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Matematika attestatsiya №7

InfoMaster

Fevral 2, 2022

44 Ko'rishlar 4 izohlar

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OMAD YOR BO'LSIN!

Matematika fanidan attestatsiya savollari №7

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Testda 76% va undan yuqori natija oling va sertifikatni yuklab oling.

1 / 40

Qavariq oltiburchakning 3 ta uchidan kesib olsak qanday shakl hosil bo’ladi ?

A) 14

B) 9

C) 6

D) 12

2 / 40

Teng yonli trapetsiya asosidagi burchakning sinusi 0,6 ga, asoslarining ayirmasi 4 ga teng bo‘lsa, trapetsiyaning yon tomonini toping.

A) 4

B) 2,5

C) 3

D) 2

3 / 40

x⁴+8x³+ax²+bx+1 ko'phadning kvadrati bo'lsa, a va b koeffitiyentlarning barcha qiymatlari yig'indisini toping.

A) 27

B) 26

C) 48

D) 32

4 / 40

Aniq integralni hisoblang:

A) 0,5

B) 2

C) 3

D) 0

5 / 40

Eldor velosipetida 13m/s bilan 3 soniyada 15m tepalikka ko`tarilgan bo`lsa AC kesma uzunligi necha metr?

A) 40

B) 36

C) 32

D) 34

6 / 40

x ni toping ?

A) 6

B) 3

C) 4

D) 5

7 / 40

a,b,c -1 dan katta musbat sonlar uchun a³=b² va a⁴=c⁵ bo'lsa, log_bc ni toping.

A) 5/6

B) 8/15

C) 1/3

D) 1/2

8 / 40

Agar x =17 bo‘lsa, quyidagi ifodaning qiymatini toping.

A) √17

B) -4

C) 4

D) -√17

9 / 40

Hisoblang:

A) 19/20

B) 20/21

C) 9/10

D) 10/11

10 / 40

Hisoblang:

A) -1/2

B) -√3/2

C) 1/32

D) 0

11 / 40

ABCD to‘rtburchak yuzini toping.

A) 28

B) 32

C) 36

D) 24

12 / 40

Teng yonli ABC (AC = BC) uchburchakning AC tomonida D nuqta olingan. K – BD kesma va AH baladlik kesishgan nuqta. Agar AD = AK bo‘lsa, DBA burchakni toping.

A) 30°

B) 45°

C) 75°

D) 60°

13 / 40

ABC uchburchakning og‘irilik markazi G nuqta. Agar BG = 1 dm, CG = 6 cm va AG ⊥ CG bo‘lsa, AG ni (cm) toping.

A) 9

B) 6

C) 8

D) 10

14 / 40

Hisoblang:

A) 16/9

B) 9/16

C) 1

D) 36/16

15 / 40

ni hisoblang.

A) 1

B) 4

C) 3

D) 2

16 / 40

x tomon uzunligini toping.

A) 1

B) √2

C) 2

D) 3

17 / 40

va ab = 2 bo‘lsa, (a +b)³ ni toping.

A) 125

B) -27

C) 64

D) -64

18 / 40

sonining oxirgi raqamini toping.

A) 1

B) 0

C) 3

D) 9

19 / 40

Ifodani soddalashtiring:

A) √3

B) 1

C) 0

D) 2

20 / 40

ABCD trapetsiyada AD va BC asoslari mos ravishda 10 va 2 ga teng hamda ∠BAD= 45° . CD yon tomonning o‘rtasi E nuqta va BE ⊥ AB bo‘lsa, S_ABCD ni toping.

A) 24

B) 36

C) 48

D) 18

21 / 40

Agar n ta juft son va m ta toq sonlarni qo‘shganda juft son chiqsa, m ning qiymati qanday?

A) 0

B) juft

C) toq

D) 1

22 / 40

Sistemani yeching:

A) (–1; –3) va (1; 3)

B) (3; 1)

C) (–1; 3)

D) (1; 3)

23 / 40

Tengsizlikning qanoatlantiruvchi barcha butun sonlar ko‘paytmasini toping.

A) 3

B) 2

C) -4

D) -2

24 / 40

ni hisoblang.

A) 30

B) 15

C) 0,3

D) 3

25 / 40

ushbu kasrning maxrajini irratsionalikdan qutqaring.

A) √10-√15+√14-√21

B) √10-√15+√21-√41

C) √10-√15+√21+√14

D) √10+√15+√14-√21

26 / 40

Ko‘pyoqning 62 ta yog‘i va 120 ta uchi bor bo‘lsa, uning qirralari sonini toping.

A) 58

B) 180

C) 182

D) 60

27 / 40

Diagonallarining soni tomonlari sonidan 1,5 marta ko‘p bo‘lgan muntazam ko‘pburchakning tomoni 4 ga teng bo‘lsa, unga ichki chizilgan doira yuzini toping.

A) 16π

B) 8π

C) 10π

D) 12π

28 / 40

Qavariq ko‘pburchakning diagonallari soni 5 ga teng bo‘lsa, burchaklari sonini toping.

A) 6

B) 4

C) 7

D) 5

29 / 40

Tenglamani yeching.

A) 9

B) 5

C) 25

D) 3

30 / 40

ABCD parallelogrammda A burchakdan o‘tkazilgan bissektrisa BC tomonni E nuqtada kesib o‘tadi. Agar AB=9 va AD=15 cm bo‘lsa, BE va EC kesmalarni (cm) toping.

A) 5; 10

B) 6; 9

C) 9; 6

D) 10;5

31 / 40

T= 1·1!+2·2!+ 3· 3!+ ...+ 19·19! T bo‘lsa, T +1 yig‘indi nechta 0 bilan tugaydi?

A) 4

B) 2

C) 1

D) 3

32 / 40

|x²-5ax|=15a tenglama bitta manfiy va 3 ta musbat ildizga ega bo‘ladigan a larni toping.

A) (0; 2,5]

B) (0;2,4]

C) (2,4;∞)

D) [2,5;∞)

33 / 40

bo‘lsa, f ''(1) ning qiymatini toping.

A) -12

B) 0

C) 12

D) -24

34 / 40

P(x)=3x⁵-2x⁴+4x³+2x²-1 ko‘phadni x²- x +1 ga bo‘lgandagi qoldiqni toping.

A) x-1

B) x-10

C) x-4

D) x+4

35 / 40

G‘orning 7 ta eshigi bor, sayyoh bu g‘orga necha usulda kirib, yana qaytib chiqishi mumkin? (Bunda har bir eshikdan faqat bir marta o‘tish mumkin)

A) 42

B) 7⁷

C) 36

D) 49

36 / 40

α burchakni toping.

A) 32°

B) 36°

C) 48°

D) 54°

37 / 40

Chizmadan foydalanib noma’lum burchakni toping.

A) 50°

B) 80°

C) 70°

D) 60°

38 / 40

4·(-5)-6·(-3)+(-5)·(-2) sonli ifoda qiymatini toping.

A) 48

B) -12

C) 12

D) 8

39 / 40

8x²+4x³-3x-7=0 tenglama ildizlari ko‘paytmasini toping.

A) -4

B) -2

C) -3

D) 1,75

40 / 40

ABCD qavariq to‘rtburchakka aylana ichki chizilgan. AB = 8 va BC = 12 bo‘lsa, CD – AD ayirmani toping?

A) 2

B) 3

C) aniqlab bo’lmaydi

D) 4

Baholash mezoni

- 75 foiz va undan yuqori ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, oliy malaka toifasi (bosh o’qituvchi lavozimi) saqlansin”;

- 75 foizdan past ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, birinchi malaka toifasi (yetakchi o’qituvchi lavozimi)ga tushirilsin”;

- 75 foiz va undan yuqori ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, oliy malaka toifasi (bosh o’qituvchi lavozimi) berilsin”;

- 74 foizdan 65 foizgacha ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, birinchi malaka toifasi (yetakchi o’qituvchi lavozimi) saqlansin”;

- 65 foizdan kam ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, ikkinchi malaka toifasi (katta o’qituvchi lavozimi)ga tushirilsin”

- 64 foizdan 60 foizgacha ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, ikkinchi malaka toifasi (katta o’qituvchi lavozimi) saqlansin”;

- 60 foizdan kam ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, mutaxassis (oliy yoki o’rta maxsus, kasb-hunar ma'lumotli o’qituvchi) lavozimiga tushirilsin”