answer:First, let's calculate the cost of labor for replacing the radiator: Labor for radiator replacement = 2 hours * 75/hour = 150 Now, let's apply the 10% discount on the labor charges: Discount on labor = 10% of 150 = 0.10 * 150 = 15 Labor cost after discount = 150 - 15 = 135 Next, let's calculate the cost of the part: Part cost = 150 Now, let's calculate the cost of labor for cleaning and flushing the cooling system: Labor for cleaning and flushing = 1 hour * 60/hour = 60 Now, let's calculate the total cost before tax: Total cost before tax = Labor cost after discount + Part cost + Labor for cleaning and flushing Total cost before tax = 135 + 150 + 60 = 345 Finally, let's calculate the tax and the total cost after tax: Tax = 8% of 345 = 0.08 * 345 = 27.60 Total cost after tax = Total cost before tax + Tax Total cost after tax = 345 + 27.60 = 372.60 So, Mark paid boxed{372.60} for the radiator replacement and cooling system cleaning and flushing.

answer:Since it is known that a > b, and the quadratic trinomial ax^{2}+2x+bgeqslant 0 always holds for all real numbers x, therefore a > 0, and Delta = 4-4ab leqslant 0, therefore ab geqslant 1. Furthermore, since there exists x_{0} in mathbb{R}, such that a x_{0}^{2}+2x_{0}+b=0 holds, we can deduce Delta = 0, therefore ab = 1, therefore a > 1, Since dfrac{a^{2}+b^{2}}{a-b} = dfrac{a^{2}+ dfrac{1}{a^{2}}}{a- dfrac{1}{a}} = dfrac{(a- dfrac{1}{a})^{2}+2}{a- dfrac{1}{a}} = (a- dfrac{1}{a})+ dfrac{2}{a- dfrac{1}{a}} geqslant 2 sqrt{(a- dfrac{1}{a})cdot dfrac{2}{a- dfrac{1}{a}}} = 2 sqrt{2}, equality holds if and only if a= dfrac{1+ sqrt{5}}{2} Therefore, the minimum value of dfrac{a^{2}+b^{2}}{a-b} is 2 sqrt{2}, Hence, the answer is: boxed{2 sqrt{2}}. Given the conditions a > 1, ab=1, we transform the expression to be evaluated into dfrac{a^{4}+1}{a^{3}-a} = dfrac{(a- dfrac{1}{a})^{2}+2}{a- dfrac{1}{a}} = (a- dfrac{1}{a})+ dfrac{2}{a- dfrac{1}{a}}, and the answer can be found using the basic inequality. This problem mainly examines the application of the basic inequality and the issue of a function always holding true. The transformation of the expression is the difficulty and key to solving the problem, making it a medium-level question.

answer:**Analysis** This question examines geometric probability problems related to area and the use of definite integrals to calculate the area of a curvilinear trapezoid, which is considered a medium-level problem. **Solution** First, calculate the area of (Ω): (S_{Ω} = 2 times sqrt{2} times sqrt{2} = 4). Next, calculate the area of (A): (S_{A} = int_{- frac{pi}{3}}^{frac{pi}{3}} cos x , dx = sin frac{pi}{3} - sin (- frac{pi}{3}) = sqrt{3} ). Therefore, the probability sought is ( boxed{frac{sqrt{3}}{4}} ). Hence, the correct answer is C.

answer:Given: The polynomial ( f(x) = x^{2007} + 17x^{2006} + 1 ) has distinct zeroes ( r_{1}, ldots, r_{2007} ). A polynomial ( P ) of degree 2007 satisfies ( Pleft(r_{j} + frac{1}{r_{j}}right) = 0 ) for ( j = 1, ldots, 2007 ). We need to determine the value of ( frac{P(1)}{P(-1)} ). 1. Since ( P ) must be zero at each ( r_{j} + frac{1}{r_{j}} ), we can write: P(z) = k prod_{j=1}^{2007} left( z - left( r_{j} + frac{1}{r_{j}} right) right) for some constant ( k ). 2. To evaluate ( frac{P(1)}{P(-1)} ), we need to consider ( z = 1 ) and ( z = -1 ): frac{P(1)}{P(-1)} = frac{k prod_{j=1}^{2007} left( 1 - left( r_{j} + frac{1}{r_{j}} right) right)}{k prod_{j=1}^{2007} left( -1 - left( r_{j} + frac{1}{r_{j}} right) right)} 3. Simplifying the product for ( frac{P(1)}{P(-1)} ): frac{P(1)}{P(-1)} = prod_{j=1}^{2007} frac{1 - left( r_{j} + frac{1}{r_{j}} right)}{-1 - left( r_{j} + frac{1}{r_{j}} right)} 4. We can rewrite the terms inside the product: frac{1 - left( r_{j} + frac{1}{r_{j}} right)}{-1 - left( r_{j} + frac{1}{r_{j}} right)} = frac{1 - r_{j} - frac{1}{r_{j}}}{-1 - r_{j} - frac{1}{r_{j}}} = frac{r_{j}^{2} - r_{j} + 1}{r_{j}^{2} + r_{j} + 1} 5. Using ( omega ) where ( omega neq 1 ) and ( omega^{3} = 1 ), we have ( omega^{2} + omega = -1 ): 6. We now need to evaluate ( f(z) ) at necessary points. Doing this we get: f(z) = z^{2007} + 17z^{2006} + 1 Continuing this evaluation for roots: prod_{j=1}^{2007} frac{r_{j}^{2} - r_{j} + 1}{r_{j}^{2} + r_{j} + 1} = prod_{j=1}^{2007} frac{ - omega - r_{j}}{ omega - r_{j}} times frac{- omega^{2}- r_{j}}{ omega^{2}- r_{j}} 7. With the evaluations: frac{P(1)}{P(-1)} = frac{f(-omega) fleft(-omega^{2}right)}{f(omega) fleft(omega^{2}right)} = frac{left(-omega^{2007} + 17omega^{2006} + 1right)left(-left(omega^{2}right)^{2007} + 17left(omega^{2}right)^{2006} + 1right)}{left(omega^{2007} + 17omega^{2006} + 1right)left(left(omega^{2}right)^{2007} + 17left(omega^{2}right)^{2006} + 1right)} 8. Simplify the numerator and the denominator: [ = frac{ left(17 omega times 17 omega^{2}right)}{left( 2 + 17 omega^{2}) times (2 + 17 omega)} ] 9. Combining the terms: = frac{289}{ (4 + 34 (omega + omega^{2}) + 289)} = frac{289}{259} # Conclusion: boxed{frac{289}{259}}