From 7f95a6cc64993bbe4ae202a40085419a87675a3c Mon Sep 17 00:00:00 2001 From: WangWeiye Date: Fri, 14 Apr 2023 15:17:06 +0800 Subject: [PATCH] =?UTF-8?q?=E5=BD=95=E5=85=A5=E9=BB=84=E6=B5=A6=E9=9D=92?= =?UTF-8?q?=E6=B5=A6=E8=99=B9=E5=8F=A32023=E5=B1=8A=E4=BA=8C=E6=A8=A1?= =?UTF-8?q?=E7=AD=94=E6=A1=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 工具/文本文件/metadata.txt | 208 ++++++++++++++++++++++++++++++------- 题库0.3/Problems.json | 124 +++++++++++----------- 2 files changed, 232 insertions(+), 100 deletions(-) diff --git a/工具/文本文件/metadata.txt b/工具/文本文件/metadata.txt index 0f6d670e..93bed4cc 100644 --- a/工具/文本文件/metadata.txt +++ b/工具/文本文件/metadata.txt @@ -1,64 +1,196 @@ ans -15038 -$0$ +15059 +$\{3,5\}$ -15039 -$(1,2)$ +15060 +$\pi$ -15040 -$y=\pm \dfrac 43 x$ +15061 +$81$ -15041 -$\dfrac\pi 4$ +15062 +$-5$ -15042 -$10$ +15063 +$(x-1)^2+y^2=4$ -15043 -$3-\mathrm{i}$ +15064 +$40$ -15044 -$-19$ +15065 +$-2$ -15045 -$\dfrac 12$ +15066 +$(300+100\sqrt{2})\pi$ -15046 -$8$ +15067 +$\dfrac{2\pi}{3}$ -15047 -$(0,\dfrac 56]$ +15068 +$\dfrac{11}{32}$ -15048 -$(2-\sqrt{2},3)$ +15069 +$4$ -15049 -$\sqrt{29}-2$ +15070 +$[-2,2]$ -15050 +15071 +B + +15072 +A + +15073 D -15051 +15074 C -15052 +15075 +(1) $\dfrac{16}{65}$; (2) 周长为$32$, 面积为$24$ + +15076 +(1) 证明略; (2) 距离为$\dfrac{\sqrt{2}}2$, 所成角为$\arcsin\dfrac{\sqrt{10}}{10}$ + +15077 +(1) \begin{tabular}{|c|c|c|c|} \hline & 生产标兵 & 非生产标兵 & 总计\\\hline +$35$周岁及以上组 & $20$ & $60$ & $80$\\\hline +$35$周岁以下组 & $30$ & $50$ & $80$ \\\hline +总计& $50$ & $110$ & $160$ \\ \hline +\end{tabular}, $\chi^2\approx 2.91$, 因此没有$95\%$的把握认为是否为生产标兵与工人所在的年龄有关; (2) 估计该厂工人中$35$周岁以下占$40\%$, 该厂生产标兵中$35$周岁以下占$50\%$ + +15078 +(1) $y=\pm 2\sqrt{2}x$; (2) 最大值为$\dfrac 14$, 此时$\angle AF_1B$的正切值为$-\dfrac{24}{7}$; (3) 证明略 + +15079 +(1) $h_1(x)$是, $h_2(x)$不是; (2) $y=ax^2+(4-2a)x+a$($0