meet+one+s+match

浦华熊5093英文歌,其中有一句And I这是什么歌? -
厍史松19341036001 ______ 这首歌: i will always love you , 是Whitney Houston 为电影《保镖》改编翻唱的一首歌曲,原唱是Dolly Parton在1973年发行的一首歌,有点老咯!!不过很好听…… 希望采纳

浦华熊5093英语写作常见句型有什么结构吗? -
厍史松19341036001 ______ 开头: When it comes to ..., some think ... There is a public debate today that ... A is a commen way of ..., but is it a wise one? Recentaly the problem has been brought into focus. 提出观点: Now there is a growing awareness that... It is time we ...

浦华熊5093从键盘输入任一小于6,大于2的正整数N,用VF编程计算:(1)s=1+(1+2)+(1+2+3)+...+(1+2+3+...+N); -
厍史松19341036001 ______ 1.input ＂请输入一个小于6,大于2的正整数:＂ to n if n<=2.or.n>=6 ?＂输入数据不符合要求,请重新运行程序再输入!＂ else s=0 k=0 do whil k<=n kk=1 m=0 do whil kk<=k m=m+kk kk=kk+1 endd s=s+m k=k+1 endd ?＂你...

浦华熊5093反义疑问句 -
厍史松19341036001 ______ 反义疑问句的疑问尾句用助动词,这一点毋庸质疑.你说的那两个例句,是正确的. 若陈述句的谓语动词为“must have + 过去分词”时,若主句强调对过去事情的推测,疑问尾句的谓语动词用“didn't + 主语”结构,若强调动作完成,疑问尾句部...

浦华熊5093用meet one's demand造句 -
厍史松19341036001 ______ i meet his demand.

浦华熊5093在There be句型中加介词造句 -
厍史松19341036001 ______ 1.can you see it?I can see the teapot.but I can't see the tea.此处can't可以换成don't么,为什么? 2.你们谁要这本书,英语怎么翻? 3.is there any tea in the bag.is there any tobacco in the tin.此处可用some代替any么 4.数字什么时候用分隔符,就...

浦华熊5093meet with sales director. sale为什么加s -
厍史松19341036001 ______ meet with sales director可以看做是一个短语,销售主管会议.sales:销售的,形容词,而不是sale+s形式 director:主管,懂事,负责人 sales director是一个形容词+名词的结构

浦华熊5093定义一个求阶乘的函数,然后计算1! - 2!+3! - 4!+5! - 6!+7! - 8!+9!.C语言问题 -
厍史松19341036001 ______ int fact(int n) { int m = 1; while(n>0) { m*=n; n--; } return m; } int main() { int i, sum=0; for(i=1;i<=9;i++) { if(i%2) sum+=fact(i); else sum-=fact(i); } printf(＂1!容-2!+3!-4!+5!-6!+7!-8!+9!=%d\n＂, sum); }

浦华熊5093(用pascal)输入一个实数m,计算并输出满足条件1+1∕2+1∕3+…+1∕n〉m的n值.(满足条件的n的最小值) -
厍史松19341036001 ______ var n:integer;s,m:real; begin s:=0; n:=0; read(m); repeat n:=n+1; s:=s+1.0/n; until (s>m); writeln('n=',n); end.

浦华熊5093用c语言求e的值,e=1+1/1,+1/2,+1/3,+1/n -
厍史松19341036001 ______ #include int main(void) { int i = 0, n = 0; float sum = 0.0; pritnf(＂please input n = ＂); scanf(＂%d＂, &n); for(i = 1; i sum += 1.0 / i; printf(＂result = %f\n＂, sum); reutrn 0; }