answer:Let's denote the total area of the original triangle as 1. The area of each of the four smaller triangles formed in the first step of division is frac{1}{4}. In subsequent steps, since only the top triangle is divided further and shaded, what remains for the shading calculation follows the series: - First top triangle: frac{1}{4} - Second top triangle within the first top one: frac{1}{4^2} - And so forth: frac{1}{4^3}, frac{1}{4^4}, ldots Thus, the series summing up the areas is: [ frac{1}{4} + frac{1}{4^2} + frac{1}{4^3} + cdots ] This is a geometric series with a common ratio r = frac{1}{4}. The sum S of this infinite series is calculated using the formula for the sum of an infinite geometric series: [ S = frac{a}{1 - r} = frac{frac{1}{4}}{1 - frac{1}{4}} = frac{frac{1}{4}}{frac{3}{4}} = frac{1}{3} ] Thus, the fractional part of the triangle that is shaded is boxed{frac{1}{3}}.

answer:Since overrightarrow{m}=(lambda+1,1) and overrightarrow{n}=(lambda+2,2), then overrightarrow{m}+ overrightarrow{n}=(2lambda+3,3) and overrightarrow{m}- overrightarrow{n}=(-1,-1). Because (overrightarrow{m}+ overrightarrow{n})perp(overrightarrow{m}- overrightarrow{n}), then (overrightarrow{m}+ overrightarrow{n})cdot (overrightarrow{m}- overrightarrow{n})=0, thus -(2lambda+3)-3=0, solving this gives lambda=-3. Therefore, the correct choice is boxed{text{B}}. This problem can be solved by using the rules of vector operations and the relationship between vector perpendicularity and the dot product. Mastering the rules of vector operations and the relationship between vector perpendicularity and the dot product is key to solving the problem.

answer:**Analysis:** First, based on the fact that the sum of an odd number and an even number is odd, and given a^2 + b = 2009 with b being an odd number, we can conclude that a^2 is even. Since a is a prime number and the only even prime number is 2, we can determine the value of a. Then, we can find the value of b, and thus calculate the value of a + b. Since the only even prime number is 2, we have a = 2. Substituting a = 2 into the equation a^2 + b = 2009, we get 4 + b = 2009. Solving for b, we find b = 2005. Therefore, the sum of a and b is a + b = 2 + 2005 = 2007. Thus, the answer is boxed{2007} .

answer:The given proposition p is a universal statement. The negation of a universal statement is an existential (or particular) statement. Therefore, the negation of the proposition lnot p: exists xinmathbb{R}, x < 0. Let's break down the reasoning: 1. Identify the type of the original proposition, which is a universal statement. 2. Apply the rule for negating a universal statement, which is to change it into an existential statement. 3. Observe that the original proposition states xgeqslant 0 for all real numbers. So, negating this would mean there exists at least one real number x for which the inequality does not hold. 4. Option C correctly represents the negation, stating that there exists an x in mathbb{R} such that x < 0. Hence, the correct answer is boxed{C}.