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question:Sandy sold half of her comic books and then bought 6 more. She began with 14 comic books. How many comic books does she have now?

answer:Sandy started with 14 comic books. She sold half of them, so she sold 14 / 2 = 7 comic books. After selling 7, she had 14 - 7 = 7 comic books left. Then she bought 6 more, so she now has 7 + 6 = boxed{13} comic books.

question:A gardener plants four maple trees, five oaks, six birch trees, and two pine trees in a row. He plants them in random order, each arrangement being equally likely. Let frac{m}{n} in lowest terms be the probability that no two birch trees are next to one another. Find m+n.

answer:Start by considering the non-birch trees (maple, oak, and pine) as identical for simplicity in calculating the probability related to birch tree arrangement. There are 4+5+2=11 non-birch trees. Now, consider the placement of six birch trees amongst these non-birch trees: - Between any two non-birch trees and at the ends, there are 11+1=12 possible slots for the birch trees. - Choose 6 slots for birch trees from these 12 slots, which can be done in binom{12}{6} ways. The total ways to arrange all 17 trees (4 maple, 5 oaks, 6 birch, 2 pine) without any restrictions are binom{17}{6} ways to place the birch trees and the remaining spots can be filled with the non-birch trees. Calculating the combinations: - Valid birch arrangements: binom{12}{6} = 924 - Total arrangements: binom{17}{6} = 12376 Thus, the probability is: frac{924}{12376} = frac{21}{283} So, the answer is 21 + 283 = boxed{304}.

question:Let x be a positive real number. Find the minimum value of 6x^6 + 8x^{3}.

answer:Using the AM-GM inequality, [ 6x^6 + 8x^{-3} = x^6 + x^6 + x^6 + x^6 + x^6 + x^6 + x^{-3} + x^{-3} + x^{-3} + x^{-3} + x^{-3} + x^{-3} + x^{-3} + x^{-3} ] [ geq 14 sqrt[14]{(x^6)^6 cdot (x^{-3})^8} ] [ = 14 sqrt[14]{x^{36-24}} ] [ = 14 sqrt[14]{x^{12}} ] [ = 14. ] Equality occurs when every term in the sum equals the geometric mean: [ x^6 = x^{-3} ] which implies [ x^9 = 1 implies x = 1. ] Thus, the terms are indeed equal when x = 1, confirming that boxed{14} is the minimum value.

question:Given the sets A={x|2leqslant 2^{x}leqslant 8}, B={x|x > 2}, with the universal set U=mathbb{R}. (1) Find (complement_{U}B)cup A; (2) Given the set C={x|1 < x < a}, if Csubseteq A, find the range of the real number a.

answer:Solution: (1) A={x|2leqslant 2^{x}leqslant 8}={x|1leqslant xleqslant 3}, B={x|x > 2}, (complement_{U}B)cup A={x|xleqslant 3} (2) When aleqslant 1, C=varnothing, in this case Csubseteq A; When a > 1, for Csubseteq A, it must be 1 < aleqslant 3 Combining the above, the range of a is boxed{(-infty,3]}.

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