answer:To find the value of ( f(1998) ) given the properties of the function ( f(x) ): 1. Consider the functional equation given: [ f(x + y) = x + f(y) ] - This relation must hold for all real numbers ( x ) and ( y ). 2. Use the initial condition ( x = 0 ): [ f(0 + y) = 0 + f(y) implies f(y) = f(y) ] - This restates the fact that the function is well-defined for all ( y ). 3. Next, substitute ( y = 0 ): [ f(x + 0) = x + f(0) implies f(x) = x + f(0) ] - This tells us that the function ( f(x) ) can be expressed in another form if we know ( f(0) ). 4. Given ( f(0) = 2 ): [ f(x) = x + 2 ] 5. We now seek ( f(1998) ): [ f(1998) = 1998 + 2 = 2000 ] Conclusion: [ boxed{2000} ] Therefore, the value of ( f(1998) ) is ( 2000 ).

answer:To solve the problem, let's take it step by step: 1. First, we identify the total number of applicants from this university, which includes undergraduate students, master's students, and doctoral students. The total number is 4400 + 400 + 200. 2. Then, we calculate the actual total to get the combined number of applicants from all categories: [ 4400 + 400 + 200 = 5000 ] 3. According to the problem, the stratified sampling ratio for selecting doctoral students is provided as frac{10}{200}. To use this ratio for calculating the total number of volunteers selected from all categories, we understand this ratio applies uniformly across all types of students. Thus, the ratio is frac{1}{20}. 4. Applying the stratified sampling ratio to the total number of applicants, we find the number of volunteers selected from the university: [ 5000 times frac{1}{20} = 250 ] 5. Therefore, the total number of volunteers selected from this university, following the given stratified sampling ratio, is 250. So, the final answer, following the rules and encapsulating it as requested, is boxed{D}.

answer:To solve for the length of side BC in triangle ABC with the given conditions, we can apply the Law of Cosines. Given that angle B = 120^{circ}, AC = sqrt{19}, and AB = 2, we aim to find the length of side BC, which we'll denote as a. The Law of Cosines states that in any triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of those two sides and the cosine of the included angle. Applying this to our triangle, with side AC as c = sqrt{19}, side AB as b = 2, and the angle at B as 120^{circ}, we get: [c^2 = a^2 + b^2 - 2abcos(B)] Substituting the given values into this equation: [sqrt{19}^2 = a^2 + 2^2 - 2 times a times 2 times cos(120^{circ})] Simplifying the equation: [19 = a^2 + 4 - 4a times cos(120^{circ})] Since cos(120^{circ}) = -frac{1}{2}, we substitute this value into the equation: [19 = a^2 + 4 + 4a] Rearranging the equation to form a quadratic equation: [a^2 + 4a - 15 = 0] Solving this quadratic equation, we look for values of a that satisfy the equation. Factoring or using the quadratic formula, we find that a = 3 or a = -5. Since a side length cannot be negative in geometry, we discard a = -5 and accept a = 3. Therefore, the length of side BC is 3. Encapsulating the final answer: [boxed{D}]

answer:To solve the equation frac{a}{1-i} = 1 - bi, we multiply the numerator and denominator on the left side by the complex conjugate of the denominator to remove the imaginary part from the denominator: frac{a}{1-i} = frac{a(1+i)}{(1-i)(1+i)} = frac{a(1+i)}{1^2 - (-i)^2} = frac{a(1+i)}{1 + 1} = frac{a}{2} + frac{a}{2}i. Now we have the equation frac{a}{2} + frac{a}{2}i = 1 - bi. By equating the real and the imaginary parts of the complex numbers, we get: frac{a}{2} = 1 text{ and } frac{a}{2}i = -bi. From the first equation, by solving for a, we get a = 2. From the second equation (and knowing the value of a), we solve for b: frac{a}{2}i = -bi Rightarrow frac{2}{2}i = -bi Rightarrow i = -bi Rightarrow b = -1. Having both a and b, we can now find the modulus of the complex number a+bi: |a + bi| = sqrt{a^2 + b^2} = sqrt{2^2 + (-1)^2} = sqrt{4 + 1} = sqrt{5}. Therefore, the modulus of a + bi is boxed{sqrt{5}}.