早教吧作业答案频道 -->英语-->
heylisten的全文在网上搜了许多关于Heylisten的全文,但是没有一篇是完全的,我搜到的都是这样的Hey,Listen,Ifyouarenotthatgreatatitnow,dontworryaboutit,justdothebestyoupossiblycan,tomorrowmorningwhen
题目详情
hey listen的全文
在网上搜了许多关于Heylisten的全文,但是没有一篇是完全的,我搜到的都是这样的Hey ,Listen,If you are not that great at it now ,dont worry about it ,just do the best you possibly can ,tomorrow morning when you get out of the bed ,do a little bit better than that ,the next day after that do a lil bit better than that ,(这一段并没有翻译) if you want to compete with someone ,compete with yourself ,be the best you can possibly be everyday .
谁能帮我把中间那一段补全
在网上搜了许多关于Heylisten的全文,但是没有一篇是完全的,我搜到的都是这样的Hey ,Listen,If you are not that great at it now ,dont worry about it ,just do the best you possibly can ,tomorrow morning when you get out of the bed ,do a little bit better than that ,the next day after that do a lil bit better than that ,(这一段并没有翻译) if you want to compete with someone ,compete with yourself ,be the best you can possibly be everyday .
谁能帮我把中间那一段补全
▼优质解答
答案和解析
《Hey listen》
Hey, listen! If you're not that great at it now.
Don't worry about it.
Just do the best you possibly can.
Tomorrow morning, when you get out of bed,
do a little bit better than that.
The next day affer that,
do a little bit better than that.
If you want to compete with someone ,compete with yourself.
Be the best you can possibly be everyday.
《Hey!请听我说》
请听我说.如果你现在还不是最棒的,
不用担心.
只要尽力去做得最好.
第二天早上,当你起床的时候,
再做的比前一天好一点点.
再过一天,
再做的比前一天好一点点.
如果你想要挑战什么人的话,那么就挑战你自己吧!
每一天,都要力争——做最好的自己!
Hey, listen! If you're not that great at it now.
Don't worry about it.
Just do the best you possibly can.
Tomorrow morning, when you get out of bed,
do a little bit better than that.
The next day affer that,
do a little bit better than that.
If you want to compete with someone ,compete with yourself.
Be the best you can possibly be everyday.
《Hey!请听我说》
请听我说.如果你现在还不是最棒的,
不用担心.
只要尽力去做得最好.
第二天早上,当你起床的时候,
再做的比前一天好一点点.
再过一天,
再做的比前一天好一点点.
如果你想要挑战什么人的话,那么就挑战你自己吧!
每一天,都要力争——做最好的自己!
看了 heylisten的全文在网...的网友还看了以下:
求下列参数方程所确定的函数的导数1.x=e∧tsinty=e∧tcost求dy/dx2.x=根号( 2020-05-12 …
matlab求微分方程,常数项比如y=dsolve("Du=((a-u-b)*e-u*d)/(e* 2020-05-14 …
∮1dx/(x^2+y^2+z^2)ds,其中,曲线x=(e^t)sinty==(e^t)cost 2020-06-03 …
老师,我的线代基础比较差,是关于您对一个题目给出的解答的.题:设A是n阶矩阵,A=E+xy^T,x 2020-06-19 …
设A是n阶矩阵,A=E+xy^T,x与y都是n*1矩阵,且x^T*y=2,求A的特征值、特征向量易 2020-06-30 …
令文法G[E]为:E→T|E+T|E-TT→F|T*F|T/FF→(E)|i证明E+T*F是它的一 2020-07-08 …
dy/dx=(e^y)/2=e/2是怎么回事?原题:dy/dx(t=0)=(e^y*cost)/( 2020-07-15 …
y=xe^kx的导数我是这么做的为什么不对设kx=t则有y=xe^tt=kx的导数是ky=xe^t 2020-07-20 …
连续随机变量X,Y相互独立试用期望值的定义来证明E{exp(t(X+Y))}=E{exp(tX)} 2020-07-23 …
由参数方程x=3(e的-t次方)y=2(e的t次方)所确定的函数y=y(x)的12阶导数麻烦求2阶的 2021-02-16 …