规范),已知:如图△ABC.求作:△ABC的内角平分线AD.作法:(2)如图2,在直角坐标系中,第一次将△OAB变换成△OA1B1,第二次将△OA1B1变换成△OA2B2,---依此类推.已知A(1,3),A1(2,3),A2(4,3),A3(8,3),…;B
答案:连接直线a上的点和直线b上的点,所得到的线 段称为“正规线段”.分类计数所得三角形个数: 第1类:三角形的三个顶点都在直线a和直线b 上,其中2个顶点在直线a上,第3个顶点在直线b 上,这类三角形有 [1/2*5*(5-1)]*5=50(↑) ;类似,其中2个顶点在直线b上,第3个顶点在直线a 上...
a-Aminocyclopropanecarboxylate a-aminocyclopropylcarboxamide as surrogate 2,3-diaminopyridine appl bradykinin B1 antagonistBock, M G
Identification of SRC3/AIB1 as a Preferred Coactivator for Hormone-activated Androgen Receptor Transcription activation by androgen receptor (AR), which depends on recruitment of coactivators, is required for the initiation and progression of prostat... XE Zhou,KM Suino-Powell,J Li,... - 《...
Suppression of tumor growth via IGFBP3 depletion as a potential treatment in glioma --最爱读文献的成老师组 作者的话: 这一篇文献主要讲述了IGFBP3在GBM中的作用,是一篇比较新的文章,发表于2019年。通过研究这篇文章我们可以发现...
解:1,,且a.b夹角为150° ∵|a|=3 |b|=4 a.b =|a|⋅|b|⋅cos150° =ase =3.4⋅(-(√3)/2)=⋅a⋅b=|a|⋅|b|⋅ =-6√3 ∴(a+b)^2 |a+b| =a^2+b^2+2ab =√(|a|^2+|b|^2+2|a||b|⋅cosθ) =|a|^2+|b|^2+2|a||b|⋅cosθ =9+16...
解:1,,且a.b夹角为150° ∵|a|=3 |b|=4 ∴a⋅b =|a|⋅|b|⋅cos150° =3.4⋅(-(√3)/2)=⋅a⋅b=|a|⋅|b|⋅cos s0 =-6√3 ∴(a+b)^2 |a+b| =a^2+b^2+2ab =√(|a|^2+|b|^2+2|a+1|+|b|) =|a|^2+|b|^2+2|a|⋅|b|⋅cosθ =...
5The abbreviations used are: AFBI, aflatoxin B,; AFB,-2,3-oxide, aflatoxin B,-2,3-oxide; AFB@,aflatoxin B@(2,3-dihydroaflatoxin B,); 2,3-dihydro-2,3- dihydroxy-AFB, (dihydrodiol), 2,3-dihydro-2,3-dihydroxy-aflatoxin B1;AFB, 2,3-dichloride, aflatoxin B,-2,3-dichloride; AFB...
live with friends and often return to living with their parents. Similarly, they fall in and out of love, quit one job and try another or even shift to a new career. So, we need to recognize this new stage, t...
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