设 [tex=3.571x1.0]1OxXb/HtxwZSBmoxMs+T5CpPiru7ngyhOI3Tn6tJCR4=[/tex] 是任意 3 个数，能否构造一个 3 阶幻方，使得其每行，每列及两条对角线上的元素都是这 3 个数的排列？

从集合 [tex=4.929x1.357]68T2qWuxh0wbv30rW0kB7X2GxxrUsnvrn/P8EN1r/GrXtY+LhfOe6/DbAwBFX9jS[/tex] 中无序选取 4 个元素，求 [tex=3.214x1.0]1OxXb/HtxwZSBmoxMs+T5CpPiru7ngyhOI3Tn6tJCR4=[/tex] 中不能同时选取 2 个但可以同时选取 3 个的方法数

设 [tex=6.214x1.0]1OxXb/HtxwZSBmoxMs+T5JDiA9dAAI97DLuHfJHQ2O6Wocs/w2tJstjHGd8PXJ1T[/tex] 是任意 5 个数，试构造两个不同的 5 阶幻方，使得其每行，每列及两条对角线上的元素都是这 5 个数的排列.

设向量组 [tex=3.571x1.0]1OxXb/HtxwZSBmoxMs+T5CpPiru7ngyhOI3Tn6tJCR4=[/tex] 线性无关，则下列向量组线性无关的是（       ）. 未知类型：{'options': ['[tex=8.5x1.143]528DONpevTuOCn6TMvx84PNZbnfT98Vx3Y/1qsTsTvz5DAdzSebBV2H0h1qQYMln[/tex]', '[tex=10.714x1.214]528DONpevTuOCn6TMvx84G5P02KM3JahBX0napkQjLD7iLYjQbytOZNacz+JnR0e[/tex]', '[tex=10.5x1.214]DsQjWcGa7L6y/HdD1czl/3aLfX7b0mlXCdN0KLJjUsHij7qDlTysAd7/mXDZeU0d[/tex][tex=17.0x1.214]tbQVIM5o4LeGu215YOqZd6H3EEVXcTrfEwzsOMeBdIyx1x6erZV3C2lFAoyI5xlit5Rh5XWBoELdUJkipo8+5AIawiBVamtoJwE/OtOvlq4=[/tex]', '[tex=17.0x1.214]tbQVIM5o4LeGu215YOqZd6H3EEVXcTrfEwzsOMeBdIyx1x6erZV3C2lFAoyI5xlit5Rh5XWBoELdUJkipo8+5AIawiBVamtoJwE/OtOvlq4=[/tex]'], 'type': 102}

下表 是某求极大化线性规划问题计算得到的单纯形表。表中无人工变量，[tex=6.143x1.214]1OxXb/HtxwZSBmoxMs+T5IY9Ogpu9JnzsV95xreV1eaGP5mEU4UsUZsvENxt2Lu1[/tex] 为待定常数。试说明这些常数分别取何值时,以下结论成立。[img=913x256]179454acb1d69fe.png[/img](1)表中解为惟一最优解;(2)表中解为最优解,但存在无穷多最优解;(3)该线性规划问题具有无界解;(4)表中解非最优,为对解改进,换人变量为[tex=0.929x1.0]gli8UB1bTl2bTdA8x5IHiA==[/tex], 换出变量为 [tex=0.929x1.0]6VsMDUb1yWJbYWDftxBBSQ==[/tex] 。

for i in range(b.max_row): for j in range(b.max_column): print(b.cell(row=i,column=j).value)上面语句运行的结果是：__________。 A: 1 1 1 1 1 1 1 1 1 1 B: 1111111111 C: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 D: 出现异常

【单选题】CDMA通信的基站,假定基站A的码片序列是+1 +1 +1 -1 -1 +1 -1 -1,基站发射bit为101时,实际发射的信号是 A. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 B. +1 +1 +1 -1 -1 +1 -1 -1 –1 –1 –1 +1 +1 –1 +1 +1 +1 +1 +1 -1 -1 +1 -1 -1 C. +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 +1 +1 +1 -1 -1 +1 -1 -1 D. –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1 –1 –1 –1 +1 +1 –1 +1 +1

for i in range(1,11): for j in range(1,11): b.cell(row=i,column=j).value=1 #b是一个工作表对象for i in range(1,11): for j in range(1,11): print(b.cell(row=i,column=j).value,end=" ") print()上面程序代码运行的结果是（）。 A: 1 B: 1 1 1 1 1 1 1 1 1 1 C: 1111111111 D: 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1

下列哪个码片序列不能用于CDMA通信 A: ( -1 -1 -1 +1 +1 -1 +1 +1 ) B: ( -1 -1 +1 -1 +1 +1 +1 -1 ) C: ( -1 +1 -1 +1 +1 -1 -1 -1 ) D: ( -1 +1 -1 -1 -1 -1 +1 -1 )

请仔细观察下面行列式的计算过程，如果四个等号都理解了，请选择A，否则请选择B。 | a 1 1 1 1 | |a＋4 a＋4 a＋4 a＋4 a＋4| | 1 a 1 1 1 | | 1 a 1 1 1 | | 1 1 a 1 1 |= | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 a | | 1 1 1 1 1 | | 1 a 1 1 1 | =(a＋4) | 1 1 a 1 1 | | 1 1 1 a 1 | | 1 1 1 1 a | | 1 1 1 1 1 | | 0 a–1 0 0 0 | =(a＋4) | 0 0 a–1 0 0 | | 0 0 0 a–1 0 | | 0 0 0 0 a–1 | =(a＋4)(a–1)^4