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数学参考答案第1页（共4页）2024届南宁市初中毕业班适应性测试数学参考答案一、选择题（本大题共12小题，每小题3分，共36分）题号123456789101112答案ABACABDACBDC二、填空题（本大题共6小题，每小题2分，共12分）13．x(x-5

)；14．x≥3；15．31；16．（0，6）；17．25；18．2nm．三、解答题（本大题共72分）19．（本题满分6分）解：原式＝）（19+2，·····························3分＝-9+2，··································

···4分＝-7．········································6分20．（本题满分6分）解：原式＝22222babbaba，·············2分＝aba42

．································4分当a＝2，b＝-41时，原式＝）（412422，·····················5分＝4-2＝2．···········

·······························6分21．（本题满分10分）（1）如图所示，CE即为所求；·····················5分（2）证明：∵在Rt△ABC中，∠ACB=90°，∠A=40°

，∴∠B=90°-40°=50°．··················6分又点D为AB中点，∴CD=BD．·······························7分∴∠B=∠DCB=50°．·············

····8分∵CE平分∠BCD，∴∠BCE=21∠BCD=25°．···········9分∴∠AEC=∠B+∠BCE=75°．········10分（第21题图）{#{QQABAYCUggAoAIJAARhCQQ2QCAMQkBEACAoGhF

AEIAAByAFABAA=}#}数学参考答案第2页（共4页）22．（本题满分10分）（1）m＝40，a＝12，n%＝40%；·················3分（2）他的说法是错误的．··························

··4分理由如下：∵在参加测试的40名学生测试成绩中，排在最中间的两个分数都是85，∴中位数为8528585．·····················6分∵84＜85，∴有一半以上的同学成绩超过了84分．···7分∴小邕的说法是错误的．（3）解：500×(40%+30%)＝350（人）·

···········9分答：估计本年级中食品安全意识良好的学生人数为350人．····················10分23．（本题满分10分）（1）证明：∵AB∥DE，∴∠B＝∠E．······································1分又BF

＝CE，∴BF+FC＝CE+FC，即BC=EF．·············2分在△ABC和△DEF中ABDEBEBCEF，·····································

3分∴△ABC≌△DEF（SAS）．···················4分（2）解：连接AD交FC于点O，∵△ABC≌△DEF，∴AC＝DF，∠ACB＝∠DFE．················5分∴AC∥DF．∴四边形ACDF是平行四边形．又AF＝FD，∴四边形ACDF是菱

形．·······················6分∴AD⊥CF，OF＝OC＝12FC＝2．在Rt△ACO中∴2222(13)23AOACOC，··7分∴AD＝2AO＝6．··································8分∴S四边形ACDF

=11641222FCAD．·····10分（第23题图）{#{QQABAYCUggAoAIJAARhCQQ2QCAMQkBEACAoGhFAEIAAByAFABAA=}#}数学参考答案第3页（共4页）24．（本题满分10分）（1）解：

设团购群1《儒林外史》和《简爱》的单价分别是x元、y元；············1分由题意得：288342201223yxyx，·····3分解得3248yx．···························5分答：团购群1《儒林外史》和《简爱》的单价分别是48

元、32元．············6分（2）解：团购群1：（元））（8407.0153248，·······································7分团购群2：（元）105015701050-（元）930340，····8分∵840＜930，··

·······························9分∴选择团购群1购买更合算．············10分25．（本题满分10分）（1）证明：如图1，连接OC，·····················1分∵OA＝OB，CA＝CB，∴OC⊥AB.···········

·························3分又OC为半径，∴AB是⊙O的切线.························4分（2）解：设半径为R，在Rt△OCB中，∠OCB=90°，BC＝12AB＝

4，222OBOCBC，即22224RR，解得R＝3，···································5分∴OB＝OE+BE＝5.··························6分方法一：如图2，过O作OM

⊥EG于点M，又OA＝OB，AC＝BC，∴∠A＝∠B.又OF＝OE，∴∠F＝∠OEF＝∠BEG.∴△AFG∽△BEG.··························7分∴∠AGF＝∠BGE＝90°，842FGAGAFEGBGBE.∴FG＝4EG，AG＝4GB.∴AB＝4GB+GB

＝8.解得GB=85.··································8分∴222286255EGBEBG.··········································

···9分∴6183355EFFGEGEG.················································10分（第25题图1）（第25题图2）{#{QQABAYCUggAoAIJAARhCQQ2QCAMQkBEACAoGh

FAEIAAByAFABAA=}#}数学参考答案第4页（共4页）方法二：如图3，连接DE，∵∠DOE＝∠AOB，35ODOEOAOB，∴△DOE∽△AOB.·························7分∴35DEOEABOB.∴DE=245.···········

·························8分∵DF是⊙O直径，∴∠DEF＝90°.·····························9分∴22222418655EFDFDE.···

·······································10分26．（本题满分10分）（1）解：由题意可知，抛物线顶点为（2，2），经过B（0，1.2）设抛物线的解析式为222yax····················

···························1分将点B（0，1.2）代入，得21.2022a，解得15a.·············································································2分∴抛物

线的解析式为21225yx.············································3分（2）队伍排列时，最高的队员在正中间其位置的横坐标为2，其余同学按照从高到低的顺序在其两侧对称排列.身高为1.

68与1.73的队员所在位置的横坐标分别为1.5与2.5，···············4分身高为1.60的队员所在位置的横坐标分别为1与3；∵1.8＜2，∴身高最高的队员可以通过.············································

·················5分由题意，把1.5x代入21225yx中，得211.5221.955y.∵1.95＞1.73＞1.68，∴可以通过.·················

·································································6分由题意，把1x代入21225yx中，得211221.85y.∵1.8＞1.60，∴可以通过.····················

······························································7分综上所述，所有队员都可以通过.（3）由题意，把1.60y代入21225yx中，则211.6225x.················

·····················································8分解得1122,22xx＋（舍去）.··································

·················9分∵220.5＞，∴最左边的跳绳队员与离他最近的甩绳队员之间距离的取值范围：221x＜≤.·········································

·····································10分（第25题图3）{#{QQABAYCUggAoAIJAARhCQQ2QCAMQkBEACAoGhFAEIAAByAFABAA=}#}