answer：First, I need to determine the distance Sarah walked one way. Let's denote this distance as ( d ) miles.Sarah walks north at a rate of 3 mph, so the time it takes her to walk one way is ( frac{d}{3} ) hours.On her return trip, she walks at a rate of 4 mph, so the time it takes her to return is ( frac{d}{4} ) hours.The total time for the round trip is the sum of the times for each leg of the trip, which is given as 3.5 hours. Therefore, I can write the equation:[frac{d}{3} + frac{d}{4} = 3.5]To solve for ( d ), I first find a common denominator for the fractions, which is 12. I rewrite the equation as:[frac{4d}{12} + frac{3d}{12} = 3.5]Combining the fractions, I get:[frac{7d}{12} = 3.5]To solve for ( d ), I multiply both sides of the equation by 12:[7d = 3.5 times 12][7d = 42]Now, I divide both sides by 7 to find ( d ):[d = frac{42}{7}][d = 6]So, Sarah walked 6 miles one way. Since the round trip is twice the one-way distance, the total distance she walked is:[6 text{ miles} times 2 = 12 text{ miles}]

answer：First, I need to determine the size of the car's gasoline tank. I know that the car travels 336 miles in the city and achieves 32 miles per gallon. By dividing the total miles by the miles per gallon, I can find the tank size.Next, I'll use the tank size to calculate the car's miles per gallon on the highway. The car travels 462 miles on the highway, so I'll divide this distance by the tank size to find the highway mileage.Finally, I'll subtract the city miles per gallon from the highway miles per gallon to determine how many fewer miles per gallon the car achieves in the city compared to the highway.

answer：Okay, so I have this problem about triangle ABC. The sides opposite angles A, B, and C are a, b, and c respectively. There are two vectors given: e₁ = (2cos C, c/2 - b) and e₂ = (a/2, 1). It says that e₁ is perpendicular to e₂. Part I asks for the value of cos 2A. Hmm, okay. Since the vectors are perpendicular, their dot product should be zero. So, I can write the dot product of e₁ and e₂ and set it equal to zero. Let me write that out:e₁ • e₂ = (2cos C)(a/2) + (c/2 - b)(1) = 0.Simplifying that, I get:(2cos C)(a/2) is just a cos C, and then plus (c/2 - b). So:a cos C + (c/2 - b) = 0.So, a cos C + c/2 - b = 0.Hmm, okay. So, that's the equation I get from the perpendicular vectors. Now, I need to relate this to angle A. Maybe I can use the Law of Cosines or the Law of Sines here.Let me recall the Law of Cosines: c² = a² + b² - 2ab cos C. But I'm not sure if that's directly helpful here. Maybe the Law of Sines would be better. The Law of Sines says that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle.So, perhaps I can express a, b, c in terms of the sines of their opposite angles. Let me denote 2R as a common term. So, a = 2R sin A, b = 2R sin B, c = 2R sin C.Substituting these into the equation a cos C + c/2 - b = 0:(2R sin A) cos C + (2R sin C)/2 - 2R sin B = 0.Simplify each term:First term: 2R sin A cos C.Second term: (2R sin C)/2 = R sin C.Third term: -2R sin B.So, putting it all together:2R sin A cos C + R sin C - 2R sin B = 0.I can factor out R:R [2 sin A cos C + sin C - 2 sin B] = 0.Since R is not zero (it's the radius of the circumscribed circle), the expression inside the brackets must be zero:2 sin A cos C + sin C - 2 sin B = 0.Hmm, okay. Let me see if I can express sin B in terms of A and C. Since in a triangle, A + B + C = π, so B = π - A - C. Therefore, sin B = sin (A + C).So, sin B = sin (A + C) = sin A cos C + cos A sin C.So, substituting back into the equation:2 sin A cos C + sin C - 2 [sin A cos C + cos A sin C] = 0.Let me expand the last term:2 sin A cos C + sin C - 2 sin A cos C - 2 cos A sin C = 0.Simplify term by term:2 sin A cos C - 2 sin A cos C cancels out.Then, sin C - 2 cos A sin C remains.So, sin C (1 - 2 cos A) = 0.Hmm, so either sin C = 0 or 1 - 2 cos A = 0.But in a triangle, angle C can't be 0 or π, so sin C ≠ 0. Therefore, 1 - 2 cos A = 0.So, 1 - 2 cos A = 0 => 2 cos A = 1 => cos A = 1/2.Therefore, angle A is π/3, since cos π/3 = 1/2.Now, we need to find cos 2A. Using the double angle formula:cos 2A = 2 cos² A - 1.Since cos A = 1/2, cos² A = 1/4.So, cos 2A = 2*(1/4) - 1 = 1/2 - 1 = -1/2.So, cos 2A is -1/2.Alright, that seems solid. Let me recap to make sure I didn't make any mistakes.1. Vectors are perpendicular, so their dot product is zero.2. Substituted the expressions for a, b, c using the Law of Sines.3. Simplified the equation, leading to an expression involving sin C and cos A.4. Found that cos A must be 1/2, so A is 60 degrees or π/3 radians.5. Applied the double angle formula to find cos 2A.Seems good. I think part I is done.Now, part II: If a = 2, find the range of the perimeter L of triangle ABC.Perimeter L = a + b + c. Since a = 2, L = 2 + b + c. So, I need to find the range of b + c.Given that a = 2, and from part I, we know angle A is π/3.So, we can use the Law of Cosines again. For angle A:a² = b² + c² - 2bc cos A.Since a = 2 and cos A = 1/2,4 = b² + c² - 2bc*(1/2) => 4 = b² + c² - bc.So, 4 = b² + c² - bc.I need to find the range of b + c.Let me denote s = b + c. Then, I can express b² + c² in terms of s and bc.Recall that (b + c)² = b² + 2bc + c². So, b² + c² = (b + c)² - 2bc = s² - 2bc.Substituting into the equation:4 = (s² - 2bc) - bc => 4 = s² - 3bc.So, 4 = s² - 3bc => 3bc = s² - 4 => bc = (s² - 4)/3.Now, I need to relate bc to s. Maybe using the AM-GM inequality or something.But since b and c are sides of a triangle, they must satisfy the triangle inequalities. Also, since angle A is fixed, perhaps we can use some other relations.Alternatively, maybe express bc in terms of s and then find constraints on s.Wait, from the equation bc = (s² - 4)/3.But also, from the Law of Cosines, we have 4 = b² + c² - bc.But b² + c² can also be written as (b + c)^2 - 2bc = s² - 2bc.So, 4 = s² - 2bc - bc = s² - 3bc, which is consistent with what I had before.So, 4 = s² - 3bc => bc = (s² - 4)/3.Now, to find the range of s, we can use the fact that in any triangle, the sum of two sides must be greater than the third side.So, b + c > a = 2. So, s > 2.Also, since a = 2, and the triangle inequality, b + c > 2.But what's the upper limit? Hmm.We can use the AM-GM inequality on b and c. The maximum value of bc occurs when b = c, perhaps?Wait, but let's think about it. From the equation bc = (s² - 4)/3.We can also note that for fixed s, bc is maximized when b = c. Because for a given sum, the product is maximized when the numbers are equal.So, if b = c, then s = 2b, so b = s/2.Then, bc = (s/2)^2 = s²/4.But from above, bc = (s² - 4)/3.So, setting them equal:s²/4 = (s² - 4)/3.Multiply both sides by 12:3s² = 4(s² - 4)3s² = 4s² - 163s² - 4s² = -16-s² = -16 => s² = 16 => s = 4.So, when b = c, s = 4.Therefore, the maximum value of s is 4, since when b and c are equal, the product bc is maximized, and s cannot be larger than 4.Wait, but let me verify that. If s = 4, then b = c = 2.Then, plugging back into the equation 4 = b² + c² - bc:4 = 4 + 4 - 4 => 4 = 4. So, that works.So, s can be at most 4.But wait, earlier I thought s > 2, but actually, in a triangle, the sum of two sides must be greater than the third side. So, b + c > a = 2, so s > 2.But also, the other triangle inequalities: a + b > c and a + c > b.Since a = 2, and s = b + c, then 2 + b > c and 2 + c > b.But since s = b + c, and b and c are positive, these inequalities are automatically satisfied if s > 2.Wait, no. Let me think.If s = b + c, then for 2 + b > c, since c = s - b, then 2 + b > s - b => 2 + 2b > s.Similarly, 2 + c > b => 2 + (s - b) > b => 2 + s - b > b => 2 + s > 2b => b < (2 + s)/2.But since s = b + c, and c = s - b, these inequalities might not add much beyond s > 2.So, perhaps the only constraints are s > 2 and s ≤ 4.But wait, when s approaches 2, what happens?If s approaches 2, then b + c approaches 2. But since a = 2, the triangle becomes degenerate when s = 2, so s must be greater than 2.So, the perimeter L = 2 + s, so L ranges from just above 4 to 6.Wait, s is between 2 and 4, so L = 2 + s is between 4 and 6.But wait, when s = 4, L = 6, which is the maximum perimeter.When s approaches 2, L approaches 4, but cannot be equal to 4.So, the perimeter L is in the interval (4, 6].Wait, but let me check if s can actually reach 4.When s = 4, b = c = 2, so triangle is equilateral? Wait, no, because angle A is 60 degrees, but sides b and c are 2, and a is 2, so it is an equilateral triangle.Yes, in that case, all sides are 2, so perimeter is 6.But if s approaches 2, then b + c approaches 2, but a is 2, so the triangle becomes very "flat", almost a straight line, but not quite, so the perimeter approaches 4.But can s actually reach 4? Yes, when b = c = 2, as above.So, the range of L is (4, 6].Wait, but let me think again. Is there a lower bound higher than 4?Suppose b approaches 0, then c approaches 2, but then angle A would have to adjust. Wait, but angle A is fixed at 60 degrees.Wait, no, if a = 2 and angle A is 60 degrees, then sides b and c can't be arbitrary.Wait, perhaps I need to find the minimum value of b + c given that a = 2 and angle A = 60 degrees.So, maybe using the Law of Cosines again.We have 4 = b² + c² - bc.We can consider this as a quadratic in terms of b or c.Alternatively, we can use the method of Lagrange multipliers or something, but maybe it's simpler.Let me consider b and c as variables satisfying 4 = b² + c² - bc.We can write this as b² - bc + c² = 4.Let me denote t = b + c, and u = bc.Then, b² + c² = t² - 2u.So, substituting into the equation:t² - 2u - u = 4 => t² - 3u = 4 => u = (t² - 4)/3.But we also know that for real numbers b and c, the discriminant of the quadratic equation must be non-negative.Assume b and c are roots of x² - tx + u = 0.Then, discriminant D = t² - 4u ≥ 0.Substituting u:D = t² - 4*(t² - 4)/3 = t² - (4t² - 16)/3 = (3t² - 4t² + 16)/3 = (-t² + 16)/3 ≥ 0.So, (-t² + 16)/3 ≥ 0 => -t² + 16 ≥ 0 => t² ≤ 16 => t ≤ 4.Which is consistent with earlier result.Also, since t = b + c > 2, as per triangle inequality.So, t ∈ (2, 4].Therefore, perimeter L = 2 + t ∈ (4, 6].So, the range of the perimeter is (4, 6].Wait, but can t actually reach 4? Yes, when b = c = 2, as before.And t approaches 2 when one of the sides approaches 0, but in reality, sides can't be zero, so t approaches 2 from above.Therefore, the perimeter L is greater than 4 and less than or equal to 6.So, the range is (4, 6].I think that's the answer.**Final Answer**(I) boxed{-dfrac{1}{2}} (II) boxed{(4, 6]}

answer：First, I need to calculate the total cost of the chapatis, plates of rice, and plates of mixed vegetables that Alok ordered.The cost of 16 chapatis at Rs. 6 each is:16 chapatis * Rs. 6/chapati = Rs. 96The cost of 5 plates of rice at Rs. 45 each is:5 plates * Rs. 45/plate = Rs. 225The cost of 7 plates of mixed vegetables at Rs. 70 each is:7 plates * Rs. 70/plate = Rs. 490Next, I'll add up these costs to find the total cost of these items:Total cost of chapatis, rice, and mixed vegetables = Rs. 96 + Rs. 225 + Rs. 490 = Rs. 811Alok paid a total of Rs. 961, so the total cost of the ice-cream cups is:Total cost of ice-cream cups = Rs. 961 - Rs. 811 = Rs. 150Alok ordered 6 ice-cream cups, so the cost of each ice-cream cup is:Cost per ice-cream cup = Total cost of ice-cream cups / Number of ice-cream cups = Rs. 150 / 6 = Rs. 25Therefore, the cost of each ice-cream cup is Rs. 25.

answer：Alright, so I have this problem here about a variable X that follows a binomial distribution. It says that the probability of X being 1 is p, and the probability of X being 0 is q, where 0 < p < 1. I need to find the variance of X, denoted as D(X), and express it in terms of p.Okay, let me start by recalling what a binomial distribution is. I remember that a binomial distribution is used to model the number of successes in a fixed number of independent trials, where each trial has the same probability of success. In this case, it seems like we're dealing with a single trial because X can only be 0 or 1. So, this is actually a Bernoulli trial, which is a special case of the binomial distribution where the number of trials n is 1.Now, the variance of a random variable measures how spread out its values are. For a binomial distribution, the variance is given by the formula D(X) = n * p * q, where n is the number of trials, p is the probability of success, and q is the probability of failure, which is 1 - p. Since in this case, n = 1, the formula simplifies to D(X) = p * q.But wait, let me make sure I'm not missing anything here. Is this really a binomial distribution with n = 1, or is it something else? The problem states that X follows a binomial distribution with P(X=1) = p and P(X=0) = q. Since there are only two possible outcomes, 0 and 1, and the probabilities sum up to 1 (because p + q = p + (1 - p) = 1), this does indeed fit the definition of a Bernoulli distribution, which is a special case of the binomial distribution with n = 1.So, if n = 1, then the variance D(X) should be p * q. Let me double-check this formula. I recall that for a Bernoulli random variable, the variance is p * (1 - p), which is the same as p * q since q = 1 - p. That seems consistent.Just to be thorough, let me derive the variance from scratch to ensure I understand where the formula comes from. The variance of a random variable X is defined as D(X) = E[X^2] - (E[X])^2, where E[X] is the expected value of X.First, let's find E[X]. Since X can take on two values, 0 and 1, with probabilities q and p respectively, the expected value E[X] is:E[X] = 0 * q + 1 * p = p.Next, let's find E[X^2]. Since X is either 0 or 1, squaring it doesn't change its value:E[X^2] = 0^2 * q + 1^2 * p = 0 * q + 1 * p = p.Now, plug these into the variance formula:D(X) = E[X^2] - (E[X])^2 = p - p^2 = p(1 - p) = p * q.So, that confirms it. The variance D(X) is indeed p * q, which is equal to p(1 - p).I think I've got it. The variance of X is p times (1 - p). This makes sense because when p is 0 or 1, the variance is 0, which means there's no spread since the outcome is certain. As p increases from 0 to 0.5, the variance increases, reaching a maximum at p = 0.5, and then decreases again as p approaches 1. This behavior aligns with my understanding of variance for a Bernoulli distribution.Just to recap, the steps I took were:1. Recognize that X follows a Bernoulli distribution since it has two possible outcomes.2. Recall the formula for variance in a binomial distribution and simplify it for n = 1.3. Derive the variance from the definition using expected values to confirm the formula.4. Verify the logic by considering the behavior of variance as p changes.I think I've covered all the necessary steps and verified the result through multiple methods. I feel confident that the variance D(X) is p(1 - p).**Final Answer**The variance of ( X ) is boxed{p(1-p)}.

answer：First, I identify the relationship between the distances of Sky Falls and Rock Mist Mountains from the city. Rock Mist Mountains are 50 times farther than Sky Falls.Next, I set up an equation where the distance of Rock Mist Mountains (400 miles) equals 50 times the distance of Sky Falls (x).Finally, I solve for x by dividing 400 miles by 50, which gives me the distance of Sky Falls from the city.