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25．如圖1.正方形和正三角形的邊長(zhǎng)都為1.點(diǎn)分別在線段上滑動(dòng).設(shè)點(diǎn)到的距離為.到的距離為.記為(當(dāng)點(diǎn)分別與重合時(shí).記)． (1)當(dāng)時(shí).求的值, (2)當(dāng)為何值時(shí).點(diǎn)落在對(duì)角形上?請(qǐng)說(shuō)出你的理由.并求出此時(shí)的值, (3)請(qǐng)你補(bǔ)充完成下表: 0.03 0 0.29 0.29 0.13 0.03 (4)若將“點(diǎn)分別在線段上滑動(dòng) 改為“點(diǎn)分別在正方形邊上滑動(dòng) ．當(dāng)滑動(dòng)一周時(shí).請(qǐng)使用(3)的結(jié)果.在圖4中描出部分點(diǎn)后.勾畫出點(diǎn)運(yùn)動(dòng)所形成的大致圖形． (參考數(shù)據(jù):．) 25．解:(1)過(guò)作于交于.于． .. .．······················································································ 2分 .．························································································· 3分 (2)當(dāng)時(shí).點(diǎn)在對(duì)角線上.其理由是:·········································· 4分過(guò)作交于. 過(guò)作交于．平分..． ..． .． .．即時(shí).點(diǎn)落在對(duì)角線上．································································ 6分 (以下給出兩種求的解法) 方法一:.．在中.. ．·········································································· 7分．························································································ 8分方法二:當(dāng)點(diǎn)在對(duì)角線上時(shí).有 .··························································································· 7分解得．························································································ 8分 (3) 0.13 0.03 0 0.03 0.13 0.29 0.50 0.50 0.29 0.13 0.03 0 0.03 0.13 ································································ 10分 (4)由點(diǎn)所得到的大致圖形如圖所示: ··························································································· 12分說(shuō)明:1．第(2)問(wèn)回答正確的得1分.證明正確的得2分.求出的值各得1分, 【查看更多】

正在江西南昌舉行的全國(guó)第七屆城市運(yùn)動(dòng)會(huì)共有10350人參加決賽階段的比賽，10350用科學(xué)記數(shù)法表示應(yīng)為（　�。�

A．10.35×10³

B．1.035×10⁴

C．0.1035×10⁵

D．1035×10

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（2011江西南昌，4，3分）下列運(yùn)算正確的是（）.

A.a+b=ab B.a2·a3=a5 C.a2+2ab－b2=(a－b)2 D.3a－2a=1

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（2011江西南昌，12，3分）方程組的解是 .