几道初三上学期的数学选择题,希望能帮忙解释一下正确答案,一定要,谢谢了
推荐回答(6个)
1.D(点可能在左也可能在右啊)
2.D(画直线后 已有一角相等 再有一角相等便由角角可相似)
3.C(把k乘过去 让后慢慢做 注意别消去 a+b+c)
4.B(30与50、60分别相对应)
5.A(这个套一下也可以啊 或者设长为一 1/宽=宽/二分之一 的宽为二分之根号二)
6题错了把 没直角啊
1 解:由于C为线段AB=2的黄金分割点,
则AC=2× 二分之根号5-1= 根号5-1或AC=2除以二分之根号5-1=.3减根号5
故选D.
2 直线可能平行AB,可能平行AC,可能过A点,也可能既不过A点也不平行.故选D
3 解:分情况进行:当a+b+c≠0时,根据等比性质,得k= 2c+2a+2bc+a+b=2;
当a+b+c=0时,则a+b=-c,k=-1,故选C.
4 解:取30cm为一边,另两边设为x、y;
(1)30cm与20cm对应,即 x50=y60=3020
解得x=75,y=90;
75+90>50,不可以.
(2)30cm与50cm对应,即 x20=3050=y60
解得x=12,y=36;
12+36=48<50,可以.
(3)30cm与60cm对应,即 x20=y50=3060
解得x=10,y=25;
10+25<50,可以.
所以有两种不同的截法.选B
5解:设报纸的长为2x,报纸的宽为y,又整张报纸与半张报纸相似,则 y2x= xy;即 y2x2=2; yx= 2:1.故选C.
6 由已知,∠AOB=∠COD=90°,0A:OB=1:2,由于对应顶点不确定,那么OC:OD应等于1:2,或2:1.点D的坐标可能为(-6,0),(-1.5,0),(1.5,0),(6,0)那么过CD的直线最多可做4条.选C
我也是初三的 讲过 打了这么多字 评个最佳吧
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