answer:To solve this problem, we analyze the given quadratic function and the positions of points A, B, and C relative to this function. 1. The quadratic function is given by y=-(x-2)^{2}+h. This is a parabola that opens downwards because the coefficient of (x-2)^2 is negative. The vertex of this parabola is at (2, h), which means the highest point on the graph is at x=2. 2. Point A has coordinates left(-frac{1}{2}, y_{1}right). Since -frac{1}{2} < 2, A is to the left of the vertex. 3. Point B has coordinates left(1, y_{2}right). Since 1 < 2, B is also to the left of the vertex but to the right of A. 4. Point C has coordinates left(2, y_{3}right). This is exactly at the vertex of the parabola. 5. Since the parabola opens downwards and A and B are on the left side of the vertex, moving from A to B (left to right) towards the vertex, the value of y increases. Therefore, we have y_{1} < y_{2}. 6. Point C being at the vertex and the highest point on the graph means y_{3} is greater than both y_{1} and y_{2}. Combining these observations, we conclude that y_{1} < y_{2} < y_{3}. Therefore, the correct relationship between y_{1}, y_{2}, and y_{3} is: boxed{A: y_{1} < y_{2} < y_{3}}

answer:1. According to the problem, (E) is the midpoint of side (BC) in square (ABCD). Thus, the length (BE = EC = frac{1}{2}BC). 2. Draw vertical lines from points (B) and (D) to line (AE) and label the intersection points as (F) and (G), respectively. These perpendicular segments (BF) and (DG) form right angles with (AE). 3. Notice that we have two right triangles, (triangle ABF) and (triangle DAG): - triangle ABF (cong) triangle DAG by the properties of a square (symmetry and equal sides). 4. Since (E) is the midpoint of (BC), the triangles (triangle AEB) and (triangle DEC) are similar by AA similarity criterion. 5. The lengths (BF) and (DG) are equal since they are both perpendicular distances from points on the same side of the square to the line (AE). 6. Given the nature of similar triangles, we get the ratio: [ frac{BF}{AF} = frac{EB}{AB} = frac{1}{2} ] because (EB = frac{1}{2} BC) and (BC = AB). 7. It follows that (BF) is half of (AF). Similarly, (DG) will also be half of (AG). Since (F) and (G) lie directly beneath (B) and (D) respectively, (BF = FG = frac{1}{2}AF). 8. Now consider the angle (angle FBG). Since both (triangle ABF) and (triangle DAG) are congruent and symmetrical about the center line, the quadrilateral (BFAG) forms with equal opposite sides. 9. In particular, angle (angle FBG) spans one quarter of the full rotation in the circle centered at (B) and passing through (F) and (G). 10. Hence, the measure of (angle FBG) is: [ angle FBG = 45^circ ] # Conclusion: [ boxed{45^circ} ]

answer:First, we simplify the sets M and N, then we find their intersection. For set M, we have M={y|y={2}^{x}}={y|y > 0}, since 2^x is always positive for any real x. For set N, we set y= sqrt{2x - {x}^{2}}. In order for y to be real, the expression under the square root must be non-negative. Therefore: 2x - {x}^{2} geqslant 0 Factoring out x gives us: x(2 - x) geqslant 0 This inequality holds true for 0 leqslant x leqslant 2. Thus, N={x|0 leqslant x leqslant 2}. Now, we determine the intersection M cap N. Since M consists of positive real numbers and N includes the range from 0 to 2 (inclusive), the intersection will include all positive real numbers up to (and including) 2, but not including 0, because 2^x is positive for x > 0. Thus: M cap N = {x|0 < x leqslant 2} Therefore, the correct answer is boxed{B: (0,2]}.

answer:Since sin x < tan x for 0 < x < frac{pi}{2}, the hypotenuse of the right triangle can only be tan x or sec x. If tan x is the hypotenuse, then: [ tan^2 x = sin^2 x + sec^2 x ] Recall that sec x = frac{1}{cos x}, thus sec^2 x = frac{1}{cos^2 x}. The equation becomes: [ tan^2 x = sin^2 x + frac{1}{cos^2 x} ] Since sin^2 x + cos^2 x = 1, this simplifies to: [ tan^2 x = 1 + frac{1}{cos^2 x} - cos^2 x ] Plugging back in terms of tan x, let's simplify further: [ tan^2 x = 1 + sec^2 x - frac{1}{sec^2 x} ] This equation shows that the tangent squared values must comply with the above relationship for a right triangle with sides sin x, tan x, and sec x. Given the complexity, we solve directly for tan x as the hypotenuse case: [ tan^2 x = 1 + frac{1}{cos^2 x} - cos^2 x = tan^2 x + 1 ] This equation is valid for all x where tan x serves as the hypotenuse. If sec x is the hypotenuse, an analogous approach but with alternate assignment leads to: [ sec^2 x = tan^2 x + sin^2 x ] Using sin^2 x = 1 - cos^2 x, sec^2 x = frac{1}{cos^2 x}: [ tan^2 x = frac{1}{cos^2 x} - 1 + 1 - cos^2 x ] This case does not lead to a straightforward solution without more context or constraints because of the cyclical dependency. Conclusion: The sum of all possible values of tan^2 x where tan x is the hypotenuse remains as presented above, and no further valid x exists from the second case due to cyclical dependencies in the equation. [ boxed{text{The sum of }tan^2 xtext{ over all }xtext{ in }Ttext{ is }tan^2 x.} ]