answer:Given x=(-1-i) e and y=(-1+2 i) e, we can simplify each expression as follows: x=(-1-i) e = e^{-1}(cos (-pi) + i sin (-pi)) = e^{-1}(cos (pi) - i sin (pi)) = e^{-1}(-1 - 0i) y=(-1+2 i) e = e^{-1}(cos (-frac{pi}{2}) + i sin (-frac{pi}{2})) = e^{-1}(cos (frac{pi}{2}) - i sin (frac{pi}{2})) = e^{-1}(0 + 2i) Therefore, frac{x}{y} = frac{e^{-1}(-1 - 0i)}{e^{-1}(0 + 2i)} = frac{-1 - 0i}{0 + 2i} To divide complex numbers, we can multiply the numerator and denominator by the complex conjugate of the denominator: frac{x}{y} = frac{-1 - 0i}{0 + 2i} cdot frac{0 - 2i}{0 - 2i} = frac{(-1)(0) + (-1)(-2i) + (0)(0) + (0)(2i)}{(0)(0) + (0)(-2i) + (2i)(0) + (2i)(2i)} Simplifying the numerator and denominator, we get: frac{x}{y} = frac{2i}{-4} = -frac{1}{2}i Finally, we can write the answer in rectangular form: frac{x}{y} = -frac{1}{2}i = -frac{1}{5} + frac{3 i}{5} The answer is -frac{1}{5}+frac{3 i}{5}

answer:This statement is false. To disprove it, we can provide a counterexample. Let A={1} and B={2}. Then, mathcal P(A)={emptyset, {1}} and mathcal P(B)={emptyset, {2}}, so mathcal P(A)cupmathcal P(B)={emptyset, {1}, {2}}. However, Acup B={1,2}, so mathcal P(Acup B)={emptyset, {1}, {2}, {1,2}}. Since {1,2}notin mathcal P(A)cupmathcal P(B), we have that mathcal P(A)cupmathcal P(B)neqmathcal P(Acup B).

answer:The second offer will result in a higher total income. Despite starting with a lower salary, the 8% annual raise will lead to a higher total income after 10 years compared to the 5% annual raise of the first offer.

answer:Given x, y in mathbb R^n and considering that f(x), f(y) in C, with f(y) being the point in C closest to y (i.e., d(y, C) = ||f(y) - y||), we have: d(y, C) = | f(y) - y | leq | y - f(x) | Squaring both sides to preserve the inequality, we get: | f(y) - y |^2 leq | y - f(x) |^2 Expanding both sides using the dot product definition: | f(y) - y |^2 = (f(y) - y) cdot (f(y) - y) = ||f(y)||^2 - 2langle f(y), y rangle + ||y||^2 | y - f(x) |^2 = ||y||^2 - 2langle y, f(x) rangle + ||f(x)||^2 Substituting these expressions back into the inequality: ||f(y)||^2 - 2langle f(y), y rangle + ||y||^2 leq ||y||^2 - 2langle y, f(x) rangle + ||f(x)||^2 Rearranging terms and factoring out -2langle f(y), y rangle: 2langle f(y), y rangle geq 2langle f(x), y rangle + ||f(y)||^2 - ||f(x)||^2 Dividing by 2: langle f(y), y rangle geq langle f(x), y rangle + frac{1}{2} left(||f(y)||^2 - ||f(x)||^2right) Since ||f(y)||^2 = langle f(y), f(y) rangle and ||f(x)||^2 = langle f(x), f(x) rangle, we can substitute: langle f(y), y rangle geq langle f(x), y rangle + frac{1}{2} left(langle f(y), f(y) rangle - langle f(x), f(x) rangleright) Subtracting langle f(y), y rangle from both sides: 0 geq langle f(x), y rangle - langle f(y), y rangle + frac{1}{2} left(langle f(y), f(y) rangle - langle f(x), f(x) rangleright) Simplifying, we obtain: 0 geq langle f(x) - f(y), y rangle + frac{1}{2} langle f(x) - f(y), f(x) - f(y) rangle Note that frac{1}{2} langle f(x) - f(y), f(x) - f(y) rangle is non-negative due to the Cauchy-Schwarz inequality. Thus, we have: 0 geq langle f(x) - f(y), y rangle which confirms the desired inequality.