answer:To solve for the conjugate of z, we first simplify z: z=frac{{5-i}}{{1+i}} To eliminate the imaginary unit i in the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator: =frac{(5-i)(1-i)}{(1+i)(1-i)} Expanding both the numerator and the denominator: =frac{5(1)-5(i)-i(1)+i(i)}{1^2-(i)^2} Since i^2 = -1, we simplify further: =frac{5-5i-i-1}{1+1} =frac{4-6i}{2} Simplifying the fraction: =2-3i Therefore, the conjugate of z, denoted as overline{z}, is obtained by changing the sign of the imaginary part of z: overline{z}=2+3i Thus, the correct answer is: boxed{A}

answer:The domain of the function f(x)=a+frac{1}{2^{x}+1} is mathbb{R}, and it is given to be an odd function. For an odd function, f(0)=0. So, we have: a+frac{1}{2^{0}+1}=0 a+frac{1}{2}=0 a=-frac{1}{2} Let's verify that a=-frac{1}{2} makes the function f(x) an odd function: f(x)=-frac{1}{2}+frac{1}{2^{x}+1}=frac{1-2^{x}}{2(2^{x}+1)} f(-x)=frac{1-2^{-x}}{2(2^{-x}+1)}=-frac{1-2^{x}}{2(2^{x}+1)}=-f(x) Since f(-x)=-f(x), the function is indeed an odd function, which is consistent with the problem statement. Therefore, the answer is boxed{a=-frac{1}{2}}. Note that if an odd function has a domain of mathbb{R} (meaning x=0 is defined), then f(0)=0. Alternatively, you can use the definition of odd functions: f(-x)=-f(x) to find a. If the domain of a function f(x) includes 0 and the function is odd, then f(0)=0. However, this is a sufficient but not necessary condition, so it's essential to verify the results in this type of problem. You could also directly use the definition of odd and even functions: f(-x)=-f(x) to solve for a.

answer:Let's denote the number of pieces of paper Jessica used as P. Each piece of paper weighs 1/5 of an ounce, so the total weight of the paper is P * (1/5) ounces. The envelope weighs 2/5 of an ounce. The total weight of the letter is the weight of the paper plus the weight of the envelope, which is P * (1/5) + 2/5 ounces. Jessica needed 2 stamps to mail her letter, which means the total weight must be more than 1 ounce but less than or equal to 2 ounces (since she needs one stamp per ounce, and you can't use a fraction of a stamp). So, we have the inequality: 1 < P * (1/5) + 2/5 ≤ 2 To find the number of pieces of paper (P), we need to solve for P: First, let's multiply the entire inequality by 5 to get rid of the fractions: 5 < P + 2 ≤ 10 Now, subtract 2 from all parts of the inequality to isolate P: 5 - 2 < P ≤ 10 - 2 3 < P ≤ 8 Since P must be a whole number (you can't use a fraction of a piece of paper), the possible values for P are 4, 5, 6, 7, or 8. However, since Jessica used 2 stamps, the weight must be more than 1 ounce but less than or equal to 2 ounces. If P were 8, the total weight would be 8 * (1/5) + 2/5 = 1.6 + 0.4 = 2 ounces, which is the maximum weight for 2 stamps. Therefore, Jessica used boxed{8} pieces of paper.

answer:Part 1: 1. Let the midpoint of line segment AB be G(a, b). Since points A and B are on the lines x pm 2y = 0, we express their coordinates as follows: [ A = (a + t cos alpha, b + t sin alpha), quad B = (a - t cos alpha, b - t sin alpha) ] 2. Since A and `B` are on the lines `x = 2y` and `x = -2y`, their coordinates must satisfy: [ (a + t cos alpha)^2 - 4 (b + t sin alpha)^2 = 0 quad text{and} quad (a - t cos alpha)^2 - 4 (b - t sin alpha)^2 = 0 ] 3. By combining equations and simplifying, we get: [ left( cos^2 alpha - 4 sin^2 alpha right) t^2 + left(2a cos alpha - 8b sin alpha right) t + a^2 - 4b^2 = 0 ] 4. Given that t_1 t_2 = -4, the product of roots for the quadratic equation derived from the above expression must satisfy: [ t_1 t_2 = -frac{a^2 - 4b^2}{cos^2 alpha - 4 sin^2 alpha} = -4 ] 5. Thus, rearranging and solving for `a` and `b`, we get: [ -frac{a^2 - 4b^2}{cos^2 alpha - 4 sin^2 alpha} = -4 Rightarrow a^2 - 4b^2 = frac{16(a^2 - 4b^2)}{a^2 + 16b^2} ] [ a^2 + 16b^2 = 16 Rightarrow frac{a^2}{16} + b^2 = 1 ] 6. Therefore, the equation of the locus for G, or C is: [ C: frac{x^2}{16} + y^2 = 1 ] Part 2: 1. Let point P be (m, n). The equation of the tangent to the curve at `P` is: [ y - n = k(x - m) ] 2. Substituting this into the equation of the ellipse: [ x^2 + 16y^2 = 16 ] results in the following quadratic equation in x: [ (1 + 16k^2)x^2 - 32k(km - n)x + 16(km - n)^2 - 16 = 0 ] 3. For the lines to be perpendicular, the slopes k_1 and k_2 must satisfy: [ k_1k_2 = -1 ] This implies that the discriminant Delta must equal zero: [ Delta = 32^2 k^2 (km - n)^2 - 4 times 16 (1 + 16k^2)[(km - n)^2 - 1] = 0 ] 4. Rearranging and simplifying gives: [ (m^2 - 16)k^2 - 2mnk + n^2 - 1 = 0 ] 5. Therefore, further analysis shows: [ m^2 + n^2 - 17 = 0 Rightarrow x^2 + y^2 = 17 ] This implies that point P lies on a circle with equation: (boxed{x^2 + y^2 = 17})