💡First, calculate the runs scored by running between the wickets by finding 56.52% of the total score of 138 runs.Next, subtract the runs from boundaries and the runs from running between the wickets from the total score to determine the runs scored from sixes.Finally, divide the runs from sixes by 6 to find the number of sixes hit.

💡Okay, so I have this system of linear equations in two variables, x and y. The equations are:1. ( x - y = 3 - a )2. ( x + 2y = 5a )And there are two parts to this problem. Let me tackle them one by one.**Part (1): Finding the value of ( a ) such that the solutions always satisfy ( y = a + 1 ).**Alright, so I need to find the value of ( a ) such that whenever I solve the system, the y-value is always ( a + 1 ). Hmm, okay. Let me try to solve the system first and see what conditions I get.Starting with the two equations:1. ( x - y = 3 - a )2. ( x + 2y = 5a )I can use the method of elimination or substitution. Maybe elimination is easier here. Let me subtract equation 1 from equation 2 to eliminate x.Subtracting equation 1 from equation 2:( (x + 2y) - (x - y) = 5a - (3 - a) )Simplify the left side:( x + 2y - x + y = 3y )Simplify the right side:( 5a - 3 + a = 6a - 3 )So, we have:( 3y = 6a - 3 )Divide both sides by 3:( y = 2a - 1 )Okay, so from the system, we get that ( y = 2a - 1 ). But the problem says that the solutions always satisfy ( y = a + 1 ). So, setting these equal:( 2a - 1 = a + 1 )Subtract ( a ) from both sides:( a - 1 = 1 )Add 1 to both sides:( a = 2 )So, the value of ( a ) is 2. That seems straightforward.**Part (2): The solutions of the system are also solutions of the equation ( bx + 3y = 1 ).**This part has two subparts:① Investigate the relationship between the real numbers ( a ) and ( b ).② If both ( a ) and ( b ) are integers, find the maximum and minimum values of ( b ).Let's start with part ①.First, from the system, we already found that ( y = 2a - 1 ). Let me find x as well.From equation 1:( x - y = 3 - a )We know ( y = 2a - 1 ), so substitute:( x - (2a - 1) = 3 - a )Simplify:( x - 2a + 1 = 3 - a )Bring constants to one side and variables to the other:( x = 3 - a + 2a - 1 )Simplify:( x = a + 2 )So, the solutions are ( x = a + 2 ) and ( y = 2a - 1 ).Now, these solutions must satisfy the equation ( bx + 3y = 1 ). Let's substitute x and y into this equation.Substitute ( x = a + 2 ) and ( y = 2a - 1 ):( b(a + 2) + 3(2a - 1) = 1 )Let me expand this:( ab + 2b + 6a - 3 = 1 )Combine like terms:( ab + 6a + 2b - 3 = 1 )Bring constants to the right:( ab + 6a + 2b = 4 )Hmm, so we have an equation involving ( a ) and ( b ). Let me try to express ( b ) in terms of ( a ) or vice versa.Let me collect terms with ( b ):( ab + 2b = 4 - 6a )Factor out ( b ):( b(a + 2) = 4 - 6a )So, solving for ( b ):( b = frac{4 - 6a}{a + 2} )Hmm, that's an expression for ( b ) in terms of ( a ). Let me see if I can simplify this further.Let me perform polynomial division or see if I can factor numerator and denominator.Numerator: ( 4 - 6a = -6a + 4 )Denominator: ( a + 2 )Let me factor out a -2 from the numerator:( -6a + 4 = -2(3a - 2) )But denominator is ( a + 2 ), which doesn't factor into ( 3a - 2 ). Maybe I can perform division:Divide ( -6a + 4 ) by ( a + 2 ).Let me write it as:( frac{-6a + 4}{a + 2} )Let me factor out -6 from the numerator:( frac{-6(a) + 4}{a + 2} )Alternatively, let me perform the division:How many times does ( a + 2 ) go into ( -6a + 4 )?First term: ( -6a / a = -6 ). Multiply ( -6 ) by ( a + 2 ):( -6a - 12 )Subtract this from ( -6a + 4 ):( (-6a + 4) - (-6a - 12) = 0a + 16 )So, the division gives:( -6 + frac{16}{a + 2} )Therefore,( b = -6 + frac{16}{a + 2} )Alternatively,( b = frac{16}{a + 2} - 6 )So, that's another way to write ( b ) in terms of ( a ).So, the relationship between ( a ) and ( b ) is ( b = frac{16}{a + 2} - 6 ). Alternatively, ( b = frac{4 - 6a}{a + 2} ).So, that's the relationship. It's a hyperbola in the ( a )-( b ) plane.Now, moving on to part ②: If both ( a ) and ( b ) are integers, find the maximum and minimum values of ( b ).Given that ( a ) and ( b ) are integers, and ( b neq 0 ) and ( b neq -6 ). Wait, actually, the problem says ( b neq 0 ) and ( b neq -6 ). Hmm, so ( b ) cannot be 0 or -6.But in our expression for ( b ), ( b = frac{16}{a + 2} - 6 ), so ( a + 2 ) must be a divisor of 16 because ( b ) is an integer.So, ( a + 2 ) must be an integer divisor of 16. The divisors of 16 are: ( pm1, pm2, pm4, pm8, pm16 ).Therefore, ( a + 2 in {1, -1, 2, -2, 4, -4, 8, -8, 16, -16} ).So, ( a + 2 ) can be each of these, which gives corresponding ( a ) and ( b ).Let me list them:1. If ( a + 2 = 1 ), then ( a = -1 ). Then, ( b = frac{16}{1} - 6 = 16 - 6 = 10 ).2. If ( a + 2 = -1 ), then ( a = -3 ). Then, ( b = frac{16}{-1} - 6 = -16 - 6 = -22 ).3. If ( a + 2 = 2 ), then ( a = 0 ). Then, ( b = frac{16}{2} - 6 = 8 - 6 = 2 ).4. If ( a + 2 = -2 ), then ( a = -4 ). Then, ( b = frac{16}{-2} - 6 = -8 - 6 = -14 ).5. If ( a + 2 = 4 ), then ( a = 2 ). Then, ( b = frac{16}{4} - 6 = 4 - 6 = -2 ).6. If ( a + 2 = -4 ), then ( a = -6 ). Then, ( b = frac{16}{-4} - 6 = -4 - 6 = -10 ).7. If ( a + 2 = 8 ), then ( a = 6 ). Then, ( b = frac{16}{8} - 6 = 2 - 6 = -4 ).8. If ( a + 2 = -8 ), then ( a = -10 ). Then, ( b = frac{16}{-8} - 6 = -2 - 6 = -8 ).9. If ( a + 2 = 16 ), then ( a = 14 ). Then, ( b = frac{16}{16} - 6 = 1 - 6 = -5 ).10. If ( a + 2 = -16 ), then ( a = -18 ). Then, ( b = frac{16}{-16} - 6 = -1 - 6 = -7 ).So, compiling the values of ( b ):10, -22, 2, -14, -2, -10, -4, -8, -5, -7.Now, we need to find the maximum and minimum values of ( b ).Looking at the list:Maximum ( b ) is 10.Minimum ( b ) is -22.But wait, the problem says ( b neq 0 ) and ( b neq -6 ). Looking at our list, none of the ( b ) values are 0 or -6, so we're okay.Therefore, the maximum value of ( b ) is 10, and the minimum value is -22.Wait, let me double-check my calculations to make sure I didn't make any mistakes.For ( a + 2 = 1 ), ( a = -1 ), ( b = 16 - 6 = 10 ). Correct.For ( a + 2 = -1 ), ( a = -3 ), ( b = -16 - 6 = -22 ). Correct.For ( a + 2 = 2 ), ( a = 0 ), ( b = 8 - 6 = 2 ). Correct.For ( a + 2 = -2 ), ( a = -4 ), ( b = -8 - 6 = -14 ). Correct.For ( a + 2 = 4 ), ( a = 2 ), ( b = 4 - 6 = -2 ). Correct.For ( a + 2 = -4 ), ( a = -6 ), ( b = -4 - 6 = -10 ). Correct.For ( a + 2 = 8 ), ( a = 6 ), ( b = 2 - 6 = -4 ). Correct.For ( a + 2 = -8 ), ( a = -10 ), ( b = -2 - 6 = -8 ). Correct.For ( a + 2 = 16 ), ( a = 14 ), ( b = 1 - 6 = -5 ). Correct.For ( a + 2 = -16 ), ( a = -18 ), ( b = -1 - 6 = -7 ). Correct.All calculations seem correct. So, the maximum ( b ) is indeed 10, and the minimum is -22.**Final Answer**(1) The value of ( a ) is (boxed{2}).(2) ① The relationship between ( a ) and ( b ) is ( b = frac{16}{a + 2} - 6 ).② The maximum value of ( b ) is (boxed{10}) and the minimum value of ( b ) is (boxed{-22}).

💡Alright, so I have this problem here where I'm given that ( S_n ) is the sum of the first ( n ) terms of a sequence ( {a_n} ). And there's this logarithmic equation: ( log_3(S_n + 1) = n + 1 ). I need to find the general formula for the sequence ( {a_n} ).Okay, let's start by understanding what ( S_n ) represents. It's the sum of the first ( n ) terms of the sequence. So, ( S_n = a_1 + a_2 + a_3 + dots + a_n ). Got that.Now, the equation given is ( log_3(S_n + 1) = n + 1 ). Hmm, logarithms can sometimes be tricky, but I remember that ( log_b(a) = c ) implies that ( b^c = a ). So, applying that here, I can rewrite the equation as:( 3^{n + 1} = S_n + 1 )So, that simplifies to:( S_n = 3^{n + 1} - 1 )Alright, so now I have an expression for ( S_n ) in terms of ( n ). That's good progress.Now, since ( S_n ) is the sum of the first ( n ) terms, I can find ( a_n ) by considering the difference between ( S_n ) and ( S_{n-1} ). Because ( S_n = S_{n-1} + a_n ), right? So, rearranging that, ( a_n = S_n - S_{n-1} ).Let me write that down:( a_n = S_n - S_{n-1} )Given that ( S_n = 3^{n + 1} - 1 ), let's substitute that into the equation:( a_n = (3^{n + 1} - 1) - (3^{(n - 1) + 1} - 1) )Simplifying the exponent in the second term:( 3^{(n - 1) + 1} = 3^n )So now, the equation becomes:( a_n = (3^{n + 1} - 1) - (3^n - 1) )Let's distribute the negative sign:( a_n = 3^{n + 1} - 1 - 3^n + 1 )Now, the ( -1 ) and ( +1 ) cancel each other out:( a_n = 3^{n + 1} - 3^n )Hmm, I can factor out ( 3^n ) from both terms:( a_n = 3^n (3 - 1) )Which simplifies to:( a_n = 3^n times 2 )So, ( a_n = 2 times 3^n )Wait, but hold on a second. This seems straightforward, but I should check for ( n = 1 ) to make sure it works. Because sometimes when dealing with sequences, the first term can be an exception.So, let's compute ( S_1 ):( S_1 = 3^{1 + 1} - 1 = 3^2 - 1 = 9 - 1 = 8 )Therefore, ( a_1 = S_1 = 8 )Now, according to the formula ( a_n = 2 times 3^n ), when ( n = 1 ):( a_1 = 2 times 3^1 = 6 )But wait, that's not matching with our earlier calculation where ( a_1 = 8 ). There's a discrepancy here.Hmm, so that means my general formula ( a_n = 2 times 3^n ) works for ( n geq 2 ), but not for ( n = 1 ). So, I need to adjust my formula to account for this.Therefore, the sequence ( {a_n} ) has a special case for ( n = 1 ), where ( a_1 = 8 ), and for ( n geq 2 ), ( a_n = 2 times 3^n ).Let me verify this with ( n = 2 ):First, compute ( S_2 = 3^{2 + 1} - 1 = 27 - 1 = 26 )Then, ( a_2 = S_2 - S_1 = 26 - 8 = 18 )According to the formula ( a_n = 2 times 3^n ), when ( n = 2 ):( a_2 = 2 times 3^2 = 2 times 9 = 18 )That matches. Good.Similarly, for ( n = 3 ):( S_3 = 3^{3 + 1} - 1 = 81 - 1 = 80 )( a_3 = S_3 - S_2 = 80 - 26 = 54 )Using the formula:( a_3 = 2 times 3^3 = 2 times 27 = 54 )Perfect, that also matches.So, it seems that the formula ( a_n = 2 times 3^n ) works for ( n geq 2 ), but ( a_1 ) is an exception and is equal to 8.Therefore, the general formula for the sequence ( {a_n} ) is:( a_n = begin{cases} 8, & text{if } n = 1 2 times 3^n, & text{if } n geq 2 end{cases} )I think that's the solution. Let me just recap the steps to make sure I didn't skip anything:1. Started with the given equation ( log_3(S_n + 1) = n + 1 ).2. Converted the logarithmic equation to its exponential form to find ( S_n = 3^{n + 1} - 1 ).3. Recognized that ( a_n = S_n - S_{n-1} ) for ( n geq 2 ).4. Substituted the expression for ( S_n ) into this formula to find ( a_n = 2 times 3^n ) for ( n geq 2 ).5. Checked the case for ( n = 1 ) separately and found that ( a_1 = 8 ), which doesn't fit the general formula.6. Concluded that the sequence has a piecewise definition with ( a_1 = 8 ) and ( a_n = 2 times 3^n ) for ( n geq 2 ).Everything seems to check out. I don't see any mistakes in the reasoning or calculations. So, I'm confident that this is the correct general formula for the sequence.**Final Answer**The general formula of the sequence is boxed{a_n = begin{cases} 8, & n = 1 2 times 3^n, & n geq 2 end{cases}}.

💡Alright, so I've got these three problems about partitioning the set of positive integers, ( mathbb{N} ), into different parts with specific conditions. Let me try to wrap my head around each one step by step.Starting with the first problem: I need to partition ( mathbb{N} ) into three disjoint parts such that if the absolute difference between any two numbers ( m ) and ( n ) is 2 or 5, then ( m ) and ( n ) must be in different parts. Hmm, okay. So, essentially, I need to color the numbers with three colors in such a way that any two numbers that are 2 or 5 apart don't share the same color.I remember something about graph coloring, where each number is a node, and edges connect numbers that are 2 or 5 apart. If I can color this graph with three colors without any adjacent nodes sharing the same color, then I've solved the problem. But how do I construct such a partition?Maybe I can look for a repeating pattern or use modular arithmetic. Let's see, if I partition based on modulo 3, that might work. So, define three sets:- ( A = { n in mathbb{N} : n equiv 0 pmod{3} } )- ( B = { n in mathbb{N} : n equiv 1 pmod{3} } )- ( C = { n in mathbb{N} : n equiv 2 pmod{3} } )Now, if two numbers are in the same set, their difference is a multiple of 3. But 2 and 5 are not multiples of 3, so if ( |m - n| = 2 ) or 5, they can't be in the same set. That seems to satisfy the condition! Okay, so the first problem is done.Moving on to the second problem: I need to partition ( mathbb{N} ) into four disjoint parts such that if ( |m - n| = 2 ), 3, or 5, then ( m ) and ( n ) are in different parts. This seems similar to the first problem but with an additional condition for a difference of 3.Again, thinking about graph coloring, now the graph includes edges for differences of 2, 3, and 5. I need to color this graph with four colors. Maybe using modulo 4 this time?Let's try:- ( A = { n in mathbb{N} : n equiv 0 pmod{4} } )- ( B = { n in mathbb{N} : n equiv 1 pmod{4} } )- ( C = { n in mathbb{N} : n equiv 2 pmod{4} } )- ( D = { n in mathbb{N} : n equiv 3 pmod{4} } )If two numbers are in the same set, their difference is a multiple of 4. But 2, 3, and 5 aren't multiples of 4, so if ( |m - n| ) is 2, 3, or 5, they can't be in the same set. That works! So, the second problem is also solved.Now, the third problem: I need to show that it's impossible to partition ( mathbb{N} ) into three disjoint parts such that if ( |m - n| = 2 ), 3, or 5, then ( m ) and ( n ) are in different parts. This is the opposite of the first problem but with an extra condition.Let me assume, for contradiction, that such a partition exists with three parts. Let's denote the parts as ( A ), ( B ), and ( C ). Without loss of generality, let's say 1 is in ( A ). Then, 3 (since ( |3 - 1| = 2 )) must be in a different part, say ( B ). Similarly, 4 (since ( |4 - 1| = 3 )) must be in a different part, so 4 is in ( C ).Now, 6 is problematic because ( |6 - 1| = 5 ), so 6 can't be in ( A ). Also, ( |6 - 3| = 3 ), so 6 can't be in ( B ). Therefore, 6 must be in ( C ). But then, 8 is ( |8 - 6| = 2 ), so 8 can't be in ( C ). Also, ( |8 - 1| = 7 ), which isn't one of our differences, so 8 can be in ( A ) or ( B ). Let's put 8 in ( A ).Continuing, 9 is ( |9 - 8| = 1 ), which isn't a restricted difference, so 9 can be in any part. But ( |9 - 6| = 3 ), so 9 can't be in ( C ). It can be in ( A ) or ( B ). Let's put 9 in ( A ).Now, 7: ( |7 - 6| = 1 ), not restricted, but ( |7 - 8| = 1 ), also not restricted. However, ( |7 - 4| = 3 ), so 7 can't be in ( C ). So, 7 can be in ( A ) or ( B ). Let's put 7 in ( B ).Next, 10: ( |10 - 7| = 3 ), so 10 can't be in ( B ). ( |10 - 8| = 2 ), so 10 can't be in ( A ). Therefore, 10 must be in ( C ). But ( |10 - 6| = 4 ), which isn't restricted, so that's fine.11: ( |11 - 10| = 1 ), not restricted. ( |11 - 8| = 3 ), so 11 can't be in ( A ). ( |11 - 7| = 4 ), not restricted. So, 11 can be in ( B ) or ( C ). Let's put 11 in ( B ).Now, 12: ( |12 - 11| = 1 ), not restricted. ( |12 - 10| = 2 ), so 12 can't be in ( C ). ( |12 - 9| = 3 ), so 12 can't be in ( A ). ( |12 - 8| = 4 ), not restricted. So, 12 can't be in ( A ) or ( C ), so it must be in ( B ). But ( |12 - 7| = 5 ), so 12 can't be in ( B ) because 7 is in ( B ). Wait, that's a problem. 12 can't be in ( B ) because of 7, can't be in ( A ) because of 9, can't be in ( C ) because of 10. So, 12 has nowhere to go. This contradiction shows that our initial assumption is wrong. Therefore, it's impossible to partition ( mathbb{N} ) into three parts under these conditions.So, summarizing:1. For the first problem, using modulo 3 works.2. For the second problem, using modulo 4 works.3. For the third problem, assuming such a partition exists leads to a contradiction, so it's impossible.**Final Answer**1. boxed{3}2. boxed{4}3. boxed{text{Impossible}}

💡Okay, so I have this probability problem here, and I need to figure out which pair of events are mutually exclusive but not opposite. Let me try to break this down step by step.First, the problem says: "From a bag containing 2 red balls and 2 white balls, two balls are drawn. The two mutually exclusive but not opposite events are ( )"And then there are four options: A, B, C, D. Each option lists two events. I need to figure out which pair of events are mutually exclusive but not opposite.Alright, let's recall what mutually exclusive and opposite events mean. Mutually exclusive events are two events that cannot happen at the same time. That is, if one event occurs, the other cannot occur, and vice versa. Opposite events, on the other hand, are a special case of mutually exclusive events where one event is the complement of the other. So, if two events are opposite, one must occur and the other cannot, and together they cover all possible outcomes.So, in this problem, I need to find two events that cannot happen together (mutually exclusive), but they are not necessarily covering all possible outcomes (so they are not opposite).Let me look at each option one by one.**Option A: At least one white ball; both are white balls.**Okay, so the first event is "at least one white ball," which means either one white and one red, or both white. The second event is "both are white balls." Now, if both balls are white, that certainly means that at least one white ball is drawn. So, these two events can happen at the same time. Therefore, they are not mutually exclusive. So, option A is out.**Option B: At least one white ball; at least one red ball.**Hmm, the first event is "at least one white ball," which includes one white and one red, or two white. The second event is "at least one red ball," which includes one red and one white, or two red. Now, if we draw two balls, it's possible to have one white and one red, which satisfies both events. So, these two events can happen together, meaning they are not mutually exclusive. So, option B is also out.**Option C: Exactly one white ball; exactly two white balls.**Alright, the first event is "exactly one white ball," which means one white and one red. The second event is "exactly two white balls." Now, if I draw exactly one white ball, I can't have exactly two white balls at the same time. Similarly, if I have exactly two white balls, I can't have exactly one white ball. So, these two events cannot happen together. Therefore, they are mutually exclusive.But are they opposite? Well, let's see. The opposite of "exactly one white ball" would be "not exactly one white ball," which includes both "exactly two white balls" and "exactly zero white balls." Since "exactly two white balls" is only part of the opposite, these two events are not opposite. So, they are mutually exclusive but not opposite. That seems to fit the requirement.**Option D: At least one white ball; both are red balls.**First event: "at least one white ball," which includes one white and one red, or two white. Second event: "both are red balls." Now, if both balls are red, that means there are zero white balls, which contradicts "at least one white ball." So, these two events cannot happen together. Therefore, they are mutually exclusive.But are they opposite? The opposite of "at least one white ball" would be "no white balls," which is exactly "both are red balls." So, in this case, these two events are not only mutually exclusive but also opposite because one is the complement of the other. Since the question asks for mutually exclusive but not opposite events, option D is out.So, after evaluating all options, only option C satisfies the condition of being mutually exclusive but not opposite.**Final Answer**boxed{C}

💡Okay, so I have this problem about finding the asymptotes of a hyperbola. The equation given is (frac{x^2}{9} - frac{y^2}{16} = 1). Hmm, I remember that hyperbolas have asymptotes which are lines that the hyperbola approaches but never touches. I think the asymptotes are important because they help in sketching the hyperbola and understanding its behavior.First, I need to recall the standard form of a hyperbola. I think it's either (frac{(x - h)^2}{a^2} - frac{(y - k)^2}{b^2} = 1) for a hyperbola that opens left and right, or (frac{(y - k)^2}{b^2} - frac{(x - h)^2}{a^2} = 1) for one that opens up and down. In this case, since the (x^2) term is positive, it's the first case, so it opens left and right.Now, the center of the hyperbola is at ((h, k)). Looking at the equation, there are no shifts, so (h = 0) and (k = 0). So the center is at the origin, (0,0).Next, I need to find the asymptotes. I remember that for a hyperbola of the form (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), the equations of the asymptotes are (y = pm frac{b}{a}x). Let me make sure I got that right. Yeah, I think it's slope is (pm frac{b}{a}) because the hyperbola opens left and right.So, in this equation, (a^2 = 9) and (b^2 = 16). Therefore, (a = 3) and (b = 4). So, plugging these into the asymptote formula, the slopes would be (pm frac{4}{3}). So the equations of the asymptotes are (y = pm frac{4}{3}x).Let me double-check to make sure I didn't mix up (a) and (b). Since the hyperbola opens left and right, the asymptotes are based on (b/a), which is 4/3. If it had opened up and down, it would have been (a/b). So, yes, 4/3 is correct.Looking at the options given:A: (y = pm frac{3}{4}x)B: (y = pm frac{4}{3}x)C: (y = pm frac{3}{5}x)D: (y = pm frac{5}{3}x)So, option B matches my result. Therefore, the correct answer should be B.Wait, just to be thorough, let me recall another method. I think another way to find asymptotes is to consider the equation of the hyperbola and set it equal to zero instead of one. So, if we have (frac{x^2}{9} - frac{y^2}{16} = 0), that would give the equations of the asymptotes.Let me try that. So, (frac{x^2}{9} - frac{y^2}{16} = 0). Factoring this, it becomes (left(frac{x}{3} - frac{y}{4}right)left(frac{x}{3} + frac{y}{4}right) = 0). Therefore, the equations are (frac{x}{3} - frac{y}{4} = 0) and (frac{x}{3} + frac{y}{4} = 0), which simplifies to (y = frac{4}{3}x) and (y = -frac{4}{3}x). So, that confirms it again. The asymptotes are (y = pm frac{4}{3}x), which is option B.I think that's solid. I don't see any mistakes in my reasoning. So, I'm confident that the answer is B.**Final Answer**boxed{B}