早教吧作业答案频道 -->英语-->
改英语语法和句子顺序Changingfromdirectspeechtofreedirectspeechindicatesthegovernessisnotareasonableperson.Sheimmersesinherfantasyfromtimetotime.Fromthedirectspeechtothefreedirectspeech,theatmosphereofmenacek
题目详情
改英语语法和句子顺序
Changing from direct speech to free direct speech indicates the governess
is not a reasonable person.She immerses in her fantasy from time to time.From
the direct speech to the free direct speech,the atmosphere of menace keeps
increasing.Passages in direct speech
can only deliver the fact.Readers can feel the menace
in an indirect way.Using the free direct speech allows the narrator adds more personal
emotions.Readers can deal with terror face to face though the words of the narrator.
Large number of monologues and psychological descriptions create atmosphere of
menace tremendously.
Changing from direct speech to free direct speech indicates the governess
is not a reasonable person.She immerses in her fantasy from time to time.From
the direct speech to the free direct speech,the atmosphere of menace keeps
increasing.Passages in direct speech
can only deliver the fact.Readers can feel the menace
in an indirect way.Using the free direct speech allows the narrator adds more personal
emotions.Readers can deal with terror face to face though the words of the narrator.
Large number of monologues and psychological descriptions create atmosphere of
menace tremendously.
▼优质解答
答案和解析
Changing from direct speech to free direct speech 【indicated】 the governess is not a reasonable person.She immerses in her fantasy 【all the time】.From the direct speech to the free direct speech,t...
看了改英语语法和句子顺序Chang...的网友还看了以下:
设a>0,f(x)=e^x/a+a/e^x是R上的偶函数,求a值.∵f(x)=e^x/a+a/e^ 2020-05-17 …
main(){unionEXAMPLE{struct{intx,y;}in;inta,b;}e;e 2020-06-12 …
用以下英文宇母填在上a,a,a,a,a,a,b,e,e,d,e,e,e,e,e,e,f,g,g用以 2020-06-24 …
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 …
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 …
若函数f(x)在R上可导,且f(x)>f'(x),当a>b时,下列不等式成立的是A.e^af(若函 2020-07-29 …
已知向量a≠e,|e|=1,对任意t∈R,恒有|a-te|≥|a-e|,则(a,e皆为向量)A.a⊥ 2020-11-02 …
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 …
人染色体上的基因E突变为e,导致编码的蛋白质中段一个氨基酸改变,下列叙述正确的是()A.E基因突变为 2020-12-28 …
试求矩阵B!设A,B为n阶矩阵,2A-B-AB=E,A^2=A,其中E为n阶单位矩阵.已知A=100 2021-02-05 …