💡First, I recognize that the two triangles are similar, which means their corresponding linear dimensions are proportional.The ratio of their areas is given as 1:9. To find the ratio of their corresponding linear dimensions, I take the square root of the area ratio.Calculating the square root of 9 gives me 3, so the ratio of their heights is 1:3.Given that the height of the smaller triangle is 5 cm, I multiply this by the scaling factor of 3 to find the height of the larger triangle.5 cm multiplied by 3 equals 15 cm.Therefore, the height of the larger triangle is 15 centimeters.

💡Okay, so I have this probability problem here about a college student who needs to finish both a chemistry and a math problem set by February 24, 2006. He has the night of February 23 to work on them, but he can only work in one library because he doesn't have a bicycle. The math library and the chemistry library have different probabilities for him finishing each problem set. I need to find the probability that he worked in the math library given that he finished both problem sets that night.Alright, let's break this down step by step. First, I need to understand the given probabilities and what we're trying to find. The student can choose between two libraries: the math library or the chemistry library. He chooses the math library with a probability of 60%, which means he chooses the chemistry library with a probability of 40%. If he works in the math library, the probability of finishing the math problem set is 95%, and the probability of finishing the chemistry problem set is 75%. On the other hand, if he works in the chemistry library, the probability of finishing the chemistry problem set is 90%, and the probability of finishing the math problem set is 80%. We need to find the probability that he worked in the math library given that he finished both problem sets. This sounds like a conditional probability problem, so I think I should use Bayes' Theorem here. Bayes' Theorem relates the conditional and marginal probabilities of random events, which seems perfect for this situation.Let me recall Bayes' Theorem formula:[ P(A|B) = frac{P(B|A) cdot P(A)}{P(B)} ]In this context, event A is "worked in the math library," and event B is "finished both problem sets." So, we need to find ( P(M|A) ), where M is working in the math library and A is finishing both problem sets.First, I need to define all the probabilities:- ( P(M) = 0.60 ) (probability of choosing the math library)- ( P(C) = 0.40 ) (probability of choosing the chemistry library)- ( P(text{Math finished} | M) = 0.95 )- ( P(text{Chemistry finished} | M) = 0.75 )- ( P(text{Chemistry finished} | C) = 0.90 )- ( P(text{Math finished} | C) = 0.80 )Now, to find ( P(A|B) ), I need to calculate the joint probabilities of finishing both problem sets in each library and then apply Bayes' Theorem.First, let's find the probability of finishing both problem sets given that he worked in the math library:[ P(text{Math finished} cap text{Chemistry finished} | M) = P(text{Math finished} | M) times P(text{Chemistry finished} | M) ][ = 0.95 times 0.75 = 0.7125 ]Similarly, the probability of finishing both problem sets given that he worked in the chemistry library:[ P(text{Math finished} cap text{Chemistry finished} | C) = P(text{Math finished} | C) times P(text{Chemistry finished} | C) ][ = 0.80 times 0.90 = 0.72 ]Now, to find the total probability of finishing both problem sets regardless of the library, we use the law of total probability:[ P(text{Both finished}) = P(text{Both finished} | M) times P(M) + P(text{Both finished} | C) times P(C) ][ = 0.7125 times 0.60 + 0.72 times 0.40 ][ = 0.4275 + 0.288 = 0.7155 ]Now, applying Bayes' Theorem:[ P(M | text{Both finished}) = frac{P(text{Both finished} | M) times P(M)}{P(text{Both finished})} ][ = frac{0.7125 times 0.60}{0.7155} ][ = frac{0.4275}{0.7155} ]Let me compute this division:[ frac{0.4275}{0.7155} approx 0.597 ]So, approximately 59.7%, which is roughly 60%. Wait, that seems interesting because the prior probability of choosing the math library was 60%, and the posterior probability is also around 60%. Does that mean that the evidence (finishing both problem sets) doesn't change the probability much? Or maybe I made a mistake in my calculations.Let me double-check the calculations:First, ( 0.95 times 0.75 = 0.7125 ) – that's correct.Then, ( 0.80 times 0.90 = 0.72 ) – that's also correct.Next, ( 0.7125 times 0.60 = 0.4275 ) and ( 0.72 times 0.40 = 0.288 ). Adding those together: ( 0.4275 + 0.288 = 0.7155 ) – correct.Then, ( 0.4275 / 0.7155 ). Let me compute this more precisely:Divide numerator and denominator by 0.0005 to eliminate decimals:Numerator: 0.4275 / 0.0005 = 855Denominator: 0.7155 / 0.0005 = 1431So, ( 855 / 1431 ). Let's simplify this fraction.Divide numerator and denominator by 3:855 ÷ 3 = 2851431 ÷ 3 = 477Again, divide by 3:285 ÷ 3 = 95477 ÷ 3 = 159So, the simplified fraction is ( frac{95}{159} ).Calculating ( 95 ÷ 159 ) gives approximately 0.597, which is about 59.7%, as I had before.So, the probability that he worked in the math library given that he finished both problem sets is ( frac{95}{159} ) or approximately 59.7%.Wait, but the prior probability was 60%, and the posterior is almost the same. That suggests that the evidence (finishing both problem sets) doesn't significantly change the probability. Is that correct?Let me think about it. If both libraries have similar probabilities of finishing both problem sets, then the posterior probability would be close to the prior. In this case, the math library has a slightly lower probability (0.7125) compared to the chemistry library (0.72). So, the chemistry library is slightly better for finishing both problem sets. Therefore, if he finished both, it's slightly less likely he was in the math library. But since the prior was 60%, the posterior is slightly less, around 59.7%, which is almost the same.So, it seems my calculations are consistent. Therefore, the probability is ( frac{95}{159} ).**Final Answer**The probability that he worked in the math library given that he finished both problem sets is boxed{dfrac{95}{159}}.

💡First, I need to determine the total amount of work involved. The original plan was to complete the work in 8 days with 30 men. Therefore, the total work can be calculated as:[text{Total Work} = 30 text{ men} times 8 text{ days} = 240 text{ man-days}]Next, the work was finished 3 days earlier, which means it was completed in 5 days. Let ( x ) represent the number of additional men who joined. The total number of men working on the project then becomes ( 30 + x ).The work done with the additional men is:[text{Work Done} = (30 + x) text{ men} times 5 text{ days}]Since the total work remains the same, I can set up the equation:[(30 + x) times 5 = 240]Solving for ( x ):[150 + 5x = 240][5x = 240 - 150][5x = 90][x = frac{90}{5}][x = 18]Therefore, 18 additional men joined to finish the work earlier.

💡Okay, so I have this problem about a cyclic quadrilateral (ABCD) with sides (AB = a), (BC = b), (CD = c), (DA = d), and diagonals (AC = e), (BD = f). The equation I need to solve is:[ a sqrt[3]{x + c^3} + b sqrt[3]{x + d^3} = e sqrt[3]{x + f^3}.]Hmm, cyclic quadrilaterals have some special properties, like the sum of opposite angles being 180 degrees, and there are formulas like Ptolemy's theorem which relates the sides and diagonals. Maybe that could be useful here.First, let me understand the equation. It involves cube roots, which can sometimes be tricky. The variable (x) is inside these cube roots, so I need to find all non-negative real (x) that satisfy this equation.I wonder if there's a substitution or transformation that can simplify this equation. Maybe if I let (y = sqrt[3]{x}), but that might not help directly because the cube roots are of (x + c^3), not just (x). Alternatively, perhaps I can cube both sides to eliminate the cube roots, but that might complicate things because of the addition on the left side.Let me think about the structure of the equation. It's a linear combination of two cube roots on the left equal to another cube root on the right. Maybe I can consider each term separately and see if there's a relationship between them.Since (ABCD) is cyclic, Ptolemy's theorem tells us that (AC cdot BD = AB cdot CD + BC cdot DA), which translates to:[ e cdot f = a cdot c + b cdot d. ]That's interesting because the coefficients (a), (b), and (e) are related through the sides and diagonals. Maybe this relationship can be used in the equation.Let me write down the equation again:[ a sqrt[3]{x + c^3} + b sqrt[3]{x + d^3} = e sqrt[3]{x + f^3}.]I notice that each term has a similar structure: a coefficient multiplied by a cube root of (x) plus another cube. Maybe I can factor out something or find a common term.What if I consider each cube root as a function of (x)? Let me define a function (F(y) = sqrt[3]{x + y^3}). Then the equation becomes:[ a F(c) + b F(d) = e F(f). ]Hmm, that might be a useful perspective. Now, I can think about the properties of this function (F(y)).Let me compute the derivative of (F(y)) with respect to (y) to understand its behavior. The derivative (F'(y)) is:[ F'(y) = frac{d}{dy} sqrt[3]{x + y^3} = frac{1}{3}(x + y^3)^{-2/3} cdot 3y^2 = frac{y^2}{(x + y^3)^{2/3}}. ]So, (F'(y)) is positive because both numerator and denominator are positive for real (y) and (x geq 0). That means (F(y)) is an increasing function of (y).Wait, but in the equation, we have (a F(c) + b F(d) = e F(f)). Since (F) is increasing, larger (y) will give larger (F(y)). So, if (f) is larger than both (c) and (d), then (F(f)) would be larger than both (F(c)) and (F(d)). But the left side is a combination of (F(c)) and (F(d)) scaled by (a) and (b), respectively.I wonder if there's a way to relate (F(f)) to (F(c)) and (F(d)) using the properties of cyclic quadrilaterals. From Ptolemy's theorem, we have (e f = a c + b d). Maybe I can use this to relate the terms in the equation.Let me think about the function (F(y)) again. Since it's increasing, maybe I can apply some inequality here. For example, if I can express (F(f)) in terms of (F(c)) and (F(d)), perhaps using the fact that (e f = a c + b d).Alternatively, maybe I can consider the function (G(y) = F(y) - y). Let me define that:[ G(y) = sqrt[3]{x + y^3} - y. ]Then, the original equation becomes:[ a G(c) + b G(d) = e G(f). ]Hmm, interesting. Let me compute the derivative of (G(y)):[ G'(y) = frac{y^2}{(x + y^3)^{2/3}} - 1. ]So, (G'(y)) is equal to (F'(y) - 1). Since (F'(y) = frac{y^2}{(x + y^3)^{2/3}}), which is less than 1 because (x + y^3 > y^3), so the denominator is larger, making the fraction smaller than 1. Therefore, (G'(y) < 0), which means (G(y)) is a decreasing function.So, (G(y)) is decreasing. That means as (y) increases, (G(y)) decreases. So, if (f > c) and (f > d), then (G(f) < G(c)) and (G(f) < G(d)). But in our equation, we have (a G(c) + b G(d) = e G(f)). Since (G(f)) is less than both (G(c)) and (G(d)), the right-hand side is smaller than each term on the left. But since (a) and (b) are positive, the left-hand side is a sum of positive terms, while the right-hand side is a single term scaled by (e). Wait, but from Ptolemy's theorem, (e f = a c + b d). So, (e = frac{a c + b d}{f}). Maybe I can substitute that into the equation:[ a G(c) + b G(d) = frac{a c + b d}{f} G(f). ]Hmm, not sure if that helps directly. Maybe I can think about the behavior of (G(y)) as (x) changes.If (x = 0), then (G(y) = sqrt[3]{y^3} - y = y - y = 0). So, at (x = 0), all terms (G(c)), (G(d)), and (G(f)) are zero, making the equation (0 = 0). So, (x = 0) is a solution.But the problem asks for all non-negative real solutions, so maybe (x = 0) is the only solution. Let me check for (x > 0).Suppose (x > 0). Then, (G(y) = sqrt[3]{x + y^3} - y). Since (x > 0), (sqrt[3]{x + y^3} > y), so (G(y) > 0). Therefore, all terms (G(c)), (G(d)), and (G(f)) are positive.But since (G(y)) is decreasing, as (y) increases, (G(y)) decreases. So, (G(f)) is less than both (G(c)) and (G(d)) if (f > c) and (f > d). But in a cyclic quadrilateral, the diagonals are usually longer than the sides, so (f > c) and (f > d) might hold. Let me assume that (f > c) and (f > d).Therefore, (G(f) < G(c)) and (G(f) < G(d)). So, the right-hand side (e G(f)) is less than both (a G(c)) and (b G(d)), but the left-hand side is a sum of (a G(c)) and (b G(d)), which are both larger than (e G(f)). So, the left-hand side is greater than the right-hand side. Therefore, the equation cannot hold for (x > 0).Wait, but this seems to suggest that (x = 0) is the only solution. Let me verify this.At (x = 0), as I saw earlier, all (G(y)) terms are zero, so the equation holds. For (x > 0), the left-hand side is greater than the right-hand side, so the equation doesn't hold. For (x < 0), but the problem specifies non-negative solutions, so we don't need to consider negative (x).Therefore, the only non-negative real solution is (x = 0).But wait, let me double-check. Suppose (x = 0), then the equation becomes:[ a sqrt[3]{c^3} + b sqrt[3]{d^3} = e sqrt[3]{f^3}, ]which simplifies to:[ a c + b d = e f. ]But from Ptolemy's theorem, we have (e f = a c + b d). So, indeed, (x = 0) satisfies the equation.Is there any other (x) that could satisfy the equation? From the earlier reasoning, for (x > 0), the left-hand side is greater than the right-hand side, so no solution exists there. For (x = 0), it's exactly equal. So, (x = 0) is the only solution.Therefore, the only non-negative real solution is (x = 0).

💡First, I recognize that 90% of the study group members are women. This means that out of the total number of members, 90% are women.Next, I need to find out what percentage of these women are lawyers. Let's denote this unknown percentage as ( L ).The probability that a randomly selected member is a woman lawyer is given as 0.54. This probability can be broken down into two parts: the probability of selecting a woman and the probability that the selected woman is a lawyer.Mathematically, this relationship can be expressed as:[0.90 times L = 0.54]To find ( L ), I divide both sides of the equation by 0.90:[L = frac{0.54}{0.90} = 0.60]Converting this decimal to a percentage, I find that 60% of the women in the study group are lawyers.

💡To determine the degree of ( h(x) ), we start by analyzing the given polynomial ( f(x) = -5x^5 + 2x^4 + 7x - 8 ). The goal is to find a polynomial ( h(x) ) such that the degree of ( f(x) - h(x) ) is 3.1. **Identify the highest degree terms in ( f(x) ):** - The term with the highest degree in ( f(x) ) is ( -5x^5 ), which is of degree 5. - The next highest term is ( 2x^4 ), which is of degree 4.2. **Determine the terms in ( h(x) ) needed to cancel the higher-degree terms in ( f(x) ):** - To cancel the ( x^5 ) term in ( f(x) ), ( h(x) ) must include a term ( 5x^5 ). - To cancel the ( x^4 ) term in ( f(x) ), ( h(x) ) must include a term ( -2x^4 ).3. **Construct ( h(x) ) based on the cancellation requirements:** - ( h(x) ) must have at least the terms ( 5x^5 ) and ( -2x^4 ) to cancel the corresponding terms in ( f(x) ). - The remaining terms in ( h(x) ) (if any) do not affect the highest degree of ( f(x) - h(x) ), which is 3.4. **Conclude the degree of ( h(x) ):** - Since ( h(x) ) contains an ( x^5 ) term, the degree of ( h(x) ) is 5.Thus, the degree of ( h(x) ) is (boxed{5}).