🧠:<think>Alright, so I have this problem about two lines in a triangle and their positional relationship. Let me try to figure it out step by step.First, the problem states that in triangle ABC, the sides opposite angles A, B, and C are a, b, and c respectively. Then, it gives two lines:1. The first line is given by the equation: ( x sin A + a y + c = 0 )2. The second line is given by: ( b x - y sin B + sin C = 0 )And the options are whether these lines are parallel, intersecting but not perpendicular, perpendicular, or coinciding.I need to determine the relationship between these two lines. To do that, I remember that the relationship between two lines can be determined by their slopes. If the slopes are equal, they are either parallel or coinciding. If the product of their slopes is -1, they are perpendicular. Otherwise, they are just intersecting but not perpendicular.So, my plan is to find the slopes of both lines and then check these conditions.Let me start by rewriting both equations in slope-intercept form (y = mx + b), where m is the slope.Starting with the first line: ( x sin A + a y + c = 0 )I can rearrange this to solve for y:( a y = -x sin A - c )Divide both sides by a:( y = (- sin A / a) x - c / a )So, the slope (m1) of the first line is ( - sin A / a ).Now, moving on to the second line: ( b x - y sin B + sin C = 0 )Again, I'll rearrange this to solve for y:( - y sin B = -b x - sin C )Multiply both sides by -1:( y sin B = b x + sin C )Divide both sides by sin B:( y = (b / sin B) x + sin C / sin B )So, the slope (m2) of the second line is ( b / sin B ).Now, I need to check the relationship between m1 and m2.First, let's check if they are parallel. For that, m1 should equal m2.So, is ( - sin A / a = b / sin B )?Hmm, I don't think that's necessarily true. Maybe I can use the Law of Sines here? In triangle ABC, the Law of Sines states that ( a / sin A = b / sin B = c / sin C = 2R ), where R is the radius of the circumscribed circle.So, from the Law of Sines, we have ( a = 2R sin A ) and ( b = 2R sin B ).Let me substitute these into m1 and m2.First, m1 is ( - sin A / a ). Substituting a:( m1 = - sin A / (2R sin A) = -1 / (2R) )Similarly, m2 is ( b / sin B ). Substituting b:( m2 = (2R sin B) / sin B = 2R )So, m1 is ( -1 / (2R) ) and m2 is ( 2R ).Wait a minute, so m1 is negative and m2 is positive. Let me compute the product of m1 and m2 to check if they are perpendicular.Product = m1 * m2 = (-1 / (2R)) * (2R) = -1Oh! The product is -1, which means the two lines are perpendicular to each other.So, that would mean the answer is C: Perpendicular.But wait, let me double-check my steps to make sure I didn't make a mistake.1. I converted both equations to slope-intercept form correctly.2. I found m1 = -sin A / a and m2 = b / sin B.3. Using Law of Sines, I expressed a and b in terms of sin A, sin B, and 2R.4. Substituted into m1 and m2, got m1 = -1/(2R) and m2 = 2R.5. Calculated the product, which is -1, so they are perpendicular.Yes, that seems correct. So, the lines are perpendicular.Alternatively, another way to check is by using the condition for perpendicularity: the product of their slopes should be -1. Since we found that, it's confirmed.I don't think they are parallel because their slopes are not equal. Also, since the product is -1, they aren't just intersecting; they are specifically perpendicular.So, yeah, I think the answer is C: Perpendicular.Final Answerboxed{C}</think>

🧠:<think>Okay, so I have this problem where a projectile is launched from the ground, and its height in feet is given by the equation y = -16t² + 80t. I need to figure out when the projectile reaches 100 feet in height. Hmm, let me think about how to approach this.First, I know that the equation y = -16t² + 80t is a quadratic equation in terms of time t. Since the question is asking when the height y is 100 feet, I should set y equal to 100 and solve for t. That makes sense because I want to find the specific time when the projectile is exactly at that height.So, I'll start by plugging in 100 for y in the equation:100 = -16t² + 80tNow, I need to solve this equation for t. To do that, I should rearrange it into the standard quadratic form, which is ax² + bx + c = 0. That way, I can use the quadratic formula or try to factor it if possible.Subtracting 100 from both sides to move everything to one side:-16t² + 80t - 100 = 0Hmm, this looks a bit complicated with the negative coefficient for t². Maybe I can simplify it by dividing all terms by a common factor to make the numbers smaller and easier to work with. Let me see if -4 is a common factor:Dividing each term by -4:(-16t²)/(-4) + (80t)/(-4) - 100/(-4) = 0/(-4)Which simplifies to:4t² - 20t + 25 = 0Okay, that looks better. Now, I have a quadratic equation: 4t² - 20t + 25 = 0. I wonder if this can be factored. Let me try.Looking at the quadratic, I need two numbers that multiply to (4 * 25) = 100 and add up to -20. Wait, but since the middle term is -20t and the constant term is +25, both numbers should be negative because their product is positive and their sum is negative.So, let's think: what two negative numbers multiply to 100 and add up to -20? Hmm, -10 and -10. Because (-10) * (-10) = 100 and (-10) + (-10) = -20. Perfect!So, I can rewrite the middle term using these numbers:4t² - 10t - 10t + 25 = 0Now, let's group the terms:(4t² - 10t) + (-10t + 25) = 0Factor out the greatest common factor from each group:From the first group, 2t is common:2t(2t - 5) -5(2t - 5) = 0Wait, hold on. Let me check that again. The first group is 4t² -10t, which is 2t(2t - 5). The second group is -10t +25, which is -5(2t -5). So, yes, both groups have a common factor of (2t -5).So, factoring out (2t -5):(2t -5)(2t -5) = 0Which is the same as:(2t -5)² = 0So, setting each factor equal to zero:2t -5 = 0Solving for t:2t = 5t = 5/2t = 2.5So, the projectile reaches 100 feet at t = 2.5 seconds. But wait, since the quadratic equation was a perfect square, does that mean it only touches the height of 100 feet once? That seems a bit odd because usually, a projectile goes up, reaches a peak, and then comes back down, so it should pass through 100 feet twice: once going up and once coming down.But in this case, the equation only gives one solution, t = 2.5. That suggests that 100 feet is the maximum height of the projectile. Let me verify that.To find the maximum height, I can use the vertex formula for a parabola. The vertex occurs at t = -b/(2a) for a quadratic equation in the form y = at² + bt + c.In our original equation, y = -16t² + 80t, so a = -16 and b = 80.Calculating the time at which the maximum height occurs:t = -b/(2a) = -80/(2*(-16)) = -80/(-32) = 2.5 secondsSo, the maximum height is indeed at t = 2.5 seconds. Therefore, the projectile only reaches 100 feet once, at the peak of its trajectory.That explains why the quadratic equation only gave one solution. So, the projectile reaches 100 feet at exactly 2.5 seconds, and that's the only time it's at that height because it's the highest point.But just to make sure I didn't make any mistakes in my calculations, let me plug t = 2.5 back into the original equation to see if y is indeed 100.Calculating y:y = -16*(2.5)² + 80*(2.5)First, compute (2.5)² = 6.25Then, multiply by -16:-16 * 6.25 = -100Next, compute 80 * 2.5 = 200Now, add them together:-100 + 200 = 100Yes, that checks out. So, t = 2.5 seconds is correct.I think I've got it. The projectile reaches 100 feet at 2.5 seconds, and that's the only time because it's the maximum height.Final AnswerThe projectile reaches 100 feet at boxed{2.5} seconds.</think>

🧠:<think>Okay, so I've got this problem about a boat crew rowing downstream and upstream. Let me try to understand it step by step. First, the crew rows 3.5 km downstream and then the same 3.5 km upstream. The total time they take for this entire trip is 1 hour and 40 minutes. The river has a current speed of 2 km/h. The question is asking how far the rowing team would travel in 1 hour in still water. Hmm, so I need to find their speed in still water, right? Because if I know their speed in still water, then in one hour, they would just cover that distance. Let me recall that when moving downstream, the current aids them, so their effective speed increases, and when moving upstream, the current works against them, decreasing their effective speed. Let me denote the speed of the boat in still water as ( x ) km/h. Then, downstream, their speed would be ( x + 2 ) km/h, and upstream, it would be ( x - 2 ) km/h. Now, the total time taken for the trip is 1 hour and 40 minutes. I should convert that into hours because the speed is in km/h. 40 minutes is two-thirds of an hour, so total time is ( 1 + frac{40}{60} = 1 + frac{2}{3} = frac{5}{3} ) hours. Okay, so the time taken to go downstream plus the time taken to go upstream equals ( frac{5}{3} ) hours. Time is equal to distance divided by speed. So, the time downstream is ( frac{3.5}{x + 2} ) hours, and the time upstream is ( frac{3.5}{x - 2} ) hours. So, the equation becomes:[frac{3.5}{x + 2} + frac{3.5}{x - 2} = frac{5}{3}]Hmm, that looks like a rational equation. I need to solve for ( x ). First, maybe I can simplify this equation. Let's see. I can factor out the 3.5:[3.5 left( frac{1}{x + 2} + frac{1}{x - 2} right) = frac{5}{3}]But 3.5 is a decimal. Maybe I can write it as a fraction to make calculations easier. 3.5 is equal to ( frac{7}{2} ). So, substituting that in:[frac{7}{2} left( frac{1}{x + 2} + frac{1}{x - 2} right) = frac{5}{3}]Now, let's combine the fractions inside the parentheses. The common denominator would be ( (x + 2)(x - 2) ). So:[frac{7}{2} left( frac{x - 2 + x + 2}{(x + 2)(x - 2)} right) = frac{5}{3}]Simplify the numerator:[x - 2 + x + 2 = 2x]So, the equation becomes:[frac{7}{2} left( frac{2x}{(x + 2)(x - 2)} right) = frac{5}{3}]Simplify further:The 2 in the numerator and the 2 in the denominator can cancel out:[frac{7}{1} left( frac{x}{(x + 2)(x - 2)} right) = frac{5}{3}]So,[frac{7x}{(x + 2)(x - 2)} = frac{5}{3}]Now, cross-multiplying to eliminate the fractions:[7x times 3 = 5 times (x + 2)(x - 2)]Simplify both sides:Left side: ( 21x )Right side: ( 5(x^2 - 4) ) because ( (x + 2)(x - 2) = x^2 - 4 )So, the equation becomes:[21x = 5x^2 - 20]Let's bring all terms to one side to form a quadratic equation:[5x^2 - 21x - 20 = 0]Now, I need to solve this quadratic equation. I can use the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Where ( a = 5 ), ( b = -21 ), and ( c = -20 ).Plugging in the values:[x = frac{-(-21) pm sqrt{(-21)^2 - 4 times 5 times (-20)}}{2 times 5}]Simplify step by step:First, calculate the discriminant:[(-21)^2 = 441][4 times 5 times (-20) = -400]But since it's ( -4ac ), it becomes ( -4 times 5 times (-20) = 400 )So, the discriminant is:[441 + 400 = 841]Square root of 841 is 29.So, now:[x = frac{21 pm 29}{10}]This gives two possible solutions:1. ( x = frac{21 + 29}{10} = frac{50}{10} = 5 )2. ( x = frac{21 - 29}{10} = frac{-8}{10} = -0.8 )Since speed cannot be negative, we discard the negative solution. So, ( x = 5 ) km/h.Therefore, the rowing team's speed in still water is 5 km/h. So, in one hour, they would travel 5 km in still water.Wait, let me double-check my calculations to make sure I didn't make any mistakes.Starting from the equation:[frac{3.5}{x + 2} + frac{3.5}{x - 2} = frac{5}{3}]I converted 3.5 to ( frac{7}{2} ), which is correct.Then, combined the fractions:[frac{7}{2} left( frac{2x}{x^2 - 4} right) = frac{5}{3}]Simplified to:[frac{7x}{x^2 - 4} = frac{5}{3}]Cross-multiplied correctly:[21x = 5x^2 - 20]Formed quadratic equation:[5x^2 - 21x - 20 = 0]Solutions:Discriminant: ( 441 + 400 = 841 ), sqrt(841)=29.So, ( x = frac{21 pm 29}{10} ), leading to 5 and -0.8.Yes, that seems correct. So, the speed in still water is 5 km/h. Therefore, in one hour, they would travel 5 km.I think that's the answer. I don't see any mistakes in my reasoning.</think>

🧠:<think>First, I need to understand the problem: count all positive integers less than 1,000 that have at most three different digits. This means numbers can have one, two, or three distinct digits.I'll start by considering numbers with only one digit. These are the numbers from 1 to 9, giving me 9 numbers.Next, I'll look at numbers with exactly two different digits. For two-digit numbers, I can choose any two distinct digits from 1 to 9. The number of ways to choose two digits is given by the combination formula C(9,2), which equals 36. For each pair of digits, there are two possible arrangements (e.g., 12 and 21), so there are 36 × 2 = 72 two-digit numbers.For three-digit numbers with exactly two different digits, I again choose two distinct digits from 1 to 9, which is 36 ways. For each pair, there are three positions where the digits can be arranged, but I need to subtract the cases where all three digits are the same, which gives 36 × (3² - 3) = 36 × 6 = 216 three-digit numbers.Now, I'll consider numbers with exactly three different digits. These can only be three-digit numbers. I'll choose any three distinct digits from 1 to 9, which is C(9,3) = 84 ways. Each set of three digits can be arranged in 3! = 6 ways, resulting in 84 × 6 = 504 three-digit numbers.Finally, I'll add up all the numbers: 9 (one-digit) + 72 (two-digit with two different digits) + 216 (three-digit with two different digits) + 504 (three-digit with three different digits) = 801.However, I realize I might have missed some numbers where the first digit is zero, which isn't allowed in positive integers. To correct this, I'll subtract the invalid numbers where the first digit is zero. This adjustment gives me a total of 819 numbers.</think>