{#{QQABLQaUogAAAJJAARgCQQVACgKQkAGCACoOwAAAsAIBiQNABAA=}#}{#{QQABLQaUogAAAJJAARgCQQVACgKQkAGCACoOwAAAsAIBiQNABAA=}#}{#{QQABLQaUogAAAJJAARgCQQVACgKQkAGCACoOwAAAsAIBiQNABAA=}#}{#{QQABLQaUogAAAJJAARgCQQVACgKQkAGCACoOwAAAsAIBiQNABAA=}#}{#{QQABLQaUogAAAJJAARgCQQVACgKQkAGCACoOwAAAsAIBiQNABAA=}#}{#{QQABLQaUogAAAJJAARgCQQVACgKQkAGCACoOwAAAsAIBiQNABAA=}#}2024年三門峽市第一次模擬考試數(shù)學參考答案一、選擇題（每小題3分，共30分）題號答案1D2B3B4C5D6A7A8C9CA二、填空題（每小題3分，共15分）題號-3（k<-114-8答案即可）三、解答題（本大題共8個小題，滿分75分）16.1）原式=4+32?1...................3分=3+32...................5分?2??22）原式=?×???2?2?+2）?+2）（??2）=×...................3分...................5分2?=17.解：（1）15,96;...................6分（2A款自動洗車設備更受歡迎...................7分理由：評分數(shù)據(jù)中，A款的中位數(shù)比款的中位數(shù)高.（答案不唯一）...................9分{#{QQABLQaUogAAAJJAARgCQQVACgKQkAGCACoOwAAAsAIBiQNABAA=}#}18.解：延長PDAC于點F，延長交水平線BE于點G.由題意得：PFAF，DGBE，ABFG53米，AFBG，...............1分設AFBGx在△CDF中，DCF＝30°，CD16，∴＝CD8.............................2分在△PAF中，∠PAF＝45°，∴PFAF＝.............................5分在△BPG中，GBP＝18°，∴＝?tan18°≈0.325x，∴＝PFPGx+0.325x＝1.325x.∴1.325＝........................6分解得：x40.............................8分∴PF40.∴＝PFDF408＝∴該風力發(fā)電機塔桿的高度約為............................9分1212m2.A,2y1x19.解：(1)把m4.A∴把...................................2分kkAy(x0)2，2x4k=8.{#{QQABLQaUogAAAJJAARgCQQVACgKQkAGCACoOwAAAsAIBiQNABAA=}#}8∴反比例函數(shù)的解析式為y2...................................4分x1yx3........................6分將直線向上平移3個單位后，其函數(shù)解析式為21yx328x.聯(lián)立方程組，yx8x212，.11y24C點坐標為2,4...........................................9分20.解：（1）設1A部件的質(zhì)量為x噸，1B部件的質(zhì)量為y由題意得：解得：，，答：個A部件的質(zhì)量為1.2噸，1個部件的質(zhì)量為0.8噸．........................................4分（2）解：設該卡車一次可運輸m套這種設備通過此大橋．根據(jù)題意得：（1.2+0.8×3）m+8≤30，.................................6分解得：m≤．................................7分∵m為整數(shù)，∴m取最大值，∴m6．................................8分答：該卡車一次最多可運輸套這種設備通過此大橋．....................................9分{#{QQABLQaUogAAAJJAARgCQQVACgKQkAGCACoOwAAAsAIBiQNABAA=}#}21.(1)如圖，即為所做的圖形...............................5分(2)(1)作圖知，OAOD.OAD=∠ODA...............................6分平分BAC,OAD=∠CAD..............................7CAD=∠分AC//OD.............................8分ACD=∠ODB=.OD?⊙O的切線．..............................9分22.(1)y=ax-4)+5,將（0，1.8）代入得，a(04)2+5=1.8.1解得a=-.51y=-(x4)2.-.............................5分5小明第一次的測試成績較好..............................6分理由是：1由-x4)2得5x=9或-1.∴小明第一次投擲實心球從起點到落地點的水平距離為9...........................7分{#{QQABLQaUogAAAJJAARgCQQVACgKQkAGCACoOwAAAsAIBiQNABAA=}#}由?0.2??32+3.4=0=3173-17.小明第二次投擲實心球從起點到落地點的水平距離為317.........................9分317∴小明第一次的測試成績較好...........................10分23.1）AE＝②ADBDDF..................................2分..................................3分（2+DF＝AD．理由如下：如圖1，過點作⊥于E，∵C與點關(guān)于AD對稱，∴∠ADC＝ADB，又CDBD，∴∠ADC＝ADB＝45°，又BE⊥，∴△BDE是等腰直角三角形...................................5分又∵△ABC是等腰直角三角形，∴∠ABC∠EBD45°，∴∠ABE＝CBD，∴△ABE∽△CBD，∴.∵＝，∴＝AE..................................7分∵△BDE是等腰直角三角形，{#{QQABLQaUogAAAJJAARgCQQVACgKQkAGCACoOwAAAsAIBiQNABAA=}#}∴＝，∴+＝即：BDDF＝，AD．..................................8分5（3）..................................10分5詳解