answer：Okay, so I have this problem where I need to find the value of ( fleft(frac{2015}{4}right) + log_2 5 ). The function ( f(x) ) is given as an odd function with a period of 2. When ( x ) is in the interval ([0,1)), ( f(x) = log_2(x + 1) ). Hmm, let me try to break this down step by step.First, since ( f(x) ) has a period of 2, that means ( f(x + 2) = f(x) ) for all ( x ). So, if I can find an equivalent value of ( frac{2015}{4} ) within one period, say between 0 and 2, that would simplify things. Let me calculate ( frac{2015}{4} ) to see where it falls.Calculating ( frac{2015}{4} ), I get ( 503.75 ). That's a pretty large number, but since the function has a period of 2, I can subtract multiples of 2 until I get a value within the interval ([0, 2)). Let me figure out how many times 2 goes into 503.75.Dividing 503.75 by 2, I get 251.875. So, that means 2 goes into 503.75 exactly 251 times with a remainder. To find the remainder, I can calculate ( 503.75 - 2 times 251 ). Let me compute that:( 2 times 251 = 502 ), so ( 503.75 - 502 = 1.75 ). Therefore, ( frac{2015}{4} ) is equivalent to 1.75 in terms of the function's period. So, ( fleft(frac{2015}{4}right) = f(1.75) ).Now, 1.75 is within the interval ([1, 2)). But the function is defined differently on ([0,1)) and we know it's an odd function. Since it's an odd function, ( f(-x) = -f(x) ). But how does that help me here?Wait, maybe I should consider the periodicity first. Since 1.75 is in ([1,2)), I can subtract 2 to bring it into the interval ([-1,0)). So, ( f(1.75) = f(1.75 - 2) = f(-0.25) ).Now, ( f(-0.25) ) can be expressed using the odd function property: ( f(-0.25) = -f(0.25) ). So, ( f(1.75) = -f(0.25) ).Given that ( f(x) = log_2(x + 1) ) for ( x in [0,1) ), I can compute ( f(0.25) ):( f(0.25) = log_2(0.25 + 1) = log_2(1.25) ).So, ( f(1.75) = -log_2(1.25) ).Therefore, ( fleft(frac{2015}{4}right) = -log_2(1.25) ).Now, the expression I need to evaluate is ( fleft(frac{2015}{4}right) + log_2 5 ), which becomes:( -log_2(1.25) + log_2 5 ).Let me simplify this expression. First, I can rewrite ( 1.25 ) as ( frac{5}{4} ), so:( -log_2left(frac{5}{4}right) + log_2 5 ).Using the logarithm property ( log_bleft(frac{a}{c}right) = log_b a - log_b c ), I can rewrite ( log_2left(frac{5}{4}right) ) as ( log_2 5 - log_2 4 ). Since ( log_2 4 = 2 ), this becomes:( -(log_2 5 - 2) + log_2 5 ).Distributing the negative sign:( -log_2 5 + 2 + log_2 5 ).Now, the ( -log_2 5 ) and ( log_2 5 ) terms cancel each other out, leaving:( 2 ).So, the value of ( fleft(frac{2015}{4}right) + log_2 5 ) is 2.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I found that ( frac{2015}{4} = 503.75 ).2. Subtracted 2 multiplied by 251 (which is 502) to get 1.75.3. Recognized that 1.75 is in ([1,2)), so subtracted 2 to get -0.25.4. Applied the odd function property: ( f(-0.25) = -f(0.25) ).5. Calculated ( f(0.25) = log_2(1.25) ).6. Expressed the original expression as ( -log_2(1.25) + log_2 5 ).7. Rewrote ( 1.25 ) as ( frac{5}{4} ) and applied logarithm properties.8. Simplified to get 2.Everything seems to check out. I don't see any errors in my reasoning.**Final Answer**boxed{2}

answer：To solve the equation ( |5x + 9| = 34 ), I recognize that the absolute value equation can be split into two separate linear equations: one where the expression inside the absolute value is positive and another where it is negative.First, I consider the case where ( 5x + 9 = 34 ). Solving for ( x ) involves subtracting 9 from both sides to isolate the term with ( x ), resulting in ( 5x = 25 ). Dividing both sides by 5 gives ( x = 5 ).Next, I examine the case where ( 5x + 9 = -34 ). Here, subtracting 9 from both sides yields ( 5x = -43 ). Dividing both sides by 5 gives ( x = -8.6 ).Comparing the two solutions, ( x = 5 ) and ( x = -8.6 ), the smallest value of ( x ) is ( -8.6 ).

answer：Okay, so I need to prove that for any real numbers a, b, and c, the sum of their squares is greater than or equal to the sum of their pairwise products. That is, a² + b² + c² ≥ ab + ac + bc. Hmm, I remember something about inequalities involving squares, maybe the Cauchy-Schwarz inequality or something similar? Or perhaps it's related to expanding some expression.Let me think. I recall that sometimes you can manipulate an expression by completing the square or using known inequalities like AM-GM. Wait, AM-GM is the Arithmetic Mean-Geometric Mean inequality, which states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. But here, we have squares and products, so maybe that's not directly applicable.Alternatively, I remember that sometimes you can express the difference between the two sides of the inequality and show that it's always non-negative. Let me try that. So, let's compute a² + b² + c² - ab - ac - bc. If I can show this is always greater than or equal to zero, then the original inequality holds.How can I rewrite this expression? Maybe by grouping terms or factoring. Let me see. If I factor out 1/2, I can write it as (1/2)(2a² + 2b² + 2c² - 2ab - 2ac - 2bc). Hmm, that might help because now I can try to express this as a sum of squares, which are always non-negative.Let me try to break it down. 2a² - 2ab - 2ac can be written as 2(a² - ab - ac). Similarly, 2b² - 2bc would be 2(b² - bc), and 2c² is just 2c². Wait, maybe that's not the most straightforward way. Alternatively, perhaps I can pair the terms differently.I remember that (a - b)² = a² - 2ab + b². So, if I consider (a - b)² + (a - c)² + (b - c)², let's expand that:(a - b)² = a² - 2ab + b² (a - c)² = a² - 2ac + c² (b - c)² = b² - 2bc + c² Adding these together: (a - b)² + (a - c)² + (b - c)² = (a² - 2ab + b²) + (a² - 2ac + c²) + (b² - 2bc + c²) = 2a² + 2b² + 2c² - 2ab - 2ac - 2bcHey, that's exactly the numerator I had earlier when I factored out 1/2! So, going back, a² + b² + c² - ab - ac - bc = (1/2)[(a - b)² + (a - c)² + (b - c)²]. Since squares are always non-negative, the entire expression is non-negative. Therefore, a² + b² + c² - ab - ac - bc ≥ 0, which implies a² + b² + c² ≥ ab + ac + bc.Wait, that seems to work. But let me double-check if I did everything correctly. I started by expressing the difference as a sum of squares, which are non-negative, so the entire expression is non-negative. Therefore, the original inequality must hold. Also, equality occurs when each squared term is zero, meaning a = b = c.Is there another way to approach this? Maybe using vectors or something more advanced? But since the problem is about real numbers, perhaps the method I used is sufficient. Alternatively, I could consider using the Cauchy-Schwarz inequality, which in one of its forms states that (a₁² + a₂² + ... + aₙ²)(b₁² + b₂² + ... + bₙ²) ≥ (a₁b₁ + a₂b₂ + ... + aₙbₙ)². But I'm not sure if that directly applies here because we have three variables and their products.Wait, maybe if I set up vectors appropriately. Let me think. If I take two vectors, say, (a, b, c) and (b, c, a), then their dot product is ab + bc + ca, and the product of their magnitudes is sqrt(a² + b² + c²) * sqrt(b² + c² + a²) = (a² + b² + c²). So, by Cauchy-Schwarz, (a² + b² + c²)² ≥ (ab + bc + ca)². Taking square roots on both sides, we get a² + b² + c² ≥ |ab + bc + ca|. But since a² + b² + c² is always non-negative and ab + bc + ca can be positive or negative, this might not directly give the desired inequality because if ab + bc + ca is negative, the inequality would still hold, but if it's positive, we need to ensure that a² + b² + c² is greater than or equal to it.Hmm, so maybe the Cauchy-Schwarz approach gives a slightly different inequality, but it still supports the original statement. Alternatively, perhaps I can use the fact that for any real numbers, (a - b)² + (b - c)² + (c - a)² is non-negative, which again leads to the same conclusion as before.Another thought: Maybe using the AM-GM inequality on pairs of terms. For example, a² + b² ≥ 2ab, which is a direct result of AM-GM. Similarly, a² + c² ≥ 2ac and b² + c² ≥ 2bc. If I add these three inequalities together, I get 2a² + 2b² + 2c² ≥ 2ab + 2ac + 2bc. Dividing both sides by 2, we get a² + b² + c² ≥ ab + ac + bc. That's another way to prove it!So, that's actually a simpler approach. Using the basic AM-GM inequality on each pair of variables and then summing them up gives the desired result. This method also shows that equality holds when a = b = c, since that's when each of the AM-GM inequalities becomes an equality.Let me summarize both methods to make sure I understand them:1. Expressing the difference as a sum of squares: a² + b² + c² - ab - ac - bc = ½[(a - b)² + (a - c)² + (b - c)²] ≥ 0.2. Using AM-GM on each pair: a² + b² ≥ 2ab, a² + c² ≥ 2ac, b² + c² ≥ 2bc. Adding these gives 2a² + 2b² + 2c² ≥ 2ab + 2ac + 2bc, which simplifies to a² + b² + c² ≥ ab + ac + bc.Both methods seem solid. The first method directly shows the non-negativity of the difference, while the second method uses a known inequality and combines them. I think either method is acceptable, but the second one might be more straightforward for someone just learning inequalities.I wonder if there are any special cases or if the inequality can be tight. For example, if a = b = c, then both sides are equal. Let's test with specific numbers. Let a = 1, b = 1, c = 1. Then LHS = 1 + 1 + 1 = 3, and RHS = 1 + 1 + 1 = 3. So equality holds.What if a, b, c are different? Let's say a = 2, b = 3, c = 4. Then LHS = 4 + 9 + 16 = 29, and RHS = 6 + 8 + 12 = 26. So 29 ≥ 26, which holds.What if one of them is negative? Let's say a = -1, b = 2, c = 3. Then LHS = 1 + 4 + 9 = 14, and RHS = (-1)(2) + (-1)(3) + (2)(3) = -2 -3 + 6 = 1. So 14 ≥ 1, which is true.Another case: a = 0, b = 0, c = 0. Then both sides are zero, so equality holds.What if two variables are equal and the third is different? Let a = b = 1, c = 2. Then LHS = 1 + 1 + 4 = 6, and RHS = 1 + 2 + 2 = 5. So 6 ≥ 5, which is true.It seems like the inequality holds in all these cases, which gives me more confidence in the proof.I also recall that this inequality is a specific case of the more general inequality that the sum of squares is greater than or equal to the sum of products, which is related to the variance in statistics. The expression a² + b² + c² - ab - ac - bc is actually equal to ½[(a - b)² + (a - c)² + (b - c)²], which is the same as the variance formula scaled by a factor. Since variance is always non-negative, this again confirms the inequality.Another angle: Maybe using Lagrange multipliers or calculus to find minima, but that seems overkill for this problem. The algebraic methods are sufficient.Wait, could I also use induction? Probably not straightforward, since we have three variables. Induction is more for sequences or expressions with a variable number of terms. Here, it's fixed at three variables, so induction might not be the best approach.Alternatively, could I use symmetry? The inequality is symmetric in a, b, c, so perhaps assuming without loss of generality that a ≥ b ≥ c or something like that, but I don't see an immediate way to exploit that for the proof.I think the two methods I have are sufficient. The first method is more insightful because it shows exactly when equality holds, and it breaks down the expression into squares, which are fundamental in showing non-negativity. The second method is more of a cookbook approach, using known inequalities and adding them up, which is also valid but might be less revealing about why the inequality holds.In conclusion, both methods are correct, and they complement each other. The first method gives a deeper understanding by expressing the difference as a sum of squares, while the second method reinforces the result by leveraging a well-known inequality. Therefore, I can confidently say that the inequality a² + b² + c² ≥ ab + ac + bc holds for all real numbers a, b, and c, with equality if and only if a = b = c.

answer：Okay, so I have this problem where Sam invested Rs. 12,000 at a 10% annual interest rate for one year. The interest is compounded at a certain frequency, and at the end of the year, Sam received Rs. 13,230. I need to figure out how often the interest was compounded—like, is it compounded annually, semi-annually, quarterly, or maybe even monthly?Alright, first, I remember that compound interest means that the interest is calculated on the initial principal and also on the accumulated interest from previous periods. The formula for compound interest is:A = P (1 + r/n)^(nt)Where:- A is the amount of money accumulated after n years, including interest.- P is the principal amount (the initial amount of money).- r is the annual interest rate (decimal).- n is the number of times that interest is compounded per year.- t is the time the money is invested for in years.In this problem:- A = Rs. 13,230- P = Rs. 12,000- r = 10% per annum = 0.10- t = 1 yearSo, plugging these values into the formula:13,230 = 12,000 (1 + 0.10/n)^(n*1)First, I'll divide both sides by 12,000 to simplify:13,230 / 12,000 = (1 + 0.10/n)^nLet me calculate 13,230 divided by 12,000:13,230 ÷ 12,000 = 1.1025So, now the equation is:1.1025 = (1 + 0.10/n)^nNow, I need to find the value of n that satisfies this equation. Since n is both in the base and the exponent, it's a bit tricky to solve algebraically. Maybe I can try plugging in common values of n to see which one works.Let's start with n=1, which means compounded annually.(1 + 0.10/1)^1 = 1.10But 1.10 is less than 1.1025, so n=1 doesn't work.Next, let's try n=2, which means compounded semi-annually.(1 + 0.10/2)^2 = (1 + 0.05)^2 = 1.1025Hey, that's exactly what we have on the left side of the equation! So, n=2 works.Just to be thorough, let's check n=4, which is quarterly.(1 + 0.10/4)^4 = (1 + 0.025)^4 ≈ 1.1038That's a bit higher than 1.1025, so n=4 doesn't fit.What about n=12, which is monthly?(1 + 0.10/12)^12 ≈ 1.1047That's even higher. So, n=12 doesn't work either.Since n=2 gives us exactly 1.1025, which matches our equation, the compounding frequency must be semi-annually.Wait, just to make sure I didn't make any calculation errors, let me double-check the semi-annual calculation:(1 + 0.10/2)^2 = (1.05)^2 = 1.1025Yes, that's correct. So, n=2 is indeed the right answer.I think that's it. The interest is compounded semi-annually.

answer：First, I need to determine the total rainfall for the months of March to June by adding the given amounts.Next, I'll calculate the average rainfall for these four months by dividing the total rainfall by the number of months.Since the overall average rainfall for Virginia is 4 inches per month over five months, I'll find the total rainfall required for this average by multiplying the average by the number of months.Finally, I'll subtract the total rainfall from March to June from the total rainfall needed for five months to find the rainfall for July.

answer：Alright, so I have this problem here about averages and variances, and I need to figure out the average and standard deviation of a transformed dataset. Let me try to break it down step by step.First, the problem says that the average of a set of data ( x_1, x_2, ldots, x_n ) is 5. That means if I add up all these ( x ) values and divide by the number of data points ( n ), I get 5. So, mathematically, that's:[frac{x_1 + x_2 + ldots + x_n}{n} = 5]Which implies:[x_1 + x_2 + ldots + x_n = 5n]Okay, got that part. Now, the variance of this original dataset is given as 16. Variance is a measure of how spread out the data is. The formula for variance is:[sigma^2 = frac{1}{n} sum_{i=1}^{n} (x_i - mu)^2]Where ( mu ) is the mean, which is 5 in this case. So, plugging in the numbers:[16 = frac{1}{n} sum_{i=1}^{n} (x_i - 5)^2]Alright, now the problem asks about another set of data: ( 3x_1 + 2, 3x_2 + 2, ldots, 3x_n + 2 ). I need to find the average and standard deviation of this new dataset.Let me tackle the average first. If each data point is transformed by multiplying by 3 and then adding 2, how does that affect the average? Well, I remember that when you apply a linear transformation to each data point, the average also undergoes the same transformation. So, if the original average is 5, then the new average should be:[3 times 5 + 2 = 15 + 2 = 17]That seems straightforward. So, the new average is 17.Now, onto the variance and standard deviation. I recall that variance is affected by the scaling factor but not by the shift. Specifically, if you multiply each data point by a constant ( a ), the variance gets multiplied by ( a^2 ). Adding a constant ( b ) doesn't change the variance because it just shifts all the data points by the same amount, not affecting their spread.So, in this case, since we're multiplying each ( x_i ) by 3, the variance should be multiplied by ( 3^2 = 9 ). The original variance is 16, so the new variance is:[9 times 16 = 144]Therefore, the variance of the new dataset is 144. But the question asks for the standard deviation, which is just the square root of the variance. So, the standard deviation is:[sqrt{144} = 12]Putting it all together, the average of the new dataset is 17, and the standard deviation is 12. Looking back at the options provided:A: 15, 144 B: 17, 144 C: 17, 12 D: 7, 16Option C matches our calculations: average 17 and standard deviation 12. So, I think that's the correct answer.Wait, just to make sure I didn't make any mistakes. Let me double-check the variance part. The original variance is 16, multiplying each term by 3 increases the spread by a factor of 9, so 16 times 9 is indeed 144. Then, the standard deviation is the square root, which is 12. Yep, that seems right.And for the average, since we're scaling by 3 and shifting by 2, the average scales and shifts the same way. 3 times 5 is 15, plus 2 is 17. That checks out too.So, I feel confident that option C is the correct answer.**Final Answer**boxed{C}