推荐回答(2个)
英语
一、听力测试题(1-25小题,每小题1分,共25分)
1. A 2. B 3. A 4. B 5. C 6. B 7. A 8. C 9. B 10. B
11. C 12. A 13. B 14. C 15. B 16. B 17. A 18. C 19. A 20. B
21. A 22. A 23. C 24. B 25. B
二、单选(26-40小题,每小题1分,共15分)
26. B 27. B 28. C 29. D 30. C 31. C 32. C 33. A 34. B 35. D
36. C 37. B 38. C 39. B 40. B
三、完形填空(41-55小题,每小题1分,共15分)
41. A 42. C 43. B 44. D 45. D 46. D 47. C 48. C 49. D 50. C
51. B 52. D 53. D 54. A 55. A
四、阅读理解(56-70小题,每小题2分,共30分)
56. C 57. B 58. A 59. C 60. D 61. C 62. A 63. A 64. B 65. A
66. A 67. B 68. B 69. C 70. A
五、词与短语填空 (71-75小题,每小题2分,共10分)
71. imagine 72. got out 73. followed 74. went into 75. amazing
六、阅读理解填词(76-85小题,每小题1分,共10分)
76. proud 77. exactly 78. wait 79. studied 80. truth
81. lazy 82. watched 83. family 84. strange 85. page
数学
参考答案:
题号 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
答案 C B A D D C C B
C B C C ;-9a6 ;3; 60或110
-2<x≤-1 6
17.解:a=1,b=2,c=-2.
b2-4ac=22-4×1×(-2)=4+8=12.
x= . ∴ x= .∴ x1= ,x2= .
18. 解:原式= = =2—x.
当 时,原式= .
19. 证明:∵AB‖DE,AC‖DF,∴∠B=∠DEF,∠ACB=∠F.
在△ABC和△DEF中,
∵∠B=∠DEF, BC=EF, ∠ACB=∠F.
∴△ABC≌△DEF.
∴AB=DE.
20.解:列表如下:
红 黄 白 黑
红 红,红 黄,红 白,红 黑,红
黄 红,黄 黄,黄 白,黄 黑,黄
白 红,白 黄,白 白,白 黑,白
黑 红,黑 黄,黑 白,黑 黑,黑
由表或图可知,共有16种可能的结果,其中小菲两次都能摸到同色球出现4次,
故P(小菲两次都能摸到白球)= =
21.解:(1)3 —6;
(2)答案不唯一,以下提供两种图案.
22.(1)证明:
∵AB是⊙O的直径, ∴∠ACB=90°.
∵CD平分∠ACB,∴∠ACD=∠FCB=45°.
∵AE⊥CD,∴∠CAE=45°=∠FCB.
在△ACE与△BCF中,
∠CAE=∠FCB,∠E=∠B,∴△ACE∽△CFB.
(2)解:延长AE、CB交于点M.
∵∠FCB=45°,∠CHM=90°,
∴∠M=45°=∠CAE.
∴HA=HC=HM,CM=CA=6.
∵CB=4 ,∴BM=2.
∵OA=OB,∴OH= BM=1.
23.解:(1)当50≤ ≤60时, ;
当60< ≤80时, ;
∴ (50≤ ≤60且 为整数)
=
(60< ≤80且 为整数)
(2)当50≤ ≤60时, ;
∵a=-1<0,且x的取值在对称轴的左侧,∴y随x的增大而增大,
∴当x=60时, 有最大值2000;
当60< ≤80时, ;
∵a=-2<0,∴当 =75时, 有最大值2450.
综上所述,每件商品的售价定为75元时,每个月可获得最大利润,最大的月利润是2450元.
(3)当60< ≤80时, .
当y=2250元时, ,解得:
其中,x=85不符合题意,舍去.
∴当每件商品的售价为65元时,每个月的利润恰为2250元.
24.(1)证明:∵AC=BC,∠ACB=90°,
∴∠CAB=∠ABC=45°.
∵∠CAD=∠CBD=15°,
∴∠BAD=∠ABD=30°.
∴ AD=BD.
(2)证明:在DE上截取DM=DC,连接CM.
∵AD=BD,AC=BC,DC=DC,
∴△ACD≌△BCD,
∴∠ACD=∠BCD= 45°.
∵∠CAD=15°,
∴∠EDC=60°.
∵DM=DC,
∴△CMD是等边三角形.
∴∠CDA=∠CME=120°,
∵CE=CA,
∴∠E=∠CAD.
∴△CAD≌△CEM,
∴ME=AD.
∴DA+DC=ME+MD=DE.
即AD+CD=DE.
(3) 延长CD交AB于点H.则CH⊥AB.
∵∠HBD=30°,BD=2,
∴BH=BD?cos30°= .
∴AC=BC=BH÷sin45°= .
25解:(1)∵ M为抛物线 的顶点,
∴M(2,c).∴OH=2,MH=|c|.∵a<0,且抛物线与x轴有交点,∴c>0,∴MH=c.
∵sin∠MOH= ,∴ .∴OM= ,∵ ,∴MH=c=4.∴M(2,4).∴抛物线的函数表达式为: .
(2)如图1,∵OE⊥PH,MF⊥PH,MH⊥OH.
∴∠EHO=∠FMH,∠OEH=∠HFM.∴△OEH∽△HFM.
∴HEMF =HOMH =12 .∵HEHF =12 ,∴MF=HF.
∴∠OHP=∠FHM=45°.∴OP=OH=2,∴P(0,2).
如图2,同理可得,P(0,-2).
(3)∵A(-1,0),∴D(1,0).
∵M(2,4),D(1,0),∴MD: .∵ON‖MH,∴△AON∽△AHM,∴ ,∴AN= ,ON= ,N(0, ).
如图3,若△ANG ∽ △AMD,可得NG‖MD,∴QG: .
如图4,若△ANG ∽ △ADM,可得, .
∴AG= ,∴G( ,0),∴QG: ;
综上所述,符合条件的所有直线QG的解析式为:Y=4X+4/3或y=-8/19X
+4/3
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