💡Alright, so I have this problem here where I need to compare three values: a, b, and c. They're defined as follows:- ( a = 0.3^{0.6} )- ( b = 0.5^{0.6} )- ( c = log_{frac{pi}{3}} frac{pi}{4} )And I need to figure out the order of these values from greatest to least. The options are:A: ( b > a > c )B: ( a > b > c )C: ( c > b > a )D: ( c > b > a )Hmm, okay. Let me try to break this down step by step.First, let's look at ( a ) and ( b ). Both are exponential expressions with exponents of 0.6. The bases are 0.3 and 0.5, respectively. Since both bases are between 0 and 1, I remember that when you raise a number between 0 and 1 to a positive exponent, the result is larger than the original number but still less than 1. Also, the larger the base, the larger the result when the exponent is the same. So, since 0.5 is larger than 0.3, ( b ) should be larger than ( a ). That gives me ( b > a ).Now, let's think about ( c ). It's a logarithm with base ( frac{pi}{3} ) and argument ( frac{pi}{4} ). I know that ( pi ) is approximately 3.1416, so ( frac{pi}{3} ) is roughly 1.0472, and ( frac{pi}{4} ) is about 0.7854. So, the logarithm is ( log_{1.0472} 0.7854 ).I remember that logarithms can be tricky, especially when the base is greater than 1 but the argument is less than 1. Let me recall the logarithm properties. If the base is greater than 1 and the argument is less than 1, the logarithm is negative. That's because ( log_b a ) is negative when ( a < 1 ) and ( b > 1 ). So, ( c ) is negative.So, putting it all together, I have:- ( b > a ) because 0.5 > 0.3 and both raised to 0.6.- ( c ) is negative, so it's less than both ( a ) and ( b ).Therefore, the order should be ( b > a > c ), which is option A.Wait, but let me double-check my reasoning about the logarithm. Maybe I made a mistake there. Let me calculate ( c ) more precisely.Given ( c = log_{frac{pi}{3}} frac{pi}{4} ), I can use the change of base formula to convert this logarithm to a natural logarithm or base 10. The formula is:( log_b a = frac{ln a}{ln b} )So, plugging in the values:( c = frac{ln frac{pi}{4}}{ln frac{pi}{3}} )Calculating the natural logs:- ( ln frac{pi}{4} approx ln 0.7854 approx -0.241 )- ( ln frac{pi}{3} approx ln 1.0472 approx 0.046 )So, ( c approx frac{-0.241}{0.046} approx -5.24 )That's definitely negative, so my initial thought was correct. ( c ) is negative, and both ( a ) and ( b ) are positive numbers less than 1.Now, let's get back to ( a ) and ( b ). I said ( b > a ) because 0.5 > 0.3, but let me verify that with actual calculations.Calculating ( a = 0.3^{0.6} ):First, take the natural logarithm of 0.3: ( ln 0.3 approx -1.20397 )Multiply by 0.6: ( -1.20397 times 0.6 approx -0.72238 )Exponentiate: ( e^{-0.72238} approx 0.485 )So, ( a approx 0.485 )Calculating ( b = 0.5^{0.6} ):Take the natural logarithm of 0.5: ( ln 0.5 approx -0.69315 )Multiply by 0.6: ( -0.69315 times 0.6 approx -0.41589 )Exponentiate: ( e^{-0.41589} approx 0.658 )So, ( b approx 0.658 )Therefore, ( b approx 0.658 ) and ( a approx 0.485 ), which confirms that ( b > a ).Since ( c approx -5.24 ), it's much less than both ( a ) and ( b ). So, the order is ( b > a > c ), which is option A.Wait a second, but looking back at the options, both C and D are ( c > b > a ). That seems contradictory to my conclusion. Did I misread the options?Looking back:A: ( b > a > c )B: ( a > b > c )C: ( c > b > a )D: ( c > b > a )So, both C and D are the same? That might be a typo, but assuming that's correct, both C and D are the same option. So, the correct answer should be A.But just to be thorough, let me check my calculations again.For ( a = 0.3^{0.6} ):Using a calculator, 0.3^0.6 ≈ e^{0.6 * ln(0.3)} ≈ e^{0.6 * (-1.20397)} ≈ e^{-0.72238} ≈ 0.485. That seems correct.For ( b = 0.5^{0.6} ):0.5^0.6 ≈ e^{0.6 * ln(0.5)} ≈ e^{0.6 * (-0.69315)} ≈ e^{-0.41589} ≈ 0.658. That also seems correct.For ( c = log_{frac{pi}{3}} frac{pi}{4} ):Using change of base: ( frac{ln (pi/4)}{ln (pi/3)} ≈ frac{ln 0.7854}{ln 1.0472} ≈ frac{-0.241}{0.046} ≈ -5.24 ). That's definitely negative.So, ( c ) is negative, ( a ) is about 0.485, ( b ) is about 0.658. So, ( b > a > c ), which is option A.Therefore, the correct answer is A.

💡First, I need to determine the daily cost of Claire's coffee and pastry. She buys a latte for 3.75 and a croissant for 3.50 every morning.Next, I'll calculate the total cost for one day by adding the price of the latte and the croissant:3.75 + 3.50 = 7.25Since she plans to do this every morning for a week, I'll multiply the daily cost by 7 to find the weekly total:7.25 * 7 = 50.75Claire also wants to buy 5 cookies, each costing 1.25. I'll calculate the total cost for the cookies:1.25 * 5 = 6.25Now, I'll add the cost of the cookies to the weekly total for the coffee and pastry:50.75 + 6.25 = 57.00Finally, I'll subtract the total expenditure from the 100 gift card to find out how much money will be left:100 - 57.00 = 43.00Therefore, Claire will have 43.00 left on her gift card after a week of coffee and pastry.

💡Okay, so I've got this trigonometry problem here. It says that (cosα + sinα)/(cosα - sinα) equals 2, and then I need to find the value of 1 + 3 sinα cosα - 2 cos²α. Hmm, let me see how to approach this.First, the given equation is a ratio of two expressions involving cosα and sinα. Maybe I can simplify this ratio somehow. I remember that sometimes dividing both numerator and denominator by cosα can be helpful because it can turn the expression into something involving tanα, which might be easier to handle.So, let me try that. If I divide both the numerator and the denominator by cosα, the equation becomes:(cosα + sinα)/cosα divided by (cosα - sinα)/cosα.That simplifies to (1 + tanα)/(1 - tanα) = 2.Alright, so now I have (1 + tanα)/(1 - tanα) = 2. Let me set tanα = t to make it easier to write. So, the equation becomes (1 + t)/(1 - t) = 2.Now, I can solve for t. Multiply both sides by (1 - t):1 + t = 2(1 - t).Expanding the right side:1 + t = 2 - 2t.Now, let's get all the t terms on one side and constants on the other:t + 2t = 2 - 1.That simplifies to:3t = 1.So, t = 1/3. Since t is tanα, that means tanα = 1/3.Alright, so now I know that tanα is 1/3. I need to find the value of 1 + 3 sinα cosα - 2 cos²α. Hmm, let's think about how to express this in terms of tanα because I already know tanα.I remember that sinα = tanα / sqrt(1 + tan²α) and cosα = 1 / sqrt(1 + tan²α). Maybe I can use these expressions to rewrite the given expression.Alternatively, I can use the identity sin²α + cos²α = 1. Since I have terms involving sinα cosα and cos²α, perhaps I can express everything in terms of sinα and cosα and then substitute using tanα.Wait, another approach is to express everything in terms of tanα by dividing numerator and denominator by cos²α. Let me try that.The expression is 1 + 3 sinα cosα - 2 cos²α. Let me factor out cos²α from all terms:1 + 3 sinα cosα - 2 cos²α = (1/cos²α) * (cos²α + 3 sinα cosα - 2 cos²α).Wait, that might complicate things. Alternatively, I can divide numerator and denominator by cos²α, but since it's not a fraction, maybe that's not the right approach.Wait, perhaps I can write the expression as:1 + 3 sinα cosα - 2 cos²α = 1 - 2 cos²α + 3 sinα cosα.I know that 1 - 2 cos²α is equal to -cos(2α), but I'm not sure if that helps here. Alternatively, maybe I can factor something out.Alternatively, since I know tanα = 1/3, I can find sinα and cosα in terms of t.Let me set tanα = 1/3, so sinα = (1/3)/sqrt(1 + (1/3)²) = (1/3)/sqrt(10/9) = (1/3)/(√10/3) = 1/√10.Similarly, cosα = 1/sqrt(1 + (1/3)²) = 1/sqrt(10/9) = 3/√10.So, sinα = 1/√10 and cosα = 3/√10.Now, let's compute each term in the expression:First term: 1.Second term: 3 sinα cosα = 3*(1/√10)*(3/√10) = 3*(3/10) = 9/10.Third term: -2 cos²α = -2*(9/10) = -18/10.So, adding them all together:1 + 9/10 - 18/10.Convert 1 to 10/10 to have a common denominator:10/10 + 9/10 - 18/10 = (10 + 9 - 18)/10 = (19 - 18)/10 = 1/10.So, the value of the expression is 1/10.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I started with (cosα + sinα)/(cosα - sinα) = 2.2. Divided numerator and denominator by cosα to get (1 + tanα)/(1 - tanα) = 2.3. Let tanα = t, so (1 + t)/(1 - t) = 2. Solved for t and got t = 1/3.4. Then, found sinα and cosα using tanα = 1/3, so sinα = 1/√10 and cosα = 3/√10.5. Plugged into the expression: 1 + 3 sinα cosα - 2 cos²α.6. Calculated each term: - 3 sinα cosα = 3*(1/√10)*(3/√10) = 9/10. - -2 cos²α = -2*(9/10) = -18/10.7. Added them up: 1 + 9/10 - 18/10 = 1 - 9/10 = 1/10.Everything seems to check out. I think the answer is 1/10.

💡Alright, so I've got this problem about the curve ( y = x^{n+1} ) and its tangent at the point (1,1). The x-coordinate where this tangent intersects the x-axis is ( x_n ). There are two parts to this problem: part (I) asks for the sum ( a_1 + a_2 + ldots + a_9 ) where ( a_n = log{x_n} ), and part (II) is about determining whether there's a largest positive integer ( m ) such that the inequality ( f(n) + f(n+1) + ldots + f(2n-1) > frac{m}{24} ) holds for all positive integers ( n ), given that ( nf(n) = x_n ).Starting with part (I). I need to find ( x_n ), the x-intercept of the tangent to the curve at (1,1). To find the equation of the tangent, I need the derivative of ( y = x^{n+1} ). The derivative ( y' ) is ( (n+1)x^n ). At the point (1,1), the slope of the tangent is ( (n+1)(1)^n = n+1 ). So, the equation of the tangent line is ( y - 1 = (n+1)(x - 1) ).To find the x-intercept, set ( y = 0 ):[0 - 1 = (n+1)(x_n - 1)][-1 = (n+1)(x_n - 1)][x_n - 1 = -frac{1}{n+1}][x_n = 1 - frac{1}{n+1} = frac{n}{n+1}]So, ( x_n = frac{n}{n+1} ).Given ( a_n = log{x_n} ), then ( a_n = log{left(frac{n}{n+1}right)} ). Therefore, the sum ( a_1 + a_2 + ldots + a_9 ) is:[sum_{k=1}^{9} log{left(frac{k}{k+1}right)} = log{left(frac{1}{2}right)} + log{left(frac{2}{3}right)} + ldots + log{left(frac{9}{10}right)}]This is a telescoping series because when you add the logs, the terms cancel out:[log{left(frac{1}{2} times frac{2}{3} times ldots times frac{9}{10}right)} = log{left(frac{1}{10}right)} = log{1} - log{10} = 0 - log{10} = -log{10}]But wait, in the original problem, it's just ( log{x_n} ), so if ( x_n = frac{n}{n+1} ), then ( a_n = log{left(frac{n}{n+1}right)} ). So, the sum is ( log{left(frac{1}{10}right)} ), which is ( -log{10} ). But in the problem statement, is the logarithm base 10 or natural? It doesn't specify, but in math problems like this, it's usually natural logarithm, but the answer is a simple number, so maybe it's base 10? Wait, no, the answer is -1, which would make sense if it's base 10 because ( log_{10}{(1/10)} = -1 ). So, if it's base 10, the answer is -1. If it's natural logarithm, it would be ( -ln{10} ), which is approximately -2.3026, but the answer is given as -1, so it must be base 10. So, the sum is -1.Moving on to part (II). We're told that ( nf(n) = x_n ), so ( f(n) = frac{x_n}{n} = frac{1}{n(n+1)} times n = frac{1}{n+1} ). Wait, no, ( x_n = frac{n}{n+1} ), so ( f(n) = frac{x_n}{n} = frac{n}{n+1} times frac{1}{n} = frac{1}{n+1} ). So, ( f(n) = frac{1}{n+1} ).We need to determine if there's a largest positive integer ( m ) such that the inequality ( f(n) + f(n+1) + ldots + f(2n-1) > frac{m}{24} ) holds for all positive integers ( n ). So, let's write out what ( f(n) + f(n+1) + ldots + f(2n-1) ) is.Since ( f(k) = frac{1}{k+1} ), the sum becomes:[sum_{k=n}^{2n-1} frac{1}{k+1} = sum_{k=n+1}^{2n} frac{1}{k}]So, it's the sum of reciprocals from ( n+1 ) to ( 2n ). This is a harmonic series segment. I know that the harmonic series grows like ( ln{n} + gamma ), but here it's a partial sum.We need to find the maximum ( m ) such that:[sum_{k=n+1}^{2n} frac{1}{k} > frac{m}{24}]for all positive integers ( n ).So, we need to find the infimum of ( sum_{k=n+1}^{2n} frac{1}{k} ) over all ( n ), then set ( frac{m}{24} ) to be less than that infimum. The largest integer ( m ) such that ( frac{m}{24} ) is less than the infimum.First, let's compute this sum for small ( n ) to get an idea.For ( n = 1 ):[sum_{k=2}^{2} frac{1}{k} = frac{1}{2} = 0.5]So, ( 0.5 > frac{m}{24} ) implies ( m < 12 ).For ( n = 2 ):[sum_{k=3}^{4} frac{1}{k} = frac{1}{3} + frac{1}{4} approx 0.3333 + 0.25 = 0.5833]So, ( 0.5833 > frac{m}{24} ) implies ( m < 14 ).For ( n = 3 ):[sum_{k=4}^{6} frac{1}{k} = frac{1}{4} + frac{1}{5} + frac{1}{6} approx 0.25 + 0.2 + 0.1667 = 0.6167]So, ( 0.6167 > frac{m}{24} ) implies ( m < 14.8 ).For ( n = 4 ):[sum_{k=5}^{8} frac{1}{k} = frac{1}{5} + frac{1}{6} + frac{1}{7} + frac{1}{8} approx 0.2 + 0.1667 + 0.1429 + 0.125 = 0.6346]So, ( 0.6346 > frac{m}{24} ) implies ( m < 15.23 ).For ( n = 5 ):[sum_{k=6}^{10} frac{1}{k} approx 0.1667 + 0.1429 + 0.125 + 0.1111 + 0.1 = 0.6457]So, ( m < 15.5 ).As ( n ) increases, the sum ( sum_{k=n+1}^{2n} frac{1}{k} ) approaches ( ln{2} approx 0.6931 ). So, the infimum is actually ( ln{2} ), but for finite ( n ), the sum is slightly less than ( ln{2} ). Wait, no, as ( n ) increases, the sum approaches ( ln{2} ) from below because the integral ( int_{n}^{2n} frac{1}{x} dx = ln{2n} - ln{n} = ln{2} ). But the sum is less than the integral because the function ( 1/x ) is decreasing. So, the sum ( sum_{k=n+1}^{2n} frac{1}{k} ) is less than ( ln{2} ), but approaches ( ln{2} ) as ( n ) increases.Wait, but for ( n = 1 ), the sum is 0.5, which is less than ( ln{2} approx 0.6931 ). For ( n = 2 ), it's 0.5833, still less than ( ln{2} ). For ( n = 3 ), 0.6167, still less. For ( n = 4 ), 0.6346, and so on, approaching ( ln{2} ). So, the infimum is actually 0.5, achieved at ( n = 1 ).Wait, but for ( n = 1 ), the sum is 0.5, which is the smallest value. For larger ( n ), the sum increases towards ( ln{2} ). So, the minimum value of the sum is 0.5, and it increases from there. Therefore, the inequality ( sum_{k=n+1}^{2n} frac{1}{k} > frac{m}{24} ) must hold for all ( n ), including ( n = 1 ). So, the most restrictive case is ( n = 1 ), where the sum is 0.5. Therefore, ( 0.5 > frac{m}{24} ) implies ( m < 12 ). So, the largest integer ( m ) is 11.But wait, let's check for ( n = 1 ), ( m = 11 ):[0.5 > frac{11}{24} approx 0.4583]Yes, 0.5 > 0.4583.For ( n = 2 ):[0.5833 > frac{11}{24} approx 0.4583]True.For ( n = 3 ):[0.6167 > 0.4583]True.And as ( n ) increases, the sum increases towards ( ln{2} approx 0.6931 ), which is still greater than ( 0.4583 ).Therefore, the largest integer ( m ) such that ( sum_{k=n+1}^{2n} frac{1}{k} > frac{m}{24} ) for all positive integers ( n ) is ( m = 11 ).But wait, let's make sure that ( m = 11 ) is indeed the largest. If we try ( m = 12 ), then ( frac{12}{24} = 0.5 ). For ( n = 1 ), the sum is exactly 0.5, so ( 0.5 > 0.5 ) is false. Therefore, ( m = 12 ) doesn't work. Hence, ( m = 11 ) is indeed the largest integer satisfying the condition.So, summarizing:(I) The sum ( a_1 + a_2 + ldots + a_9 = -1 ).(II) The largest positive integer ( m ) is 11.

💡First, I need to find the derivative of the function ( f(x) = 5sin x ). I recall that the derivative of ( sin x ) is ( cos x ). Multiplying by the constant 5, the derivative of ( 5sin x ) is ( 5cos x ). So, ( f'(x) = 5cos x ).Next, I need to evaluate this derivative at ( x = frac{pi}{2} ).Substituting ( frac{pi}{2} ) into the derivative, I get ( f'left(frac{pi}{2}right) = 5cosleft(frac{pi}{2}right) ).I know that ( cosleft(frac{pi}{2}right) = 0 ).Therefore, ( f'left(frac{pi}{2}right) = 5 times 0 = 0 ).

💡First, I need to determine the speed of the train. The train passes a platform that is 450 meters long, and it takes 105 seconds to do so. When passing the platform, the train covers its own length plus the length of the platform. So, the total distance covered while passing the platform is:600 meters (train length) + 450 meters (platform length) = 1050 meters.Using the formula for speed:Speed = Distance / Time,the speed of the train is:1050 meters / 105 seconds = 10 meters per second.Now, to find out how long it takes for the train to cross a tree, I only need to consider the length of the train, since the tree has negligible length compared to the train.Using the formula for time:Time = Distance / Speed,the time it takes to cross the tree is:600 meters / 10 meters per second = 60 seconds.