answer:Let's denote the cost of purchasing the new shoes as ( P ). The cost per year of repairing the used shoes is ( frac{13.50}{1} = 13.50 ) dollars per year. The average cost per year of the new shoes is 18.52% greater than the cost of repairing the used shoes. Therefore, the cost per year of the new shoes is ( 13.50 + 0.1852 times 13.50 ). Let's calculate that: ( 0.1852 times 13.50 = 2.5002 ) So the cost per year of the new shoes is: ( 13.50 + 2.5002 = 16.0002 ) Since the new shoes last for 2 years, the total cost of the new shoes is twice the annual cost: ( P = 2 times 16.0002 ) ( P = 32.0004 ) Therefore, the cost to purchase the new shoes is approximately boxed{32.00} .

answer:To solve this problem, first, we convert the equation of the circle into its standard form: (x-2)^2+(y-2)^2=9. Therefore, the center of the circle is (2, 2), and the radius r=3. When the slope of the tangent line is undefined, the line x=5 satisfies the condition. When the slope of the tangent line exists, let it be k. The equation of the tangent line is y+2=k(x-5), which can be rewritten as kx-y-5k-2=0. Since the distance d from the center of the circle to the tangent line equals the radius r, we have frac{|2k-2-5k-2|}{sqrt{k^2+1}}=3. Solving this equation, we find k=-frac{7}{24}. Thus, the equation of the tangent line at this slope is -frac{7}{24}x-y+frac{35}{24}-2=0, which simplifies to 7x+24y+13=0. In summary, the equations of the tangent lines are 7x+24y+13=0 or x=5. Therefore, the answer is: boxed{7x+24y+13=0 text{ or } x=5}. This problem involves finding the equation of a tangent line to a circle, which includes converting the circle's equation to its standard form, using the point-slope form of a line, calculating the distance from a point to a line, and considering different cases based on the slope of the tangent line. Understanding the property that the distance from the center of the circle to the tangent line equals the radius of the circle is key to solving this problem.

answer:1. **Claim:** For every ( i ), (gcd(x_{i+1}, x_i) geq 2). - To prove this, assume there exists an ( i ) such that ( x_i ) is divisible by a prime ( p ). If no such ( i ) exists, then ( x_i = 1 ) for all ( i ), which does not satisfy the given equation. - Given the equation ( x_{i+2}^2 = text{lcm}(x_{i+1}, x_i) + text{lcm}(x_i, x_{i-1}) ), we can deduce that ( p mid x_i ) implies ( p mid x_{i+2} ). - From the equation ( x_{i+3}^2 = text{lcm}(x_{i+2}, x_{i+1}) + text{lcm}(x_{i+1}, x_i) ), we get ( (x_{i+3} - x_{i+2})(x_{i+3} + x_{i+2}) = text{lcm}(x_{i+2}, x_{i+1}) - text{lcm}(x_i, x_{i-1}) ). - Since ( p mid x_i ), it follows that ( p mid x_{i+3} ). By induction, every ( x_i ) is divisible by ( p ). Therefore, for every ( i ), (gcd(x_{i+1}, x_i) geq 2). 2. **Summing the Equations:** - Summing all the given equations, we obtain: [ 2 sum_{i=1}^{20} text{lcm}(x_{i+1}, x_i) = sum_{i=1}^{20} x_i^2 ] - Using the inequality (sum_{i=1}^{20} x_i^2 geq sum_{i=1}^{20} x_i x_{i+1}), where equality holds if and only if ( x_i = x_{i+1} ) for all ( i ). 3. **Rewriting the LCM:** - Rewrite (text{lcm}(x_{i+1}, x_i) = frac{x_i x_{i+1}}{gcd(x_{i+1}, x_i)}). - Since (gcd(x_{i+1}, x_i) geq 2), we conclude that the inequality must hold, and therefore all integers must be equal. 4. **Solving for ( x_i ):** - Given ( x_i = x_{i+1} ) for all ( i ), we substitute into the original equation: [ x_i^2 = 2x_i ] - Solving ( x_i^2 = 2x_i ) gives ( x_i (x_i - 2) = 0 ), so ( x_i = 2 ). Conclusion: [ boxed{x_i = 2 text{ for all } i in {1, 2, ldots, 20}} ]

answer:1. **Calculate the amount of paint remaining after the first day:** Initially, there is 1 gallon of paint. On the first day, one fourth of the paint is used. The amount of paint used on the first day is: [ frac{1}{4} times 1 = frac{1}{4} text{ gallon} ] Therefore, the amount of paint remaining after the first day is: [ 1 - frac{1}{4} = frac{4}{4} - frac{1}{4} = frac{3}{4} text{ gallons} ] 2. **Calculate the amount of paint remaining after the second day:** On the second day, half of the remaining paint is used. The amount of paint used on the second day is: [ frac{1}{2} times frac{3}{4} = frac{3}{8} text{ gallons} ] Therefore, the amount of paint remaining after the second day is: [ frac{3}{4} - frac{3}{8} = frac{6}{8} - frac{3}{8} = frac{3}{8} text{ gallons} ] 3. **Conclusion:** The fraction of the original amount of paint that is left to use on the third day is frac{3}{8}. Thus, the answer is: [ frac{3{8}} ] The final answer is boxed{textbf{(C) } frac{3}{8}}