2.
如图,在矩形ABCD中,E是AD边的中点,BE⊥AC于点,连接DF,分析下列四个结论:①△AEF∽△CAB;②CF=2AF;③DF=DC;④S四边形CDEF=$\frac{5}{2}$S△ABF其中正确的结论有( )
| A. | 4个 | B. | 3个 | C. | 2个 | D. | 1个 |
1.下列各组数中是同类项的是( )
0 290996 291004 291010 291014 291020 291022 291026 291032 291034 291040 291046 291050 291052 291056 291062 291064 291070 291074 291076 291080 291082 291086 291088 291090 291091 291092 291094 291095 291096 291098 291100 291104 291106 291110 291112 291116 291122 291124 291130 291134 291136 291140 291146 291152 291154 291160 291164 291166 291172 291176 291182 291190 366461
| A. | 7x和7y | B. | 7xy2和7xy | C. | 3xy2和-7x2y | D. | -3xy2和3y2x |