💡Okay, so I have this problem about a regular hexagon with a side length of 3. Inside this hexagon, there are three semicircles inscribed such that their diameters coincide with alternating sides of the hexagon. I need to find the area of the shaded region, which is inside the hexagon but outside all of the semicircles. The answer choices are given, so I need to figure out which one is correct.First, I should recall some properties of a regular hexagon. A regular hexagon has all sides equal and all internal angles equal to 120 degrees. The area of a regular hexagon can be calculated using the formula:[text{Area} = frac{3sqrt{3}}{2} s^2]where ( s ) is the side length. Since the side length here is 3, plugging that in:[text{Area} = frac{3sqrt{3}}{2} times 3^2 = frac{3sqrt{3}}{2} times 9 = frac{27sqrt{3}}{2} = 13.5sqrt{3}]So the area of the hexagon is ( 13.5sqrt{3} ). That seems straightforward.Now, the problem mentions three semicircles inscribed in the hexagon, with their diameters on alternating sides. Since the hexagon has six sides, alternating sides would mean every other side. So, if I imagine the hexagon, the semicircles are drawn on three non-adjacent sides.Each semicircle has a diameter equal to the side length of the hexagon, which is 3. Therefore, the radius of each semicircle is half of that, so ( r = frac{3}{2} = 1.5 ).The area of a full circle is ( pi r^2 ), so the area of a semicircle would be half of that:[text{Area of one semicircle} = frac{1}{2} pi r^2 = frac{1}{2} pi (1.5)^2 = frac{1}{2} pi times 2.25 = frac{2.25pi}{2} = frac{9pi}{8}]Wait, let me double-check that calculation. ( (1.5)^2 = 2.25 ), and half of that is 1.125, which is ( frac{9}{8} ) because ( 1.125 = frac{9}{8} ). So, yes, the area of one semicircle is ( frac{9pi}{8} ).Since there are three such semicircles, the total area covered by the semicircles is:[text{Total area of semicircles} = 3 times frac{9pi}{8} = frac{27pi}{8}]So, the total area of the three semicircles is ( frac{27pi}{8} ).Now, the shaded region is the area inside the hexagon but outside all of the semicircles. Therefore, to find the area of the shaded region, I need to subtract the total area of the semicircles from the area of the hexagon.So, the area of the shaded region is:[text{Area of shaded region} = text{Area of hexagon} - text{Total area of semicircles} = 13.5sqrt{3} - frac{27pi}{8}]Looking back at the answer choices:A. ( 12sqrt{3} - frac{25pi}{8} )B. ( 13.5sqrt{3} - frac{26pi}{8} )C. ( 13.5sqrt{3} - frac{27pi}{8} )D. ( 14.5sqrt{3} - frac{27pi}{8} )E. ( 14sqrt{3} - frac{28pi}{8} )My calculation gives ( 13.5sqrt{3} - frac{27pi}{8} ), which is option C. So, I think the correct answer is C.But wait, let me just make sure I didn't make any mistakes in my calculations. Let me go through each step again.1. Area of the hexagon: Formula is correct, ( frac{3sqrt{3}}{2} s^2 ). Plugging in 3, I get ( 13.5sqrt{3} ). That seems right.2. Radius of each semicircle: Since the diameter is 3, radius is 1.5. Correct.3. Area of one semicircle: ( frac{1}{2} pi r^2 = frac{1}{2} pi (1.5)^2 = frac{9pi}{8} ). Yes, that's correct.4. Total area of three semicircles: 3 times ( frac{9pi}{8} ) is ( frac{27pi}{8} ). Correct.5. Subtracting the total semicircle area from the hexagon area: ( 13.5sqrt{3} - frac{27pi}{8} ). That's exactly option C.Hmm, seems solid. I don't see any errors in my reasoning. So, I think I can confidently go with option C.**Final Answer**boxed{C}

💡First, I need to calculate the value of ( frac{9!}{168} ). I know that ( 9! ) is the product of all integers from 1 to 9, which is ( 362880 ).Next, I divide ( 362880 ) by ( 168 ) to simplify the expression:[frac{362880}{168} = 2160]Now, I need to find the square root of ( 2160 ). To do this, I'll factorize ( 2160 ) into its prime factors:[2160 = 2^4 times 3^3 times 5]Taking the square root involves taking half of each exponent in the prime factorization:[sqrt{2160} = sqrt{2^4 times 3^3 times 5} = 2^2 times 3^{1.5} times sqrt{5} = 4 times 3 sqrt{3 times 5} = 12 sqrt{15}]Therefore, the square root of ( frac{9!}{168} ) is ( 12sqrt{15} ).

💡Alright, so I need to figure out how many two-digit prime numbers have a ones digit of 3. Hmm, okay, let's break this down step by step. First, I should recall what a prime number is. A prime number is a number greater than 1 that has no positive divisors other than 1 and itself. So, for example, 2 is prime because it's only divisible by 1 and 2, but 4 isn't prime because it's divisible by 1, 2, and 4.Now, the question specifically asks about two-digit primes. That means we're looking at numbers from 10 up to 99. Within this range, we need to find all the numbers where the ones digit is 3. The ones digit is the digit in the units place, so numbers like 13, 23, 33, and so on.Let me list out all the two-digit numbers that end with 3. Starting from the smallest two-digit number ending with 3, which is 13, and going up by 10 each time: 13, 23, 33, 43, 53, 63, 73, 83, and 93. So, that's nine numbers in total.Now, I need to check which of these numbers are prime. I'll go through each one and see if it's prime or not.Starting with 13: I know that 13 is a prime number because its only divisors are 1 and 13. There's no number between 2 and the square root of 13 (which is approximately 3.6) that divides 13 evenly. So, 13 is prime.Next is 23: Similarly, 23 is a prime number. The square root of 23 is about 4.8, so I check divisibility by 2, 3, and 4. 23 divided by 2 is 11.5, which isn't an integer. 23 divided by 3 is approximately 7.666, also not an integer. And 23 divided by 4 is 5.75, still not an integer. So, 23 is prime.Moving on to 33: This one might be tricky. I think 33 is divisible by 3 because 3 times 11 is 33. Let me check: 3 times 11 is indeed 33, so 33 is not a prime number. It has divisors other than 1 and itself.Next is 43: I believe 43 is a prime number. The square root of 43 is approximately 6.557, so I need to check divisibility by primes less than or equal to 6, which are 2, 3, and 5. 43 divided by 2 is 21.5, not an integer. 43 divided by 3 is about 14.333, not an integer. 43 divided by 5 is 8.6, also not an integer. So, 43 is prime.Now, 53: I think 53 is a prime number. The square root of 53 is roughly 7.28, so I'll check divisibility by primes up to 7, which are 2, 3, 5, and 7. 53 divided by 2 is 26.5, not an integer. 53 divided by 3 is approximately 17.666, not an integer. 53 divided by 5 is 10.6, not an integer. 53 divided by 7 is about 7.571, still not an integer. So, 53 is prime.Next is 63: I'm pretty sure 63 isn't a prime number. 63 divided by 3 is 21, so it's divisible by 3. Therefore, 63 is not prime.Then, 73: I think 73 is a prime number. The square root of 73 is approximately 8.544, so I'll check divisibility by primes up to 8, which are 2, 3, 5, 7. 73 divided by 2 is 36.5, not an integer. 73 divided by 3 is about 24.333, not an integer. 73 divided by 5 is 14.6, not an integer. 73 divided by 7 is approximately 10.428, not an integer. So, 73 is prime.Next up is 83: I believe 83 is a prime number. The square root of 83 is roughly 9.11, so I'll check divisibility by primes up to 9, which are 2, 3, 5, 7. 83 divided by 2 is 41.5, not an integer. 83 divided by 3 is about 27.666, not an integer. 83 divided by 5 is 16.6, not an integer. 83 divided by 7 is approximately 11.857, not an integer. So, 83 is prime.Finally, 93: I think 93 isn't a prime number. 93 divided by 3 is 31, so it's divisible by 3. Therefore, 93 is not prime.So, out of the nine numbers I listed, 13, 23, 33, 43, 53, 63, 73, 83, and 93, the prime numbers are 13, 23, 43, 53, 73, and 83. That makes six prime numbers in total.Wait, did I miss any? Let me double-check. 13 is prime, 23 is prime, 33 isn't, 43 is prime, 53 is prime, 63 isn't, 73 is prime, 83 is prime, and 93 isn't. So, that's six primes. Hmm, but earlier I thought it was seven. Did I miscount?Let me recount: 13, 23, 43, 53, 73, 83. That's six numbers. Wait, maybe I missed one. Let me check each number again.13: Prime.23: Prime.33: Not prime.43: Prime.53: Prime.63: Not prime.73: Prime.83: Prime.93: Not prime.So, that's six primes. But earlier, I thought it was seven. Maybe I made a mistake in my initial count. Let me see the list again: 13, 23, 33, 43, 53, 63, 73, 83, 93.Wait, 13, 23, 43, 53, 73, 83. That's six. So, the answer should be six, not seven. Did I make a mistake in my initial thought process? It seems like I did. So, the correct number is six.But wait, let me check again to be sure. Maybe I missed a prime number.13: Prime.23: Prime.33: Not prime.43: Prime.53: Prime.63: Not prime.73: Prime.83: Prime.93: Not prime.Yes, that's six primes. So, the correct answer is six.

💡First, Ben thinks of the number 8. According to the teacher's instructions, he multiplies it by 2, adds 1, and then doubles the result. This gives Ben a value of 34 to give to Sue.Sue receives the number 34. Following the teacher's instructions, she subtracts 1 from this number and then doubles the result. This calculation results in 66.Therefore, Sue's final answer is 66.

💡Okay, so I have this function f(x) that's defined piecewise. For x less than or equal to 1, it's a quadratic function: -x² - a x - 5. And for x greater than 1, it's a reciprocal function: a/x. The problem says that for all x1 not equal to x2, the difference quotient (f(x2) - f(x1))/(x2 - x1) is greater than 0. That means the function is strictly increasing everywhere. So, I need to find the range of real numbers a that makes this function strictly increasing on the entire real line.First, I remember that if a function is strictly increasing, its derivative should be positive wherever it's differentiable. But since this is a piecewise function, I have to check the derivatives on each piece and also ensure that the function doesn't have any jumps that would violate the increasing nature at the boundary point x=1.Let me start by analyzing the derivative of each piece.For x ≤ 1, f(x) = -x² - a x - 5. The derivative f’(x) would be the slope of this quadratic function. Taking the derivative, f’(x) = -2x - a. Since the function needs to be increasing, this derivative must be positive for all x ≤ 1.So, -2x - a > 0 for all x ≤ 1.I need to find the condition on a such that this inequality holds for all x ≤ 1. Since x can be any number less than or equal to 1, the most restrictive case will be when x is at its maximum, which is x=1. Because as x decreases, -2x becomes more positive, so the derivative becomes larger. Therefore, the minimum value of the derivative on x ≤ 1 occurs at x=1.So, plugging x=1 into the derivative: f’(1) = -2(1) - a = -2 - a. For the function to be increasing at x=1, we need -2 - a > 0, which simplifies to a < -2.Okay, so that's one condition: a must be less than -2.Next, let's look at the other piece of the function for x > 1, which is f(x) = a/x. The derivative here is f’(x) = -a/x². Again, since the function needs to be increasing, the derivative must be positive. So, -a/x² > 0.Since x² is always positive for x ≠ 0, the sign of the derivative depends on -a. So, -a must be positive, which implies that a must be negative. So, a < 0.But from the first piece, we already have a < -2, which is a subset of a < 0. So, the condition from the second piece is automatically satisfied if a < -2.However, I also need to ensure that the function is continuous at x=1. Wait, actually, the problem doesn't explicitly state that the function is continuous, but if the function is strictly increasing, it must be continuous except possibly at points where it has jump discontinuities. But if there's a jump discontinuity, the function can't be strictly increasing across that point because the left limit and right limit would have to maintain the increasing property.So, let me check the continuity at x=1.For x approaching 1 from the left, the limit of f(x) is f(1) = -(1)^2 - a(1) - 5 = -1 - a - 5 = -6 - a.For x approaching 1 from the right, the limit of f(x) is a/1 = a.Since the function is strictly increasing, the value at x=1 from the right must be greater than or equal to the value at x=1 from the left. But actually, since it's strictly increasing, it must be strictly greater. So, we have:a > -6 - aSolving this inequality:a + a > -62a > -6a > -3So, from the continuity condition, we get that a must be greater than -3.But earlier, from the first piece, we had a < -2. So, combining these two conditions, we have:-3 < a < -2Wait, but I should also check if the function is strictly increasing across the entire domain. So, not only do the derivatives have to be positive on each interval, but also the function must not decrease when moving from the left side to the right side at x=1.I think I already considered that by ensuring that the right-hand limit at x=1 is greater than the left-hand limit. So, that should take care of the increasing nature across the boundary.Let me double-check by considering two points, one on each side of x=1.Suppose x1 < 1 and x2 > 1. Then, f(x2) - f(x1) must be positive if x2 > x1.But since x2 > x1, and the function is increasing, f(x2) > f(x1). So, the difference quotient is positive.But to ensure that, we need to make sure that f(x) is increasing across the entire real line, so the function doesn't have any decreasing parts.Wait, another thing to consider is whether the function is increasing on both sides, which we've already checked via the derivatives, and also ensuring that the function doesn't have a drop at x=1.So, the key points are:1. The derivative on x ≤ 1 is positive, leading to a < -2.2. The derivative on x > 1 is positive, leading to a < 0, which is already satisfied by a < -2.3. The function is increasing across x=1, leading to a > -3.Therefore, combining these, the range of a is -3 < a < -2.But wait, should I include the endpoints? If a = -3, then at x=1, f(1) from the left is -6 - (-3) = -3, and from the right, it's -3. So, f(x) would be equal at x=1 from both sides, but since the function is strictly increasing, it can't have equal values at x=1. So, a cannot be equal to -3.Similarly, if a = -2, then f’(1) = -2 - (-2) = 0, which would make the derivative zero at x=1, meaning the function isn't strictly increasing there. So, a cannot be equal to -2 either.Therefore, the range is -3 < a < -2.I think that's the final answer.

💡First, I need to determine the total number of ways to choose any two books from the available genres. There are 4 mystery novels, 4 fantasy novels, and 4 biographies, making a total of 12 books.The total number of ways to choose any two books from these 12 is calculated using the combination formula ( C(n, k) = frac{n!}{k!(n - k)!} ), where ( n = 12 ) and ( k = 2 ):[C(12, 2) = frac{12!}{2!(12 - 2)!} = frac{12 times 11}{2 times 1} = 66]Next, I need to subtract the number of ways to choose two books from the same genre. For each genre:- Mystery: ( C(4, 2) = 6 )- Fantasy: ( C(4, 2) = 6 )- Biography: ( C(4, 2) = 6 )The total number of ways to choose two books from the same genre is:[6 + 6 + 6 = 18]Finally, to find the number of ways to choose two books from different genres, I subtract the same-genre combinations from the total combinations:[66 - 18 = 48]