answer:To find the annual interest rate, we can use the compound interest formula: A = P(1 + r/n)^(nt) Where: A = the future value of the investment/loan, including interest P = the principal investment amount (the initial deposit or loan amount) r = the annual interest rate (decimal) n = the number of times that interest is compounded per year t = the number of years the money is invested or borrowed for In this case, we are given: A = Rs. 1183 (future value) P = Rs. 1093.75 (present worth) t = 2 years n = 1 (since it's not mentioned, we'll assume the interest is compounded annually) We need to find r. Let's rearrange the formula to solve for r: 1183 = 1093.75(1 + r/1)^(1*2) 1183 = 1093.75(1 + r)^2 Now, let's solve for (1 + r): (1 + r)^2 = 1183 / 1093.75 (1 + r)^2 = 1.0815 (rounded to 4 decimal places) Now, take the square root of both sides to solve for (1 + r): 1 + r = sqrt(1.0815) 1 + r = 1.0399 (rounded to 4 decimal places) Now, subtract 1 from both sides to solve for r: r = 1.0399 - 1 r = 0.0399 To express r as a percentage, we multiply by 100: r = 0.0399 * 100 r = 3.99% So, the annual interest rate is approximately boxed{3.99%} .

answer:Let's denote the number of students in section B as ( x ). The total weight of students in section A is ( 40 times 50 ) kg. The total weight of students in section B is ( x times 40 ) kg. The average weight of the whole class is given as 46.67 kg, and the total number of students in the class is ( 40 + x ). The total weight of the whole class is the sum of the total weights of sections A and B, which is ( (40 times 50) + (x times 40) ) kg. The average weight of the class is the total weight divided by the total number of students, so we have: [ frac{(40 times 50) + (x times 40)}{40 + x} = 46.67 ] Now we can solve for ( x ): [ 2000 + 40x = 46.67 times (40 + x) ] [ 2000 + 40x = 1866.8 + 46.67x ] Subtract ( 40x ) from both sides: [ 2000 = 1866.8 + 6.67x ] Subtract ( 1866.8 ) from both sides: [ 133.2 = 6.67x ] Divide both sides by ( 6.67 ) to solve for ( x ): [ x = frac{133.2}{6.67} ] [ x approx 19.97 ] Since the number of students cannot be a fraction, we round to the nearest whole number. Therefore, section B has approximately boxed{20} students.

answer:Using the vector triple product formula: [mathbf{a} times (mathbf{b} times mathbf{c}) = (mathbf{a} cdot mathbf{c}) mathbf{b} - (mathbf{a} cdot mathbf{b}) mathbf{c},] we calculate, [mathbf{i} times (kmathbf{w} times mathbf{i}) = (mathbf{i} cdot mathbf{i}) kmathbf{w} - (mathbf{i} cdot kmathbf{w}) mathbf{i} = kmathbf{w} - k(mathbf{i} cdot mathbf{w}) mathbf{i},] [mathbf{j} times (kmathbf{w} times mathbf{j}) = (mathbf{j} cdot mathbf{j}) kmathbf{w} - (mathbf{j} cdot kmathbf{w}) mathbf{j} = kmathbf{w} - k(mathbf{j} cdot mathbf{w}) mathbf{j},] [mathbf{k} times (kmathbf{w} times mathbf{k}) = (mathbf{k} cdot mathbf{k}) kmathbf{w} - (mathbf{k} cdot kmathbf{w}) mathbf{k} = kmathbf{w} - k(mathbf{k} cdot mathbf{w}) mathbf{k}.] Adding these gives: begin{align*} &mathbf{i} times (kmathbf{w} times mathbf{i}) + mathbf{j} times (kmathbf{w} times mathbf{j}) + mathbf{k} times (kmathbf{w} times mathbf{k}) &= 3kmathbf{w} - k((mathbf{i} cdot mathbf{w}) mathbf{i} + (mathbf{j} cdot mathbf{w}) mathbf{j} + (mathbf{k} cdot mathbf{w}) mathbf{k}) &= 3kmathbf{w} - kmathbf{w} = 2kmathbf{w}. end{align*} Thus, comparing with c mathbf{w}, we have c = 2k. But since this must hold for all k, and the problem defines independent of k, it must be that c = 2. Therefore, c = boxed{2}.

answer:1. **Express the equation using logarithm properties**: [ A log_{180} 3 + B log_{180} 5 = C ] Use the change of base formula: [ A frac{log 3}{log 180} + B frac{log 5}{log 180} = C ] Since (log 180 = log (2^2 cdot 3^2 cdot 5) = 2 log 2 + 2 log 3 + log 5), the equation becomes: [ A frac{log 3}{2 log 2 + 2 log 3 + log 5} + B frac{log 5}{2 log 2 + 2 log 3 + log 5} = C ] 2. **Simplify the equation**: Multiply through by (2 log 2 + 2 log 3 + log 5) to clear the denominators: [ A log 3 + B log 5 = C (2 log 2 + 2 log 3 + log 5) ] This can be rewritten using the properties of logarithms: [ log(3^A) + log(5^B) = log(180^C) ] [ log(3^A cdot 5^B) = log(180^C) ] Hence: [ 3^A cdot 5^B = 180^C ] Expanding (180^C): [ 3^A cdot 5^B = (2^2 cdot 3^2 cdot 5)^C = 2^{2C} cdot 3^{2C} cdot 5^C ] 3. **Equating the powers of the same bases**: [ 3^A = 3^{2C} quad text{and} quad 5^B = 5^{C} ] Hence, (A = 2C) and (B = C). 4. **Determine the values of (A), (B), (C)**: To keep (A), (B), and (C) relatively prime, the smallest value for (C) that makes (A) and (B) integers and maintains the gcd condition is (C = 1). Therefore, (A = 2) and (B = 1). 5. **Calculate (A + B + C)**: [ A + B + C = 2 + 1 + 1 = 4 ] Thus, the answer is 4. The final answer is boxed{4} (Choice A).