💡First, I need to determine Tully's current age. Since Tully was 60 years old a year ago, Tully is now 61 years old.Kate is currently 29 years old.Let ( x ) represent the number of years from now until Tully is twice as old as Kate.In ( x ) years, Tully will be ( 61 + x ) years old, and Kate will be ( 29 + x ) years old.According to the problem, Tully's age in ( x ) years will be twice Kate's age at that time. This gives the equation:[61 + x = 2(29 + x)]Simplifying the equation:[61 + x = 58 + 2x]Subtracting ( 58 ) and ( x ) from both sides:[3 = x]Therefore, in 3 years, Tully will be twice as old as Kate.

💡Okay, so I need to figure out on which interval the function ( y = log(-x^2 + 2x) ) is monotonically increasing. Hmm, let's break this down step by step. First, I remember that for a function to be defined, especially when dealing with logarithms, the argument inside the log must be positive. So, I need to find where ( -x^2 + 2x > 0 ). Let me solve that inequality.Starting with:[ -x^2 + 2x > 0 ]I can factor this expression. Let me factor out a negative sign to make it easier:[ -(x^2 - 2x) > 0 ]Now, factor the quadratic inside the parentheses:[ -(x(x - 2)) > 0 ]So, the inequality becomes:[ -x(x - 2) > 0 ]To solve this, I can find the critical points where the expression equals zero. Setting each factor equal to zero:[ -x = 0 Rightarrow x = 0 ][ x - 2 = 0 Rightarrow x = 2 ]So, the critical points are at ( x = 0 ) and ( x = 2 ). These points divide the number line into intervals. I'll test each interval to see where the inequality holds.The intervals are:1. ( (-infty, 0) )2. ( (0, 2) )3. ( (2, infty) )Let's pick test points from each interval.For ( (-infty, 0) ), let's choose ( x = -1 ):[ -(-1)(-1 - 2) = -(-1)(-3) = -3 ]Which is negative, so the inequality ( -x(x - 2) > 0 ) is not satisfied here.For ( (0, 2) ), let's choose ( x = 1 ):[ -(1)(1 - 2) = -(1)(-1) = 1 ]Which is positive, so the inequality holds here.For ( (2, infty) ), let's choose ( x = 3 ):[ -(3)(3 - 2) = -(3)(1) = -3 ]Which is negative, so the inequality doesn't hold here.Therefore, the domain of the function ( y = log(-x^2 + 2x) ) is ( (0, 2) ).Now, to determine where the function is monotonically increasing, I need to look at its derivative. If the derivative is positive on an interval, the function is increasing there.Let me find the derivative ( y' ).Given:[ y = log(-x^2 + 2x) ]Using the chain rule, the derivative of ( log(u) ) with respect to ( x ) is ( frac{1}{u} cdot u' ), where ( u = -x^2 + 2x ).So, first, find ( u' ):[ u = -x^2 + 2x ][ u' = -2x + 2 ]Now, the derivative ( y' ) is:[ y' = frac{1}{-x^2 + 2x} cdot (-2x + 2) ]Simplify this expression:[ y' = frac{-2x + 2}{-x^2 + 2x} ]I can factor out a -2 from the numerator:[ y' = frac{-2(x - 1)}{-x^2 + 2x} ]Notice that the denominator ( -x^2 + 2x ) can be factored as ( -x(x - 2) ), which is the same as ( x(2 - x) ).So, rewriting the derivative:[ y' = frac{-2(x - 1)}{x(2 - x)} ]I can also factor out a negative from the denominator:[ y' = frac{-2(x - 1)}{-x(x - 2)} ]Which simplifies to:[ y' = frac{2(x - 1)}{x(x - 2)} ]Wait, let me check that again. When I factor out a negative from the denominator, it becomes:[ -x(x - 2) = x(2 - x) ]So, the denominator is ( x(2 - x) ), which is positive in the interval ( (0, 2) ) since both ( x ) and ( 2 - x ) are positive there.Now, looking at the numerator ( 2(x - 1) ), it changes sign at ( x = 1 ). So, let's analyze the sign of the derivative in the intervals ( (0, 1) ) and ( (1, 2) ).In ( (0, 1) ):- ( x - 1 ) is negative, so the numerator is negative.- The denominator ( x(2 - x) ) is positive.- So, overall, ( y' ) is negative divided by positive, which is negative.In ( (1, 2) ):- ( x - 1 ) is positive, so the numerator is positive.- The denominator ( x(2 - x) ) is still positive.- So, overall, ( y' ) is positive divided by positive, which is positive.Wait, that seems contradictory to my initial thought. Let me double-check.Wait, I think I made a mistake in simplifying the derivative. Let me go back.Original derivative:[ y' = frac{-2x + 2}{-x^2 + 2x} ]Let me factor numerator and denominator:Numerator: ( -2x + 2 = -2(x - 1) )Denominator: ( -x^2 + 2x = -x(x - 2) )So, the derivative is:[ y' = frac{-2(x - 1)}{-x(x - 2)} ]Simplify the negatives:[ y' = frac{-2(x - 1)}{-x(x - 2)} = frac{2(x - 1)}{x(x - 2)} ]Wait, that's correct. So, in the interval ( (0, 1) ), ( x - 1 ) is negative, so numerator is negative, denominator ( x(x - 2) ) is positive because ( x ) is positive and ( x - 2 ) is negative, so ( x(x - 2) ) is negative. Wait, hold on, denominator is ( x(x - 2) ). In ( (0, 2) ), ( x ) is positive, ( x - 2 ) is negative, so ( x(x - 2) ) is negative.Wait, so denominator is negative in ( (0, 2) ). So, let's correct that.So, denominator ( x(x - 2) ) is negative in ( (0, 2) ).So, in ( (0, 1) ):- Numerator ( 2(x - 1) ) is negative (since ( x - 1 < 0 ))- Denominator ( x(x - 2) ) is negative- So, negative divided by negative is positive.In ( (1, 2) ):- Numerator ( 2(x - 1) ) is positive (since ( x - 1 > 0 ))- Denominator ( x(x - 2) ) is negative- So, positive divided by negative is negative.Wait, that's different from what I thought earlier. So, actually, the derivative is positive in ( (0, 1) ) and negative in ( (1, 2) ).Therefore, the function is increasing on ( (0, 1) ) and decreasing on ( (1, 2) ).So, the interval where the function is monotonically increasing is ( (0, 1) ).Let me just confirm this by considering the behavior of the function.The function inside the log is a quadratic ( -x^2 + 2x ), which opens downward. Its vertex is at ( x = 1 ), which is the maximum point. So, the quadratic increases from ( x = 0 ) to ( x = 1 ), and then decreases from ( x = 1 ) to ( x = 2 ).Since the logarithm function is monotonically increasing, the composition ( log(-x^2 + 2x) ) will increase when the quadratic inside increases, and decrease when the quadratic inside decreases.Therefore, the function ( y = log(-x^2 + 2x) ) is increasing on ( (0, 1) ) and decreasing on ( (1, 2) ).So, the interval of monotonic increase is ( (0, 1) ).

💡First, I need to determine the total number of silver and bronze coins. The pirate has 500 silver coins and twice as many bronze coins, which means there are 1000 bronze coins.Next, I'll calculate how these coins are distributed across the 5 chests. Each chest will receive 100 silver coins and 200 bronze coins, totaling 300 coins per chest.Since each chest must contain 1000 coins in total, the remaining coins must be gold. This means each chest will have 700 gold coins.Finally, to find the total number of gold coins, I'll multiply the number of gold coins per chest by the total number of chests, resulting in 3500 gold coins.

💡Okay, so I need to figure out the smallest natural number ( n ) such that for every integer ( p geq n ), I can partition a square into ( p ) smaller squares. These smaller squares don't have to be equal in size, which is important. First, I should start by understanding what it means to partition a square into smaller squares. If ( p = 1 ), then the square itself is the only square, so that's trivial. For ( p = 2 ), is it possible? Hmm, if I try to divide a square into two smaller squares, I realize that it's not possible because if I split the square along a straight line, I'll end up with rectangles, not squares. So, ( p = 2 ) is impossible.What about ( p = 3 )? Let me visualize this. If I try to divide the square into three smaller squares, I might think of arranging them in some way, but I can't seem to find a configuration where all three are squares without overlapping or leaving gaps. So, ( p = 3 ) is also impossible.Moving on to ( p = 4 ). This is straightforward because I can simply divide the square into four equal smaller squares by making two cuts along the midpoints of the sides. So, ( p = 4 ) is possible.Now, ( p = 5 ). This is trickier. I know that ( p = 4 ) is possible, but adding one more square without breaking the rules is challenging. I remember hearing that it's impossible to partition a square into five smaller squares, but I'm not entirely sure. Let me think. If I try to divide one of the four equal squares into smaller squares, say into four even smaller squares, that would give me ( 4 + 3 = 7 ) squares, which skips ( p = 5 ). Alternatively, if I try to divide one square into two smaller squares, but that would result in rectangles again, not squares. So, it seems like ( p = 5 ) is indeed impossible.So far, I have that ( p = 1 ) is possible, ( p = 2 ) and ( p = 3 ) are impossible, ( p = 4 ) is possible, and ( p = 5 ) is impossible. What about ( p = 6 )? Let's see. If I take the four equal squares and then divide one of them into four smaller squares, that would give me ( 4 + 3 = 7 ) squares, which is more than 6. Alternatively, if I divide one of the four squares into two smaller squares, but that would result in rectangles, not squares. Wait, maybe there's another way. If I arrange the squares differently, perhaps not all starting from equal divisions. Maybe I can have a larger square and then smaller squares around it. Let me try to visualize this. Suppose I have a large square taking up a portion of the original square, and then fill the remaining space with smaller squares. This might work, but I need to ensure all are squares. I think this is possible, but I'm not entirely certain. Maybe I should look for a specific configuration.Upon further thought, I recall that it's possible to partition a square into six smaller squares by having one large square and five smaller squares arranged around it. This configuration ensures that all are squares without overlapping or leaving gaps. So, ( p = 6 ) is possible.Now, what about ( p = 7 )? Using a similar approach, I can take the six-square partition and further subdivide one of the smaller squares into four even smaller squares. This would increase the total number of squares by three, resulting in ( 6 + 3 = 9 ) squares, which skips ( p = 7 ). Alternatively, maybe there's a different way to partition into seven squares. Perhaps by adjusting the sizes of the squares in the six-square configuration. I think it's possible, but I'm not entirely sure. Maybe I should consider another method.Wait, another approach is to use induction. If I can partition a square into ( k ) squares, then I can partition it into ( k + 3 ) squares by subdividing one of the squares into four smaller squares. This is because subdividing one square into four adds three more squares. So, if I can get to ( p = 6 ), then I can get to ( p = 9 ), and so on. But this skips some numbers. However, if I can also find partitions for ( p = 7 ) and ( p = 8 ), then I can cover all numbers beyond a certain point.Let me try ( p = 7 ). Starting from ( p = 4 ), if I divide one square into four smaller squares, I get ( 4 + 3 = 7 ) squares. Wait, that works! So, ( p = 7 ) is possible by taking the four equal squares and subdividing one into four, resulting in seven squares. Similarly, for ( p = 8 ), I can take the four equal squares and divide two of them into four smaller squares each, resulting in ( 4 + 3 + 3 = 10 ) squares, which is more than 8. Alternatively, maybe I can find a different configuration. Perhaps by dividing one square into two smaller squares, but that would result in rectangles. Hmm, maybe I need to think differently. Alternatively, starting from ( p = 6 ), if I divide one of the squares into two smaller squares, but again, that would result in rectangles. Wait, no, if I divide a square into four smaller squares, I add three more squares. So, from ( p = 6 ), dividing one square into four would give me ( 6 + 3 = 9 ). But I need ( p = 8 ). Maybe there's another way. Perhaps by dividing one square into three smaller squares, but that would require some unequal sizes. I think it's possible, but I'm not entirely sure. Alternatively, maybe I can use a different initial partition.Wait, another idea. If I have a square divided into four equal squares, and then divide one of those into four smaller squares, I get seven squares. Then, if I divide another one of the original four into four smaller squares, I get ( 7 + 3 = 10 ) squares. But I need ( p = 8 ). Maybe I can divide one of the original four into two smaller squares, but that would result in rectangles. Hmm, this is tricky. Maybe I need to consider a different initial partition.Alternatively, perhaps I can start with a different configuration for ( p = 6 ). Instead of dividing one square into four, maybe I can have a different arrangement where I can more easily add squares. For example, having a large square and then arranging smaller squares around it in a way that allows for easier subdivision. I think this might work, but I'm not entirely sure. Maybe I should look for a specific example.Upon further research, I find that it's indeed possible to partition a square into eight smaller squares. One way is to have a 2x2 grid of squares, and then divide one of those into four smaller squares, resulting in a total of ( 4 + 3 = 7 ) squares, but that's not eight. Wait, maybe I need to divide two of the original four squares into four smaller squares each, resulting in ( 4 + 3 + 3 = 10 ) squares. Hmm, that's more than eight. Alternatively, maybe I can divide one square into two smaller squares, but that would result in rectangles. I'm stuck here.Wait, perhaps I'm overcomplicating this. Maybe I can use a different approach. If I can partition a square into six squares, and then further subdivide one of those into four, I get nine squares. But how do I get eight? Maybe by subdividing one square into three smaller squares instead of four. But is that possible? Dividing a square into three smaller squares without overlapping or leaving gaps. I think it is possible, but I'm not entirely sure. Let me try to visualize it.If I have a square and I want to divide it into three smaller squares, I can have one large square taking up a portion of the original square, and then two smaller squares filling the remaining space. This might work, but I need to ensure all are squares. I think this configuration is possible, but I'm not entirely certain. If it is, then from ( p = 6 ), subdividing one square into three would give me ( 6 + 2 = 8 ) squares. So, ( p = 8 ) is possible.Now, with ( p = 6 ), ( p = 7 ), and ( p = 8 ) all possible, I can use induction. If I can partition a square into ( k ) squares, then I can partition it into ( k + 3 ) squares by subdividing one square into four smaller squares. This means that once I have partitions for ( p = 6 ), ( p = 7 ), and ( p = 8 ), I can cover all numbers beyond that by adding three each time. Therefore, the smallest ( n ) should be 6, as every number ( p geq 6 ) can be achieved.But wait, let me double-check. If ( p = 6 ) is possible, then ( p = 9 ) is possible by adding three. Similarly, ( p = 7 ) leads to ( p = 10 ), and ( p = 8 ) leads to ( p = 11 ). So, starting from 6, 7, and 8, we can cover all numbers beyond by adding three each time. Therefore, the smallest ( n ) is indeed 6.However, I should also confirm that there are no numbers between 6 and the next possible partitions that are skipped. For example, after 6, we have 7, 8, then 9, 10, 11, etc. So, no numbers are skipped beyond 6. Therefore, ( n = 6 ) is the smallest natural number with the desired property.

💡To compare the sizes of (-9) and (-sqrt{80}), I start by examining their absolute values. First, I calculate the square of each absolute value:[9^2 = 81 quad text{and} quad (sqrt{80})^2 = 80]Since (81 > 80), it means that (9) is greater than (sqrt{80}). However, because both numbers are negative, the inequality reverses when considering their negative counterparts. Therefore, (-9) is less than (-sqrt{80}).

💡Okay, so I need to figure out if the construction company "Roga and Kopyta" will be able to complete building a 100 km road from Arbatova to Chernomorsk. The plan is that in the first month, they build 1 km, and then each subsequent month, they build an additional 1 divided by A to the power of 10 km, where A is the total length completed by the beginning of that month.Alright, let's break this down. In the first month, they build 1 km. So, by the start of the second month, they've completed 1 km. Then, in the second month, they build 1 divided by 1^10, which is just 1 km again. So, by the start of the third month, they've completed 2 km.Wait a minute, so in the third month, they build 1 divided by 2^10 km. Let me calculate that. 2^10 is 1024, so 1/1024 is approximately 0.0009766 km. That's a very small amount. So, in the third month, they only add about 0.0009766 km to the road.Hmm, so the amount they build each month is decreasing rapidly because it's 1 divided by A^10, and A is increasing each month. That means each subsequent month, the additional length they build becomes smaller and smaller.Let me see if I can model this mathematically. Let's denote L_n as the total length built by the end of the nth month. Then, according to the problem, L_1 = 1 km. For each subsequent month, L_n = L_{n-1} + 1/(L_{n-1})^10.So, this is a recursive sequence where each term depends on the previous term. I need to determine if this sequence will ever reach or exceed 100 km.This seems like a problem that could be approached using series and sequences. Maybe I can think of it as an infinite series where each term is 1/(L_{n-1})^10. But since L_{n-1} is increasing each time, the terms of the series are decreasing.Wait, but how does L_n behave as n increases? Does it converge to a finite limit, or does it diverge to infinity? If it converges, then the total length built will be finite, and they won't complete the 100 km. If it diverges, then eventually, L_n will exceed 100 km.To analyze this, I can consider the behavior of the sequence L_n. Since each term is L_{n-1} + 1/(L_{n-1})^10, and L_{n-1} is positive and increasing, the sequence is monotonically increasing. The question is whether it is bounded above or not.If the sequence is bounded above, it will converge to some finite limit L. If it's unbounded, it will diverge to infinity.Assume that the sequence converges to some limit L. Then, taking the limit on both sides of the recursive formula:L = L + 1/L^10.Subtracting L from both sides gives 0 = 1/L^10, which implies that L must be infinite. But that's a contradiction because we assumed L is finite. Therefore, the sequence cannot converge to a finite limit and must diverge to infinity.Wait, does that mean that L_n will eventually exceed any finite bound, including 100 km? So, the company will complete the road, but it will take an extremely long time.But how long exactly? The problem doesn't specify a time limit, just asks if they will complete it. Since the sequence diverges, they will eventually reach 100 km, but it might take an unimaginably large number of months.Let me try to estimate how many months it would take. Starting from L_1 = 1 km, each subsequent month adds 1/(L_{n-1})^10 km.In the first few months, the additions are significant:- Month 1: 1 km- Month 2: 1 km (total 2 km)- Month 3: ~0.0009766 km (total ~2.0009766 km)- Month 4: 1/(2.0009766)^10 ≈ 1/1024 ≈ 0.0009766 km (total ~2.0019532 km)Wait, actually, after the second month, the additions become extremely small. It seems like after the second month, each additional km added is negligible. So, does that mean that the total length will never reach 100 km?But earlier, I thought that the sequence diverges to infinity because the recursive formula suggests that L_n increases without bound. However, the increments after the second month are so minuscule that it might take an impractically long time, but theoretically, it should still reach 100 km.Let me think about it differently. The total length built after n months is the sum from k=1 to n of 1/(L_{k-1})^10, with L_0 = 0 and L_1 = 1.But this is a recursive series, so it's not straightforward to sum it up. Maybe I can approximate it using integrals or some other method.Alternatively, consider that after the first two months, the total length is 2 km. From then on, each month adds 1/(A)^10 km, where A is at least 2 km.So, the amount added each month is at most 1/2^10 = 1/1024 km ≈ 0.0009766 km.To reach 100 km, they need to add 98 more km after the first two months.If each month adds at least 1/100^10 km, which is a very small number, then the number of months required would be on the order of 98 / (1/100^10) = 98 * 100^10 months.But 100^10 is 10^20, so 98 * 10^20 months is an astronomically large number, way beyond any practical time frame.However, theoretically, since each month adds a positive amount, no matter how small, the total length will eventually reach 100 km. It just takes an unimaginably long time.So, in conclusion, yes, the company will complete the road, but it will take an extremely long time, far beyond any realistic expectation.