answer:To solve the given problem, we start by adding the fractions frac{3}{a} and frac{2}{a}. Since they have the same denominator, we can simply add the numerators: [ frac{3}{a} + frac{2}{a} = frac{3 + 2}{a} = frac{5}{a}. ] Thus, the result of frac{3}{a} + frac{2}{a} is frac{5}{a}. Therefore, the correct answer is boxed{C}.

answer:Given: - Length of edge AB = a - Length of edge CD = b - Angle between AB and CD is theta - Distance between lines AB and CD is d The goal is to find the volume of the tetrahedron ABCD. Here's how we approach the solution: 1. Consider a perpendicular distance EF between the skew lines AB and CD. 2. Let C drop a perpendicular CG to the plane ABF, where G is the foot of the perpendicular. 3. Decomposing the volume of the tetrahedron into simpler parts: - Volume of smaller tetrahedron C-ABF: [ V_{C-ABF} = frac{1}{3} times left(frac{1}{2} times EF times AB times CG right) ] - As CG = CF sin theta, we replace CG: [ V_{C-ABF} = frac{1}{3} times frac{1}{2} times EF times AB times CF sin theta ] 4. Analogously, considering another smaller tetrahedron D-ABF: - Volume [ V_{D-ABF} = frac{1}{3} times frac{1}{2} times EF times AB times FD sin theta ] 5. Adding the volumes of the two parts together to find the total volume V_{ABCD}: [ V_{ABCD} = V_{C-ABF} + V_{D-ABF} ] Summing up, using E = EF, AB = a, CD = b, and d, we get: [ V_{ABCD} = frac{1}{6} times a times b times d times sin theta ] Thus, the volume of tetrahedron ABCD is (boxed{frac{1}{6} a b d sin theta}).

answer:The graph has vertical asymptotes at ( x = -frac{2pi}{5}, 0, frac{2pi}{5} ). This implies the function's periodicity is ( frac{2pi}{5} ), as the period of ( y = a tan bx ) is ( frac{pi}{b} ), we infer ( b = frac{5}{2} ). Given the form of the function now is [ y = a tan left(frac{5x}{2}right). ] The graph passes through ( left(frac{pi}{10}, 3right) ), which can be plugged into the formula to get: [ 3 = a tan left(frac{5 cdot frac{pi}{10}}{2}right) = a tan left(frac{pi}{4}right). ] Since ( tan left(frac{pi}{4}right) = 1 ), we deduce that ( a = 3 ). Thus, the product ( ab ) is: [ ab = 3 cdot frac{5}{2} = boxed{7.5}. ] Conclusion: All conditions are satisfied according to the initial hypothesis: the period corresponds to the asymptotes, the point matches the graph, and the product ( ab ) is determined accurately.

answer:To solve this problem, we need to calculate the probability of picking at least 2 red balls and the third ball not being blue. We can break this down into two separate events: 1. Picking exactly 2 red balls and 1 non-blue ball. 2. Picking all 3 red balls. Let's calculate the probability for each event: 1. Picking exactly 2 red balls and 1 non-blue ball: The total number of ways to pick 3 balls out of 22 (5 red + 6 blue + 7 green + 4 yellow) is: Total ways = C(22, 3) The number of ways to pick exactly 2 red balls out of 5 is: Ways to pick 2 red balls = C(5, 2) The number of ways to pick 1 non-blue ball out of the remaining (7 green + 4 yellow) is: Ways to pick 1 non-blue ball = C(11, 1) So, the number of ways to pick exactly 2 red balls and 1 non-blue ball is: Ways to pick 2 red and 1 non-blue = C(5, 2) * C(11, 1) 2. Picking all 3 red balls: The number of ways to pick 3 red balls out of 5 is: Ways to pick 3 red balls = C(5, 3) Now, let's calculate the probabilities using combinations (C(n, k) = n! / (k! * (n - k)!)): Total ways = C(22, 3) = 22! / (3! * (22 - 3)!) = 22! / (3! * 19!) = (22 * 21 * 20) / (3 * 2 * 1) = 1540 Ways to pick 2 red and 1 non-blue = C(5, 2) * C(11, 1) = (5! / (2! * (5 - 2)!)) * (11! / (1! * (11 - 1)!)) = (5 * 4 / (2 * 1)) * 11 = 10 * 11 = 110 Ways to pick 3 red balls = C(5, 3) = 5! / (3! * (5 - 3)!) = (5 * 4 * 3) / (3 * 2 * 1) = 10 Now, let's add the probabilities for the two events: Probability of picking exactly 2 red and 1 non-blue = 110 / 1540 Probability of picking 3 red balls = 10 / 1540 Total probability = (110 + 10) / 1540 = 120 / 1540 Now, we simplify the fraction: Total probability = 120 / 1540 = 12 / 154 The probability that at least 2 of the picked balls are red and the other is not blue is boxed{12/154} .