版權(quán)說明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請(qǐng)進(jìn)行舉報(bào)或認(rèn)領(lǐng)
文檔簡介
1、注:名字標(biāo)紅色者為A路線人物。標(biāo)藍(lán)色者為B路線人物。角色友好度 初期友好度/友好度進(jìn)行數(shù)值 羅伊炎瑪卡斯氷(30/+2亞倫炎(20/+2蘭斯理(20/+2沃爾特氷(30/+2風(fēng)(1/+1風(fēng)(1/+1莉莉娜光(56/+4雷(1/+1理(20/+2闇(1/+1瑪卡斯冰羅伊炎(30/+2亞倫炎(20/+2蘭斯理(20/+2沃爾特氷(20/+2莉莉娜光(1/+1亞倫炎羅伊炎(20/+2瑪卡斯氷(20/+2蘭斯理(30/+2沃爾特氷(20/+2炎(1/+1氷(1/+1 波魯斯風(fēng)闇(1/+2莉莉娜光(30/+2炎(35/+3巴斯氷(20/+2風(fēng)(1/+2沃爾特
2、冰羅伊炎(30/+2瑪卡斯氷(20/+2亞倫炎(20/+2蘭斯理(20/+2風(fēng)(1/+1蘭斯理羅伊炎(20/+2瑪卡斯氷(20/+2亞倫炎(30/+2沃爾特氷(20/+2洛特理(1/+1雷(1/+1 愛玲風(fēng)查特雷(1/+1阿魯炎(30/+1薩沃爾氷(1/+1闇(20/+2雷(10/+2瓦德炎亞倫炎(1/+1雷(10/+2洛特理(40/+2風(fēng)(10/+2氷(1/+1洛特理蘭斯理(1/+1雷(10/+2炎(40/+2風(fēng)(10/+2氷(1/+1迪克雷炎(10/+2洛特理(10/+2風(fēng)(10/+2雷(1/+1闇(10/+1氷(40/+1
3、0; 夏妮風(fēng)羅伊炎(1/+1雷(10/+2炎(10/+2洛特理(10/+2澤洛特闇(30/+1氷(30/+3光(30/+3 查特雷愛玲風(fēng)(1/+1阿魯炎(30/+2氷(1/+2理(1/+1風(fēng)(1/+1阿魯炎愛玲風(fēng)(30/+1查特雷(30/+2氷(20/+3闇(1/+1風(fēng)(5/+1庫拉莉娜雷蘭斯理(1/+1雷(1/+1闇(10/+2炎(1/+1氷(40/+3魯?shù)兰影道?10/+1雷(10/+2炎(1/+1理(1/+1光(1/+1薩沃爾冰愛玲風(fēng)(1/+1炎(20/+2理(1/+1氷(1/+1光(40/+2多羅茜炎雷(1/+1薩沃爾氷(20/+
4、2氷(1/+1闇(1/+1光(40/+2絲風(fēng)羅伊炎(1/+1沃爾特氷(1/+1氷(30/+2光(1/+1理(40/+3澤洛特暗風(fēng)(30/+1特雷克風(fēng)(10/+2諾亞理(10/+2氷(30/+1光(40/+3特雷克風(fēng)澤洛特闇(10/+2諾亞理(10/+2雷(1/+1闇(1/+1光(20/+1諾亞理澤洛特闇(10/+2特雷克風(fēng)(10/+2炎(10/+2光(20/+1光(1/+1艾斯多爾暗風(fēng)(1/+2莉莉娜光(30/+2炎(1/+2巴斯氷(10/+2氷(10/+1莉莉娜光羅伊炎(56/+4瑪卡斯氷(1/+1風(fēng)(30/+2闇(30/+2炎(30/+2巴斯氷(30/+2風(fēng)(10/+2雷(30/+2理(5
5、0/+1風(fēng)(10/+2巴斯冰風(fēng)(20/+2闇(10/+2莉莉娜光(30/+2炎(20/+2風(fēng)(10/+2炎風(fēng)(35/+3闇(1/+2莉莉娜光(30/+2巴斯氷(20/+2風(fēng)(10/+2-風(fēng)風(fēng)(1/+2莉莉娜光(10/+2炎(10/+2巴斯氷(10/+2雷(1/+1炎闇(1/+1諾亞理(10/+2氷(1/+2雷(40/+3光(30/+3冰炎(1/+1風(fēng)(30/+2炎(1/+2雷(1/+1理(40/+2雷羅伊炎(1/+1風(fēng)(1/+1炎(1/+1氷(20/+2闇(1/+2風(fēng)(1/+1雷(30/+3 光氷(10/+2闇(20/+2理(20/+2光(1/+1雷(30/+
6、2-炎雷(1/+1氷(1/+1理(20/+1風(fēng)(1/+1雷(1/+1雷特雷克風(fēng)(1/+1莉莉娜光(30/+2氷(1/+1雷(1/+1風(fēng)(20/+1理(1/+1 冰雷(40/+1雷(40/+3氷(30/+2光(10/+2闇(20/+2冰亞倫炎(1/+1風(fēng)(30/+3澤洛特闇(30/+1氷(30/+2光(30/+3冰炎(1/+1洛特理(1/+1雷(1/+1炎(1/+1雷(20/+2雷炎(40/+3雷(1/+1理(1/+1雷(1/+1光(20/+2冰查特雷(1/+2阿魯炎(20/+3闇(1/+1風(fēng)(5/+1理(1/+1理查特雷(1/+1炎(20/+1雷(
7、1/+1風(fēng)(1/+1風(fēng)(1/+1暗愛玲風(fēng)(20/+2阿魯炎(1/+1特雷克風(fēng)(1/+1雷(20/+3光(1/+1-暗炎(1/+1氷(20/+2雷(1/+2光(20/+2理(30/+2雷(30/+2 理羅伊炎(20/+2薩沃爾氷(1/+1莉莉娜光(50/+1光(20/+2闇(30/+2雷(30/+2 -暗羅伊炎(1/+1氷(1/+1氷(30/+2光(30/+3理(1/+1-冰薩沃爾氷(1/+1闇(10/+1闇(30/+2光(30/+3雷(1/+1風(fēng)莉莉娜光(10/+2雷(20/+1炎(1/+1雷(1/+1
8、理(1/+1光風(fēng)(1/+1光(1/+1闇(30/+3氷(30/+3理(1/+1風(fēng)查特雷(1/+1阿魯炎(5/+1氷(5/+1理(1/+1理(40/+1雷愛玲風(fēng)(10/+2氷(1/+1雷(1/+1闇(20/+3光(1/+1雷炎(1/+1雷(30/+3光(30/+2闇(30/+2理(30/+2氷(1/+1 理闇(1/+1風(fēng)(40/+3氷(40/+2雷(1/+1光(1/+1-光風(fēng)(30/+3澤洛特闇(40/+3特雷克風(fēng)(20/+1諾亞理(20/+1氷(30/+3理氷(1/+1闇(1/+1光(1/+1風(fēng)(40/+1光(40/+1-光薩沃爾氷(40/+2炎(
9、40/+2闇(1/+1理(40/+1理(1/+1光闇(1/+1諾亞理(1/+1炎(30/+3雷(20/+2雷(1/+1支援效果一覽1:3 D2:3思路點(diǎn)撥 添加輔助線,要探求兩半徑之間的關(guān)系,必須求出C,PA交O2于點(diǎn)D,CD的延長線交Ol于點(diǎn)N(1過點(diǎn)A作AECN+5.0%外切于A,PA是內(nèi)公切線,BC是外公切線,自身的C是切點(diǎn)PB=AB;PBA=PAB;PABOlAB;PBA上述結(jié)論,正確結(jié)論的個(gè)數(shù)是( A1 B2 C3 D回避率、F,求證:(1CD是Ol的直徑;(2試判斷線段BC、BE、BF的大小關(guān)系,并證明你的結(jié)論 11如圖,已知A是Ol、O2的一個(gè)交點(diǎn),點(diǎn)M是 OlO2的中點(diǎn),過點(diǎn)A
10、的直線BC1,2 B1,3 C1,2,3 D1,2,3,4 16如圖,相等兩圓交于A、B兩點(diǎn),過B任作一直線交兩圓于M、N,過M、N各引所在圓的切線相交于C,則四邊形AMCN有下面關(guān)系成立( 于點(diǎn)D,E,過點(diǎn)E作EFCE交CB的延長線于F(C自身的屬性對(duì)手的屬性支援等級(jí)攻撃力防御力命中率回避率必殺率必殺回避率炎炎C+1.00.0+5.0%+5.0%+5.0%0.0%炎炎B+2.00.0+10.0%+10.0%+10.0%0.0%炎炎A+3.00.0+15.0%+15.0%+15.0%0.0%炎雷C+0.5+0.5+2.5%+5.0%+5.0%+2.5%炎雷B+1.0+1.0+5.0%+10.0
11、%+10.0%+5.0%炎雷A+1.5+1.5+7.5%+15.0%+15.0%+7.5%炎風(fēng)C+1.00.0+5.0%+2.5%+5.0%+2.5%炎風(fēng)B+2.00.0+10.0%+5.0%+10.0%+5.0%炎風(fēng)A+3.00.0+15.0%+7.5%+15.0%+7.5%炎氷C+0.5+0.5+5.0%+5.0%+2.5%+2.5%炎氷B+1.0+1.0+10.0%+10.0%+5.0%+5.0%炎氷A+1.5+1.5+15.0%熟悉以下基本圖形、基本結(jié)論:【例題求解】【例1】 如圖,Ol與半徑為40.0A,Ol+5.0%2的直徑BC交Ol于點(diǎn)D,EF為過點(diǎn)A的公切線,若O2D=,那么B
12、AF= 思路點(diǎn)撥 直徑、公切線、O2的特殊位置等,隱含豐富的信息,而連O2Ol必過A點(diǎn),先求出D O2A+15.0%+15.0%+15.0%(2涉及兩圓位置關(guān)系的計(jì)算題,常作半徑、連心線,結(jié)合切線性質(zhì)等構(gòu)造直角三角形,將分散的條件集中,通過解直角三角形求解【例2】 如圖,Ol與+1.02外切于點(diǎn)A,兩圓的一條外公切線與O1相切于點(diǎn)B,若AB與兩圓的另一條外公切線平行,則Ol 與O2的半徑之比為( 炎光B0.0O2 (或DO2Ol的度數(shù),為此需尋求CO1B、CO1A、BO1A的關(guān)系炎理C與O2相交于A、B兩點(diǎn),P是+5.0%+2.5%+2.5%炎,求證:PA=PE;(2連結(jié)PN,若PB=4,BC
13、=2,求PN的長思路點(diǎn)撥 (1+7.5%+7.5%(2PB·PC=PD·PA,探尋PN、PD、PA對(duì)應(yīng)三角形的聯(lián)系【例4】命中率回避率D,連結(jié)OD并延長交大圓于點(diǎn)E,連結(jié)BE交AC于點(diǎn)F,已知AC=,大、小兩圓半徑差為2(1求大圓半徑長;(2求線段BF的長;(3求證:EC與過B、F、C三點(diǎn)的圓相切思路點(diǎn)撥 (1設(shè)大圓半徑為R,則小圓半徑為R-2,建立R的方程;(2證明EBCECF;(3過B、F、C三點(diǎn)的圓的圓心O,必在BF上,連OC,證明OCE=90°注:本例以同心圓為背景,綜合了垂徑定理、直徑所對(duì)的圓周角為直角、切線的判定、勾股定理、相似三角形等豐富的知識(shí)作出圓
14、中基本輔助線、運(yùn)用與圓相關(guān)的角是解本例的關(guān)鍵 【例5】 如圖,AOB雷C0.0OA上,并與弧AB內(nèi)切于點(diǎn)A,半圓O2的圓心O2在OB上,并與弧AB內(nèi)切于點(diǎn)B,半圓O1與半圓O2相切,設(shè)兩半圓的半徑之和為+2.00.0試建立以為自變量的函數(shù)的解析式;(2求函數(shù)的最小值思路點(diǎn)撥 設(shè)兩圓半徑分別為R、r,對(duì)于(1,通過變形把R2+r2用“=R+r”的代數(shù)式表示,作出基本輔助線;對(duì)于(2,因=R+r,故是在約束條件下求的最小值,解題的關(guān)鍵是求出風(fēng)注:如圖,半徑分別為r、R的Ol 、O2外切于C,AB,CM分別為兩圓的公切線,OlO2與AB交于P點(diǎn),則: (1AB=2;(2 ACB=Ol M O2=90
15、°;(3PC2=PA·PB;(4sinP=;(5設(shè)C到AB的距離為d,則學(xué)力訓(xùn)練1已知:Ol和O2交于A、B兩點(diǎn),且Ol經(jīng)過點(diǎn)O+1.0AOlB=90°,則A O2B的度數(shù)是 2矩形ABCD中,雷氷BD0.0+2.0+5.0%+10.0%+10.0%3雷氷相交于點(diǎn)A、B,現(xiàn)給出+3.0+7.5%AC+15.0%2的切線且交Ol于點(diǎn)C,AD是Ol的切線且交O2于點(diǎn)D,則AB2=BC·BD;(2連結(jié)AB+5.0%+5.0%雷A=15cm,O2A=20cm,AB=24cm,則OlO2=25cm;(3若CA是Ol的直徑,DA是O2 的一條非直徑的弦,且點(diǎn)D、B不
16、重合,則C、B、D三點(diǎn)不在同一條直線上,(4若過點(diǎn)A作O+15.0%+15.0%于點(diǎn)D,直線DB交Ol于點(diǎn)C,直線CA 交O2+0.5+1.0,則DE2=DB·DC,則正確命題的序號(hào)是 (寫出所有正確命題的序號(hào) 4如圖,半圓O的直徑AB=4,與半圓O內(nèi)切的動(dòng)圓Ol與AB切于點(diǎn)M,設(shè)Ol的半徑為,AM的長為,則A+1.5,自變量的取值范圍是 +7.5% +7.5%5如圖,施工工地的水平地面上,有三根外徑都是1米的水泥管兩兩相切摞在一起,則其最高點(diǎn)到地面的距離是( A2 B C D0.0%+5.0%+2.5%+5.0%、B兩點(diǎn),且點(diǎn)Ol在O2上,過A作Oll的切線AC交B Ol的延長線于
17、點(diǎn)P,交+10.0%+5.0%+10.0%雷理AA B C D+15.0%攻撃力防御力風(fēng)r (R>r,圓心距為d,若關(guān)于的方程有兩個(gè)相等的實(shí)數(shù)根,則兩圓的位置關(guān)系是( A一定內(nèi)切 B一定外切 C相交 D內(nèi)切或外切9如圖,Ol和O2內(nèi)切于點(diǎn)P,過點(diǎn)P的直線交Ol于點(diǎn)D,交O2于點(diǎn)E,DA與2相切,切點(diǎn)為C(1)求證:PC平分APD;0.0+15.0%+7.5%+15.0%(3若PE=3,PA=6,求PC的長10如圖,已知Ol和O2外切于A,BC是Ol和O2的公切線,切點(diǎn)為B、C,連結(jié)BA并延長交Ol于+1.0+1.0+5.0%+1.02于B、C(1求證:AB=AC;+5.0%風(fēng)風(fēng)B+2.0
18、0.0+10.0%,求證:dl%1O2;(3在(2的條件下,若dld2=1,設(shè)Ol、O2的半徑分別為R、r,求證:R2+r2= R2r2+15.0%氷C+0.5+0.5+5.0%+2.5%+2.5%+5.0%根圓形筷子的橫截面圓半徑為r,則捆扎這7根筷子一周的繩子的長度為 14如圖,Ol和O+10.0%+5.0%2的弦AB經(jīng)過Ol風(fēng)l氷AD,若AC:CD:DB=3:4:2,則Ol與O2的直徑之比為( 2:7 BC5 C2:3 D 1:3 15如圖,Ol與+5.0%風(fēng)闇B風(fēng)光既有內(nèi)切圓,也有外接圓 D以上情況都不對(duì) 17已知:如圖,O與相交于A,B兩點(diǎn),點(diǎn)P在O上,O的弦風(fēng)光B+5.0%風(fēng)光A+
19、3.0+1.5+15.0%0.0%+15.0%+7.5%風(fēng)理C+1.0+0.5+2.5%+2.5%+2.5%+5.0%風(fēng)理B+2.0+1.0+5.0%+5.0%+5.0%+10.0%風(fēng)理A+3.0+1.5+7.5%+7.5%+7.5%+15.0%自身的屬性對(duì)手的屬性支援等級(jí) 攻撃力防御力命中率回避率必殺率必殺回避率氷炎C+0.5+0.5+5.0%+5.0%+2.5%+2.5%氷炎B+1.0+1.0+10.0%+10.0%+5.0%+5.0%氷炎A+1.5+1.5+15.0%+15.0%+7.5%+7.5%氷雷C0.0+1.0+2.5%+5.0%+2.5%+5.0%氷雷B0.0+2.0+5.0%
20、+10.0%+5.0%+10.0%氷雷A0.0+3.0+7.5%+15.0%+7.5%+15.0%氷風(fēng)C+0.5+0.5+5.0%+2.5%+2.5%+5.0%氷風(fēng)B+1.0+1.0+10.0%+5.0%+5.0%+10.0%氷風(fēng)A+1.5+1.5+15.0%+7.5%+7.5%+15.0%氷氷C0.0+1.0+5.0%+5.0%0.0%+5.0%氷氷B0.0+2.0+10.0%+10.0%0.0%+10.0%氷氷A0.0+3.0+15.0%+15.0%0.0%+15.0%氷闇C0.0+0.5+5.0%+5.0%+2.5%+5.0%氷闇B0.0+1.0+10.0%+10.0%+5.0%+10.
21、0%氷闇A0.0+1.5+15.0%+15.0%+7.5%+15.0%氷光C+0.5+1.0+5.0%+2.5%+2.5%+2.5%氷光B+1.0+2.0+10.0%+5.0%+5.0%+5.0%氷光A+1.5+3.0+15.0%+7.5%+7.5%+7.5%氷理C+0.5+1.0+2.5%+5.0%0.0%+5.0%氷理B+1.0+2.0+5.0%+10.0%0.0%+10.0%氷理A+1.5+3.0+7.5%+15.0%0.0%+15.0%自身的屬性對(duì)手的屬性支援等級(jí)攻撃力防御力命中率回避率必殺率必殺回避率闇炎C+0.50.0+5.0%+5.0%+5.0%+2.5%闇炎B+1.00.0+1
22、0.0%+10.0%+10.0%+5.0%闇炎A+1.50.0+15.0%+15.0%+15.0%+7.5%闇雷C0.0+0.5+2.5%+5.0%+5.0%+5.0%闇雷B0.0+1.0+5.0%+10.0%+10.0%+10.0%闇雷A0.0+1.5+7.5%+15.0%+15.0%+15.0%闇風(fēng)C+0.50.0+5.0%+2.5%+5.0%+5.0%闇風(fēng)B+1.00.0+10.0%+5.0%+10.0%+10.0%闇風(fēng)A+1.50.0+15.0%+7.5%+15.0%+15.0%闇氷C0.0+0.5+5.0%+5.0%+2.5%+5.0%闇氷B0.0+1.0+10.0%+10.0%+5
23、.0%+10.0%闇氷A0.0+1.5+15.0%+15.0%+7.5%+15.0%闇闇C0.00.0+5.0%+5.0%+5.0%+5.0%闇闇B0.00.0+10.0%+10.0%+10.0%+10.0%闇闇A0.00.0+15.0%+15.0%+15.0%+15.0%闇光C+0.5+0.5+5.0%+2.5%+5.0%+2.5%闇光B+1.0+1.0+10.0%+5.0%+10.0%+5.0%闇光A+1.5+1.5+15.0%+7.5%+15.0%+7.5%闇理C+0.5+0.5+2.5%+5.0%+2.5%+5.0%闇理B+1.0+1.0+5.0%+10.0%+5.0%+10.0%闇理
24、A+1.5+1.5+7.5%+15.0%+7.5%+15.0%自身的屬性對(duì)手的屬性支援等級(jí)攻撃力防御力命中率回避率必殺率必殺回避率光炎C+1.0+0.5+5.0%+2.5%+5.0%0.0%光炎B+2.0+1.0+10.0%+5.0%+10.0%0.0%p 光炎A+3.0+1.5+15.0%+7.5%+15.0%0.0%p 光雷C+0.5+1.0+2.5%+2.5%+5.0%+2.5%p 光雷B+1.0+2.0+5.0%+5.0%+10.0%+5.0%p 光雷A+1.5+3.0+7.5%+7.5%+15.0%+7.5%p 光風(fēng)C+1.0+0.5+5.0%0.0%+5.0%+2.5%p 光風(fēng)B+
25、2.0+1.0+10.0%0.0%+10.0%+5.0%p 光風(fēng)A+3.0+1.5+15.0%0.0%+15.0%+7.5%p 光氷C+0.5+1.0+5.0%+2.5%+2.5%+2.5%p 光氷B+1.0+2.0+10.0%+5.0%+5.0%+5.0%p 光氷A+1.5+3.0+15.0%+7.5%+7.5%+7.5%p 光闇C+0.5+0.5+5.0%+2.5%+5.0%+2.5%p 光闇B+1.0+1.0+10.0%+5.0%+10.0%+5.0%p 光闇A+1.5+1.5+15.0%+7.5%+15.0%+7.5%p 光光C+1.0+1.0+5.0%0.0%+5.0%0.0%p 光光B+2.0+2.0+10.0%0.0%+10.0%0.0%p 光光A+3.0+3.0+15.0%0.0%+15.0%0
溫馨提示
- 1. 本站所有資源如無特殊說明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請(qǐng)下載最新的WinRAR軟件解壓。
- 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請(qǐng)聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
- 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁內(nèi)容里面會(huì)有圖紙預(yù)覽,若沒有圖紙預(yù)覽就沒有圖紙。
- 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
- 5. 人人文庫網(wǎng)僅提供信息存儲(chǔ)空間,僅對(duì)用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對(duì)用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對(duì)任何下載內(nèi)容負(fù)責(zé)。
- 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請(qǐng)與我們聯(lián)系,我們立即糾正。
- 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對(duì)自己和他人造成任何形式的傷害或損失。
最新文檔
- 廣場(chǎng)物業(yè)管理保密合同
- 保證書承諾文書的寫作要點(diǎn)
- 遼寧省大連市高中化學(xué) 第三章 金屬及其化合物 3.2.2 鈉的重要化合物習(xí)題課教案 新人教版必修1
- 2024秋一年級(jí)語文上冊(cè) 漢語拼音 11 ie üe er教案 新人教版
- 2024秋六年級(jí)英語上冊(cè) Unit 4 I have a pen pal說課稿 人教PEP
- 2024六年級(jí)英語上冊(cè) Module 2 Unit 2 There are lots of beautiful lakes in China教案 外研版(三起)
- 2023九年級(jí)物理上冊(cè) 第一章 分子動(dòng)理論與內(nèi)能1.3 比熱容教案 (新版)教科版
- 河北省工程大學(xué)附屬中學(xué)初中體育《第一課 技巧 跳躍練習(xí) 》教案
- 2024學(xué)年八年級(jí)英語上冊(cè) Module 9 Population Unit 1 The population of China is about 137 billion教案 (新版)外研版
- 2024-2025版高中物理 第二章 恒定電流 7 閉合電路的歐姆定律教案 新人教版選修3-1
- 血標(biāo)本采集法并發(fā)癥
- 2024天津港保稅區(qū)管委會(huì)雇員公開招聘6人高頻500題難、易錯(cuò)點(diǎn)模擬試題附帶答案詳解
- 上海離職協(xié)議書模板
- TGDNAS 056-2024 胚胎移植婦女圍術(shù)期護(hù)理
- 第十五屆全國交通運(yùn)輸行業(yè)職業(yè)技能大賽(公路收費(fèi)及監(jiān)控員賽項(xiàng))考試題庫-下(簡答題)
- 2024年中考語文復(fù)習(xí)分類必刷:非連續(xù)性文本閱讀(含答案解析)
- 項(xiàng)目經(jīng)理或管理招聘面試題與參考回答(某大型國企)
- 《進(jìn)一步規(guī)范管理燃煤自備電廠工作方案》發(fā)改體改〔2021〕1624號(hào)
- 2024年國際貿(mào)易實(shí)務(wù)試題及答案
- 血透進(jìn)修總結(jié)匯報(bào)
- 冀少版(2024)七年級(jí)上冊(cè)生物單元+期中+期末共6套學(xué)情評(píng)估測(cè)試卷匯編(含答案)
評(píng)論
0/150
提交評(píng)論