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解:(1)設藥物燃燒階段函數(shù)解析式為.由題意得: ························································································································ 2分 .此階段函數(shù)解析式為······································································· 3分 (2)設藥物燃燒結束后的函數(shù)解析式為.由題意得: ·························································································································· 5分 .此階段函數(shù)解析式為······································································ 6分 (3)當時.得···················································································· 7分 ························································································································· 8分 ·························································································································· 9分 從消毒開始經(jīng)過50分鐘后學生才可回教室.···························································· 10分 查看更多

 

題目列表(包括答案和解析)

方程
x+2
x
+
x
3x+6
=2,用換元法解,若設
x+2
x
=y,則此方程化為整式方程的是(  )

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七年級學生參加了社會實踐調(diào)查活動,到生態(tài)果園調(diào)查后得到如下信息:今年收獲了15噸李子和8噸桃子,要租用甲、乙兩種貨車共6輛,及時運往外地,經(jīng)詢問,甲種貨車可裝李子4噸和桃子1噸,乙種貨車可裝李子1噸和桃子3噸.根據(jù)同學們帶回的信息,試探究以下問題:
(1)共有幾種租車方案?
(2)經(jīng)咨詢運輸公司,甲種貨車每輛需付運費1000元,乙種貨車每輛需付運費700元,試幫助選出最佳方案,并求出此方案運費是多少.
請同學們補充完成下列部分解題過程:
(1)解:
①若設租用甲車x輛,則租用乙車
(6-x)
(6-x)
輛,
②由題意可知:甲車一共可裝
x
x
噸桃子,乙車一共可裝
3(6-x)
3(6-x)
噸桃子,則甲,乙兩種車一共可裝
x+3(6-x)
x+3(6-x)
噸桃子.(用含有x的代數(shù)式表示)
請列出不等式
x+3(6-x)≥8
x+3(6-x)≥8

③甲車一共可裝
4x
4x
噸李子,乙車一共可裝
(6-x)
(6-x)
噸李子,則甲,乙兩種車一共可裝
4x+(6-x)
4x+(6-x)
噸李子.(用含有x的代數(shù)式表示)
請列出不等式
4x+(6-x)≥15
4x+(6-x)≥15

④請列出不等式組,并求出滿足不等組的整數(shù)解,寫出相應的方案
(2)解:

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25、已知:兩個正整數(shù)的和與積相等,求這兩個正整數(shù).
解:不妨設這兩個正整數(shù)為a、b,且a≤b.
由題意,得ab=a+b,(*)
則ab=a+b≤b+b=2b,所以a≤2,
因為a為正整數(shù),所以a=1或2,
①當a=1時,代入等式(*),得1•b=1+b,b不存在;
②當a=2時,代入等式(*),得2•b=2+b,b=2.
所以這兩個正整數(shù)為2和2.
仔細閱讀以上材料,根據(jù)閱讀材料的啟示,思考是否存在三個正整數(shù),它們的和與積相等試說明你的理由.

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用換元法解方程,若設,則原方程化為關于的整式方程是

A、            B、       

C、                D、

 

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用換元法解方程,設,那么原方程可化為   

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