💡First, I need to identify the first ten prime numbers. These are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.Next, I consider the conditions under which the sum of two prime numbers is also a prime number. Since all prime numbers except for 2 are odd, adding two odd numbers results in an even number, which cannot be prime (except for 2). Therefore, one of the primes in the sum must be 2, the only even prime, and the other must be an odd prime.I then examine each possible pair involving the prime number 2 and check if their sum is a prime number:- 2 + 3 = 5 (prime)- 2 + 5 = 7 (prime)- 2 + 11 = 13 (prime)- 2 + 17 = 19 (prime)- 2 + 23 = 25 (not prime)- 2 + 29 = 31 (prime)From these, there are 5 valid pairs where the sum is a prime number: (2, 3), (2, 5), (2, 11), (2, 17), and (2, 29).The total number of possible pairs that can be drawn from the ten prime numbers is calculated using the combination formula ( binom{10}{2} ), which equals 45.Finally, the probability is the number of valid pairs divided by the total number of pairs, resulting in ( frac{5}{45} = frac{1}{9} ).

💡To find the value of ( a ) in the sequence ([..., a, b, c, d, 0, 2, 2, 4, 6, 10, 16, ...]) where each term is the sum of the two terms to its left, we can work backwards from the known terms.1. **Finding ( d ):** The term after ( d ) is ( 0 ) and the term following ( 0 ) is ( 2 ). Therefore: [ d + 0 = 2 implies d = 2 ]2. **Finding ( c ):** The term after ( c ) is ( d = 2 ) and the term following ( 2 ) is ( 2 ). Therefore: [ c + 2 = 2 implies c = 0 ]3. **Finding ( b ):** The term after ( b ) is ( c = 0 ) and the term following ( 0 ) is ( 4 ). Therefore: [ b + 0 = 4 implies b = 4 ]4. **Finding ( a ):** The term after ( a ) is ( b = 4 ) and the term following ( 4 ) is ( 0 ). Therefore: [ a + 4 = 0 implies a = -4 ]Thus, the value of ( a ) is (-4).

💡Okay, so I have this problem where I need to find f(1) given that f is a smooth function from R to R, f'(x) = f(1 - x) for all x, and f(0) = 1. Hmm, interesting. Let me try to figure this out step by step.First, I know that f is smooth, which means it's infinitely differentiable. That's good because it means I can take derivatives as needed without worrying about the function not being differentiable. The condition given is f'(x) = f(1 - x). So, the derivative of f at any point x is equal to the function evaluated at 1 - x. That seems like a functional equation involving derivatives. I think I can approach this by taking derivatives multiple times and seeing if I can find a pattern or maybe a differential equation that I can solve.Let me write down what I know:1. f'(x) = f(1 - x) for all x.2. f(0) = 1.I need to find f(1). Maybe if I can express f(1) in terms of other values or derivatives, I can find an equation to solve for it.Let me try taking the derivative of both sides of the equation f'(x) = f(1 - x). If I differentiate both sides with respect to x, I get:f''(x) = d/dx [f(1 - x)].Using the chain rule on the right side, the derivative of f(1 - x) with respect to x is f'(1 - x) * (-1). So,f''(x) = -f'(1 - x).But from the original equation, f'(1 - x) = f(1 - (1 - x)) = f(x). So substituting that in,f''(x) = -f(x).Wait, that's interesting. So, f''(x) = -f(x). That's a second-order linear differential equation. The general solution to f''(x) = -f(x) is f(x) = A cos(x) + B sin(x), where A and B are constants. So, f(x) must be a combination of sine and cosine functions.So, f(x) = A cos(x) + B sin(x). Now, let's find A and B using the given conditions.First, we know that f(0) = 1. Let's plug x = 0 into the general solution:f(0) = A cos(0) + B sin(0) = A * 1 + B * 0 = A.So, A = 1.Now, we need another condition to find B. Let's use the original equation f'(x) = f(1 - x). Let's compute f'(x) using the general solution.f'(x) = -A sin(x) + B cos(x).But from the original condition, f'(x) = f(1 - x). So,f'(x) = f(1 - x) = A cos(1 - x) + B sin(1 - x).So, we have:-A sin(x) + B cos(x) = A cos(1 - x) + B sin(1 - x).Hmm, this equation must hold for all x. Let me see if I can simplify this or find a relationship between A and B.First, let's plug in x = 0 into this equation to see if that helps:Left side: -A sin(0) + B cos(0) = 0 + B * 1 = B.Right side: A cos(1 - 0) + B sin(1 - 0) = A cos(1) + B sin(1).So, we have:B = A cos(1) + B sin(1).But we already know that A = 1, so substituting that in:B = cos(1) + B sin(1).Let's solve for B:B - B sin(1) = cos(1)B (1 - sin(1)) = cos(1)So,B = cos(1) / (1 - sin(1)).Alright, so now we have both A and B. Therefore, the function f(x) is:f(x) = cos(x) + [cos(1) / (1 - sin(1))] sin(x).Now, we need to find f(1). Let's plug x = 1 into the function:f(1) = cos(1) + [cos(1) / (1 - sin(1))] sin(1).Let me compute this:f(1) = cos(1) + [cos(1) sin(1)] / (1 - sin(1)).I can factor out cos(1):f(1) = cos(1) [1 + sin(1) / (1 - sin(1))].Let me combine the terms inside the brackets:1 + sin(1)/(1 - sin(1)) = [ (1 - sin(1)) + sin(1) ] / (1 - sin(1)) = 1 / (1 - sin(1)).So, f(1) = cos(1) * [1 / (1 - sin(1))] = cos(1) / (1 - sin(1)).Therefore, f(1) is equal to cos(1) divided by (1 - sin(1)).Wait, let me double-check my steps to make sure I didn't make any mistakes.1. I started with f'(x) = f(1 - x) and took the second derivative to get f''(x) = -f(x). That seems correct.2. The general solution f(x) = A cos(x) + B sin(x) is correct for f''(x) = -f(x).3. Using f(0) = 1, I found A = 1. That seems right.4. Then, using f'(x) = f(1 - x), I computed f'(x) as -A sin(x) + B cos(x) and set it equal to f(1 - x) = A cos(1 - x) + B sin(1 - x). That's correct.5. Plugging x = 0 into that equation gave me B = A cos(1) + B sin(1). Substituting A = 1, I solved for B and got B = cos(1)/(1 - sin(1)). That seems correct.6. Then, plugging x = 1 into f(x), I substituted into the general solution and simplified. The steps seem correct, and I ended up with f(1) = cos(1)/(1 - sin(1)).I think that's the correct answer. Let me just verify if this makes sense. If I plug x = 1 into f'(x) = f(1 - x), then f'(1) = f(0) = 1. Let's check if that's consistent with our function.From f(x) = cos(x) + [cos(1)/(1 - sin(1))] sin(x), f'(x) = -sin(x) + [cos(1)/(1 - sin(1))] cos(x).So, f'(1) = -sin(1) + [cos(1)/(1 - sin(1))] cos(1).Let me compute that:f'(1) = -sin(1) + [cos^2(1)] / (1 - sin(1)).I can write this as:f'(1) = [ -sin(1)(1 - sin(1)) + cos^2(1) ] / (1 - sin(1)).Expanding the numerator:-sin(1) + sin^2(1) + cos^2(1).But sin^2(1) + cos^2(1) = 1, so numerator becomes:-sin(1) + 1.Therefore,f'(1) = (1 - sin(1)) / (1 - sin(1)) = 1.Which matches the condition f'(1) = f(0) = 1. So, that checks out. Therefore, my answer seems consistent.I think I've got it right. So, f(1) is cos(1) divided by (1 - sin(1)).**Final Answer**The value of ( f(1) ) is boxed{dfrac{cos 1}{1 - sin 1}}.

💡Okay, so I have this problem where I need to find the maximum value of y/x for real numbers x and y that satisfy the equation x² + y² - 6x - 6y + 12 = 0. Hmm, let me try to figure this out step by step.First, the equation given is x² + y² - 6x - 6y + 12 = 0. I remember that equations of circles can be written in the form (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. Maybe I can rewrite the given equation in that form.Let me complete the squares for both x and y terms. Starting with the x terms: x² - 6x. To complete the square, I take half of -6, which is -3, and square it to get 9. So, x² - 6x becomes (x - 3)² - 9.Similarly, for the y terms: y² - 6y. Half of -6 is -3, squared is 9. So, y² - 6y becomes (y - 3)² - 9.Now, substituting these back into the original equation:(x - 3)² - 9 + (y - 3)² - 9 + 12 = 0Simplify the constants: -9 -9 +12 = -6So, the equation becomes:(x - 3)² + (y - 3)² - 6 = 0Which means:(x - 3)² + (y - 3)² = 6Alright, so this is a circle with center at (3, 3) and radius √6. Got that.Now, I need to find the maximum value of y/x for points (x, y) on this circle. Hmm, y/x is the slope of the line connecting the origin (0,0) to the point (x, y). So, essentially, I need to find the line through the origin that touches the circle and has the maximum slope.I remember that the maximum slope will occur where the line is tangent to the circle. So, I need to find the tangent lines from the origin to the circle and then determine their slopes.Let me denote the slope as k. So, the equation of the line through the origin with slope k is y = kx.This line should be tangent to the circle (x - 3)² + (y - 3)² = 6. The condition for a line to be tangent to a circle is that the distance from the center of the circle to the line equals the radius.The formula for the distance from a point (h, k) to the line ax + by + c = 0 is |ah + bk + c| / sqrt(a² + b²). In this case, the line is y = kx, which can be rewritten as kx - y = 0. So, a = k, b = -1, c = 0.The center of the circle is (3, 3). Plugging into the distance formula:|k*3 - 1*3 + 0| / sqrt(k² + (-1)²) = |3k - 3| / sqrt(k² + 1)This distance should equal the radius, which is √6. So, we have:|3k - 3| / sqrt(k² + 1) = √6Let me drop the absolute value by squaring both sides to eliminate the square roots:(3k - 3)² / (k² + 1) = 6Expanding the numerator:(9k² - 18k + 9) / (k² + 1) = 6Multiply both sides by (k² + 1):9k² - 18k + 9 = 6k² + 6Bring all terms to one side:9k² - 18k + 9 - 6k² - 6 = 0Simplify:3k² - 18k + 3 = 0Divide the entire equation by 3:k² - 6k + 1 = 0Now, solve this quadratic equation for k. Using the quadratic formula:k = [6 ± sqrt(36 - 4*1*1)] / 2 = [6 ± sqrt(32)] / 2 = [6 ± 4√2] / 2 = 3 ± 2√2So, the two possible slopes are k = 3 + 2√2 and k = 3 - 2√2. Since we are looking for the maximum value of y/x, which corresponds to the maximum slope, we take the larger value, which is 3 + 2√2.Let me just verify if this makes sense. The circle is centered at (3,3) with radius √6. The origin is at (0,0), which is outside the circle because the distance from (0,0) to (3,3) is √(9 + 9) = √18 ≈ 4.24, which is greater than √6 ≈ 2.45. So, there should be two tangent lines from the origin to the circle, and the slopes of these lines are 3 ± 2√2. Since 3 + 2√2 is approximately 3 + 2.828 ≈ 5.828, which is positive, and 3 - 2√2 is approximately 0.172, which is also positive. So, both slopes are positive, which makes sense because the circle is in the first quadrant.Therefore, the maximum value of y/x is 3 + 2√2.**Final Answer**The maximum value of frac{y}{x} is boxed{3 + 2sqrt{2}}.

💡First, calculate the total number of connections if each of the 30 switches connects to exactly 4 other switches. This gives 30 × 4 = 120 potential connections. Since each connection is shared between two switches, divide by 2 to get 60 unique connections.Next, account for the additional connections required for the 5 crucial switches. Each crucial switch needs one extra connection, adding 5 more connections.Finally, add the initial 60 connections to the 5 extra connections to determine the total number of connections needed.

💡First, I notice that the expression contains multiple factorial terms and coefficients. To simplify the calculation, I can look for common factors or patterns that allow me to combine terms.I start by examining the terms (6 cdot 6!), (5 cdot 5!), and another (5 cdot 5!). I observe that (6! = 6 cdot 5!), which means I can express (6 cdot 6!) as (6 cdot 6 cdot 5! = 36 cdot 5!).Next, I combine the two (5 cdot 5!) terms, which adds up to (10 cdot 5!).Now, the expression becomes (36 cdot 5! + 10 cdot 5!). Since both terms have a common factor of (5!), I can factor that out:[(36 + 10) cdot 5! = 46 cdot 5!]Finally, I calculate (5! = 120) and multiply it by 46:[46 cdot 120 = 5520]Therefore, the final result is 5520.