💡Okay, so I have this function f(x) = log₂(|x+1| + |x-2| - m). There are two parts to the problem. Let me tackle them one by one.Starting with part (1): When m = 7, find the domain of f(x). Hmm, the domain of a logarithmic function is where the argument is positive. So, I need to find all x such that |x+1| + |x-2| - 7 > 0. That simplifies to |x+1| + |x-2| > 7.Alright, to solve this inequality, I should consider the expression |x+1| + |x-2|. This is the sum of two absolute values, which can be tricky because absolute value functions have different behaviors depending on the value of x. I remember that for expressions like |x - a| + |x - b|, the minimum value occurs at some point between a and b, depending on the specific values.In this case, the critical points are at x = -1 and x = 2. So, I should break down the problem into intervals based on these critical points: x < -1, -1 ≤ x < 2, and x ≥ 2.Let me handle each interval separately.1. **Case 1: x < -1** In this interval, both x + 1 and x - 2 are negative. So, |x+1| = -(x + 1) = -x - 1, and |x-2| = -(x - 2) = -x + 2. Adding them together: (-x - 1) + (-x + 2) = -2x + 1. So, the inequality becomes -2x + 1 > 7. Subtracting 1 from both sides: -2x > 6. Dividing by -2 (and remembering to reverse the inequality sign): x < -3. So, in this interval, x must be less than -3. Since we're already considering x < -1, the solution here is x < -3.2. **Case 2: -1 ≤ x < 2** In this interval, x + 1 is non-negative, so |x+1| = x + 1. However, x - 2 is still negative, so |x-2| = -(x - 2) = -x + 2. Adding them together: (x + 1) + (-x + 2) = 3. So, the inequality becomes 3 > 7, which simplifies to 3 > 7. That's obviously false. So, there are no solutions in this interval.3. **Case 3: x ≥ 2** In this interval, both x + 1 and x - 2 are non-negative. So, |x+1| = x + 1, and |x-2| = x - 2. Adding them together: (x + 1) + (x - 2) = 2x - 1. So, the inequality becomes 2x - 1 > 7. Adding 1 to both sides: 2x > 8. Dividing by 2: x > 4. So, in this interval, x must be greater than 4. Since we're considering x ≥ 2, the solution here is x > 4.Putting it all together, the solutions are x < -3 or x > 4. Therefore, the domain of f(x) when m = 7 is (-∞, -3) ∪ (4, ∞).Wait, let me double-check that. When x is between -1 and 2, the sum of the absolute values is always 3, which is less than 7, so indeed, there are no solutions there. For x < -1, we found x < -3, and for x ≥ 2, x > 4. That seems correct.Moving on to part (2): If the solution set of the inequality f(x) ≥ 2 is ℝ, find the range of values for m.So, f(x) ≥ 2 implies log₂(|x+1| + |x-2| - m) ≥ 2. Since log₂ is an increasing function, this inequality holds when the argument is at least 2² = 4. Therefore, |x+1| + |x-2| - m ≥ 4, which simplifies to |x+1| + |x-2| ≥ m + 4.The problem states that the solution set is ℝ, meaning that for all real numbers x, the inequality |x+1| + |x-2| ≥ m + 4 must hold. Therefore, the minimum value of |x+1| + |x-2| must be greater than or equal to m + 4.So, I need to find the minimum value of |x+1| + |x-2|. Let me think about how to find this minimum.I recall that the expression |x - a| + |x - b| represents the sum of distances from x to a and x to b on the real line. The minimum of this sum occurs when x is between a and b, and the minimum value is |a - b|.In this case, a = -1 and b = 2. So, the minimum value of |x+1| + |x-2| is |(-1) - 2| = | -3 | = 3.Therefore, the minimum value is 3. So, for the inequality |x+1| + |x-2| ≥ m + 4 to hold for all x, we must have 3 ≥ m + 4.Solving for m: 3 ≥ m + 4 ⇒ m ≤ 3 - 4 ⇒ m ≤ -1.So, the range of values for m is all real numbers less than or equal to -1.Wait, let me verify that. The minimum of |x+1| + |x-2| is 3, so to have 3 ≥ m + 4, m must be ≤ -1. That seems correct.Therefore, the range of m is (-∞, -1].I think that's all. Let me just recap:1. For m = 7, the domain is where |x+1| + |x-2| > 7, which gives x < -3 or x > 4.2. For f(x) ≥ 2 to hold for all x, the minimum of |x+1| + |x-2| must be ≥ m + 4, which gives m ≤ -1.Yeah, that makes sense.

💡Okay, so I have this problem with three equations, and I need to figure out the pattern and then prove it using mathematical induction. Let me try to understand what's going on here.First, let me write down the given equations again to see the pattern clearly:1. (dfrac{1}{1 times 3} = dfrac{1}{2 times 1 + 1})2. (dfrac{1}{1 times 3} + dfrac{1}{3 times 5} = dfrac{2}{2 times 2 + 1})3. (dfrac{1}{1 times 3} + dfrac{1}{3 times 5} + dfrac{1}{5 times 7} = dfrac{3}{2 times 3 + 1})Hmm, okay. So each equation is adding up fractions where the denominators are products of consecutive odd numbers. The first term is (1 times 3), the next is (3 times 5), then (5 times 7), and so on. On the right side, each equation equals a fraction where the numerator is the number of terms added, and the denominator is (2n + 1), where (n) is the number of terms.Let me check:- For the first equation, there's 1 term, and the right side is (dfrac{1}{2 times 1 + 1} = dfrac{1}{3}), which matches the left side.- For the second equation, there are 2 terms, and the right side is (dfrac{2}{2 times 2 + 1} = dfrac{2}{5}). Let me compute the left side: (dfrac{1}{3} + dfrac{1}{15} = dfrac{5}{15} + dfrac{1}{15} = dfrac{6}{15} = dfrac{2}{5}). That works.- For the third equation, 3 terms, right side is (dfrac{3}{7}). Left side: (dfrac{1}{3} + dfrac{1}{15} + dfrac{1}{35}). Let me compute that:Convert to common denominator, which is 105:(dfrac{35}{105} + dfrac{7}{105} + dfrac{3}{105} = dfrac{45}{105} = dfrac{9}{21} = dfrac{3}{7}). Perfect.So, the pattern seems to be that if we sum the reciprocals of the products of consecutive odd numbers starting from 1×3 up to (2n-1)(2n+1), the sum equals (dfrac{n}{2n + 1}).So, for part (1), the conjecture is:[dfrac{1}{1 times 3} + dfrac{1}{3 times 5} + cdots + dfrac{1}{(2n - 1)(2n + 1)} = dfrac{n}{2n + 1}]Now, moving on to part (2), which is to prove this conjecture using mathematical induction.I remember that mathematical induction has two main steps: the base case and the inductive step.**Base Case (n = 1):**Let's verify if the equation holds for n = 1.Left side: (dfrac{1}{1 times 3} = dfrac{1}{3})Right side: (dfrac{1}{2 times 1 + 1} = dfrac{1}{3})They are equal, so the base case holds.**Inductive Step:**Assume that the equation holds for some arbitrary positive integer k. That is, assume:[dfrac{1}{1 times 3} + dfrac{1}{3 times 5} + cdots + dfrac{1}{(2k - 1)(2k + 1)} = dfrac{k}{2k + 1}]Now, we need to show that the equation holds for n = k + 1, i.e.,[dfrac{1}{1 times 3} + dfrac{1}{3 times 5} + cdots + dfrac{1}{(2k - 1)(2k + 1)} + dfrac{1}{(2(k + 1) - 1)(2(k + 1) + 1)} = dfrac{k + 1}{2(k + 1) + 1}]Simplify the last term:[dfrac{1}{(2(k + 1) - 1)(2(k + 1) + 1)} = dfrac{1}{(2k + 1)(2k + 3)}]So, the left side for n = k + 1 is:[left( dfrac{1}{1 times 3} + dfrac{1}{3 times 5} + cdots + dfrac{1}{(2k - 1)(2k + 1)} right) + dfrac{1}{(2k + 1)(2k + 3)}]By the inductive hypothesis, the sum up to k terms is (dfrac{k}{2k + 1}). So, substitute that in:[dfrac{k}{2k + 1} + dfrac{1}{(2k + 1)(2k + 3)}]Now, let's combine these two fractions. The common denominator is ((2k + 1)(2k + 3)):[dfrac{k(2k + 3)}{(2k + 1)(2k + 3)} + dfrac{1}{(2k + 1)(2k + 3)} = dfrac{k(2k + 3) + 1}{(2k + 1)(2k + 3)}]Expand the numerator:[k(2k + 3) + 1 = 2k^2 + 3k + 1]Factor the numerator if possible. Let me see:Looking at (2k^2 + 3k + 1), can this be factored?Yes, because 2k^2 + 3k + 1 = (2k + 1)(k + 1). Let me check:[(2k + 1)(k + 1) = 2k(k) + 2k(1) + 1(k) + 1(1) = 2k^2 + 2k + k + 1 = 2k^2 + 3k + 1]Perfect, so the numerator factors as (2k + 1)(k + 1). Therefore, the fraction becomes:[dfrac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)} = dfrac{k + 1}{2k + 3}]Simplify the denominator:[2k + 3 = 2(k + 1) + 1]So, the right side is:[dfrac{k + 1}{2(k + 1) + 1}]Which is exactly the form we wanted for n = k + 1.Therefore, if the equation holds for n = k, it also holds for n = k + 1.**Conclusion:**Since the base case holds and the inductive step has been proven, by mathematical induction, the conjectured equation is valid for all positive integers n.**Final Answer**The conjectured equation is valid for all positive integers ( n ), so the final answer is boxed{dfrac{n}{2n + 1}}.

💡Okay, so I have this equation: 30x² + 13 = 47x - 2. And I know that x = 3/5 is a solution to this equation. I need to find the other value of x that solves it. Hmm, let me think about how to approach this.First, I remember that quadratic equations usually have two solutions. Since I already know one solution, maybe I can use that to find the other one. But let me start by writing the equation in standard form, which is ax² + bx + c = 0. That might make it easier to work with.So, starting with 30x² + 13 = 47x - 2. I need to move all terms to one side. Let me subtract 47x and add 2 to both sides to get everything on the left side. That would look like:30x² + 13 - 47x + 2 = 0Wait, that doesn't seem right. Let me double-check. If I subtract 47x and add 2, it should be:30x² + 13 - 47x + 2 = 0But 13 + 2 is 15, so combining like terms, I get:30x² - 47x + 15 = 0Okay, that looks better. So the equation in standard form is 30x² - 47x + 15 = 0.Now, since I know that x = 3/5 is a solution, I can use this information to factor the quadratic equation. Factoring quadratics can sometimes be tricky, but knowing one root can help me find the other.I remember that if a quadratic equation has roots r and s, it can be written as a(x - r)(x - s) = 0, where a is the coefficient of x². In this case, a is 30. So, the equation can be expressed as 30(x - 3/5)(x - s) = 0, where s is the other root I'm trying to find.But maybe it's easier to factor it without the 30 first and then adjust. Let me think. If x = 3/5 is a root, then (x - 3/5) is a factor. However, since the coefficient of x² is 30, which is not 1, I need to consider how to factor it properly.Alternatively, I can use the fact that for a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a and the product of the roots is c/a. Maybe that's a quicker way since I already know one root.Let me recall: if the roots are r and s, then r + s = -b/a and r * s = c/a.Given that, let's compute the sum and product of the roots.Here, a = 30, b = -47, c = 15.So, the sum of the roots is -b/a = -(-47)/30 = 47/30.The product of the roots is c/a = 15/30 = 1/2.I know one root is 3/5, so let's denote r = 3/5 and s as the other root.So, r + s = 47/30Therefore, s = 47/30 - r = 47/30 - 3/5But I need to subtract these fractions. Let me find a common denominator. 30 is a common denominator.3/5 is equal to 18/30.So, s = 47/30 - 18/30 = (47 - 18)/30 = 29/30Wait, that doesn't seem right because when I multiply the roots, it should be 1/2.Let me check: r * s = (3/5) * (29/30) = (87)/150 = 29/50, which is not equal to 1/2.Hmm, that's a problem. That means I made a mistake somewhere.Wait, maybe I messed up the sum of the roots. Let me double-check.The sum of the roots is -b/a. In the equation 30x² - 47x + 15 = 0, b is -47, so -b is 47. Therefore, the sum is 47/30.Yes, that's correct.But then, if r = 3/5 = 18/30, then s = 47/30 - 18/30 = 29/30.But 29/30 is approximately 0.9667, and 3/5 is 0.6. Multiplying them gives approximately 0.58, which is close to 1/2 (0.5), but not exactly. Wait, maybe I did the multiplication wrong.Let me compute (3/5)*(29/30):3 * 29 = 875 * 30 = 150So, 87/150 simplifies to 29/50, which is 0.58, not 0.5. So, that's not equal to 1/2. Therefore, my calculation must be wrong.Wait, maybe I made a mistake in the sum of the roots. Let me re-examine the formula.The sum of the roots is -b/a. In the equation 30x² - 47x + 15 = 0, a = 30, b = -47.So, sum = -(-47)/30 = 47/30. That's correct.Product of the roots is c/a = 15/30 = 1/2. That's correct.So, if r = 3/5, then s = 47/30 - 3/5.But 3/5 is 18/30, so s = 47/30 - 18/30 = 29/30.But 29/30 * 3/5 = 87/150 = 29/50, which is not 1/2.Hmm, that's a contradiction. That means either my initial assumption is wrong, or I made a mistake in calculations.Wait, maybe I should try factoring the quadratic equation instead of using the sum and product. Let's see.The equation is 30x² - 47x + 15 = 0.I need to factor this into two binomials. Since the coefficient of x² is 30, which factors into 5 and 6, or 3 and 10, etc.Let me try to factor it.Looking for two numbers that multiply to 30*15 = 450 and add up to -47.Wait, but actually, when factoring, I need to find two numbers that multiply to a*c = 30*15 = 450 and add up to b = -47.So, I need two numbers that multiply to 450 and add up to -47.Hmm, 450 is a large number. Let me think about factors of 450.450 divided by 25 is 18. 25 and 18 multiply to 450. 25 + 18 = 43, which is not 47.Wait, 450 divided by 27 is 16.666, not an integer.Wait, 450 divided by 15 is 30. 15 and 30 multiply to 450. 15 + 30 = 45, which is close to 47, but not quite.Wait, 450 divided by 18 is 25. 18 and 25 multiply to 450, and 18 + 25 = 43.Wait, maybe negative numbers. Since the middle term is negative, both numbers should be negative.So, looking for two negative numbers that multiply to 450 and add up to -47.So, let's think: -25 and -18 multiply to 450 and add up to -43.-27 and -16.666, not integer.-30 and -15 multiply to 450 and add up to -45.Still not -47.Wait, maybe -25 and -18.2, but that's not integer.Hmm, maybe I'm approaching this wrong. Perhaps I should use the quadratic formula instead.The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a).Given the equation 30x² - 47x + 15 = 0, a = 30, b = -47, c = 15.So, discriminant D = b² - 4ac = (-47)² - 4*30*15.Calculating that:(-47)² = 22094*30*15 = 1800So, D = 2209 - 1800 = 409Wait, 409 is a prime number, I think. So, sqrt(409) is irrational.But wait, I know that x = 3/5 is a solution, so the quadratic should factor nicely, meaning the discriminant should be a perfect square. But 409 is not a perfect square. That suggests that maybe I made a mistake in forming the equation.Wait, let me go back to the original equation: 30x² + 13 = 47x - 2.When I moved all terms to the left, I got 30x² - 47x + 15 = 0.Let me double-check that:30x² + 13 = 47x - 2Subtract 47x and add 2 to both sides:30x² - 47x + 15 = 0Yes, that's correct.But then, using the quadratic formula, the solutions are:x = [47 ± sqrt(47² - 4*30*15)] / (2*30)Which is x = [47 ± sqrt(2209 - 1800)] / 60sqrt(409) is approximately 20.223, so:x ≈ (47 + 20.223)/60 ≈ 67.223/60 ≈ 1.120x ≈ (47 - 20.223)/60 ≈ 26.777/60 ≈ 0.446Wait, 3/5 is 0.6, but 0.446 is approximately 5/11 or something. Hmm, but I was told that x = 3/5 is a solution, but according to this, it's not matching.Wait, maybe I made a mistake in calculating the discriminant.Wait, 47 squared is 2209, correct.4*30*15 is 1800, correct.2209 - 1800 is 409, correct.So, sqrt(409) is indeed irrational, approximately 20.223.But if x = 3/5 is a solution, then plugging it into the equation should satisfy it.Let me check:30*(3/5)^2 + 13 = 47*(3/5) - 2Calculate left side:30*(9/25) + 13 = (270/25) + 13 = 10.8 + 13 = 23.8Right side:47*(3/5) - 2 = (141/5) - 2 = 28.2 - 2 = 26.2Wait, 23.8 ≠ 26.2. That means x = 3/5 is not a solution to this equation. But the problem statement says it is. So, I must have made a mistake in forming the equation.Wait, let me go back to the original problem: "Given that x = 3/5 is a solution to the equation 30x² + 13 = 47x - 2, find the other value of x that will solve the equation."Wait, maybe I misread the equation. Let me check again.It says 30x² + 13 = 47x - 2.So, moving all terms to the left:30x² + 13 - 47x + 2 = 0Wait, that's 30x² - 47x + 15 = 0, which is what I had before.But when I plug x = 3/5 into this equation, it doesn't satisfy it. So, either the problem is incorrect, or I made a mistake.Wait, maybe I miscalculated when plugging in x = 3/5.Let me do it again carefully.Left side: 30x² + 13x = 3/5, so x² = 9/2530*(9/25) = 270/25 = 10.810.8 + 13 = 23.8Right side: 47x - 247*(3/5) = 141/5 = 28.228.2 - 2 = 26.223.8 ≠ 26.2, so x = 3/5 is not a solution. But the problem says it is. That's confusing.Wait, maybe I misread the equation. Let me check again.It says 30x² + 13 = 47x - 2.Wait, could it be 30x² + 13x = 47x - 2? Because that would make more sense, perhaps.If that's the case, then moving all terms to the left:30x² + 13x - 47x + 2 = 0Which simplifies to 30x² - 34x + 2 = 0Then, let's check if x = 3/5 is a solution.Left side: 30*(9/25) + 13*(3/5) + 2Wait, no, if the equation is 30x² - 34x + 2 = 0, then plugging x = 3/5:30*(9/25) - 34*(3/5) + 2= 270/25 - 102/5 + 2= 10.8 - 20.4 + 2= (10.8 + 2) - 20.4 = 12.8 - 20.4 = -7.6 ≠ 0Still not zero. Hmm.Wait, maybe the original equation was 30x² + 13 = 47x - 2, but perhaps I need to check my arithmetic when plugging in x = 3/5.Wait, let's do it step by step.Left side: 30x² + 13x = 3/5, so x² = 9/2530*(9/25) = (30/1)*(9/25) = (270)/25 = 10.810.8 + 13 = 23.8Right side: 47x - 247*(3/5) = (47*3)/5 = 141/5 = 28.228.2 - 2 = 26.223.8 ≠ 26.2, so x = 3/5 is not a solution. Therefore, either the problem is incorrect, or I misread it.Wait, maybe the equation is 30x² + 13x = 47x - 2. Let me try that.So, 30x² + 13x = 47x - 2Move all terms to the left:30x² + 13x - 47x + 2 = 0Simplify:30x² - 34x + 2 = 0Now, let's check x = 3/5:30*(9/25) - 34*(3/5) + 2= 270/25 - 102/5 + 2= 10.8 - 20.4 + 2= (10.8 + 2) - 20.4 = 12.8 - 20.4 = -7.6 ≠ 0Still not zero. Hmm.Wait, maybe the equation is 30x² + 13 = 47x - 2, but I need to check if x = 3/5 is a solution.Wait, perhaps I made a mistake in the calculation. Let me use fractions instead of decimals to be precise.Left side: 30x² + 13x = 3/5, so x² = 9/2530*(9/25) = (30*9)/25 = 270/25 = 54/554/5 + 13 = 54/5 + 65/5 = 119/5Right side: 47x - 247*(3/5) = 141/5141/5 - 2 = 141/5 - 10/5 = 131/5So, left side is 119/5, right side is 131/5. They are not equal. Therefore, x = 3/5 is not a solution to the equation as given.This is confusing because the problem states that x = 3/5 is a solution. Maybe there's a typo in the problem, or I misread it.Alternatively, perhaps the equation is 30x² + 13x = 47x - 2, which would make more sense because then x = 3/5 could be a solution.Let me try that equation: 30x² + 13x = 47x - 2Move all terms to the left:30x² + 13x - 47x + 2 = 0Simplify:30x² - 34x + 2 = 0Now, let's check x = 3/5:30*(9/25) - 34*(3/5) + 2= 270/25 - 102/5 + 2= 10.8 - 20.4 + 2= -7.6 ≠ 0Still not zero. Hmm.Wait, maybe the equation is 30x² + 13 = 47x + 2. Let me try that.So, 30x² + 13 = 47x + 2Move all terms to the left:30x² - 47x + 11 = 0Check x = 3/5:30*(9/25) - 47*(3/5) + 11= 270/25 - 141/5 + 11= 10.8 - 28.2 + 11= (10.8 + 11) - 28.2 = 21.8 - 28.2 = -6.4 ≠ 0Still not zero.Wait, maybe the equation is 30x² + 13 = 47x - 2, but I need to check again.Alternatively, perhaps the equation is 30x² + 13x = 47x - 2, but that didn't work either.Wait, maybe I should try to factor the original equation 30x² - 47x + 15 = 0, even though the discriminant is not a perfect square, just to see if x = 3/5 is a solution.Wait, if x = 3/5 is a solution, then (x - 3/5) is a factor. Let me try to factor it.So, 30x² - 47x + 15 = (x - 3/5)(something)Let me write it as (x - 3/5)(ax + b) = 0Expanding this, we get:x*(ax + b) - 3/5*(ax + b) = ax² + bx - (3a/5)x - 3b/5Combine like terms:ax² + (b - 3a/5)x - 3b/5This should equal 30x² - 47x + 15So, setting coefficients equal:a = 30b - 3a/5 = -47-3b/5 = 15Let's solve these equations.From the third equation: -3b/5 = 15Multiply both sides by 5: -3b = 75Divide by -3: b = -25Now, from the second equation: b - 3a/5 = -47We know a = 30, b = -25So, -25 - 3*30/5 = -25 - 18 = -43But this should equal -47, which it doesn't. Therefore, x = 3/5 is not a root of this quadratic equation.This is perplexing. The problem states that x = 3/5 is a solution, but according to my calculations, it's not. Maybe I made a mistake in the problem statement.Wait, let me check the original problem again: "Given that x = 3/5 is a solution to the equation 30x² + 13 = 47x - 2, find the other value of x that will solve the equation."Hmm, perhaps the equation is written incorrectly. Maybe it's 30x² + 13x = 47x - 2, but even then, x = 3/5 doesn't satisfy it.Alternatively, maybe the equation is 30x² + 13 = 47x + 2, which would give 30x² - 47x + 11 = 0. Let's check x = 3/5:30*(9/25) - 47*(3/5) + 11 = 270/25 - 141/5 + 11 = 10.8 - 28.2 + 11 = -6.4 ≠ 0Still not zero.Wait, maybe the equation is 30x² + 13 = 47x - 2, but I need to check if x = 3/5 is a solution.Wait, perhaps I made a mistake in the calculation. Let me use fractions:Left side: 30x² + 13x = 3/5, so x² = 9/2530*(9/25) = 270/25 = 54/554/5 + 13 = 54/5 + 65/5 = 119/5Right side: 47x - 247*(3/5) = 141/5141/5 - 2 = 141/5 - 10/5 = 131/5So, 119/5 ≠ 131/5. Therefore, x = 3/5 is not a solution.This is very confusing. Maybe the problem has a typo, or I misread it. Alternatively, perhaps I need to consider that x = 3/5 is a solution, so I can use polynomial division or synthetic division to factor the quadratic.Wait, if x = 3/5 is a root, then (5x - 3) is a factor, because 5x - 3 = 0 when x = 3/5.So, let's try to factor 30x² - 47x + 15 by dividing it by (5x - 3).Using polynomial long division:Divide 30x² - 47x + 15 by (5x - 3).First term: 30x² ÷ 5x = 6xMultiply (5x - 3) by 6x: 30x² - 18xSubtract this from the original polynomial:(30x² - 47x + 15) - (30x² - 18x) = (-47x + 18x) + 15 = -29x + 15Now, divide -29x by 5x: -29/5Multiply (5x - 3) by -29/5: -29x + 87/5Subtract this from the previous remainder:(-29x + 15) - (-29x + 87/5) = 15 - 87/5 = (75/5 - 87/5) = -12/5So, the division gives a quotient of 6x - 29/5 with a remainder of -12/5. Since the remainder is not zero, (5x - 3) is not a factor of 30x² - 47x + 15, meaning x = 3/5 is not a root.This confirms my earlier conclusion that x = 3/5 is not a solution to the equation as given. Therefore, there must be an error in the problem statement.Alternatively, perhaps the equation is different. Maybe it's 30x² + 13x = 47x - 2, which would make the equation 30x² - 34x + 2 = 0. Let's check if x = 3/5 is a solution here.30*(9/25) - 34*(3/5) + 2 = 270/25 - 102/5 + 2 = 10.8 - 20.4 + 2 = -7.6 ≠ 0Still not zero.Wait, maybe the equation is 30x² + 13 = 47x + 2, which would be 30x² - 47x + 11 = 0.Check x = 3/5:30*(9/25) - 47*(3/5) + 11 = 270/25 - 141/5 + 11 = 10.8 - 28.2 + 11 = -6.4 ≠ 0Still not zero.I'm really stuck here. The problem states that x = 3/5 is a solution, but according to my calculations, it's not. Maybe I need to consider that the equation is written differently, or perhaps I made a mistake in the arithmetic.Wait, let me try to solve the equation 30x² + 13 = 47x - 2 again.30x² + 13 = 47x - 2Subtract 47x and add 2 to both sides:30x² - 47x + 15 = 0Now, let's use the quadratic formula:x = [47 ± sqrt(47² - 4*30*15)] / (2*30)Calculate discriminant:47² = 22094*30*15 = 1800So, sqrt(2209 - 1800) = sqrt(409) ≈ 20.223Thus, x ≈ [47 ± 20.223]/60So, x ≈ (47 + 20.223)/60 ≈ 67.223/60 ≈ 1.120x ≈ (47 - 20.223)/60 ≈ 26.777/60 ≈ 0.446Neither of these is 3/5 (0.6). So, x = 3/5 is not a solution.Therefore, I must conclude that there's an error in the problem statement. Either the equation is incorrect, or the given solution is wrong.Alternatively, perhaps the equation is 30x² + 13x = 47x - 2, which would make it 30x² - 34x + 2 = 0. Let's solve this:x = [34 ± sqrt(34² - 4*30*2)] / (2*30)= [34 ± sqrt(1156 - 240)] / 60= [34 ± sqrt(916)] / 60sqrt(916) ≈ 30.265So, x ≈ (34 + 30.265)/60 ≈ 64.265/60 ≈ 1.071x ≈ (34 - 30.265)/60 ≈ 3.735/60 ≈ 0.06225Still, x = 3/5 is not a solution.Wait, maybe the equation is 30x² + 13 = 47x + 2, which is 30x² - 47x + 11 = 0.Solutions:x = [47 ± sqrt(47² - 4*30*11)] / (2*30)= [47 ± sqrt(2209 - 1320)] / 60= [47 ± sqrt(889)] / 60sqrt(889) ≈ 29.816So, x ≈ (47 + 29.816)/60 ≈ 76.816/60 ≈ 1.280x ≈ (47 - 29.816)/60 ≈ 17.184/60 ≈ 0.2864Still not 3/5.I'm really stuck here. Maybe I need to consider that the equation is different, or perhaps the given solution is incorrect.Alternatively, perhaps the equation is 30x² + 13 = 47x - 2, and x = 3/5 is a solution, but I need to find the other solution despite the inconsistency.Wait, maybe I should proceed with the assumption that x = 3/5 is a solution, even though my calculations suggest otherwise. Perhaps I made a mistake in the arithmetic.Let me try to factor the quadratic equation 30x² - 47x + 15 = 0, assuming that x = 3/5 is a root.If x = 3/5 is a root, then (5x - 3) is a factor.So, let's write 30x² - 47x + 15 = (5x - 3)(ax + b)Expanding the right side:(5x - 3)(ax + b) = 5a x² + (5b - 3a)x - 3bSet this equal to 30x² - 47x + 15So, we have:5a x² + (5b - 3a)x - 3b = 30x² - 47x + 15Now, equate coefficients:5a = 30 ⇒ a = 65b - 3a = -47-3b = 15 ⇒ b = -5Now, substitute a = 6 into the second equation:5b - 3*6 = -475b - 18 = -475b = -47 + 18 = -29b = -29/5But earlier, from -3b = 15, we got b = -5. This is a contradiction because b cannot be both -5 and -29/5.Therefore, x = 3/5 is not a root of the quadratic equation 30x² - 47x + 15 = 0.This confirms that there's an inconsistency in the problem statement. Either the equation is incorrect, or the given solution is wrong.Given this, I cannot proceed further because the problem as stated contains conflicting information. It's possible that there was a typo in the equation or the given solution.Alternatively, perhaps I need to consider that the equation is different, such as 30x² + 13x = 47x - 2, but even then, x = 3/5 is not a solution.In conclusion, based on the given equation and the provided solution, there seems to be an inconsistency. Therefore, I cannot accurately determine the other solution without resolving this discrepancy.

💡Okay, so I need to figure out how long it took Rich to run a mile on average during his 50-mile ultra-marathon. He ran this race in 8 hours and 37 minutes, and the terrain varied, meaning he ran faster on flat sections but slower on uphill parts. First, I think I need to convert the total time he took into minutes because the question asks for the average time per mile in minutes. Rich ran for 8 hours and 37 minutes. I know that 1 hour is 60 minutes, so 8 hours would be 8 times 60 minutes. Let me calculate that: 8 * 60 = 480 minutes. Then, I need to add the remaining 37 minutes to this. So, 480 minutes + 37 minutes equals 517 minutes in total.Now, Rich ran 50 miles in 517 minutes. To find the average time per mile, I think I need to divide the total time by the number of miles. So, that would be 517 minutes divided by 50 miles. Let me do that division: 517 / 50. Hmm, 50 goes into 517 ten times because 50 * 10 = 500, and that leaves a remainder of 17. So, 17 divided by 50 is 0.34. Therefore, the average time per mile is 10.34 minutes.Wait, but the problem mentions that Rich ran faster on flat terrain and slower on uphill sections. Does that affect the average time per mile? I think the average time already accounts for the varying speeds because it's based on the total time divided by the total distance. So, even though he ran some parts faster and some slower, the overall average time per mile should still be 10.34 minutes.I'm not sure if I need to consider any other factors, like elevation changes or the specific distribution of flat and uphill sections. But since the problem doesn't provide details about how much of the race was flat versus uphill, I think it's safe to assume that the average time per mile is simply the total time divided by the total distance.So, after converting the total time into minutes and dividing by the number of miles, I conclude that Rich took approximately 10.34 minutes per mile on average during the race.

💡First, I need to determine the leak rate of the tank. Given that the leak can empty the full tank in 6 hours, the leak rate ( L ) can be calculated by dividing the tank's capacity by the time it takes to empty:[L = frac{5760}{6} = 960 text{ litres per hour}]Next, when the inlet pipe is opened, the tank empties in 8 hours. This means that the net rate at which the tank is being emptied is the leak rate minus the fill rate of the inlet pipe. I can express this as:[L - F = frac{5760}{8} = 720 text{ litres per hour}]Now, I can solve for the fill rate ( F ) by rearranging the equation:[F = L - 720 = 960 - 720 = 240 text{ litres per hour}]Therefore, the rate at which the inlet pipe fills the tank is 240 litres per hour.

💡Okay, so I have this problem where I need to find the sum of the reciprocals of (p-2), (q-2), and (r-2), where p, q, and r are the roots of the cubic equation x³ - 2x² - x + 3 = 0. Hmm, that seems a bit tricky, but I think I can figure it out by using some algebraic techniques I've learned before.First, I remember that for a cubic equation like ax³ + bx² + cx + d = 0, the sum of the roots is -b/a, the sum of the products of the roots two at a time is c/a, and the product of all three roots is -d/a. Maybe I can use Vieta's formulas here somehow.But wait, the problem isn't directly asking for the sum of the roots or their products. It's asking for the sum of 1/(p-2) + 1/(q-2) + 1/(r-2). That seems a bit different. I think I need to find a way to relate this expression to the coefficients of the given polynomial.Let me think. If I can find a polynomial whose roots are (p-2), (q-2), and (r-2), then maybe I can use Vieta's formulas on that polynomial to find the sum of the reciprocals of its roots. That might work.So, how do I find a polynomial whose roots are shifted by 2? I remember that if you have a polynomial with roots p, q, r, then replacing x with (x + 2) will shift the roots to the left by 2 units. Wait, actually, if I substitute x = y + 2 into the original polynomial, then the new variable y will correspond to x - 2, so the roots in terms of y will be p - 2, q - 2, r - 2. That sounds right.Let me try that substitution. Let me set x = y + 2. Then, substitute this into the original equation:(y + 2)³ - 2(y + 2)² - (y + 2) + 3 = 0.Now, I need to expand this expression. Let me do that step by step.First, expand (y + 2)³:(y + 2)³ = y³ + 3* y²*2 + 3*y*4 + 8 = y³ + 6y² + 12y + 8.Next, expand -2(y + 2)²:(y + 2)² = y² + 4y + 4, so -2(y² + 4y + 4) = -2y² - 8y - 8.Then, expand -(y + 2):-(y + 2) = -y - 2.Now, add all these together along with the constant term +3:(y³ + 6y² + 12y + 8) + (-2y² - 8y - 8) + (-y - 2) + 3 = 0.Let me combine like terms:- y³ term: y³.- y² terms: 6y² - 2y² = 4y².- y terms: 12y - 8y - y = 3y.- Constant terms: 8 - 8 - 2 + 3 = 1.So, the transformed polynomial is y³ + 4y² + 3y + 1 = 0.Okay, so now I have a new polynomial in terms of y, where the roots are (p - 2), (q - 2), and (r - 2). Let me denote these roots as y₁ = p - 2, y₂ = q - 2, y₃ = r - 2.So, the polynomial is y³ + 4y² + 3y + 1 = 0. Now, I need to find 1/y₁ + 1/y₂ + 1/y₃.I recall that for a polynomial ay³ + by² + cy + d = 0, the sum of the reciprocals of the roots is -c/d, provided that none of the roots are zero. Let me verify that.Wait, if the polynomial is y³ + 4y² + 3y + 1 = 0, then a = 1, b = 4, c = 3, d = 1.So, the sum of the reciprocals would be -c/d = -3/1 = -3.But wait, let me make sure I'm applying Vieta's correctly for reciprocals. Vieta's formulas give us:For a cubic equation y³ + a y² + b y + c = 0,Sum of roots: y₁ + y₂ + y₃ = -a.Sum of products two at a time: y₁y₂ + y₁y₃ + y₂y₃ = b.Product of roots: y₁y₂y₃ = -c.So, the sum of reciprocals is (1/y₁ + 1/y₂ + 1/y₃) = (y₂y₃ + y₁y₃ + y₁y₂)/(y₁y₂y₃) = (sum of products two at a time)/(product of roots).From Vieta's, that would be b/(-c). In our case, b = 3, c = 1, so it's 3/(-1) = -3.Yes, that matches what I thought earlier. So, the sum of the reciprocals is -3.Therefore, the value of 1/(p - 2) + 1/(q - 2) + 1/(r - 2) is -3.I think that makes sense. Let me just recap what I did:1. Recognized that shifting the variable x by 2 would transform the roots p, q, r into p - 2, q - 2, r - 2.2. Substituted x = y + 2 into the original polynomial to find the new polynomial in terms of y.3. Expanded and simplified the substituted polynomial to get the new cubic equation.4. Applied Vieta's formulas to the new polynomial to find the sum of the reciprocals of its roots, which corresponds to the desired expression.I don't see any mistakes in my calculations, so I feel confident that the answer is correct.**Final Answer**The value of frac{1}{p-2} + frac{1}{q-2} + frac{1}{r-2} is boxed{-3}.