answer:The triangle's angles sum up to 180^{circ}, and since angle B is a right angle (90^{circ}), we have 180 = A + 90 + C. Thus, A + C = 90^{circ}. Given that C = x + 30^{circ}, we can express angle A as A = 90 - (x + 30) = 60 - x. The reason angle C is written as x + 30^{circ} instead of just 30^{circ} is to indicate that it's a variable angle, not a fixed value. This format, (x pm theta^{circ}), is used to represent any angle in terms of an unknown angle x plus or minus a specific measure (theta^{circ}). It allows the problem to be generalized for different values of x and not just a particular angle. Answer a) is correct because when considering angles A and C, their relationship is complementary, meaning they add up to 90^{circ}. With A = 60 - x and C = x + 30^{circ}, if A and C were complementary, their sum would equal 90^{circ}, which is not the case for all values of x. This rules out d) as a valid choice.

answer:Using the trigonometric identities cos left(frac{19 pi }{90}right) = cos left(2 pi - frac{19 pi }{90}right) = cos left(frac{2 pi }{15}right) and sin left(frac{19 pi }{90}right) = sin left(2 pi - frac{19 pi }{90}right) = sin left(frac{2 pi }{15}right), we can rewrite the expression as: left(9 left(cos left(frac{2 pi }{15}right)+i sin left(frac{2 pi }{15}right)right)right)^3 Expanding the cube, we get: 9^3 left(cos left(frac{2 pi }{15}right)+i sin left(frac{2 pi }{15}right)right)^3 =729 left(cos left(frac{2 pi }{15}right)+i sin left(frac{2 pi }{15}right)right)^3 Using De Moivre's theorem, we can simplify further: =729 left(cos left(3 cdot frac{2 pi }{15}right)+i sin left(3 cdot frac{2 pi }{15}right)right) =729 left(cos left(frac{2 pi }{5}right)+i sin left(frac{2 pi }{5}right)right) Therefore, the simplified expression is 729 left(cos left(frac{2 pi }{15}right)+i sin left(frac{2 pi }{15}right)right). The answer is 729 left(cos left(frac{2 pi }{15}right)+i sin left(frac{2 pi }{15}right)right)

answer:Given the ellipse: {eq}dfrac{{{x^2}}}{{{a^2}}} + dfrac{{{y^2}}}{{{b^2}}} = 1. {/eq} To find the dimensions and area of the largest rectangle that can be inscribed in the upper half of the ellipse: Let PQRS be a rectangle inscribed in the ellipse in an anticlockwise direction. Let point {eq}Pleft( {acos theta ,bsin theta } right) {/eq} be in the first quadrant. Calculate the area of the rectangle: {eq}begin{align*} {rm{Length}}:PQ &= 2acos theta {rm{Width}}:RS &= bsin theta {rm{Area}} &= PQ times RS fleft( theta right) &= 2absin theta cos theta &= absin 2theta end{align*} {/eq} Find the derivatives of the area with respect to {eq}theta . {/eq} {eq}begin{align*} f'left( theta right) &= 2abcos 2theta f''left( theta right) &= - 4absin 2theta end{align*} {/eq} To maximize the area, equate the first derivative to 0: {eq}begin{align*} f'left( theta right) &= 2abcos 2theta = 0 cos 2theta &= 0 theta &= dfrac{pi }{4} end{align*} {/eq} Substitute this value into the second derivative: {eq}begin{align*} f''left( {dfrac{pi }{4}} right) &= - 4absin 2left( {dfrac{pi }{4}} right) &= - 4ab end{align*} {/eq} Since the second derivative is negative, the area is maximum at this point. Find the dimensions and area of the rectangle: {eq}begin{align*} PQ &= 2acos dfrac{pi }{4} = sqrt 2 a RS &= bsin dfrac{pi }{4} = dfrac{b}{{sqrt 2 }} {rm{Area}} &= sqrt 2 a times dfrac{b}{{sqrt 2 }} &= ab end{align*} {/eq} Therefore, the dimensions of the rectangle are {eq}left( {sqrt 2 a,dfrac{b}{{sqrt 2 }}} right) {/eq} and the area is ab.

answer:To achieve the highest degree of precision, the quadrature formula should exactly reproduce all polynomials up to a certain degree. For a single-point formula like this one, the highest possible degree of precision is 1, meaning it will exactly integrate all linear functions. By the mean value theorem for integrals, there exists a point xi in (0,1) such that f(xi) is equal to the average value of the function over the interval [0,1], which is also the exact result of the integral for linear functions: f(xi) = frac{1}{b-a} int_a^b f(x), mathrm{d}x = int_0^1 f(x), mathrm{d}x Choosing c_0 = 0, c_1 = 1, and x_1 = xi ensures that the formula correctly integrates all linear functions. However, without further information, the exact value of xi cannot be determined. In practice, selecting x_1 as the midpoint of the interval, x_1 = frac{1}{2}, would provide a good approximation for many functions.