dredging+of+an+anchor
作者&投稿：孛苏 （若有异议请与网页底部的电邮联系）
color edging的意思
color edging 英 [ˈkʌlə(r) ˈedʒɪŋ] 美 [ˈkʌlər ˈedʒɪŋ]网络 彩色镶边
求新加坡足协的详细资料
Four years later in 1925, the Malaya Cup was played for the first time in Singapore and the Lion City duly celebrated by edging Selangor 2-1 at the Anson Road Stadium. Over the next o decades, names such as inside forward Chia Keng Hock and full back Abdul Rahman who appeared in nine ...
how did princess diana die?
while being chased by paparazzi. But how did she die the condition of Diana was at first considered serious until the surgeons discovered that her coronary artery was ruptured. The surgeons then changed her condition to grave. Princess Diana died after o hours in surgery at 4:00 a...
石屏县葆利回答： a piece of meat a loaf of bread a cup of tea a glass of water a bowl of rice a bottle of milk
田肃18678693036问： by+the+end+of+和at+the+end+of+的区别及用法？
石屏县葆利回答： 1. ①一般说来,at the end of用于表示具体事物或场所的场合,它也可以用来表示比喻意.例: The school is situated at the end of the street. 该校位于这条街的尽头. They were at the end of their patience. 他们忍无可忍. ②at the end of亦可用于...
田肃18678693036问： 数学—数列前N项和,Sn - (Sn - 1)与S(n+1) - Sn,求{an}有事么不同?为什么求出的{an}结果不同? - ？
石屏县葆利回答：[答案] 你需要注意两点 第一,Sn-S(n-1)=an中,n大于等于2.(当n=1时,n-1=0) 第二,Sn-S(n-1)=an S(n+1)-Sn=a(n+1) 两者的下标不一样,结果当然不一样.
田肃18678693036问： 数列An+1=An+1/An,A1=1,求An的通项 - ？
石屏县葆利回答： 依我看,应该是An+(1/An),此递推式没有通项,因为特征根不存在. A(n+1)=(An+1)/An的递推式也没有通项. A(n+1)=A(n+1)/An的通项就是An=1,无探讨的意义. 总之,原题试着用数学归纳法解吧,实在不懂问老师.求通项的路无法走.
田肃18678693036问： 等差中项和等比中项的问题等差和等比中项的公式可以逆用吗?an=an - 1+an+1 那么an是 等差数列对吗 那 an=an - 2+an+2 的、an是等差数列对吗 an²=an - 1*... - ？
石屏县葆利回答：[答案] 如果2a[n]=a[n-1]+a[n+1],那么a[n]就是等差数列 但是后一个不正确,因为还有a[n]=0的情况.当一数列为常数列0时,由于没有公比而不成等比数列.
田肃18678693036问： 已知数列{an}的各项均为正数,且满足an+1=an+(二倍的根号下an)+1,a1=2.(1)求证:数列{根号下an}是等差数...已知数列{an}的各项均为正数,且满足an+1... - ？
石屏县葆利回答：[答案] a(n+1)=a(n)+2√[a(n)]+1 a(n+1)=[√a(n)+1]² 因为数列{an}都是正项,则: √[a(n+1)]=√[a(n)]+1 即: √[a(n+1)]-√[a(n)]=1=常数 则数列{√[a(n)]}是以√a1=√2为首项、以d=1为公差的等差数列,则: √[a(n)]=√2+(n-1)d √[a(n)]=√2+n-1 两边平方...
田肃18678693036问： 如果数列{an}是等比数列,那么下列数列中不是等比数列的是_____. - ？
石屏县葆利回答：[选项] A. {1/an}. B. {3√an}. C. { an•an+1 } D. {an+an+1} 为什么!
田肃18678693036问： 设数列{an}的前n项和Sn满足:Sn+an=(n - 1)/n(n+1),n1,2……,求an - ？
石屏县葆利回答：[答案] n=1时,S1+a1=2a1=(1-1)/(1*2)=0 a1=0 n≥2时, Sn+an=(n-1)/[n(n+1)]=n/[n(n+1)]-1/[n(n+1)]=1/(n+1)-[1/n -1/(n+1)]=2/(n+1) -1/n S(n-1)+a(n-1)=2/n -1/(n-1) Sn+an-S(n-1)-a(n-1)=2an-a(n-1)=2/(n+1)-1/n-2/n +1/(n-1)=[2/(n+1)-2/n]-[1/n -1/(n-1)] 2an+[2/n-2/...
田肃18678693036问： 数列{an}的通项公式为an=an^2+n,若a1 - ？
石屏县葆利回答：[答案] a(n+1)-an=a*(n+1)^2+n+1-an^2-n=2na+a+1 当n≤4时,2na+a+1>0a>-1/(2n+1)≥-1/9 当n≥8时,2na+a+1因此,-1/9解析看不懂?免费查看同类题视频解析查看解答
田肃18678693036问： 设an=(1/n+1)+(1/n+2)+(1/n+3)+...+1/2n,则an+1 - an等于? - ？
石屏县葆利回答：[答案] An=1/(n+1)+1/(n+2)+…+1/(2n-1)+1/(2n) 则An+1=1/(n+2)+1/(n+3)+…+1/(2n-1)+1/(2n)+ 1/(2n+1)+1/(2n+2)则An+1-An=1/(2n+1)+1/(2n+2)-1/(n+1)=1/(2n+1)-1/(2n+2)=1/[(2n+1)(2n+2)]
