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電子課本網(wǎng) 第28頁(yè)

第28頁(yè)

信息發(fā)布者:
解:?$(1)$?由題意得,甲三項(xiàng)成績(jī)之和為:?$9+5+9=23($?分),
乙三項(xiàng)成績(jī)之和為:?$8+9+5=22($?分),
?$∵23>22,$?
∴會(huì)錄用甲.
?$(2)$?由題意得,甲三項(xiàng)成績(jī)之加權(quán)平均數(shù)為:?$9×\frac {120}{360}+5×\frac {360-120-60}{360}+9×\frac {60}{360}$?
?$=3+2.5+1.5$?
?$=7($?分),
乙三項(xiàng)成績(jī)之加權(quán)平均數(shù)為:?$8×\frac {120}{360}+9×\frac {360-120-60}{360}+5×\frac {60}{360}$?
?$=\frac {8}{3}+4.5+\frac {5}{6}$?
?$=8($?分),
?$∵7<8,$?
∴會(huì)改變?$(1)$?的錄用結(jié)果.

證明?$:(1) ∵ $?在?$⊙O$?中?$,\widehat{AD}=\widehat{AD},∠ABD=45°, $?
?$∴ ∠ACD=∠ABD=45°,$?
即?$∠FCE=45°. $?
?$∵ △FCE$?的內(nèi)角和為?$180°,∠CFE=45°, $?
?$∴ ∠CEF=180°-∠CFE-∠FCE=180°-45°-45°=90°, $?
?$∴ l⊥CE$?
?$(2)∵ $?四邊形?$ABCD$?是?$⊙O$?的內(nèi)接四邊形,
?$∴ ∠ABC+∠ADC=180°. $?
?$∵ ∠GDE+∠ADC=180°,$?
?$∴∠ABC=∠GDE. $?
?$∵ AB$?為?$⊙O$?的直徑,
?$∴∠ACB=90°$?
由?$(1),$?知?$∠CEF=90°,$?
即?$∠GED=90°,$?
?$∴ ∠ACB=∠GED.$?
在?$△ABC$?和?$△GDE$?中, 
?$\begin{cases}{∠ACB=∠GED,}\\{∠ABC=∠GDE,}\\{AB=GD,}\end{cases}$?
?$∴△ABC≌△GDE$?