answer:1. **Identification of the Given Information**: The problem provides a circle with center (O) and a point (D) on the circle. We need to determine the value of the angle ( x ). 2. **Identification of the Key Angles**: We start by noting that (angle ACD) is a right angle. This is because it is subtended by the diameter (AC) of the circle. According to the theorem: [ text{An angle subtended by the diameter of a circle is a right angle}. ] Therefore, (angle ACD = 90^circ). 3. **Calculation of Complementary Angles**: As given in the problem, (angle CDA = 48^circ). Thus, (angle CAD) can be calculated using the fact that the sum of angles in a right triangle is (90^circ): [ angle CAD = 90^circ - angle CDA = 90^circ - 48^circ = 42^circ. ] 4. **Use Symmetry and Radii Properties**: Since (O) is the center of the circle, all radii are equal: [ OA = OB = OC. ] Hence, (triangle OAC) and (triangle OBC) are isosceles triangles. Consequently, the base angles are equal: [ angle OCA = angle CAO = angle OCB ] Since (angle OCA = 42^circ), it follows that: [ angle CAO = 42^circ. ] 5. **Formation of Another Angle**: We need to determine the angle ( x ). Given that: [ angle OBC = angle OCA + 10^circ. ] Utilizing the values obtained: [ x = angle OBC = 42^circ + 10^circ = 52^circ. ] 6. **Conclusion**: Thus, the value of ( x ) is [ boxed{58^circ}. ]

answer:First, calculate a: a= int_{0}^{1}2xdx=x^{2} |_{0}^{1}=1, Then, for the binomial (ax^{2}- frac{1}{x})^{6}=(x^{2}- frac{1}{x})^{6}, the general term formula in its expansion is: T_{r+1}= binom{6}{r}(x^{2})^{6-r}(- frac{1}{x})^{r}=(-1)^{r} binom{6}{r}x^{12-3r}, Let 12-3r=0, solving this gives r=4. Therefore, the constant term = binom{6}{4}=15. Hence, the answer is: boxed{15}. By using the Fundamental Theorem of Calculus, we obtain: a= int_{0}^{1}2xdx=x^{2} |_{0}^{1}=1, and by applying the general term formula of the binomial expansion (ax^{2}- frac{1}{x})^{6}=(x^{2}- frac{1}{x})^{6}, we can derive the solution. This problem tests the understanding of the binomial theorem's general term formula, reasoning, and computational skills, and is considered a basic question.

answer:Since overrightarrow{OP_1} + overrightarrow{OP_2} + overrightarrow{OP_3} = overrightarrow{0}, we can write overrightarrow{OP_1} + overrightarrow{OP_2} = -overrightarrow{OP_3}. Given that |overrightarrow{OP_1}| = |overrightarrow{OP_2}| = |overrightarrow{OP_3}| = 1, squaring both sides and simplifying, we obtain overrightarrow{OP_1} cdot overrightarrow{OP_2} = -frac{1}{2}. By the definition of the dot product, we can deduce that angle P_1OP_2 = 120^circ. Thus, |overrightarrow{P_1P_2}| = sqrt{1 + 1 - 2 times 1 times 1 times (-frac{1}{2})} = sqrt{3}. Therefore, the magnitude of vector overrightarrow{P_1P_2} is boxed{sqrt{3}}. First, we prove that overrightarrow{OP_1} cdot overrightarrow{OP_2} = -frac{1}{2}, and then by the definition of the dot product, we can deduce that angle P_1OP_2 = 120^circ, leading to the final result. This problem involves understanding the dot product definition and applying the cosine rule. It is a moderate-level question.

answer:From the given, we have f(1)=2^{1}-4=-2, Therefore, f(a) > f(1) can be transformed into begin{cases} 2^{a}-4 > -2, & a > 0 -a-3 > -2, & a < 0end{cases}, Solving these sets of inequalities respectively, we get a > 1 or a < -1 Hence, the answer is: a > 1 or a < -1. By converting the inequality into two sets of inequalities and solving them, we can find the solution. This question tests the method of solving piecewise inequalities, and converting it into a set of inequalities is key to solving the problem. It is a basic question. boxed{a > 1 text{ or } a < -1}