answer:We need to show that the expression: [ 1 + frac{x}{2} - frac{x^2}{2^3} + frac{x^3}{2^4} - frac{5 x^4}{2^7} ] can be derived by modifying the procedure mentioned in the 1029th practice problem. We will also extend the expression from (1) by adding two more terms, and apply this to approximate (sqrt{10}) up to 7 decimal places. We will compare this result with the usual numerical approximation for (sqrt{10}). I. In the 1029th exercise, the fourth approximating value to (sqrt{a^2 + b}) is given by: [ sqrt{a^2 + b} approx a + frac{b}{2a} - frac{b^2}{4a(2a^2 + b)} - frac{b^4}{8a(2a^2 + b)(8a^4 + 8a^2b + b^2)} ] To make (a^2 + b) transform into (1 + x), we set (a = 1) and (b = x). Thus, (2) becomes: [ sqrt{1 + x} approx 1 + frac{x}{2} - frac{x^2}{4(2 + x)} - frac{x^4}{8(2 + x)(8 + 8 x + x^2)} ] Initially, both this expression and (1) match in the first two terms. To generate the third term in (1) from the third term in (3), we omit the (x) term in the denominator of ( frac{x^2}{4(2 + x)} ), assuming that (x) is small compared to 2. So we set: [ d_2 = -frac{x^2}{4(2 + x)} ] to [ d_2' = -frac{x^2}{2^3} ] We apply the same step to the fourth term, i.e., removing (x) from the larger term in the denominator of the fourth term in (3). Simplifying, we get: [ -frac{x^4}{128} = -frac{x^4}{2^7} ] which resembles the fifth term in (1), but it is five times larger. The query arises: where does the fourth term in (1) originate from? Moving to our next steps, supposing that [ m_3' = m_2 - (2c_2 + d_2') cdot d_2 ] Here, (m_2 = -frac{x^2}{4}), and (c_2 = 1 + frac{x}{2}). Thus, we compute: [ begin{aligned} m_3' &= -frac{x^2}{4} - left(2 + x - frac{x^2}{2^3}right)left(-frac{x^2}{2^3}right) &= frac{x^3}{2^3} - frac{x^4}{2^6} end{aligned} ] For the next term, we compute with (m_{3}^prime) first term: [ d_3' = frac{m_3'}{2c_1} = left(frac{x^3}{2^3}right) div 2 = frac{x^3}{2^4} ] As such, this explains the origin of the fourth term in (1). To summarize: [ m_4' = m_3' - left(2c_3' + d_3'right) cdot d_3' ] where [ begin{aligned} m_4' &= frac{x^3}{2^3} - frac{x^4}{2^6} - left(2 + x - frac{x^2}{2^3} + frac{x^3}{2^4}right) cdot frac{x^3}{2^4} &= -frac{5x^4}{2^6} + frac{x^5}{2^6} - frac{x^6}{2^8} end{aligned} ] Finally, [ d_4' = left(-frac{5x^4}{2^6}right) div 2 = -frac{5x^4}{2^7} ] which gives the fifth term in (1). # II. Adding two more terms, we proceed as follows: [ begin{aligned} m_5' &= m_4' - left(2c_4' + d_4'right) cdot d_4' &= -frac{5x^4}{2^6} + frac{x^5}{2^6} - frac{x^6}{2^8} - left(2 + x - frac{x^2}{2^2} + frac{x^3}{2^3} - frac{5x^4}{2^7}right) cdot left(-frac{5x^4}{2^7}right) &= frac{7x^5}{2^7} - frac{7x^6}{2^9} + text{higher degree terms} end{aligned} ] Thus, [ d_5' = frac{7x^5}{2^7} div 2 = frac{7x^5}{2^8}, quad m_6' = ldots = -frac{21x^6}{2^9} + text{higher degree terms}, quad d_6' = frac{21x^6}{2^{10}} ] Finally, [ sqrt{1 + x} approx 1 + frac{x}{2} - frac{x^2}{2^3} + frac{x^3}{2^4} - frac{5x^4}{2^7} + frac{7x^5}{2^8} - frac{21x^6}{2^{10}} ] III. To apply this to (sqrt{10}) and approximate it to 7 decimal places, we set (x = -frac{1}{10}) and (x = frac{1}{9}) respectively, giving: [ sqrt{10} approx frac{10}{3} sqrt{1 - frac{1}{10}} = frac{10}{3} - frac{1}{6} - frac{1}{12 cdot 20} - frac{1}{12 cdot 20^2} - frac{5}{48 cdot 20^3} - frac{7}{48 cdot 20^4} - frac{21}{32 cdot 20^5} ] and [ sqrt{10} approx 3 sqrt{1 + frac{1}{9}} = 3 + frac{1}{6} - frac{3}{2 cdot 18^2} + frac{3}{2 cdot 18^3} - frac{15}{2 cdot 18^4} + frac{21}{8 cdot 18^5} - frac{63}{16 cdot 18^6} ] Calculating these terms up to 7 decimal places: [ begin{aligned} & 3.16666666 -& 0.60416666 -& 0.04120833 -& 0.00325042 -& 0.00027084 -& 0.00002161 & approx 3.16227768 end{aligned} ] and similarly: [ begin{aligned} & 3.16666666 -& 0.03703703 & 0.00154321 -& 0.00006420 & 0.00000257 -& 0.00000011 & approx 3.16227759 end{aligned} ] The usual calculation for (sqrt{10}) results in approximately: [ 3.16227766016838... ] Thus: Conclusion: [ boxed{3.1622776} ]

answer:To approach this problem, we will use some principles of combinatorics and geometry to prove the statement. 1. **Consider a Regular Pentagon:** - First, let's inscribe a regular pentagon into the given circle. A regular pentagon is a five-sided polygon with all sides and angles equal. - As per the properties of a regular pentagon, any three vertices form an isosceles triangle. This is because the distances between non-adjacent vertices of a regular pentagon are equal. 2. **Coloring Vertices:** - Since every point on the circle is colored in one of two colors, and there are five vertices in the pentagon, consider these five vertices and their colorings. - According to the pigeonhole principle, if we distribute more than (n) objects (in this case, vertices) into (n) groups (in this case, the two colors), at least one group must contain more than (leftlfloor frac{n}{2} rightrfloor) objects. Hence, out of the five vertices, at least three must be of the same color. 3. **Forming an Isosceles Triangle:** - Let's denote the color that appears at least three times as (C). - Select any three vertices among those that are colored (C). We denote these vertices as (A), (B), and (C). - Because (A), (B), and (C) are among the vertices of the regular pentagon, the triangle (ABC) must be an isosceles triangle, as explained in the first step. This is because, in a regular pentagon, the lengths of the sides connecting any two vertices are consistent, making at least one pair of sides equal. 4. **Conclusion:** - Therefore, these three vertices (A), (B), and (C) form an isosceles triangle with vertices of the same color. - Thus, we have proved that no matter how the circle is colored in two colors, it is always possible to find an isosceles triangle whose vertices are all the same color, inscribed within the circle. [ boxed{} ]

answer:According to the universal set I={1,2,3,4,5}, A={1,3}, therefore C_I A={2,4,5}, and B={2}; then Bcup C_I A={2,4,5} Therefore, the correct option is boxed{text{C}}.

answer:First, replace ( cos^2 x ) with ( 1 - sin^2 x ) and simplify: [ g(x) = frac{sin^3 x + 7 sin^2 x + 2 sin x + 3(1 - sin^2 x) - 10}{sin x - 1} = frac{sin^3 x + 4 sin^2 x + 2 sin x - 7}{sin x - 1}. ] Factor the numerator: [ sin^3 x + 4 sin^2 x + 2 sin x - 7 = (sin x - 1)(sin^2 x + 5 sin x + 7). ] This simplifies to: [ g(x) = sin^2 x + 5 sin x + 7. ] Let ( y = sin x ). Then: [ g(x) = y^2 + 5y + 7 = left( y + frac{5}{2} right)^2 - frac{25}{4} + 7 = left( y + frac{5}{2} right)^2 + frac{3}{4}. ] Since ( y = sin x ) and ( -1 leq y leq 1 ), ( y + frac{5}{2} ) ranges from ( frac{3}{2} ) to ( frac{7}{2} ). Thus, [ left( frac{3}{2} right)^2 + frac{3}{4} leq g(x) leq left( frac{7}{2} right)^2 + frac{3}{4}. ] Calculating these values: [ left( frac{3}{2} right)^2 + frac{3}{4} = frac{9}{4} + frac{3}{4} = 3, quad left( frac{7}{2} right)^2 + frac{3}{4} = frac{49}{4} + frac{3}{4} = 13. ] Thus, the range of ( g(x) ) is ( boxed{[3, 13)} ), excluding the point where ( sin x = 1 ).