answer:Let's denote the speed of the current as ( c ) km/hr. When the man is rowing downstream, the current assists him, so his effective speed is ( (4.2 + c) ) km/hr. When he is rowing upstream, the current is against him, so his effective speed is ( (4.2 - c) ) km/hr. According to the problem, it takes him twice as long to row upstream as to row downstream. This means that the distance he covers upstream in a certain amount of time is half the distance he covers downstream in the same amount of time. Let's denote the time it takes to row downstream as ( t ) hours. Then the time it takes to row upstream is ( 2t ) hours. The distance covered downstream in time ( t ) is ( (4.2 + c) cdot t ) km, and the distance covered upstream in time ( 2t ) is ( (4.2 - c) cdot 2t ) km. Since the distances are the same, we can set these two expressions equal to each other: [ (4.2 + c) cdot t = (4.2 - c) cdot 2t ] Now we can solve for ( c ): [ 4.2t + ct = 8.4t - 2ct ] Combine like terms: [ 3ct = 4.2t ] Divide both sides by ( 3t ) to solve for ( c ): [ c = frac{4.2t}{3t} ] [ c = frac{4.2}{3} ] [ c = 1.4 ] So the rate of the current is ( boxed{1.4} ) km/hr.

answer:Suppose we have a pentagon inscribed around a circle, with angles at each vertex denoted as (A, B, C, D, ) and ( E). 1. **Initial Setup and Property Usage**: Considering the property of a cyclic polygon, the angle bisectors of a polygon inscribed in a circle meet at the circle's center. In this case, the line segments connecting each vertex to the center of the inscribed circle would be the angle bisectors of each interior angle. Let (O) be the center of the inscribed circle. 2. **Triangles and Symmetry**: Draw the angle bisectors from each vertex to the center (O): - (OA) - (OB) - (OC) - (OD) - (OE) These bisectors divide the pentagon into five isosceles triangles (OAB, OBC, OCD, ODE, OEA). 3. **Triangles Properties**: In each isosceles triangle, we have two equal angles. For instance, in ( triangle OAB ): [ text{Angle at } A = text{Angle at } B ] 4. **Sum of Interior Angles in a Pentagn**: The sum of the interior angles of a pentagon is given by: [ (n-2) times 180^{circ}, text{ where } n = 5 ] Therefore, [ (5-2) times 180^{circ} = 3 times 180^{circ} = 540^{circ} ] 5. **Equal Angles**: Since all the interior angles are equal due to the symmetry and properties of the angles bisectors meeting at the center, [ 5 times text{Angle at each vertex} = 540^{circ} ] Hence, each angle at each vertex is: [ text{Angle at each vertex} = frac{540^{circ}}{5} = 108^{circ} ] 6. **Conclusion**: Each angle of the pentagon is (108^{circ}). (boxed{108^{circ}})

answer:1. **Identify Key Points and Lines:** - Given an isosceles triangle ( triangle ABC ) with ( AB ) as the base. - ( M ) is a point on ( BC ). - ( O ) is the center of the circumscribed circle (circumcenter). - ( S ) is the center of the inscribed circle (incenter). - ( SM parallel AC ). - We need to prove that ( OM perp BS ). 2. **Construct Additional Points and Lines:** - Let ( D ) be the point where ( CS ) intersects ( AB ). - Let ( OM ) intersect ( BS ) at ( F ). 3. **Consider Circumcircles:** - Let ( omega_1 ) be the circumcircle of ( triangle BSD ). - Let ( omega_2 ) be the circumcircle of ( triangle SOF ). - These circles intersect again at point ( E ). 4. **Spiral Similarity:** - Consider the spiral similarity ( mathcal{S} ) centered at ( E ) that maps ( omega_1 ) to ( omega_2 ). - Since ( B, S, F ) are collinear and ( CA = CB ), the points ( O, S, D ) are also collinear. 5. **Mapping Points:** - The spiral similarity ( mathcal{S} ) maps ( D ) to ( F ) and ( B ) to ( O ). 6. **Angle Analysis:** - Since ( mathcal{S} ) maps ( D ) to ( F ) and ( B ) to ( O ), we have: [ angle OFS = angle SDB ] - Given that ( angle SDB = 90^circ ) (since ( S ) is the incenter and ( D ) lies on ( AB )), it follows that: [ angle OFS = 90^circ ] 7. **Conclusion:** - Since ( angle OFS = 90^circ ), it implies that ( OM perp BS ). (blacksquare)

answer:**Analysis** This question examines the operation of derivatives, the zeros of functions, and the application of new concepts. According to the new definition, setting the second derivative of the original function to 0 yields f(x_0)=3x_0, which means point M is on the line y=3x. **Solution** Since f(x)=3x+4sin x-cos x, Then f'(x)=3+4cos x+sin x, And f''(x)=-4sin x+cos x, From f''(x_0)=0, we get -4sin x_0+cos x_0=0, Which simplifies to 4sin x_0+cos x_0=0, Therefore, f(x_0)=3x_0, Hence, M(x_0,f(x_0)) is on the line y=3x. Therefore, the correct choice is boxed{text{A}}.