💡First, I need to find all the positive factors of 84. To do this, I'll start by determining its prime factorization.84 can be factored as:[84 = 2^2 times 3^1 times 7^1]Using the formula for finding the number of divisors, which is ((e_1 + 1)(e_2 + 1) ldots (e_n + 1)) where (e_i) are the exponents in the prime factorization, I can calculate the total number of positive factors:[(2+1)(1+1)(1+1) = 3 times 2 times 2 = 12]Next, I'll list all the factors of 84:[1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84]Now, I need to identify which of these factors are less than 8:[1, 2, 3, 4, 6, 7]There are 6 factors less than 8. To find the probability that a randomly drawn positive factor of 84 is less than 8, I'll divide the number of favorable outcomes by the total number of possible outcomes:[text{Probability} = frac{6}{12} = frac{1}{2}]Therefore, the probability that a randomly drawn positive factor of 84 is less than 8 is (frac{1}{2}).

💡First, I will multiply 8 by 1/15, which simplifies to 8/15.Next, I will multiply 8/15 by 30. Dividing 30 by 15 gives 2, so the expression becomes 8 multiplied by 2, which equals 16.Finally, I will multiply 16 by 3 to get the final result of 48.

💡First, I need to understand the given ratio ( A:B:C = 3:2:6 ). This means that ( A ) is to ( B ) as 3 is to 2, and ( B ) is to ( C ) as 2 is to 6. To simplify calculations, I'll introduce a variable ( x ) such that ( A = 3x ), ( B = 2x ), and ( C = 6x ).Next, I'll substitute these expressions into the given equation ( (4A + 3B) div (5C - 2A) ).For the numerator:[4A + 3B = 4(3x) + 3(2x) = 12x + 6x = 18x]For the denominator:[5C - 2A = 5(6x) - 2(3x) = 30x - 6x = 24x]Now, I'll divide the two results:[frac{4A + 3B}{5C - 2A} = frac{18x}{24x} = frac{18}{24}]Finally, I'll simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 6:[frac{18}{24} = frac{3}{4}]Therefore, the value of ( (4A + 3B) div (5C - 2A) ) is ( frac{3}{4} ).

💡Okay, so I need to find the slope of the tangent line to the function y = x e^x at x = 1. Hmm, I remember that the slope of the tangent line at a particular point is just the derivative of the function evaluated at that point. So, first things first, I need to find the derivative of y with respect to x.Alright, the function is y = x e^x. That looks like a product of two functions: one is x and the other is e^x. I think I need to use the product rule here. The product rule says that if you have a function that's the product of two functions, say u(x) and v(x), then the derivative is u'(x)v(x) + u(x)v'(x). So, in this case, u(x) is x and v(x) is e^x.Let me write that down. So, u(x) = x, which means u'(x) is just 1, because the derivative of x with respect to x is 1. And v(x) = e^x, whose derivative is also e^x. So, v'(x) = e^x.Putting that into the product rule formula, we have y' = u'(x)v(x) + u(x)v'(x) = 1 * e^x + x * e^x. So, simplifying that, y' = e^x + x e^x. Wait, that can be factored as y' = e^x (1 + x). Hmm, that might be useful later.But for now, I just need to evaluate this derivative at x = 1. So, plugging x = 1 into y', we get y'(1) = e^1 (1 + 1) = e * 2 = 2e. So, the slope of the tangent line at x = 1 is 2e.Let me double-check my work to make sure I didn't make any mistakes. So, the function is y = x e^x. The derivative using the product rule is correct: derivative of x is 1, derivative of e^x is e^x, so 1*e^x + x*e^x. That seems right. Then, factoring out e^x gives e^x (1 + x), which is correct. Plugging in x = 1, we get e*(1 + 1) = 2e. Yeah, that seems correct.Wait, just to be thorough, maybe I can compute the derivative another way to confirm. Let me try expanding the function or using another method. Hmm, y = x e^x. If I think of it as y = e^x + x e^x, but that doesn't really help. Alternatively, maybe I can use logarithmic differentiation, but that seems unnecessary here since the product rule worked fine.Alternatively, I can compute the derivative from first principles, using the limit definition. The derivative of y at x is the limit as h approaches 0 of [y(x + h) - y(x)] / h. So, let's try that.Compute y(x + h) = (x + h) e^{x + h} = (x + h) e^x e^h. Then, y(x) = x e^x. So, y(x + h) - y(x) = (x + h) e^x e^h - x e^x. Factor out e^x: e^x [ (x + h) e^h - x ].So, [y(x + h) - y(x)] / h = e^x [ (x + h) e^h - x ] / h. Let's expand that: e^x [ x e^h + h e^h - x ] / h = e^x [ x (e^h - 1) + h e^h ] / h.Now, split the fraction: e^x [ x (e^h - 1)/h + e^h ]. Then, take the limit as h approaches 0. We know that lim_{h->0} (e^h - 1)/h = 1, and lim_{h->0} e^h = 1. So, the expression becomes e^x [ x * 1 + 1 ] = e^x (x + 1). Which is the same as before, so that confirms that the derivative is indeed y' = e^x (x + 1). So, evaluating at x = 1, we get 2e.Okay, so both methods give me the same result, so I feel confident that the slope is 2e. Therefore, the slope of the tangent line to the function y = x e^x at x = 1 is 2e.**Final Answer**The slope of the tangent line to the function ( y = xe^{x} ) at ( x = 1 ) is boxed{2e}.

💡Okay, so I have this problem about four identical point charges at the corners of a square, storing 20 Joules of energy. Then, one of the charges is moved to the center of the square, and I need to find out how much more energy is stored in this new configuration. Hmm, let me try to break this down step by step.First, I remember that the energy stored between two point charges is given by Coulomb's law, which is inversely proportional to the distance between them and directly proportional to the product of their charges. Since all charges are identical, let's denote each charge as ( q ). The formula for the potential energy between two charges is ( U = k frac{q_1 q_2}{r} ), where ( k ) is Coulomb's constant, ( q_1 ) and ( q_2 ) are the charges, and ( r ) is the distance between them.In the initial configuration, the four charges are at the corners of a square. Let's assume the side length of the square is ( s ). So, each charge has three neighboring charges: two adjacent ones at a distance ( s ) and one diagonal charge at a distance ( ssqrt{2} ). Wait, but in a square, each charge is connected to three others, right? So, for four charges, how many unique pairs are there? That's a combination problem. The number of unique pairs is ( binom{4}{2} = 6 ). So, there are six pairs of charges interacting with each other.Now, out of these six pairs, four pairs are adjacent, each separated by distance ( s ), and two pairs are diagonal, each separated by distance ( ssqrt{2} ). So, the total potential energy ( U_{text{total}} ) is the sum of the potential energies of all these pairs.Let me write that down:[U_{text{total}} = 4 times left( k frac{q^2}{s} right) + 2 times left( k frac{q^2}{ssqrt{2}} right)]Simplifying this:[U_{text{total}} = 4k frac{q^2}{s} + 2k frac{q^2}{ssqrt{2}} = 4k frac{q^2}{s} + sqrt{2}k frac{q^2}{s}]So, combining the terms:[U_{text{total}} = left(4 + sqrt{2}right)k frac{q^2}{s}]And we know that this total energy is 20 Joules:[left(4 + sqrt{2}right)k frac{q^2}{s} = 20 , text{J}]Okay, so that's the initial configuration. Now, when one charge is moved to the center of the square, the configuration changes. Let's visualize this: one charge is at the center, and the other three are still at the corners. So, now, each of the three corner charges will interact with the center charge, and also with each other.Wait, so in the new configuration, how many unique pairs are there? We still have four charges, so again ( binom{4}{2} = 6 ) pairs. But now, the distances have changed for some of these pairs.Specifically, the three corner charges are each at a distance of ( frac{s}{sqrt{2}} ) from the center charge because the distance from the center of a square to a corner is half the length of the diagonal, which is ( frac{ssqrt{2}}{2} = frac{s}{sqrt{2}} ).Additionally, the three corner charges still interact with each other. Among these, two are adjacent (distance ( s )) and one is diagonal (distance ( ssqrt{2} )). Wait, no, actually, if one charge is moved to the center, the remaining three corner charges form a sort of triangle. Let me think: originally, the square had four corners, but now one is missing, so the remaining three form a right triangle with sides ( s ), ( s ), and ( ssqrt{2} ). So, among these three, how many unique pairs are there? Again, ( binom{3}{2} = 3 ) pairs. Two of these pairs are adjacent (distance ( s )), and one pair is diagonal (distance ( ssqrt{2} )).So, in the new configuration, the total potential energy will consist of:1. The interactions between the center charge and each of the three corner charges.2. The interactions between the three corner charges themselves.Let's compute each part.First, the interactions between the center charge and each corner charge. There are three such pairs, each with distance ( frac{s}{sqrt{2}} ). So, the potential energy for each of these is:[U_{text{center-corner}} = k frac{q^2}{frac{s}{sqrt{2}}} = k frac{q^2 sqrt{2}}{s}]Since there are three such pairs, the total energy for these interactions is:[3 times k frac{q^2 sqrt{2}}{s} = 3sqrt{2}k frac{q^2}{s}]Next, the interactions between the three corner charges. As I mentioned, there are three pairs: two adjacent (distance ( s )) and one diagonal (distance ( ssqrt{2} )). So, the potential energies are:- For the two adjacent pairs: each has energy ( k frac{q^2}{s} ), so total for two is ( 2k frac{q^2}{s} ).- For the diagonal pair: energy ( k frac{q^2}{ssqrt{2}} ).So, total energy for the corner-corner interactions is:[2k frac{q^2}{s} + k frac{q^2}{ssqrt{2}} = 2k frac{q^2}{s} + frac{sqrt{2}}{2}k frac{q^2}{s}]Wait, hold on, ( frac{1}{sqrt{2}} = frac{sqrt{2}}{2} ), so that term is ( frac{sqrt{2}}{2}k frac{q^2}{s} ).Therefore, the total energy for the corner-corner interactions is:[2k frac{q^2}{s} + frac{sqrt{2}}{2}k frac{q^2}{s} = left(2 + frac{sqrt{2}}{2}right)k frac{q^2}{s}]Now, adding the two parts together: the center-corner interactions and the corner-corner interactions.Total potential energy in the new configuration:[U'_{text{total}} = 3sqrt{2}k frac{q^2}{s} + left(2 + frac{sqrt{2}}{2}right)k frac{q^2}{s}]Let me combine these terms:First, let's factor out ( k frac{q^2}{s} ):[U'_{text{total}} = left(3sqrt{2} + 2 + frac{sqrt{2}}{2}right)k frac{q^2}{s}]Combine the ( sqrt{2} ) terms:[3sqrt{2} + frac{sqrt{2}}{2} = left(3 + frac{1}{2}right)sqrt{2} = frac{7}{2}sqrt{2}]So, now:[U'_{text{total}} = left(frac{7}{2}sqrt{2} + 2right)k frac{q^2}{s}]Now, let's recall from the initial configuration that:[left(4 + sqrt{2}right)k frac{q^2}{s} = 20 , text{J}]So, we can express ( k frac{q^2}{s} ) in terms of Joules:[k frac{q^2}{s} = frac{20}{4 + sqrt{2}} , text{J}]Let me rationalize the denominator:[frac{20}{4 + sqrt{2}} = frac{20(4 - sqrt{2})}{(4 + sqrt{2})(4 - sqrt{2})} = frac{20(4 - sqrt{2})}{16 - 2} = frac{20(4 - sqrt{2})}{14} = frac{10(4 - sqrt{2})}{7}]So,[k frac{q^2}{s} = frac{10(4 - sqrt{2})}{7} , text{J}]Now, plug this back into the expression for ( U'_{text{total}} ):[U'_{text{total}} = left(frac{7}{2}sqrt{2} + 2right) times frac{10(4 - sqrt{2})}{7}]Simplify this expression step by step.First, let's distribute the multiplication:[U'_{text{total}} = left(frac{7}{2}sqrt{2} times frac{10(4 - sqrt{2})}{7}right) + left(2 times frac{10(4 - sqrt{2})}{7}right)]Simplify each term separately.First term:[frac{7}{2}sqrt{2} times frac{10(4 - sqrt{2})}{7} = frac{7}{2} times frac{10}{7} times sqrt{2}(4 - sqrt{2}) = frac{10}{2} times sqrt{2}(4 - sqrt{2}) = 5 times sqrt{2}(4 - sqrt{2})]Multiply this out:[5 times (4sqrt{2} - (sqrt{2})^2) = 5 times (4sqrt{2} - 2) = 20sqrt{2} - 10]Second term:[2 times frac{10(4 - sqrt{2})}{7} = frac{20(4 - sqrt{2})}{7}]So, now, combining both terms:[U'_{text{total}} = (20sqrt{2} - 10) + frac{20(4 - sqrt{2})}{7}]Let me write both terms with a common denominator to combine them:First term: ( 20sqrt{2} - 10 = frac{140sqrt{2} - 70}{7} )Second term: ( frac{20(4 - sqrt{2})}{7} = frac{80 - 20sqrt{2}}{7} )Now, add them together:[U'_{text{total}} = frac{140sqrt{2} - 70 + 80 - 20sqrt{2}}{7} = frac{(140sqrt{2} - 20sqrt{2}) + (-70 + 80)}{7} = frac{120sqrt{2} + 10}{7}]So,[U'_{text{total}} = frac{120sqrt{2} + 10}{7} , text{J}]Now, let's compute the numerical value to understand how much more energy is stored. But before that, let's also compute the initial energy in terms of the same denominator to see the difference.From earlier, we have:[U_{text{total}} = left(4 + sqrt{2}right)k frac{q^2}{s} = 20 , text{J}]But we also expressed ( k frac{q^2}{s} = frac{10(4 - sqrt{2})}{7} ). So, let's compute ( U'_{text{total}} ) and then subtract the initial 20 J to find the increase.But wait, actually, the initial total energy is 20 J, and the new total energy is ( frac{120sqrt{2} + 10}{7} ) J. So, the increase in energy is:[Delta U = U'_{text{total}} - U_{text{total}} = frac{120sqrt{2} + 10}{7} - 20]Convert 20 J to a fraction with denominator 7:[20 = frac{140}{7}]So,[Delta U = frac{120sqrt{2} + 10 - 140}{7} = frac{120sqrt{2} - 130}{7}]Simplify this:[Delta U = frac{120sqrt{2} - 130}{7} = frac{10(12sqrt{2} - 13)}{7}]But let's see if this can be simplified further or if there's a better way to express it. Alternatively, we can factor out a 10:[Delta U = frac{10(12sqrt{2} - 13)}{7}]But perhaps it's better to leave it as ( frac{120sqrt{2} - 130}{7} ).However, let me double-check my calculations to make sure I didn't make a mistake somewhere.Starting from the new total energy:[U'_{text{total}} = left(frac{7}{2}sqrt{2} + 2right) times frac{10(4 - sqrt{2})}{7}]Breaking it down:First term: ( frac{7}{2}sqrt{2} times frac{10(4 - sqrt{2})}{7} = 5sqrt{2}(4 - sqrt{2}) = 20sqrt{2} - 10 )Second term: ( 2 times frac{10(4 - sqrt{2})}{7} = frac{20(4 - sqrt{2})}{7} )Adding them: ( 20sqrt{2} - 10 + frac{80 - 20sqrt{2}}{7} )Convert ( 20sqrt{2} - 10 ) to sevenths:( 20sqrt{2} = frac{140sqrt{2}}{7} ), ( -10 = frac{-70}{7} )So, ( frac{140sqrt{2} - 70}{7} + frac{80 - 20sqrt{2}}{7} = frac{140sqrt{2} - 70 + 80 - 20sqrt{2}}{7} = frac{120sqrt{2} + 10}{7} )Yes, that seems correct.So, the increase in energy is:[Delta U = frac{120sqrt{2} + 10}{7} - 20 = frac{120sqrt{2} + 10 - 140}{7} = frac{120sqrt{2} - 130}{7}]Which is approximately:Compute ( 120sqrt{2} approx 120 times 1.4142 approx 169.704 )So,[Delta U approx frac{169.704 - 130}{7} = frac{39.704}{7} approx 5.672 , text{J}]So, approximately 5.67 Joules more energy is stored.But let's see if we can express this in a simplified exact form. Let's factor numerator:( 120sqrt{2} - 130 = 10(12sqrt{2} - 13) )So,[Delta U = frac{10(12sqrt{2} - 13)}{7}]Alternatively, we can write it as:[Delta U = frac{10}{7}(12sqrt{2} - 13)]But perhaps the problem expects an exact form, so we can leave it as ( frac{120sqrt{2} - 130}{7} ) or factor out 10 as above.Alternatively, let's see if the initial approach was correct. Wait, in the initial problem, the user mentioned that the energy stored by any pair is inversely proportional to the distance and directly proportional to the charges. Since all charges are identical, the energy per pair is proportional to ( frac{1}{r} ).But in my calculation, I used Coulomb's law which includes ( k ), but since the problem states that energy is inversely proportional to distance and directly proportional to charges, perhaps we can consider the energy per pair as ( U = frac{q^2}{r} ), ignoring the constant ( k ) because it's just a proportionality constant and will cancel out in the ratio.Wait, that might be a simpler approach. Let me try that.Let’s denote the energy per pair as ( U = frac{q^2}{r} ), where ( q ) is the charge and ( r ) is the distance. Since all charges are identical, the total energy is the sum over all pairs.In the initial configuration, we have six pairs: four with distance ( s ) and two with distance ( ssqrt{2} ). So, total energy:[U_{text{initial}} = 4 times frac{q^2}{s} + 2 times frac{q^2}{ssqrt{2}} = left(4 + frac{2}{sqrt{2}}right)frac{q^2}{s} = left(4 + sqrt{2}right)frac{q^2}{s}]Given that ( U_{text{initial}} = 20 , text{J} ), so:[left(4 + sqrt{2}right)frac{q^2}{s} = 20]Thus,[frac{q^2}{s} = frac{20}{4 + sqrt{2}} = frac{20(4 - sqrt{2})}{(4 + sqrt{2})(4 - sqrt{2})} = frac{20(4 - sqrt{2})}{14} = frac{10(4 - sqrt{2})}{7}]So, ( frac{q^2}{s} = frac{10(4 - sqrt{2})}{7} )Now, in the new configuration, we have six pairs again, but with different distances.Three pairs are between the center charge and the corner charges, each at distance ( frac{s}{sqrt{2}} ). So, energy for each is ( frac{q^2}{frac{s}{sqrt{2}}} = frac{q^2 sqrt{2}}{s} ). Three such pairs give:[3 times frac{q^2 sqrt{2}}{s} = 3sqrt{2} times frac{q^2}{s}]Then, the remaining three pairs are between the corner charges. Two of these are adjacent (distance ( s )) and one is diagonal (distance ( ssqrt{2} )). So, their energies are:- Two pairs: ( 2 times frac{q^2}{s} )- One pair: ( frac{q^2}{ssqrt{2}} )So, total energy for corner-corner pairs:[2 times frac{q^2}{s} + frac{q^2}{ssqrt{2}} = left(2 + frac{1}{sqrt{2}}right)frac{q^2}{s} = left(2 + frac{sqrt{2}}{2}right)frac{q^2}{s}]Therefore, total energy in the new configuration:[U'_{text{total}} = 3sqrt{2} times frac{q^2}{s} + left(2 + frac{sqrt{2}}{2}right)frac{q^2}{s} = left(3sqrt{2} + 2 + frac{sqrt{2}}{2}right)frac{q^2}{s}]Combine the ( sqrt{2} ) terms:[3sqrt{2} + frac{sqrt{2}}{2} = frac{6sqrt{2} + sqrt{2}}{2} = frac{7sqrt{2}}{2}]So,[U'_{text{total}} = left(frac{7sqrt{2}}{2} + 2right)frac{q^2}{s}]Now, substitute ( frac{q^2}{s} = frac{10(4 - sqrt{2})}{7} ):[U'_{text{total}} = left(frac{7sqrt{2}}{2} + 2right) times frac{10(4 - sqrt{2})}{7}]Simplify:First, factor out the 10 and 7:[U'_{text{total}} = frac{10}{7} times left(frac{7sqrt{2}}{2} + 2right) times (4 - sqrt{2})]Simplify inside the parentheses:[frac{7sqrt{2}}{2} + 2 = frac{7sqrt{2} + 4}{2}]So,[U'_{text{total}} = frac{10}{7} times frac{7sqrt{2} + 4}{2} times (4 - sqrt{2})]The 7 in the denominator cancels with the 7 in the numerator:[U'_{text{total}} = frac{10}{1} times frac{sqrt{2} + frac{4}{7}}{2} times (4 - sqrt{2})]Wait, that might complicate things. Alternatively, let's multiply it out step by step.First, multiply ( frac{7sqrt{2}}{2} + 2 ) with ( 4 - sqrt{2} ):[left(frac{7sqrt{2}}{2} + 2right)(4 - sqrt{2}) = frac{7sqrt{2}}{2} times 4 + frac{7sqrt{2}}{2} times (-sqrt{2}) + 2 times 4 + 2 times (-sqrt{2})]Compute each term:1. ( frac{7sqrt{2}}{2} times 4 = 14sqrt{2} )2. ( frac{7sqrt{2}}{2} times (-sqrt{2}) = -frac{7 times 2}{2} = -7 )3. ( 2 times 4 = 8 )4. ( 2 times (-sqrt{2}) = -2sqrt{2} )Now, add all these together:[14sqrt{2} - 7 + 8 - 2sqrt{2} = (14sqrt{2} - 2sqrt{2}) + (-7 + 8) = 12sqrt{2} + 1]So, the product is ( 12sqrt{2} + 1 ).Now, multiply this by ( frac{10}{7} ):[U'_{text{total}} = frac{10}{7} times (12sqrt{2} + 1) = frac{10(12sqrt{2} + 1)}{7}]So,[U'_{text{total}} = frac{120sqrt{2} + 10}{7} , text{J}]Which matches what I got earlier. So, the increase in energy is:[Delta U = U'_{text{total}} - U_{text{initial}} = frac{120sqrt{2} + 10}{7} - 20]Convert 20 to sevenths:[20 = frac{140}{7}]So,[Delta U = frac{120sqrt{2} + 10 - 140}{7} = frac{120sqrt{2} - 130}{7}]Factor out 10:[Delta U = frac{10(12sqrt{2} - 13)}{7}]Alternatively, we can write this as:[Delta U = frac{10}{7}(12sqrt{2} - 13)]But perhaps the problem expects the answer in a simplified radical form without factoring out 10. So, ( frac{120sqrt{2} - 130}{7} ) is also acceptable.Alternatively, we can rationalize or approximate it numerically, but since the problem asks for the exact value, we'll keep it in terms of ( sqrt{2} ).So, the increase in energy is ( frac{120sqrt{2} - 130}{7} ) Joules.But let me check if this can be simplified further or if I made a miscalculation somewhere.Wait, earlier I thought the increase was approximately 5.67 J, but let's compute ( frac{120sqrt{2} - 130}{7} ):Compute ( 120sqrt{2} approx 120 times 1.4142 approx 169.704 )So,[169.704 - 130 = 39.704]Then,[frac{39.704}{7} approx 5.672 , text{J}]So, approximately 5.67 J increase.But let's see if the exact form can be expressed differently. Alternatively, perhaps the initial approach of considering energy per side was flawed because in the initial configuration, the total energy is not simply four times the energy per side, but rather six pairs. So, maybe the initial assumption that energy per side is 5 J is incorrect.Wait, in the initial problem, the user wrote:"Four identical point charges start at the vertices of a square, and this configuration stores 20 Joules of energy."So, the total energy is 20 J, which includes all six pairs. So, my initial approach was correct in considering all six pairs.But in the initial problem, the user's assistant had a different approach, assuming that the energy per side was 5 J, which is incorrect because in a square, each charge interacts with three others, leading to six pairs, not four.So, the correct approach is to consider all six pairs, which I did.Therefore, the increase in energy is ( frac{120sqrt{2} - 130}{7} ) J, which is approximately 5.67 J.But let's see if this can be simplified or expressed in a more elegant form. Alternatively, perhaps we can factor out a common term.Wait, ( 120sqrt{2} - 130 = 10(12sqrt{2} - 13) ), so:[Delta U = frac{10(12sqrt{2} - 13)}{7}]Alternatively, we can write this as:[Delta U = frac{10}{7}(12sqrt{2} - 13)]But I don't think this simplifies further. So, the exact increase in energy is ( frac{10(12sqrt{2} - 13)}{7} ) Joules.Alternatively, we can rationalize or present it differently, but I think this is as simplified as it gets.So, to summarize:- Initial total energy: 20 J- New total energy: ( frac{120sqrt{2} + 10}{7} ) J- Increase in energy: ( frac{120sqrt{2} - 130}{7} ) J ≈ 5.67 JTherefore, the energy stored increases by approximately 5.67 Joules when one charge is moved to the center of the square.But since the problem asks for the exact value, we'll present it as ( frac{120sqrt{2} - 130}{7} ) Joules.Alternatively, we can factor out a 10:[Delta U = frac{10(12sqrt{2} - 13)}{7}]Which is also acceptable.So, the final answer is ( frac{10(12sqrt{2} - 13)}{7} ) Joules, or approximately 5.67 Joules.

💡Okay, so I have this trigonometry problem here. It says that if sin α plus sin γ equals 2 sin β, then I need to prove that tan((α + β)/2) plus tan((β + γ)/2) equals 2 tan((γ + α)/2). Hmm, that seems a bit involved, but let me try to break it down step by step.First, I remember that there are some trigonometric identities that relate sums of sines and cosines to products. Maybe I can use those to simplify the given equation. The equation is sin α + sin γ = 2 sin β. I think the sum-to-product identity for sine is sin A + sin B = 2 sin((A + B)/2) cos((A - B)/2). Let me apply that here.So, applying the identity to sin α + sin γ, I get:sin α + sin γ = 2 sin((α + γ)/2) cos((α - γ)/2).According to the problem, this equals 2 sin β. So, I can write:2 sin((α + γ)/2) cos((α - γ)/2) = 2 sin β.Dividing both sides by 2 to simplify:sin((α + γ)/2) cos((α - γ)/2) = sin β.Okay, so that's one equation. Now, I need to relate this to the tangent expressions given in the problem. The target equation involves tangents of half-angles, so maybe I should express the sines and cosines in terms of tangents.I recall that tan θ = sin θ / cos θ, but I also remember the tangent half-angle formulas. Let me write those down:tan(θ/2) = sin θ / (1 + cos θ) = (1 - cos θ) / sin θ.Hmm, maybe I can use these to express the tangents in terms of sines and cosines. Let me denote the angles involved in the tangent terms:Let’s define:A = (α + β)/2,B = (β + γ)/2,C = (γ + α)/2.So, the equation I need to prove is:tan A + tan B = 2 tan C.Given that, and knowing that A, B, and C are related to α, β, and γ, maybe I can find a relationship between them.Wait, let's see:A = (α + β)/2,B = (β + γ)/2,C = (γ + α)/2.If I add A and B, I get:A + B = (α + β)/2 + (β + γ)/2 = (α + 2β + γ)/2.Similarly, C = (γ + α)/2.So, A + B = C + β.Interesting. Maybe that can help later.Going back to the equation we had earlier:sin((α + γ)/2) cos((α - γ)/2) = sin β.Let me denote θ = (α + γ)/2 and φ = (α - γ)/2. Then, the equation becomes:sin θ cos φ = sin β.Hmm, so sin θ cos φ = sin β. Maybe I can express sin β in terms of θ and φ.Alternatively, perhaps I can relate θ and φ to the angles A, B, and C.Wait, θ is (α + γ)/2, which is exactly C. So, θ = C.And φ is (α - γ)/2. Hmm, I don't know if that relates directly to A or B.But maybe I can express sin β in terms of θ and φ.Wait, from the equation sin θ cos φ = sin β, so sin β = sin θ cos φ.Is there a way to express β in terms of θ and φ? Maybe not directly, but perhaps using some identities.Alternatively, maybe I can express tan A and tan B in terms of θ and φ.Let me try that.First, let's express tan A:tan A = tan((α + β)/2).Similarly, tan B = tan((β + γ)/2).And tan C = tan((γ + α)/2) = tan θ.So, I need to express tan A + tan B in terms of tan C.Let me recall that tan(A + B) = (tan A + tan B) / (1 - tan A tan B). But in our case, A + B = C + β, so tan(A + B) = tan(C + β).But I'm not sure if that helps directly. Maybe I can find expressions for tan A and tan B in terms of tan C and other terms.Alternatively, perhaps I can use the sine and cosine of A, B, and C to express the tangents.Let me think about expressing tan A and tan B in terms of sine and cosine.tan A = sin A / cos A,tan B = sin B / cos B,tan C = sin C / cos C.So, tan A + tan B = (sin A / cos A) + (sin B / cos B).To combine these, I need a common denominator:= (sin A cos B + sin B cos A) / (cos A cos B).Hmm, the numerator looks familiar. That's the sine of (A + B):sin(A + B) = sin A cos B + cos A sin B.So, tan A + tan B = sin(A + B) / (cos A cos B).Similarly, 2 tan C = 2 sin C / cos C.So, to prove that tan A + tan B = 2 tan C, we need:sin(A + B) / (cos A cos B) = 2 sin C / cos C.From earlier, we have A + B = C + β, so sin(A + B) = sin(C + β).Therefore, sin(C + β) / (cos A cos B) = 2 sin C / cos C.Let me write that:sin(C + β) / (cos A cos B) = 2 sin C / cos C.Cross-multiplying:sin(C + β) cos C = 2 sin C cos A cos B.Hmm, that's a bit complicated. Maybe I can find expressions for cos A and cos B in terms of other variables.Alternatively, let's recall that we have an earlier equation:sin θ cos φ = sin β, where θ = C and φ = (α - γ)/2.So, sin C cos φ = sin β.Maybe I can express sin β in terms of sin C and cos φ.Also, perhaps I can relate cos A and cos B to cos C and other terms.Let me try to express cos A and cos B.cos A = cos((α + β)/2),cos B = cos((β + γ)/2).Hmm, can I express these in terms of θ and φ?Wait, θ = (α + γ)/2, so α = 2θ - γ.Similarly, φ = (α - γ)/2, so α = γ + 2φ.Wait, that might not be helpful directly.Alternatively, maybe I can use the sum and difference formulas.Let me think about cos A:cos A = cos((α + β)/2).Similarly, cos B = cos((β + γ)/2).I wonder if I can express these in terms of θ and φ.Alternatively, perhaps I can use the fact that A + B = C + β, so maybe express cos A and cos B in terms of cos(C + β) or something.Wait, maybe I can use the identity for cos(A + B):cos(A + B) = cos C cos β - sin C sin β.But A + B = C + β, so cos(A + B) = cos(C + β).Wait, that might not help directly.Alternatively, perhaps I can express cos A and cos B in terms of θ and φ.Wait, let me try to express α and γ in terms of θ and φ.We have θ = (α + γ)/2,and φ = (α - γ)/2.So, solving for α and γ:α = θ + φ,γ = θ - φ.So, α = θ + φ,γ = θ - φ.Therefore, β can be expressed in terms of θ and φ from the equation sin θ cos φ = sin β.So, sin β = sin θ cos φ.Hmm, so β is such that sin β = sin θ cos φ.But θ is (α + γ)/2, which is C.So, sin β = sin C cos φ.I wonder if I can express β in terms of C and φ.Alternatively, perhaps I can use the sine addition formula.Wait, sin β = sin θ cos φ.But θ is C, so sin β = sin C cos φ.Hmm, maybe I can write sin β = sin C cos φ.Is there a way to express β in terms of C and φ? Maybe not directly, but perhaps using the sine of sum or difference.Alternatively, perhaps I can use the identity sin β = sin C cos φ to find a relationship between β and C.Wait, if sin β = sin C cos φ, then maybe β is equal to C + something or C - something, but I'm not sure.Alternatively, perhaps I can use the identity sin(A) = sin(B) implies A = B + 2πn or A = π - B + 2πn, but since we're dealing with general angles, maybe that's not directly helpful.Hmm, maybe I need a different approach.Let me go back to the target equation:tan A + tan B = 2 tan C.Expressed as:sin(A + B) / (cos A cos B) = 2 sin C / cos C.From earlier, we have A + B = C + β, so sin(A + B) = sin(C + β).Therefore, the equation becomes:sin(C + β) / (cos A cos B) = 2 sin C / cos C.So, cross-multiplying:sin(C + β) cos C = 2 sin C cos A cos B.Now, let's express sin(C + β) using the sine addition formula:sin(C + β) = sin C cos β + cos C sin β.So, substituting back:(sin C cos β + cos C sin β) cos C = 2 sin C cos A cos B.Expanding the left side:sin C cos β cos C + cos^2 C sin β = 2 sin C cos A cos B.Now, let's recall that from the given equation, sin θ cos φ = sin β, where θ = C and φ = (α - γ)/2.So, sin β = sin C cos φ.Therefore, we can substitute sin β in the equation:sin C cos β cos C + cos^2 C (sin C cos φ) = 2 sin C cos A cos B.Factor out sin C:sin C [cos β cos C + cos^2 C cos φ] = 2 sin C cos A cos B.Assuming sin C ≠ 0, we can divide both sides by sin C:cos β cos C + cos^2 C cos φ = 2 cos A cos B.Hmm, now I need to express cos A and cos B in terms of other variables.Recall that A = (α + β)/2,and B = (β + γ)/2.So, let's express cos A and cos B.Using the cosine of sum formula:cos A = cos((α + β)/2) = cos(α/2 + β/2) = cos(α/2) cos(β/2) - sin(α/2) sin(β/2).Similarly,cos B = cos((β + γ)/2) = cos(β/2 + γ/2) = cos(β/2) cos(γ/2) - sin(β/2) sin(γ/2).Hmm, that seems complicated. Maybe there's a better way.Alternatively, perhaps I can use the product-to-sum identities for cos A cos B.Wait, cos A cos B = [cos(A + B) + cos(A - B)] / 2.We know that A + B = C + β, so:cos A cos B = [cos(C + β) + cos(A - B)] / 2.Similarly, cos(A - B) = cos((α + β)/2 - (β + γ)/2) = cos((α - γ)/2) = cos φ.So, cos A cos B = [cos(C + β) + cos φ] / 2.Therefore, 2 cos A cos B = cos(C + β) + cos φ.So, going back to the equation:cos β cos C + cos^2 C cos φ = 2 cos A cos B = cos(C + β) + cos φ.Therefore, we have:cos β cos C + cos^2 C cos φ = cos(C + β) + cos φ.Let me write that:cos β cos C + cos^2 C cos φ = cos(C + β) + cos φ.Now, let's expand cos(C + β) using the cosine addition formula:cos(C + β) = cos C cos β - sin C sin β.So, substituting back:cos β cos C + cos^2 C cos φ = cos C cos β - sin C sin β + cos φ.Subtract cos C cos β from both sides:cos^2 C cos φ = - sin C sin β + cos φ.Bring all terms to one side:cos^2 C cos φ + sin C sin β - cos φ = 0.Factor out cos φ:cos φ (cos^2 C - 1) + sin C sin β = 0.But cos^2 C - 1 = -sin^2 C, so:cos φ (-sin^2 C) + sin C sin β = 0.Which simplifies to:- sin^2 C cos φ + sin C sin β = 0.Factor out sin C:sin C (-sin C cos φ + sin β) = 0.So, either sin C = 0 or -sin C cos φ + sin β = 0.If sin C = 0, then C = 0 or π, which would mean α + γ = 0 or 2π, which might not be generally true. So, assuming sin C ≠ 0, we have:-sin C cos φ + sin β = 0.Which implies:sin β = sin C cos φ.But wait, that's exactly the equation we had earlier from the given condition! So, this checks out.Therefore, the equation holds true, which means that our initial assumption leads to a valid conclusion.Hence, we have proven that:tan((α + β)/2) + tan((β + γ)/2) = 2 tan((γ + α)/2).**Final Answer**boxed{tan frac{alpha + beta}{2} + tan frac{beta + gamma}{2} = 2 tan frac{gamma + alpha}{2}}