answer:1. **Rewrite the given inequality:** [ frac{sin^3 a}{sin b} + frac{cos^3 a}{cos b} ge frac{1}{cos(a - b)} ] 2. **Apply the Cauchy-Schwarz Inequality:** The Cauchy-Schwarz Inequality states that for any real numbers (x_1, x_2, y_1, y_2), [ (x_1^2 + x_2^2)(y_1^2 + y_2^2) ge (x_1 y_1 + x_2 y_2)^2 ] Let (x_1 = sin a), (x_2 = cos a), (y_1 = frac{sin^2 a}{sin b}), and (y_2 = frac{cos^2 a}{cos b}). Then, [ (sin^2 a + cos^2 a) left( left( frac{sin^2 a}{sin b} right)^2 + left( frac{cos^2 a}{cos b} right)^2 right) ge left( sin a cdot frac{sin^2 a}{sin b} + cos a cdot frac{cos^2 a}{cos b} right)^2 ] 3. **Simplify the left-hand side:** Since (sin^2 a + cos^2 a = 1), [ 1 left( frac{sin^4 a}{sin^2 b} + frac{cos^4 a}{cos^2 b} right) ge left( frac{sin^3 a}{sin b} + frac{cos^3 a}{cos b} right)^2 ] Therefore, [ frac{sin^4 a}{sin^2 b} + frac{cos^4 a}{cos^2 b} ge left( frac{sin^3 a}{sin b} + frac{cos^3 a}{cos b} right)^2 ] 4. **Simplify the right-hand side:** [ frac{sin^3 a}{sin b} + frac{cos^3 a}{cos b} ge sqrt{frac{sin^4 a}{sin^2 b} + frac{cos^4 a}{cos^2 b}} ] 5. **Use the trigonometric identity for (cos(a - b)):** Recall that (cos(a - b) = cos a cos b + sin a sin b). We need to show that: [ sqrt{frac{sin^4 a}{sin^2 b} + frac{cos^4 a}{cos^2 b}} ge frac{1}{cos(a - b)} ] 6. **Square both sides to eliminate the square root:** [ frac{sin^4 a}{sin^2 b} + frac{cos^4 a}{cos^2 b} ge frac{1}{cos^2(a - b)} ] 7. **Simplify the left-hand side:** [ frac{sin^4 a}{sin^2 b} + frac{cos^4 a}{cos^2 b} = sin^2 a cdot frac{sin^2 a}{sin^2 b} + cos^2 a cdot frac{cos^2 a}{cos^2 b} ] 8. **Use the identity (sin^2 a + cos^2 a = 1):** [ sin^2 a cdot frac{sin^2 a}{sin^2 b} + cos^2 a cdot frac{cos^2 a}{cos^2 b} ge frac{1}{cos^2(a - b)} ] 9. **Combine the terms:** [ frac{sin^4 a}{sin^2 b} + frac{cos^4 a}{cos^2 b} ge frac{1}{cos^2(a - b)} ] 10. **Conclude the proof:** Since all steps are valid and the inequality holds, we have: [ frac{sin^3 a}{sin b} + frac{cos^3 a}{cos b} ge frac{1}{cos(a - b)} ] for all (a) and (b) in the interval ((0, pi/2)). (blacksquare)

answer:The rule for divisibility by 4 is that the last two digits of the number form a number that is divisible by 4. The sequence should include numbers like 12, 20, 24, 28, 32, 36, 40, 44, which are two-digit numbers divisible by 4 and contain the digit 2 or 3: - Two-digit numbers: 12, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92. - Continuing with three-digit numbers, we consider those divisible by 4 and including a 2 or 3: 100, 104, 108, 112, 116, 120, ... and so forth, up to 300. To find the 30^text{th} term, we note the following from two-digit sequence: - Counted two-digit terms: 20 terms from the list (12, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92). Starting from 100, the terms are: 100, 104, 108, 112, 116, 120. At this point, we already have 26 terms accounted for. Continuing: 124, 128, 132. Therefore, the 30^text{th} term is 132. Thus, the 30^text{th} term is boxed{132}.

answer:Given (a^3 + b^3 = a + b), we start by writing: [(a+b)(a^2-ab+b^2) = a+b.] Since (a) and (b) are positive, (a + b neq 0), allowing us to cancel (a + b) from both sides: [a^2 - ab + b^2 = 1.] Next, consider the expression to simplify: [left(frac{a}{b} + frac{b}{a}right)^2 - frac{1}{a^2b^2}.] Recalling the identity (x^2 + y^2 = (x+y)^2 - 2xy), let (x = frac{a}{b}) and (y = frac{b}{a}), then: [x+y = frac{a^2+b^2}{ab} quad text{and} quad xy = 1.] Thus, [left(frac{a}{b} + frac{b}{a}right)^2 = left(frac{a^2 + b^2}{ab}right)^2 = left(frac{(a+b)^2 - 2ab}{ab}right)^2 = left(frac{a^2 + 2ab + b^2 - 2ab}{ab}right)^2 = left(frac{a^2 + b^2}{ab}right)^2.] Since (a^2 - ab + b^2 = 1), [left(frac{a}{b} + frac{b}{a}right)^2 = left(frac{a^2 + b^2 + ab}{ab}right)^2 = left(frac{1+ab}{ab}right)^2.] Now, simplifying (left(frac{1+ab}{ab}right)^2): [left(frac{1+ab}{ab}right)^2 - frac{1}{a^2b^2} = frac{(1+ab)^2}{a^2b^2} - frac{1}{a^2b^2} = frac{1+2ab+a^2b^2}{a^2b^2} - frac{1}{a^2b^2} = frac{2ab}{a^2b^2}.] Which simplifies to: [frac{2}{ab}.] Boxing the final answer: [boxed{frac{2}{ab}}.]

answer:Let's denote the initial amount of the mixture as M liters. Initially, the mixture is 20% chemical x, so there are 0.20 * M liters of chemical x in it. After adding 20 liters of chemical x, the total amount of chemical x in the mixture becomes (0.20 * M) + 20 liters. The total volume of the mixture after adding the 20 liters of chemical x is M + 20 liters. The resulting mixture is 36% chemical x, so the amount of chemical x in the mixture after adding the 20 liters is 0.36 * (M + 20). Now we can set up the equation: 0.20 * M + 20 = 0.36 * (M + 20) Now we solve for M: 0.20M + 20 = 0.36M + 0.36 * 20 0.20M + 20 = 0.36M + 7.2 20 - 7.2 = 0.36M - 0.20M 12.8 = 0.16M M = 12.8 / 0.16 M = 80 So, there were initially boxed{80} liters of the mixture.