cost+a+lot+of+money

宗珊吴2639Cost smokers a lot of money是个句子吗 Educate people on the dangers of smoking是个句子吗 谢谢 -
栾饺季13644576065 ______ 你好,同学,才看到你向我发来的求助题,现在为你解答 Cost smokers a lot of money 这不是一个句子,因为没有主语,cost 一般是表示物的名词或动名词作主语,所以前面少了一个主语,前面最好加上Smoking cost smokers a lot of money. 吸...

宗珊吴2639不规则动词复数形式和不规则过去式 -
栾饺季13644576065 ______ 哥们 动词没有复数 谢谢 (1) AAA型(动词原形、过去式、过去分词同形) cost(花费) cost cost cut(割) cut cut hit(打) hit hit hurt 伤害) hurt hurt let(让) let let put(放) put put read (读) read read (2) AAB型(动词原形与过去式同形...

宗珊吴2639求圆x=a(cost+tsint),y=a(sint - tcost)的渐伸线的弧长 -
栾饺季13644576065 ______ ^x'(t)=a(-sint+sint+tcost)=atcost,y'(t)=a(cost-cost+tsint)=atsint,x'(t)^2+y'(t)^2=a^2t^2,s=∫ [0,2π] √ [x'(t)^2+y'(t)^2]dt=∫ [0,2π] (at)dt=a(t^2)/2[0,2π]=2π^2a.∴圆的渐开线一周弧长为2π^2a.

宗珊吴2639化简1+cos(t)+cos(2t)+...+cos(nt) -
栾饺季13644576065 ______ 解 法一 cost+cos2t+...+cosnt=1/2[(cost+cosnt)+(cos2t+cos(n-1)t)+...+(cosnt+cost)]=[cos(n+1)t/2][cos((n-1)t/2)+cos(((n-3)t/2)+...+cos((n-(2n-1))t/2)=[cos(n+1)t/2/sin(t/2)]*[sin(t/2)*cos((n-1)t/2)+sin(t/2)*cos(((n-3)t/2)+...+sin(t/2)*cos((n-(2n-1))t/2)=1/2[cos...

宗珊吴2639the watch costs a lot.句型转换 the - - - --- - the watch is very - --- -
栾饺季13644576065 ______ 【The price of 】the watch is very 【high.】 祝你学习进步,更上一层楼! (*^__^*) 不明白的再问哟,请及时采纳,多谢!

宗珊吴2639cost a lot的同义词是什么 -
栾饺季13644576065 ______ expensive

宗珊吴2639英语中有几个表示花费的词语分别是什么?用法有什么区别吗? -
栾饺季13644576065 ______ take cost spend pay take 一般主语是物,一般是it.spend主语是人一般后面跟on sth 或者ving sth.pay一般后面跟钱数或者for sth.cost主语也是人. Spend . Expend . Cost . Pay 以上四个词和“消费”、“花费”和“消耗”有关. ●首先,...

宗珊吴2639.Reading such a thick book must take a lot of time. -
栾饺季13644576065 ______ take正确, 意思是花费时间.cost 花费 钱,付出代价 (不能专门用来指花时间) 当然这里并非说cost不对,用cost翻译成: 读这么厚的一本书一定会付出很多时间的代价.但肯定没take好.

宗珊吴2639初二英语问题 -
栾饺季13644576065 ______ 正选的话:it takes somebody sometime to do something这是个固定搭配,当然时态是可以随语境变化的.排除法的话:spend的主语必须是人,所以D排除了.B项的pay的主语也是人,所以排除.A项的cost的主...

宗珊吴2639化简1+cos(t)+cos(2t)+...+cos(nt) -
栾饺季13644576065 ______[答案] 解 法一 cost+cos2t+...+cosnt=1/2[(cost+cosnt)+(cos2t+cos(n-1)t)+...+(cosnt+cost)]=[cos(n+1)t/2][cos((n-1)t/2)+cos(((n-3)t/2)+...+cos((n-(2n-1))t/2)=[cos(n+1)t/2/sin(t/2)]*[sin(t/2)*cos((n-1)t/2)+sin(t/2)*cos(((n-3)t/2)+...+sin(t/2)*cos((n-(2n-1))t/2)=1/2[cos(n+1...