2k}{x - k}) behaves like (frac{2k}{x}) for large x, so as x approaches infinity or negative infinity, each term approaches zero. Therefore, f(x) approaches zero as x approaches both positive and negative infinity.So, putting this together, f(x) has vertical asymptotes at each integer from 1 to 100, and it approaches zero as x goes to positive or negative infinity. Between each pair of consecutive asymptotes, the function goes from negative infinity to positive infinity or vice versa.Wait, actually, let's clarify that. For each interval between two asymptotes, say between k and k+1, what is the behavior of f(x)?Let's take the interval (1, 2). As x approaches 1 from the right, f(x) approaches positive infinity because of the term (frac{2}{x - 1}). As x approaches 2 from the left, f(x) approaches negative infinity because of the term (frac{4}{x - 2}). So, in the interval (1, 2), f(x) decreases from positive infinity to negative infinity.Similarly, in the interval (2, 3), as x approaches 2 from the right, f(x) approaches positive infinity because of the term (frac{4}{x - 2}), and as x approaches 3 from the left, f(x) approaches negative infinity because of the term (frac{6}{x - 3}). So, again, f(x) decreases from positive infinity to negative infinity in that interval.Wait, is that the case for all intervals? Let me check another one. Take the interval (k, k+1). As x approaches k from the right, the term (frac{2k}{x - k}) dominates and approaches positive infinity. As x approaches (k+1) from the left, the term (frac{2(k+1)}{x - (k+1)}) dominates and approaches negative infinity. So, yes, in each interval (k, k+1), f(x) goes from positive infinity to negative infinity.Similarly, in the interval (-infty, 1), as x approaches 1 from the left, f(x) approaches negative infinity, and as x approaches negative infinity, f(x) approaches zero. So, in (-infty, 1), f(x) goes from zero to negative infinity.Wait, but hold on. Let me think about the behavior in (-infty, 1). As x approaches 1 from the left, f(x) approaches negative infinity because of the term (frac{2}{x - 1}). As x approaches negative infinity, each term (frac{2k}{x - k}) behaves like (frac{2k}{x}), so the sum behaves like (sum_{k=1}^{100} frac{2k}{x}), which is (frac{2}{x} sum_{k=1}^{100} k). The sum of k from 1 to 100 is (frac{100 cdot 101}{2} = 5050). So, as x approaches negative infinity, f(x) behaves like (frac{2 cdot 5050}{x} = frac{10100}{x}), which approaches zero from above because x is negative and large in magnitude, so 10100/x approaches zero from below? Wait, no, 10100 is positive, x is negative, so 10100/x is negative and approaches zero.Wait, but as x approaches negative infinity, each term (frac{2k}{x - k}) is approximately (frac{2k}{x}), which is a small negative number because x is negative. So, the sum of all these terms is a sum of negative numbers, each approaching zero. So, f(x) approaches zero from below as x approaches negative infinity.Therefore, in the interval (-infty, 1), f(x) goes from approaching zero from below as x approaches negative infinity to approaching negative infinity as x approaches 1 from the left.Similarly, in each interval (k, k+1), f(x) goes from positive infinity to negative infinity.Now, the function y = x is a straight line with a slope of 1, passing through the origin.To find the number of real solutions to f(x) = x, we need to find how many times the graph of f(x) intersects the graph of y = x.Let me analyze each interval one by one.1. Interval (-infty, 1): - As x approaches negative infinity, f(x) approaches zero from below. - As x approaches 1 from the left, f(x) approaches negative infinity. - The line y = x is a straight line increasing with x. - At x = 0, y = 0. f(0) is the sum of all terms with x = 0: (frac{2}{-1} + frac{4}{-2} + frac{6}{-3} + dots + frac{200}{-100}). Each term is negative, so f(0) is negative. - At x approaching negative infinity, f(x) approaches zero from below, which is slightly negative, while y = x approaches negative infinity. - So, in this interval, f(x) starts near zero (slightly negative) and decreases to negative infinity, while y = x starts at negative infinity and increases to 1. - Therefore, somewhere in this interval, f(x) must cross y = x exactly once because f(x) is decreasing from near zero to negative infinity, while y = x is increasing from negative infinity to 1. They must intersect once.2. Interval (1, 2): - As x approaches 1 from the right, f(x) approaches positive infinity. - As x approaches 2 from the left, f(x) approaches negative infinity. - The line y = x is increasing from 1 to 2. - So, f(x) starts at positive infinity and decreases to negative infinity, while y = x increases from 1 to 2. - Since f(x) is continuous in this interval and goes from positive infinity to negative infinity, and y = x is a straight line, they must cross exactly once in this interval.3. Interval (2, 3): - Similar to (1, 2), f(x) goes from positive infinity to negative infinity. - y = x increases from 2 to 3. - Again, f(x) is continuous and spans from positive infinity to negative infinity, while y = x is increasing. So, they must cross exactly once.This pattern continues for each interval (k, k+1) where k ranges from 1 to 99. In each such interval, f(x) goes from positive infinity to negative infinity, while y = x is increasing from k to k+1. Therefore, in each of these intervals, f(x) and y = x must intersect exactly once.4. Interval (100, ∞): - As x approaches 100 from the right, f(x) approaches positive infinity because of the term (frac{200}{x - 100}). - As x approaches infinity, f(x) approaches zero from above because each term (frac{2k}{x - k}) approaches zero, and the sum of positive terms approaches zero. - The line y = x is increasing from 100 to infinity. - So, f(x) starts at positive infinity and decreases to zero, while y = x increases from 100 to infinity. - At some point, f(x) will be less than y = x. Since f(x) starts at positive infinity and decreases to zero, and y = x starts at 100 and increases, they must intersect exactly once in this interval.Now, let's count the number of intervals where f(x) and y = x intersect:- Interval (-infty, 1): 1 solution- Intervals (1, 2), (2, 3), ..., (99, 100): 99 solutions- Interval (100, ∞): 1 solutionTotal number of solutions: 1 + 99 + 1 = 101.Wait, but let me double-check. Each interval between the asymptotes (from 1 to 100) gives one solution, and the intervals before 1 and after 100 each give one solution. So, that's 100 intervals between 1 to 100, but actually, the intervals are (1,2), (2,3), ..., (99,100), which is 99 intervals, each giving one solution. Then, the intervals (-infty,1) and (100, ∞) each give one solution. So, 99 + 2 = 101 solutions.Yes, that seems correct.I should also consider whether there could be more than one solution in any interval, but given that f(x) is strictly decreasing in each interval (from positive infinity to negative infinity or from negative infinity to positive infinity depending on the interval), and y = x is strictly increasing, they can intersect at most once in each interval. Since f(x) spans all real numbers in each interval, it must cross y = x exactly once.Therefore, the total number of real solutions is 101.

💡Okay, so I need to figure out how many perfect cubes lie between (2^8 + 1) and (2^{18} + 1), inclusive. Hmm, let me break this down step by step.First, I should calculate what (2^8 + 1) and (2^{18} + 1) are. I know that (2^8) is 256, so adding 1 gives me 257. For (2^{18}), that's a bit bigger. I remember that (2^{10}) is 1024, so (2^{18}) is (2^{10} times 2^8), which is 1024 multiplied by 256. Let me calculate that: 1024 times 256. Hmm, 1024 times 200 is 204,800, and 1024 times 56 is 57,344. Adding those together gives me 204,800 + 57,344 = 262,144. So, (2^{18} + 1) is 262,145.Now, I need to find all the perfect cubes between 257 and 262,145, inclusive. A perfect cube is a number that can be written as (n^3) where (n) is an integer. So, I need to find the smallest integer (n) such that (n^3) is greater than or equal to 257, and the largest integer (n) such that (n^3) is less than or equal to 262,145. Then, the number of perfect cubes will be the count of integers from the smallest (n) to the largest (n), inclusive.Let me start with the lower bound. I need to find the smallest (n) where (n^3 geq 257). I know that (6^3 = 216) and (7^3 = 343). Since 216 is less than 257 and 343 is greater than 257, the smallest (n) is 7.Now, for the upper bound, I need the largest (n) where (n^3 leq 262,145). I recall that (64^3 = 262,144). Let me verify that: (64 times 64 = 4096), and (4096 times 64). Let me compute 4096 times 60, which is 245,760, and 4096 times 4, which is 16,384. Adding those together gives 245,760 + 16,384 = 262,144. So, (64^3 = 262,144), which is just one less than 262,145. Therefore, the largest (n) is 64.So, the perfect cubes we're considering are from (7^3) up to (64^3). To find how many perfect cubes there are, I need to count the integers from 7 to 64, inclusive. The number of integers from (a) to (b) inclusive is (b - a + 1). Plugging in the values, that's (64 - 7 + 1 = 58).Wait, let me double-check that. If I have numbers from 1 to 64, that's 64 numbers. From 1 to 6, that's 6 numbers. So, subtracting, 64 - 6 = 58. Yep, that seems right.So, there are 58 perfect cubes between (2^8 + 1) and (2^{18} + 1), inclusive.**Final Answer**boxed{E}

💡First, I recognize that the exponential function is given by ( f(x) = a^x ), where ( a ) is the base we need to find.The graph passes through the point ( (2, frac{1}{16}) ), which means that when ( x = 2 ), ( f(x) = frac{1}{16} ).Substituting these values into the equation, I get:[frac{1}{16} = a^2]To solve for ( a ), I take the square root of both sides:[a = sqrt{frac{1}{16}} = frac{1}{4}]Since the base ( a ) of an exponential function must be positive, ( a = frac{1}{4} ) is the valid solution.

💡First, I need to determine the amount of the service charge by subtracting the original order amount from the total amount paid.Next, I will calculate the service charge percentage by dividing the service charge by the original order amount and then multiplying by 100 to convert it into a percentage.Finally, I will present the calculated service charge percentage as the solution.

💡First, determine the total number of boxes Seth bought initially. He bought 3 boxes at the market and twice as many at the farm, which is 6 boxes. So, he initially has 3 + 6 = 9 boxes.Next, Seth gives 1 box to his mother, leaving him with 9 - 1 = 8 boxes.He donates a quarter of the remaining boxes to charity. A quarter of 8 is 2, so he donates 2 boxes, leaving him with 8 - 2 = 6 boxes.Seth then distributes an equal number of boxes to his 5 friends. Let F be the number of boxes each friend receives. The total distributed is 5F, and he has 4 boxes left. Therefore, 6 - 5F = 4.Solving for F:5F = 6 - 45F = 2F = 2/5Since F must be a whole number, this implies that Seth did not distribute any boxes to his friends, meaning he kept all 6 boxes. Thus, the total number of boxes he bought initially is 9.

💡Okay, so I need to figure out the value of ( m ) that makes the polynomial ( 4x^2 - 3x + m ) divisible by ( x - 2 ). Hmm, divisibility of polynomials... I remember something about the Remainder Theorem from algebra class. Let me try to recall how that works.The Remainder Theorem states that if you divide a polynomial ( f(x) ) by ( x - a ), the remainder is ( f(a) ). So, if ( x - 2 ) is a divisor of ( 4x^2 - 3x + m ), that means when I divide ( 4x^2 - 3x + m ) by ( x - 2 ), the remainder should be zero. Therefore, according to the Remainder Theorem, substituting ( x = 2 ) into the polynomial should give me zero.Let me write that down:[f(2) = 4(2)^2 - 3(2) + m = 0]Calculating each term step by step:First, ( 4(2)^2 ). The exponent comes first, so ( 2^2 = 4 ). Then multiply by 4: ( 4 times 4 = 16 ).Next, ( -3(2) ). That's straightforward: ( -3 times 2 = -6 ).So now, putting it all together:[16 - 6 + m = 0]Simplifying ( 16 - 6 ):[10 + m = 0]To solve for ( m ), I subtract 10 from both sides:[m = -10]Wait, let me double-check my calculations to make sure I didn't make a mistake. So, plugging ( x = 2 ) into ( 4x^2 - 3x + m ):( 4(2)^2 = 16 ), ( -3(2) = -6 ), so ( 16 - 6 = 10 ). Then ( 10 + m = 0 ), so ( m = -10 ). Yeah, that seems right.Looking at the answer choices, option B is -10, so that must be the correct answer. Just to be thorough, maybe I should also try polynomial long division to confirm. Let's see.Dividing ( 4x^2 - 3x + m ) by ( x - 2 ):1. Divide ( 4x^2 ) by ( x ) to get ( 4x ).2. Multiply ( 4x ) by ( x - 2 ) to get ( 4x^2 - 8x ).3. Subtract that from the original polynomial: ( (4x^2 - 3x + m) - (4x^2 - 8x) = 5x + m ).4. Now, divide ( 5x ) by ( x ) to get ( 5 ).5. Multiply ( 5 ) by ( x - 2 ) to get ( 5x - 10 ).6. Subtract that from the previous remainder: ( (5x + m) - (5x - 10) = m + 10 ).For the polynomial to be divisible by ( x - 2 ), the remainder must be zero. So, ( m + 10 = 0 ), which again gives ( m = -10 ). Yep, that confirms it.I think I'm confident now that ( m = -10 ) is the correct answer. It's option B.