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22、(2009年長(zhǎng)春市)26.甲船從A港出發(fā)順流勻速駛向B港,行至某處,發(fā)現(xiàn)船上一救生圈不知何時(shí)落入水中,立刻原路返回,找到救生圈后,繼續(xù)順流駛向B港.乙船從B港出發(fā)逆流勻速駛向A港.已知救生圈漂流的速度和水流速度相同;甲、乙兩船在靜水中的速度相同.甲、乙兩船到A港的距離y1y2(km)與行駛時(shí)間x(h)之間的函數(shù)圖象如圖所示.

(1)寫(xiě)出乙船在逆流中行駛的速度.(2分)

(2)求甲船在逆流中行駛的路程.(2分)

(3)求甲船到A港的距離y1與行駛時(shí)間x之間的函數(shù)關(guān)系式.(4分)

(4)求救生圈落入水中時(shí),甲船到A港的距離.(2分)

[參考公式:船順流航行的速度船在靜水中航行的速度+水流速度,船逆流航行的速度船在靜水中航行的速度水流速度.]

試題詳情

20、(德州市二○○九年)20。 (本題滿分9分)為了貫徹落實(shí)國(guó)務(wù)院關(guān)于促進(jìn)家電下鄉(xiāng)的指示精神,有關(guān)部門(mén)自2007年12月底起進(jìn)行了家電下鄉(xiāng)試點(diǎn),對(duì)彩電、冰箱(含冰柜)、手機(jī)三大類(lèi)產(chǎn)品給予產(chǎn)品銷(xiāo)售價(jià)格13%的財(cái)政資金直補(bǔ).企業(yè)數(shù)據(jù)顯示,截至2008年12月底,試點(diǎn)產(chǎn)品已銷(xiāo)售350萬(wàn)臺(tái)(部),銷(xiāo)售額達(dá)50億元,與上年同期相比,試點(diǎn)產(chǎn)品家電銷(xiāo)售量增長(zhǎng)了40%.

(1)求2007年同期試點(diǎn)產(chǎn)品類(lèi)家電銷(xiāo)售量為多少萬(wàn)臺(tái)(部)?

(2)如果銷(xiāo)售家電的平均價(jià)格為:彩電每臺(tái)1500元,冰箱每臺(tái)2000元,手機(jī)每部800元,已知銷(xiāo)售的冰箱(含冰柜)數(shù)量是彩電數(shù)量的倍,求彩電、冰箱、手機(jī)三大類(lèi)產(chǎn)品分別銷(xiāo)售多少萬(wàn)臺(tái)(部),并計(jì)算獲得的政府補(bǔ)貼分別為多少萬(wàn)元?  

解:(1)2007年銷(xiāo)量為a萬(wàn)臺(tái),則a(1+40%)=350,a =250(萬(wàn)臺(tái)). ……3分

(2)設(shè)銷(xiāo)售彩電x萬(wàn)臺(tái),則銷(xiāo)售冰箱x萬(wàn)臺(tái),銷(xiāo)售手機(jī)(350-x)萬(wàn)臺(tái).由題意得:1500x+2000×+800(350x)=500000. ……………6分

解得x=88.  ………………………………………………………7分

,

所以,彩電、冰箱(含冰柜)、手機(jī)三大類(lèi)產(chǎn)品分別銷(xiāo)售88萬(wàn)臺(tái)、132萬(wàn)臺(tái)、130萬(wàn)部.………………………………………………………………8分

∴ 88×1500×13%=17160(萬(wàn)元),132×2000×13%=34320(萬(wàn)元),

130×800×13%=13520(萬(wàn)元). 

獲得的政府補(bǔ)貼分別是17160萬(wàn)元、34320萬(wàn)元、13520萬(wàn)元. ……9分

試題詳情

19、(濟(jì)南市2009年)21.(本小題滿分8分)自2008年爆發(fā)全球金融危機(jī)以來(lái),部分企業(yè)受到了不同程度的影響,為落實(shí)“促民生、促經(jīng)濟(jì)”政策,濟(jì)南市某玻璃制品銷(xiāo)售公司今年1月份調(diào)整了職工的月工資分配方案,調(diào)整后月工資由基本保障工資和計(jì)件獎(jiǎng)勵(lì)工資兩部分組成(計(jì)件獎(jiǎng)勵(lì)工資=銷(xiāo)售每件的獎(jiǎng)勵(lì)金額×銷(xiāo)售的件數(shù)).下表是甲、乙兩位職工今年五月份的工資情況信息:

職工


月銷(xiāo)售件數(shù)(件)
200
180
月工資(元)
1800
1700

(1)試求工資分配方案調(diào)整后職工的月基本保障工資和銷(xiāo)售每件產(chǎn)品的獎(jiǎng)勵(lì)金額各多少元?

(2)若職工丙今年六月份的工資不低于2000元,那么丙該月至少應(yīng)銷(xiāo)售多少件產(chǎn)品?

解:(1)設(shè)職工的月基本保障工資為元,銷(xiāo)售每件產(chǎn)品的獎(jiǎng)勵(lì)金額為元·················· 1分

由題意得··················································································· 3分

解這個(gè)方程組得······················································································ 4分

答:職工月基本保障工資為800元,銷(xiāo)售每件產(chǎn)品的獎(jiǎng)勵(lì)金額5元.······························ 5分

(2)設(shè)該公司職工丙六月份生產(chǎn)件產(chǎn)品····································································· 6分

由題意得··················································································· 7分

解這個(gè)不等式得

答:該公司職工丙六月至少生產(chǎn)240件產(chǎn)品   8分

試題詳情

18、(濟(jì)寧市二○○九年)25.(9分)某體育用品商店購(gòu)進(jìn)一批滑板,每件進(jìn)價(jià)為100元,售價(jià)為130元,每星期可賣(mài)出80件.商家決定降價(jià)促銷(xiāo),根據(jù)市場(chǎng)調(diào)查,每降價(jià)5元,每星期可多賣(mài)出20件.

(1)求商家降價(jià)前每星期的銷(xiāo)售利潤(rùn)為多少元?

(2)降價(jià)后,商家要使每星期的銷(xiāo)售利潤(rùn)最大,應(yīng)將售價(jià)定為多少元?最大銷(xiāo)售利潤(rùn)是多少?

解:(1) (130-100)×80=2400(元);…………………………………4分

(2)設(shè)應(yīng)將售價(jià)定為元,則銷(xiāo)售利潤(rùn)

……………………………………6分

.……………………………………………8分

當(dāng)時(shí),有最大值2500.

∴應(yīng)將售價(jià)定為125元,最大銷(xiāo)售利潤(rùn)是2500元.  ……………9分

試題詳情

17、(2009年臨沂市)24.(本小題滿分10分)在全市中學(xué)運(yùn)動(dòng)會(huì)800m比賽中,甲乙兩名運(yùn)動(dòng)員同時(shí)起跑,剛跑出200m后,甲不慎摔倒,他又迅速地爬起來(lái)繼續(xù)投入比賽,并取得了優(yōu)異的成績(jī).圖中分別表示甲、乙兩名運(yùn)動(dòng)員所跑的路程y(m)與比賽時(shí)間x(s)之間的關(guān)系,根據(jù)圖像解答下列問(wèn)題:

(1)甲摔倒前,________的速度快(填甲或乙);

(2)甲再次投入比賽后,在距離終點(diǎn)多遠(yuǎn)處追上乙?

解:(1)甲.········································································································· (3分)

(2)設(shè)線段的解析式為

代入,得

線段的解析式為().··············································· (5分)

設(shè)線段的解析式為

,分別代入

   解得

線段的解析式為().····································· (7分)

解方程組······························································· (9分)

答:甲再次投入比賽后,在距離終點(diǎn)處追上了乙.    (10分)

試題詳情

16、(泰安市2009年)23(本小題滿分10分)某旅游商品經(jīng)銷(xiāo)店欲購(gòu)進(jìn)A、B兩種紀(jì)念品,若用380元購(gòu)進(jìn)A種紀(jì)念品7件,B種紀(jì)念品8件;也可以用380元購(gòu)進(jìn)A種紀(jì)念品10件,B種紀(jì)念品6件。

(1) 求A、B兩種紀(jì)念品的進(jìn)價(jià)分別為多少?

(2) 若該商店每銷(xiāo)售1件A種紀(jì)念品可獲利5元,每銷(xiāo)售1件B種紀(jì)念品可獲利7元,該商店準(zhǔn)備用不超過(guò)900元購(gòu)進(jìn)A、B兩種紀(jì)念品40件,且這兩種紀(jì)念品全部售出候總獲利不低于216元,問(wèn)應(yīng)該怎樣進(jìn)貨,才能使總獲利最大,最大為多少?

解:(1)設(shè)A、B兩種紀(jì)念品的進(jìn)價(jià)分別為x元、y元。

                                                        

由題意,       得 …………  2分                                                               

解之,得… …4分

答:A、B兩種紀(jì)念品的進(jìn)價(jià)分別為20元、30元… …… …5分

(2)設(shè)上點(diǎn)準(zhǔn)備購(gòu)進(jìn)A種紀(jì)念品a件,則購(gòu)進(jìn)B種紀(jì)念品(40-x)件,

由題意,得

… …… …… ……7分

解之,得:… ………………………………………………8分

∵總獲利是a的一次函數(shù),且w隨a的增大而減小

∴當(dāng)a=30時(shí),w最大,最大值w=-2×30+280=220.

∴40-a=10

∴應(yīng)進(jìn)A種紀(jì)念品30件,B種紀(jì)念品10件,在能是獲得利潤(rùn)最大,最大值是220元!10分

試題詳情

15、(威海市2009年)22.(10分)響應(yīng)“家電下鄉(xiāng)”的惠農(nóng)政策,某商場(chǎng)決定從廠家購(gòu)進(jìn)甲、乙、丙三種不同型號(hào)的電冰箱80臺(tái),其中甲種電冰箱的臺(tái)數(shù)是乙種電冰箱臺(tái)數(shù)的2倍,購(gòu)買(mǎi)三種電冰箱的總金額不超過(guò)132 000元.已知甲、乙、丙三種電冰箱的出廠價(jià)格分別為:1 200元/臺(tái)、1 600元/臺(tái)、2 000元/臺(tái).

(1)至少購(gòu)進(jìn)乙種電冰箱多少臺(tái)?

(2)若要求甲種電冰箱的臺(tái)數(shù)不超過(guò)丙種電冰箱的臺(tái)數(shù),則有哪些購(gòu)買(mǎi)方案?

解:(1)設(shè)購(gòu)買(mǎi)乙種電冰箱臺(tái),則購(gòu)買(mǎi)甲種電冰箱臺(tái),

丙種電冰箱臺(tái),根據(jù)題意,列不等式:·························································· 1分

.······················································· 3分

解這個(gè)不等式,得.·························································································· 4分

至少購(gòu)進(jìn)乙種電冰箱14臺(tái).······················································································ 5分

(2)根據(jù)題意,得.·············································································· 6分

解這個(gè)不等式,得.·························································································· 7分

由(1)知.    .    又為正整數(shù),

.··········································································································· 8分

所以,有三種購(gòu)買(mǎi)方案:

方案一:甲種電冰箱為28臺(tái),乙種電冰箱為14臺(tái),丙種電冰箱為38臺(tái);

方案二:甲種電冰箱為30臺(tái),乙種電冰箱為15臺(tái),丙種電冰箱為35臺(tái);

方案三:甲種電冰箱為32臺(tái),乙種電冰箱為16臺(tái),丙種電冰箱為32臺(tái).    10分

試題詳情

14、(2009年煙臺(tái)市)23.(本題滿分10分)某商場(chǎng)將進(jìn)價(jià)為2000元的冰箱以2400元售出,平均每天能售出8臺(tái),為了配合國(guó)家“家電下鄉(xiāng)”政策的實(shí)施,商場(chǎng)決定采取適當(dāng)?shù)慕祪r(jià)措施.調(diào)查表明:這種冰箱的售價(jià)每降低50元,平均每天就能多售出4臺(tái).

  (1)假設(shè)每臺(tái)冰箱降價(jià)x元,商場(chǎng)每天銷(xiāo)售這種冰箱的利潤(rùn)是y元,請(qǐng)寫(xiě)出yx之間的函數(shù)表達(dá)式;(不要求寫(xiě)自變量的取值范圍)

  (2)商場(chǎng)要想在這種冰箱銷(xiāo)售中每天盈利4800元,同時(shí)又要使百姓得到實(shí)惠,每臺(tái)冰箱應(yīng)降價(jià)多少元?

  (3)每臺(tái)冰箱降價(jià)多少元時(shí),商場(chǎng)每天銷(xiāo)售這種冰箱的利潤(rùn)最高?最高利潤(rùn)是多少?

解:(1)根據(jù)題意,得

.······················································································ 2分

(2)由題意,得

整理,得.················································································ 4分

解這個(gè)方程,得.·········································································· 5分

要使百姓得到實(shí)惠,取.所以,每臺(tái)冰箱應(yīng)降價(jià)200元.································· 6分

(3)對(duì)于

當(dāng)時(shí),······················································································ 8分

所以,每臺(tái)冰箱的售價(jià)降價(jià)150元時(shí),商場(chǎng)的利潤(rùn)最大,最大利潤(rùn)是5000元.    10分

試題詳情

13、(2009年山西省太原市)28.(本小題滿分9分)、兩座城市之間有一條高速公路,甲、乙兩輛汽車(chē)同時(shí)分別從這條路兩端的入口處駛?cè)耄⑹冀K在高速公路上正常行駛.甲車(chē)駛往城,乙車(chē)駛往城,甲車(chē)在行駛過(guò)程中速度始終不變.甲車(chē)距城高速公路入口處的距離(千米)與行駛時(shí)間(時(shí))之間的關(guān)系如圖.

(1)求關(guān)于的表達(dá)式;

(2)已知乙車(chē)以60千米/時(shí)的速度勻速行駛,設(shè)行駛過(guò)程中,兩車(chē)相距的路程為(千米).請(qǐng)直接寫(xiě)出關(guān)于的表達(dá)式;

(3)當(dāng)乙車(chē)按(2)中的狀態(tài)行駛與甲車(chē)相遇后,速度隨即改為(千米/時(shí))并保持勻速行駛,結(jié)果比甲車(chē)晚40分鐘到達(dá)終點(diǎn),求乙車(chē)變化后的速度.在下圖中畫(huà)出乙車(chē)離開(kāi)城高速公路入口處的距離(千米)與行駛時(shí)間(時(shí))之間的函數(shù)圖象.

解:(1)方法一:由圖知的一次函數(shù),設(shè)········································ 1分

      圖象經(jīng)過(guò)點(diǎn)(0,300),(2,120),∴··························· 2分

      解得························································································ 3分

      ∴關(guān)于的表達(dá)式為····················· 4分

方法二:由圖知,當(dāng)時(shí),;時(shí),

      所以,這條高速公路長(zhǎng)為300千米.

   甲車(chē)2小時(shí)的行程為300-120=180(千米).

      ∴甲車(chē)的行駛速度為180÷2=90(千米/時(shí)).··········································· 3分

      ∴關(guān)于的表達(dá)式為().······················ 4分

(2)······················································································ 5分

(3)在中.當(dāng)時(shí),

即甲乙兩車(chē)經(jīng)過(guò)2小時(shí)相遇.············································································· 6分

中,當(dāng).所以,相遇后乙車(chē)到達(dá)終點(diǎn)所用的時(shí)間為(小時(shí)).

乙車(chē)與甲車(chē)相遇后的速度

 (千米/時(shí)).

    ∴(千米/時(shí)).··································· 7分

   乙車(chē)離開(kāi)城高速公路入口處的距離(千米)與行

駛時(shí)間(時(shí))之間的函數(shù)圖象如圖所示.·············· 9分

試題詳情

12、(2009年山西省)(24.(本題8分)某批發(fā)市場(chǎng)批發(fā)甲、乙兩種水果,根據(jù)以往經(jīng)驗(yàn)和市場(chǎng)行情,預(yù)計(jì)夏季某一段時(shí)間內(nèi),甲種水果的銷(xiāo)售利潤(rùn)(萬(wàn)元)與進(jìn)貨量(噸)近似滿足函數(shù)關(guān)系;乙種水果的銷(xiāo)售利潤(rùn)(萬(wàn)元)與進(jìn)貨量(噸)近似滿足函數(shù)關(guān)系(其中為常數(shù)),且進(jìn)貨量為1噸時(shí),銷(xiāo)售利潤(rùn)為1.4萬(wàn)元;進(jìn)貨量為2噸時(shí),銷(xiāo)售利潤(rùn)為2.6萬(wàn)元.

(1)求(萬(wàn)元)與(噸)之間的函數(shù)關(guān)系式.

(2)如果市場(chǎng)準(zhǔn)備進(jìn)甲、乙兩種水果共10噸,設(shè)乙種水果的進(jìn)貨量為噸,請(qǐng)你寫(xiě)出這兩種水果所獲得的銷(xiāo)售利潤(rùn)之和(萬(wàn)元)與(噸)之間的函數(shù)關(guān)系式.并求出這兩種水果各進(jìn)多少噸時(shí)獲得的銷(xiāo)售利潤(rùn)之和最大,最大利潤(rùn)是多少?

解:(1)由題意,得:解得······················································· (2分)

       ∴····················································································· (3分)

(2)

   ∴··················································································· (5分)

    時(shí),有最大值為6.6. ····························· (7分)

(噸).

答:甲、乙兩種水果的進(jìn)貨量分別為4噸和6噸時(shí),獲得的銷(xiāo)售利潤(rùn)之和最大,最大利潤(rùn)是6.6萬(wàn)元.     (8分)

試題詳情


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