answer:The minimum number of people who like both tea and coffee is found when the number of people liking only tea is maximized. There are 150 - 100 = 50 people who do not like coffee. Assuming all these people like tea, then the number of people who like both tea and coffee would be 120 - 50 = boxed{70}.

answer:1. **Determine the relative speed difference between Chris and Sam:** - Chris's speed: ( 24 , text{km/h} ) - Sam's speed: ( 16 , text{km/h} ) - Relative speed (Chris gains on Sam): [ 24 , text{km/h} - 16 , text{km/h} = 8 , text{km/h} ] 2. **Calculate the initial distance gap between Sam and Chris:** - Initial distance: ( 1 , text{km} ) 3. **Calculate the time (in hours) it takes for Chris to catch up to Sam using the relative speed:** - Formula: [ text{Time} = frac{text{Distance}}{text{Speed}} ] Plugging in the values: [ text{Time} = frac{1 , text{km}}{8 , text{km/h}} = frac{1}{8} , text{hours} ] 4. **Convert the time from hours to minutes:** [ frac{1}{8} , text{hours} times 60 , text{minutes/hour} = frac{60}{8} , text{minutes} ] 5. **Simplify the fraction:** [ frac{60}{8} = frac{15}{2} , text{minutes} = 7 frac{1}{2} , text{minutes} ] # Conclusion: [ boxed{7 frac{1}{2} , text{minutes}} ]

answer:1. The given elipse in the problem is (Gamma: x^2 + frac{y^2}{4} = 1) constrained by (0 leq x leq 1) and (0 leq y leq 2). We denote a segment of ellipse located in the first quadrant by (overparen{AB}). 2. Point (P(x, y)) lies on the arc (overparen{AB}). We need to find the function (f(P) = t(P) cdot overrightarrow{OP}), where (t(P)) represents the unit tangent vector at point (P) on the ellipse (Gamma), with the direction specified as counter-clockwise (clockwise). 3. Using the general tangent vector at any point ((x, y)) on the ellipse: [ text{For the ellipse } x^2 + frac{y^2}{4} = 1, quad text{the ellipse is in a general form:} quad frac{x^2}{a^2} + frac{y^2}{b^2} = 1 quad text{with} quad a = 1 text{ and } b = 2. ] Thus, points on the ellipse comply with ( overrightarrow{OP} = (x, y) ) such that ( t(P) ) is essentially the direction of the tangent line as described. 4. To find ( f(P) = t(P) cdot overrightarrow{OP} ): The unit tangent vector ( t(P) ) at ( P(x, y)) is orthogonal to the normal vector. The gradient vector of the ellipse is given by: [ nabla left( x^2 + frac{y^2}{4} right) = (2x, frac{y}{2}). ] Therefore, the normal vector ( N_P = (2x, frac{y}{2}) ). 5. The unit tangent vector ( t(P) ) can be found via rotating the normal vector by (90^circ): [ t(P) = left( frac{y}{2}, -2x right). ] 6. To normalize ( t(P) ), calculate its magnitude: [ left| t(P) right| = sqrt{ left( frac{y}{2} right)^2 + ( -2x )^2 } = sqrt{ frac{y^2}{4} + 4x^2 }. ] Therefore, the unit tangent vector: [ t(P) = left( frac{frac{y}{2}}{sqrt{ 4x^2 + frac{y^2}{4} }}, frac{-2x}{sqrt{ 4x^2 + frac{y^2}{4} }} right). ] 7. Now, finding ( f(P) = t(P) cdot (x, y) ): [ f(P) = left( frac{y}{2 sqrt{ 4x^2 + frac{y^2}{4}} }, frac{-2x}{sqrt{4x^2 + frac{y^2}{4}}} right) cdot (x, y) = frac{ xy }{ 2 sqrt{ 4x^2 + frac{y^2}{4} } } - frac{ 2xy }{ sqrt{ 4x^2 + frac{y^2}{4} } } = frac{ xy - 4xy }{ 2 sqrt{ 4x^2 + frac{y^2}{4} } } = frac{-3xy}{2 sqrt{ 4x^2 + frac{y^2}{4} } } = -frac{3xy}{2sqrt{ 4x^2 + frac{y^2}{4} } }. ] 8. The expression ( f(P) = f(x, y) ) is thus: [ f(x) = -frac{3xy}{2sqrt{ 4x^2 + y^2/4 } }. ] To find the maximum value of (f(P)), we need to identify its constraints. 9. As noted earlier (referencing max functions): [ max left{ x, y right} = frac{ left| x - y right| + x + y }{ 2 }. ] 10. Setting ( x = frac{ left| b - a right| }{ left| ab right| } + frac{ b + a }{ ab }, y = frac{2}{c} ): [ f(a, b, c) = 2 max left{ frac{ left| b - a right| }{ left| ab right| } + frac{ b + a }{ ab }, frac{2}{c} right} ] Simplifying this expression, we find: [ frac{left| b - a right|}{ left| ab right| } + frac{ b + a }{ ab } = 2 max left{ frac{1}{a}, frac{1}{b} right} Rightarrow f(a, b, c) = 4 max left{ frac{1}{a}, frac{1}{b}, frac{1}{c} right} ] Hence, we obtain: [ frac{ max left{ frac{1}{a}, frac{1}{b}, frac{1}{c} right} }{ f(a, b, c) } = frac{1}{4}. ] # Conclusion: [ boxed{frac{1}{4}} ]

answer:To find the area of the shaded figure described by a semicircle rotated about one end by an angle alpha = 20^circ, we proceed as follows: 1. **Identify the area of the semicircle:** The area of the semicircle S_0 with radius R is given by: [ S_0 = frac{pi R^2}{2} ] 2. **Identify the areas of individual segments and sectors:** - Let the area of the lune AB_1C be x. - Let the area of the lune CB_1B be y. - Let the area of the sector ACB be a. - Let the area of the segment AC be b. 3. **Establish relationships among the areas:** Given: [ x + b = S_0 ] and, [ b + a = S_0 - text{(since AC and S_0 are parts of the same figure)} ] This implies: [ x = a ] 4. **Calculate the total shaded area:** The total shaded area is: [ x + y = a + y ] Since a + y is the area of the composite sector ABB_1 with radius 2R, we need to calculate this: 5. **Calculate the area of sector ABB_1:** The radius of the sector ABB_1 is 2R, and the angle subtended at the center is alpha = 20^circ, which in radians is: [ alpha = frac{20 pi}{180} = frac{pi}{9} ] The area of a sector of a circle is given by: [ text{Area of sector} = frac{1}{2} cdot (text{radius})^2 cdot alpha ] Substituting the values: [ text{Area of sector } ABB_1 = frac{1}{2} cdot (2R)^2 cdot frac{pi}{9} = frac{1}{2} cdot 4R^2 cdot frac{pi}{9} = frac{2 pi R^2}{9} ] 6. **Conclusion:** The area of the shaded figure is: [ boxed{frac{2 pi R^2}{9}} ]