💡First, I need to determine the total area of the four walls of the room. The room has two pairs of walls with the same dimensions. For the walls with the width of 20 feet:2 × (Height × Width) = 2 × (15 feet × 20 feet) = 600 square feetFor the walls with the length of 40 feet:2 × (Height × Length) = 2 × (15 feet × 40 feet) = 1200 square feetAdding these together gives the total area of the four walls:600 square feet + 1200 square feet = 1800 square feetNext, I'll calculate the total area of the doors and windows that will not be whitewashed.Area of the first door (7 feet × 4 feet):7 × 4 = 28 square feetArea of the second door (5 feet × 3 feet):5 × 3 = 15 square feetArea of the first window (5 feet × 4 feet):5 × 4 = 20 square feetArea of the second window (4 feet × 3 feet):4 × 3 = 12 square feetArea of the third window (3 feet × 3 feet):3 × 3 = 9 square feetArea of the fourth window (4 feet × 2.5 feet):4 × 2.5 = 10 square feetArea of the fifth window (6 feet × 4 feet):6 × 4 = 24 square feetAdding these areas together gives the total area of the doors and windows:28 + 15 + 20 + 12 + 9 + 10 + 24 = 118 square feetSubtracting the total area of the doors and windows from the total area of the walls gives the area to be whitewashed:1800 square feet - 118 square feet = 1682 square feetFinally, calculating the cost of whitewashing at Rs. 10 per square foot:1682 × 10 = Rs. 16820Therefore, the total cost of whitewashing the four walls, excluding the doors and windows, is Rs. 16,820.

💡To determine the domain of the function ( f(x) = frac{1}{sqrt{2x - 3}} ), we need to ensure that the expression inside the square root is positive and that the denominator is not zero.1. **Condition for the Square Root:** The expression inside the square root, ( 2x - 3 ), must be greater than zero because we cannot take the square root of a negative number or zero. [ 2x - 3 > 0 ] Solving for ( x ): [ 2x > 3 implies x > frac{3}{2} ]2. **Condition for the Denominator:** The denominator ( sqrt{2x - 3} ) must not be zero because division by zero is undefined. However, since ( 2x - 3 > 0 ) already ensures that ( sqrt{2x - 3} ) is not zero, this condition is automatically satisfied.3. **Combining Conditions:** The only condition we need to consider is ( x > frac{3}{2} ). Therefore, the domain of the function is all real numbers greater than ( frac{3}{2} ).4. **Conclusion:** The domain of ( f(x) ) is ( left( frac{3}{2}, +infty right) ).Thus, the correct option is (boxed{D}).

💡First, I need to determine the total number of balls in the bag. I know there are 8 black balls and 4 white balls, so together that's 12 balls.Next, I'm given that the frequency of drawing a red ball stabilizes at 0.4. This means that 40% of the time, a red ball is drawn.Since the total probability of drawing any ball must add up to 1, the probability of drawing a black or white ball combined is 1 - 0.4 = 0.6.Using the number of black and white balls, I can set up the equation 12 / Total = 0.6 to find the total number of balls.Solving for Total, I get Total = 12 / 0.6 = 20.Finally, to find the number of red balls, I subtract the number of black and white balls from the total: 20 - 12 = 8.Therefore, there are approximately 8 red balls in the bag.

💡Alright, so I have this problem about the function f(x) = |x + 1|. It has two parts: part (I) is to solve the inequality f(x + 8) ≥ 10 - f(x), and part (II) is to prove an inequality involving f(y) and f(y/x²) given some conditions on x and y. Let me try to tackle each part step by step.Starting with part (I). The function f(x) is given as |x + 1|. So, f(x + 8) would be |(x + 8) + 1|, which simplifies to |x + 9|. Similarly, f(x) is |x + 1|. So, the inequality f(x + 8) ≥ 10 - f(x) becomes |x + 9| ≥ 10 - |x + 1|.Hmm, okay. So, I need to solve |x + 9| + |x + 1| ≥ 10. Wait, is that right? Because if I move |x + 1| to the left side, it becomes |x + 9| + |x + 1| ≥ 10. That might be a better way to think about it because I can analyze the sum of two absolute values.I remember that the sum of absolute values can be tricky, but maybe I can consider different cases based on the critical points where the expressions inside the absolute values change their signs. The expressions inside the absolute values are x + 9 and x + 1. These expressions change sign at x = -9 and x = -1, respectively. So, I can break down the real line into intervals based on these critical points: x < -9, -9 ≤ x ≤ -1, and x > -1.Let me handle each case separately.**Case 1: x < -9**In this interval, x + 9 is negative because x is less than -9, so |x + 9| = -(x + 9) = -x - 9. Similarly, x + 1 is also negative because x is less than -9, which is certainly less than -1, so |x + 1| = -(x + 1) = -x - 1.So, the inequality becomes:(-x - 9) + (-x - 1) ≥ 10Simplify:-2x - 10 ≥ 10Add 10 to both sides:-2x ≥ 20Divide both sides by -2, remembering to reverse the inequality sign:x ≤ -10But in this case, we're considering x < -9, so the solution here is x ≤ -10.**Case 2: -9 ≤ x ≤ -1**In this interval, x + 9 is non-negative because x is at least -9, so |x + 9| = x + 9. However, x + 1 is still negative because x is less than or equal to -1, so |x + 1| = -(x + 1) = -x - 1.So, the inequality becomes:(x + 9) + (-x - 1) ≥ 10Simplify:x + 9 - x - 1 ≥ 10Which simplifies to:8 ≥ 10Wait, that's not true. 8 is not greater than or equal to 10. So, in this interval, there are no solutions.**Case 3: x > -1**In this interval, both x + 9 and x + 1 are positive because x is greater than -1. So, |x + 9| = x + 9 and |x + 1| = x + 1.So, the inequality becomes:(x + 9) + (x + 1) ≥ 10Simplify:2x + 10 ≥ 10Subtract 10 from both sides:2x ≥ 0Divide by 2:x ≥ 0But in this case, we're considering x > -1, so the solution here is x ≥ 0.**Combining the solutions from all cases:**From Case 1: x ≤ -10From Case 2: No solutionFrom Case 3: x ≥ 0So, the solution set is all real numbers x such that x ≤ -10 or x ≥ 0.Okay, that seems solid. I think I covered all the cases and checked each interval properly. Let me just verify with a test value in each interval.For x < -10, say x = -11:f(x + 8) = f(-11 + 8) = f(-3) = |-3 + 1| = | -2 | = 210 - f(x) = 10 - | -11 + 1 | = 10 - | -10 | = 10 - 10 = 0So, 2 ≥ 0, which is true.For x between -10 and -9, say x = -9.5:f(x + 8) = f(-9.5 + 8) = f(-1.5) = |-1.5 + 1| = | -0.5 | = 0.510 - f(x) = 10 - | -9.5 + 1 | = 10 - | -8.5 | = 10 - 8.5 = 1.5So, 0.5 ≥ 1.5? No, that's false, which is consistent with our solution since x = -9.5 is not in the solution set.For x between -9 and -1, say x = -5:f(x + 8) = f(3) = |3 + 1| = 410 - f(x) = 10 - | -5 + 1 | = 10 - 4 = 6So, 4 ≥ 6? No, again consistent with our solution.For x between -1 and 0, say x = -0.5:f(x + 8) = f(7.5) = |7.5 + 1| = 8.510 - f(x) = 10 - | -0.5 + 1 | = 10 - 0.5 = 9.5So, 8.5 ≥ 9.5? No, which is consistent because x = -0.5 is not in the solution set.For x = 0:f(x + 8) = f(8) = |8 + 1| = 910 - f(x) = 10 - |0 + 1| = 10 - 1 = 9So, 9 ≥ 9, which is true.For x > 0, say x = 1:f(x + 8) = f(9) = |9 + 1| = 1010 - f(x) = 10 - |1 + 1| = 10 - 2 = 8So, 10 ≥ 8, which is true.Alright, all test cases seem to confirm the solution set is x ≤ -10 or x ≥ 0.Moving on to part (II). We need to prove that if |x| > 1 and |y| < 1, then f(y) < |x| * f(y / x²).Given f(x) = |x + 1|, so f(y) = |y + 1| and f(y / x²) = |(y / x²) + 1|.So, the inequality to prove is |y + 1| < |x| * |(y / x²) + 1|.Let me write that down:|y + 1| < |x| * |(y / x²) + 1|Hmm, okay. Let me see if I can manipulate this inequality to see if it's true under the given conditions.First, note that |x| > 1, so |x| is at least greater than 1, and |y| < 1, so y is between -1 and 1.Let me square both sides of the inequality to eliminate the absolute values, since both sides are non-negative.So, squaring both sides:(y + 1)² < |x|² * (y / x² + 1)²Let me compute both sides.Left side: (y + 1)² = y² + 2y + 1Right side: |x|² * (y / x² + 1)²First, note that |x|² = x², so:Right side = x² * (y / x² + 1)²Let me compute the expression inside the square:(y / x² + 1) = (1 + y / x²) = (x² + y) / x²So, (y / x² + 1)² = (x² + y)² / x⁴Therefore, right side becomes:x² * (x² + y)² / x⁴ = (x² + y)² / x²So, the inequality becomes:y² + 2y + 1 < (x² + y)² / x²Multiply both sides by x² (since x² is positive, the inequality sign doesn't change):x²(y² + 2y + 1) < (x² + y)²Let me expand both sides.Left side: x²y² + 2x²y + x²Right side: (x² + y)² = x⁴ + 2x²y + y²So, the inequality is:x²y² + 2x²y + x² < x⁴ + 2x²y + y²Subtract 2x²y from both sides:x²y² + x² < x⁴ + y²Bring all terms to the left side:x²y² + x² - x⁴ - y² < 0Factor terms:Let me group terms:(x²y² - y²) + (x² - x⁴) < 0Factor y² from the first group and x² from the second group:y²(x² - 1) + x²(1 - x²) < 0Notice that (1 - x²) is the negative of (x² - 1), so:y²(x² - 1) - x²(x² - 1) < 0Factor out (x² - 1):(x² - 1)(y² - x²) < 0So, the inequality simplifies to:(x² - 1)(y² - x²) < 0Now, let's analyze this expression.Given that |x| > 1, so x² > 1, which means (x² - 1) is positive.Also, |y| < 1, so y² < 1. Since x² > 1, y² - x² is negative because y² < 1 < x².Therefore, (x² - 1) is positive, and (y² - x²) is negative, so their product is negative.Hence, (x² - 1)(y² - x²) < 0 is true.Therefore, the original inequality |y + 1| < |x| * |(y / x²) + 1| holds under the given conditions.Wait, let me just double-check my steps to make sure I didn't make a mistake.Starting from |y + 1| < |x| * |(y / x²) + 1|, I squared both sides correctly, expanded both sides, and simplified to (x² - 1)(y² - x²) < 0. Then, since x² > 1 and y² < 1, (x² - 1) is positive and (y² - x²) is negative, so their product is negative, which satisfies the inequality.Yes, that seems correct. I don't see any errors in the algebraic manipulations or the reasoning.So, both parts (I) and (II) are solved.**Final Answer**Ⅰ. The solution set is boxed{{x mid x leq -10 text{ or } x geq 0}}.Ⅱ. The inequality is proven as required.

💡Okay, so I have this problem here: If ( a ) and ( b ) are positive integers such that ( gcd(a, b) = 10 ), what is the smallest possible value of ( gcd(12a, 18b) )? Hmm, let me try to figure this out step by step.First, I know that the greatest common divisor (gcd) of two numbers is the largest number that divides both of them without leaving a remainder. So, if ( gcd(a, b) = 10 ), that means both ( a ) and ( b ) are multiples of 10. In other words, I can write ( a = 10m ) and ( b = 10n ) where ( m ) and ( n ) are positive integers that are coprime, meaning ( gcd(m, n) = 1 ). That makes sense because if ( m ) and ( n ) had a common divisor greater than 1, then the gcd of ( a ) and ( b ) would be greater than 10, which contradicts the given information.Now, the problem asks for the smallest possible value of ( gcd(12a, 18b) ). Let me substitute ( a ) and ( b ) with ( 10m ) and ( 10n ) respectively. So, ( 12a = 12 times 10m = 120m ) and ( 18b = 18 times 10n = 180n ). Therefore, I need to find ( gcd(120m, 180n) ).To find the gcd of these two numbers, I should probably factorize them into their prime factors. Let's do that:- ( 120 = 2^3 times 3 times 5 )- ( 180 = 2^2 times 3^2 times 5 )So, ( 120m = 2^3 times 3 times 5 times m ) and ( 180n = 2^2 times 3^2 times 5 times n ).The gcd of two numbers is the product of the smallest powers of all common prime factors. So, let's identify the common prime factors between ( 120m ) and ( 180n ):- The prime number 2: The smallest power is ( 2^2 ) because ( 120m ) has ( 2^3 ) and ( 180n ) has ( 2^2 ).- The prime number 3: The smallest power is ( 3^1 ) because ( 120m ) has ( 3^1 ) and ( 180n ) has ( 3^2 ).- The prime number 5: The smallest power is ( 5^1 ) because both have ( 5^1 ).So, the gcd so far is ( 2^2 times 3 times 5 = 4 times 3 times 5 = 60 ).But wait, I also have ( m ) and ( n ) in the factors. Since ( m ) and ( n ) are coprime, they don't share any common prime factors. Therefore, the gcd of ( 120m ) and ( 180n ) cannot be affected by ( m ) and ( n ) because any common factors would have to come from the parts that are already in 120 and 180. So, the gcd remains 60 regardless of the values of ( m ) and ( n ), as long as they are coprime.But hold on, is 60 really the smallest possible value? Let me test this with specific numbers to make sure. If I set ( m = 1 ) and ( n = 1 ), then ( a = 10 times 1 = 10 ) and ( b = 10 times 1 = 10 ). Then, ( 12a = 120 ) and ( 18b = 180 ). The gcd of 120 and 180 is indeed 60. What if I choose different values for ( m ) and ( n )? Let's say ( m = 2 ) and ( n = 3 ). Then, ( a = 10 times 2 = 20 ) and ( b = 10 times 3 = 30 ). Now, ( 12a = 240 ) and ( 18b = 540 ). The gcd of 240 and 540 is still 60 because:- ( 240 = 2^4 times 3 times 5 )- ( 540 = 2^2 times 3^3 times 5 )The gcd is ( 2^2 times 3 times 5 = 60 ). So, even with different ( m ) and ( n ), the gcd remains 60.Wait, but the question asks for the smallest possible value. Is there a way to make the gcd smaller than 60? Let me think. If I can somehow reduce the common factors, maybe by choosing ( m ) and ( n ) such that they don't contribute to the gcd. But since ( m ) and ( n ) are coprime, they don't share any prime factors, so they can't add any new common factors to ( 120m ) and ( 180n ). Therefore, the gcd is fixed at 60 regardless of ( m ) and ( n ).But let me double-check. Suppose ( m ) is a multiple of 3, say ( m = 3 ), and ( n ) is not a multiple of 3, say ( n = 1 ). Then, ( a = 10 times 3 = 30 ) and ( b = 10 times 1 = 10 ). Then, ( 12a = 360 ) and ( 18b = 180 ). The gcd of 360 and 180 is 180, which is larger than 60. So, in this case, the gcd increased. Alternatively, if ( m ) is a multiple of 2, say ( m = 2 ), and ( n ) is not a multiple of 2, say ( n = 1 ). Then, ( a = 20 ) and ( b = 10 ). ( 12a = 240 ) and ( 18b = 180 ). The gcd is still 60. So, even if ( m ) is a multiple of 2, the gcd doesn't decrease because 2 is already a common factor in both 120 and 180.Wait, but if ( m ) is a multiple of 5, say ( m = 5 ), and ( n ) is not a multiple of 5, say ( n = 1 ). Then, ( a = 50 ) and ( b = 10 ). ( 12a = 600 ) and ( 18b = 180 ). The gcd of 600 and 180 is still 60 because:- ( 600 = 2^3 times 3 times 5^2 )- ( 180 = 2^2 times 3^2 times 5 )The gcd is ( 2^2 times 3 times 5 = 60 ). So, even if ( m ) is a multiple of 5, the gcd remains 60.Hmm, so it seems like no matter what ( m ) and ( n ) I choose (as long as they are coprime), the gcd of ( 12a ) and ( 18b ) is always 60. Therefore, 60 is the only possible value, and hence it's also the smallest possible value.But just to be thorough, let me consider another example where ( m ) and ( n ) are different. Let’s say ( m = 7 ) and ( n = 3 ). Then, ( a = 70 ) and ( b = 30 ). ( 12a = 840 ) and ( 18b = 540 ). The gcd of 840 and 540 is:- ( 840 = 2^3 times 3 times 5 times 7 )- ( 540 = 2^2 times 3^3 times 5 )The gcd is ( 2^2 times 3 times 5 = 60 ). Again, it's 60.Wait, so in all these cases, the gcd is 60. So, is 60 the only possible value? It seems like it. Therefore, the smallest possible value is 60 because it can't be smaller than that.But let me think again. Is there a way to have a smaller gcd? For the gcd to be smaller, there must be a way to reduce the common factors. But since ( a ) and ( b ) are both multiples of 10, and 12 and 18 both have factors of 2, 3, and 6, respectively, the common factors are already fixed.Let me break it down:- ( 12a = 12 times 10m = 120m )- ( 18b = 18 times 10n = 180n )So, ( gcd(120m, 180n) ). Let's factor out the common parts:- ( 120 = 2^3 times 3 times 5 )- ( 180 = 2^2 times 3^2 times 5 )So, the gcd is the product of the minimum exponents of the common primes:- For 2: min(3, 2) = 2- For 3: min(1, 2) = 1- For 5: min(1, 1) = 1So, ( 2^2 times 3 times 5 = 4 times 3 times 5 = 60 ).Therefore, regardless of ( m ) and ( n ), as long as ( gcd(a, b) = 10 ), the gcd of ( 12a ) and ( 18b ) will always be 60. Hence, the smallest possible value is 60.I think I've considered multiple angles and examples, and in each case, the gcd remains 60. So, I'm confident that 60 is indeed the smallest possible value.

💡First, I need to determine the ratio of A's profit share to the total profit. A's share is Rs. 3780 out of a total profit of Rs. 12600.Next, I'll set up a proportion where A's investment ratio equals the profit ratio. Let ( x ) represent the amount A invested.The total investment is the sum of A's investment, B's investment (Rs. 4200), and C's investment (Rs. 10500).Setting up the equation based on the ratio:[frac{3780}{12600} = frac{x}{x + 4200 + 10500}]Simplifying the left side of the equation:[frac{3780}{12600} = frac{3}{10}]Now, the equation becomes:[frac{3}{10} = frac{x}{x + 14700}]Cross-multiplying to solve for ( x ):[3(x + 14700) = 10x]Expanding and simplifying:[3x + 44100 = 10x][44100 = 7x]Finally, solving for ( x ):[x = frac{44100}{7} = 6300]Therefore, A invested Rs. 6300 in the business.