2015中考题3答案 (2)

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1、试题答案一、填空题1、;2、4.1×107；3、；4.350；5、1；6、；7、是正数就可以8、9、5和6；10、2或8；11、－2或2或4二、12、A13、D14、C15、C16、C17、D18、D19、B20、C三、21、化简：原式=·························3分当x=2007时，原式=·························2分22、答案：（1）·························3分（2）由························2分时，原方程=－7，舍去当时，原方程，，························1分2

2、3、解：过D作DM⊥AE于M，过C作CN⊥AE于N，则：MN=CD=3米，设AM=x，则AN=x+3，由题意：∠ADM=30ｏ，∠ACN=45ｏ，·························3分在Rt△ADM中，DM=AM·cot30ｏ=x，在Rt△ANC中，CN=AN=x+3，又DM=CN=MB，∴x=x+3，解之得，x=（+1），···························2分∴AB=AM+MB=x+x+3=2×（+1）+3=3+6·······················1分24、（1）频数：18　频数：3，频率：0.075（2）略（3）三（4）10025．（1

3、）甲队在0≤x≤6的时段内：y甲=10x·································2分乙队在2≤x≤6的时段内：y乙=5x+20································2分(2)由10x=5x+20有x=4，开挖4小时后甲队所挖掘河渠的长度开始超过乙队···2分（3）设长度为W米，则，有W=110，甲队从开挖到完工所挖河渠的长度为110米······················································1分26、（1）证明：，，，．又，．　······························

4、·3分．．．　　　······················································2分（2）直线与⊙O相切．　················································1分理由如下：连接．为⊙O的直径，．．．，．．直线与⊙O相切．·····································2分27．（1）y=3x+2(20－x)=x+40·····································3分(2)·····································

5、2分∴12≤x≤14∵x为整数∴x=12或13或14······························1分∴有三种修建方案：A型12个，B型8个；A型13个，B型7个；A型14个，B型6个；·······························1分（3）∵y=x+40，y随x的增大而增大，∴修建A型12个，B型8个时费用最少为52万元；····································2分∵0。05×360+34=52（万元）∴平均每户村民筹集500元钱，能满足所需费用最少的修建方案·································

6、···1分28．（1）由△=－2(n－2)2≥0有n=2，····························2分方程为x2—8x+64=0,故OA=OC=4····························2分(2)证△CDO∽△BED····························2分即得BE=则AE=因此点E的坐标为（4，）···2分(2)存在S的最大值．····································1分由△CDO∽△BED，所以，即，BE=t－t2，×4×(4＋t－t2)．故当t=2时，S有最大值10．··············1分