answer:Given vectors overrightarrow {a}=(2,1) and overrightarrow {b}=(1,3), we are asked to find the angle between the vectors 2overrightarrow {a}-overrightarrow {b} and overrightarrow {a}.First, let's calculate 2overrightarrow {a}-overrightarrow {b}:begin{align*}2overrightarrow {a}-overrightarrow {b} &= 2(2,1)-(1,3) &= (4,2)-(1,3) &= (4-1, 2-3) &= (3,-1).end{align*}Next, we find the magnitudes of 2overrightarrow {a}-overrightarrow {b} and overrightarrow {a}:begin{align*}|2overrightarrow {a}-overrightarrow {b}| &= sqrt{3^2+(-1)^2} = sqrt{9+1} = sqrt{10}, |overrightarrow {a}| &= sqrt{2^2+1^2} = sqrt{4+1} = sqrt{5}.end{align*}Now, we calculate the dot product of 2overrightarrow {a}-overrightarrow {b} and overrightarrow {a}:begin{align*}(2overrightarrow {a}-overrightarrow {b})cdot overrightarrow {a} &= (3,-1)cdot(2,1) &= 3cdot2 + (-1)cdot1 &= 6 - 1 &= 5.end{align*}Let the angle between vectors 2overrightarrow {a}-overrightarrow {b} and overrightarrow {a} be theta. The cosine of the angle is given by:begin{align*}cos theta &= dfrac{(2overrightarrow {a}-overrightarrow {b})cdot overrightarrow {a}}{|2overrightarrow {a}-overrightarrow {b}|cdot |overrightarrow {a}|} &= dfrac{5}{sqrt{10}cdot sqrt{5}} &= dfrac{5}{sqrt{50}} &= dfrac{5}{5sqrt{2}} &= dfrac{sqrt{2}}{2}.end{align*}Since cos theta = dfrac{sqrt{2}}{2} corresponds to an angle of 45^{circ} within the range 0^{circ}leqslant theta leqslant 180^{circ}, we conclude that theta = 45^{circ}.Thus, the answer is boxed{C}.

answer:Analysis: The difficulty ratio of this problem has significantly increased from 1, mainly due to its "apparent" asymmetry (the number of rows and columns are inconsistent). Therefore, the key to solving the problem is to find the symmetry hidden behind this asymmetry.Inspired by Example 1, we will also start from the local perspective.Solution: In each 1 times 5 rectangle, at least one chess piece must be taken. Following Example 1, as shown in Figure 5, the area outside the shaded part must have at least 10 chess pieces taken.If at least one chess piece is taken from the shaded part in Figure 5, then the total number of chess pieces taken is no less than 11.If no chess pieces are taken from the shaded part in Figure 5, then at least 2 chess pieces must be taken from each of the middle two columns, and at least 2 chess pieces must be taken from each of the middle three rows. Therefore, at least 10 chess pieces have been taken.Considering the four squares (1), (2), (3), and (4) in Figure 6.Looking diagonally, since the shaded area in Figure 6 indicates that no chess pieces have been taken, to avoid a line of five, the four squares (1), (2), (3), and (4) must all have chess pieces taken. Therefore, in Figure 6, 2 chess pieces are taken from the 3rd column and 2 chess pieces are taken from the 6th column; at least 1 chess piece is taken from the white area of each of the 1st, 2nd, 7th, and 8th columns, totaling 4 chess pieces; at least 1 chess piece is taken from the white area of each of the 1st, 2nd, 6th, and 7th rows, totaling 4 chess pieces. Therefore, 12 chess pieces are taken.In summary, at least 11 chess pieces must be taken.Figure 7 provides one possible method.

answer:Method 1: According to the problem, all possible values of xi are 2, 4, 6. Let each two games be a round, then the probability that the match stops at the end of this round is left(frac{2}{3}right)^{2}+left(frac{1}{3}right)^{2}=frac{5}{9}. If the match continues after this round, then A and B must each have scored one point in this round, and the result of this round has no effect on whether the match stops in the next round. Therefore, we havebegin{array}{l}P(xi=2)=frac{5}{9} P(xi=4)=left(frac{4}{9}right)left(frac{5}{9}right)=frac{20}{81} P(xi=6)=left(frac{4}{9}right)^{2}=frac{16}{81}end{array}Thus,E xi=2 times frac{5}{9}+4 times frac{20}{81}+6 times frac{16}{81}=frac{266}{81} .Therefore, the answer is B.Method 2: According to the problem, all possible values of xi are 2, 4, 6. Let A_{k} represent A winning the k-th game, then bar{A}_{k} represents B winning the k-th game. By independence and mutual exclusivity, we getbegin{aligned}P(xi=2)= & Pleft(A_{1} A_{2}right)+Pleft(bar{A}_{1} bar{A}_{2}right)=frac{5}{9} P(xi=4)= & Pleft(A_{1} bar{A}_{2} A_{3} A_{4}right)+Pleft(A_{1} bar{A}_{2} bar{A}_{3} bar{A}_{4}right) & +Pleft(bar{A}_{1} A_{2} A_{3} A_{4}right)+Pleft(bar{A}_{1} A_{2} bar{A}_{3} bar{A}_{4}right) = & 2left[left(frac{2}{3}right)^{3}left(frac{1}{3}right)+left(frac{1}{3}right)^{3}left(frac{2}{3}right)right] P(xi=6)= & Pleft(A_{1} bar{A}_{2} A_{3} bar{A}_{4}right)+Pleft(A_{1} bar{A}_{2} bar{A}_{3} A_{4}right) & +Pleft(bar{A}_{1} A_{2} A_{3} bar{A}_{4}right)+Pleft(bar{A}_{1} A_{2} bar{A}_{3} A_{4}right) = & frac{20}{81}, = & 4left(frac{2}{3}right)^{2}left(frac{1}{3}right)^{2} = & frac{16}{81}end{aligned}Therefore,E xi=2 times frac{5}{9}+4 times frac{20}{81}+6 times frac{16}{81}=frac{266}{81} .Therefore, the answer is B.

answer:7. 1202.Notice that, 52-7=97-52=45,begin{array}{l}7=45 times 1-38,52=45 times 2-38, 97=45 times 3-38 .end{array}Let Q(x)=P(x)-45 x+38. Then Q(x) is a quartic polynomial with leading coefficient 1, andQ(1)=Q(2)=Q(3)=0 text {. }Thus, for some r,begin{array}{l}Q(x)=(x-1)(x-2)(x-3)(x-r) . text { Therefore, } frac{1}{4}(P(9)+P(-5)) =frac{1}{4}(Q(9)+Q(-5))+26 =frac{1}{4}(8 times 7 times 6(9-r)+6 times 7 times 8(5+r))+26 =frac{1}{4}(6 times 7 times 8 times 14)+26=1202 .end{array}

answer:Let the equation of the desired line be x-2y+c=0. Substituting point (3,0) into the line equation, we get 3+c=0,therefore c=-3Hence, the equation of the desired line is: x-2y-3=0,So the answer is: boxed{x-2y-3=0}.Based on the perpendicular relationship, set the equation of the desired line as x-2y+c=0. Substitute point (3,0) into the line equation to find the value of c, and thus obtain the equation of the desired line.This problem primarily examines the properties of perpendicular lines, where the product of the slopes of two perpendicular lines equals (-1). Use the method of undetermined coefficients to find the equation of the line.

answer:Horner's Rule is a method to efficiently evaluate polynomial functions at a given value of x. It restructures the polynomial to lessen the number of multiplications required.For the polynomial f(x) = 2x^4 + 3x^3 + 5x - 4 and x = 2, we proceed as follows:[f(x) = 2x^4 + 3x^3 + 5x - 4][= x(2x^3 + 3x^2) + 5x - 4][= x[x(2x^2 + 3x) + 5] - 4][= x{x[x(2x + 3)] + 5} - 4]Now we substitute x = 2 and evaluate step by step:[f(2) = 2{2[2(2 times 2 + 3)] + 5} - 4][= 2{2[2(4 + 3)] + 5} - 4][= 2{2[2 times 7] + 5} - 4][= 2{2[14] + 5} - 4][= 2{28 + 5} - 4][= 2 times 33 - 4][= 66 - 4][= boxed{62}]