answer:To find the 2023rd monomial in the given sequence, we observe the pattern in the sequence of monomials: -2x, 3x^{2}, -4x^{3}, 5x^{4}, ldots. From this pattern, we can deduce the following rules for any nth monomial in the sequence: 1. The absolute value of the coefficient is the ordinal number (n) plus 1. This gives us the coefficient as n + 1. 2. The sign of the coefficient alternates, which can be represented mathematically as left(-1right)^{n}. This means the sign is negative for odd n and positive for even n. 3. The exponent of x in the monomial is exactly n. Applying these rules to find the 2023rd monomial: - The coefficient's absolute value is 2023 + 1 = 2024. - Since 2023 is odd, the sign of the coefficient will be negative, which is represented by left(-1right)^{2023} = -1. - The exponent of x will be 2023. Combining these observations, the 2023rd monomial in the sequence is given by left(-1right)^{2023} cdot 2024 cdot x^{2023} = -2024x^{2023}. Therefore, the 2023rd monomial is boxed{-2024x^{2023}}.

answer:The line l: x+2y+c=0 can be rewritten as y=-frac{1}{2}x-frac{1}{2}c. Translating it 1 unit to the left, we get y=-frac{1}{2}(x+1)-frac{1}{2}c. Translating it further 2 units down, we obtain y=-frac{1}{2}(x+1)-frac{1}{2}c-2. After rearranging, the equation of line l' becomes x+2y+c+5=0. The circle C: x^2+y^2+2x-4y=0 can be rewritten as (x+1)^2+(y-2)^2=5, thus, the center of the circle is (-1, 2), and its radius is sqrt{5}. Since the line and the circle are tangent, we have frac{|-1+2times2+c+5|}{sqrt{1^2+2^2}}=sqrt{5}, Solving this equation, we find c=-3 or c=-13. Therefore, the answer is: boxed{{-3, -13}}. By translating the line l' and rewriting the equation of the circle into its standard form, we can find the center and radius of the circle. Using the formula for the distance from a point to a line, we can solve for c. This problem examines the relationship between a line and a circle, involving translations and the formula for the distance from a point to a line, and is considered a medium-level question.

answer:To simplify 100r - 48r + 10, follow these steps: 1. **Combine like terms**: Simplify the terms involving r. - 100r - 48r = 52r 2. **Incorporate the constant term**: As there is no other constant to combine with, the constant term remains as is. - Simplified expression becomes 52r + 10. Thus, the simplified form of the expression 100r - 48r + 10 is boxed{52r + 10}.

answer:Let z = x + yi. Then (frac{z}{50} = frac{x}{50} + frac{y}{50}i), so [0 leq frac{x}{50} leq 1] [0 leq frac{y}{50} leq 1] which implies (0 leq x leq 50) and (0 leq y leq 50). Also, [frac{50}{overline{z}} = frac{50}{x - yi} = frac{50(x + yi)}{x^2 + y^2} = frac{50x}{x^2 + y^2} + frac{50y}{x^2 + y^2}i,] which implies [0 leq frac{50x}{x^2 + y^2} leq 1] [0 leq frac{50y}{x^2 + y^2} leq 1] For (x geq 0), the first condition is ( 50x leq x^2 + y^2), or [(x - 25)^2 + y^2 geq 25^2.] For (y geq 0), the second condition is (50y leq x^2 + y^2), or [x^2 + (y - 25)^2 geq 25^2.] Thus, B is the region inside the square with vertices at 0, 50, 50 + 50i, and 50i, but outside the circles centered at 25 and 25i with radius 25. The area inside the square but outside the circle is simpler calculated in one quarter and then doubled because of symmetry. The area of one quarter outside the circle of radius 25 starting at center 25 (or 25i) is: [625 - frac{1}{4} cdot pi cdot 625 = 625 - frac{625pi}{4}.] Thus, the total area in B is: [2(625 - frac{625pi}{4}) + 625 = boxed{1875 - frac{625pi}{2}}.]