[tex=2.286x1.286]1zGb/3i3ElmwBKoVatn0Cw==[/tex]曲线和[tex=1.857x1.286]CAc16G0Xd4lVk++G23HiXg==[/tex]曲线的关系

[tex=2.286x1.286]1zGb/3i3ElmwBKoVatn0Cw==[/tex]曲线与[tex=2.286x1.286]snW3q/lmQgJrLk4mbqpImg==[/tex]曲线的关系

证明一般行业短期边际成本曲线[tex=1.857x1.286]CAc16G0Xd4lVk++G23HiXg==[/tex]与平均总成本曲线[tex=2.214x1.286]J827CAqERC4rxVpm+yGxAg==[/tex]、平均可变成本曲线[tex=2.286x1.286]1zGb/3i3ElmwBKoVatn0Cw==[/tex]相交，且交点为[tex=2.214x1.286]J827CAqERC4rxVpm+yGxAg==[/tex]和[tex=2.286x1.286]1zGb/3i3ElmwBKoVatn0Cw==[/tex]的最低点。

表[tex=1.786x1.143]XHxmx7LhfLulASYY0gvGAA==[/tex]是一张关于短期生产函数 [tex=5.429x1.429]qe8/S0+VBwYXZRYkCFp8a2WXUxpnPQtetAvkc6AO5Vg=[/tex] 的产量表:[img=731x236]17b101808fc9a76.png[/img]根据上表，在一张坐标图上作出[tex=2.286x1.286]snW3q/lmQgJrLk4mbqpImg==[/tex]曲线，在另一张坐标图上作出 [tex=2.286x1.286]1zGb/3i3ElmwBKoVatn0Cw==[/tex]曲线和[tex=1.857x1.286]CAc16G0Xd4lVk++G23HiXg==[/tex]曲线。(提示:为了便于作图与比较，[tex=2.286x1.286]snW3q/lmQgJrLk4mbqpImg==[/tex]曲线图的纵坐标的单位刻度大于[tex=2.286x1.286]1zGb/3i3ElmwBKoVatn0Cw==[/tex]曲线和曲[tex=1.857x1.286]CAc16G0Xd4lVk++G23HiXg==[/tex]线图。)

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。