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Matematika attestatsiya №5

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Yanvar 24, 2022

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Matematika fanidan attestatsiya savollari №5

1 / 40

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

A) 2

B) 2,5

C) 3

D) 4

2 / 40

ni qanoatlantiruvchi yechimlar nechta?

A) 3

B) 2

C) 0

D) 1

3 / 40

Muntazam oltiburchakning tomoni 2√6 ga teng. Shu ko‘pburchakka tengdosh bo‘lgan teng tomonli uchburchakning tomonini toping.

A) 24

B) 12

C) 18

D) 30

4 / 40

Uchlari A(4;5;1), B(2;3;0) va C(2;–1;–3) nuqtalarda joylashgan uchburchakning BD medianasi uzunligini toping.

A) 1

B) √2

C) 2

D) √3

5 / 40

Aniq integralni hisoblang:

A) 3

B) 0,5

C) 2

D) 0

6 / 40

Umumiy hadi x_n=3+5n-n² formula bo'yicha berilgan sonli ketma-ketlikning eng katta hadi 15 dan qanchaga kam?

A) 5,75

B) 9,25

C) 13

D) 6

7 / 40

Quyida berilgan ko`rinishda mashinadan yuknitushurish uchun necha metr taxta kerak ?

A) 13

B) 12

C) 11

D) 14

8 / 40

Shaharlar orasidagi masofa xaritada 5 sm ga teng bo`lsava xaritaning masshtabi 1:4000000 bo`lsa , shaharlar orasidagi haqiqiy masofa qanchaga teng ?

A) 20 km

B) 2 km

C) 200 km

D) 0,2km

9 / 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) 10

B) 11

C) 12

D) 13

10 / 40

Agar bo'lsa, ni m orqali ifodalang.

A) (m+4)/4

B) 2/m

C) (m+1)/2

D) (4-m)/4

11 / 40

1, 1, 2, 3, 5, 8, 13, … ketma-ketlikning umumiy hadi an bo‘lsa, quyidagilardan qaysi biri to‘g‘ri?

A) 2

B) 4

C) 1

D) 3

12 / 40

Ayniyatdan foydalanib x + y + z ni toping:

A) 6

B) 12

C) 3

D) 9

13 / 40

ABC to‘g‘ri burchakli uchburchakning og‘irlik markazi G nuqta. Bunda AB ⊥ BC, AG ⊥ GB va AG = 8 bo‘lsa, BG ni toping.

A) 4√2

B) 3√2

C) 6

D) 4√3

14 / 40

Hisoblang:

A) 10/11

B) 19/20

C) 20/21

D) 9/10

15 / 40

tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping.

A) -2

B) 5

C) 3

D) 8

16 / 40

a+b+c=10, bo‘lsa, ni toping.

A) 6

B) 5

C) 4

D) 11

17 / 40

3x3 o‘lchamli kvadratning tugunlarida 16 ta nuqta belgilanib, ularning o‘ng tomondan eng yuqorisidagi A bilan belgilangan. Bir uchi A nuqtada, qolgan uchlari qolgan 15 ta nuqtada orasidan tanlanadigan uchburchaklarning sonini toping.

A) 100

B) 25

C) 96

D) 105

18 / 40

Sonlarining o‘rta geometrik qiymatini toping.

A) 4√3

B) 2√2

C) 3√2

D) 2√3

19 / 40

Tenglamani yeching:

A) 5

B) -6

C) -6; 5

D) 6; -5

20 / 40

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

A) 13

B) 12

C) 15

D) 14

21 / 40

bo‘lsa,

A) 2

B) 3/2

C) 1

D) 2/3

22 / 40

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

A) 1

B) bc/d

C) bd/c

D) cd/b

23 / 40

Ifodaning qiymatini quyidagi sonlardan qaysi biriga teng.

3¹⁹·2⁵
3¹⁶·2⁸
3¹³·2⁸
32⁹·2⁸

A) 2

B) 3

C) 4

D) 1

24 / 40

Tenglamalar sistemasining ildizlari yig‘indisini toping.

A) 20

B) 14

C) 10

D) 12

25 / 40

x²-√11x+1=0 0 bo‘lsa,

A) 12

B) 10

C) 11

D) 9

26 / 40

sonlarini taqqolsang.

A) c

B) b

C) c

D) a

27 / 40

Ifodaning qiymatini toping.

A) 0,04

B) 0,0(4)

C) 0,(04)

D) 0,0(2)

28 / 40

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

A) 32

B) 16(π-2)

C) 64

D) 8(π+2)

29 / 40

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

A) 32

B) 48

C) 42

D) 64

30 / 40

vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping.

A) 4

B) 3

C) 1

D) 2

31 / 40

Kvadratlarning yuzlari yig‘indisini toping.

A) 11

B) berilganlar yetarli emas

C) 22

D) 121

32 / 40

Hisoblang:

A) cos10°

B) sin10°

C) cos50°

D) 1

33 / 40

1+sin 2x = 7(sin x + cos x) tenglamani yeching.

A) 0

B) Ø

C) π/4+πk, k∈Z

D) -π/4+πk, k∈Z

34 / 40

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

A) 4

B) 2

C) 3

D) 1

35 / 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

36 / 40

P(x)=x¹ºº ko‘phadni x³-3x+2 ga bo‘lganda qoladigan qoldiqni toping.

2¹ºº-1
(2¹ºº-1)x+2(2⁹⁹-1)
(2¹ºº-1)x-2(2⁹⁹-1)
2¹ººx-3·2¹⁰⁰

A) 3

B) 1

C) 2

D) 4

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) 12

C) 8

D) 10

38 / 40

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