answer:1. For proposition p to be false and q to be true, we have: - The negation of p: There exists an x in [-1, 0] such that log_2(x + 2) geq 2m. The smallest value of log_2(x + 2) is log_2(1) = 0, so we have 2m > 1 Rightarrow m > frac{1}{2}. - Proposition q: x^2 - 2x + m^2 = 0 has two distinct real roots, so its discriminant Delta = 4 - 4m^2 > 0. This implies m^2 < 1, or -1 < m < 1. Combining both conditions, we have -1 < m leq frac{1}{2}. boxed{text{The range of } m text{ is } (-1, frac{1}{2}].} 2. For the proposition (p lor q) to be true and (p land q) to be false, either p or q must be true, but not both. - If p is true and q is false, we have m > frac{1}{2} (from p) and m notin (-1, 1) (as q is false, meaning the discriminant is non-positive). This gives m geq 1. - If p is false and q is true, from part 1 we have -1 < m leq frac{1}{2}. Combining these cases, the range of m is (-1, frac{1}{2}] cup [1, +infty). boxed{text{The range of } m text{ is } (-1, frac{1}{2}] cup [1, +infty).}

answer:(1) According to the problem, the equation of line l_1 is y=frac{y_1}{x_1}x. Using the point-to-line distance formula, the distance d from point C to line l_1 is [ d = frac{|y_1x_2 - x_1y_2|}{sqrt{x_1^2 + y_1^2}} ] Since |AB|=2|AO|=2sqrt{x_1^2+y_1^2}, the area of the parallelogram S is [ S = |AB| cdot d = 2|x_1y_2-x_2y_1| ] This relationship still holds when the slope of either l_1 or l_2 is undefined. (2) Method 1: Let the slope of line l_1 be k, then the slope of line l_2 is -frac{1}{2k}. The equation of line l_1 is y=kx. Solving the system [ begin{cases} y = kx x^2 + 2y^2 = 1 end{cases} ] by eliminating y, we obtain x=pmfrac{1}{sqrt{1+2k^2}}. By symmetry, let x_1=frac{1}{sqrt{1+2k^2}} and y_1=frac{k}{sqrt{1+2k^2}}. Similarly, we can find that x_2=frac{sqrt{2}k}{sqrt{1+2k^2}} and y_2=-frac{sqrt{2}}{2sqrt{1+2k^2}}. Therefore, [ S = 2|x_1y_2-x_2y_1| = boxed{sqrt{2}} ] Method 2: Let the slopes of l_1 and l_2 be frac{y_1}{x_1} and frac{y_2}{x_2} respectively; then frac{y_1y_2}{x_1x_2}=-frac{1}{2}. Thus, x_1x_2=-2y_1y_2 and x_1^2x_2^2=4y_1^2y_2^2=-2x_1x_2y_1y_2. Since A(x_1, y_1) and C(x_2, y_2) lie on the ellipse x^2+2y^2=1, [ (x_1^2+2y_1^2)(x_2^2+2y_2^2) = x_1^2x_2^2+4y_1^2y_2^2+2(x_1^2y_2^2+x_2^2y_1^2) = 1 ] which gives -4x_1x_2y_1y_2+2(x_1^2y_2^2+x_2^2y_1^2)=1. Therefore, [ (x_1y_2-x_2y_1)^2=frac{1}{2} ] and |x_1y_2-x_2y_1|=frac{sqrt{2}}{2}. Finally, [ S = 2|x_1y_2-x_2y_1| = boxed{sqrt{2}} ]

answer:1. Identify the lengths of the pieces of string: - First string: 2 inches - Second string: 5 inches - Third string: 3 inches 2. Sum the lengths of the three pieces: [ 2 + 5 + 3 = 10 text{ inches} ] 3. Divide the total length by the number of pieces to find the average length: [ frac{10}{3} = boxed{3.33} text{ inches (recurring)} ] Conclusion: The average length of the three pieces of string is boxed{3.33} inches.

answer:For a line y = kx + 3 to be tangent to the circle x^2+y^2=1, the distance from the center of the circle, O(0,0), to the line should equal the radius of the circle, which is 1. The distance d from a point (x_0, y_0) to a line Ax + By + C = 0 is given by the formula: d = frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}} For the line kx - y + 3=0 and the point O(0,0), substituting the values gives us: 1 = frac{|0cdot k - 0 + 3|}{sqrt{k^2 + (-1)^2}} Simplifying, we get: 1 = frac{3}{sqrt{k^2 + 1}} Squaring both sides to clear the square root, we obtain: 1^2 = left(frac{3}{sqrt{k^2 + 1}}right)^2 1 = frac{9}{k^2 + 1} Multiplying both sides by k^2 + 1 to get rid of the denominator: k^2 + 1 = 9 Subtracting 1 from both sides: k^2 = 8 Taking the square root of both sides: k = pmsqrt{8} Since sqrt{8} = 2sqrt{2}, we get: k = pm2sqrt{2} Thus, the value of k that satisfies the given condition is boxed{pm2sqrt{2}}.