answer:Given (f(x)= begin{cases} -2x-2, & xin (-infty ,0) x^{2}-2x-1, & xin [0,+infty ) end{cases}), When (x < 0), (y > -2); When (xgeqslant 0), (y=(x-1)^{2}-2geqslant -2), (f(0)=f(2)=-1), Given (x_{1} < x_{2} < x_{3}) and (f(x_{1})=f(x_{2})=f(x_{3})), Then (x_{2}+x_{3}=2), which implies (x_{1}+x_{2}+x_{3}=x_{1}+2), When (f(x_{1})=-1), i.e., (-2x_{1}-2=-1), solving gives (x_{1}=- dfrac {1}{2}), Since (- dfrac {1}{2}leqslant x_{1} < 0), We can derive that ( dfrac {3}{2}leqslant x_{1}+2 < 2), Therefore, the correct choice is: boxed{A}. By the symmetry of the quadratic function, we have (x_{2}+x_{3}=2), which implies (x_{1}+x_{2}+x_{3}=x_{1}+2). From the graph, we solve (- dfrac {1}{2}leqslant x_{1} < 0), and thus obtain the desired range. This question examines the graph and application of piecewise functions, the symmetry of quadratic functions, and the method of combining numerical and geometric thinking. It is considered a medium difficulty and commonly mistaken question.

answer:To solve this, we start with frac {9}{18} = 0.5 = 27 div 54 = 50% = 10 : 20 = text{five out of ten}. Therefore, the answers are: 9, 54, 50, 10, five. Converting 0.5 into a fraction and simplifying gives frac {1}{2}, and by applying the basic property of fractions, multiplying both the numerator and denominator by 9 yields frac {9}{18}. According to the relationship between fractions and division, frac {1}{2} = 1 div 2, and by applying the property of constant quotient, multiplying both the dividend and divisor by 27 gives 27 div 54. Based on the relationship between ratios and fractions, frac {1}{2} = 1 : 2, and by applying the basic property of ratios, multiplying both terms of the ratio by 10 gives 10 : 20. Moving the decimal point of 0.5 two places to the right and adding a percent sign gives 50%. According to the meaning of percentages, 50% is equivalent to five out of ten. The key to solving this problem is 0.5, and by understanding the relationships among decimals, fractions, percentages, division, ratios, and the basic properties of fractions, constant quotient, and ratios, the conversion can be made. Thus, the final answers are: boxed{9, 54, 50, 10, five}.

answer:Solution: A, A triangle is determined by three non-collinear points and lies on a plane, so it is a planar figure. Hence, A is incorrect. B, When all sides of a spatial quadrilateral are equal, it is a spatial solid rather than a planar figure. Hence, B is correct. C, A trapezoid has one pair of parallel sides, and two parallel lines determine a plane. Therefore, a trapezoid is a planar figure. Hence, C is incorrect. D, A parallelogram has its opposite sides parallel, and two parallel lines determine a plane. Therefore, a parallelogram is a planar figure. Hence, D is incorrect. Thus, the correct answer is B. According to the axioms and theorems for determining a plane, the shapes in options A, C, and D are planar figures. However, a quadrilateral with all sides equal in spatial geometry is not necessarily a planar figure. This question tests the application of axioms and theorems for determining a plane. Note that in solid geometry, a quadrilateral does not have to be a planar figure and can form a solid, such as a triangular pyramid. The final answer is boxed{text{B}}.

answer:If Zachary did 19 push-ups, and David did 3 times more than Zachary, then David did: 3 times 19 = 57 push-ups. So, David did boxed{57} push-ups.