job+qualification
@奚媚3732：求一篇英文作文,题目是跳槽(JOB - HOPPING) -
师梵19287396421…… My View on Job-Hopping Some people tend to stick to their positions all tile time, as theythink the longer one works in a particular field, the more skillful one will be at it. Some have been teachers all their lives. Some devote their life energy to ...
@奚媚3732：设A和B是n阶实方阵,B是正定矩阵,b是B的伴随矩阵,证明:对于任一正数t,方阵tb+aA是正定矩 -
师梵19287396421…… 【解答】 因为B正定矩阵,所以BT=B,bT=b.(tb+aA)T=tbT+ATaT=tb+aA,所以tb+aA是实对称矩阵 因为B正定矩阵,所以对于任意x≠0,xTbx>0 当x≠0时,xT(tb+aA)x=txTbx+xTaAx=txTbx+(Ax)TAx t>0,txTbx>0,(Ax)TAx>0 所以xT(tb+aA)x=txTbx+(Ax)TAx>0 即tb+aA正定 【评注】 证明矩阵正定,首先判定矩阵是对称矩阵 此题根据正定定义:x≠0时,xTAx>0,则A正定.满足正定定义.newmanhero 2015年3月26日20:15:26 希望对你有所帮助,望采纳.
@奚媚3732：已知△ABC的三边分别为abc,满足a=2b,a+b+c=t,求最短边的取值范围 -
师梵19287396421…… 2b+b+3b>a+b+c=t, 所以t/4,bt/, b+c>a或者a+b>c 那么b
@奚媚3732：二次函数y=ax2+bx+1 a≠0 的图象的顶点在第一象限设t=a+b+1 则t 二次函数y=ax2+bx+1(a≠0)的图象的顶点在第一象限,且过点( - 1,0).设t=a+b+1,则t值的变... - 作业帮
师梵19287396421…… [选项] A. 0
@奚媚3732：a high diploma and a good job - ----英语作文 -
师梵19287396421…… A high diploma and a good job 正文:Some people think that one must find a good job as long as one has a high diploma.Maybe you are a very capable person,but you can't find a good job unless you meet a person who know how to make use of ...
@奚媚3732：已知a、b、c属于r a+b+c=0,abc大于0,T=1/a+1/b+1/c则T大于0还是小于0还是=0还是无解? - 作业帮
师梵19287396421…… [答案] 已知等式平方得 a^2+b^2+c^2+2(ab+bc+ca)=0 , 因此 ab+bc+ca= -(a^2+b^2+c^2)/2而 abc>0 , 所以 T=1/a+1/b+1/c=(ab+bc+ca) / (abc)解析看不懂?免费查看同类题视频解析查看解答更多答案(2)
@奚媚3732：如图1所示,物体A放在足够长的木板B上,木板B静止于水平面.已知A的质量mA和B的质量mB均为2.0kg,A、B之 -
师梵19287396421…… (1)根据牛顿第二定律得 对A:μ1mAg=mAaA 解得:aA=μ1g=0.2*10m/s2=2 m/s2 对B:F1-μ2(mA+mB) g-μ1mAg=mBaB 代入数据得 aB=4 m/s2 (2)t1=1s时,A、B分别为vA、vB vA=aA t1=2*1 m/s=2 m/s vB=aB t1=4*1 m/s=4 m/s F1改为F2=4N后,...
@奚媚3732：(lgb+lgt)/lg9=lgb/lg12,lgb/lg12=[lgb+lg(t+1)]/lg16,求t - 作业帮
师梵19287396421…… [答案] 令(lgb+lgt)/lg9=lgb/lg12=[lgb+lg(t+1)]/lg16=k~ 好了,自己做去吧
@奚媚3732：若关于x的不等式|x - 2|+|x - 3|<t,(t∈T)的解集非空.(Ⅰ)求集合T;(Ⅱ)若a,b∈T,求证:ab+1>a+b -
师梵19287396421…… ∴(b-1)(a-1)>0,b>1,即(ab+1)-(a+b)>0(Ⅰ)∵关于x的不等式|x-2|+|x-3|1,即得T={t|t>1 }. (Ⅱ)(ab+1)-(a+b)=(ab-a)+(1-b)=a(b-1)-(b-1)=(b-1)(a-1), ∵a,b∈T,∴a>1
@奚媚3732：Everyone konws his job-------; Everyone konws their job-------.反问疑问句是什么? - 作业帮
师梵19287396421…… [答案] Everyone knows his job,doesn't he? Everyone knows their job,don't they? everyone 作主语,反义疑问句可以用he或they
师梵19287396421…… My View on Job-Hopping Some people tend to stick to their positions all tile time, as theythink the longer one works in a particular field, the more skillful one will be at it. Some have been teachers all their lives. Some devote their life energy to ...
@奚媚3732：设A和B是n阶实方阵,B是正定矩阵,b是B的伴随矩阵,证明:对于任一正数t,方阵tb+aA是正定矩 -
师梵19287396421…… 【解答】 因为B正定矩阵,所以BT=B,bT=b.(tb+aA)T=tbT+ATaT=tb+aA,所以tb+aA是实对称矩阵 因为B正定矩阵,所以对于任意x≠0,xTbx>0 当x≠0时,xT(tb+aA)x=txTbx+xTaAx=txTbx+(Ax)TAx t>0,txTbx>0,(Ax)TAx>0 所以xT(tb+aA)x=txTbx+(Ax)TAx>0 即tb+aA正定 【评注】 证明矩阵正定,首先判定矩阵是对称矩阵 此题根据正定定义:x≠0时,xTAx>0,则A正定.满足正定定义.newmanhero 2015年3月26日20:15:26 希望对你有所帮助,望采纳.
@奚媚3732：已知△ABC的三边分别为abc,满足a=2b,a+b+c=t,求最短边的取值范围 -
师梵19287396421…… 2b+b+3b>a+b+c=t, 所以t/4,bt/, b+c>a或者a+b>c 那么b
@奚媚3732：二次函数y=ax2+bx+1 a≠0 的图象的顶点在第一象限设t=a+b+1 则t 二次函数y=ax2+bx+1(a≠0)的图象的顶点在第一象限,且过点( - 1,0).设t=a+b+1,则t值的变... - 作业帮
师梵19287396421…… [选项] A. 0
@奚媚3732：a high diploma and a good job - ----英语作文 -
师梵19287396421…… A high diploma and a good job 正文:Some people think that one must find a good job as long as one has a high diploma.Maybe you are a very capable person,but you can't find a good job unless you meet a person who know how to make use of ...
@奚媚3732：已知a、b、c属于r a+b+c=0,abc大于0,T=1/a+1/b+1/c则T大于0还是小于0还是=0还是无解? - 作业帮
师梵19287396421…… [答案] 已知等式平方得 a^2+b^2+c^2+2(ab+bc+ca)=0 , 因此 ab+bc+ca= -(a^2+b^2+c^2)/2而 abc>0 , 所以 T=1/a+1/b+1/c=(ab+bc+ca) / (abc)解析看不懂?免费查看同类题视频解析查看解答更多答案(2)
@奚媚3732：如图1所示,物体A放在足够长的木板B上,木板B静止于水平面.已知A的质量mA和B的质量mB均为2.0kg,A、B之 -
师梵19287396421…… (1)根据牛顿第二定律得 对A:μ1mAg=mAaA 解得:aA=μ1g=0.2*10m/s2=2 m/s2 对B:F1-μ2(mA+mB) g-μ1mAg=mBaB 代入数据得 aB=4 m/s2 (2)t1=1s时,A、B分别为vA、vB vA=aA t1=2*1 m/s=2 m/s vB=aB t1=4*1 m/s=4 m/s F1改为F2=4N后,...
@奚媚3732：(lgb+lgt)/lg9=lgb/lg12,lgb/lg12=[lgb+lg(t+1)]/lg16,求t - 作业帮
师梵19287396421…… [答案] 令(lgb+lgt)/lg9=lgb/lg12=[lgb+lg(t+1)]/lg16=k~ 好了,自己做去吧
@奚媚3732：若关于x的不等式|x - 2|+|x - 3|<t,(t∈T)的解集非空.(Ⅰ)求集合T;(Ⅱ)若a,b∈T,求证:ab+1>a+b -
师梵19287396421…… ∴(b-1)(a-1)>0,b>1,即(ab+1)-(a+b)>0(Ⅰ)∵关于x的不等式|x-2|+|x-3|1,即得T={t|t>1 }. (Ⅱ)(ab+1)-(a+b)=(ab-a)+(1-b)=a(b-1)-(b-1)=(b-1)(a-1), ∵a,b∈T,∴a>1
@奚媚3732：Everyone konws his job-------; Everyone konws their job-------.反问疑问句是什么? - 作业帮
师梵19287396421…… [答案] Everyone knows his job,doesn't he? Everyone knows their job,don't they? everyone 作主语,反义疑问句可以用he或they
