answer:Let s_3 be the side length of the square inscribed in the quarter circle of radius r. The square fits into the quarter circle such that its diagonal from the origin (center of the full circle) to the opposite vertex equals r. Applying the Pythagorean theorem to the right triangle formed by the diagonal, we have: [ left(frac{s_3}{sqrt{2}}right)^2 + left(frac{s_3}{sqrt{2}}right)^2 = r^2, ] [ 2 left(frac{s_3}{sqrt{2}}right)^2 = r^2, ] [ s_3^2 = r^2. ] Let s_2 be the side length of the square inscribed in the circle of radius r, as previously defined. From the initial solution, we know s_2^2 = 2r^2. The ratio of the areas of the square in the quarter circle to the square in the circle is: [ frac{s_3^2}{s_2^2} = frac{r^2}{2r^2} = boxed{frac{1}{2}}. ]

answer:(1) Since x in mathbb{R}, the first derivative of f(x) is f'(x) = (x-1)(e^{x} + a). When a geqslant 0, for x in (-infty, 1), we have f'(x) < 0. When x in (1, +infty), we have f'(x) > 0. Therefore, f(x) is monotonically decreasing in (-infty, 1) and monotonically increasing in (1, +infty). When a < 0, setting f'(x) = 0 yields x = 1 and x = ln (-a). (i) If a < -e, for x in (-infty, 1), we have f'(x) > 0. For x in (1, ln (-a)), we have f'(x) < 0, and for x in (ln (-a), +infty), we have f'(x) > 0. Thus, f(x) is monotonically increasing in (-infty, 1) and (ln (-a), +infty), and monotonically decreasing in (1, ln (-a)). (ii) If a = -e, since f'(x) geqslant 0, f(x) is monotonically increasing on mathbb{R}. (iii) If -e < a < 0, for x in (-infty, ln (-a)), we have f'(x) > 0. For x in (ln (-a), 1), we have f'(x) < 0, and for x in (1, +infty), we have f'(x) > 0. Therefore, f(x) is monotonically increasing in (-infty, ln (-a)) and (1, +infty), and monotonically decreasing in (ln (-a), 1). (2) Let g(x) = f(x) - kx + 2 = (x-2)e^{x} + frac{1}{2}x^2 - x -kx + 2, and consequently, we find g'(x) = (x-1)e^{x} + x - 1 - k. Define h(x) = (x-1)e^{x} + x - 1 - k, and its first derivative is h'(x) = xe^{x} + 1. For x geqslant 0, since h'(x) = xe^{x} + 1 > 0, h(x) is monotonically increasing. Therefore, h(x) geqslant h(0) = -2-k, which implies g'(x) geqslant -2 - k. (i) For k leqslant -2, it holds that g'(x) geqslant 0, and so g(x) is monotonically increasing on (0, +infty). Since g(0) = 0, the inequality f(x) geqslant kx - 2 is always satisfied. (ii) For k > -2, g'(x) = 0 has a solution at some x_0. This implies that g'(x) < 0 for x in (0, x_0), making g(x) decrease in this interval. As x enters (x_0, +infty), g'(x) > 0, making g(x) increase again. However, g(x_0) < g(0) = 0, which means, for x geqslant 0, f(x) geqslant kx - 2 does not always hold. Summarizing the above, the range of values for k is boxed{(-infty, -2]}.

answer:Baker made a total of 62 + 149 = 211 cakes. If he still has 67 cakes, then he must have sold 211 - 67 = boxed{144} cakes.

answer:Since f(x) is an even function, we have f(-x)=f(x). Therefore, we can rewrite f(-4) as f(4). Now, we have: begin{align*} f(-4)+f(9) &= f(4)+f(9) &= log_{6}4+log_{6}9 &= log_{6}(4times9) quad text{(using the product rule for logarithms)} &= log_{6}36 &= 2 quad text{(since 6^2=36)} end{align*} Therefore, the answer is: boxed{2}. This problem primarily tests the computation of function values and requires an understanding of the properties of even functions to solve it.