answer:Given the equation left(2x-3right)^{5}=a_{0}+a_{1}(x-1)+a_{2}(x-1)^{2}+a_{3}(x-1)^{3}+a_{4}(x-1)^{4}+a_{5}(x-1)^{5}, we're asked to find the sum a_{0}+a_{2}+a_{4}. **Step 1:** Substitute x = 0 into the given equation. left(2(0)-3right)^{5} = a_{0}+a_{1}(0-1)+a_{2}(0-1)^{2}+a_{3}(0-1)^{3}+a_{4}(0-1)^{4}+a_{5}(0-1)^{5} Simplifying, we get: -243 = a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - a_{5} (1) **Step 2:** Substitute x = 2 into the given equation. left(2(2)-3right)^{5} = a_{0}+a_{1}(2-1)+a_{2}(2-1)^{2}+a_{3}(2-1)^{3}+a_{4}(2-1)^{4}+a_{5}(2-1)^{5} Simplifying, we find: 1 = a_{0} + a_{1} + a_{2} + a_{3} + a_{4} + a_{5} (2) **Step 3:** Add equations (1) and (2) together to eliminate the terms with odd powers of (x-1), which are a_{1}, a_{3}, and a_{5}: -243 + 1 = (a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - a_{5}) + (a_{0} + a_{1} + a_{2} + a_{3} + a_{4} + a_{5}) Simplifying, we get: -242 = 2a_{0} + 2a_{2} + 2a_{4} Dividing through by 2: -121 = a_{0} + a_{2} + a_{4} Therefore, the sum a_{0}+a_{2}+a_{4} equals -121, which means the correct answer is boxed{text{D}}.

answer:We start by identifying the essential properties of the problem: a circle with 2019 points divides the circle into 2019 arcs. Specifically, we have: - 673 arcs of length 1. - 673 arcs of length 2. - 673 arcs of length 3. Step-by-step Proof: 1. **Divide the Circle**: Consider a circular arrangement of 2019 points with arcs of different lengths as described: 673 arcs of length 1, 673 arcs of length 2, and 673 arcs of length 3. 2. **Introduce Midpoints**: By placing blue points at the midpoints of the arcs of length 2 and trisection points on arcs of length 3, we now effectively divide the circle into arcs of length 1. 3. **Assignnment of Points**: Take one of the arcs of length 2, denoted as ( overparen{AC} ), where (A) and (C) are red points. Let (B) be the midpoint of ( overparen{AC} ) which we colored blue. 4. **Consider Polarity**: Now, consider the diametrically opposite points ( A', B', C' ) to (A, B, C), respectively: - If ( A' ) is a red point, then (AA') is a diameter, satisfying the problem condition. - Similarly, if ( C' ) is a red point, then (CC') is a diameter. - If both (A') and (C') are blue points, ( B' ) must be a red point (since three consecutive blue points are not possible). 5. **Contradiction Approach**: Suppose there does not exist a pair of diametric red points. Now consider the minor arc ( overparen{AB'} ) subtending ( angle ACB' ) with length 2018. 6. **Arc Analysis**: For ( overparen{AB'} ), we determine the number of segments of each category: Let ( n_{1} ) be the number of segments of length 1, ( n_{2} ) be the number of segments of length 2, ( n_{3} ) be the number of segments of length 3: [ 2018 = n_{1} + 2n_{2} + 3n_{3} ] Similarly, for arc ( overparen{CB'} ): [ 2018 = n_{1}' + 2n_{2}' + 3n_{3}' ] 7. **Opposite Arc Consideration**: Each arc of length 1 on ( overparen{AB'} ) has endpoints opposed to the arcs of adjacent points opposite on ( overparen{CB'}): Resulting in: [ n_{1} = n_{3}', quad n_{1}' = n_{3} ] 8. **Deriving an Equation**: The total number of arcs of length 1 is 673: [ n_{1} + n_{1}'= 673 Rightarrow n_{1} + n_{3} = 673 ] Plugging into our equations: [ 2018 = n_{1} + 2n_{2} + 3(673 - n_{1}) ] Simplifying we find: [ 2018 = n_{1} + 2n_{2} + 2019 - 3n_{1} ] [ 2n_{2} = 2n_{1} - 1 ] Which gives an inherent contradiction since (2n_{2}) must be even. Conclusion: We conclude that our assumption that no two diametrically opposite red points exist is incorrect—proving that there must exist two diametrically opposite red points. (blacksquare)

answer:To find out how many games the soccer team won, you need to calculate 40 percent of 158 games. 40% of 158 = (40/100) * 158 = 0.4 * 158 = 63.2 Since a team cannot win a fraction of a game, we round to the nearest whole number. Therefore, the soccer team won boxed{63} games.

answer:Let's assume the man's total income is 100 units (it could be dollars, euros, etc., but we'll use units for simplicity). He spends: - 35% on food, which is 35 units. - 25% on children's education, which is 25 units. So, the total percentage spent on food and children's education is 35% + 25% = 60%. This means he has 100% - 60% = 40% of his income remaining. Out of this remaining income, he is left with 8% of his total income. Therefore, he spends the rest of the remaining income on house rent. The amount spent on house rent is 40% (remaining income) - 8% (left with him) = 32% of his total income. However, the question asks for the percentage of the remaining income (not the total income) that he spends on house rent. Since the remaining income is 40% of the total, and he spends 32% of his total income on rent, we need to find out what percentage 32% is of 40%. To find this, we use the formula: Percentage of remaining income spent on house rent = (Amount spent on house rent / Remaining income) * 100 Plugging in the values, we get: Percentage of remaining income spent on house rent = (32 / 40) * 100 Percentage of remaining income spent on house rent = 0.8 * 100 Percentage of remaining income spent on house rent = 80% So, the man spends boxed{80%} of the remaining income on house rent.