answer:Let U be an open neighborhood of f^{-1}(C) in X. Since f is continuous, f^{-1}(C) is closed in X. Since X is compact, f^{-1}(C) is also compact. Therefore, f^{-1}(C) is a closed and compact subset of X. Since C is closed in Y, its complement Y-C is open in Y. Therefore, there exists an open neighborhood V of C in Y such that Vsubset Y-C. Now, let xin f^{-1}(V). Then f(x)in V. Since Vsubset Y-C, we have f(x)notin C. Therefore, xnotin f^{-1}(C). Hence, f^{-1}(V)cap f^{-1}(C)=emptyset. Since f^{-1}(C) is closed and f^{-1}(V) is open, we have f^{-1}(V)subset X-f^{-1}(C). Since U is an open neighborhood of f^{-1}(C), we have X-f^{-1}(C)subset U. Therefore, f^{-1}(V)subset U. Hence, for any open neighborhood U of f^{-1}(C) in X, there exists an open neighborhood V of C in Y such that f^{-1}(V)⊂U.

answer:Upon multiplication and expansion, we get: [ p(x)q(x) = left(-10sqrt{2}x^2 - frac{3x}{sqrt{2}} - frac{19}{sqrt{2}}right)left(-frac{13x^2}{sqrt{2}} - 10sqrt{2}x - frac{11}{sqrt{2}}right) ] After simplifying the expression, the result is: [ p(x)q(x) = 130x^4 + frac{439x^3}{2} + frac{527x^2}{2} + frac{413x}{2} + frac{209}{2} ]

answer:Variable costs are crucial to consider in short-term decision-making because they fluctuate with production levels and have a significant impact on the profitability of short-term decisions.

answer:To find x cdot y, we can multiply the complex numbers x and y directly: x cdot y = left(frac{16+7 i}{sqrt{3}}right) cdot left(frac{10+i}{sqrt{3}}right) Multiply the numerators and the denominators separately: = frac{(16+7 i)(10+i)}{(sqrt{3})(sqrt{3})} = frac{160 + 16i + 70i + 7i^2}{3} Remember that i^2 = -1, so we can substitute 7i^2 with -7: = frac{160 + 86i - 7}{3} = frac{153 + 86i}{3} To combine the real and imaginary parts, multiply each by 3: = 51 + frac{86}{3}i Thus, x cdot y = 51 + frac{86 i}{3}.