💡First, calculate the cost of raising the child until the age of 18.For the first 8 years, the annual cost is 10,000:8 years * 10,000/year = 80,000From age 9 to 18, the annual cost is 20,000:10 years * 20,000/year = 200,000Total cost to raise the child until age 18:80,000 + 200,000 = 280,000John pays half of this cost:280,000 / 2 = 140,000The total cost including university tuition is 265,000. Subtract John's contribution to find the university tuition cost:265,000 - 140,000 = 125,000

💡First, I need to determine the percentage of the budget allocated to basic astrophysics using the given 39.6 degrees. Since a full circle is 360 degrees, I can set up a proportion to find the percentage for basic astrophysics.Next, I'll calculate the percentage for basic astrophysics by solving the proportion.After finding the percentage for basic astrophysics, I'll add up all the known percentages to find the total allocated so far.Finally, I'll subtract the total known percentages from 100% to determine the percentage allocated to food additives.

💡Okay, so I have this question here about functions and whether they represent the same function. There are four options: A, B, C, and D. Let me go through each one step by step to figure out which ones are the same.Starting with option A: f(x) = t + 1 and g(x) = (x² + x)/x. Hmm, wait, f(x) is given in terms of t, but g(x) is in terms of x. That seems a bit confusing. Maybe that's a typo? Or maybe t is supposed to be x? If f(x) is actually supposed to be x + 1, then let me consider that. So f(x) = x + 1, and g(x) = (x² + x)/x. Let me simplify g(x). If I factor the numerator, it's x(x + 1), so g(x) = x(x + 1)/x. The x cancels out, so g(x) = x + 1. That seems like f(x) and g(x) are the same, right? But wait, hold on. The domain of f(x) is all real numbers because it's a linear function. However, for g(x), the original expression is (x² + x)/x. That means x cannot be zero because division by zero is undefined. So the domain of g(x) is all real numbers except x = 0. Since the domains are different, even though the simplified expressions are the same, they aren't the same function. So option A is not correct.Moving on to option B: f(x) = t + 1 and g(x) = (x² + x)/x. Wait, this is the same as option A. Did I miss something? Let me double-check. If f(x) is t + 1, that still doesn't make much sense. Maybe it's another typo. If f(x) is supposed to be x²/(√x)², that simplifies to x²/x = x. So f(x) = x, and g(x) is (x² + x)/x, which simplifies to x + 1. So f(x) = x and g(x) = x + 1. These are different functions because their outputs are different for any x. So option B is also not correct.Now, option C: f(x) = |x| and g(x) = nth root of xⁿ. Let me think about this. The nth root of xⁿ depends on whether n is even or odd. If n is even, then the nth root of xⁿ is |x| because even roots can't be negative. If n is odd, then the nth root of xⁿ is just x because odd roots can handle negative numbers. So g(x) is equal to |x| only when n is even. But since n isn't specified, we can't be sure. Therefore, f(x) and g(x) might not always be the same. So option C is not necessarily the same function.Finally, option D: f(x) = x and g(t) = (t³ + t)/(t² + 1). Let me simplify g(t). The numerator is t³ + t, which can be factored as t(t² + 1). So g(t) = t(t² + 1)/(t² + 1). The (t² + 1) terms cancel out, leaving g(t) = t. So g(t) simplifies to t, which is the same as f(x) = x. Now, checking the domains: f(x) = x is defined for all real numbers, and g(t) is also defined for all real numbers because the denominator t² + 1 is never zero. So both the expressions and the domains are the same. Therefore, f(x) and g(t) are the same function.So after going through all the options, only option D represents the same function.

💡Okay, so I have this function f(x) = √3 sin x - cos x, and I need to find its range when x is between -π/2 and π/2. Hmm, I remember that functions involving sine and cosine can often be rewritten using some kind of identity to make them easier to analyze. Maybe I can express this as a single sine or cosine function with a phase shift. Let me think about how to do that.First, I recall that any expression of the form a sin x + b cos x can be written as R sin(x + φ) or R cos(x + φ), where R is the amplitude and φ is the phase shift. I think the formula for R is √(a² + b²). Let me check that. If I have a sin x + b cos x, then R = √(a² + b²) and tan φ = b/a or something like that. Yeah, I think that's right.In this case, my function is √3 sin x - cos x. So, comparing to a sin x + b cos x, a is √3 and b is -1. So, R should be √[(√3)² + (-1)²] = √[3 + 1] = √4 = 2. Okay, so R is 2. That means I can write f(x) as 2 sin(x + φ) or 2 cos(x + φ). Let me figure out which one it is and what φ is.Wait, actually, it can be either sine or cosine, but I think sine might be easier here. Let me try writing it as 2 sin(x + φ). The expansion of 2 sin(x + φ) is 2 sin x cos φ + 2 cos x sin φ. Comparing this to √3 sin x - cos x, I can set up the equations:2 cos φ = √3 (coefficient of sin x)2 sin φ = -1 (coefficient of cos x)So, from the first equation, cos φ = √3 / 2, and from the second equation, sin φ = -1/2. Hmm, what angle φ has cosine √3/2 and sine -1/2? That should be φ = -π/6 because cos(-π/6) = √3/2 and sin(-π/6) = -1/2. So, φ is -π/6.Therefore, f(x) can be rewritten as 2 sin(x - π/6). Let me verify that:2 sin(x - π/6) = 2 [sin x cos(π/6) - cos x sin(π/6)] = 2 [sin x (√3/2) - cos x (1/2)] = √3 sin x - cos x. Yep, that matches the original function. Great, so f(x) = 2 sin(x - π/6).Now, I need to find the range of f(x) when x is in [-π/2, π/2]. Since f(x) is a sine function with amplitude 2, its maximum value is 2 and its minimum is -2, but I have to consider the domain of x to see if these extremes are actually achieved within the given interval.Let me think about the argument of the sine function, which is x - π/6. When x is in [-π/2, π/2], then x - π/6 is in [-π/2 - π/6, π/2 - π/6] = [-2π/3, π/3]. So, the argument of the sine function ranges from -2π/3 to π/3.Now, let's consider the sine function over this interval. The sine function reaches its maximum at π/2 and its minimum at -π/2. But our interval is from -2π/3 to π/3. Let me see where the maximum and minimum occur within this interval.First, the maximum value of sin θ is 1, which occurs at θ = π/2. But π/2 is approximately 1.5708, and our upper limit is π/3, which is approximately 1.0472. So, π/2 is outside our interval. Therefore, the maximum value of sin θ in our interval will occur at the upper limit, θ = π/3. Let me compute sin(π/3): that's √3/2 ≈ 0.866. So, the maximum value of sin θ in this interval is √3/2, and thus the maximum value of f(x) is 2*(√3/2) = √3.Now, for the minimum value. The sine function reaches its minimum at θ = -π/2, which is -1. But our lower limit is θ = -2π/3, which is approximately -2.0944. Since -π/2 is approximately -1.5708, which is greater than -2π/3, the minimum of sin θ in our interval will occur at θ = -2π/3. Let me compute sin(-2π/3): that's sin(2π/3) with a negative sign, so sin(2π/3) is √3/2, so sin(-2π/3) is -√3/2 ≈ -0.866. Therefore, the minimum value of sin θ is -√3/2, and thus the minimum value of f(x) is 2*(-√3/2) = -√3.Wait, but hold on. The function f(x) is 2 sin(x - π/6). So, if the maximum of sin is √3/2, then f(x) max is √3, and the minimum is -√3. But wait, I thought earlier that the amplitude is 2, so shouldn't the range be from -2 to 2? But apparently, within the given domain, it doesn't reach -2 or 2.Wait, maybe I made a mistake. Let me double-check. The function is 2 sin(x - π/6). The amplitude is 2, so theoretically, the range should be [-2, 2]. But because x is restricted to [-π/2, π/2], the argument x - π/6 is restricted to [-2π/3, π/3]. So, sin(x - π/6) will not reach its full range of [-1, 1] within this interval.Wait, let me compute sin(-2π/3). That's sin(2π/3) with a negative sign, which is -√3/2 ≈ -0.866. So, the minimum value of sin(x - π/6) is -√3/2, and the maximum is sin(π/3) = √3/2. Therefore, f(x) ranges from 2*(-√3/2) = -√3 to 2*(√3/2) = √3. So, the range is [-√3, √3].But wait, that can't be right because when x is π/2, x - π/6 is π/2 - π/6 = π/3, and sin(π/3) is √3/2, so f(x) is √3. When x is -π/2, x - π/6 is -π/2 - π/6 = -2π/3, and sin(-2π/3) is -√3/2, so f(x) is -√3. So, the range is indeed [-√3, √3].But wait, the options given are A: [-1,1], B: [-2,1], C: [-2,√3], D: [-1,√3]. Hmm, none of these options is [-√3, √3]. So, maybe I did something wrong.Wait, let me check my steps again. Maybe I made a mistake in rewriting the function. Let me go back.Original function: f(x) = √3 sin x - cos x.I rewrote it as 2 sin(x - π/6). Let me verify that:2 sin(x - π/6) = 2 [sin x cos(π/6) - cos x sin(π/6)] = 2 [sin x (√3/2) - cos x (1/2)] = √3 sin x - cos x. Yes, that's correct.So, f(x) = 2 sin(x - π/6). The amplitude is 2, but due to the domain restriction, the range is limited.Wait, but when x is in [-π/2, π/2], x - π/6 is in [-2π/3, π/3]. So, the sine function over this interval.Wait, maybe I need to consider the critical points within this interval. Let me find the maximum and minimum values by taking the derivative.f(x) = 2 sin(x - π/6). The derivative f’(x) = 2 cos(x - π/6). Setting this equal to zero, cos(x - π/6) = 0. So, x - π/6 = π/2 + kπ. Solving for x, x = π/2 + π/6 + kπ = 2π/3 + kπ. But within our domain x ∈ [-π/2, π/2], 2π/3 is approximately 2.094, which is outside the domain. Similarly, x = -π/2 + π/6 = -π/3, which is within the domain.So, the critical point is at x = -π/3. Let's evaluate f(x) at x = -π/3:f(-π/3) = 2 sin(-π/3 - π/6) = 2 sin(-π/2) = 2*(-1) = -2.Wait, so at x = -π/3, f(x) = -2. That's lower than -√3, which is approximately -1.732. So, that's a lower value.Wait, but earlier I thought that the minimum was at x = -π/2, giving f(x) = -√3. But now, by taking the derivative, I find that at x = -π/3, f(x) = -2, which is lower. So, that must be the minimum.Similarly, let's check the endpoints:At x = π/2, f(x) = 2 sin(π/2 - π/6) = 2 sin(π/3) = 2*(√3/2) = √3.At x = -π/2, f(x) = 2 sin(-π/2 - π/6) = 2 sin(-2π/3) = 2*(-√3/2) = -√3.But wait, at x = -π/3, f(x) = -2, which is less than -√3. So, the minimum is -2, and the maximum is √3.Therefore, the range of f(x) is [-2, √3], which corresponds to option C.Wait, but earlier when I thought the range was [-√3, √3], that was incorrect because I didn't consider the critical point inside the interval. By taking the derivative, I found that the function reaches -2 at x = -π/3, which is within the domain. So, the correct range is from -2 to √3.Therefore, the answer should be option C: [-2, √3].I think that's it. So, the range is from -2 to √3.

💡To find the smallest positive prime number that is 20 less than a perfect square, I'll start by considering perfect squares and subtracting 20 from them.First, I'll list some perfect squares:- (5^2 = 25)- (7^2 = 49)- (9^2 = 81)Next, I'll subtract 20 from each of these perfect squares to see if the result is a prime number:- (25 - 20 = 5), which is prime.- (49 - 20 = 29), which is also prime.- (81 - 20 = 61), another prime.Among these results, the smallest prime number is 5. Therefore, the smallest positive prime number that is 20 less than a perfect square is 5.

💡First, I need to determine the interval ((a, b)) where the equation (x^3 - x + 1 = 0) has a root. Since (a) and (b) are integers and (b - a = 1), the interval is between two consecutive integers.I'll evaluate the function (f(x) = x^3 - x + 1) at several integer points to identify where the function changes sign, indicating a root in that interval.1. **Evaluate at (x = -2):** [ f(-2) = (-2)^3 - (-2) + 1 = -8 + 2 + 1 = -5 ] The function value is negative.2. **Evaluate at (x = -1):** [ f(-1) = (-1)^3 - (-1) + 1 = -1 + 1 + 1 = 1 ] The function value is positive.Since (f(-2) = -5 < 0) and (f(-1) = 1 > 0), the function changes sign between (x = -2) and (x = -1). By the Intermediate Value Theorem, there is at least one root in the interval ((-2, -1)).Thus, (a = -2) and (b = -1). Adding these together:[a + b = -2 + (-1) = -3]