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广东省东莞市石碣新民学校2023——2024学年上学期九年级数学期中教学检测(PDF版含答案)

日期:2024-06-16 科目:数学 类型:初中试卷 查看:60次 大小:1152836Byte 来源:二一课件通
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广东省,数学,答案,版含,PDF,检测
    2023-2024学年第一学期期中教学检测 九年级数学答案 1-10:DDDBA ADCCB 11.k<1 12.1 13. 14.(-1,3) 15.> 16.(1)法1:解:2x2﹣4x﹣1=0 x2﹣2x﹣=0 ..........................................(1分) x2﹣2x+1=+1.......................................(2分) (x﹣1)2= ................................... (3分) ∴x1=1+,x2=1﹣................. (4分) 法2:........................(1分) 1=24>0...........(2分) .................................(3分) x1=1+,x2=1﹣..................................(4分) 解:(x﹣2)(x﹣5)=0..............(2分) ∴x1=2,x2=5. .....................(4分) 解:y=x2﹣6x+4 =x2﹣6x++4.....................(1分) =(x﹣3)2.............................(2分) 开口方向:向上,...........................(4分) 对称轴:..................................(6分) 顶点坐标:..........................(8分) 解:将...................(1分) .........................................................(3分) 解得:..........................................................................(7分) ∴.....................................................................(8分) 19.解:............................... ............(1分) =16+4>0...........(4分) ∴该方程有两个不相等的实数根...............(5分) ====-4.........................................................(7分) ∴2.............................................................................(9分) 20,解:(1)如图; ..........................................................................................................................................(3分) (2)A1(4,﹣3).....................................(5分) B1(1,﹣2)...............................................(7分) C1(2,﹣1)..............................................(9分) 解:(1)设全省5G基站数量的年平均增长率为x...........(1分) 有:..............................................(3分) 解得:x1=0.1,x2=-2.1(舍)...................................(5分) ∴全省5G基站数量的年平均增长率为10%........(6分) (2)................................(7分) ................................................................(8分) ∴能.................................................................................(9分) 22.解:(1)2000............................................................................. (2分) (2)①依题意得:(100﹣80﹣x)(100+10x)=2160...........(4分) 解得:x1=2,x2=8......................................................................(6分) 答:每件商品应降价2元或8元............................................(7分) ②依题意得:y=(100﹣80﹣x)(100+10x),.....................(8分) =﹣10x2+100x+2000......................................(9分) =﹣10(x﹣5)2+2250..................................(11分) ∴当x=5时,商店所获利润最大...........................................(12分) 23解:(1)将A(1,0),C(﹣2,3)代入y=﹣x2+bx+c,得: ,解得:, ∴抛物线的函数关系式为y=﹣x2﹣2x+3;................................................. ... ...

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