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take+good+care

来源:baiyundou.net   日期:2024-08-06

武辰婵3211Let's Trade!Will you take a piece of cake for an apple?a good book for a game?a toy boat for a ca我要整篇文章 谢谢快点 急~~~ -
武陆威14730998449 ______[答案] Let's Trade!Will you take a piece of cake for an apple?a good book for a game?a toy boat for a bike? Long ago, people had to get what they wanted by trading. Some people farmed the land. Other people made tools. Tolls were traded for food. All ...

武辰婵3211七年级上册英语重点句型 列如:be good +at擅长于这样类型的急用 -
武陆威14730998449 ______[答案] I.重点句型 Good morning/afternoon/evening. Good morning/afternoon/evening. How are you? I'm fine,/OK,thanks. Fine,thanks. What's this in English? It's a map. It's V. Spell it please. K-E-Y. What color is it/the key? It's blue. The key is yellow. ...

武辰婵3211Good night's sleep early late to take good care of yourself Da[月亮]翻译成中文什么意思 -
武陆威14730998449 ______ Good night's sleep early late to take good care of yourself Da 良好的睡眠早晚照顾好自己哒

武辰婵3211that's++a+good+exercise的句式错在哪儿 -
武陆威14730998449 ______[答案] 应改为That's good exercise.那是很好的锻炼.exercise作为“锻炼\”讲时,是不可数名词,前面不能用a修饰.

武辰婵3211CH3COOH+Ca的反应方程式和离子反应方程式 -
武陆威14730998449 ______ 反应方程式:2 CH3COOH+Ca=(CH3COO)2Ca+H2↑ 离子方程式:2 CH3COOH+Ca=2 CH3COOH- +Ca+H2↑

武辰婵3211thank+good+the+a+all+things+have+you+we+怎样连词成句
武陆威14730998449 ______ 最相近的句子:thank you (for) all the good things we have.缺1个for,多了1个a

武辰婵3211已知:△ABC的三边分别为a,b,c,且a方+b方+c方=ab+bc+ca,求证:此三角形为等边三角形 -
武陆威14730998449 ______[答案] 已知a^2+b^2+c^2=ab+bc+ac 同乘2 2a^2+2b^2+2c^2=2ab+2bc+2ac (a-b)^2+(b-c)^2+(a-c)^2=0 推出a=b=c 等边三角形

武辰婵3211碳酸氢钠与氢氧化钙反应离子方程式是HCO3+Ca+OH=CaCO3+H2O还是HCO3 - +Ca2++2OH - =CaCO3+H2O+OH - ?哪个方程式对 -
武陆威14730998449 ______[答案] 第一个对(加上离子符号和沉淀符号后)第二个的OH-多出来了……过量氢氧化钙(少量碳酸氢钠)NaHCO3 +Ca(OH)2+ =CaCO3+H2O+NaOH即HCO3- +Ca2+ +OH-=CaCO3+H2O过量碳酸氢钠(少量氢氧化钙)2NaHCO3-+Ca(OH)2=CaCO3+Na...

武辰婵3211已知正整数a、b、c满足:a武陆威14730998449 ______[答案] 解法一:由1≤a

武辰婵3211因式分解abc+ab+bc+ca+a+b+c+1= -
武陆威14730998449 ______[答案] abc+ab+bc+ca+a+b+c+1 =(abc+ab)+(bc+b)+(ca+a)+(c+1) =ab(c+1)+b(c+1)+a(c+1)+(c+1) =(c+1)(ab+a+b+1) =(c+1)[(ab+a)+(b+1)] =(c+1)[a(b+1)+(b+1)] =(c+1)(b+1)(a+1)

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