💡Okay, so I have this problem here: Given a constant ( a > 0 ), the expansion of ( left(ax^{2} + frac{1}{sqrt{x}}right)^{4} ) contains a term with ( x^{3} ) whose coefficient is ( frac{3}{2} ). I need to find the value of ( lim_{n rightarrow infty} (a + a^{2} + cdots + a^{n}) ).Hmm, let me break this down step by step. First, I remember that when you expand a binomial like ( (A + B)^n ), each term is given by the binomial theorem: ( T_{k+1} = binom{n}{k} A^{n-k} B^{k} ). So in this case, ( A = ax^2 ) and ( B = frac{1}{sqrt{x}} ), and ( n = 4 ).So, the general term in the expansion should be ( T_{r+1} = binom{4}{r} (ax^2)^{4 - r} left(frac{1}{sqrt{x}}right)^r ). Let me write that out:( T_{r+1} = binom{4}{r} cdot a^{4 - r} cdot x^{2(4 - r)} cdot x^{-frac{r}{2}} ).Simplifying the exponents of ( x ), we have ( x^{8 - 2r - frac{r}{2}} = x^{8 - frac{5r}{2}} ).Okay, so each term has an exponent of ( x ) equal to ( 8 - frac{5r}{2} ). The problem states that there's a term with ( x^3 ). So I need to find the value of ( r ) such that:( 8 - frac{5r}{2} = 3 ).Let me solve for ( r ):( 8 - 3 = frac{5r}{2} )( 5 = frac{5r}{2} )Multiply both sides by 2:( 10 = 5r )Divide both sides by 5:( r = 2 ).So, the term corresponding to ( r = 2 ) is the one with ( x^3 ). Now, let's find the coefficient of this term.The coefficient is ( binom{4}{2} cdot a^{4 - 2} ).Calculating ( binom{4}{2} ), which is 6. So, the coefficient is ( 6 cdot a^{2} ).The problem says this coefficient is ( frac{3}{2} ). So:( 6a^{2} = frac{3}{2} ).Let me solve for ( a ):Divide both sides by 6:( a^{2} = frac{3}{2} div 6 = frac{3}{12} = frac{1}{4} ).Taking the square root of both sides:( a = sqrt{frac{1}{4}} = frac{1}{2} ).Since ( a > 0 ), we don't consider the negative root.Alright, so ( a = frac{1}{2} ). Now, the next part is to find the limit ( lim_{n rightarrow infty} (a + a^{2} + cdots + a^{n}) ).I recognize this as an infinite geometric series. The sum of an infinite geometric series with first term ( a ) and common ratio ( r ) is ( frac{a}{1 - r} ), provided ( |r| < 1 ).In this case, the first term is ( a = frac{1}{2} ), and the common ratio is also ( a = frac{1}{2} ), since each term is multiplied by ( a ) to get the next term.So, plugging into the formula:Sum = ( frac{frac{1}{2}}{1 - frac{1}{2}} = frac{frac{1}{2}}{frac{1}{2}} = 1 ).Therefore, the limit is 1.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Expanded the binomial correctly using the binomial theorem. Yes, that seems right.2. Calculated the exponent of ( x ) correctly: ( 8 - frac{5r}{2} ). Yes, because ( ax^2 ) contributes ( 2(4 - r) ) and ( frac{1}{sqrt{x}} ) contributes ( -frac{r}{2} ). So, total exponent is ( 8 - 2r - frac{r}{2} = 8 - frac{5r}{2} ). Correct.3. Solved for ( r ) when exponent is 3: ( 8 - frac{5r}{2} = 3 ) leading to ( r = 2 ). That seems correct.4. Calculated the coefficient: ( binom{4}{2} a^{2} = 6a^{2} ). That's right.5. Set ( 6a^{2} = frac{3}{2} ) and solved for ( a ): ( a^{2} = frac{1}{4} ), so ( a = frac{1}{2} ). Correct.6. Recognized the series as a geometric series with first term ( a = frac{1}{2} ) and ratio ( r = frac{1}{2} ). Sum is ( frac{frac{1}{2}}{1 - frac{1}{2}} = 1 ). Correct.So, all steps seem solid. I don't see any mistakes here.**Final Answer**boxed{1}

💡First, determine Charlie's floor number by calculating one-fourth of Frank's floor number.Charlie's floor number = 1/4 * Frank's floor numberCharlie's floor number = 1/4 * 16Charlie's floor number = 4Next, find the difference between Dennis's floor number and Charlie's floor number to determine how many floors above Charlie Dennis lives.Dennis's floor number - Charlie's floor number = Number of floors above Charlie6 - 4 = 2Dennis lives 2 floors above Charlie.

💡First, I need to determine the initial number of men in the group. Let's denote this number as ( M ) and the initial average age of the men as ( A ).When two men, aged 20 and 24 years, are replaced by two women, each aged 30 years, the average age of the group increases by 2 years. This means that the total age of the group increases by ( 2M ) years.The total age of the two women is ( 2 times 30 = 60 ) years. The total age of the two men being replaced is ( 20 + 24 = 44 ) years. The net increase in total age due to the replacement is ( 60 - 44 = 16 ) years.Since this net increase equals the total age increase for the group, we have the equation:[ 2M = 16 ]Solving for ( M ):[ M = frac{16}{2} = 8 ]Therefore, there were initially 8 men in the group.

💡Okay, so I need to find the value of (frac{p(1)}{q(1)}) given the graph of (y = frac{p(x)}{q(x)}), where both (p(x)) and (q(x)) are quadratic polynomials. The graph has vertical asymptotes at (x = -4) and (x = 1), passes through the points ((0,0)) and ((2,-2)), and has a horizontal asymptote at (y = 1).First, I remember that vertical asymptotes occur where the denominator is zero (provided the numerator isn't also zero there). So, since there are vertical asymptotes at (x = -4) and (x = 1), the denominator (q(x)) must have factors of ((x + 4)) and ((x - 1)). Therefore, I can write (q(x)) as:[q(x) = a(x + 4)(x - 1)]where (a) is a constant.Next, the horizontal asymptote of a rational function where the degrees of the numerator and denominator are equal is given by the ratio of the leading coefficients. Since the horizontal asymptote is (y = 1), the leading coefficients of (p(x)) and (q(x)) must be equal. So, if (q(x)) has a leading coefficient of (a), then (p(x)) must also have a leading coefficient of (a).Now, let's consider the numerator (p(x)). Since the graph passes through the point ((0,0)), when (x = 0), (y = 0). Plugging this into the equation:[0 = frac{p(0)}{q(0)}]This implies that (p(0) = 0), so (x = 0) is a root of (p(x)). Therefore, (p(x)) must have a factor of (x). So, I can write (p(x)) as:[p(x) = x(bx + c)]where (b) and (c) are constants to be determined.Now, the graph also passes through the point ((2, -2)). Plugging (x = 2) and (y = -2) into the equation:[-2 = frac{p(2)}{q(2)}]Substituting the expressions for (p(x)) and (q(x)):[-2 = frac{2(b cdot 2 + c)}{a(2 + 4)(2 - 1)}]Simplifying the denominator:[a(6)(1) = 6a]So, the equation becomes:[-2 = frac{2(2b + c)}{6a}]Simplify the numerator:[2(2b + c) = 4b + 2c]So, the equation is:[-2 = frac{4b + 2c}{6a}]Multiply both sides by (6a):[-12a = 4b + 2c]Divide both sides by 2:[-6a = 2b + c]So, that's one equation.Now, since the horizontal asymptote is (y = 1), the ratio of the leading coefficients of (p(x)) and (q(x)) must be 1. The leading term of (q(x)) is (a x^2), and the leading term of (p(x)) is (b x^2). Therefore:[frac{b}{a} = 1 implies b = a]So, we can substitute (b = a) into the previous equation:[-6a = 2a + c]Subtract (2a) from both sides:[-8a = c]So, (c = -8a).Now, let's write (p(x)) with these values:[p(x) = x(a x - 8a) = a x(x - 8)]And (q(x)) is:[q(x) = a(x + 4)(x - 1)]So, the function is:[y = frac{a x(x - 8)}{a(x + 4)(x - 1)}]We can cancel out the (a) since it's a common factor:[y = frac{x(x - 8)}{(x + 4)(x - 1)}]Now, we need to find (frac{p(1)}{q(1)}). Let's compute (p(1)) and (q(1)).First, (p(1)):[p(1) = 1(1 - 8) = 1(-7) = -7]Next, (q(1)):[q(1) = (1 + 4)(1 - 1) = 5 times 0 = 0]Wait, that's a problem. (q(1)) is zero, which would make (frac{p(1)}{q(1)}) undefined. But the question is asking for (frac{p(1)}{q(1)}). Hmm, maybe I made a mistake somewhere.Let me double-check. The vertical asymptotes are at (x = -4) and (x = 1), so (q(1) = 0), which is correct. But the function (y = frac{p(x)}{q(x)}) is undefined at (x = 1), so (frac{p(1)}{q(1)}) is undefined. But the question is asking for it, so perhaps I need to interpret it differently.Wait, maybe I misapplied the horizontal asymptote condition. Let me check that again. The horizontal asymptote is the ratio of the leading coefficients, which for both (p(x)) and (q(x)) are quadratic. So, leading term of (p(x)) is (a x^2) and leading term of (q(x)) is (a x^2), so their ratio is 1, which matches the horizontal asymptote (y = 1). So that seems correct.But then, when I plug in (x = 1), (q(1) = 0), so the function is undefined there. So, perhaps the question is a trick question, and the answer is undefined. But the problem didn't specify that (x = 1) is a hole instead of an asymptote, so it's an asymptote, meaning the function isn't defined there.Wait, but the problem says "the graph of (y = frac{p(x)}{q(x)})", so maybe I need to consider the limit as (x) approaches 1? But the question specifically asks for (frac{p(1)}{q(1)}), which is undefined because (q(1) = 0). Hmm.Wait, maybe I made a mistake in setting up (p(x)). Let me go back. I assumed (p(x) = x(bx + c)), but maybe I should have included another root. Since (p(x)) is quadratic, it has two roots. One is at (x = 0), as we saw from the point ((0,0)). The other root is at (x = c), so (p(x) = x(x - c)). Then, we can write (p(x) = x(x - c)), and (q(x) = (x + 4)(x - 1)), with leading coefficients matching for the horizontal asymptote.Wait, but I included a constant (a) earlier. Maybe I should set (a = 1) for simplicity, since the ratio of leading coefficients is 1. Let me try that.So, let (q(x) = (x + 4)(x - 1)), and (p(x) = x(x - c)). Then, the function is:[y = frac{x(x - c)}{(x + 4)(x - 1)}]Now, using the point ((2, -2)):[-2 = frac{2(2 - c)}{(2 + 4)(2 - 1)} = frac{2(2 - c)}{6 times 1} = frac{2(2 - c)}{6} = frac{2 - c}{3}]So, set up the equation:[frac{2 - c}{3} = -2]Multiply both sides by 3:[2 - c = -6]Subtract 2 from both sides:[-c = -8]Multiply both sides by -1:[c = 8]So, (p(x) = x(x - 8)), and (q(x) = (x + 4)(x - 1)).Now, let's compute (frac{p(1)}{q(1)}):[p(1) = 1(1 - 8) = 1 times (-7) = -7][q(1) = (1 + 4)(1 - 1) = 5 times 0 = 0]So, (frac{p(1)}{q(1)} = frac{-7}{0}), which is undefined. But the problem is asking for this value, so perhaps I need to consider the limit as (x) approaches 1? Or maybe there's a mistake in my setup.Wait, perhaps I should not have set (a = 1). Let me go back to the earlier step where I had:[-6a = 2b + c]And since (b = a), then:[-6a = 2a + c implies c = -8a]So, (p(x) = a x(x - 8)), and (q(x) = a(x + 4)(x - 1)). So, the function is:[y = frac{a x(x - 8)}{a(x + 4)(x - 1)} = frac{x(x - 8)}{(x + 4)(x - 1)}]So, the (a) cancels out, and we're back to the same situation. Therefore, (q(1) = 0), so (frac{p(1)}{q(1)}) is undefined.But the problem is asking for (frac{p(1)}{q(1)}), so maybe I need to consider the limit as (x) approaches 1. Let's compute that:[lim_{x to 1} frac{p(x)}{q(x)} = lim_{x to 1} frac{x(x - 8)}{(x + 4)(x - 1)}]As (x) approaches 1 from the right, the denominator approaches 0 from the positive side, and the numerator approaches (1 times (-7) = -7), so the limit is (-infty). As (x) approaches 1 from the left, the denominator approaches 0 from the negative side, so the limit is (+infty). Therefore, the limit does not exist, but the function approaches positive and negative infinity from either side.But the problem is asking for (frac{p(1)}{q(1)}), which is undefined because (q(1) = 0). So, perhaps the answer is undefined, but the problem might expect the limit or something else. Alternatively, maybe I made a mistake in setting up (p(x)) and (q(x)).Wait, another thought: maybe (x = 1) is a hole instead of an asymptote. But for that, both (p(x)) and (q(x)) would have to have a common factor of ((x - 1)). But in our case, (p(x) = x(x - 8)) and (q(x) = (x + 4)(x - 1)), so there's no common factor. Therefore, (x = 1) is indeed a vertical asymptote, not a hole.So, given that, (frac{p(1)}{q(1)}) is undefined. But the problem is asking for it, so maybe I need to reconsider my approach.Wait, perhaps I should not have set (a = 1). Let me try keeping (a) as a variable and see if that helps.So, (p(x) = a x(x - 8)), (q(x) = a(x + 4)(x - 1)). Then, (frac{p(1)}{q(1)} = frac{a times 1 times (1 - 8)}{a times (1 + 4) times (1 - 1)} = frac{a times (-7)}{a times 5 times 0}). The (a) cancels out, but we still have division by zero, so it's undefined.Therefore, regardless of the value of (a), (frac{p(1)}{q(1)}) is undefined. So, perhaps the answer is that it's undefined, but the problem might expect a specific value. Maybe I made a mistake in determining (p(x)) and (q(x)).Wait, let me check the point ((2, -2)) again. With (p(x) = x(x - 8)) and (q(x) = (x + 4)(x - 1)), let's compute (y) at (x = 2):[y = frac{2(2 - 8)}{(2 + 4)(2 - 1)} = frac{2(-6)}{6 times 1} = frac{-12}{6} = -2]Which matches the given point. So, my (p(x)) and (q(x)) are correct.Therefore, (frac{p(1)}{q(1)}) is indeed undefined because (q(1) = 0). But the problem is asking for it, so maybe I need to interpret it differently. Alternatively, perhaps the problem expects the value of the function at (x = 1) in some extended sense, but since it's an asymptote, it's not defined.Wait, another thought: maybe the problem is expecting the value of the limit as (x) approaches 1, but as I computed earlier, the limit does not exist because it approaches positive and negative infinity from either side. So, perhaps the answer is that it's undefined or does not exist.But the problem is specifically asking for (frac{p(1)}{q(1)}), which is undefined. So, maybe the answer is undefined, but in the context of the problem, perhaps it's expecting a specific value. Alternatively, maybe I made a mistake in setting up (p(x)) and (q(x)).Wait, perhaps I should consider that (p(x)) and (q(x)) could have a common factor, but in this case, they don't because (p(x) = x(x - 8)) and (q(x) = (x + 4)(x - 1)). So, no common factors, hence no holes.Therefore, I think the conclusion is that (frac{p(1)}{q(1)}) is undefined because (q(1) = 0). However, the problem is asking for it, so maybe I need to express it in terms of limits or something else. Alternatively, perhaps I made a mistake in the setup.Wait, let me try a different approach. Let me assume that (p(x)) and (q(x)) are both quadratics, so let me write them as:[p(x) = ax^2 + bx + c][q(x) = dx^2 + ex + f]Given that the vertical asymptotes are at (x = -4) and (x = 1), so (q(-4) = 0) and (q(1) = 0). Therefore, (q(x)) can be written as:[q(x) = d(x + 4)(x - 1) = d(x^2 + 3x - 4)]So, (q(x) = d x^2 + 3d x - 4d).Similarly, since the horizontal asymptote is (y = 1), the ratio of the leading coefficients (a/d = 1), so (a = d).Now, the graph passes through ((0,0)), so:[0 = frac{p(0)}{q(0)} = frac{c}{-4d}]Therefore, (c = 0). So, (p(x) = a x^2 + b x).Now, the graph also passes through ((2, -2)), so:[-2 = frac{p(2)}{q(2)} = frac{a(4) + b(2)}{d(4) + 3d(2) - 4d}]Simplify the denominator:[4d + 6d - 4d = 6d]So, the equation becomes:[-2 = frac{4a + 2b}{6d}]But since (a = d), substitute (a) for (d):[-2 = frac{4a + 2b}{6a}]Multiply both sides by (6a):[-12a = 4a + 2b]Subtract (4a) from both sides:[-16a = 2b]Divide both sides by 2:[-8a = b]So, (b = -8a).Therefore, (p(x) = a x^2 - 8a x = a(x^2 - 8x)).Now, let's write (p(x)) and (q(x)):[p(x) = a(x^2 - 8x)][q(x) = a(x^2 + 3x - 4)]So, the function is:[y = frac{a(x^2 - 8x)}{a(x^2 + 3x - 4)} = frac{x^2 - 8x}{x^2 + 3x - 4}]Now, let's compute (frac{p(1)}{q(1)}):[p(1) = 1^2 - 8(1) = 1 - 8 = -7][q(1) = 1^2 + 3(1) - 4 = 1 + 3 - 4 = 0]So, (frac{p(1)}{q(1)} = frac{-7}{0}), which is undefined.Therefore, regardless of the approach, (frac{p(1)}{q(1)}) is undefined because (q(1) = 0). So, the answer is undefined.But the problem is asking for (frac{p(1)}{q(1)}), so perhaps the answer is undefined. However, in the context of the problem, maybe it's expecting a specific value, but I think it's correct to say it's undefined.Wait, but in the initial setup, I considered (p(x)) as (x(x - 8)) and (q(x)) as ((x + 4)(x - 1)), which led to the same conclusion. So, I think the answer is indeed undefined.But let me check if there's any other way to interpret the problem. Maybe the function is defined at (x = 1) despite the denominator being zero, but that would require (p(1) = 0) as well, which it's not. So, no, that's not possible.Therefore, the conclusion is that (frac{p(1)}{q(1)}) is undefined because (q(1) = 0). So, the answer is undefined.

💡Okay, so I have this function f(x) = x³ - 3x + 5, and I need to find the range of values for a such that the equation f(x) = a has at least two distinct real roots. Hmm, okay. Let me try to figure this out step by step.First, I remember that for a cubic function like this, the graph is going to have a certain shape. It might have one or two turning points, right? So, maybe I should find the critical points of the function to understand where it's increasing or decreasing. That might help me figure out how many times it crosses the horizontal line y = a.To find the critical points, I need to take the derivative of f(x). The derivative, f'(x), will tell me where the function has local maxima or minima. Let me compute that:f'(x) = d/dx (x³ - 3x + 5) = 3x² - 3.Okay, so f'(x) = 3x² - 3. To find the critical points, I set this equal to zero:3x² - 3 = 0.Dividing both sides by 3 gives x² - 1 = 0, so x² = 1. Taking the square root of both sides, I get x = ±1. So, the critical points are at x = 1 and x = -1.Now, I should figure out whether these critical points are maxima or minima. I can use the second derivative test for that. Let me compute the second derivative:f''(x) = d/dx (3x² - 3) = 6x.Now, plug in x = 1 into f''(x):f''(1) = 6(1) = 6, which is positive. So, at x = 1, the function has a local minimum.Similarly, plug in x = -1:f''(-1) = 6(-1) = -6, which is negative. So, at x = -1, the function has a local maximum.Alright, so now I know that the function has a local maximum at x = -1 and a local minimum at x = 1. Let me find the corresponding y-values for these points because that will help me understand the range of a where the equation f(x) = a has at least two real roots.First, compute f(-1):f(-1) = (-1)³ - 3(-1) + 5 = -1 + 3 + 5 = 7.So, the local maximum is at (-1, 7).Next, compute f(1):f(1) = (1)³ - 3(1) + 5 = 1 - 3 + 5 = 3.So, the local minimum is at (1, 3).Now, thinking about the graph of this function, it's a cubic, so it goes from negative infinity to positive infinity. But because it has a local maximum at y = 7 and a local minimum at y = 3, the function will cross the y-axis somewhere between these points.Wait, actually, the function will have a "hill" at y = 7 and a "valley" at y = 3. So, if I draw a horizontal line y = a, the number of times this line intersects the graph of f(x) depends on where a is.If a is greater than 7, the line y = a will intersect the graph only once because the function only goes up to 7 and then continues to infinity. Similarly, if a is less than 3, the line y = a will intersect the graph only once because the function only goes down to 3 and then continues to negative infinity.But if a is between 3 and 7, the line y = a will intersect the graph three times: once before the local maximum, once between the local maximum and local minimum, and once after the local minimum. Wait, but the question says "at least two distinct real roots." So, does that mean two or three?Hmm, actually, when a is exactly 7 or exactly 3, the line y = a will be tangent to the graph at the local maximum or minimum, respectively. That means it will intersect at exactly two points: one at the tangent point and one other point. So, in those cases, there are exactly two real roots.But when a is strictly between 3 and 7, the line y = a will intersect the graph three times, meaning three real roots. So, does the question consider three real roots as "at least two"? I think yes, because three is more than two. So, the range of a where f(x) = a has at least two distinct real roots would be from 3 to 7, inclusive.Wait, let me double-check that. If a is exactly 7, then f(x) = a will have two real roots: one at x = -1 (since that's the local maximum) and another somewhere else. Similarly, if a is exactly 3, it will have two real roots: one at x = 1 (the local minimum) and another somewhere else. For a between 3 and 7, it will have three real roots. So, in all cases, a in [3,7] will give at least two real roots.But just to make sure, let me think about the behavior of the function. As x approaches negative infinity, x³ dominates, so f(x) approaches negative infinity. As x approaches positive infinity, f(x) approaches positive infinity. So, the function is continuous and smooth.Since it has a local maximum at y = 7 and a local minimum at y = 3, the function must cross the y-axis somewhere in between. So, for a between 3 and 7, the horizontal line will intersect the graph three times. For a equal to 3 or 7, it will intersect twice, and for a outside of [3,7], it will intersect once.Therefore, the range of a for which f(x) = a has at least two distinct real roots is all real numbers a such that 3 ≤ a ≤ 7.Wait, but let me think again. If a is exactly 3, the equation f(x) = 3 will have a double root at x = 1 and another root somewhere else. Similarly, for a = 7, it will have a double root at x = -1 and another root somewhere else. So, in both cases, there are two distinct real roots, one of which is a double root.But the question says "at least two distinct real roots." Does a double root count as two distinct roots? Hmm, no, a double root is still just one distinct root with multiplicity two. So, actually, when a is exactly 3 or 7, the equation f(x) = a has exactly one distinct real root with multiplicity two and another distinct real root. So, in total, two distinct real roots.Wait, no. If a = 3, f(x) = 3, which is f(x) - 3 = x³ - 3x + 2 = 0. Let me factor that. Maybe it factors as (x - 1)^2 (x + 2). Let me check:(x - 1)^2 (x + 2) = (x² - 2x + 1)(x + 2) = x³ + 2x² - 2x² - 4x + x + 2 = x³ - 3x + 2. Yes, that's correct. So, f(x) - 3 = (x - 1)^2 (x + 2). So, the roots are x = 1 (double root) and x = -2. So, there are two distinct real roots: x = 1 and x = -2. Similarly, for a = 7, f(x) - 7 = x³ - 3x - 2. Let me factor that. Maybe (x + 1)^2 (x - 2):(x + 1)^2 (x - 2) = (x² + 2x + 1)(x - 2) = x³ - 2x² + 2x² - 4x + x - 2 = x³ - 3x - 2. Yes, that's correct. So, the roots are x = -1 (double root) and x = 2. So, two distinct real roots.Therefore, when a is exactly 3 or 7, the equation f(x) = a has exactly two distinct real roots, one of which is a double root. For a between 3 and 7, it has three distinct real roots. So, in all cases, a in [3,7] gives at least two distinct real roots.Therefore, the range of a is [3,7].**Final Answer**The range of values for ( a ) is boxed{[3, 7]}.

💡First, determine the amount of paint needed by dividing the total cost by the cost per quart:16 / 3.20 = 5 quarts.Next, calculate the total area covered by multiplying the number of quarts by the coverage per quart:5 quarts * 120 square feet per quart = 600 square feet.The surface area of a cube is given by 6 times the square of its edge length. Set up the equation:6 * (edge length)^2 = 600 square feet.Solve for the edge length by dividing both sides by 6:(edge length)^2 = 100 square feet.Finally, take the square root to find the edge length:edge length = √100 = 10 feet.