answer:Let's denote the initial amount of coolant in the car as ( x ) liters. The initial coolant is 30% antifreeze, so the amount of antifreeze in the initial coolant is ( 0.30x ) liters. Ann drained some of the coolant and replaced it with 80% antifreeze. Let's denote the amount of coolant drained and replaced as ( y ) liters. The amount of antifreeze in the replaced coolant is ( 0.80y ) liters. After the replacement, there are ( 7.6 ) liters of the original coolant left in the car, which means ( y = x - 7.6 ) liters were drained and replaced. The final mixture is 50% antifreeze, so the total amount of antifreeze in the final mixture is ( 0.50x ) liters. The amount of antifreeze in the final mixture comes from two sources: the original coolant that was not drained and the new 80% antifreeze coolant that was added. So we can write the equation: [ 0.30x + 0.80y = 0.50x ] Substitute ( y ) with ( x - 7.6 ): [ 0.30x + 0.80(x - 7.6) = 0.50x ] Expand the equation: [ 0.30x + 0.80x - 6.08 = 0.50x ] Combine like terms: [ 1.10x - 6.08 = 0.50x ] Subtract ( 0.50x ) from both sides: [ 0.60x - 6.08 = 0 ] Add ( 6.08 ) to both sides: [ 0.60x = 6.08 ] Divide both sides by ( 0.60 ): [ x = frac{6.08}{0.60} ] [ x = 10.13 ] So the initial amount of coolant in the car was approximately ( boxed{10.13} ) liters.

answer:Given that U={0,1,2,3,4,5,6}, M={1,3,5}, and N={2,4,6}, First, we find the complements of sets M and N in U. The complement of M in U, denoted by complement_U M, consists of the elements that are in U but not in M. So, complement_U M = {0,2,4,6}. Similarly, the complement of N in U, denoted by complement_U N, consists of the elements that are in U but not in N. So, complement_U N = {0,1,3,5}. Now, we take the union of complement_U M and complement_U N. (complement_U M) cup (complement_U N) = {0,2,4,6} cup {0,1,3,5}. To find the union, we combine the elements of both sets without repeating any elements. Thus, (complement_U M) cup (complement_U N) = {0,1,2,3,4,5,6}. So the correct answer is boxed{D: {0,1,2,3,4,5,6}}.

answer:Given the right-angled triangle triangle ABC with angle C = 90^circ , where the internal angle bisectors of angle A and angle B intersect at point P , and PE perp AB meeting at point E . Also given BC = 2 and AC = 3 , we need to find AE cdot EB . 1. **Identify segments**: - Construct PD perp AC at point D and PF perp BC at point F . 2. **Determine length of segments CD and CF**: [ CD = CF = frac{1}{2}(AC + BC - AB) quad text{(by the Apollonius Circle Theorem)} ] Next step is to find AB . 3. **Use Pythagorean Theorem in triangle ABC to find AB **: [ AB = sqrt{AC^2 + BC^2} = sqrt{3^2 + 2^2} = sqrt{9 + 4} = sqrt{13} ] 4. **Calculate CD and CF **: [ CD = CF = frac{1}{2}(AC + BC - AB) = frac{1}{2}(3 + 2 - sqrt{13}) = frac{1}{2}(5 - sqrt{13}) ] 5. **Find segment AD **: [ AD = AC - CD = 3 - frac{1}{2}(5 - sqrt{13}) = 3 - frac{5}{2} + frac{sqrt{13}}{2} = frac{6}{2} - frac{5}{2} + frac{sqrt{13}}{2} = frac{1 + sqrt{13}}{2} ] 6. **Find segment BF **: [ BF = BC - CF = 2 - frac{1}{2}(5 - sqrt{13}) = 2 - frac{5}{2} + frac{sqrt{13}}{2} = frac{4}{2} - frac{5}{2} + frac{sqrt{13}}{2} = frac{sqrt{13} - 1}{2} ] 7. **Find AE cdot EB **: [ AE cdot EB = AD cdot BF = left(frac{1 + sqrt{13}}{2}right) left(frac{sqrt{13} - 1}{2}right) ] Multiply the fractions: [ AE cdot EB = frac{(1 + sqrt{13})(sqrt{13} - 1)}{4} ] 8. **Simplify the product**: [ AE cdot EB = frac{(1 cdot sqrt{13} - 1 + sqrt{13} cdot sqrt{13} - sqrt{13})}{4} = frac{sqrt{13} + 13 - 1 - sqrt{13}}{4} = frac{13 - 1}{4} = frac{12}{4} = 3 ] # Conclusion: [ boxed{3} ]

answer:Solution: (Ⅰ) The polar coordinates of point A are (4 sqrt {2}, frac {pi}{4}), the polar equation of line l is rhocos (theta- frac {pi}{4})=a, therefore a=4 sqrt {2}cos ( frac {pi}{4}- frac {pi}{4})=4 sqrt {2}. therefore The polar equation of line l is rhocos (theta- frac {pi}{4})=4 sqrt {2}, which expands to: frac { sqrt {2}}{2}rho(cos theta+sin theta)=4 sqrt {2}, converting to Cartesian coordinates equation: x+y-8=0. therefore The distance d from points on curve C to line l is d= frac {|2cos theta+ sqrt {3}sin theta-8|}{ sqrt {2}}= frac {| sqrt {7}sin (theta+phi)-8|}{ sqrt {2}}geqslant frac {8- sqrt {7}}{ sqrt {2}}= frac {8 sqrt {2}- sqrt {14}}{2}, equality holds when sin (theta+phi)=1. (Ⅱ) Let the equation of l_{1} be: x+y+m=0, substituting B(1,1) into the above equation yields: m=-2. therefore The equation of line l_{1} is: x+y-2=0. The parametric equation is: begin{cases} x=1- frac { sqrt {2}}{2}t y=1+ frac { sqrt {2}}{2}t end{cases} (t is the parameter), substituting into the general equation of curve C frac {x^{2}}{4}+ frac {y^{2}}{3}=1. This leads to: 7t^{2}+2 sqrt {2}t-10=0, therefore t_{1}+t_{2}=- frac {2 sqrt {2}}{7}, t_{1}⋅t_{2}=- frac {10}{7}, therefore|AB|=|t_{1}-t_{2}|= sqrt {(t_{1}+t_{2})^{2}-4t_{1}t_{2}}= sqrt {(- frac {2 sqrt {2}}{7})^{2}-4×(- frac {10}{7})}= frac {12 sqrt {2}}{7}. Thus, the answers are: (Ⅰ) The minimum distance from points on curve C to line l is boxed{frac {8 sqrt {2}- sqrt {14}}{2}}. (Ⅱ) The value of |AB| is boxed{frac {12 sqrt {2}}{7}}.