answer:To solve, we first find the derivative of the function f(x)=x^{3}-3x^{2}+ax+2,f′(x)=3x^{2}-6x+a.Therefore, the slope of the tangent line to the curve y=f(x) at (0,2) is a.This means the equation of the tangent line at (0,2) is: y=ax+2.Setting y=0, we get x=- dfrac {2}{a}. From -2=- dfrac {2}{a}, we find a=1.Thus, the correct choice is: boxed{A}.The solution involves finding the derivative to determine the slope of the tangent line, formulating the equation of the tangent line, setting y=0 to solve for the equation, and solving for a. This problem tests the application of derivatives: finding the equation of a tangent line, understanding the geometric meaning of derivatives, and the correct application of the equation of a line are key to solving the problem. It is a basic question that tests computational skills.

answer:Solution 1 because a_{1}=1, a_{n+1}=3+frac{1}{2} a_{n}, (n geqslant 1) thenbegin{array}{l} a_{1}=1, a_{2}=frac{1}{2}+3, a_{3}=frac{1}{2^{frac{2}{2}}}+3 times frac{1}{2}+3, a_{4}=frac{1}{2^{3}}+3 times frac{1}{2^{2}}+3 times frac{1}{2}+3, cdots cdots a_{n}=frac{1}{2^{n-1}}+3 times frac{1}{2^{n-2}}+3 times frac{1}{2^{n-3}}+cdots +3 times frac{1}{2}+3 =frac{1}{2^{n-1}}+3 timesleft(frac{1}{2^{n-2}}+frac{1}{2^{n-3}}+cdots+right. frac{1}{2}+1left.=frac{1}{2^{n-1}}+frac{1}{3left(1-frac{1}{2}right.}right) =6-frac{5}{2^{n-1}} cdotend{array}(Proof omitted)Solution 2 because a_{n+1}=3+frac{1}{2} a_{n}begin{array}{l}-) quad a_{n}=3+frac{1}{2} a_{n-1} a_{n+1}-a_{n}=frac{1}{2}left(a_{n}-a_{n-1}right) . (n geqslant 2)end{array}Let a_{n+1}-a_{n}=b_{n}, then we get a new sequence {b_{n}}, where b_{1}=a_{2}-a_{1}=left(3+frac{1}{2} a_{1}right)-a_{1}=left(3+frac{1}{2}right)-1=frac{5}{2} text {, }and b_{n}=frac{1}{2} b_{n-1}, i.e., frac{b_{n}}{b_{n-1}}=frac{1}{2}.therefore The sequence {b_{n}} is a geometric sequence with the first term b_{1}=frac{5}{2}.begin{aligned}& text { Also } b_{n-1}+b_{n-2}+cdots+b_{2}+b_{1} & =a_{n}-a_{n-1}+a_{n-1}-a_{n-2}+cdots+a_{3} -a_{2} & +a_{2}-a_{1} & =a_{n}-a_{1} text {. }end{aligned}That is, a_{n}-a_{1}=b_{1}+b_{2}+cdots+b_{n-}+{ }_{2} b_{n-1}begin{aligned}& =frac{frac{5}{2}left(1-frac{1}{2^{n-1}}right)}{1-frac{1}{2}}=5-frac{5}{2^{n^{-1}}} . therefore quad a_{n} & =6-frac{5}{2^{n-1}} .end{aligned}

answer:Answer: 284160. Solution. We will add the numbers in a column. Each last digit appears in the units place as many times as there are three-digit numbers ending with this digit. Therefore, it will appear 8 cdot 8=64 times (since a total of 8 digits are used for the hundreds and tens places). Thus, the sum of the digits in the last place is 64 cdot(1+2+3+4+6+7+8+9)=2560. Similarly, in the tens and hundreds places, we get the same sum. In the end, we get 2560 cdot 100+2560 cdot 10+2560=2560 cdot 111=284160.

answer:-1 . frac{1}{10} text {. }Since left(frac{1}{f(x)}right)^{2}=frac{1}{x^{2}}+1, then,left(frac{1}{f^{(n)}(x)}right)^{2}=frac{1}{x^{2}}+n text {. }Therefore, f^{(99)}(1)=frac{1}{10}.

answer:To solve the equation x^{2}-2x-5=0 using the completing the square method, we follow these steps:1. Move the constant term to the other side of the equation:[x^{2}-2x = 5]2. To complete the square, we need to add (frac{-2}{2})^{2} = 1 to both sides of the equation. This is because the coefficient of x is -2, and we use the formula left(frac{b}{2}right)^{2} where b is the coefficient of x:[x^{2}-2x + 1 = 5 + 1]3. Simplify the equation:[x^{2}-2x+1 = 6]4. Now, the left side of the equation is a perfect square trinomial, which can be written as:[(x-1)^{2} = 6]Therefore, the original equation x^{2}-2x-5=0 when transformed using the completing the square method becomes:[boxed{(x-1)^{2} = 6}]Hence, the correct answer is boxed{B}.

answer:[Analysis]This problem mainly tests the application of vectors in geometry, vector dot product, vector collinearity, and the formula for the area of a triangle. By using the vector triangle theorem and determining the positional relationship between points P and Q, we can solve the problem.From the given conditions, we can deduce that P is the midpoint of AC, and Q is the trisection point closer to B on AB. Using the ratio of the areas of triangles APQ and ABC and the given condition (S_{Delta APQ} = frac{2}{3}), we can find the area of triangle ABC. With the formula for the area of a triangle, we can find the value of sin A and subsequently cos A. Finally, using the formula for the dot product of vectors, we can find the value of overrightarrow{AB} cdot overrightarrow{AC} + {overrightarrow{BC}}^{2}.[Solution]Since overrightarrow{PA} + overrightarrow{PC} = overrightarrow{0}, P is the midpoint of AC.Also, overrightarrow{QA} + overrightarrow{QB} + overrightarrow{QC} = overrightarrow{BC} implies overrightarrow{QA} + overrightarrow{QB} = overrightarrow{BC} - overrightarrow{QC} = overrightarrow{BQ}, so overrightarrow{QA} = 2overrightarrow{BQ}. Thus, Q is the trisection point closer to B on AB.Consequently, S_{Delta APQ} = frac{1}{3}S_{Delta ABC}, so S_{Delta ABC} = frac{1}{2}|overrightarrow{AB}||overrightarrow{AC}|sin A = 2,and thus sin A = frac{1}{2}, cos A = frac{sqrt{3}}{2} or -frac{sqrt{3}}{2}.Therefore, overrightarrow{AB} cdot overrightarrow{AC} = |overrightarrow{AB}||overrightarrow{AC}|cos A = pm 4sqrt{3},and {overrightarrow{BC}}^{2} = (overrightarrow{AC} - overrightarrow{AB})^{2} = 20 - 2overrightarrow{AB} cdot overrightarrow{AC},so overrightarrow{AB} cdot overrightarrow{AC} + {overrightarrow{BC}}^{2} = 20 - overrightarrow{AB} cdot overrightarrow{AC} = 20 pm 4sqrt{3}.Thus, the answer is boxed{D}.