💡First, identify the given information:- Total number of students: 880- Number of students who play cricket: 500- Number of students who play both cricket and basketball: 220Let ( B ) represent the number of students who play basketball.Using the principle of inclusion-exclusion:[text{Total} = |C| + |B| - |C cap B|]Plugging in the known values:[880 = 500 + B - 220]Simplify the equation:[880 = 280 + B]Solve for ( B ):[B = 880 - 280 = 600]

💡Okay, so I have this problem about a sequence with 8 terms. The first term is 2014, and the eighth term is 2015. The difference between consecutive terms can be -1, 1/3, or 1. I need to find how many such sequences exist. The options are A:7, B:252, C:210, D:35.Hmm, let me break this down. The sequence has 8 terms, so there are 7 differences between them. Each difference can be -1, 1/3, or 1. The total change from the first term to the eighth term is 2015 - 2014 = 1. So, the sum of these 7 differences must be 1.Let me denote the differences as ( d_1, d_2, ldots, d_7 ), where each ( d_i in {-1, frac{1}{3}, 1} ). So, we have:[d_1 + d_2 + ldots + d_7 = 1]I need to find the number of sequences of differences that satisfy this equation. Each difference can be one of three values, so it's a combinatorial problem where I have to count the number of ways to assign these differences such that their sum is 1.Let me think about how to model this. Let's say I have ( x ) differences of -1, ( y ) differences of 1/3, and ( z ) differences of 1. Since there are 7 differences in total, we have:[x + y + z = 7]And the sum of the differences is:[(-1)x + left(frac{1}{3}right)y + 1 cdot z = 1]So, substituting ( z = 7 - x - y ) into the sum equation:[-x + frac{y}{3} + (7 - x - y) = 1]Simplify this:[-x + frac{y}{3} + 7 - x - y = 1][-2x - frac{2y}{3} + 7 = 1][-2x - frac{2y}{3} = -6]Multiply both sides by 3 to eliminate the fraction:[-6x - 2y = -18]Divide both sides by -2:[3x + y = 9]So, now I have the equation ( 3x + y = 9 ) with the constraints that ( x, y, z ) are non-negative integers and ( x + y leq 7 ) because ( z = 7 - x - y ) must also be non-negative.Let me find all possible integer solutions for ( x ) and ( y ):1. If ( x = 0 ), then ( y = 9 ). But ( x + y = 9 ), which is more than 7. So, this is invalid.2. If ( x = 1 ), then ( y = 6 ). ( x + y = 7 ), which is okay because ( z = 0 ).3. If ( x = 2 ), then ( y = 3 ). ( x + y = 5 ), so ( z = 2 ).4. If ( x = 3 ), then ( y = 0 ). ( x + y = 3 ), so ( z = 4 ).5. If ( x = 4 ), then ( y = -3 ). Negative y is invalid.So, the valid solutions are:- ( x = 1, y = 6, z = 0 )- ( x = 2, y = 3, z = 2 )- ( x = 3, y = 0, z = 4 )Now, for each of these solutions, I need to calculate the number of sequences. Since the differences can be arranged in any order, the number of sequences is the multinomial coefficient for each case.1. For ( x = 1, y = 6, z = 0 ):We have 7 differences, with 1 of them being -1, 6 being 1/3, and 0 being 1. The number of such sequences is:[frac{7!}{1!6!0!} = frac{5040}{1 times 720 times 1} = 7]2. For ( x = 2, y = 3, z = 2 ):We have 2 of -1, 3 of 1/3, and 2 of 1. The number of sequences is:[frac{7!}{2!3!2!} = frac{5040}{2 times 6 times 2} = frac{5040}{24} = 210]3. For ( x = 3, y = 0, z = 4 ):We have 3 of -1, 0 of 1/3, and 4 of 1. The number of sequences is:[frac{7!}{3!0!4!} = frac{5040}{6 times 1 times 24} = frac{5040}{144} = 35]Now, adding up all these possibilities:[7 + 210 + 35 = 252]So, the total number of sequences is 252, which is option B.Wait, let me just double-check my calculations to make sure I didn't make any mistakes.First, for the case ( x = 1, y = 6, z = 0 ):Yes, 7 choose 1 is 7, and the rest are fixed as 1/3. That seems correct.For ( x = 2, y = 3, z = 2 ):7! divided by (2!3!2!) is indeed 210. That's a standard multinomial coefficient.For ( x = 3, y = 0, z = 4 ):7! divided by (3!4!) is 35, which is correct.Adding them up: 7 + 210 is 217, plus 35 is 252. Yep, that's right.I think that's solid. So, the answer should be 252, which is option B.**Final Answer**boxed{B}

💡Okay, so I need to figure out how to express the square root of 10 and 1/9 as a common fraction in its simplest form. Hmm, let me think about this step by step.First, I remember that when dealing with square roots of mixed numbers, it's often easier to convert the mixed number into an improper fraction. That way, I can work with a single fraction instead of dealing with a whole number and a fraction separately. So, let me try that.10 and 1/9 is the mixed number. To convert this into an improper fraction, I need to multiply the whole number part (which is 10) by the denominator of the fractional part (which is 9), and then add the numerator of the fractional part (which is 1). So, that would be:10 * 9 = 90Then, adding the numerator:90 + 1 = 91So, the improper fraction is 91/9. That makes sense because 91 divided by 9 is approximately 10.111..., which is the same as 10 and 1/9.Now, the expression we're dealing with is the square root of this improper fraction. So, we have:√(91/9)I remember that the square root of a fraction can be expressed as the square root of the numerator divided by the square root of the denominator. So, that would be:√91 / √9I know that √9 is 3 because 3 times 3 is 9. So, that simplifies the denominator to 3. So now, the expression becomes:√91 / 3Now, I need to check if √91 can be simplified further. To do this, I should factor 91 and see if any of its factors are perfect squares. Let's factor 91.91 divided by 7 is 13, so 91 = 7 * 13. Both 7 and 13 are prime numbers, which means they don't have any square factors other than 1. Therefore, √91 cannot be simplified any further because there are no perfect square factors in 91.So, putting it all together, the square root of 10 and 1/9 is equal to √91 divided by 3. Since √91 is already in its simplest form, the final answer is √91 over 3.I think that's it. Let me just recap to make sure I didn't miss anything. I converted the mixed number to an improper fraction, took the square root of both the numerator and the denominator, simplified the denominator, checked if the numerator's square root can be simplified, and it couldn't. So, the answer should be √91 over 3.

💡First, recognize that in an isosceles right triangle, the two legs are congruent, and the hypotenuse is √2 times longer than each leg. Given that side XY is longer than side YZ, XY must be the hypotenuse.Let the length of each leg be "a". The area of the triangle is given by the formula:Area = (1/2) * base * heightSince both the base and height are "a", we have:49 = (1/2) * a * aSolving for "a":98 = a²a = √98a = 7√2Now, calculate the hypotenuse XY:XY = a * √2XY = 7√2 * √2XY = 7 * 2XY = 14Therefore, the measure of side XY is 14.

💡First, identify the total number of baseball cards Nell originally had, which is 242.Next, determine how many cards Nell has left after giving some to Jeff, which is 106.To find out how many cards Nell gave to Jeff, subtract the number of cards she has left from the original number of cards.So, 242 minus 106 equals 136.Therefore, Nell gave Jeff 136 cards.

💡Okay, so I've got this problem here with two parts. Let me try to work through each step carefully.First, part (1): I need to find the range of x when the function f(x) reaches its minimum value. The function f(x) is defined as the dot product of vectors m and n. Vector m is (2 cos x, sin x + cos x) and vector n is (√3 sin x, sin x - cos x). So, f(x) = m · n.Alright, let's compute the dot product. The dot product of two vectors (a, b) and (c, d) is ac + bd. So applying that here:f(x) = (2 cos x)(√3 sin x) + (sin x + cos x)(sin x - cos x)Let me compute each part separately.First term: (2 cos x)(√3 sin x) = 2√3 cos x sin xSecond term: (sin x + cos x)(sin x - cos x). Hmm, that looks like a difference of squares. So, (sin x)^2 - (cos x)^2.So putting it all together:f(x) = 2√3 cos x sin x + sin²x - cos²xHmm, I remember that 2 sin x cos x is sin 2x, so 2√3 cos x sin x is √3 sin 2x. Also, sin²x - cos²x is -cos 2x because cos 2x = cos²x - sin²x, so sin²x - cos²x = -cos 2x.So substituting these identities in:f(x) = √3 sin 2x - cos 2xNow, this looks like a single sinusoidal function. Maybe I can write it in the form A sin(2x + φ) or something similar. Let me recall that a sin θ + b cos θ can be written as R sin(θ + φ), where R = √(a² + b²) and tan φ = b/a.Wait, actually, in this case, it's √3 sin 2x - cos 2x. So, a = √3, b = -1. So, R = √( (√3)² + (-1)^2 ) = √(3 + 1) = √4 = 2.So, f(x) can be written as 2 sin(2x - π/6). Let me check that.Using the identity: a sin θ + b cos θ = R sin(θ + φ), where R = √(a² + b²) and φ = arctan(b/a). Wait, actually, the formula is a bit different. Let me recall:If I have A sin θ + B cos θ, it can be written as R sin(θ + φ), where R = √(A² + B²) and φ = arctan(B/A) if A ≠ 0.Wait, but in our case, it's √3 sin 2x - cos 2x. So, A = √3, B = -1. So, R = √( (√3)^2 + (-1)^2 ) = √(3 + 1) = 2.Then, φ is such that sin φ = B/R = (-1)/2 and cos φ = A/R = √3/2. So, φ is -π/6 because sin(-π/6) = -1/2 and cos(-π/6) = √3/2.Therefore, √3 sin 2x - cos 2x = 2 sin(2x - π/6). So, f(x) = 2 sin(2x - π/6).Okay, so f(x) is a sine function with amplitude 2, frequency doubled, and phase shift of π/6.Now, the minimum value of f(x) occurs when sin(2x - π/6) = -1. Because the sine function ranges between -1 and 1, so the minimum value is -2.So, when does sin(2x - π/6) = -1? That happens when 2x - π/6 = 3π/2 + 2πk, where k is any integer, because sine is -1 at 3π/2 plus multiples of 2π.So, solving for x:2x - π/6 = 3π/2 + 2πkAdd π/6 to both sides:2x = 3π/2 + π/6 + 2πkConvert 3π/2 to 9π/6 to add:2x = 9π/6 + π/6 + 2πk = 10π/6 + 2πk = 5π/3 + 2πkDivide both sides by 2:x = 5π/6 + πkSo, the solutions are x = 5π/6 + πk, where k is any integer.Wait, let me check that again. If 2x - π/6 = 3π/2 + 2πk, then 2x = 3π/2 + π/6 + 2πk.3π/2 is 9π/6, so 9π/6 + π/6 is 10π/6, which simplifies to 5π/3. So, 2x = 5π/3 + 2πk, so x = 5π/6 + πk.Yes, that seems correct.So, the range of x when f(x) reaches its minimum is x = 5π/6 + πk, where k is any integer.Wait, but the problem says "the range of x". Hmm, maybe they want all x such that f(x) is at its minimum. So, the set of x is {x | x = 5π/6 + πk, k ∈ Z}.Alternatively, in interval notation, but since it's periodic, it's better to express it as x = 5π/6 + kπ, k ∈ Z.So, that's part (1).Now, moving on to part (2). It says: Let a, b, c be the sides opposite to angles A, B, C in triangle ABC, respectively. If f(C) = 2 and c = √3, find the maximum value of the area S of triangle ABC.Alright, so f(C) = 2. From part (1), f(x) = 2 sin(2x - π/6). So, f(C) = 2 implies 2 sin(2C - π/6) = 2.Divide both sides by 2: sin(2C - π/6) = 1.So, when does sin θ = 1? At θ = π/2 + 2πk, where k is integer.So, 2C - π/6 = π/2 + 2πk.Solving for C:2C = π/2 + π/6 + 2πk = (3π/6 + π/6) + 2πk = 4π/6 + 2πk = 2π/3 + 2πk.Divide by 2:C = π/3 + πk.But since C is an angle in a triangle, it must be between 0 and π. So, let's find possible C.If k = 0: C = π/3.If k = 1: C = π/3 + π = 4π/3, which is greater than π, so not possible.k = -1: C = π/3 - π = -2π/3, which is negative, so not possible.Thus, the only possible solution is C = π/3.So, angle C is π/3 radians, which is 60 degrees.Now, we have triangle ABC with side c opposite angle C, and c = √3.We need to find the maximum area S of triangle ABC.Recall that the area S of a triangle can be given by (1/2)ab sin C, where a and b are the sides adjacent to angle C.So, S = (1/2)ab sin(π/3) = (√3/4)ab.So, to maximize S, we need to maximize ab.But we also know from the Law of Cosines that c² = a² + b² - 2ab cos C.Given that c = √3 and C = π/3, so cos C = 1/2.So, substituting:(√3)² = a² + b² - 2ab*(1/2)Simplify:3 = a² + b² - abSo, we have the equation a² + b² - ab = 3.We need to maximize ab given this constraint.Hmm, this is an optimization problem with constraint. Let me think about how to approach this.One way is to use the method of Lagrange multipliers, but maybe there's a simpler way using algebra.Let me consider that a² + b² - ab = 3.I can rewrite this as a² - ab + b² = 3.Notice that a² - ab + b² is equal to (a - b/2)^2 + (3b²)/4, which is always positive.Alternatively, maybe I can express this in terms of variables u = a + b and v = ab, but I'm not sure if that helps.Wait, another approach: Let's consider a and b as variables, and we can use the AM-GM inequality or other inequalities to find the maximum of ab.But let's see:We have a² + b² - ab = 3.We can write this as a² + b² = ab + 3.We know that a² + b² ≥ 2ab by AM-GM inequality.So, substituting:2ab ≤ ab + 3Subtract ab from both sides:ab ≤ 3So, ab is at most 3.Therefore, the maximum value of ab is 3.Thus, the maximum area S is (√3/4)*3 = (3√3)/4.Wait, but is this achievable? Let me check.If ab = 3, then from a² + b² - ab = 3, we have a² + b² = 3 + ab = 3 + 3 = 6.So, a² + b² = 6 and ab = 3.Let me see if such a and b exist.We can solve for a and b.Let me set a = b, then from ab = 3, a² = 3, so a = √3, b = √3.Then, a² + b² = 3 + 3 = 6, which matches. So, yes, when a = b = √3, we have ab = 3 and a² + b² = 6.Therefore, the maximum value of ab is indeed 3, and thus the maximum area is (3√3)/4.So, the maximum area S is 3√3/4.Wait, let me just double-check everything.We have f(C) = 2, which led us to C = π/3.Then, using the Law of Cosines, we got a² + b² - ab = 3.We need to maximize ab, which we found to be 3, leading to the area being (√3/4)*3 = 3√3/4.Yes, that seems correct.So, the maximum area is 3√3/4.**Final Answer**(1) The range of ( x ) when ( f(x) ) reaches its minimum value is ( boxed{x = frac{5pi}{6} + kpi text{ for any integer } k} ).(2) The maximum value of the area ( S ) of ( triangle ABC ) is ( boxed{dfrac{3sqrt{3}}{4}} ).