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12. 解：(1)如圖：，；(2)　 (b，a) ；

(3)由(2)得，D(1,-3) 關(guān)于直線l的對(duì)稱點(diǎn)的坐標(biāo)為(-3,1)，連接E交直線l于點(diǎn)Q，此時(shí)點(diǎn)Q到D、E兩點(diǎn)的距離之和最小　

設(shè)過(-3,1) 、E(-1,-4)的設(shè)直線的解析式為，則

，∴，∴．由得 ，∴所求Q點(diǎn)的坐標(biāo)為(，)

11. 解：由圖象可知，點(diǎn)在直線上，．解得．直線的解析式為．令，可得．直線與軸的交點(diǎn)坐標(biāo)為．令，可得．直線與軸的交點(diǎn)坐標(biāo)為．

10. 解：(1)由直角三角形紙板的兩直角邊的長為1和2，

知兩點(diǎn)的坐標(biāo)分別為．

設(shè)直線所對(duì)應(yīng)的函數(shù)關(guān)系式為．···························································· 2分

有解得

所以，直線所對(duì)應(yīng)的函數(shù)關(guān)系式為．····················································· 4分

(2)①點(diǎn)到軸距離與線段的長總相等．

因?yàn)辄c(diǎn)的坐標(biāo)為，

所以，直線所對(duì)應(yīng)的函數(shù)關(guān)系式為．

又因?yàn)辄c(diǎn)在直線上，

所以可設(shè)點(diǎn)的坐標(biāo)為．

過點(diǎn)作軸的垂線，設(shè)垂足為點(diǎn)，則有．

因?yàn)辄c(diǎn)在直線上，所以有．······················· 6分

因?yàn)榧埌鍨槠叫幸苿?dòng)，故有，即．

又，所以．

法一：故，

從而有．

得，．

所以．

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

所以，得，而，

從而總有．····································································································· 10分

法二：故，可得．

故．

所以．

故點(diǎn)坐標(biāo)為．

設(shè)直線所對(duì)應(yīng)的函數(shù)關(guān)系式為，

則有解得

所以，直線所對(duì)的函數(shù)關(guān)系式為．·············································· 8分

將點(diǎn)的坐標(biāo)代入，可得．解得．

而，從而總有．························································· 10分

②由①知，點(diǎn)的坐標(biāo)為，點(diǎn)的坐標(biāo)為．

．····································································· 12分

當(dāng)時(shí)，有最大值，最大值為．

取最大值時(shí)點(diǎn)的坐標(biāo)為．　　 14分

9.

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

(2)由題意，可得：．

．·········································································································· (5分)

當(dāng)時(shí)，．

造這片林的總費(fèi)用需45 000元．········································································· (8分)

8.

解：(1)；

(2)，

(天)

答：乙隊(duì)單獨(dú)完成這項(xiàng)工程要60天.

(3)(天)

答：圖中的值是28.

7. 解：(1)設(shè)與之間的關(guān)系為一次函數(shù)，其函數(shù)表達(dá)式為··················· 1分

將，代入上式得，

　 解得

········································································································· 4分

驗(yàn)證：當(dāng)時(shí)，，符合一次函數(shù)；

當(dāng)時(shí)，，也符合一次函數(shù)．

可用一次函數(shù)表示其變化規(guī)律，

而不用反比例函數(shù)、二次函數(shù)表示其變化規(guī)律．···························································· 5分

與之間的關(guān)系是一次函數(shù)，其函數(shù)表達(dá)式為································ 6分

(2)當(dāng)時(shí)，由可得

即貨車行駛到處時(shí)油箱內(nèi)余油16升．········································································ 8分

(3)方法不唯一，如：

方法一：由(1)得，貨車行駛中每小時(shí)耗油20升，····················································· 9分

設(shè)在處至少加油升，貨車才能到達(dá)地．

依題意得，，···························································· 11分

解得，(升)··································································································· 12分

方法二：由(1)得，貨車行駛中每小時(shí)耗油20升，····················································· 9分

汽車行駛18千米的耗油量：(升)

之間路程為：(千米)

汽車行駛282千米的耗油量：

(升)······························································································· 11分

(升)················································································ 12分

方法三：由(1)得，貨車行駛中每小時(shí)耗油20升，····················································· 9分

設(shè)在處加油升，貨車才能到達(dá)地．

依題意得，，

解得，············································································································· 11分

在處至少加油升，貨車才能到達(dá)地．·························································· 12分

6. 解：(1)特征數(shù)為的一次函數(shù)為，

，

．

(2)拋物線與軸的交點(diǎn)為，

與軸的交點(diǎn)為．

若，則，；

若，則，．

當(dāng)時(shí)，滿足題設(shè)條件．

此時(shí)拋物線為．

它與軸的交點(diǎn)為，

與軸的交點(diǎn)為，

一次函數(shù)為或，

特征數(shù)為或．

5. 解⑴y＝(63－55)x+(40－35)(500－x)……………3分

＝2x+2500。即y＝2x+2500(0≤x≤500)，………………4分

⑵由題意，得55x+35(500－x)≤20000，………………6分

解這個(gè)不等式，得x≤125，………………………………7分

∴當(dāng)x＝125時(shí),y_最大值＝3×12+2500＝2875(元),…………9分

∴該商場(chǎng)購進(jìn)A、B兩種品牌的飲料分別為125箱、375箱時(shí)，能獲得最大利潤2875元.………………………………………………………………10分

4. 解：設(shè)這個(gè)一次函數(shù)的解析式為y=kx+b.

則解得，函數(shù)的解析式為y=-2x+3.

由題意，得得，所以使函數(shù)為正值的x的范圍為

3. 解：(1)，所以不能在60天內(nèi)售完這些椪柑，

　　　 (千克)

　　　 即60天后還有庫存5000千克，總毛利潤為

　　　 W=；

　 (2)

　　 要在2月份售完這些椪柑，售價(jià)x必須滿足不等式

　　 解得

　　 所以要在2月份售完這些椪柑，銷售價(jià)最高可定為1.4元/千克。