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Matematika attestatsiya №5

InfoMaster

Yanvar 24, 2022

33 Ko'rishlar 1 izoh

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2 ovozlar, 1 avg

Matematika fanidan attestatsiya savollari №5

1 / 40

Kamayish tartibida joylashtiring.

a)5√6
b)12
c)4√10

A) b>a>c

B) a>c>b

C) c>a>b

D) a>b>c

2 / 40

Agar bo‘lsa, x ni toping.

A) 2

B) -5

C) 5

D) -2

3 / 40

|x+1|=2|x–2| tenglamaning ildizlari yig’indisini toping.

A) 6

B) 7

C) 5

D) 0

4 / 40

funksiyaning [2; 3] kesmadagi eng katta qiymatini

A) 3

B) 7,5

C) 4,5

D) 4

5 / 40

Tengsizlikni yeching.

A) [-3;2]∪{1}

B) (-∞;-2]

C) [-3;-2]

D) [-2;1]∪{-3}

6 / 40

Hisoblang:

A) 4

B) 7

C) 9

D) 27

7 / 40

Quyida berilgan holatda stol va uni atrofida qo’yiladigan stullarning holati berilgan va raqamlangan k-holatda stullarning raqamlari yig`indisi 351 taga teng bo`lsa, k ni toping ?

A) 11

B) 13

C) 12

D) 10

8 / 40

Hisoblang:

A) 4

B) 1

C) 3

D) 2

9 / 40

DC||AB , ∠DCE=45°,∠CEA=x va ∠EAB=115° ga tengbo`lsa x ni toping ?

A) 130°

B) 120°

C) 100°

D) 110°

10 / 40

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

A) 1/2

B) 5/6

C) 8/15

D) 1/3

11 / 40

sin a = x va 1+cos a= y bo‘lsa, x va y o‘zaro qanday bog‘langan?

A) 2

B) 1

C) 4

D) 3

12 / 40

Uchburchakning 3 va 4 ga teng bo‘lgan tomonlariga o‘tkazilgan medianalar o‘zaro perpendikulyar bo‘lsa, bu uchburchakning uchinchi tomonini toping.

A) 2,5

B) 2,4

C) √6

D) √5

13 / 40

Tenglikdan foydalanib a ni toping. (a+b+c+d)·(a-b-c+d)=(a-b+c-d)·(a+b-c-d)

A) cd/b

B) bc/d

C) bd/c

D) 1

14 / 40

sonlari geometrik progressiyaning ketma-ket hadlari bo‘ladigan barcha n larning yig‘indisini (agar bitta qiymati bo‘lsa, o‘zini) toping.

A) 24

B) 23

C) 35

D) 14

15 / 40

√3

A) 7/3

B) 5/5

C) 3/2

D) 21/10

16 / 40

P(x+1)=x³+3x²-2x+a+3 ko‘phadi berilgan. P(x+2) ko‘phadining koeffitsiyentlari yig‘indisini 8 ga teng bo‘lsa, a nechaga teng?

A) -6

B) -3

C) -4

D) 5

17 / 40

Sonlarining o‘rta geometrik qiymatini toping.

A) 2√3

B) 3√2

C) 4√3

D) 2√2

18 / 40

6-(5:3)·(-3)²-(-3)³+15:(-3) hisoblang.

A) 14

B) 12

C) 15

D) 13

19 / 40

Tengsizlik nechta butun juft yechimga
ega?

A) 116

B) 112

C) 110

D) 115

20 / 40

x²-√11x+1=0 0 bo‘lsa,

A) 12

B) 9

C) 10

D) 11

21 / 40

vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping.

A) 4

B) 2

C) 3

D) 1

22 / 40

kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping.

A) 5

B) 11

C) 10

D) 7

23 / 40

Agar α=60°, β=70°, γ=50° bo‘lsa, tgα+ tgβ+ tgγ yig‘indi quyidagilardan qaysi biriga teng?

A) 4 √3tg50° tg70°

B) 2√3tg50° tg70°

C) √3ctg40° tg70°

D) √3tg50° tg70°

24 / 40

Hisoblang:

A) 19/20

B) 20/21

C) 9/10

D) 10/11

25 / 40

Agar geometrik progressiyaning umumiy hadi b_n= 3·2ⁿ bo‘lsa, ni toping.

A) 1

B) 2

C) 4

D) 3

26 / 40

a_n - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping:

A) 7

B) 8

C) 6

D) 5

27 / 40

bo‘lsa,

A) 2

B) 2/3

C) 1

D) 3/2

28 / 40

A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping.

A) 15

B) 8

C) 20

D) 10

29 / 40

Tenglamalar sistemasining ildizlari yig‘indisini toping.

A) 10

B) 12

C) 14

D) 20

30 / 40

Rasmdagi shakl perimetrini toping.

A) 32

B) 24

C) 28

D) 30

31 / 40

Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping.

A) 64

B) 8(π+2)

C) 32

D) 16(π-2)

32 / 40

a+b+c=10, bo‘lsa, ni toping.

A) 4

B) 5

C) 6

D) 11

33 / 40

Tenglamalar sistemani yeching:

A) (7; 2), (28; -1)

B) (9; 0), (2; 7)

C) (9; 0), (28; -1)

D) (2; 3)

34 / 40

Asoslari 5 va 5√7 ga teng bo‘lgan trapetsiyaning yuzini teng ikkiga bo‘luvchi kesma asoslarga parallel. Shu kesma uzunligini toping.

A) 8

B) 10

C) 6√7

D) 4√7

35 / 40

ABC uchburchakning A burchgi 30° ga, B burchagi 75° ga teng. B uchidan AC tomonga BD kesma o‘tkazilgan. ABD burchak 45° ga teng bo‘lsa, quyidagilardan qaysi biri noto‘g‘ri?

A) DC < AD

B) BC > AD

C) AB = BC

D) BD = BC

36 / 40

To‘g‘ri to‘rtburchakning 16 ga teng diagonali yon tomoni bilan 15° li burchak tashkil etadi. To‘rtburchak yuzini toping.

A) 64

B) 42

C) 32

D) 48

37 / 40

ABC o‘tkir burchakli uchburchakning BC asosiga AD balandlik, AC yon tomoniga BE balandlik o‘tkazilgan. Bunda CE =2AE = 8, DC = 6 bo‘lsa, BD ni toping.

A) 6

B) 10

C) 12

D) 8

38 / 40

Hisoblang: