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31553155年上市中業(yè)統(tǒng)學(xué)考數(shù)學(xué)卷(分150分,試間分)考注:.試含個(gè)題共題.題,生必答要在題規(guī)的置作,草稿、試上題律效.第、大外其各如特說,必在題的相位上出明計(jì)的要驟一選題本題題,每4分,分24分)【列題四選中有只一選是確,擇確項(xiàng)代并涂答紙相位上.計(jì)算()的果是(
)A
5
B
6
C.
8
D.
9.不等式組
x,的解集是(C)xA
x
B.
D.
x.用換元法解分式方程
3時(shí),如果設(shè)x
,將原方程化為關(guān)于
的整式方程,那么這個(gè)整式方程是(A)A
y
B
yyC.
y
2
y
D.
y
2
.拋物線
y2(x)
2
(
m,
是常數(shù))的頂點(diǎn)坐標(biāo)是(
B)A
(m,n)
B
()
C.
(m
D.
(.下列正多邊形中,中心角等于內(nèi)角的是()A正六邊形B.正五邊形正四邊形
C.三邊形.如圖,已知
EF
,那么下列結(jié)論正確的是A
)
A
BBCDFABDFCECDCDC.D.EFAF二填題本題12題,題分滿分
E
C
圖
D
F【將果線入題的應(yīng)置.分母有理化.方程
x
的根是x=2
.
2y2y.如果關(guān)于
的方程
(
為常數(shù))有兩個(gè)相等的實(shí)數(shù)根,那么
.10已知函數(shù)
f(x)
11
,那么
f
—1/2
..反比例函數(shù)
2
圖像的兩支分別在第IIII象..將拋物線y向上平移一個(gè)單位后,得以新的拋物線,那么新的拋物的表達(dá)式是.13如果從小明等6名生中任選名為“世博會(huì)志愿者,那么小明被選中的概率是.1/614某商品的原價(jià)為元如果經(jīng)過兩次降價(jià),且每次降價(jià)的百分率都是
,那么該商品現(xiàn)在的價(jià)格是100*(—果用含m的代數(shù)式表示15如圖,在ABC中AD是BC上中線,設(shè)向量,BC如果用向量,b表示量,么=a(b/2
A16在圓徑OA
O
中,弦.
的長為6它對(duì)應(yīng)的弦心距為,那么半
B
A
D圖
C.在四邊形
ABCD
中,對(duì)角線
AC
與
BD
互相平分,交點(diǎn)為O
.在不添加任何輔助線的前提下,要使四邊形
成為矩形,還需添加一個(gè)條件,這個(gè)條件可以是AC=BD或有個(gè)內(nèi)角
B等于度.18在Rt中,M為BC上點(diǎn),聯(lián)結(jié)AM(圖3所示果將△ABM沿線AM翻后,點(diǎn)好落在邊AC的點(diǎn)處,那么點(diǎn)MAC的離是.三解題本題題,滿分19題滿分)
M圖
C計(jì)算:
2aaaa2a
.=20題滿分)解方程組:22xy.
(X=2)(x=-1y=0)21題滿分,每小題滿分各)如圖,在梯形ABCD,ADBC,60聯(lián)結(jié)AC.(1求
tanACB
的值;
(2若M、N分別是、DC的中點(diǎn),聯(lián)結(jié)MN,線段MN的.(1二分之根號(hào)(2
A
DB
圖
C22題滿分10分第()小題滿分2分,第)小題滿分分第3小題滿分分,第()小題滿分分為了了解某校初中男生的身體素質(zhì)狀況該六年級(jí)至九年級(jí)共四個(gè)年級(jí)的男生中別抽取部分學(xué)生進(jìn)行“引體向上”測(cè)試.所有被測(cè)試者的“引體向上”次數(shù)情況如表一所示;各年級(jí)的被測(cè)試人數(shù)占所有被測(cè)試人數(shù)的百分率如圖5所中六年級(jí)相關(guān)數(shù)據(jù)未標(biāo)出次數(shù)人數(shù)
表一根據(jù)上述信息,回答下列問題(直接寫出結(jié)果()年級(jí)的被試人數(shù)占有被測(cè)試人數(shù)的百分率是20%;(2在所有被測(cè)試者中,九年級(jí)的人數(shù)是6;(3在所有被測(cè)試者中體向上”次數(shù)小于的數(shù)所占的百分率是;
八年級(jí)25%七年級(jí)25%
九年級(jí)30%六年級(jí)(所有被測(cè)試者引向上數(shù)數(shù)是.
圖523題滿分,每小題滿分各)已知線段AC與相交于點(diǎn)O,結(jié)、,E
A
D的中點(diǎn),F(xiàn)OC的點(diǎn),聯(lián)結(jié)EF(圖示
O(1添加條件
OFE
,求證:.證明:由已知條件得又
B
E
圖6
F
C角角所以三角形全于三角形AB所以(2分別將“
記為①OEFOFE
”記為②ABDC記為③,添加條件①、③,以②為結(jié)論構(gòu)成命題1添加條件②、③,以①為結(jié)論構(gòu)成命題.命題真命,命題假命(選擇“真”或“假”填入空格24題滿分,每小題滿分各)
BCQBCQ在直角坐標(biāo)平面內(nèi),
O
為原點(diǎn),點(diǎn)
A
的坐標(biāo)為
,C
的坐標(biāo)為
(0,線CMx軸(如圖7所
B
與點(diǎn)
A
關(guān)于原點(diǎn)對(duì)稱,直線
x
(
為常數(shù))經(jīng)過點(diǎn)
B
,且與直線相交于點(diǎn)D,結(jié)OD(1求的值和點(diǎn)D的坐標(biāo);
.
y
yx(2點(diǎn)
P
在
軸的正半軸上若
POD
是等腰三角形,求點(diǎn)P的坐標(biāo);(3在()的條件下,如以PD為半徑的圓與圓O外,求圓的半.
D
M解1b=1直BD:y=x+1Y=4代入點(diǎn)D3)(21、則P(,0)
B
AO1
x、則則(,0)、PD=PO設(shè)(x,0D(3,)則由勾股定理解則P(,)(3由,D兩坐標(biāo)可以算出:
圖7、
5
—2
5
、PD=5r=1、PD=25/6r=025題滿分14分第()小題滿分4分,第)小題滿分分第3小題滿分分)已知
90,,AD,
為線段
BD
上的動(dòng)點(diǎn),點(diǎn)
Q
在射線AB
上,且滿足
PQAB
(如圖所示(1當(dāng),點(diǎn)Q與B重合時(shí)(如圖9所示線段PC的;(2在圖8中聯(lián)結(jié)AP.AD
32
,且點(diǎn)在段上,設(shè)點(diǎn)B、之的距離為,
SAPQSPBC
y,其中
△APQ
表示
△的面,
△
表示
PBC
的面積,求
y
關(guān)于
的函數(shù)解析式,并寫出函數(shù)定義域;(3當(dāng)
AD
,且點(diǎn)
Q
在線段
AB
的延長線上時(shí)(如圖所示
QPC
的大小.A
P
DA
P
DA
DPQB
C(Q圖圖92021年上海市初中業(yè)統(tǒng)一學(xué)業(yè)考試
圖10
C
5.115125.11512數(shù)學(xué)卷答案要與評(píng)分標(biāo)準(zhǔn)說明:.解答只列出試題一種或幾種解法.如果考生的解法與所列解法不同,可參照解中評(píng)分標(biāo)準(zhǔn)相應(yīng)評(píng)分;.第一、二大題若特別說明,每題評(píng)分只有滿分或零分;.第三大題中各題端所注分?jǐn)?shù),表示考生正確做對(duì)這一步應(yīng)得分?jǐn)?shù);.評(píng)閱試卷,要堅(jiān)每題評(píng)閱到底,不能因考生解答中出現(xiàn)錯(cuò)誤而中斷對(duì)本題的評(píng).如果考生的解答在某一步出現(xiàn)錯(cuò)誤影響后繼部分而未改變本題的內(nèi)容和難度,視影響的程度決定后繼部分的給分,但原上不超過后繼部分應(yīng)得分?jǐn)?shù)的一半;.評(píng)分時(shí),給分或分均以為基本單位.一選題本題題,滿分.B;2C;3.A;.5C;6A.二填題本題12題,分48分);.;9;10;4212x2;13;.100(1);
11一、;115a2165;17AC(或
90
等
。
。三解題本題題,滿分.:原式=
2(a(aaa(a
···········································(7)=
2a
·······································································()=
1a
··············································································(1分=
.················································································().:由方程①得
x
,③·······················································1分將③代入②,得
2x
x(
,··········································()整理,得
,······························································()解得x,··································································3分1分別將,入③,得y,0,··························()121,所以,原方程組的解為·····································(1分y;y0.12.)過點(diǎn)A作AEBC,足為E.···········································(1分在
Rt
△
ABE
中,∵
,
AB
,∴
BE
,··············································(分AEAB860
.··················································1)∵12,EC.
·······························································(分在Rt△中tanACB
AE43EC
.··································(1分(2在梯形
中,∵
ABDC
,
,
∴
DCB
.········································································()過點(diǎn)D作,足為F,DFCAE//.∵AD,四邊形是平四邊形.∴AD.···················(1分在
△
DCF
中,
60
,····················(1)∴
EFEC
.∴
AD
.∵M(jìn)、N分是、DC中點(diǎn),∴MN
2
.······(2分)
;·················································································
(2分(2
;···················································································(3分)(3
35%
;···············································································(2分(45.·····················································································())證明
OEF
,∴OE.···································································()∵
E
為
OB
的中點(diǎn),
為
OC
的中點(diǎn),∴
OB
,
OCOF
.·············································(1分)∴
OB
.···································································(1分∵
,DOC
,∴△
AOB
≌△
.························································(2分AB
.···································································()(2真························································································3)假.···························································································(3).)∵點(diǎn)的標(biāo)為,
B
與點(diǎn)
A
關(guān)于原點(diǎn)對(duì)稱,∴點(diǎn)
B
的坐標(biāo)為(································································(1)∵直線
x
經(jīng)過點(diǎn)
B
,∴
,得
.···························()∵點(diǎn)
的坐標(biāo)為
(0,線
軸,∴設(shè)點(diǎn)
D
的坐標(biāo)為
(.·······(1分)∵直線
y
與直線
相交于點(diǎn)
D
∴
3
∴
D
的坐標(biāo)為
(1分(2∵
D
的坐標(biāo)為
,OD5
.··············································)當(dāng)當(dāng)
PDODPO5
時(shí),點(diǎn)時(shí),點(diǎn)
PP
的坐標(biāo)為的坐標(biāo)為
(6····································(1分····································)當(dāng)
PD
時(shí),設(shè)點(diǎn)
P
的坐標(biāo)為
((x
,∴
(2
,得
x
256
,∴點(diǎn)
P
的坐標(biāo)為
25(6
.··········()綜上所述,所求點(diǎn)P坐標(biāo)是(6、或(
256
.(3當(dāng)以為半徑的圓與外時(shí),若點(diǎn)
P
的坐標(biāo)為
,則圓
P
的半徑
PD
,圓心距
PO
,∴圓
O
的半徑
r
.·····································································2分若點(diǎn)
P
的坐標(biāo)為
,圓P
的半徑
2
,圓心距
,∴圓
O
的半徑
r5
.··························································(2分)綜上所述,所求圓
O
的半徑等于
或
..)∵//BC∴DBC
.
△△PBCAP△△PBCAP∵
ADAB,ABDADB∴DBC
.∵
ABC45
.···············································(1)∵
PQADPCAB
,
AD
,點(diǎn)
與點(diǎn)
B
重合,∴
PQPC
.∴
PCBPBC45
.······························································()∴
.········································································()在
△
中,
322
.···················(1分(2過點(diǎn)作,PFAB,足分別為、F.····················(1)∴
PFBFBEBEP90
.∴四邊形
是矩形.∴
PF//BC
,
PEBF
.PFAD∵AD//,PF//..BFAB∵
3PF3,,2
.···············································()∵
232,BC,PF,S2
.S∴SPBC函數(shù)的定義域是0
,即≤x
2.·················································()47≤.···························································(1分)8(3過點(diǎn)作PM,PNAB,足分別為M、.易得四邊形
PNBM
為矩形,∴
,
PMBN
,
90
.∵
AD//
,∴
PN//
.∴
ADPNAD.∴AB
.··············(1分∵
PQADPNPQ,.······················································(1分)PCABPMPC又∵
,∴
Rt
△
PCM
∽
Rt
△
PQN
.···············(1分)∴
QPN
.···································································()∵
MPN90
,∴
QPMQPMMPN
,即
90
.·········································································()2021年考備考指1中考最后20天,用有限的時(shí)間把學(xué)習(xí)效率最大化一分鐘學(xué)一分鐘,不要30秒是看書,另外30秒是發(fā)呆。2、加強(qiáng)你的接受能力和專注程度中考最后20天是攻堅(jiān)戰(zhàn),拼的不只是學(xué)
習(xí)知識(shí)。3、如果感覺很多知識(shí)“跟不上”,回過頭把初二知識(shí)理一理。同時(shí)在這里告誡初二學(xué)生,初二基本是分水嶺,一定要重視初二知識(shí)的學(xué)習(xí)。4、中考馬上就到,學(xué)校里一些學(xué)生會(huì)說“對(duì)數(shù)學(xué)這門科目沒興趣……怎么辦”,我只想說還有30天就中考了你卻說你對(duì)數(shù)學(xué)沒興趣所以要擺正學(xué)習(xí)態(tài)度,沒興趣不是理由!5、如果文科的秘籍是多聽、多背、多,那么數(shù)學(xué)就是要多練、多整理錯(cuò)題就不用多說了,為什么整理錯(cuò)題這么重要?因?yàn)槌踔袛?shù)學(xué)題目你是做不完的,關(guān)注題型、關(guān)注你不會(huì)的,把錯(cuò)的做對(duì),那么你的數(shù)學(xué)成績就沒有問題。6如果你平常只能考一般分?jǐn)?shù)那么你掌握的基礎(chǔ)知識(shí)還可以,是考試不僅考基礎(chǔ)題,還考綜合題壓軸題所以最后一定要加強(qiáng)綜合訓(xùn)練尤其是要給自己營造出一種緊張的考試氛圍,在規(guī)定時(shí)間內(nèi)進(jìn)行綜合訓(xùn)練。7、中考最20要克服粗心的毛病,培養(yǎng)堅(jiān)持到底的毅力。最后的關(guān)鍵時(shí)刻,誰堅(jiān)持到最后,誰就是贏家,考完以后再回首你會(huì)覺得幸虧自己懂得及時(shí)。8最后這段時(shí)間學(xué)習(xí)計(jì)劃更重要每天列出需要完成的任務(wù)不要只會(huì)“刷題”,這樣學(xué)習(xí)效率會(huì)更高,你也會(huì)在完成任務(wù)的成就感中更加喜歡學(xué)習(xí)。9、要明白到底什么是“會(huì)”和“不會(huì)”。很多同學(xué)拿到試卷后看到錯(cuò)題第一反應(yīng)就是“我粗心”如果問“1加1等于幾?”,最差的初三學(xué)生都知道等于2,這跟知識(shí)點(diǎn)的熟練度相關(guān)以要明白“懂”不代表會(huì)分?jǐn)?shù)拿不到就是不會(huì)。粗心只是因?yàn)槟阕龅眠€不夠,熟練程度還沒達(dá)到!10、中考實(shí)際上是對(duì)你學(xué)習(xí)能力、心理素質(zhì)、抗壓能力、協(xié)調(diào)能力等綜合能力的考查,所以一定要注意綜合發(fā)展,別只會(huì)傻傻“刷題”。
11、學(xué)習(xí)是一個(gè)連續(xù)的過程。即使明天中,也別忘了學(xué)習(xí)計(jì)劃的實(shí)施。到現(xiàn)在還沒有一個(gè)屬于自己的計(jì)劃?更要好好反思,可以跟老師好好討論給自己制訂一個(gè)科學(xué)的復(fù)習(xí)計(jì)劃!12、不久后你會(huì)參加中,后你還會(huì)面對(duì)高,會(huì)上也有各種考試等著你,要想取得好成績,先要武裝好自己括堅(jiān)韌不拔的意志、不怕輸?shù)挠職?、勇往直前的沖勁等,具備了這些精神品質(zhì),你將一往無前。語文備建議語文最容易得分的是理解性默寫的題15分的題只要背下來記下來對(duì)字,就不成問題?;A(chǔ)題靠積累:中考,每天早自習(xí)抽出10分鐘看一下易錯(cuò)字、易錯(cuò)讀音、病句修改、文化常識(shí),30天足夠你對(duì)這些知識(shí)了然于胸考場上信手拈來。古詩詞:一般情況下考一個(gè)選擇,個(gè)分析題。中考古詩詞都是課內(nèi),所以你對(duì)這些不會(huì)陌生。但是分析題不僅需要你有一定的語言組織能,還需要你把平常上課的語文筆記都背得滾瓜爛熟。這個(gè)需要時(shí),但是如果你仍然不太熟悉的話,同理,每天復(fù)習(xí)兩篇古詩詞的筆記足夠了?,F(xiàn)代文閱讀和作文這里不必多說因?yàn)橐粋€(gè)月時(shí)間不僅不能提高你的作文和閱讀能力,還會(huì)適得其反。數(shù)學(xué)備建議建議各位在這天里,備60道二次函數(shù)壓軸題和60道幾何證明的大題。每天分析一道,做一道。那些一遍做對(duì),析一下是哪種類型,幾道同類型的,如果都能成功地pass掉,恭喜你,這個(gè)類型暫時(shí)沒問題了!
如果沒做對(duì),找一張A4,首行寫題目,下面一步一步寫過程。一道題用一張紙,不夠可以改用八開紙每一步都寫出來每個(gè)細(xì)節(jié)都不要放過每一步過程旁邊用紅筆寫出這一步的知識(shí)點(diǎn),不懂的標(biāo)注一下。這樣做的目的,你可以把自己的思維理順,知道中考不比平常的考,思路清晰、過程條理就會(huì)得到過程分??偨Y(jié)出你所有不會(huì)的題里面涉及到的卡住你的知識(shí),之后,多看看這些知識(shí)點(diǎn)。英語備建議一、聽力部分在做題之,應(yīng)利用間隙時(shí)間審題,據(jù)題干預(yù)測(cè)即將聽到的內(nèi)容,做到心中有數(shù);做題的時(shí)候,手中握筆,對(duì)關(guān)鍵信息點(diǎn)做簡要記錄,并通過對(duì)話的重音語氣等判斷人物關(guān)系
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