answer:【Answer】Solution: According to the analysis, if there is only one point in the repeating section, then it can only be above the digit "6";if there are 2 points in the repeating section, then one of them must be above the digit "6"; the other may be above " 1, 2, 3, 4, 5 ",so there are 1+5=6 possibilities in total.Answer: The original decimal has 6 possible forms.Therefore, the answer is: 6.

answer:9. Let the equation of the hyperbola be (frac{x^{2}}{a^{2}}-frac{y^{2}}{b^{2}}=1).Let (Pleft(x_{0}, y_{0}right), Q(-c, 0), R(c, 0), S(x, y)), and (P R) intersects (Q S) at point (T). Let the complex numbers corresponding to points (P, Q, R, S, T) be (p, q, r, s, t), respectively.It is easy to see that (q = -r).Let (|P Q| = u, |P R| = v). Then[u = e x_{0} + a, quad v = e x_{0} - a.]By the properties of internal and external angle bisectors, we have[begin{array}{l}frac{|P T|}{|T R|} = frac{|P Q|}{|Q R|} Rightarrow frac{p - t}{t - r} = frac{u}{2c} Rightarrow t = frac{u r + 2c p}{u + 2c}, frac{|S Q|}{|S T|} = frac{|P Q|}{|P T|} = frac{|R Q|}{|R T|} Rightarrow frac{|S Q|}{|S T|} = frac{|P Q| + |R Q|}{|P R|} Rightarrow frac{s - q}{s - t} = frac{u + 2c}{v} Rightarrow s = frac{(u + 2c) t - q v}{u + 2c - v}.end{array}]Substituting equation (1) into equation (2), we get[begin{aligned}s & = frac{u r + 2c p - q v}{u + 2c - v} = frac{(u + v) r + 2c p}{2(a + c)} = frac{e x_{0} r + c p}{a + c} & Rightarrow (x, y) = left(frac{c(e + 1) x_{0}}{a + c}, frac{c y_{0}}{a + c}right) = left(frac{c x_{0}}{a}, frac{c y_{0}}{a + c}right) Rightarrow & x_{0} = frac{a x}{c}, quad y_{0} = frac{(a + c) y}{c}.end{aligned}]Substituting into the hyperbola equation, we get[frac{x^{2}}{c^{2}} - frac{(a + c)^{2} y^{2}}{b^{2} c^{2}} = 1.]Thus, the trajectory equation of point (S) is[frac{x^{2}}{25} - frac{9 y^{2}}{25} = 1.]Let the circle with center (D) and radius (|D S| = r_{0}) be[x^{2} + (y - 1)^{2} = r_{0}^{2}.]Combining equations (3) and (4), we get[begin{array}{l}r_{0}^{2} - (y - 1)^{2} - 9 y^{2} = 25 Rightarrow 10 y^{2} - 2 y + 26 - r_{0}^{2} = 0.end{array}]By the minimality of (|D S|), the circle (odot D) is tangent to the hyperbola (3), so[Delta = 4 - 40(26 - r_{0}^{2}) = 0 Rightarrow r_{0} = sqrt{25.9}.]

answer:1. Define the coordinates: - Let the coordinates of the square ABCD be A(0, 0), B(5, 0), C(5, 5), and D(0, 5). - The midpoint E of CD has coordinates Eleft(frac{5+0}{2}, frac{5+5}{2}right) = E(2.5, 5).2. Equation of line AE: - The line AE passes through points A(0, 0) and E(2.5, 5). The slope of AE is: [ text{slope of } AE = frac{5 - 0}{2.5 - 0} = 2 ] - The equation of the line AE in slope-intercept form is: [ y = 2x ]3. Find point F on AE such that CF = 5: - Let F have coordinates (x, 2x) since it lies on AE. - The distance CF is given by: [ CF = sqrt{(x - 5)^2 + (2x - 5)^2} ] - Set CF = 5: [ sqrt{(x - 5)^2 + (2x - 5)^2} = 5 ] - Square both sides to eliminate the square root: [ (x - 5)^2 + (2x - 5)^2 = 25 ] - Expand and simplify: [ (x - 5)^2 = x^2 - 10x + 25 ] [ (2x - 5)^2 = 4x^2 - 20x + 25 ] [ x^2 - 10x + 25 + 4x^2 - 20x + 25 = 25 ] [ 5x^2 - 30x + 50 = 25 ] [ 5x^2 - 30x + 25 = 0 ] [ x^2 - 6x + 5 = 0 ]4. Solve the quadratic equation: - Factorize the quadratic equation: [ (x - 1)(x - 5) = 0 ] - Thus, x = 1 or x = 5.5. Determine the correct x value: - If x = 5, then F would be at (5, 10), which is not on the line segment AE. - Therefore, x = 1.6. Find AF: - The coordinates of F are (1, 2). - The distance AF is: [ AF = sqrt{(1 - 0)^2 + (2 - 0)^2} = sqrt{1 + 4} = sqrt{5} ]The final answer is boxed{sqrt{5}}.

answer:From the given, we have f(x)=2^{2x}-2^{x+1}+2=(2^x-1)^2+1 in [1,2], thus 2^x-1 in (-1,1], which means 2^x in (0,2]. Therefore, x in (-infty,1], which means the domain of the function f(x)=2^{2x}-2^{x+1}+2 is (-infty,1], i.e., M=(-infty,1]. Combining the given options, the numbers of the conclusions that are certainly correct are ③④⑤⑥. Hence, the answer is boxed{C}. By analyzing the range of f(x), we can determine the range of 2^x-1, which allows us to find that 2^x in (0,2], and from this, we can deduce the domain of the function M=(-infty,1], thus identifying the correct conclusions. This question mainly examines the knowledge of the domain of functions, the judgment of the relationship between elements and sets, the judgment of the inclusion relationship between sets and its application, and other basic knowledge. It involves the integration of numerical and geometric thinking, reduction, and transformation thinking, and is considered a medium-level problem.

answer:Let the acute angle at vertex D of the rectangular trapezoid A B C D with bases A D and B C be bisected by the diagonal B D, and A B=8 andC D=10. Triangle B C D is isosceles because angle C B D=angle B D A=angle C D B, so B C=C D=10.Drop the height C H to the base A D. Since A B C H is a rectangle, C H=A B=8 and A H=B C=10. From the right triangle C D H, we find that D H=6, soA D=A H+D H=16 . text { Therefore, } S_{A B C D}=1 / 2(A D+B C) cdot C H=1 / 2(16+10) cdot 8=104 .## Answer104.

answer: Solution:# Part (Ⅰ) - Finding the Measure of Angle AGiven that b^{2}+c^{2}-a^{2}=accos C+c^{2}cos A, we can manipulate this equation as follows:1. Start with the given equation: [b^{2}+c^{2}-a^{2}=accos C+c^{2}cos A]2. Rearrange the terms to isolate terms involving cos A: [b^{2}+c^{2}-a^{2}-c^{2}cos A=accos C]3. Notice that the left side can be rewritten using the cosine rule for triangle ABC: [2bccos A = accos C + c^{2}cos A]4. Simplify to isolate terms with cos A: [2bcos A = acos C + ccos A]5. Apply the sine rule, frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C}, to convert to sines: [2sin Bcos A = sin Acos C + sin Ccos A]6. Use the sum-to-product identities to simplify further: [2sin Bcos A = sin(A+C)]7. Since A + B + C = pi, we have sin(A+C) = sin(pi - B) = sin B: [2sin Bcos A = sin B]8. Divide both sides by sin B (noting sin B neq 0 since 0 < B < pi): [2cos A = 1]9. Solve for A: [cos A = frac{1}{2}]10. Given 0 < A < pi, we find: [A = frac{pi}{3}]Therefore, the measure of angle A is boxed{frac{pi}{3}}.# Part (Ⅱ) - Finding sin B + sin CGiven S_{triangle ABC} = frac{25sqrt{3}}{4} and a = 5, we proceed as follows:1. Use the formula for the area of a triangle S_{triangle ABC} = frac{1}{2}bcsin A: [frac{25sqrt{3}}{4} = frac{sqrt{3}}{4}bc]2. Solve for bc: [bc = 25]3. Use the cosine rule with cos A = frac{1}{2}: [frac{1}{2} = frac{b^{2} + c^{2} - a^{2}}{2bc} = frac{b^{2} + c^{2} - 25}{50}]4. Solve for b^{2} + c^{2}: [b^{2} + c^{2} = 50]5. Find (b + c)^{2}: [(b + c)^{2} = b^{2} + 2bc + c^{2} = 50 + 2 times 25 = 100]6. Solve for b + c: [b + c = 10]7. Calculate sin B + sin C using sin A = frac{sqrt{3}}{2} and a = 5: [sin B + sin C = (b + c) frac{sin A}{a} = 10 cdot frac{frac{sqrt{3}}{2}}{5} = sqrt{3}]Therefore, sin B + sin C = boxed{sqrt{3}}.