💡Okay, so I have this problem about a function ( f(x) ) that satisfies the property ( f(x_1 cdot x_2) = f(x_1) + f(x_2) ) for any nonzero real numbers ( x_1 ) and ( x_2 ). There are three parts to this problem, and I need to tackle each one step by step.Starting with part (1): Find the value of ( f(-1) ).Hmm, the function has this additive property when multiplying inputs. I remember that functions with such properties are often logarithmic or something similar, but since it's defined for all nonzero real numbers, it might be a logarithm-like function. But let me not get ahead of myself.First, maybe I can find ( f(1) ). Let me set ( x_1 = x_2 = 1 ). Then, according to the given property:( f(1 cdot 1) = f(1) + f(1) )( f(1) = 2f(1) )Subtracting ( f(1) ) from both sides:( 0 = f(1) )So, ( f(1) = 0 ). That's a useful piece of information.Now, to find ( f(-1) ), maybe I can set ( x_1 = x_2 = -1 ). Let's try that:( f((-1) cdot (-1)) = f(-1) + f(-1) )( f(1) = 2f(-1) )But we already know that ( f(1) = 0 ), so:( 0 = 2f(-1) )( f(-1) = 0 )Alright, so ( f(-1) = 0 ). That seems straightforward.Moving on to part (2): Prove that ( f(x) ) is an even function.An even function satisfies ( f(-x) = f(x) ) for all ( x ). Let me see how I can use the given property to show this.Let me set ( x_1 = x ) and ( x_2 = -1 ). Then:( f(x cdot (-1)) = f(x) + f(-1) )( f(-x) = f(x) + f(-1) )But from part (1), we know that ( f(-1) = 0 ). So:( f(-x) = f(x) + 0 )( f(-x) = f(x) )Therefore, ( f(x) ) is indeed an even function. That wasn't too bad.Now, part (3): Given that ( f(x) ) is increasing on the interval ( (0, +infty) ) and ( f(2x - 1) < f(x) ), find the range of values for ( x ).Okay, so ( f ) is increasing on ( (0, +infty) ). Since ( f ) is even, it must be decreasing on ( (-infty, 0) ) because it mirrors the behavior on the positive side. But let me focus on the inequality ( f(2x - 1) < f(x) ).Since ( f ) is increasing on ( (0, +infty) ), if both ( 2x - 1 ) and ( x ) are positive, then the inequality ( f(2x - 1) < f(x) ) would imply that ( 2x - 1 < x ). Let me write that down:If ( 2x - 1 > 0 ) and ( x > 0 ), then ( 2x - 1 < x ).Solving ( 2x - 1 < x ):Subtract ( x ) from both sides:( x - 1 < 0 )( x < 1 )But we also have the conditions that ( 2x - 1 > 0 ) and ( x > 0 ). Let's solve those:( 2x - 1 > 0 )( 2x > 1 )( x > frac{1}{2} )And ( x > 0 ) is already satisfied if ( x > frac{1}{2} ).So, combining these, if ( frac{1}{2} < x < 1 ), then ( f(2x - 1) < f(x) ).But wait, what if ( 2x - 1 ) is negative? Since ( f ) is even, ( f(2x - 1) = f(|2x - 1|) ). So, if ( 2x - 1 ) is negative, then ( |2x - 1| = -(2x - 1) = 1 - 2x ). So, in that case, ( f(2x - 1) = f(1 - 2x) ).But since ( f ) is increasing on ( (0, +infty) ), ( f(1 - 2x) < f(x) ) would imply ( 1 - 2x < x ) because both ( 1 - 2x ) and ( x ) would be positive in this case.Let me check when ( 2x - 1 < 0 ):( 2x - 1 < 0 )( 2x < 1 )( x < frac{1}{2} )So, if ( x < frac{1}{2} ), then ( f(2x - 1) = f(1 - 2x) ), and since ( f ) is increasing on ( (0, +infty) ), ( f(1 - 2x) < f(x) ) implies:( 1 - 2x < x )( 1 < 3x )( x > frac{1}{3} )So, combining this with ( x < frac{1}{2} ), we get ( frac{1}{3} < x < frac{1}{2} ).Therefore, combining both cases:- When ( frac{1}{2} < x < 1 ), the inequality holds.- When ( frac{1}{3} < x < frac{1}{2} ), the inequality also holds.So, overall, the solution is ( frac{1}{3} < x < 1 ).But wait, I should also consider the domain of ( f ). The function is defined for nonzero real numbers, so ( 2x - 1 ) and ( x ) must be nonzero.So, ( 2x - 1 neq 0 ) implies ( x neq frac{1}{2} ).And ( x neq 0 ), but since ( x > frac{1}{3} ), we don't have to worry about ( x = 0 ).So, the range of ( x ) is ( frac{1}{3} < x < 1 ), excluding ( x = frac{1}{2} ). But wait, in our earlier analysis, when ( x = frac{1}{2} ), ( 2x - 1 = 0 ), which is not allowed since ( f ) is defined only for nonzero real numbers. Therefore, ( x = frac{1}{2} ) is excluded.But in our solution intervals, ( frac{1}{3} < x < 1 ) already excludes ( x = frac{1}{2} ) because ( frac{1}{2} ) is between ( frac{1}{3} ) and 1. Wait, no, ( frac{1}{2} ) is within ( frac{1}{3} ) and 1, so actually, ( x = frac{1}{2} ) is in the interval ( frac{1}{3} < x < 1 ), but it's excluded because ( 2x - 1 = 0 ) is not allowed.Therefore, the solution is ( frac{1}{3} < x < 1 ), but ( x neq frac{1}{2} ). However, since ( x = frac{1}{2} ) is just a single point, it doesn't affect the interval, so we can write the solution as ( frac{1}{3} < x < 1 ).Wait, but actually, when ( x = frac{1}{2} ), ( 2x - 1 = 0 ), which is undefined for ( f ). Therefore, ( x = frac{1}{2} ) is not in the domain of the inequality ( f(2x - 1) < f(x) ). So, the solution is ( frac{1}{3} < x < 1 ), excluding ( x = frac{1}{2} ).But since ( x = frac{1}{2} ) is just a single point, it's often acceptable to write the interval as ( frac{1}{3} < x < 1 ) with the understanding that ( x neq frac{1}{2} ). However, to be precise, we should exclude it.But let me double-check. If ( x = frac{1}{2} ), then ( f(2x - 1) = f(0) ), but ( f ) is not defined at 0. Therefore, ( x = frac{1}{2} ) is not allowed. So, the solution is ( frac{1}{3} < x < 1 ) with ( x neq frac{1}{2} ).But in the context of inequalities, when we write ( frac{1}{3} < x < 1 ), it's understood that ( x ) is in that interval, and since ( x = frac{1}{2} ) is within that interval but excluded, we might need to write it as two separate intervals: ( frac{1}{3} < x < frac{1}{2} ) and ( frac{1}{2} < x < 1 ).Yes, that's more accurate. So, the solution is ( x ) in ( left( frac{1}{3}, frac{1}{2} right) cup left( frac{1}{2}, 1 right) ).But let me verify this. When ( x ) is between ( frac{1}{3} ) and ( frac{1}{2} ), ( 2x - 1 ) is negative, so ( f(2x - 1) = f(1 - 2x) ). Since ( f ) is increasing on ( (0, +infty) ), ( f(1 - 2x) < f(x) ) implies ( 1 - 2x < x ), which gives ( x > frac{1}{3} ). So, in this case, ( frac{1}{3} < x < frac{1}{2} ).When ( x ) is between ( frac{1}{2} ) and 1, ( 2x - 1 ) is positive, so ( f(2x - 1) < f(x) ) implies ( 2x - 1 < x ), which gives ( x < 1 ). So, in this case, ( frac{1}{2} < x < 1 ).Therefore, combining both intervals, the solution is ( frac{1}{3} < x < 1 ), excluding ( x = frac{1}{2} ). So, the range of ( x ) is ( left( frac{1}{3}, frac{1}{2} right) cup left( frac{1}{2}, 1 right) ).But in the original problem, it just says "find the range of values for ( x )", so I think writing it as ( frac{1}{3} < x < 1 ) is acceptable, with the understanding that ( x neq frac{1}{2} ). However, to be precise, it's better to write it as two separate intervals.Wait, but actually, when ( x = frac{1}{2} ), the expression ( f(2x - 1) ) is undefined, so ( x = frac{1}{2} ) is not part of the domain. Therefore, the solution is indeed ( frac{1}{3} < x < 1 ), excluding ( x = frac{1}{2} ).But in terms of intervals, it's ( left( frac{1}{3}, frac{1}{2} right) cup left( frac{1}{2}, 1 right) ).Alternatively, since ( x = frac{1}{2} ) is just a single point, sometimes it's written as ( frac{1}{3} < x < 1 ), ( x neq frac{1}{2} ).I think either way is acceptable, but to be thorough, I'll present it as two intervals.So, summarizing:1. ( f(-1) = 0 )2. ( f(x) ) is even because ( f(-x) = f(x) )3. The range of ( x ) is ( left( frac{1}{3}, frac{1}{2} right) cup left( frac{1}{2}, 1 right) )I think that covers all parts of the problem.

💡Okay, so I have this problem where there are 10 white balls and 15 black balls in a box, making a total of 25 balls. I need to find the probability that if I draw four balls from the box, all of them are white. Alright, let's break this down. Probability problems like this usually involve combinations because the order in which we draw the balls doesn't matter. So, I think I need to figure out how many ways there are to draw four white balls out of the ten available, and then divide that by the total number of ways to draw any four balls from the entire set of 25 balls.First, let's recall the combination formula. The number of ways to choose k items from a set of n items is given by the combination formula:[binom{n}{k} = frac{n!}{k!(n - k)!}]Where "!" denotes factorial, which is the product of all positive integers up to that number. So, for example, 5! = 5 × 4 × 3 × 2 × 1 = 120.Okay, so applying this to our problem:1. **Total number of ways to draw 4 balls from 25:**[binom{25}{4} = frac{25!}{4!(25 - 4)!} = frac{25!}{4! cdot 21!}]But calculating factorials for such large numbers seems complicated. Maybe there's a simpler way to compute this without calculating the full factorials. Let's see:[binom{25}{4} = frac{25 times 24 times 23 times 22}{4 times 3 times 2 times 1} = frac{25 times 24 times 23 times 22}{24} = 25 times 23 times 22 / 1]Wait, that doesn't seem right. Let me recalculate:Actually, the denominator is 4! which is 24, so:[binom{25}{4} = frac{25 times 24 times 23 times 22}{4 times 3 times 2 times 1} = frac{25 times 24 times 23 times 22}{24}]Here, the 24 in the numerator and denominator cancels out:[= 25 times 23 times 22]Now, let's compute that:25 × 23 = 575575 × 22 = Let's compute 575 × 20 = 11,500 and 575 × 2 = 1,150. So total is 11,500 + 1,150 = 12,650.So, (binom{25}{4} = 12,650).2. **Number of ways to draw 4 white balls from 10:**Similarly, using the combination formula:[binom{10}{4} = frac{10!}{4!(10 - 4)!} = frac{10!}{4! cdot 6!}]Again, simplifying without calculating full factorials:[binom{10}{4} = frac{10 times 9 times 8 times 7}{4 times 3 times 2 times 1} = frac{10 times 9 times 8 times 7}{24}]Let's compute the numerator first:10 × 9 = 9090 × 8 = 720720 × 7 = 5,040Now, divide by 24:5,040 ÷ 24 = 210So, (binom{10}{4} = 210).3. **Calculating the probability:**Now, the probability that all four balls drawn are white is the number of favorable outcomes divided by the total number of possible outcomes:[P(text{all 4 white}) = frac{binom{10}{4}}{binom{25}{4}} = frac{210}{12,650}]Simplifying this fraction:Divide numerator and denominator by 10:[frac{21}{1,265}]Can this be simplified further? Let's check if 21 and 1,265 have any common factors.21 factors: 1, 3, 7, 211,265 ÷ 5 = 253, so 5 is a factor. 253 is a prime number (since it's not divisible by 2, 3, 5, 7, 11, 13, etc.). So, 1,265 = 5 × 253.21 and 1,265 have no common factors other than 1, so the fraction is already in its simplest form.Therefore, the probability is:[frac{21}{1,265}]But just to make sure, let's double-check the calculations:- Total ways: 25 choose 4 = 12,650 (correct)- White ways: 10 choose 4 = 210 (correct)- Probability: 210 / 12,650 = 21 / 1,265 (correct)Yes, that seems right.Alternatively, if I want to express this as a decimal or percentage:[frac{21}{1,265} approx 0.0166]So, approximately 1.66%.That seems low, but considering there are more black balls than white balls, it makes sense that the probability of drawing all white balls is relatively low.I think that's it. The probability is 21/1,265, which simplifies to 3/181 if we divide numerator and denominator by 7:21 ÷ 7 = 31,265 ÷ 7 = 181So, (frac{3}{181}) is the simplified form.Yes, that's correct.

💡First, I need to determine the individual rates of pipes a, b, and c. Let’s assume the rate at which pipe a fills the tank is ( a ) tanks per hour. Since pipe b is twice as fast as pipe a, its rate is ( 2a ). Similarly, pipe c is twice as fast as pipe b, so its rate is ( 4a ).Next, I'll calculate the combined rate of all three pipes working together. Adding their individual rates gives:[a + 2a + 4a = 7a text{ tanks per hour}]Given that together they can fill the tank in 6 hours, I can set up the equation:[7a times 6 = 1 text{ tank}][42a = 1][a = frac{1}{42} text{ tanks per hour}]Finally, to find out how long it would take for pipe a to fill the tank alone, I take the reciprocal of its rate:[text{Time taken by pipe a alone} = frac{1}{a} = frac{1}{frac{1}{42}} = 42 text{ hours}]

💡Okay, so I need to figure out how many more days of good air quality Beijing needs in the second half of 2014 compared to the same period in 2013, to achieve a 20% increase in the total number of good air quality days for the entire year. Let me break this down step by step.First, I know that in 2013, Beijing had 47.9% of its days with good air quality. Since there are 365 days in a year, I can calculate the number of good air quality days in 2013 by multiplying 365 by 47.9%. Let me do that:365 * 0.479 = approximately 174.835 days.Since we can't have a fraction of a day, I'll round this to 175 days.Now, the target for 2014 is a 20% increase in the number of good air quality days compared to 2013. So, I need to calculate 20% of 175 days and add that to 175 to find the target for 2014.20% of 175 is 0.2 * 175 = 35 days.So, the target for 2014 is 175 + 35 = 210 days of good air quality.Next, I know that in the first half of 2014, there was a 30% reduction in emissions, which led to an increase of 20 days of good air quality compared to the same period in 2013. I need to figure out how many good air quality days there were in the first half of 2013 and then add 20 days to that to get the number for the first half of 2014.Assuming that the number of good air quality days is evenly distributed throughout the year, I can estimate that in 2013, the first half had about half of 175 days, which is approximately 87.5 days. Since we can't have half days, I'll round this to 88 days.So, in the first half of 2014, there were 88 + 20 = 108 days of good air quality.Now, to find out how many good air quality days are needed in the second half of 2014 to reach the target of 210 days, I subtract the number of good days already achieved in the first half from the target.210 - 108 = 102 days.So, Beijing needs 102 days of good air quality in the second half of 2014.But I need to compare this to the number of good air quality days in the second half of 2013. Again, assuming an even distribution, the second half of 2013 would have had about the same number of good days as the first half, which is approximately 87.5 days, rounded to 88 days.Therefore, the number of additional good air quality days needed in the second half of 2014 compared to the same period in 2013 is:102 - 88 = 14 days.Wait, but in my initial calculation, I had 102 - 87 = 15 days. There's a slight discrepancy here because I rounded differently. To be precise, let's use the exact numbers without rounding.In 2013, the number of good air quality days was 174.835, so half of that is approximately 87.4175 days for each half.In 2014, the first half had 87.4175 + 20 = 107.4175 days.To reach the target of 210 days, the second half needs 210 - 107.4175 = 102.5825 days.Comparing this to the second half of 2013, which had 87.4175 days, the additional days needed are:102.5825 - 87.4175 = 15.165 days.Rounding this to the nearest whole number, it's approximately 15 days.So, Beijing needs 15 more days of good air quality in the second half of 2014 compared to the same period in 2013 to achieve the target of a 20% increase in the number of good air quality days for the entire year.

💡Okay, so I need to figure out how many boxes of Trefoils Jenny sold. The problem says she sold some boxes and needs to deliver 3 cases of 8 boxes each, plus extra boxes. Hmm, okay, so first, I think I need to understand what a "case" means here. I guess a case is like a container that holds multiple boxes, right? So, if each case has 8 boxes, and she has 3 cases, then the total number of boxes in the cases would be 3 times 8.Let me write that down: 3 cases * 8 boxes per case = 24 boxes. So, she has 24 boxes in the cases. But the problem also mentions extra boxes. It says she needs to deliver 3 cases plus extra boxes. So, she has more boxes beyond the 24 in the cases. But wait, the problem doesn't specify how many extra boxes there are. It just says "plus extra boxes." That's a bit confusing because without knowing the number of extra boxes, I can't find the exact total number of boxes she sold.Maybe I'm missing something. Let me read the problem again: "Jenny sold some boxes of Trefoils. She needs to deliver 3 cases of 8 boxes, plus extra boxes. How many boxes of Trefoils did Jenny sell?" It still doesn't mention how many extra boxes there are. So, I think the problem is incomplete because it doesn't provide enough information to find the exact number of boxes she sold.Alternatively, maybe the problem is asking for the number of boxes in the cases, and the extra boxes are just additional information that isn't needed to solve the main question. If that's the case, then the answer would just be the number of boxes in the 3 cases, which is 24 boxes. But I'm not entirely sure because the problem does mention extra boxes, which suggests that they might be part of the total number she sold.Wait, maybe the extra boxes are included in the "some boxes" that she sold. So, the total number of boxes she sold would be the boxes in the cases plus the extra boxes. But since we don't know how many extra boxes there are, we can't calculate the exact total. Therefore, the problem might be incomplete or there might be a mistake in the way it's presented.Alternatively, perhaps the problem expects me to assume that the extra boxes are zero, meaning she only sold the boxes in the cases. In that case, the total number of boxes she sold would be 24. But that seems like making an assumption that isn't stated in the problem, which isn't ideal.Maybe I should look for more clues or see if there's any additional information I can infer from the problem. It says she "sold some boxes" and "needs to deliver" 3 cases plus extra boxes. So, perhaps the boxes she sold are the ones she needs to deliver, meaning the total number of boxes she sold is the sum of the boxes in the cases and the extra boxes. But again, without knowing the number of extra boxes, I can't find the exact total.Is there a way to express the total number of boxes she sold in terms of the extra boxes? For example, if I let E represent the number of extra boxes, then the total number of boxes she sold would be 24 + E. But the problem asks for a specific number, not an expression. So, unless there's more information, I can't determine the exact number.Maybe the problem is testing my understanding that without the number of extra boxes, the total can't be determined. In that case, the answer would be that there's not enough information to find the exact number of boxes Jenny sold.Alternatively, perhaps the problem is trying to trick me into thinking that the extra boxes are separate from the cases, but actually, they might be included in the cases. But that doesn't make much sense because it specifically mentions 3 cases and then extra boxes.Wait, maybe the extra boxes are part of the cases. So, each case has 8 boxes, and she has 3 cases, which is 24 boxes, and then she has some extra boxes on top of that. But again, without knowing how many extra boxes, I can't find the total.I'm going in circles here. Let me try to summarize:- Each case has 8 boxes.- She has 3 cases, so that's 3 * 8 = 24 boxes.- She also has extra boxes, but the number isn't specified.- Therefore, the total number of boxes she sold is 24 plus the number of extra boxes.- Since the number of extra boxes isn't given, I can't find the exact total.So, the problem doesn't provide enough information to determine the exact number of boxes Jenny sold. Maybe the problem expects me to recognize that and state that the information is insufficient. Alternatively, if I assume that the extra boxes are zero, then the total is 24 boxes. But that's making an assumption that isn't supported by the problem.Perhaps the problem is part of a larger context where the number of extra boxes was mentioned earlier, but in this isolated problem, it's missing. Without that information, I can't proceed further.In conclusion, based on the information provided, I can determine that Jenny has 24 boxes in the cases, but without knowing the number of extra boxes, I can't find the exact total number of boxes she sold. Therefore, the problem is incomplete or there's missing information needed to solve it accurately.

💡First, I need to find the first number, which I'll call "y".The average of y, 76, and x is given as 5. Since there are three numbers, I can set up the equation:(y + 76 + x) / 3 = 5I know that x is -63, so I'll substitute that value into the equation:(y + 76 - 63) / 3 = 5Next, I'll simplify the equation by combining like terms:(y + 13) / 3 = 5To isolate y, I'll multiply both sides of the equation by 3:y + 13 = 15Finally, I'll subtract 13 from both sides to solve for y:y = 2Therefore, the first number is 2.