answer:From the given information, we know that places A and B are 150 km apart. A person drives a car from A to B at a speed of 60 km/h, which takes 2.5 hours. The person then returns to A from B at a speed of 50 km/h, which takes 3 hours. - When 0 leqslant t leqslant 2.5, x = 60t - When 2.5 < t leqslant 3.5, x = 150 - When 3.5 < t leqslant 6.5, x = 150 - 50(t - 3.5) Thus, x= begin{cases} 60t,(0 leqslant t leqslant 2.5) 150,(2.5 leqslant t leqslant 3.5) 150-50(t-3.5),(3.5 < t leqslant 6.5) end{cases} So the answer is boxed{D}. This problem primarily focuses on the analytical expression of a piecewise function. The key to solving this problem is to discuss the analytical expression of the function for each segment separately.

answer:The given function is f(x)= begin{cases} 2x-b,x < 1 2^{x},xgeqslant 1 end{cases}, and we are given that f(f( dfrac {1}{2}))=4. First, let's find f(dfrac{1}{2}). Since dfrac{1}{2} < 1, we use the first part of the function, which gives us f(dfrac{1}{2}) = 2 cdot dfrac{1}{2} - b = 1 - b. Now, we have f(f(dfrac{1}{2})) = f(1 - b) = 4. We have two cases to consider: Case 1: If 1 - b < 1, which simplifies to b > 0, then we use the first part of the function again, giving us 2(1 - b) - b = 4. Solving this equation yields b = -dfrac{2}{3}, but this contradicts our assumption that b > 0, so we discard this solution. Case 2: If 1 - b geqslant 1, which simplifies to b leqslant 0, then we use the second part of the function, giving us 2^{1 - b} = 4. Solving this equation yields b = -1. So, the correct answer is b = boxed{-1}.

answer:1. Given information: - O B is the angle bisector of angle A O C - O E is the angle bisector of angle D O F - angle A O F = 146^{circ} - angle C O D = 42^{circ} 2. To find the measure of angle B O E, we first determine the sum of angles angle A O B and angle E O F. 3. Since O B is the angle bisector of angle A O C, we have: [ angle A O B = frac{angle A O C}{2} ] 4. Similarly, since O E is the angle bisector of angle D O F, we have: [ angle E O F = frac{angle D O F}{2} ] 5. Using the relationships given, we calculate: [ angle A O B + angle E O F = frac{angle A O C}{2} + frac{angle D O F}{2} ] 6. Next, since angle A O C and angle D O F are parts of the total angle angle A O F: [ angle A O C + angle D O F = angle A O F - angle C O D ] 7. Substitute the given angle values: [ angle A O C + angle D O F = 146^{circ} - 42^{circ} = 104^{circ} ] 8. Since O B and O E bisect these angles, we have: [ angle A O B + angle E O F = frac{104^{circ}}{2} = 52^{circ} ] 9. To find angle B O E, we subtract the sum angle A O B + angle E O F from angle A O F: [ angle B O E = angle A O F - (angle A O B + angle E O F) ] 10. Substituting the known values: [ angle B O E = 146^{circ} - 52^{circ} = 94^{circ} ] # Conclusion: [ boxed{94^{circ}} ]

answer:Part (a): 1. **Changing to Polar Coordinates**: - The joint density function of random variables (X) and (Y) is given by ( f(x, y) = g(x^2 + y^2) ). - We transition to polar coordinates where: [ X = R cos theta, quad Y = R sin theta ] where (R) is the radial distance and (theta) is the angle. 2. **Determining the Density Function in Polar Coordinates**: - The joint density in polar coordinates is: [ f(x, y) = f(R cos theta, R sin theta) ] Given that: [ x^2 + y^2 = R^2 ] the joint density becomes: [ f(x, y) = g(R^2) ] 3. **Transformation of Variables**: - The Jacobian of the transformation is ( R ), and the new density in polar coordinates (h(r, theta)) is: [ h(r, theta) = f(r cos theta, r sin theta) left| frac{partial (x, y)}{partial (r, theta)} right| = g(r^2) cdot r ] 4. **Final Joint Density in Polar Coordinates**: - Since we need to ensure non-negativity for (r): [ h(r, theta) = begin{cases} r g(r^2), & text{if } r geq 0, 0, & text{otherwise} end{cases} ] 5. **Independence of (R) and (theta)**: - Rewrite (h(r, theta)) as: [ h(r, theta) = r g(r^2) cdot 1 = p(r) q(theta) ] where (q(theta) = frac{1}{2pi}). - Here, (R) and (theta) are independent, and (theta) has the uniform distribution on ([0, 2pi)). - The independence and uniform distribution establish that (theta) is uniformly distributed on ([0, 2pi)), confirming the results. **Conclusion**: (R) and (theta) are independent, and (theta) is uniformly distributed on ([0, 2pi)). Part (b): 1. **Rotation Representation**: - Representing vector rotation by angle (alpha ): [ begin{pmatrix} X_alpha Y_alpha end{pmatrix} = begin{pmatrix} cos alpha & sin alpha -sin alpha & cos alpha end{pmatrix} begin{pmatrix} X Y end{pmatrix} ] 2. **Sufficiency**: - Given that: [ X_alpha^2 + Y_alpha^2 = X^2 + Y^2 ] the density under rotation by (alpha) remains unchanged. - Using the linear transformation density formula: [ f_zeta (cdot) = |det C| f_xi ( C^{-1} cdot ) ] where (C) is the rotation matrix, ensures density invariance, confirming the sufficiency. 3. **Necessity**: - Considering (alpha) as a uniform random variable independently of ((X, Y)): [ alpha sim U[0, 2pi), quad f(alpha, x, y) = frac{1}{2pi} f(x, y) ] 4. **Transforming the Density**: - Applying the rotation: [ f(x, y) = f(x_alpha, y_alpha) ] results in: [ f(alpha, x, y) = frac{1}{2pi} f(x_alpha, y_alpha) ] 5. **Calculating the Density of the Rotated Vector**: - Integrate over (alpha): [ int_0^{2pi} f(x_alpha, y_alpha) , dalpha = 2pi g(r^2) = 2pi g(x^2 + y^2) ] 6. **Conclusion**: - We conclude that: [ f(x, y) = g(x^2 + y^2) ] Thus, the distribution has rotational invariant density if and only if (f(x, y) = g(x^2 + y^2)). ___ Hence, the required condition is met. boxed{}