早教吧作业答案频道 -->英语-->
下面这段话中的最后一句:Thanisfourtreatmentsasecond,这里的asecondSTEPHENO’BRIEN:"Wewillincreaseourfinancialinvestmentandcumulative['kju:mju,leitiv]spendingfive-fold,fromfiftymillionpounds($79million)to245mi
题目详情
下面这段话中的最后一句:Than is four treatments a second,这里的 a second
STEPHEN O’BRIEN:"We will increase our financial investment and cumulative ['kju:mju,leitiv] spending five-fold,from fifty million pounds ($79 million) to 245 million pounds ($387 million) by twenty-fifteen.That is four treatments a second."
STEPHEN O’BRIEN:"We will increase our financial investment and cumulative ['kju:mju,leitiv] spending five-fold,from fifty million pounds ($79 million) to 245 million pounds ($387 million) by twenty-fifteen.That is four treatments a second."
▼优质解答
答案和解析
That is four treatments a second.这是四项措施中的第二项.
看了 下面这段话中的最后一句:Th...的网友还看了以下:
设A是n阶矩阵A^2=E,证明r(A+E)+r(A-E)=n,的一步证明过程不懂由A^2=E,得A 2020-05-14 …
(“*”为未知数x)e*/a+a/e*=1/ae*+ae*为什么会等于(a-1/a)(1/e*-e 2020-06-07 …
main(){unionEXAMPLE{struct{intx,y;}in;inta,b;}e;e 2020-06-12 …
矩阵平方差设方阵A满足A²-A-2E=O,求A的逆矩阵.答案是1/2(A-E).为啥不是1/2E, 2020-07-18 …
A,B均为三阶可逆矩阵,且A^3=0,则A:E-A,E+A均不可逆?B:E-A不可逆但E+A可逆? 2020-07-20 …
设A为n阶非零矩阵,E为n阶单位矩阵.若A3=0,则()A.E-A不可逆,E+A不可逆B.E-A不 2020-07-22 …
已知向量a≠e,|e|=1,满足:任意t∈R.已知向量a不等于e,|e|=1,对任意t属于R,恒有 2020-07-25 …
已知向量a≠e,|e|=1,对任意t∈R,恒有|a-te|≥|a-e|,则[]A.a⊥eB.a⊥(a 2020-11-02 …
n阶方阵A满足A^2=O,E是n阶单位阵,则A.|E-A|≠0,但|E+A|=0B|E-An阶方阵A 2020-11-02 …
线性代数题目设A为n阶矩阵,(E-A)的行列式不等于零,证明(E+A)(E-A)*=(E-A)*(E 2020-12-23 …