answer:For a = 2^j 3^k 5^l, b = 2^m 3^n 5^o and c = 2^p 3^q 5^r using nonnegative integers j, k, l, m, n, o, p, q, r, we need to satisfy: 1. max(j, m) = 4, max(k, n) = 1, max(l, o) = 2 for [a, b] = 2^4 3^1 5^2 = 1200. 2. max(m, p) = max(q, n) = max(o, r) = 4 for [b, c] = 2^4 3^1 5^2 = 2400. 3. max(p, j) = 4, max(q, k) = 1, max(r, l) = 2 for [c, a] = 2^4 3^1 5^2 = 2400. - For powers of 2: p = 4 and at least one of m, j = 4. Possible pairs (j, m, p) are (4, 4, 4), (3, 4, 4), (4, 3, 4). - For powers of 3 and 5: Both are maximized at 1 and 2 respectively, leading to simplest situation where k = n = q = 1 and l = o = r = 2. This results in 3 possible triples for powers of 2, and for powers of 3 and 5, the situations are fixed as indicated. Thus, the total number of valid triples (a, b, c) is 3 cdot 1 = boxed{3}.

answer:To solve this problem, we follow the given information and the standard solution closely, breaking it down into detailed steps. 1. **Identify the Focus of the Parabola C_1:** Given the parabola y^2 = 2px (where p > 0), its focus is at F(frac{p}{2}, 0). This is because the focus of a parabola y^2 = 4ax is at (a, 0), and in this case, 2p = 4a implies a = frac{p}{2}. 2. **Relate the Focus to the Hyperbola C_2:** Since the hyperbola and the parabola share a common focus, and the focus of a hyperbola frac{x^2}{a^2} - frac{y^2}{b^2} = 1 is at (pm c, 0) where c^2 = a^2 + b^2, we equate c = frac{p}{2}. 3. **Equation of the Parabola with Identified Focus:** Given c = frac{p}{2}, the equation of the parabola becomes y^2 = 4cx. Substituting c into this equation, we get y^2 = 4(frac{p}{2})x = 2px, which is consistent with the original equation of the parabola. 4. **Finding Points A and B on C_1:** Letting x = c and substituting into the parabola's equation, we find y = pm 2c. Thus, points A and B on C_1 are A(c, 2c) and B(c, -2c). 5. **Finding Point M on C_2:** Substituting y = pm 2c into the hyperbola's equation, we find y = pm frac{bsqrt{c^2 - a^2}}{a} = pm frac{b^2}{a}. Since M is in the first quadrant, we take the positive value, so M(c, frac{b^2}{a}). 6. **Relation Among OM, OA, and OB:** Given overrightarrow{OM} = moverrightarrow{OA} + noverrightarrow{OB}, we set up the system of equations: [ begin{cases} m + n = 1 m - n = frac{b^2}{2ac} end{cases} ] Subtracting the squares of these equations, we get: [ 4mn = 1 - left(frac{b^2}{2ac}right)^2 ] 7. **Using Given mn = frac{1}{8} to Find b^2:** Substituting mn = frac{1}{8} into the equation from step 6, we solve for b^2 and find b^2 = sqrt{2}ac. 8. **Relating b^2 to a^2 and c^2:** Given b^2 = c^2 - a^2, we substitute b^2 = sqrt{2}ac to get c^2 - sqrt{2}ac - a^2 = 0. 9. **Finding the Eccentricity e of C_2:** Since the eccentricity e = frac{c}{a}, we substitute c = ea into the equation from step 8 to get e^2 - sqrt{2}e - 1 = 0. Solving this quadratic equation for e (and considering e > 1 for a hyperbola), we find e = frac{sqrt{6} + sqrt{2}}{2}. Therefore, the eccentricity of the hyperbola C_2 is boxed{frac{sqrt{6} + sqrt{2}}{2}}, which corresponds to choice C.

answer:To find the opposite of a number, we simply multiply it by -1. Therefore, the opposite of frac{2}{3} is calculated as follows: [ -1 times frac{2}{3} = -frac{2}{3} ] Hence, the correct answer is boxed{D}.

answer:First, let's convert the dozens of berries into individual berries: - Strawberries: 2.5 dozen x 12 = 30 berries - Blueberries: 1.75 dozen x 12 = 21 berries - Raspberries: 1.25 dozen x 12 = 15 berries Now, let's subtract the berries Micah ate: - Strawberries left: 30 - 6 = 24 berries - Blueberries left: 21 - 4 = 17 berries - Raspberries left: 15 (since he didn't eat any raspberries) Finally, let's add up all the berries left for his mom: 24 strawberries + 17 blueberries + 15 raspberries = 56 berries So, there are boxed{56} berries left for Micah's mom.