💡First, I need to determine the area of the central area of the bathroom, which is a square with sides measuring 10 feet each.Next, I'll calculate the area of the hallway by subtracting the area of the central area from the total area of hardwood flooring installed.Finally, using the width of the hallway, I can find its length by dividing the area of the hallway by the width.

💡First, I need to determine the number of students enrolled in biology classes. Given that 47.5% of the 880 students are in biology classes, I can calculate this by multiplying 880 by 0.475.Next, I will subtract the number of students enrolled in biology classes from the total number of students to find out how many students are not enrolled in biology classes.Finally, I will present the final answer in a clear and concise manner.

💡Okay, so I have this problem: find all two-digit numbers that are divisible by the product of their digits. Hmm, let me try to understand what this means. A two-digit number can be represented as 10a + b, where 'a' is the tens digit and 'b' is the units digit. The product of the digits would then be a * b. So, the problem is asking for numbers where (10a + b) is divisible by (a * b). Alright, so I need to find all such numbers where this condition holds true. Let me think about how to approach this. Maybe I can start by considering each two-digit number and check if it satisfies the condition. But that might take a long time since there are 90 two-digit numbers (from 10 to 99). Maybe there's a smarter way to narrow it down.First, let's note that both 'a' and 'b' are digits, so they can range from 0 to 9. However, since it's a two-digit number, 'a' can't be 0. So 'a' is from 1 to 9, and 'b' is from 0 to 9. Now, the condition is (10a + b) divisible by (a * b). That means (10a + b) / (a * b) should be an integer. Let's denote this as:(10a + b) / (a * b) = k, where k is an integer.So, (10a + b) = k * (a * b). Let's rearrange this equation:10a + b = k * a * b10a = k * a * b - b10a = b * (k * a - 1)Hmm, so 10a must be divisible by (k * a - 1). That seems a bit complicated. Maybe I can think of it differently. Let's consider that (10a + b) must be divisible by both 'a' and 'b' individually because if it's divisible by their product, it should be divisible by each of them separately.So, first, (10a + b) must be divisible by 'a'. Let's check that:(10a + b) / a = 10 + (b / a)For this to be an integer, b must be divisible by a. So, b = m * a, where m is an integer. Since both 'a' and 'b' are digits, m can only be 0, 1, 2, 3, 4, or 5 because if m is 6 or higher, b would exceed 9.Similarly, (10a + b) must be divisible by 'b'. So:(10a + b) / b = (10a / b) + 1This must also be an integer, meaning 10a must be divisible by b. But since b = m * a, we can substitute:10a / (m * a) = 10 / mSo, 10 / m must be an integer. Therefore, m must be a divisor of 10. The divisors of 10 are 1, 2, 5, and 10. But since m is a digit, m can only be 1, 2, or 5.So, m can be 1, 2, or 5. Let's consider each case:1. **Case m = 1**: - Then b = a * 1 = a. - So, the number is 10a + a = 11a. - We need 11a to be divisible by a * a = a². - So, 11a / a² = 11 / a must be an integer. - Therefore, a must be a divisor of 11. The divisors of 11 are 1 and 11, but since a is a digit, a = 1. - So, the number is 11 * 1 = 11.2. **Case m = 2**: - Then b = a * 2. - The number is 10a + 2a = 12a. - We need 12a to be divisible by a * 2a = 2a². - So, 12a / 2a² = 6 / a must be an integer. - Therefore, a must be a divisor of 6. The divisors of 6 are 1, 2, 3, 6. - But since a is a digit, a can be 1, 2, 3, or 6. - Let's check each: - a = 1: b = 2, number = 12. Check if 12 is divisible by 1*2=2. Yes, 12/2=6. - a = 2: b = 4, number = 24. Check if 24 is divisible by 2*4=8. Yes, 24/8=3. - a = 3: b = 6, number = 36. Check if 36 is divisible by 3*6=18. Yes, 36/18=2. - a = 6: b = 12, but b must be a single digit, so this is invalid.3. **Case m = 5**: - Then b = a * 5. - The number is 10a + 5a = 15a. - We need 15a to be divisible by a * 5a = 5a². - So, 15a / 5a² = 3 / a must be an integer. - Therefore, a must be a divisor of 3. The divisors of 3 are 1 and 3. - Let's check each: - a = 1: b = 5, number = 15. Check if 15 is divisible by 1*5=5. Yes, 15/5=3. - a = 3: b = 15, but b must be a single digit, so this is invalid.So, compiling all the valid numbers we found:- From m=1: 11- From m=2: 12, 24, 36- From m=5: 15Let me double-check each of these:- 11: 1*1=1, 11/1=11 ✔️- 12: 1*2=2, 12/2=6 ✔️- 24: 2*4=8, 24/8=3 ✔️- 36: 3*6=18, 36/18=2 ✔️- 15: 1*5=5, 15/5=3 ✔️Looks like all these numbers satisfy the condition. I don't think I missed any because I considered all possible values of m that make sense given the constraints of digits. So, these should be all the two-digit numbers that are divisible by the product of their digits.

💡Alright, so I have this problem here about a function f(x) = |ax - 1|. The solution set for f(x) ≤ 2 is given as [-1, 3]. I need to find the value of a, and then use that to find the minimum value of another expression involving x, y, and z. Let me take it step by step.First, part (1): Find the value of the real number a.Okay, so f(x) = |ax - 1|, and we know that f(x) ≤ 2. That means |ax - 1| ≤ 2. I remember that absolute value inequalities can be rewritten without the absolute value by creating two separate inequalities. So, |ax - 1| ≤ 2 implies that -2 ≤ ax - 1 ≤ 2.Let me write that down:-2 ≤ ax - 1 ≤ 2.Now, I can solve for x in this inequality. Let's add 1 to all parts:-2 + 1 ≤ ax ≤ 2 + 1,which simplifies to:-1 ≤ ax ≤ 3.Now, to solve for x, I need to divide all parts by a. But wait, I don't know if a is positive or negative. If a is positive, the direction of the inequalities remains the same. If a is negative, the inequalities will reverse. Hmm, that's important.The solution set is given as [-1, 3]. So, the inequality -1 ≤ x ≤ 3 is the solution. Let me consider two cases: a positive and a negative.Case 1: a > 0.If a is positive, then dividing by a doesn't change the inequality signs. So,-1/a ≤ x ≤ 3/a.But the solution set is [-1, 3], so:-1/a = -1 and 3/a = 3.Let me solve these equations.From -1/a = -1:Multiply both sides by a: -1 = -a.So, a = 1.From 3/a = 3:Multiply both sides by a: 3 = 3a.So, a = 1.Great, both equations give a = 1. So, in this case, a is 1.Case 2: a < 0.If a is negative, dividing by a reverses the inequality signs. So,-1/a ≥ x ≥ 3/a.But the solution set is [-1, 3], which is x between -1 and 3. So, in this case, the inequality would be:3/a ≤ x ≤ -1/a.But the solution set is x between -1 and 3, so:3/a = -1 and -1/a = 3.Let me solve these.From 3/a = -1:Multiply both sides by a: 3 = -a.So, a = -3.From -1/a = 3:Multiply both sides by a: -1 = 3a.So, a = -1/3.Wait, that's a problem. In this case, a would have to be both -3 and -1/3, which is impossible. So, this case doesn't give a consistent solution. Therefore, a cannot be negative.Case 3: a = 0.If a is 0, then f(x) = |0*x - 1| = |-1| = 1. So, f(x) = 1 for all x. Then, f(x) ≤ 2 would be true for all real numbers x, not just [-1, 3]. So, a can't be 0.Therefore, the only valid solution is a = 1.Okay, so part (1) is solved, a = 1.Now, part (2): If x + y + z = a (where x, y, z are positive real numbers), find the minimum value of u = 1/(x + y) + (x + y)/z.Since we found a = 1, the equation becomes x + y + z = 1, with x, y, z > 0.We need to minimize u = 1/(x + y) + (x + y)/z.Hmm, okay. Let me think about how to approach this. It seems like an optimization problem with constraints. Maybe I can use calculus or some inequality like AM-GM.First, let's note that x + y + z = 1, so z = 1 - x - y. Since x, y, z are positive, x + y must be less than 1.Let me denote s = x + y. Then, since x, y > 0, s is between 0 and 1. So, z = 1 - s.Therefore, u can be written in terms of s:u = 1/s + s/z.But z = 1 - s, so:u = 1/s + s/(1 - s).So, u is now a function of s: u(s) = 1/s + s/(1 - s), where 0 < s < 1.We need to find the minimum of u(s) in the interval (0, 1).Okay, so this reduces the problem to a single variable optimization. Let's try to find the derivative of u with respect to s and set it to zero to find critical points.First, compute u(s):u(s) = 1/s + s/(1 - s).Compute the derivative u’(s):u’(s) = d/ds [1/s] + d/ds [s/(1 - s)].Compute each derivative separately.d/ds [1/s] = -1/s².For the second term, d/ds [s/(1 - s)]:Use the quotient rule: [ (1 - s)(1) - s(-1) ] / (1 - s)^2.Simplify numerator:(1 - s) + s = 1.So, derivative is 1/(1 - s)^2.Therefore, u’(s) = -1/s² + 1/(1 - s)^2.Set derivative equal to zero:-1/s² + 1/(1 - s)^2 = 0.Bring one term to the other side:1/(1 - s)^2 = 1/s².Take reciprocals on both sides:(1 - s)^2 = s².Take square roots (since both sides are positive):|1 - s| = |s|.But since 0 < s < 1, both 1 - s and s are positive, so:1 - s = s.Solve for s:1 = 2s => s = 1/2.So, critical point at s = 1/2.Now, we need to check if this is a minimum.Compute the second derivative or test intervals around s = 1/2.Alternatively, since it's the only critical point in (0,1), and as s approaches 0, u(s) approaches infinity, and as s approaches 1, u(s) also approaches infinity, so s = 1/2 must be a minimum.Therefore, the minimum occurs at s = 1/2.Now, compute u(1/2):u(1/2) = 1/(1/2) + (1/2)/(1 - 1/2) = 2 + (1/2)/(1/2) = 2 + 1 = 3.So, the minimum value of u is 3.Wait, but let me double-check. If s = 1/2, then z = 1 - s = 1/2. So, z = 1/2.Also, since s = x + y = 1/2, and x, y > 0, we can choose x = y = 1/4 to satisfy x + y = 1/2.So, x = y = 1/4, z = 1/2.Compute u:1/(x + y) + (x + y)/z = 1/(1/2) + (1/2)/(1/2) = 2 + 1 = 3.Yes, that's correct.Alternatively, using AM-GM inequality.Let me see if I can find another way without calculus.We have u = 1/s + s/z, with s + z = 1.Wait, s = x + y, z = 1 - s.So, u = 1/s + s/(1 - s).Let me denote t = s, so u = 1/t + t/(1 - t).We can write u as:u = 1/t + t/(1 - t) = (1 - t)/t(1 - t) + t^2/t(1 - t) = [ (1 - t) + t^2 ] / [ t(1 - t) ].Wait, maybe that's not helpful.Alternatively, let me consider the two terms: 1/t and t/(1 - t).Let me set t = 1/2, then u = 2 + 1 = 3.Is there a way to apply AM-GM here?AM-GM states that for positive numbers a and b, (a + b)/2 ≥ sqrt(ab), with equality when a = b.But u is 1/t + t/(1 - t). It's not immediately clear how to apply AM-GM here.Alternatively, let me consider substituting t = 1/2.Wait, but I already found the minimum using calculus, so maybe that's sufficient.Alternatively, let me consider writing u in terms of t and 1 - t.Wait, u = 1/t + t/(1 - t).Let me write this as:u = 1/t + t/(1 - t) = (1 - t + t^2)/(t(1 - t)).Hmm, not sure.Alternatively, let me consider u = 1/t + t/(1 - t).Let me set t = 1/2, which gives u = 2 + 1 = 3.If I set t = 1/3, then u = 3 + (1/3)/(2/3) = 3 + 1/2 = 3.5, which is larger.If I set t = 2/3, then u = 3/2 + (2/3)/(1/3) = 1.5 + 2 = 3.5, which is also larger.So, it seems that t = 1/2 gives the minimum.Alternatively, let me consider the function u(t) = 1/t + t/(1 - t).Let me compute its derivative again:u’(t) = -1/t² + [ (1 - t) - t(-1) ] / (1 - t)^2 = -1/t² + 1/(1 - t)^2.Set to zero: -1/t² + 1/(1 - t)^2 = 0 => 1/(1 - t)^2 = 1/t² => (1 - t)^2 = t² => 1 - 2t + t² = t² => 1 - 2t = 0 => t = 1/2.So, same result.Therefore, the minimum value is 3.So, the answer is 3.Wait, but let me think again. Is there a way to express u in terms of variables that can be directly applied with AM-GM?Let me consider the expression u = 1/(x + y) + (x + y)/z.Let me denote s = x + y, so u = 1/s + s/z.Given that s + z = 1, so z = 1 - s.Therefore, u = 1/s + s/(1 - s).Let me consider the two terms: 1/s and s/(1 - s).Let me denote a = 1/s and b = s/(1 - s).Then, u = a + b.I want to minimize a + b.But I don't see an immediate way to apply AM-GM here because a and b are not symmetric.Alternatively, let me consider the function u(s) = 1/s + s/(1 - s).Let me write it as:u(s) = 1/s + s/(1 - s) = (1 - s + s²)/(s(1 - s)).Hmm, not helpful.Alternatively, let me consider substituting s = t, so u(t) = 1/t + t/(1 - t).Let me compute u(t) at t = 1/2: 2 + 1 = 3.At t approaching 0: u(t) approaches infinity.At t approaching 1: u(t) approaches infinity.So, the function has a minimum at t = 1/2.Therefore, the minimum value is 3.Yes, that seems consistent.So, the minimum value of u is 3.Therefore, the answers are:(1) a = 1(2) The minimum value of u is 3.

💡Okay, so I have this problem where I need to find the value of a + b + c, given that ab = 36, ac = 72, and bc = 108. Hmm, all three variables are positive numbers, which is good because it means I don't have to worry about negative values messing things up. Let me see how I can approach this.First, I know that I have three equations with three variables. That usually means I can solve for each variable by substitution or elimination. Let me write down the equations again to make it clearer:1. ab = 362. ac = 723. bc = 108I think I can express two of the variables in terms of the third one and then substitute them into the third equation to solve for that variable. Let me try expressing b and c in terms of a.From the first equation, ab = 36, I can solve for b:b = 36 / aSimilarly, from the second equation, ac = 72, I can solve for c:c = 72 / aOkay, so now I have expressions for both b and c in terms of a. Let me plug these into the third equation, which is bc = 108.So, substituting b and c:(36 / a) * (72 / a) = 108Let me compute that:(36 * 72) / (a * a) = 10836 multiplied by 72... Let me calculate that. 36 times 70 is 2520, and 36 times 2 is 72, so total is 2520 + 72 = 2592. So,2592 / a² = 108Now, I can solve for a²:a² = 2592 / 108Let me compute 2592 divided by 108. Hmm, 108 times 24 is 2592 because 100*24=2400 and 8*24=192, so 2400+192=2592. So,a² = 24Therefore, a = sqrt(24). Since a is positive, I don't need to worry about the negative root. Simplifying sqrt(24), which is sqrt(4*6) = 2*sqrt(6). So,a = 2√6Now that I have a, I can find b and c using the expressions I had earlier.Starting with b:b = 36 / a = 36 / (2√6) = (36 / 2) / √6 = 18 / √6But it's better to rationalize the denominator. So, multiply numerator and denominator by √6:18 / √6 = (18√6) / (√6 * √6) = (18√6) / 6 = 3√6So, b = 3√6Now, finding c:c = 72 / a = 72 / (2√6) = (72 / 2) / √6 = 36 / √6Again, rationalizing the denominator:36 / √6 = (36√6) / (√6 * √6) = (36√6) / 6 = 6√6So, c = 6√6Now, I have all three variables:a = 2√6b = 3√6c = 6√6To find a + b + c, I just add them up:a + b + c = 2√6 + 3√6 + 6√6Since all terms have √6, I can factor that out:= (2 + 3 + 6)√6Adding the coefficients:2 + 3 = 5, and 5 + 6 = 11So,a + b + c = 11√6Let me double-check my steps to make sure I didn't make a mistake. Starting from the equations, expressing b and c in terms of a, substituting into the third equation, solving for a, then finding b and c. The calculations seem correct. The multiplication and division steps look accurate, and the rationalization of denominators was done properly. Adding the coefficients at the end also seems right. So, I think 11√6 is the correct answer.**Final Answer**The value of (a + b + c) is boxed{11sqrt{6}}.

💡To compute the binomial coefficient (dbinom{9}{5}), I start by using the formula:[dbinom{9}{5} = frac{9!}{5! times 4!}]Next, I notice that (9!) can be expressed as (9 times 8 times 7 times 6 times 5!), which allows me to cancel out the (5!) in the denominator:[dbinom{9}{5} = frac{9 times 8 times 7 times 6 times 5!}{5! times 4!} = frac{9 times 8 times 7 times 6}{4!}]Since (4! = 24), the expression simplifies to:[frac{9 times 8 times 7 times 6}{24}]Calculating the numerator:[9 times 8 = 72 72 times 7 = 504 504 times 6 = 3024]Finally, dividing by 24:[frac{3024}{24} = 126]Thus, the value of (dbinom{9}{5}) is 126.