💡Okay, so I have this hyperbola equation: (x²)/9 - (y²)/(b²) = 1. They told me that one of the asymptotes has an inclination angle of 150 degrees, and I need to find the value of b. Hmm, let me think about how to approach this.First, I remember that for hyperbolas of the form (x²)/(a²) - (y²)/(b²) = 1, the asymptotes are straight lines that the hyperbola approaches but never touches. The equations of these asymptotes are y = (b/a)x and y = -(b/a)x. So, in this case, since a² is 9, that means a is 3 because the square root of 9 is 3. So, a = 3.Now, the asymptotes are y = (b/3)x and y = -(b/3)x. These are straight lines with slopes of b/3 and -b/3. The problem mentions an inclination angle of 150 degrees. I think the inclination angle is the angle that the line makes with the positive x-axis, measured counterclockwise. So, 150 degrees is in the second quadrant, which means the slope of that asymptote should be negative because it's going downwards from left to right.So, if the angle is 150 degrees, the slope of that asymptote would be tan(150 degrees). Let me calculate that. I know that tan(150 degrees) is equal to tan(180 - 30) degrees, which is -tan(30 degrees). Since tan(30 degrees) is √3/3, then tan(150 degrees) is -√3/3.So, the slope of the asymptote is -√3/3. But from the hyperbola equation, the slopes of the asymptotes are ±b/a, which is ±b/3. Since we're dealing with the negative slope here, that would be -b/3. So, setting that equal to -√3/3, we have:-b/3 = -√3/3If I multiply both sides by 3, I get:-b = -√3Then, multiplying both sides by -1:b = √3So, that should be the value of b. Let me just double-check my steps to make sure I didn't make a mistake.1. Identified the hyperbola form and found a = 3.2. Remembered that asymptotes are y = ±(b/a)x.3. Noted that 150 degrees is in the second quadrant, so the slope is negative.4. Calculated tan(150 degrees) as -√3/3.5. Set -b/3 equal to -√3/3.6. Solved for b and got √3.Everything seems to check out. I don't think I made any calculation errors. So, yeah, b should be √3.**Final Answer**The value of ( b ) is boxed{sqrt{3}}.

💡First, I recognize that 75% of the students chose juice and 25% chose water. This means that for every 75 students who chose juice, 25 chose water.Given that 90 students chose juice, I can set up a proportion to find out how many students chose water. The ratio of students who chose water to those who chose juice is 25:75, which simplifies to 1:3.Using this ratio, I can calculate the number of students who chose water by dividing the number of students who chose juice by 3. Therefore, 90 divided by 3 equals 30 students who chose water.This approach ensures that the proportions of juice and water choices are maintained, leading to the correct number of students who opted for water.

💡First, I need to determine the total runs scored in the first 4 matches by multiplying the average score by the number of matches.Next, I'll calculate the total runs scored in the last 3 matches using the provided average.After obtaining both totals, I'll sum them up to find the overall total runs scored in all 7 matches.Finally, I'll divide the overall total runs by the total number of matches to find the overall average score.

💡Okay, so I have this problem where I need to find cos(x + 2y) given these two equations:1. ( x^3 + sin x - 2a = 0 )2. ( 4y^3 + frac{1}{2} sin 2y + a = 0 )And x and y are both in the interval ([- frac{pi}{4}, frac{pi}{4}]), which is good because it tells me something about the possible values of x and y. Also, a is a real number. Hmm, okay. So, I need to find cos(x + 2y). That seems like it might involve some trigonometric identities or maybe even some clever substitutions. Let me see.First, let me write down the equations again:1. ( x^3 + sin x = 2a ) (I moved the 2a to the other side)2. ( 4y^3 + frac{1}{2} sin 2y = -a ) (Same here, moved a to the other side)Hmm, so equation 1 gives me 2a in terms of x, and equation 2 gives me -a in terms of y. Maybe I can relate these two somehow.Let me see. If I take equation 2, which is ( 4y^3 + frac{1}{2} sin 2y = -a ), and multiply both sides by 2, I get:( 8y^3 + sin 2y = -2a )Wait a second, equation 1 is ( x^3 + sin x = 2a ), so if I set them equal in some way... Let me write both:From equation 1: ( x^3 + sin x = 2a )From equation 2 multiplied by 2: ( 8y^3 + sin 2y = -2a )So if I add these two equations together, I get:( x^3 + sin x + 8y^3 + sin 2y = 0 )Hmm, that's interesting. But I'm not sure if that helps me directly. Maybe I should think about the functions involved here.Looking at equation 1: ( x^3 + sin x ). That's a function of x. Similarly, equation 2, after multiplying by 2, is ( 8y^3 + sin 2y ). Wait, 8y^3 is the same as (2y)^3, right? Because (2y)^3 = 8y^3. And sin 2y is just sin(2y). So maybe I can rewrite equation 2 as:( (2y)^3 + sin(2y) = -2a )Wait, that's similar to equation 1, which is ( x^3 + sin x = 2a ). So, if I let u = x and v = 2y, then equation 1 becomes ( u^3 + sin u = 2a ) and equation 2 becomes ( v^3 + sin v = -2a ). So, ( u^3 + sin u = 2a ) and ( v^3 + sin v = -2a ). If I add these two equations together, I get ( u^3 + sin u + v^3 + sin v = 0 ). Hmm, but that might not be directly helpful.Wait, but if I consider the function ( f(u) = u^3 + sin u ), then equation 1 is ( f(u) = 2a ) and equation 2 is ( f(v) = -2a ). So, ( f(u) = -f(v) ). Is this function f(u) odd? Let me check. An odd function satisfies ( f(-u) = -f(u) ). So, let's see:( f(-u) = (-u)^3 + sin(-u) = -u^3 - sin u = - (u^3 + sin u) = -f(u) ). Yes, it is an odd function. So, if ( f(u) = -f(v) ), then ( f(u) = f(-v) ). Since f is odd, ( f(u) = -f(v) ) implies ( f(u) = f(-v) ).Now, is f(u) injective? That is, is it one-to-one? If f(u) is strictly increasing, then it's injective. Let's check the derivative of f(u):( f'(u) = 3u^2 + cos u )Since u is in ([- frac{pi}{4}, frac{pi}{4}]), cos u is positive because cosine is positive in the first and fourth quadrants. Also, ( 3u^2 ) is always non-negative. So, ( f'(u) ) is always positive in this interval. Therefore, f(u) is strictly increasing on ([- frac{pi}{4}, frac{pi}{4}]), which means it's injective. So, if ( f(u) = f(-v) ), then u must equal -v. So, u = -v. But u is x, and v is 2y. So, x = -2y. Therefore, x + 2y = 0. Wait, that's perfect because I need to find cos(x + 2y). If x + 2y = 0, then cos(0) is 1. So, the answer is 1.But let me double-check to make sure I didn't make any mistakes. So, starting from the two equations:1. ( x^3 + sin x = 2a )2. ( 4y^3 + frac{1}{2} sin 2y = -a )I multiplied equation 2 by 2 to get ( 8y^3 + sin 2y = -2a ). Then, I noticed that 8y^3 is (2y)^3, so I rewrote equation 2 as ( (2y)^3 + sin(2y) = -2a ). Then, I defined f(u) = u^3 + sin u, which is an odd function and strictly increasing on the given interval. So, from equation 1, f(x) = 2a, and from equation 2, f(2y) = -2a. Since f is odd, f(2y) = -f(-2y). Therefore, f(x) = f(-2y), and since f is injective, x = -2y. Thus, x + 2y = 0, so cos(x + 2y) = cos(0) = 1. I think that makes sense. Let me check if there are any alternative approaches or if I might have missed something. Another way to think about it is to express a from both equations and set them equal. From equation 1, a = (x^3 + sin x)/2. From equation 2, a = - (4y^3 + (1/2) sin 2y). So, setting them equal:( frac{x^3 + sin x}{2} = - left(4y^3 + frac{1}{2} sin 2y right) )Multiply both sides by 2:( x^3 + sin x = -8y^3 - sin 2y )Which is the same as:( x^3 + sin x + 8y^3 + sin 2y = 0 )Which is the same equation I had earlier. So, that's consistent. Alternatively, maybe I can consider substituting x = -2y into the equations to see if it works. Let's try that. If x = -2y, then let's substitute into equation 1:( (-2y)^3 + sin(-2y) - 2a = 0 )Which simplifies to:( -8y^3 - sin 2y - 2a = 0 )So, ( -8y^3 - sin 2y = 2a )But from equation 2, we have:( 4y^3 + frac{1}{2} sin 2y + a = 0 )Multiply equation 2 by 2:( 8y^3 + sin 2y + 2a = 0 )Which gives:( 8y^3 + sin 2y = -2a )Comparing this with the result from substituting x = -2y into equation 1:( -8y^3 - sin 2y = 2a )Which can be rewritten as:( 8y^3 + sin 2y = -2a )Which is exactly the same as equation 2 multiplied by 2. So, this substitution is consistent. Therefore, x = -2y is indeed a solution.Thus, x + 2y = 0, so cos(x + 2y) = 1.I think that's solid. I don't see any mistakes in my reasoning. The key was recognizing that the function f(u) = u^3 + sin u is odd and strictly increasing, which allows us to set x = -2y. Then, it's straightforward to compute the cosine of zero, which is 1.**Final Answer**boxed{1}

💡Okay, so I need to evaluate the infinite series (sum_{k = 1}^infty frac{k^2}{3^k}). Hmm, infinite series can be tricky, but I remember that there are some standard techniques for dealing with series involving (k) and (k^2). Let me think about how to approach this.First, I recall that for geometric series, the sum (sum_{k=0}^infty r^k = frac{1}{1 - r}) when (|r| < 1). But in this case, we have (k^2) in the numerator, so it's not a simple geometric series. Maybe I can relate it to a geometric series somehow.I also remember that there are formulas for sums like (sum_{k=1}^infty k r^k) and (sum_{k=1}^infty k^2 r^k). Let me try to recall those. I think for the first one, (sum_{k=1}^infty k r^k), the sum is (frac{r}{(1 - r)^2}). And for the second one, (sum_{k=1}^infty k^2 r^k), I believe the formula is (frac{r(1 + r)}{(1 - r)^3}). Let me verify that.If I differentiate the basic geometric series term by term, I can get expressions for sums involving (k) and (k^2). Let's try that.Starting with the geometric series:[S = sum_{k=0}^infty r^k = frac{1}{1 - r}, quad |r| < 1.]If I differentiate both sides with respect to (r), I get:[frac{dS}{dr} = sum_{k=0}^infty k r^{k - 1} = frac{1}{(1 - r)^2}.]Multiplying both sides by (r) gives:[sum_{k=0}^infty k r^k = frac{r}{(1 - r)^2}.]But since the (k=0) term is zero, this simplifies to:[sum_{k=1}^infty k r^k = frac{r}{(1 - r)^2}.]Okay, that matches what I remembered. Now, to find (sum_{k=1}^infty k^2 r^k), I can differentiate the previous result again.Starting from:[sum_{k=1}^infty k r^k = frac{r}{(1 - r)^2}.]Differentiate both sides with respect to (r):[sum_{k=1}^infty k^2 r^{k - 1} = frac{(1 - r)^2 cdot 1 - r cdot 2(1 - r)(-1)}{(1 - r)^4}.]Wait, let me compute that derivative carefully. The right-hand side is (frac{r}{(1 - r)^2}). Let me denote this as (f(r) = frac{r}{(1 - r)^2}). Then,[f'(r) = frac{(1 - r)^2 cdot 1 - r cdot 2(1 - r)(-1)}{(1 - r)^4} = frac{(1 - r)^2 + 2r(1 - r)}{(1 - r)^4}.]Simplify the numerator:[(1 - r)^2 + 2r(1 - r) = (1 - 2r + r^2) + (2r - 2r^2) = 1 - 2r + r^2 + 2r - 2r^2 = 1 - r^2.]So,[f'(r) = frac{1 - r^2}{(1 - r)^4} = frac{(1 - r)(1 + r)}{(1 - r)^4} = frac{1 + r}{(1 - r)^3}.]Therefore,[sum_{k=1}^infty k^2 r^{k - 1} = frac{1 + r}{(1 - r)^3}.]Multiplying both sides by (r) to adjust the exponent:[sum_{k=1}^infty k^2 r^k = frac{r(1 + r)}{(1 - r)^3}.]Great, so that formula is correct. Now, applying this to our problem where (r = frac{1}{3}), since each term is (frac{k^2}{3^k} = k^2 left(frac{1}{3}right)^k).So, substituting (r = frac{1}{3}) into the formula:[sum_{k=1}^infty k^2 left(frac{1}{3}right)^k = frac{frac{1}{3} left(1 + frac{1}{3}right)}{left(1 - frac{1}{3}right)^3}.]Let me compute this step by step.First, compute the numerator:[frac{1}{3} left(1 + frac{1}{3}right) = frac{1}{3} cdot frac{4}{3} = frac{4}{9}.]Next, compute the denominator:[left(1 - frac{1}{3}right)^3 = left(frac{2}{3}right)^3 = frac{8}{27}.]So, putting it all together:[frac{frac{4}{9}}{frac{8}{27}} = frac{4}{9} times frac{27}{8} = frac{4 times 27}{9 times 8}.]Simplify the fractions:- 27 divided by 9 is 3.- 4 divided by 8 is 0.5, or (frac{1}{2}).So,[frac{4 times 27}{9 times 8} = frac{4}{8} times frac{27}{9} = frac{1}{2} times 3 = frac{3}{2}.]Wait, that gives me (frac{3}{2}), but when I look back at the initial steps, I think I might have made a mistake. Let me double-check my calculations.Wait, actually, let's recompute the division:[frac{4}{9} div frac{8}{27} = frac{4}{9} times frac{27}{8} = frac{4 times 27}{9 times 8}.]Simplify numerator and denominator:- 27 divided by 9 is 3.- 4 divided by 8 is (frac{1}{2}).So, it's (frac{1}{2} times 3 = frac{3}{2}). Hmm, so the sum is (frac{3}{2}). But wait, in the initial problem statement, when I tried to solve it using a different method, I ended up with (S = 1), which contradicts this result. That means I must have made a mistake somewhere.Wait, no, actually, in the initial problem, the user provided a solution that led to (S = 1), but that solution seems incorrect because when I applied the standard formula, I got (frac{3}{2}). Let me check the initial solution again.In the initial solution, they let (S = sum_{k=1}^infty frac{k^2}{3^k}), then multiplied by 3 to get (3S), subtracted to get (2S), and then multiplied by 3 again to get (6S), subtracted, and ended up with (4S = 4), so (S = 1). But according to the standard formula, it should be (frac{3}{2}). So, there must be an error in that initial solution.Let me go through the initial solution step by step to see where the mistake is.Starting with:[S = sum_{k=1}^infty frac{k^2}{3^k} = frac{1}{3} + frac{4}{9} + frac{9}{27} + frac{16}{81} + dotsb]Then, multiplying both sides by 3:[3S = 1 + frac{4}{3} + frac{9}{9} + frac{16}{27} + frac{25}{81} + dotsb]Subtracting the original series from this:[3S - S = 2S = 1 + left(frac{4}{3} - frac{1}{3}right) + left(frac{9}{9} - frac{4}{9}right) + left(frac{16}{27} - frac{9}{27}right) + dotsb]Simplifying each term:[2S = 1 + frac{3}{3} + frac{5}{9} + frac{7}{27} + frac{9}{81} + dotsb]Which simplifies to:[2S = 1 + 1 + frac{5}{9} + frac{7}{27} + frac{9}{81} + dotsb]Wait, that doesn't seem right. Let me compute the subtraction term by term:- The first term: (1 - 0 = 1)- The second term: (frac{4}{3} - frac{1}{3} = frac{3}{3} = 1)- The third term: (frac{9}{9} - frac{4}{9} = frac{5}{9})- The fourth term: (frac{16}{27} - frac{9}{27} = frac{7}{27})- The fifth term: (frac{25}{81} - frac{16}{81} = frac{9}{81} = frac{1}{9})- And so on.So, actually, (2S = 1 + 1 + frac{5}{9} + frac{7}{27} + frac{1}{9} + dotsb). Wait, that seems inconsistent. Let me write it properly:[2S = 1 + 1 + frac{5}{9} + frac{7}{27} + frac{9}{81} + dotsb]Hmm, this seems a bit messy. Maybe I should consider another approach. Alternatively, perhaps the initial solution made a mistake in the subtraction step.Wait, in the initial solution, after subtracting, they got:[2S = 1 + frac{4}{3} + frac{5}{9} + frac{7}{27} + frac{9}{81} + dotsb]But when I did the subtraction, I got:[2S = 1 + 1 + frac{5}{9} + frac{7}{27} + frac{9}{81} + dotsb]So, the initial solution seems to have an extra (frac{4}{3}) instead of 1. That might be where the mistake is.Let me try to correct that. So, starting again:Given:[S = sum_{k=1}^infty frac{k^2}{3^k}]Multiply both sides by 3:[3S = sum_{k=1}^infty frac{k^2}{3^{k-1}} = sum_{k=0}^infty frac{(k+1)^2}{3^k}]But since the original series starts at (k=1), when we shift the index, we have:[3S = sum_{k=0}^infty frac{(k+1)^2}{3^k} = sum_{k=1}^infty frac{k^2}{3^{k-1}} + frac{1^2}{3^0} = sum_{k=1}^infty frac{k^2}{3^{k-1}} + 1]Wait, that might complicate things. Alternatively, perhaps it's better to express (3S) as:[3S = sum_{k=1}^infty frac{k^2}{3^{k-1}} = sum_{k=1}^infty frac{k^2}{3^{k-1}} = sum_{k=0}^infty frac{(k+1)^2}{3^k}]But this seems more involved. Maybe another approach is better.Alternatively, let's consider using generating functions or another method. Wait, I already have the formula from differentiating the geometric series, which gave me (frac{3}{2}). So, perhaps the initial solution was incorrect, and the correct answer is (frac{3}{2}).But let me try to reconcile the two methods. If the standard formula gives (frac{3}{2}), but the initial solution gave (1), there must be a mistake in the initial solution.Looking back at the initial solution:They had:[2S = 1 + frac{4}{3} + frac{5}{9} + frac{7}{27} + frac{9}{81} + dotsb]Then they multiplied by 3:[6S = 3 + 4 + frac{5}{3} + frac{7}{9} + frac{9}{27} + dotsb]Subtracting the original (2S) from this:[6S - 2S = 4S = 3 + left(4 - frac{4}{3}right) + left(frac{5}{3} - frac{5}{9}right) + left(frac{7}{9} - frac{7}{27}right) + dotsb]Wait, that seems off. Let me compute the subtraction term by term:- The first term: (3 - 1 = 2)- The second term: (4 - frac{4}{3} = frac{12}{3} - frac{4}{3} = frac{8}{3})- The third term: (frac{5}{3} - frac{5}{9} = frac{15}{9} - frac{5}{9} = frac{10}{9})- The fourth term: (frac{7}{9} - frac{7}{27} = frac{21}{27} - frac{7}{27} = frac{14}{27})- The fifth term: (frac{9}{27} - frac{9}{81} = frac{27}{81} - frac{9}{81} = frac{18}{81} = frac{2}{9})- And so on.So, putting it together:[4S = 2 + frac{8}{3} + frac{10}{9} + frac{14}{27} + frac{2}{9} + dotsb]Wait, this seems complicated. Maybe the initial solution tried to represent this as a geometric series, but it's not straightforward. Perhaps the initial solution incorrectly assumed that the remaining terms form a geometric series, which they don't.In the initial solution, they wrote:[4S = 3 + frac{1}{3} + frac{2}{9} + frac{2}{27} + frac{2}{81} + dotsb]But from my calculation, it's:[4S = 2 + frac{8}{3} + frac{10}{9} + frac{14}{27} + frac{2}{9} + dotsb]Which is different. So, the initial solution made a mistake in the subtraction step, leading to an incorrect result.Therefore, relying on the standard formula derived from differentiating the geometric series, the correct sum should be (frac{3}{2}).But wait, let me double-check my application of the formula. The formula is:[sum_{k=1}^infty k^2 r^k = frac{r(1 + r)}{(1 - r)^3}]Substituting (r = frac{1}{3}):[frac{frac{1}{3}(1 + frac{1}{3})}{(1 - frac{1}{3})^3} = frac{frac{1}{3} cdot frac{4}{3}}{left(frac{2}{3}right)^3} = frac{frac{4}{9}}{frac{8}{27}} = frac{4}{9} times frac{27}{8} = frac{4 times 27}{9 times 8} = frac{108}{72} = frac{3}{2}]Yes, that's correct. So, the sum is indeed (frac{3}{2}).Therefore, the initial solution was incorrect, and the correct answer is (frac{3}{2}).

💡First, I need to determine the reduced volumes of each vegetable after cooking. Spinach reduces to 20% of its initial volume, mushrooms to 65%, and onions to 50%.For the spinach:40 ounces multiplied by 0.20 equals 8 ounces.For the mushrooms:25 ounces multiplied by 0.65 equals 16.25 ounces.For the onions:15 ounces multiplied by 0.50 equals 7.5 ounces.Next, I add the volumes of the cooked vegetables together:8 ounces (spinach) plus 16.25 ounces (mushrooms) plus 7.5 ounces (onions) equals 31.75 ounces.Then, I add the volume of the cream cheese and eggs:31.75 ounces (vegetables) plus 6 ounces (cream cheese) plus 4 ounces (eggs) equals 41.75 ounces.Therefore, the total volume of the quiche is 41.75 ounces.