💡First, I need to translate the problem into a mathematical equation. Mickey's age is 8 years less than 200% of Jerry's age. This can be written as:M = 2J - 8Given that Mickey is 16 years old, I substitute M with 16:16 = 2J - 8Next, I add 8 to both sides of the equation to isolate the term with Jerry's age:16 + 8 = 2J24 = 2JFinally, I divide both sides by 2 to solve for Jerry's age:24 / 2 = J12 = JTherefore, Jerry is 12 years old.

💡Okay, so I have this function f(x) that's defined piecewise. It's 3^{x+1} when x is less than or equal to 0, and it's log base 1/2 of x when x is greater than 0. I need to find where f(x) is greater than 1. Hmm, okay.First, I should probably break this down into two separate cases because the function is piecewise. So, for x ≤ 0, f(x) is 3^{x+1}, and for x > 0, it's log_{1/2}(x). I need to solve the inequality f(x) > 1 for each case and then combine the solutions.Starting with the first case: x ≤ 0, f(x) = 3^{x+1}. I need to find when 3^{x+1} > 1. I remember that 3 to any power is always positive, and 3^0 is 1. So, 3^{x+1} > 1 would mean that the exponent x+1 has to be greater than 0 because 3^y > 1 when y > 0. So, x+1 > 0 implies x > -1. But wait, in this case, x is also less than or equal to 0. So combining these, x has to be greater than -1 and less than or equal to 0. So, the solution for this part is (-1, 0].Now, moving on to the second case: x > 0, f(x) = log_{1/2}(x). I need to solve log_{1/2}(x) > 1. I recall that log base a of x is greater than b when, if a < 1, x < a^b. Since 1/2 is less than 1, the inequality flips. So, log_{1/2}(x) > 1 implies x < (1/2)^1, which is x < 1/2. But in this case, x is greater than 0, so combining these, x has to be between 0 and 1/2. So, the solution for this part is (0, 1/2).Now, I need to combine the solutions from both cases. The first case gave me (-1, 0], and the second case gave me (0, 1/2). Combining these intervals, since they are adjacent and overlap at 0, the combined solution is (-1, 1/2). Wait, but at x=0, the first case includes it, but the second case starts just above 0. So, actually, the interval is from -1 to 1/2, not including -1 and 1/2, but including 0. Hmm, but in the first case, x=0 is included, but in the second case, x=0 is not in the domain. So, the overall solution is all x such that -1 < x < 1/2, because at x=0, f(x) is 3^{0+1}=3, which is greater than 1, so x=0 is included. Wait, no, in the first case, x ≤ 0, so x=0 is included, but in the second case, x > 0, so x=0 is not in the second case. So, actually, the combined solution is (-1, 1/2), because at x=0, it's already included in the first interval. So, the solution set is (-1, 1/2).I think that's it. Let me just double-check. For x between -1 and 0, 3^{x+1} is greater than 1 because x+1 is between 0 and 1, so 3^{x+1} is between 1 and 3. For x between 0 and 1/2, log_{1/2}(x) is greater than 1 because log base 1/2 of 1/2 is 1, and since x is less than 1/2, the log increases beyond 1. So, yes, that makes sense. So, the solution set is all real numbers between -1 and 1/2.

💡Okay, so I have this problem about white washing the walls of a room. The room's dimensions are 25 feet by 15 feet by 12 feet. I need to find the cost of white washing the four walls at Rs. 4 per square foot. There's also a door and three windows that I need to account for. The total cost is given as Rs. 3624, and I need to figure out the dimensions of the door, specifically its height since the width is given as 6 feet.Alright, let's break this down step by step. First, I think I need to calculate the total area of the four walls. Since the room is a rectangular prism, the walls are made up of two pairs of identical rectangles. The two longer walls will have the length of 25 feet and the height of 12 feet, and the two shorter walls will have the length of 15 feet and the same height of 12 feet.So, the area of one long wall is 25 feet multiplied by 12 feet, which is 300 square feet. Since there are two of these, that's 300 times 2, which equals 600 square feet. Similarly, the area of one short wall is 15 feet multiplied by 12 feet, which is 180 square feet. With two of these, that's 180 times 2, which equals 360 square feet. Adding those together, the total area of the four walls is 600 plus 360, which is 960 square feet.But wait, I need to subtract the areas of the door and the windows because those won't be white washed. The door has dimensions of 6 feet by some feet. Let's call the unknown height of the door "h" feet. So, the area of the door is 6 times h, which is 6h square feet.Now, for the windows. There are three windows, each with dimensions 4 feet by 3 feet. So, the area of one window is 4 times 3, which is 12 square feet. Since there are three windows, the total area for the windows is 12 times 3, which equals 36 square feet.Adding the area of the door and the windows together gives me the total area that won't be white washed: 6h plus 36 square feet.So, the area that actually needs to be white washed is the total wall area minus the areas of the door and windows. That would be 960 minus (6h + 36). Let's write that as 960 - 6h - 36, which simplifies to 924 - 6h square feet.Now, the cost of white washing is Rs. 4 per square foot. So, the total cost should be the area to be white washed multiplied by 4. According to the problem, this total cost is Rs. 3624. So, I can set up the equation:4 * (924 - 6h) = 3624Let me solve this equation for h. First, divide both sides by 4 to simplify:924 - 6h = 3624 / 4924 - 6h = 906Now, subtract 924 from both sides:-6h = 906 - 924-6h = -18Finally, divide both sides by -6:h = (-18) / (-6)h = 3So, the height of the door is 3 feet. That makes sense because if I plug it back into the area calculations, everything checks out. The door area would be 6 * 3 = 18 square feet, and the windows are 36 square feet, so the total area not white washed is 18 + 36 = 54 square feet. Subtracting that from the total wall area: 960 - 54 = 906 square feet. Multiplying by the cost per square foot: 906 * 4 = 3624, which matches the given total cost.I think I've got it right. The door is 6 feet wide and 3 feet tall.

💡Okay, so I have this problem about an arithmetic sequence. It says that S_n is the sum of the first n terms of an arithmetic sequence {a_n}. They give me two pieces of information: a_2 = 11 and S_{10} = 40. First, I need to find the general formula for {a_n}. Hmm, arithmetic sequences have a common difference, right? So, the nth term of an arithmetic sequence is given by a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference.They told me that a_2 = 11. So, using the formula, a_2 = a_1 + d = 11. That's one equation. Next, they gave me the sum of the first 10 terms, S_{10} = 40. The formula for the sum of the first n terms of an arithmetic sequence is S_n = n/2 * (2a_1 + (n-1)d). So, plugging in n=10, I get S_{10} = 10/2 * (2a_1 + 9d) = 40. Simplifying that, 5*(2a_1 + 9d) = 40, so 2a_1 + 9d = 8.Now I have two equations:1. a_1 + d = 112. 2a_1 + 9d = 8I can solve this system of equations to find a_1 and d. Let me use substitution or elimination. Maybe elimination is easier here.From the first equation, a_1 = 11 - d. Substitute that into the second equation:2*(11 - d) + 9d = 822 - 2d + 9d = 822 + 7d = 87d = 8 - 227d = -14d = -2Okay, so the common difference d is -2. Now, substitute back into a_1 = 11 - d:a_1 = 11 - (-2) = 11 + 2 = 13So, the first term is 13, and the common difference is -2. Therefore, the general formula for a_n is:a_n = a_1 + (n-1)d = 13 + (n-1)*(-2) = 13 - 2(n - 1) = 13 - 2n + 2 = 15 - 2nWait, that simplifies to a_n = -2n + 15. Let me check that:For n=1: a_1 = -2(1) + 15 = 13, which is correct.For n=2: a_2 = -4 + 15 = 11, which matches the given information.Good, that seems right.Now, part (2) asks for the sum T_n of the absolute values of the first n terms of the sequence {|a_n|}. So, T_n = |a_1| + |a_2| + ... + |a_n|.Since the sequence is arithmetic with a negative common difference, the terms will eventually become negative. So, I need to find when the terms change from positive to negative.Let me find the term where a_n = 0:-2n + 15 = 0-2n = -15n = 7.5Since n must be an integer, the 7th term is the last positive term, and the 8th term is the first negative term.So, for n ≤ 7, all terms are positive, and for n ≥ 8, some terms are negative. Therefore, the absolute values will affect the sum differently depending on whether n is less than or equal to 7 or greater than 7.Let me calculate T_n in two cases:Case 1: n ≤ 7In this case, all terms are positive, so T_n is just the sum of the first n terms, which is S_n.We know S_n = n/2*(2a_1 + (n-1)d). Plugging in a_1 =13 and d=-2:S_n = n/2*(26 + (n-1)*(-2)) = n/2*(26 - 2n + 2) = n/2*(28 - 2n) = n*(14 - n) = 14n - n^2So, T_n = 14n - n^2 for n ≤ 7.Case 2: n ≥ 8Here, the first 7 terms are positive, and from the 8th term onwards, the terms are negative. So, the absolute value of the negative terms will be positive, which effectively means we need to subtract twice the sum of the negative terms from the total sum.Wait, let me think. The sum of the absolute values would be the sum of the first 7 terms plus the sum of the absolute values of the terms from 8 to n.Alternatively, since the sum of the first n terms is S_n, and the sum of the first 7 terms is S_7, then the sum from 8 to n is S_n - S_7. But since these terms are negative, their absolute values would be -(S_n - S_7). Therefore, T_n = S_7 - (S_n - S_7) = 2S_7 - S_n.Let me compute S_7:S_7 = 7/2*(2*13 + 6*(-2)) = 7/2*(26 -12) = 7/2*14 = 7*7 = 49So, S_7 = 49.Then, T_n = 2*49 - S_n = 98 - S_n.But S_n = 14n - n^2, so T_n = 98 - (14n - n^2) = 98 -14n + n^2 = n^2 -14n +98.Therefore, for n ≥8, T_n = n^2 -14n +98.Let me verify this with n=8:Compute T_8. The first 7 terms are positive, the 8th term is a_8 = -2*8 +15 = -16 +15 = -1, so |a_8|=1.So, T_8 = S_7 + |a_8| = 49 +1=50.Using the formula for n=8: 8^2 -14*8 +98=64 -112 +98=50. Correct.Similarly, n=9:a_9 = -2*9 +15 = -18 +15 = -3, |a_9|=3.T_9 = T_8 + |a_9| =50 +3=53.Using the formula:9^2 -14*9 +98=81 -126 +98=53. Correct.Good, seems consistent.So, summarizing:For n ≤7, T_n = -n^2 +14nFor n ≥8, T_n = n^2 -14n +98I think that's the answer.**Final Answer**(1) The general formula is boxed{a_n = -2n + 15}.(2) The sum of absolute values is [T_n = begin{cases} -n^2 + 14n & text{if } 1 leq n leq 7, n^2 - 14n + 98 & text{if } n geq 8.end{cases}]So, the final answers are boxed{a_n = -2n + 15} and for ( T_n ), the piecewise function as stated.boxed{a_n = -2n + 15}For ( T_n ):[boxed{T_n = begin{cases} -n^2 + 14n & text{for } 1 leq n leq 7, n^2 - 14n + 98 & text{for } n geq 8.end{cases}}]**Final Answer**(1) boxed{a_n = -2n + 15}(2) boxed{T_n = begin{cases} -n^2 + 14n & text{if } 1 leq n leq 7, n^2 - 14n + 98 & text{if } n geq 8.end{cases}}

💡Okay, so I've got this problem here. It says that for a real number ( x_1 ) between -2 and 2, we define a sequence ( x_2, x_3, ldots ) by the recurrence relation ( x_{n+1} = x_n^2 - 2 ). We're told that none of the ( x_n ) are zero, and we need to define a number ( A ) between 0 and 1. The way we define ( A ) is by looking at the binary representation where the ( n )-th digit after the decimal point is 0 if the product ( x_1x_2cdots x_n ) is positive and 1 otherwise. The goal is to prove that ( A = frac{1}{pi} cos^{-1}left(frac{x_1}{2}right) ).Alright, let's break this down. First, I need to understand the sequence ( x_n ). It starts with ( x_1 ) between -2 and 2, and each subsequent term is the square of the previous term minus 2. So, ( x_2 = x_1^2 - 2 ), ( x_3 = x_2^2 - 2 ), and so on.Given that ( x_1 ) is between -2 and 2, let's see what happens to the sequence. If ( x_1 ) is positive, say 1, then ( x_2 = 1 - 2 = -1 ), ( x_3 = (-1)^2 - 2 = 1 - 2 = -1 ), and so on. So, it becomes periodic. If ( x_1 ) is negative, say -1, then ( x_2 = (-1)^2 - 2 = 1 - 2 = -1 ), so it's constant.But wait, the problem says that no ( x_n ) is zero. So, ( x_1 ) can't be such that any ( x_n ) becomes zero. Hmm, so if ( x_1 ) is 0, then ( x_2 = -2 ), but ( x_1 ) isn't zero. So, as long as ( x_1 ) isn't such that the sequence hits zero, which is given.Now, the number ( A ) is defined by the binary digits where each digit is 0 if the product up to that point is positive and 1 otherwise. So, the first digit after the decimal is 0 if ( x_1 ) is positive, 1 if negative. The second digit is 0 if ( x_1x_2 ) is positive, 1 otherwise, and so on.So, essentially, each digit is determined by the sign of the product of the first ( n ) terms. If the product is positive, it's 0; if negative, it's 1.The problem wants us to show that this binary number ( A ) is equal to ( frac{1}{pi} cos^{-1}left(frac{x_1}{2}right) ).Hmm, okay. So, I need to relate the sequence ( x_n ) and the product signs to an inverse cosine function scaled by ( frac{1}{pi} ).Let me think about the substitution ( x_n = 2 cos theta_n ). Since ( x_n ) is between -2 and 2, this substitution makes sense because cosine ranges between -1 and 1, so multiplying by 2 scales it appropriately.So, let me set ( x_n = 2 cos theta_n ). Then, the recurrence relation becomes:( x_{n+1} = x_n^2 - 2 )Substituting ( x_n = 2 cos theta_n ):( x_{n+1} = (2 cos theta_n)^2 - 2 = 4 cos^2 theta_n - 2 )But ( 4 cos^2 theta_n - 2 = 2(2 cos^2 theta_n - 1) = 2 cos 2theta_n )So, ( x_{n+1} = 2 cos 2theta_n )Therefore, ( theta_{n+1} = 2 theta_n )So, each ( theta_n ) is double the previous angle. That means ( theta_n = 2^{n-1} theta_1 )So, starting from ( theta_1 ), each subsequent angle is doubling. So, ( theta_2 = 2 theta_1 ), ( theta_3 = 4 theta_1 ), ( theta_4 = 8 theta_1 ), etc.Therefore, ( x_n = 2 cos(2^{n-1} theta_1) )So, the product ( x_1 x_2 cdots x_n ) can be written as:( prod_{k=1}^n x_k = prod_{k=1}^n 2 cos(2^{k-1} theta_1) )Which is ( 2^n prod_{k=1}^n cos(2^{k-1} theta_1) )Now, I remember there's a product formula for multiple angles in cosine. Specifically, the product ( prod_{k=1}^n cos(2^{k-1} theta) ) can be expressed as ( frac{sin(2^n theta)}{2^n sin theta} )Let me verify that. For n=1, it's ( cos theta ), and ( frac{sin(2 theta)}{2 sin theta} = frac{2 sin theta cos theta}{2 sin theta} = cos theta ). So, that works.Assume it's true for n, then for n+1:( prod_{k=1}^{n+1} cos(2^{k-1} theta) = left( prod_{k=1}^n cos(2^{k-1} theta) right) cos(2^n theta) = frac{sin(2^n theta)}{2^n sin theta} cos(2^n theta) )Using the identity ( sin(2 alpha) = 2 sin alpha cos alpha ), so ( sin(2^{n+1} theta) = 2 sin(2^n theta) cos(2^n theta) ). Therefore, the product becomes:( frac{sin(2^n theta)}{2^n sin theta} cos(2^n theta) = frac{sin(2^{n+1} theta)}{2^{n+1} sin theta} )So, by induction, the formula holds.Therefore, ( prod_{k=1}^n cos(2^{k-1} theta) = frac{sin(2^n theta)}{2^n sin theta} )Therefore, the product ( x_1 x_2 cdots x_n = 2^n times frac{sin(2^n theta_1)}{2^n sin theta_1} = frac{sin(2^n theta_1)}{sin theta_1} )So, ( x_1 x_2 cdots x_n = frac{sin(2^n theta_1)}{sin theta_1} )Now, the sign of this product is determined by the sign of ( sin(2^n theta_1) ), since ( sin theta_1 ) is positive because ( theta_1 ) is between 0 and ( pi ) (since ( x_1 ) is between -2 and 2, ( theta_1 = cos^{-1}(x_1 / 2) ), which is between 0 and ( pi )).Therefore, the sign of the product is the same as the sign of ( sin(2^n theta_1) ).So, the ( n )-th digit of ( A ) is 0 if ( sin(2^n theta_1) > 0 ), and 1 if ( sin(2^n theta_1) < 0 ).But ( sin alpha ) is positive when ( alpha ) is in the first or second quadrants (i.e., between 0 and ( pi )), and negative when in the third or fourth quadrants (i.e., between ( pi ) and ( 2pi )).So, ( sin(2^n theta_1) > 0 ) if ( 2^n theta_1 ) mod ( 2pi ) is in [0, ( pi )), and negative otherwise.Therefore, the ( n )-th digit is 0 if ( 2^n theta_1 ) mod ( 2pi ) is in [0, ( pi )), and 1 otherwise.So, let's define ( s = frac{theta_1}{pi} ). Then, ( s ) is between 0 and 1 because ( theta_1 ) is between 0 and ( pi ).So, ( 2^n theta_1 = 2^n pi s ). Therefore, ( 2^n theta_1 ) mod ( 2pi ) is equivalent to ( 2^n pi s ) mod ( 2pi ), which is ( 2pi times {2^{n-1} s} ), where ( {x} ) denotes the fractional part of ( x ).So, the fractional part of ( 2^{n-1} s ) determines whether ( 2^n theta_1 ) mod ( 2pi ) is in [0, ( pi )) or [( pi ), ( 2pi )).Specifically, if the fractional part ( {2^{n-1} s} < 0.5 ), then ( 2^n theta_1 ) mod ( 2pi ) is less than ( pi ), so the sine is positive, and the digit is 0. If ( {2^{n-1} s} geq 0.5 ), then it's in [( pi ), ( 2pi )), sine is negative, and the digit is 1.Therefore, the ( n )-th digit of ( A ) is the binary digit indicating whether the fractional part of ( 2^{n-1} s ) is less than 0.5 or not. That is, it's the binary expansion of ( s ).Wait, let me think about that. The fractional part of ( 2^{n-1} s ) is like shifting the binary expansion of ( s ) to the left by ( n-1 ) digits. So, the ( n )-th digit after the decimal in the binary expansion of ( s ) is determined by whether the fractional part of ( 2^{n-1} s ) is less than 0.5 or not.Therefore, the binary digits of ( s ) correspond exactly to the digits of ( A ). Hence, ( A = s ).But ( s = frac{theta_1}{pi} ), and ( theta_1 = cos^{-1}left(frac{x_1}{2}right) ). Therefore, ( s = frac{1}{pi} cos^{-1}left(frac{x_1}{2}right) ).Hence, ( A = frac{1}{pi} cos^{-1}left(frac{x_1}{2}right) ), which is what we wanted to prove.So, the key steps were:1. Substituting ( x_n = 2 cos theta_n ) to transform the recurrence into a doubling of angles.2. Expressing the product ( x_1 x_2 cdots x_n ) in terms of sine functions.3. Relating the sign of the product to the sine function's sign, which depends on the angle modulo ( 2pi ).4. Connecting this to the binary expansion of ( s = frac{theta_1}{pi} ), which is exactly the number ( A ) defined by the problem.5. Concluding that ( A ) is equal to ( frac{1}{pi} cos^{-1}left(frac{x_1}{2}right) ).I think that covers all the necessary steps. I might have skipped some details, but overall, this seems to be the path to the solution.

💡To find the value of ( a_0 - a_1 + a_2 - a_3 + a_4 ) given the expansion ( (2 - x)^4 = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 ), we can substitute ( x = -1 ) into the equation. This substitution transforms the expression into ( (2 - (-1))^4 = a_0 - a_1 + a_2 - a_3 + a_4 ), which simplifies to ( 3^4 = 81 ). Therefore, the value of ( a_0 - a_1 + a_2 - a_3 + a_4 ) is 81.