answer:The x-coordinate of the vertex of the parabola y = x^2 + 1 is frac{-b}{2a} = frac{0}{2(1)} = 0. The vertex is then (0, 1). The intersections of the line y = r with y = x^2 + 1 are found by setting the y values equal to each other: [ r = x^2 + 1 quad Rightarrow quad r - 1 = x^2 quad Rightarrow quad pmsqrt{r-1} = x. ] The vertices of our triangle are (0, 1), (-sqrt{r-1}, r), and (sqrt{r-1}, r). Using the horizontal segment along the line y = r as the base of the triangle, the length of the base is 2sqrt{r-1}. The height of the triangle is the vertical distance from (0, 1) to the line y = r, which is r - 1. The area of the triangle is then: [ A = frac{1}{2} times text{base} times text{height} = frac{1}{2} times (2sqrt{r-1}) times (r-1) = (r-1)sqrt{r-1}. ] This can also be expressed as (r-1)^{frac{3}{2}}. Given 10 leq A leq 50, we have [ 10 leq (r-1)^{frac{3}{2}} leq 50. ] Taking the cube root on all sides gives: [ sqrt[3]{10} leq (r-1)^{frac{1}{2}} leq sqrt[3]{50}. ] Squaring all sides leads to: [ (sqrt[3]{10})^2 leq r-1 leq (sqrt[3]{50})^2. ] Adding 1 to each side: [ (sqrt[3]{10})^2 + 1 leq r leq (sqrt[3]{50})^2 + 1. ] Therefore, the possible values of r are approximately in the interval boxed{[(sqrt[3]{10})^2 + 1, (sqrt[3]{50})^2 + 1]}.

answer:This problem involves the concept of vector projection. By definition, the projection of overrightarrow{b} onto overrightarrow{a} is given by the product of the magnitude of overrightarrow{b} and the cosine of the angle between overrightarrow{a} and overrightarrow{b}. Thus, the projection of overrightarrow{b} onto overrightarrow{a} is given by: begin{align*} |overrightarrow{b}| cos frac{3pi}{4} &= 2 cdot left(-frac{sqrt{2}}{2}right) &= -sqrt{2} end{align*} Therefore, the projection of overrightarrow{b} onto overrightarrow{a} is boxed{-sqrt{2}}.

answer:Let's call the number of people who left the line L. Initially, there were 7 people in line. After some left and 8 more joined, there were 11 people in line. So the equation would be: 7 - L + 8 = 11 Combining like terms, we get: 15 - L = 11 Subtracting 15 from both sides gives us: -L = 11 - 15 -L = -4 Multiplying both sides by -1 to solve for L gives us: L = 4 So, boxed{4} people left the line.

answer:From the problem, we can deduce that 5^8 = 390625, 5^9 = 1953125, thus the last four digits of each number change periodically, with a period of 4. Since 2011 = 4 times 502 + 3, the last four digits of 5^{2011} are 8125. Therefore, the correct choice is boxed{text{D}}.