[tex=6.071x2.643]Wu44PzNIxsIbmFN9e1BkkI8IocMVRf6lx/GlN+W5KqQ=[/tex]. 未知类型：{'options': ['[tex=3.571x0.929]1KQep/cJ4kpzQWTxXtALGw==[/tex]', '[tex=2.929x2.429]KIBk73uUhj8I+thVrHSfzi8N5OTZy+HmZKPcmgkQpGQ=[/tex]', '[tex=5.071x1.143]9beYR14NtFGYAmT+AX1ZeQ==[/tex]', '[tex=4.143x2.429]KIBk73uUhj8I+thVrHSfzkFqAvoyrzB4nbUcg1PtWuE=[/tex]'], 'type': 102}

找出函数[tex=3.571x0.929]1KQep/cJ4kpzQWTxXtALGw==[/tex]在区间[tex=1.857x1.357]J8sjPoB94GBNczqzS8+7VA==[/tex]的 11 个等距点上插值的 10 次多项式，打印出这个多项式的牛顿形式中的系数. 计算并打印这个多项式与[tex=3.571x0.929]1KQep/cJ4kpzQWTxXtALGw==[/tex]之差在区间[tex=1.857x1.357]zBn3HDCI1zB5XPvs1Jo9LQ==[/tex]的 33 个 等距点上的值. 由此能得出什么结论?

下列函数中不能在 x =0 处展开成幂级数的是 未知类型：{'options': ['[tex=10.143x3.357]ACpG7W/lXiEwdW69ASBj8/GXrp5udps+Ihl/bKO6isiKARpQ5kPUAYiP/yVzdE2l1FsYxvi5objNGw2HFBf79z8dVdKTguvJX/3+Y48i1LXcni0bNvPmriZBT1OkskSfhHfXaOm2bY730ciuubW6cg==[/tex]', '[tex=3.571x0.929]1KQep/cJ4kpzQWTxXtALGw==[/tex]', '[tex=3.286x1.357]S0tUTq0apy0tc15+GHWCdg==[/tex]', '[tex=5.0x2.643]TvFQr4opkaLDRfu2McOpct+P0DlJP5MisNdBRLZLqB4=[/tex]'], 'type': 102}

在[tex=4.786x1.357]WafKDm5071vVz9IYJgBhj8LbdrnQF2M50OcMtr5E7Yg=[/tex]内连续、有界且有零点，但没有最大值及最小值的函数是(  ) 未知类型：{'options': ['[tex=2.0x1.0]pCnw3JsRBb35dEjM0AXbDw==[/tex]\xa0[br][/br]', '[tex=2.143x0.786]7sCqPKbP66hcrOPfxoYK+Q==[/tex]', '[tex=3.571x0.929]1KQep/cJ4kpzQWTxXtALGw==[/tex]', '[tex=2.5x2.429]c4f2T/WoXP+taGNisOQut8pw7WXZJUXKEUeprVP1bmk=[/tex]'], 'type': 102}

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

共有4个站进行码分多址通信。4个站的码片序列为：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） 现收到这样的码片序列：（-1 +1 -3 +1 -1 -3 +1 +1），则（ ）发送1。