💡Alright, so I have this problem where I need to compare three logarithms: ( a = log_{0.3}0.2 ), ( b = log_{3}2 ), and ( c = log_{0.2}3 ). The options given are A: ( a < b < c ), B: ( a < c < b ), C: ( c < a < b ), and D: ( c < b < a ). I need to figure out the correct relationship among ( a ), ( b ), and ( c ).First, I remember that logarithms with bases less than 1 are decreasing functions, while those with bases greater than 1 are increasing. So, for ( a = log_{0.3}0.2 ), since the base is 0.3 (which is less than 1), the logarithm will increase as the argument decreases. Similarly, ( c = log_{0.2}3 ) also has a base less than 1, so it's a decreasing function. On the other hand, ( b = log_{3}2 ) has a base greater than 1, so it's an increasing function.Let me try to compute or estimate each of these logarithms.Starting with ( a = log_{0.3}0.2 ). I can use the change of base formula to convert this into natural logarithms or common logarithms. Let me use common logarithms (base 10) for simplicity. The change of base formula is:[log_{k}m = frac{log m}{log k}]So, applying this to ( a ):[a = frac{log 0.2}{log 0.3}]I know that ( log 0.2 ) is negative because 0.2 is less than 1, and ( log 0.3 ) is also negative for the same reason. So, dividing two negative numbers will give a positive result. Let me calculate the approximate values:( log 0.2 approx -0.69897 )( log 0.3 approx -0.52288 )So,[a approx frac{-0.69897}{-0.52288} approx 1.336]So, ( a ) is approximately 1.336.Next, ( b = log_{3}2 ). Again, using the change of base formula:[b = frac{log 2}{log 3}]I remember that ( log 2 approx 0.3010 ) and ( log 3 approx 0.4771 ). So,[b approx frac{0.3010}{0.4771} approx 0.6309]So, ( b ) is approximately 0.6309.Now, ( c = log_{0.2}3 ). Using the change of base formula again:[c = frac{log 3}{log 0.2}]We already know ( log 3 approx 0.4771 ) and ( log 0.2 approx -0.69897 ). So,[c approx frac{0.4771}{-0.69897} approx -0.6826]So, ( c ) is approximately -0.6826.Now, let's summarize the approximate values:- ( a approx 1.336 )- ( b approx 0.6309 )- ( c approx -0.6826 )Comparing these values, ( c ) is negative, ( b ) is positive but less than 1, and ( a ) is greater than 1. So, the order from smallest to largest is ( c < b < a ).Looking back at the options:A: ( a < b < c ) → Not possible since ( a ) is the largest.B: ( a < c < b ) → Not possible since ( c ) is negative.C: ( c < a < b ) → Not possible since ( a > b ).D: ( c < b < a ) → This matches our conclusion.Therefore, the correct answer should be D.But wait, let me double-check my calculations to make sure I didn't make any mistakes.For ( a ):[log 0.2 = log frac{1}{5} = -log 5 approx -0.69897][log 0.3 = log frac{3}{10} = log 3 - log 10 approx 0.4771 - 1 = -0.5229][a = frac{-0.69897}{-0.5229} approx 1.336]That seems correct.For ( b ):[log 2 approx 0.3010, log 3 approx 0.4771][b = frac{0.3010}{0.4771} approx 0.6309]That also seems correct.For ( c ):[log 3 approx 0.4771, log 0.2 approx -0.69897][c = frac{0.4771}{-0.69897} approx -0.6826]Yes, that's correct as well.So, the order is indeed ( c < b < a ), which is option D.I think that's solid. I don't see any errors in my reasoning or calculations. So, I feel confident that D is the correct answer.**Final Answer**boxed{D}

💡Okay, so I have this problem with two vectors, a and b. Vector a is given as (cos θ, -sin θ), and vector b is (3cos θ, sin θ). The question says that these vectors are perpendicular to each other, and I need to find the value of θ within the interval (0, π). The options are A: π/3, B: 2π/3, C: π/6 or 5π/6, and D: π/3 or 2π/3.Alright, first thing I remember is that if two vectors are perpendicular, their dot product is zero. So, I should calculate the dot product of vectors a and b and set it equal to zero. That should give me an equation to solve for θ.Let me write down the vectors again to make sure I have them right:- Vector a: (cos θ, -sin θ)- Vector b: (3cos θ, sin θ)So, the dot product of a and b is calculated by multiplying the corresponding components and then adding them together. That would be:(cos θ)(3cos θ) + (-sin θ)(sin θ)Let me compute that step by step:First component: cos θ multiplied by 3cos θ is 3cos²θ.Second component: -sin θ multiplied by sin θ is -sin²θ.So, adding those together, the dot product is 3cos²θ - sin²θ.Since the vectors are perpendicular, this dot product equals zero. Therefore:3cos²θ - sin²θ = 0Hmm, okay, so I have the equation 3cos²θ - sin²θ = 0. I need to solve for θ in the interval (0, π). Let me think about how to approach this.One way to solve this is to express everything in terms of either sine or cosine. Maybe I can use the Pythagorean identity, which is sin²θ + cos²θ = 1. Let me see if I can rewrite the equation using that identity.Alternatively, I could divide both sides of the equation by cos²θ to make it in terms of tangent. Let me try that.Dividing both sides by cos²θ:(3cos²θ)/cos²θ - (sin²θ)/cos²θ = 0/cos²θSimplifying:3 - tan²θ = 0So, 3 - tan²θ = 0. That simplifies to tan²θ = 3. Taking the square root of both sides, tanθ = ±√3.Okay, so tanθ equals positive or negative square root of 3. Now, I need to find θ in the interval (0, π) where tanθ is equal to √3 or -√3.Let me recall the values of tangent for common angles. I know that tan(π/3) = √3 and tan(2π/3) = -√3. So, θ can be π/3 or 2π/3.Wait, let me make sure. In the interval (0, π), tanθ is positive in the first quadrant (0 to π/2) and negative in the second quadrant (π/2 to π). So, tanθ = √3 at θ = π/3, which is in the first quadrant, and tanθ = -√3 at θ = 2π/3, which is in the second quadrant.So, both π/3 and 2π/3 are solutions within the interval (0, π). Therefore, the possible values of θ are π/3 and 2π/3.Looking back at the options, option D is π/3 or 2π/3, which matches my solutions. So, I think the correct answer is D.Just to double-check, let me plug these values back into the original equation to ensure they satisfy 3cos²θ - sin²θ = 0.First, θ = π/3:cos(π/3) = 1/2, sin(π/3) = √3/2So, 3cos²θ = 3*(1/2)² = 3*(1/4) = 3/4sin²θ = (√3/2)² = 3/4Thus, 3cos²θ - sin²θ = 3/4 - 3/4 = 0. That works.Next, θ = 2π/3:cos(2π/3) = -1/2, sin(2π/3) = √3/2So, 3cos²θ = 3*(-1/2)² = 3*(1/4) = 3/4sin²θ = (√3/2)² = 3/4Thus, 3cos²θ - sin²θ = 3/4 - 3/4 = 0. That also works.Okay, both solutions satisfy the equation. So, I'm confident that D is the correct answer.**Final Answer**boxed{D}

💡First, I need to determine the number of girls at each school. At Maplewood Middle School, there are 300 students with a boy to girl ratio of 3:2. This means for every 3 boys, there are 2 girls. To find the number of girls, I'll set up the equation:3x + 2x = 300Combining like terms gives:5x = 300Solving for x:x = 60So, the number of girls at Maplewood Middle School is:2x = 2 * 60 = 120Next, at Brookside Middle School, there are 240 students with a boy to girl ratio of 3:5. This means for every 3 boys, there are 5 girls. To find the number of girls, I'll set up the equation:3y + 5y = 240Combining like terms gives:8y = 240Solving for y:y = 30So, the number of girls at Brookside Middle School is:5y = 5 * 30 = 150Now, I'll add the number of girls from both schools to find the total number of girls:120 + 150 = 270The total number of students at the event is the sum of students from both schools:300 + 240 = 540Finally, to find the fraction of the students who are girls:270 / 540 = 1/2Therefore, the fraction of the students at the event who are girls is 1/2.

💡First, I need to determine how many bottle caps Danny has in total after finding more at the park. Danny initially had 6 bottle caps and found 22 more, so he now has 6 + 22 = 28 bottle caps.Next, I'll focus on the wrappers. Danny found 8 wrappers at the park. To find out how many wrappers he has now, I need to know how many wrappers he had before going to the park. Let's denote the number of wrappers he had initially as W.After finding 8 more wrappers, the total number of wrappers Danny has now is W + 8.However, the problem does not provide the initial number of wrappers (W). Without this information, I cannot determine the exact number of wrappers Danny has now. The total number of wrappers remains W + 8, where W is the number of wrappers he had before finding the additional 8.

💡First, I need to determine the initial sum of money, which is the principal amount (P), using the compound interest formula. The formula for compound interest is:A = P(1 + r/n)^(nt)Where:- A is the total amount after interest- P is the principal amount- r is the annual interest rate (in decimal)- n is the number of times interest is compounded per year- t is the time in yearsGiven:- Compound Interest (CI) = Rs. 1289.0625- Annual interest rate (r) = 25% = 0.25- Number of times compounded per year (n) = 4 (quarterly)- Time (t) = 6 months = 0.5 yearsFirst, I'll calculate the total amount (A) after 6 months by adding the compound interest to the principal:A = P + CINext, I'll substitute A into the compound interest formula:P + CI = P(1 + r/n)^(nt)Rearranging the formula to solve for P:P = (P + CI) / (1 + r/n)^(nt)Plugging in the known values:P = (P + 1289.0625) / (1 + 0.25/4)^(4*0.5)Simplifying the equation:P = (P + 1289.0625) / (1 + 0.0625)^2P = (P + 1289.0625) / 1.12890625To isolate P, I'll subtract P from both sides:P - P = 1289.0625 / 0.12890625Solving for P:P = 10000Therefore, the initial sum of money is Rs. 10,000.

💡First, I recognize that the cost price (CP) is 70% of the marked price (MP), so CP = 0.70 * MP.The gain percent is given as 28.57142857142858%, which I convert to a decimal for easier calculations: 0.2857142857142857.The selling price (SP) can be calculated by adding the gain to the cost price: SP = CP * (1 + Gain Percent).Substituting the value of CP, I get SP = 0.70 * MP * 1.2857142857142857.Next, I know that the selling price after applying a discount (D%) on the marked price is SP = MP * (1 - D%).By equating the two expressions for SP, I set up the equation 0.70 * MP * 1.2857142857142857 = MP * (1 - D%).Simplifying this, I find that 1 - D% = 0.8999999999999999, which leads to D% = 0.1000000000000001.Converting this decimal back to a percentage, I determine that the discount percentage is 10%.