answer:Given that ab = p, bc = q, and ca = r, the variables can be solved in terms of each other's products, and the quantity a^2 + b^2 + c^2 needs to be expressed using these products. 1. **Express a^2, b^2, and c^2:** - a^2 = frac{(bc)^2}{b^2} = frac{q^2}{b^2} - b^2 = frac{(ca)^2}{c^2} = frac{r^2}{c^2} - c^2 = frac{(ab)^2}{a^2} = frac{p^2}{a^2} 2. **Relation among a^2, b^2 and c^2 using abc:** - abc = sqrt[3]{pqr} - Substitute abc in terms of a, b, and c: - a^2 = frac{q^2}{b^2} = frac{q^2}{frac{r^2}{c^2}} = frac{q^2c^2}{r^2} - b^2 = frac{r^2}{frac{p^2}{a^2}} = frac{r^2 a^2}{p^2} - c^2 = frac{p^2}{frac{q^2}{b^2}} = frac{p^2b^2}{q^2} 3. **Simplify a^2 + b^2 + c^2:** - a^2 = frac{q^2c^2}{r^2} = frac{p^2b^2}{q^2} - b^2 = frac{q^2}{p^2/r^2} = frac{q^2r^2}{p^2} - c^2 = frac{p^2}{q^2/r^2} = frac{p^2r^2}{q^2} - Simplify a^2+b^2+c^2 = frac{q^2c^2}{r^2} + frac{r^2a^2}{p^2} + frac{p^2b^2}{q^2} - Factor out frac{1}{pqr} to get: - a^2 + b^2 + c^2 = frac{q^3c + r^3a + p^3b}{pqr} and adjust variables in terms of p, q, and r: - frac{(pqr){pqr}} = 1 The expression simplifies to 1. The final answer is A. boxed{1}

answer:1. **Calculate the total sum of the numbers**: Given that the mean of these six numbers is 87.5, the total sum is 6 times 87.5 = 525. 2. **Solve for the unknown number y**: - The sum of the known numbers 90 + 88 + 85 + 86 + 87 = 436. - Therefore, y = 525 - 436 = 89. 3. **Find the median**: - Arrange the numbers in ascending order: 85, 86, 87, 88, 89, 90. - The median is the average of the third and fourth numbers, which are 87 and 88. - Thus, the median is frac{87 + 88}{2} = 87.5. boxed{87.5}

answer:To solve the given expression, let's simplify [frac{3^{2040} + 3^{2038}}{3^{2040} - 3^{2038}}.] 1. **Factor out common terms:** [ frac{3^{2040} + 3^{2038}}{3^{2040} - 3^{2038}} = frac{3^{2038}(3^2 + 1)}{3^{2038}(3^2 - 1)} ] Here, we factor out (3^{2038}) from both the numerator and the denominator. 2. **Simplify the expression inside parentheses:** [ frac{3^{2038}(3^2 + 1)}{3^{2038}(3^2 - 1)} = frac{9 + 1}{9 - 1} = frac{10}{8} = frac{5}{4} ] After factoring out (3^{2038}), it cancels out from numerator and denominator, simplifying to (frac{5}{4}). Consequently, the value of the original expression is frac{5{4}}. The final answer is The correct choice is boxed{textbf{(C)} frac{5}{4}}.

answer:Since (8,6) is on the graph of y=g(x), we know: [ 6 = g(8). ] Substitute x = frac{8}{3} into 3y = frac{g(3x)}{3} + 3: [ 3y = frac{g(3 cdot frac{8}{3})}{3} + 3 = frac{g(8)}{3} + 3 = frac{6}{3} + 3 = 2 + 3 = 5. ] [ y = frac{5}{3}. ] Therefore, the point left(frac{8}{3}, frac{5}{3}right) is on the graph of 3y = frac{g(3x)}{3} + 3. The sum of these coordinates is: [ frac{8}{3} + frac{5}{3} = boxed{frac{13}{3}}. ]