answer:Since theta is an angle in the second quadrant and tan(theta+ frac {pi}{3})= frac {1}{2} > 0, it follows that theta+ frac {pi}{3} is an angle in the third quadrant. From frac {sin(theta+ frac {pi}{3})}{cos(theta+ frac {pi}{3})} = frac {1}{2}, we know that sin(theta+ frac {pi}{3}) < 0 and cos(theta+ frac {pi}{3}) < 0. Since sin^2(theta+ frac {pi}{3})+cos^2(theta+ frac {pi}{3})=1, we find that sin(theta+ frac {pi}{3})= - frac { sqrt {5}}{5}. Therefore, sintheta+ sqrt {3}costheta=2sin(theta+ frac {pi}{3})= - frac {2 sqrt {5}}{5}. Hence, the answer is boxed{- frac {2 sqrt {5}}{5}}. This solution involves using the fundamental relationship between trigonometric functions of the same angle and applying the formula for the sine of the difference of two angles. This problem primarily tests the understanding of these basic trigonometric relationships and the application of the sine difference formula, making it a foundational question.

answer:1. **Define Variables:** Let d be the distance between city A and city B, p the plane's speed in still air, and w the wind's speed. 2. **Equation for the trip against the wind:** The time for the trip against the wind is given as: [ frac{d}{p-w} = 120 quad Rightarrow quad d = 120(p-w) ] 3. **Equation for the trip with the wind:** The return trip with the wind takes 10 minutes less than it would in still air, so: [ frac{d}{p+w} = t - 10 ] where t is the time in still air for the return trip: [ t = frac{d}{p} ] Substituting t gives: [ frac{d}{p+w} = frac{d}{p} - 10 ] 4. **Substitute d and simplify:** Substitute d = 120(p-w) into the equation: [ frac{120(p-w)}{p+w} = frac{120(p-w)}{p} - 10 ] Clear the fractions and rearrange: [ 120p^2 - 120pw = 120p^2 + 120pw - 10p(p+w) ] [ 10p^2 - 130pw + 120w^2 = 0 ] 5. **Factorize the quadratic equation:** Simplify and factorize: [ p^2 - 13pw + 12w^2 = 0 ] Factoring: [ (p - 12w)(p - w) = 0 ] So, p = 12w or p = w. 6. **Calculate the time for each scenario:** - If p = 12w, then: [ frac{d}{p+w} = frac{120(12w - w)}{12w + w} = frac{120 times 11w}{13w} = 110 ] - If p = w, then: [ frac{d}{p+w} = frac{120(w - w)}{w + w} = frac{120 times 0}{2w} text{ which doesn't make sense (ignore)} ] 7. **Conclusion:** The time for the return journey can be 110 text{ minutes}. The final answer is boxed{**B** (110 minutes)}

answer:We start by aligning the numbers by their decimal points and subtract each corresponding digit: [ begin{array}{@{}c@{}c@{}c@{}c@{}c@{}c@{}c} & 6 & 7 & 8. & 9 & 0 - & 1 & 2 & 3. & 4 & 5 cline{1-6} & 5 & 5 & 5. & 5 & 5 end{array} ] Subtract each corresponding digit: - 0 - 5 = 5 (rightmost digit) - 9 - 4 = 5 - 8 - 3 = 5 - 7 - 2 = 5 - 6 - 1 = 5 (leftmost digit) Thus, the result is boxed{555.55}.

answer:1. **Assume Contra-positive**: We start by assuming that there does not exist an isosceles right triangle with vertices of the same color. 2. **Translate the Problem to Coloring Cells**: For convenience, instead of considering the coloring of the vertices, we consider the coloring of the cells on an infinite grid. 3. **Divide the Plane into 4 times 4 Squares**: - Consider the infinite grid and divide it into 4 times 4 squares. - Each 4 times 4 square contains 16 cells. - On the diagonal of each 4 times 4 square, there will be two cells that could potentially share the same color. 4. **Counting Unique Colorings**: - Let n be a number larger than the number of different ways to color a 4 times 4 square using three colors. - Next, consider a larger square consisting of n^2 such 4 times 4 squares. 5. **Application of the Pigeonhole Principle**: - On the diagonal of this larger square (which has sides of length 4n), there must be at least two 4 times 4 squares that are colored the same way. - This is due to the Pigeonhole Principle, since there are more squares (n^2) than possible colorings (less than n). 6. **Identifying Uniformly Colored Cells**: - Examine these two identically colored 4 times 4 squares. - By the structure, their cells on the diagonal lines must have two cells of the same color, which means finding four cells of the same color in the combined 4 times 4 squares. 7. **Formation of the Triangle**: - These four cells of the same color across the large square (with side 4n) guarantee that within this configuration we can find an isosceles right triangle with its vertices of the same color. 8. **Conclusion**: - The assumption that there does not exist an isosceles right triangle with vertices of the same color leads to a contradiction. Therefore, by contradiction, there must exist at least one isosceles right triangle with vertices of the same color in any three-colored infinite grid. Hence, the proof is complete, and it demonstrates that in any three-colored infinite grid, there is always an isosceles right triangle with vertices of the same color. blacksquare