give+sb+a+beating

终饰封1421动词不定式的应用有哪方面啊
家柴裕13598133639 ______ 除谓语不可以外其他成分均可 主语 To learn English well is not easy.(It's not easy to learn English well.) 表语 Your task is to clean the window. 宾语补足语 The police told the boys not to play football in the street. The teacher made him copy fifty times...

终饰封1421初一英语新目标第四单元总结 -
家柴裕13598133639 ______ 单词书上有,不说了 短语: shop assistant 店员 bank clerk 银行职员 TV station 电视台 work with和.....一起工作 talk to sb 和某人交谈 give sb sth/give sth to sb给某人某物 police station 警察局 school play 校园剧 go out出去 in the day在白...

终饰封1421逻辑函数Y=AB'+A'B+BC'+B'C化为最简与或形式 -
家柴裕13598133639 ______ 逻辑函数Y=AB'+A'B+BC'+B'C化为最简与或形式 Y=AB'+A'B+BC'+B'C= AB'+A'B(c+c')+BC'+(A+A')B'C = AB'+A'BC+A'BC'+BC'+AB'C +A'B'C =(AB'+AB'C)+(A'BC'+BC') +(A'B'C+A'BC) =AB'+BC'+A'C

终饰封1421设绝对值小于1的全体实数的集合为S,在S中定义一种运算*,使得a*b=(a+b)/(1+ab),求证:如果a与b属于S,那么a*b也属于S -
家柴裕13598133639 ______[答案] 依题意,S={x|-1如果a与b∈S,那么-1要证a*b∈S,就是要证明-1(a*b+1)(a*b-1) =[ (a+b)/(1+ab)+1] [ (a+b)/(1+ab)-1] =[a+b-1-ab] [a+b+1+ab]/(1+ab)² =[(a-1)(1-b)] [(a+1)( b+1)] /(1+ab)² 因为-10,a+1>0,b+1>0,所以上式小于0,故原命题得证

终饰封1421英语,急!tired,tiring,tiresome怎么用?
家柴裕13598133639 ______ tired和tiring的?^?e跟frightened和frightening是一?拥摹G罢咧饕ي亲餍稳菰~表示人的状?B(很疲倦),而后者作形容词是表示什麽?|西或人令人感到疲倦,累人. 而tiresome?boring,annoying的意思用法差不多.

终饰封1421用这些造一个句:be weak in/at ,give sb some adoice on sth/doing ,get +宾+形 ,feel like doing -
家柴裕13598133639 ______ 1 She is good at maths but weak in / at Chinese2 Last week he gave me some advice on how to drive a car./ some advice on car driving.3 Let's get our work done as early as possible4 I am tired, so I feel like going to bed now.

终饰封1421公式法化简逻辑函数Y=AB'+B+A' B,Y=AB+AC'D'+A' +B'+C+D -
家柴裕13598133639 ______ Y=AB'+B+A' B=AB'+B=A+B Y=AB+AC'D'+A' +B'+C+D=AB+AC'D'+(AB)'+C+D=AB+(AB)'+AC'D'+C+D=1+AC'D'+C+D=1

终饰封1421在△ABC中,已知a,b,c分别为内角A,B,C所对的边,S为△ABC的面积.若向量p=(S,a+b+c),q=(a+b−c,1)满足p∥q,则tanC2=( )A.14B.12C.2D.4 -
家柴裕13598133639 ______[答案] ∵向量 p=(S,a+b+c), q=(a+b−c,1), 由 p∥ q,得S=(a+b)2-c2=2ab+a2+b2-c2, 即 1 2absinC=2ab+2abcosC,也就是 1 4sinC=1+cosC, ∴ sinC 1+cosC=4. 则tan C 2= sinC 1+cosC=4. 故选:D.

终饰封1421已知三角形的三边为a、b、c,设p=1/2(a+b+c),求证三角形的面积S=√p(p - a)(p - b)(p - c) -
家柴裕13598133639 ______[答案] 证明:若ΔABC的三边长为a、b、c,则SΔABC=√((a+b+c)*(-a+b+c)*(a-b+c)*(a+b-c))/4(这是海伦公式的变形,“负号“-”从a左则向右经过a、b、c”,负号从x轴负轴向正轴扫描一个周期!我觉得这么...

终饰封1421是否存在整数a,b,c使二次方程 ax^2+bx+c=0 和 (a+1)x^2+(b+1)x+c+1=0都有两个整数根? -
家柴裕13598133639 ______[答案] 不存在 反证法 假如存在这样的a,b,c 显然a|b,a|c 设b=ma,c=na (m,n均为整数) 又(a+1)|(b+1)=>(a+1)|(ma+1)=>(a+1)|(m-1) 设m=r(a+1)+1 同理 n=s(a+1)+1(r,s均为整数) =>b=(r(a+1)+1)a c=(s(a+1)+1)a =>ax^2+bx+c=0ax^2+a(r(a+1)+1)x+a(s(a+1)...