answer：Okay, so I have this problem here where I need to find the value of 'a' given two equations involving exponents and a linear relationship between x and y. Let me try to break this down step by step.First, the problem states that (3^{x} = 4^{y}) and (2x = a y). I need to find the value of 'a'. Hmm, okay. So, both equations involve x and y, and I need to relate them somehow to solve for 'a'. Let me start with the first equation: (3^{x} = 4^{y}). This looks like an exponential equation where the bases are different, 3 and 4. I remember that when dealing with equations where the bases are different, taking logarithms can be helpful because logarithms can help bring down the exponents. So, maybe I should take the natural logarithm of both sides to make things easier. Let me try that:[ln(3^{x}) = ln(4^{y})]Using the logarithm power rule, which says that (ln(a^{b}) = b ln(a)), I can simplify both sides:[x ln(3) = y ln(4)]Okay, so now I have a linear equation relating x and y. Let me write that down:[x ln(3) = y ln(4)]From this equation, I can solve for either x or y in terms of the other. Since the second equation given is (2x = a y), which also relates x and y, maybe I can express x in terms of y from the first equation and substitute it into the second equation. That way, I can solve for 'a'.Let me solve for x from the first equation:[x = frac{y ln(4)}{ln(3)}]So, x is equal to ( frac{y ln(4)}{ln(3)} ). Now, I can substitute this expression for x into the second equation (2x = a y):[2 left( frac{y ln(4)}{ln(3)} right) = a y]Let me simplify this equation. First, multiply 2 with the fraction:[frac{2 y ln(4)}{ln(3)} = a y]Now, I notice that both sides of the equation have a 'y' term. Since y is a non-zero real number (because it's in (mathbb{R}^{*})), I can safely divide both sides by y to eliminate it:[frac{2 ln(4)}{ln(3)} = a]So, now I have an expression for 'a' in terms of logarithms. Let me write that down:[a = frac{2 ln(4)}{ln(3)}]Hmm, this looks a bit complicated, but maybe I can simplify it further. I remember that the change of base formula for logarithms says that (log_{b}a = frac{ln(a)}{ln(b)}). So, perhaps I can rewrite this expression using logarithms with base 3.Let me try that. Since (ln(4)) is in the numerator and (ln(3)) is in the denominator, this is actually the definition of (log_{3}4). So, I can rewrite the expression as:[a = 2 log_{3}4]Okay, that's a bit simpler. But I also remember that 4 is equal to (2^{2}), so maybe I can express (log_{3}4) in terms of (log_{3}2). Let me see:[log_{3}4 = log_{3}(2^{2}) = 2 log_{3}2]Ah, yes! Using the logarithm power rule again, which states that (log_{b}(a^{n}) = n log_{b}a), I can bring down the exponent 2. So, substituting this back into the expression for 'a':[a = 2 times 2 log_{3}2 = 4 log_{3}2]So, 'a' is equal to (4 log_{3}2). Let me check if this matches any of the answer choices. The options are:A: (2log_{3}2)B: (log_{3}2)C: (2log_{2}3)D: (4log_{3}2)Yes, option D is exactly what I got: (4log_{3}2). So, that must be the correct answer.Wait, just to make sure I didn't make any mistakes in my steps. Let me go through it again.Starting with (3^{x} = 4^{y}), taking natural logs gives (x ln3 = y ln4). Then, solving for x: (x = frac{y ln4}{ln3}). Plugging into (2x = a y): (2 times frac{y ln4}{ln3} = a y). Dividing both sides by y: (2 frac{ln4}{ln3} = a). Then, recognizing that (frac{ln4}{ln3} = log_{3}4), so (a = 2 log_{3}4). Since (4 = 2^2), (log_{3}4 = 2 log_{3}2), so (a = 4 log_{3}2). Yep, that all checks out.I think I did everything correctly. I used logarithms to solve for x in terms of y, substituted into the second equation, and simplified using logarithm properties. The answer seems consistent with the steps, so I'm confident that D is the right choice.**Final Answer**boxed{D}

answer：Okay, so I have this problem where I need to prove that ( P(P(x)) > 0 ) for all real ( x ), given that ( P(x) = x^2 + rx + s ) is a quadratic polynomial with real coefficients. The polynomial has two distinct real roots, both less than (-1), and the difference between the two roots is less than (2). First, I need to understand the properties of the polynomial ( P(x) ). Since it's a quadratic with real coefficients and two distinct real roots, its discriminant must be positive. The discriminant ( D ) of a quadratic ( ax^2 + bx + c ) is given by ( D = b^2 - 4ac ). In this case, ( a = 1 ), ( b = r ), and ( c = s ), so the discriminant is ( D = r^2 - 4s ). Since the roots are distinct, ( D > 0 ).Let me denote the roots by ( a ) and ( b ), where ( a < b ). Given that both roots are less than (-1), we have ( a < b < -1 ). Also, the difference between the roots is less than (2), so ( b - a < 2 ).I know that for a quadratic equation ( x^2 + rx + s ), the sum of the roots is ( -r ) and the product is ( s ). So, ( a + b = -r ) and ( ab = s ).Now, I need to analyze ( P(P(x)) ). Let's write it out:( P(P(x)) = (P(x))^2 + rP(x) + s ).Since ( P(x) ) is a quadratic, ( P(P(x)) ) will be a quartic polynomial. To show that ( P(P(x)) > 0 ) for all real ( x ), I need to ensure that this quartic polynomial is always positive. One way to approach this is to analyze the range of ( P(x) ) and see how it maps through ( P ) again. If I can show that ( P(x) ) never takes values in the interval where ( P ) is negative, then ( P(P(x)) ) will always be positive.Given that ( P(x) ) has two real roots ( a ) and ( b ), and since the leading coefficient is positive, ( P(x) ) is positive outside the interval ([a, b]) and negative inside ((a, b)). So, ( P(x) > 0 ) when ( x < a ) or ( x > b ), and ( P(x) < 0 ) when ( a < x < b ).Now, if I can show that ( P(x) ) never lies in the interval ((a, b)), then ( P(P(x)) ) will always be positive because ( P ) evaluated at any ( P(x) ) outside ([a, b]) is positive.But wait, ( P(x) ) can take values in ((a, b)) when ( x ) is between ( a ) and ( b ). So, ( P(x) ) is negative in that interval, which would mean ( P(P(x)) ) is evaluated at a negative value. Since ( P ) is positive outside ([a, b]), if ( P(x) ) is negative, it's still outside ([a, b]) because ( a ) and ( b ) are both less than (-1). Wait, hold on. If ( a ) and ( b ) are both less than (-1), then the interval ([a, b]) is entirely to the left of (-1). So, ( P(x) ) being negative occurs when ( x ) is between ( a ) and ( b ), which are both less than (-1). But if ( x ) is greater than ( b ) or less than ( a ), ( P(x) ) is positive. Since ( a ) and ( b ) are both less than (-1), the positive values of ( P(x) ) occur when ( x ) is either less than ( a ) (which is less than (-1)) or greater than ( b ) (which is also less than (-1)). Wait, this seems a bit confusing. Let me try to visualize the graph of ( P(x) ). It's a parabola opening upwards with roots at ( a ) and ( b ), both less than (-1). The vertex of the parabola is at ( x = -r/2 ). Since ( a + b = -r ), the vertex is at ( x = (a + b)/2 ). Given that ( a < b < -1 ), the vertex is somewhere between ( a ) and ( b ), which is also less than (-1). The minimum value of ( P(x) ) occurs at the vertex, which is ( P(-r/2) = (-r/2)^2 + r(-r/2) + s = r^2/4 - r^2/2 + s = -r^2/4 + s ). Since ( P(x) ) has two real roots, this minimum value is negative. So, the minimum value ( P(-r/2) = s - r^2/4 ) is negative. Now, considering ( P(P(x)) ), let's analyze its behavior. For any real ( x ), ( P(x) ) can be either positive or negative. If ( P(x) ) is positive, then ( P(P(x)) ) is evaluated at a positive value. If ( P(x) ) is negative, ( P(P(x)) ) is evaluated at a negative value.But since ( P(x) ) is positive when ( x < a ) or ( x > b ), and negative when ( a < x < b ), we need to see where ( P(x) ) maps these intervals.When ( x < a ) or ( x > b ), ( P(x) > 0 ). So, ( P(P(x)) ) is evaluated at positive arguments. Since ( P ) is positive for arguments outside ([a, b]), and since ( a, b < -1 ), any positive argument is definitely greater than ( b ), so ( P(P(x)) ) will be positive.When ( a < x < b ), ( P(x) < 0 ). So, ( P(P(x)) ) is evaluated at negative arguments. But the roots ( a ) and ( b ) are both less than (-1), so the negative arguments are less than (-1). Wait, but ( P(x) ) when ( a < x < b ) is negative, but how negative is it? The minimum value of ( P(x) ) is ( s - r^2/4 ), which is negative. So, ( P(x) ) can take values from just above negative infinity up to the minimum value.But since ( a ) and ( b ) are both less than (-1), the interval ([a, b]) is entirely to the left of (-1). So, if ( P(x) ) is negative, it's in the interval ((a, b)), which is less than (-1). But wait, ( P(x) ) is negative in ((a, b)), but ( P(x) ) evaluated at any ( x ) in ((a, b)) gives a value in ((-infty, 0)). However, ( P ) evaluated at any point less than ( a ) is positive, and since ( a < -1 ), any value less than ( a ) is also less than (-1). So, if ( P(x) ) is negative, say ( P(x) = c ) where ( c < 0 ), then ( c ) is less than ( b ) because ( b < -1 ). Therefore, ( P(c) ) would be positive because ( c < a ) or ( c > b ). But since ( c < 0 ) and ( a < b < -1 ), ( c ) could be either less than ( a ) or between ( a ) and ( b ). Wait, no. If ( c = P(x) ) and ( c < 0 ), then ( c ) is in ((a, b)) because ( P(x) ) is negative only in ((a, b)). But ( a ) and ( b ) are both less than (-1), so ( c ) is between ( a ) and ( b ), which are both less than (-1). So, ( c ) is less than (-1). Therefore, ( P(c) ) is evaluated at a point ( c < -1 ). Since ( c < -1 ), and ( P(x) ) is positive for ( x < a ) or ( x > b ), but ( c ) is between ( a ) and ( b ), which are both less than (-1). Wait, no. If ( c ) is between ( a ) and ( b ), which are both less than (-1), then ( c ) is less than (-1). But ( P(x) ) is positive for ( x < a ) and ( x > b ). So, if ( c < a ), then ( P(c) > 0 ). If ( c ) is between ( a ) and ( b ), then ( P(c) < 0 ). But ( c ) is in ((a, b)), so ( P(c) < 0 ). Wait, this seems like a contradiction. Let me think again.If ( x ) is in ((a, b)), then ( P(x) ) is negative, say ( c ). So, ( c ) is in ((a, b)), which is less than (-1). Therefore, ( c ) is less than (-1). Now, evaluating ( P(c) ), since ( c < -1 ), we need to see where ( c ) is relative to ( a ) and ( b ). Since ( a < b < -1 ), and ( c ) is in ((a, b)), which is less than (-1), ( c ) is between ( a ) and ( b ). But ( P(c) ) is negative because ( c ) is between ( a ) and ( b ). So, ( P(P(x)) = P(c) < 0 ) when ( x ) is in ((a, b)). But this contradicts the statement we need to prove, which is ( P(P(x)) > 0 ) for all real ( x ).Hmm, so maybe my initial approach is flawed. Let me try a different angle.Perhaps I should consider the composition ( P(P(x)) ) directly. Since ( P(x) ) is a quadratic, ( P(P(x)) ) will be a quartic polynomial. To show that it's always positive, I need to ensure that it has no real roots and that it opens upwards.The quartic polynomial ( P(P(x)) ) will open upwards because the leading term is ( (x^2)^2 = x^4 ), which has a positive coefficient. So, if I can show that ( P(P(x)) ) has no real roots, then it must be always positive.So, let's assume for contradiction that there exists some real ( x ) such that ( P(P(x)) = 0 ). Then, ( P(x) ) must be a root of ( P ), i.e., ( P(x) = a ) or ( P(x) = b ).Therefore, solving ( P(P(x)) = 0 ) is equivalent to solving ( P(x) = a ) or ( P(x) = b ).So, let's analyze the equations ( P(x) = a ) and ( P(x) = b ).First, consider ( P(x) = a ). This is a quadratic equation:( x^2 + rx + s = a )Which simplifies to:( x^2 + rx + (s - a) = 0 )Similarly, for ( P(x) = b ):( x^2 + rx + (s - b) = 0 )For these equations to have real solutions, their discriminants must be non-negative.The discriminant for ( P(x) = a ) is:( D_1 = r^2 - 4(s - a) )And for ( P(x) = b ):( D_2 = r^2 - 4(s - b) )So, for real solutions, we need ( D_1 geq 0 ) and ( D_2 geq 0 ).But let's compute these discriminants.Recall that ( a + b = -r ) and ( ab = s ).So, ( D_1 = r^2 - 4(s - a) = r^2 - 4s + 4a )But ( r^2 = (a + b)^2 = a^2 + 2ab + b^2 )So, ( D_1 = a^2 + 2ab + b^2 - 4ab + 4a = a^2 - 2ab + b^2 + 4a )Similarly, ( D_2 = r^2 - 4(s - b) = r^2 - 4s + 4b = a^2 - 2ab + b^2 + 4b )Now, let's analyze ( D_1 ) and ( D_2 ).Given that ( a < b < -1 ) and ( b - a < 2 ), let's see if these discriminants can be non-negative.First, consider ( D_1 = a^2 - 2ab + b^2 + 4a ).This can be written as ( (a - b)^2 + 4a ).Similarly, ( D_2 = (a - b)^2 + 4b ).Given that ( (a - b)^2 = (b - a)^2 < 4 ) because ( b - a < 2 ).So, ( D_1 = (b - a)^2 + 4a < 4 + 4a ).But since ( a < -1 ), ( 4a < -4 ). Therefore, ( D_1 < 4 - 4 = 0 ).Similarly, ( D_2 = (b - a)^2 + 4b < 4 + 4b ).Since ( b < -1 ), ( 4b < -4 ). Therefore, ( D_2 < 4 - 4 = 0 ).Thus, both discriminants ( D_1 ) and ( D_2 ) are negative, meaning the equations ( P(x) = a ) and ( P(x) = b ) have no real solutions. Therefore, ( P(P(x)) = 0 ) has no real roots.Since ( P(P(x)) ) is a quartic polynomial opening upwards with no real roots, it must be always positive. Hence, ( P(P(x)) > 0 ) for all real ( x ).I think this makes sense. By showing that the equations ( P(x) = a ) and ( P(x) = b ) have no real solutions, we ensure that ( P(P(x)) ) never crosses zero, and since it opens upwards, it must be always positive.

answer：Okay, so I have this physics problem here where two objects are thrown at the same time from two different points. One is thrown at an angle, and the other is thrown straight up. I need to figure out after how many seconds the distance between them is the smallest and what that minimum distance is, given some specific values. Hmm, let's break this down step by step.First, I need to visualize the scenario. There's point A on a horizontal plane where an object is thrown with speed c at an angle α. Then, there's point B, which is a meters away from point A horizontally. From point B, another object is thrown vertically upwards with the same speed c. So, both objects are in motion, but their paths are different—one is a projectile motion, and the other is vertical motion.I remember that for projectile motion, the horizontal and vertical components can be treated separately. The horizontal component of the velocity for the first object will be c*cos(α), and the vertical component will be c*sin(α). Since there's no air resistance, the horizontal velocity remains constant, but the vertical velocity will be affected by gravity, which causes a downward acceleration of g m/s².For the second object thrown vertically upwards from point B, its motion is simpler because it only has vertical velocity. Its position at any time t will depend only on its vertical motion, which is affected by gravity as well.I think the key here is to write the positions of both objects as functions of time and then find the distance between them as a function of time. Once I have that, I can find the time when this distance is minimized.Let me denote the position of the first object as (x1, y1) and the position of the second object as (x2, y2). Since the second object is thrown from point B, which is a meters away from point A, x2 will always be a, right? Because it's moving vertically, so its horizontal position doesn't change.So, for the first object:- x1 = c*t*cos(α)- y1 = c*t*sin(α) - 0.5*g*t²For the second object:- x2 = a- y2 = c*t - 0.5*g*t²Now, the distance between the two objects at time t is sqrt[(x1 - x2)² + (y1 - y2)²]. To find the minimum distance, I can minimize the square of the distance, which should give the same result and is easier to work with.So, let's compute (x1 - x2)² + (y1 - y2)².Substituting the expressions:(x1 - x2)² = (c*t*cos(α) - a)²(y1 - y2)² = [c*t*sin(α) - 0.5*g*t² - (c*t - 0.5*g*t²)]²Simplifying the y-component:y1 - y2 = c*t*sin(α) - 0.5*g*t² - c*t + 0.5*g*t² = c*t*(sin(α) - 1)So, (y1 - y2)² = [c*t*(sin(α) - 1)]² = c²*t²*(sin(α) - 1)²Therefore, the squared distance is:D² = (c*t*cos(α) - a)² + c²*t²*(sin(α) - 1)²Now, let's expand this:First term: (c*t*cos(α) - a)² = c²*t²*cos²(α) - 2*a*c*t*cos(α) + a²Second term: c²*t²*(sin(α) - 1)² = c²*t²*(sin²(α) - 2*sin(α) + 1)So, adding both terms:D² = c²*t²*cos²(α) - 2*a*c*t*cos(α) + a² + c²*t²*(sin²(α) - 2*sin(α) + 1)Now, let's combine like terms:c²*t²*cos²(α) + c²*t²*sin²(α) = c²*t²*(cos²(α) + sin²(α)) = c²*t²*1 = c²*t²Then, the other terms:-2*a*c*t*cos(α) + a² + c²*t²*(-2*sin(α) + 1)So, putting it all together:D² = c²*t² + a² - 2*a*c*t*cos(α) + c²*t²*(-2*sin(α) + 1)Wait, that doesn't seem right. Let me check:Wait, the second term is c²*t²*(sin²(α) - 2*sin(α) + 1). So, when I expand, it's c²*t²*sin²(α) - 2*c²*t²*sin(α) + c²*t².So, combining with the first term:c²*t²*cos²(α) + c²*t²*sin²(α) = c²*t²*(cos²(α) + sin²(α)) = c²*t²Then, the remaining terms:-2*a*c*t*cos(α) + a² - 2*c²*t²*sin(α) + c²*t²So, combining the t² terms:c²*t² - 2*c²*t²*sin(α) = c²*t²*(1 - 2*sin(α))And the linear term:-2*a*c*t*cos(α)And the constant term:a²So, overall:D² = c²*t²*(1 - 2*sin(α)) - 2*a*c*t*cos(α) + a²Hmm, that seems correct.Now, to find the minimum distance, we can treat D² as a function of t and find its minimum. Since D² is a quadratic function in t, its minimum occurs at t = -b/(2a), where the quadratic is in the form at² + bt + c.Wait, but in our case, the quadratic is:D² = [c²*(1 - 2*sin(α))] t² - [2*a*c*cos(α)] t + a²So, the coefficient of t² is A = c²*(1 - 2*sin(α))The coefficient of t is B = -2*a*c*cos(α)The constant term is C = a²So, the time t at which D² is minimized is t = -B/(2A) = [2*a*c*cos(α)] / [2*c²*(1 - 2*sin(α))] = [a*cos(α)] / [c*(1 - 2*sin(α))]Wait, but I have to be careful with the signs. Let me double-check:The quadratic is A*t² + B*t + C, so the vertex is at t = -B/(2A). Here, B is -2*a*c*cos(α), so -B is 2*a*c*cos(α). A is c²*(1 - 2*sin(α)). So, t = (2*a*c*cos(α)) / (2*c²*(1 - 2*sin(α))) = (a*cos(α)) / (c*(1 - 2*sin(α)))Wait, but 1 - 2*sin(α) in the denominator. If 1 - 2*sin(α) is positive, then t is positive. If it's negative, t would be negative, which doesn't make physical sense because time can't be negative. So, we need to ensure that 1 - 2*sin(α) is positive, otherwise, the minimum would occur at t=0 or something.But let's see, for α=30°, sin(30°)=0.5, so 1 - 2*0.5 = 0. So, in that case, the denominator becomes zero, which would be a problem. Hmm, that suggests that when α=30°, the coefficient A becomes zero, making the quadratic term disappear, and D² becomes a linear function in t. That might complicate things.Wait, maybe I made a mistake in the expansion earlier. Let me go back.Original D²:(c*t*cos(α) - a)² + [c*t*(sin(α) - 1)]²Expanding:c²*t²*cos²(α) - 2*a*c*t*cos(α) + a² + c²*t²*(sin(α) - 1)²Which is:c²*t²*cos²(α) - 2*a*c*t*cos(α) + a² + c²*t²*(sin²(α) - 2*sin(α) + 1)Now, combining terms:c²*t²*(cos²(α) + sin²(α) - 2*sin(α) + 1) - 2*a*c*t*cos(α) + a²Since cos²(α) + sin²(α) = 1, this becomes:c²*t²*(1 - 2*sin(α) + 1) - 2*a*c*t*cos(α) + a²Which simplifies to:c²*t²*(2 - 2*sin(α)) - 2*a*c*t*cos(α) + a²Factor out 2:2*c²*t²*(1 - sin(α)) - 2*a*c*t*cos(α) + a²So, D² = 2*c²*(1 - sin(α))*t² - 2*a*c*cos(α)*t + a²Ah, okay, so I had a mistake earlier in the expansion. The correct coefficient for t² is 2*c²*(1 - sin(α)). That makes more sense.So, now, the quadratic is:D² = 2*c²*(1 - sin(α))*t² - 2*a*c*cos(α)*t + a²Now, to find the minimum, we can use the vertex formula. The time t at which the minimum occurs is t = -B/(2A), where A = 2*c²*(1 - sin(α)) and B = -2*a*c*cos(α).So, t = [2*a*c*cos(α)] / [2*2*c²*(1 - sin(α))] = [a*cos(α)] / [2*c*(1 - sin(α))]That looks better. So, t = (a*cos(α)) / (2*c*(1 - sin(α)))Now, let's plug in the given values: α=30°, c=50 m/s, a=50√3 m.First, compute cos(30°) and sin(30°):cos(30°) = √3/2 ≈ 0.8660sin(30°) = 1/2 = 0.5So, 1 - sin(30°) = 1 - 0.5 = 0.5Now, plug into t:t = (50√3 * √3/2) / (2*50*0.5)Simplify numerator:50√3 * √3/2 = 50*(3)/2 = 75Denominator:2*50*0.5 = 50So, t = 75 / 50 = 1.5 secondsOkay, so the time at which the distance is minimized is 1.5 seconds.Now, to find the minimum distance, we can plug t=1.5 back into the distance formula.But since we have D², it might be easier to compute D² first and then take the square root.So, D² = 2*c²*(1 - sin(α))*t² - 2*a*c*cos(α)*t + a²Plugging in the values:c=50, α=30°, a=50√3, t=1.5First, compute each term:1. 2*c²*(1 - sin(α))*t²c² = 25001 - sin(30°) = 0.5t² = (1.5)^2 = 2.25So, 2*2500*0.5*2.25 = 2*2500*0.5 = 2500; 2500*2.25 = 56252. -2*a*c*cos(α)*ta=50√3, c=50, cos(30°)=√3/2, t=1.5So, -2*50√3*50*(√3/2)*1.5Simplify:-2*50*50*1.5*(√3*√3)/2√3*√3=3So, -2*50*50*1.5*3/2Simplify step by step:-2 and /2 cancel out, leaving -1Then, 50*50=25002500*1.5=37503750*3=11250So, the second term is -112503. a² = (50√3)^2 = 2500*3=7500So, D² = 5625 - 11250 + 7500 = (5625 + 7500) - 11250 = 13125 - 11250 = 1875Therefore, D = sqrt(1875) = sqrt(25*75) = 5*sqrt(75) = 5*sqrt(25*3) = 5*5*sqrt(3) = 25√3 metersSo, the minimum distance is 25√3 meters.Wait, let me double-check the calculations for D²:First term: 2*c²*(1 - sin(α))*t²2*(50)^2*(1 - 0.5)*(1.5)^2 = 2*2500*0.5*2.25 = 2*2500=5000; 5000*0.5=2500; 2500*2.25=5625Second term: -2*a*c*cos(α)*t = -2*50√3*50*(√3/2)*1.5Simplify:-2*50*50*1.5*(√3*√3)/2 = -2*50*50*1.5*3/2As before, -2 and /2 cancel, leaving -150*50=2500; 2500*1.5=3750; 3750*3=11250So, second term is -11250Third term: a² = (50√3)^2=2500*3=7500So, D²=5625 -11250 +7500=5625+7500=13125-11250=1875Yes, that's correct.So, D=√1875=√(25*75)=5√75=5*√(25*3)=5*5√3=25√3So, the minimum distance is 25√3 meters.Therefore, the time is 1.5 seconds, and the minimum distance is 25√3 meters.

answer：To determine how much Macey has already saved, I need to calculate the total amount she will save in the next 3 weeks and subtract that from the total cost of the shirt.First, I calculate the total savings for the next 3 weeks by multiplying the number of weeks by the weekly savings:3 weeks * 0.50/week = 1.50Next, I subtract this amount from the total cost of the shirt to find out how much she has already saved:3.00 (total cost) - 1.50 (savings in the next 3 weeks) = 1.50Therefore, Macey has already saved 1.50.

answer：First, I recognize that in an arithmetic sequence, each term can be expressed using the formula ( a_n = a_1 + (n-1)d ), where ( a_1 ) is the first term and ( d ) is the common difference.Given the conditions:1. ( a_5 + a_6 = 16 )2. ( a_8 = 12 )I can set up two equations based on the general term formula:1. ( (a_1 + 4d) + (a_1 + 5d) = 16 )2. ( a_1 + 7d = 12 )Simplifying the first equation:( 2a_1 + 9d = 16 )Now, I have a system of two equations:1. ( 2a_1 + 9d = 16 )2. ( a_1 + 7d = 12 )I can solve this system to find the values of ( a_1 ) and ( d ). Once I have these values, I can substitute them back into the general term formula to find ( a_3 ).

answer：Okay, so I have this problem where there are two sequences: an arithmetic sequence {a_n} and a geometric sequence {b_n}. The sums of their first n terms are S_n and T_n respectively. There are some conditions given: a_1 = b_1, 2a_2 = b_2, S_2 + T_2 = 13, and 2S_3 = b_3. I need to find the general term formulas for both sequences and then find the sum of the first n terms of another sequence c_n, which is defined as 2a_n divided by b_n.Alright, let's start with part (I). I know that for an arithmetic sequence, the nth term is given by a_n = a_1 + (n-1)d, where d is the common difference. The sum of the first n terms, S_n, is (n/2)(2a_1 + (n-1)d). For a geometric sequence, the nth term is b_n = b_1 * q^{n-1}, where q is the common ratio. The sum of the first n terms, T_n, is b_1*(1 - q^n)/(1 - q) if q ≠ 1.Given that a_1 = b_1, let's denote this common first term as a_1 = b_1 = k. So, both sequences start with the same number k.Next, the condition 2a_2 = b_2. Let's express a_2 and b_2 in terms of k, d, and q. For the arithmetic sequence, a_2 = a_1 + d = k + d. For the geometric sequence, b_2 = b_1 * q = k*q. So, 2a_2 = b_2 translates to 2(k + d) = k*q. Let's write that as equation (1): 2(k + d) = k*q.Moving on to S_2 + T_2 = 13. Let's compute S_2 and T_2. For S_2, the sum of the first two terms of the arithmetic sequence: S_2 = (2/2)(2k + (2-1)d) = (1)(2k + d) = 2k + d. For T_2, the sum of the first two terms of the geometric sequence: T_2 = k + k*q. So, S_2 + T_2 = (2k + d) + (k + k*q) = 3k + d + k*q. This equals 13, so equation (2): 3k + d + k*q = 13.Next condition: 2S_3 = b_3. Let's compute S_3 and b_3. S_3 is the sum of the first three terms of the arithmetic sequence: S_3 = (3/2)(2k + 2d) = (3/2)(2k + 2d) = 3(k + d). So, 2S_3 = 2*3(k + d) = 6(k + d). On the other hand, b_3 is the third term of the geometric sequence: b_3 = k*q^{2}. So, equation (3): 6(k + d) = k*q^{2}.So, to recap, we have three equations:1. 2(k + d) = k*q2. 3k + d + k*q = 133. 6(k + d) = k*q^{2}We need to solve for k, d, and q.Let me try to express d from equation (1). From equation (1): 2(k + d) = k*q => 2k + 2d = k*q => 2d = k*q - 2k => d = (k*q - 2k)/2 = k*(q - 2)/2.So, d is expressed in terms of k and q. Let's substitute this into equation (2). Equation (2): 3k + d + k*q = 13. Substituting d:3k + [k*(q - 2)/2] + k*q = 13.Let me compute each term:3k is 3k.[k*(q - 2)/2] is (kq - 2k)/2.k*q is kq.So, adding them up:3k + (kq - 2k)/2 + kq = 13.Let me combine these terms. Let's get a common denominator of 2:(6k)/2 + (kq - 2k)/2 + (2kq)/2 = 13.Combine the numerators:[6k + kq - 2k + 2kq]/2 = 13.Simplify numerator:6k - 2k = 4k.kq + 2kq = 3kq.So, numerator is 4k + 3kq.Thus, (4k + 3kq)/2 = 13.Multiply both sides by 2:4k + 3kq = 26.Factor out k:k*(4 + 3q) = 26. Let's call this equation (2a): k*(4 + 3q) = 26.Now, let's look at equation (3): 6(k + d) = k*q^{2}.We already have d = k*(q - 2)/2, so k + d = k + [k*(q - 2)/2] = [2k + kq - 2k]/2 = [kq]/2.So, k + d = (kq)/2.Thus, equation (3): 6*(kq/2) = k*q^{2} => 3kq = k*q^{2}.Assuming k ≠ 0 (since if k were 0, the sequences would be trivial and the sums would be zero, which doesn't fit S_2 + T_2 = 13), we can divide both sides by k:3q = q^{2} => q^{2} - 3q = 0 => q(q - 3) = 0.Thus, q = 0 or q = 3. But q = 0 would make the geometric sequence collapse after the first term, which doesn't make sense because b_2 would be zero, and from equation (1): 2(k + d) = k*q, if q=0, then 2(k + d)=0, which would imply k + d = 0. But if k is positive, d would have to be negative, but let's see.Wait, if q=0, then from equation (1): 2(k + d) = 0 => k + d = 0. So, d = -k. Then, from equation (2a): k*(4 + 3*0) = 26 => 4k = 26 => k = 26/4 = 13/2. Then d = -13/2.But let's check equation (3): 6(k + d) = k*q^{2}. If q=0, then RHS is 0. LHS: 6(k + d) = 6*(13/2 -13/2) = 6*0 = 0. So, it holds. But let's check equation (2): 3k + d + k*q = 13. Substituting k=13/2, d=-13/2, q=0: 3*(13/2) + (-13/2) + (13/2)*0 = (39/2) - (13/2) + 0 = 26/2 =13. So, it holds.But if q=0, the geometric sequence would be 13/2, 0, 0, 0,... which seems a bit odd, but mathematically, it's a valid geometric sequence with ratio 0. However, in the context of the problem, it's more likely that q=3 is the intended solution because q=0 would make the sequence collapse after the second term, which might not be meaningful. So, let's consider q=3.So, q=3. Then, from equation (2a): k*(4 + 3*3) = 26 => k*(4 +9) =26 => k*13=26 => k=2.So, k=2. Then, d = k*(q - 2)/2 = 2*(3 - 2)/2 = 2*(1)/2=1.So, d=1.So, now, we have a_1 = b_1 =2, d=1, q=3.Thus, the arithmetic sequence is a_n = a_1 + (n-1)d =2 + (n-1)*1 = n +1.The geometric sequence is b_n = b_1*q^{n-1}=2*3^{n-1}.So, that's part (I) done.Now, part (II): Let c_n = (2a_n)/b_n. Find the sum of the first n terms of {c_n}, denoted as C_n.First, let's write c_n in terms of n.From part (I), a_n =n +1, b_n=2*3^{n-1}.So, c_n = 2*(n +1)/(2*3^{n-1})= (n +1)/3^{n-1}.So, c_n = (n +1)/3^{n-1}.We need to find C_n = c_1 + c_2 + ... + c_n.So, C_n = sum_{k=1}^n (k +1)/3^{k-1}.Let me write out the terms:c_1 = (1 +1)/3^{0}=2/1=2.c_2 = (2 +1)/3^{1}=3/3=1.c_3 = (3 +1)/3^{2}=4/9.c_4 =5/27, and so on.So, C_n = 2 + 1 + 4/9 + 5/27 + ... + (n +1)/3^{n-1}.This seems like a series where each term is (k +1)/3^{k-1} for k from 1 to n.To find the sum, we can consider using the method for summing such series, which often involves manipulating the series and subtracting a shifted version.Let me denote S = C_n = sum_{k=1}^n (k +1)/3^{k-1}.Let me write S as:S = 2 + 1 + 4/9 + 5/27 + ... + (n +1)/3^{n-1}.Let me multiply both sides by 3:3S = 6 + 3 + 4/3 + 5/9 + ... + (n +1)/3^{n-2}.Now, subtract the original S from this equation:3S - S = 2S = [6 + 3 + 4/3 + 5/9 + ... + (n +1)/3^{n-2}] - [2 + 1 + 4/9 + 5/27 + ... + (n +1)/3^{n-1}].Let me compute term by term:First term: 6 - 2 =4.Second term: 3 -1=2.Third term: 4/3 - 4/9 = (12/9 -4/9)=8/9.Fourth term:5/9 -5/27= (15/27 -5/27)=10/27.And so on, until the last few terms.Wait, actually, let's see:The subtraction will result in most terms canceling out except for the first few and the last term.Let me write it more carefully.When we subtract, the terms from k=1 to k=n-1 will partially cancel.Let me index the terms:In 3S, the terms are:Term 1:6 (which is 3*2)Term 2:3 (which is 3*1)Term 3:4/3 (which is 3*(4/9))Term 4:5/9 (which is 3*(5/27))...Term n: (n +1)/3^{n-2} (which is 3*(n +1)/3^{n-1})In S, the terms are:Term 1:2Term 2:1Term 3:4/9Term 4:5/27...Term n: (n +1)/3^{n-1}So, when subtracting S from 3S, term by term:Term 1:6 -2=4Term 2:3 -1=2Term 3:4/3 -4/9= (12/9 -4/9)=8/9Term 4:5/9 -5/27= (15/27 -5/27)=10/27Term 5:6/27 -6/81= (18/81 -6/81)=12/81=4/27Wait, but actually, let's see a pattern.Alternatively, perhaps it's better to consider the general approach.Let me denote S = sum_{k=1}^n (k +1)/3^{k-1}.Let me shift the index to make it easier. Let m = k -1, so when k=1, m=0, and when k=n, m=n-1.Thus, S = sum_{m=0}^{n-1} (m +2)/3^{m}.So, S = sum_{m=0}^{n-1} (m +2)/3^{m}.This can be split into two sums:S = sum_{m=0}^{n-1} m/3^{m} + 2*sum_{m=0}^{n-1} 1/3^{m}.We know that sum_{m=0}^{n-1} 1/3^{m} is a finite geometric series with ratio 1/3, so it's equal to (1 - (1/3)^n)/(1 -1/3)= (1 - (1/3)^n)/(2/3)= (3/2)(1 - (1/3)^n).Now, the other sum is sum_{m=0}^{n-1} m/3^{m}.This is a standard sum which can be computed using the formula for sum_{m=0}^{n-1} m x^m.The formula is x(1 - (n)x^{n-1} + (n-1)x^n)/(1 -x)^2.In our case, x=1/3.So, sum_{m=0}^{n-1} m*(1/3)^m = (1/3)(1 - n*(1/3)^{n-1} + (n -1)*(1/3)^n)/(1 -1/3)^2.Compute denominator: (1 -1/3)^2=(2/3)^2=4/9.So, numerator: (1/3)[1 - n*(1/3)^{n-1} + (n -1)*(1/3)^n].Let me factor out (1/3)^{n-1} from the last two terms:= (1/3)[1 - (1/3)^{n-1}(n - (n -1)/3)].Simplify inside the brackets:n - (n -1)/3 = (3n -n +1)/3 = (2n +1)/3.Thus, numerator becomes:(1/3)[1 - (2n +1)/3 * (1/3)^{n-1}] = (1/3)[1 - (2n +1)/3^{n}].Thus, the entire sum is:[(1/3)(1 - (2n +1)/3^{n})]/(4/9) = (1/3)/(4/9) * [1 - (2n +1)/3^{n}] = (3/4)[1 - (2n +1)/3^{n}].So, sum_{m=0}^{n-1} m/3^{m} = (3/4)[1 - (2n +1)/3^{n}].Therefore, going back to S:S = sum_{m=0}^{n-1} m/3^{m} + 2*sum_{m=0}^{n-1} 1/3^{m} = (3/4)[1 - (2n +1)/3^{n}] + 2*(3/2)(1 - (1/3)^n).Simplify:First term: (3/4)[1 - (2n +1)/3^{n}].Second term: 2*(3/2)(1 - (1/3)^n)= 3*(1 - (1/3)^n).So, S = (3/4)[1 - (2n +1)/3^{n}] + 3[1 - (1/3)^n].Let me expand both terms:First term: 3/4 - (3/4)(2n +1)/3^{n}.Second term: 3 - 3*(1/3)^n.Combine them:3/4 + 3 - (3/4)(2n +1)/3^{n} -3*(1/3)^n.Convert 3 to 12/4 to have a common denominator:3/4 + 12/4 = 15/4.So, S = 15/4 - (3/4)(2n +1)/3^{n} -3*(1/3)^n.Now, let's combine the last two terms:Factor out (1/3)^n:=15/4 - [ (3/4)(2n +1) +3 ]*(1/3)^n.Compute inside the brackets:(3/4)(2n +1) +3 = (6n +3)/4 +12/4 = (6n +3 +12)/4 = (6n +15)/4.Thus, S =15/4 - (6n +15)/4*(1/3)^n.Factor out 3 from numerator:=15/4 - 3*(2n +5)/4*(1/3)^n.But (1/3)^n = 3^{-n}, so:=15/4 - 3*(2n +5)/4*3^{-n}.Simplify the constants:3*3^{-n}=3^{1 -n}=3^{-(n -1)}.But perhaps it's better to write it as:=15/4 - (2n +5)/(4*3^{n -1}).Wait, let's see:3*(2n +5)/4*3^{-n}= (2n +5)/4*3^{-(n -1)}.Because 3*3^{-n}=3^{-(n -1)}.So, yes, S=15/4 - (2n +5)/(4*3^{n -1}).But let's write 15/4 as 15/2 divided by 2, but perhaps it's better to leave it as is.Alternatively, factor out 1/4:S= (15 - (2n +5)/3^{n -1}) /4.But the question asks for the sum C_n, which is S.So, C_n=15/4 - (2n +5)/(4*3^{n -1}).We can factor out 1/4:C_n= (15 - (2n +5)/3^{n -1}) /4.Alternatively, write it as:C_n= (15/4) - (2n +5)/(4*3^{n -1}).This is the sum of the first n terms of c_n.So, to recap, after all that algebra, we find that C_n=15/4 - (2n +5)/(4*3^{n -1}).Alternatively, we can write it as (15/2 - (2n +5)/3^{n -1}) /2, but the form I have is fine.So, that's the answer for part (II).**Final Answer**(I) The general term formulas are boxed{a_n = n + 1} and boxed{b_n = 2 cdot 3^{n-1}}.(II) The sum of the first n terms of the sequence {c_n} is boxed{frac{15}{2} - frac{2n + 5}{2 cdot 3^{n-1}}}.