人教版九年级数学上册期末考试卷带答案【必考】

右转 直行 (左转,右转) (左转,直行) (右转,右转) (右转,直行) (直行,右转) (直行,直行) (2)(3分)由上表知:两辆汽车都向左转的概率是:

1. 922.(本小题满分8分)

解:解法一,选用①②④,...............................................................................3分

∵AB⊥FC,CD⊥FC, ∴∠ABF=∠DCE=90°,..................................................................................4分 又∵AF∥DE,

∴∠AFB=∠DEC,.........................................................................................5分 ∴△ABF∽△DCE,........................................................................................6分 ∴

ABFB,...............................................................................................7分 ?DCCE又∵DC=1.5m,FB=7.6m,EC=1.7m, ∴AB=6.7m.

即旗杆高度是6.7m.......................................................................................8分 解法二,选①③⑤.............................................................................................3分 过点D作DG⊥AB于点G. ∵AB⊥FC,DC⊥FC,

∴四边形BCDG是矩形,................................................................................4分 ∴CD=BG=1.5m,DG=BC=9m,.....................................................................5分 在直角△AGD中,∠ADG=30°,

AG,................................................................................................6分 DG∴AG=33,.....................................................................................................7分

∴tan30°=又∵AB=AG+GB,

∴AB=33?1.5≈6.7m.

即旗杆高度是6.7m..........................................................................................8分 23.(本小题满分9分)

解:(1)(4分)由题意的点A的坐标是(1,3),....................2分

把A(1,3)代入y=

k, x3;.......................................4分 x得k=1×3=3,.............................................................. ...3分 ∴反比例函数的解析式为y=

(2)(5分)点B在此反比例函数的图象上..

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