Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №6 Matematika attestatsiya Matematika attestatsiya №6 InfoMaster Yanvar 30, 2022 46 Ko'rishlar 2 izohlar SaqlashSaqlanganOlib tashlandi 0 0% 0 ovozlar, 0 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №6 1 / 40 Quyidagi javoblardan qaysi biri tengsizlikni yechimi bo'la oladi? A) 3 B) 4 C) 2 D) 1 2 / 40 Abdulla, Samandar va Jamshid 3 ta stulga necha xil usulda utirishi mumkin? A) 6 B) 7 C) 9 D) 27 3 / 40 Muntazam uchburchak ichidan olingan nuqtadan uchburchak tomonlarigacha bo'lgan masofalar mos holda c(2;3;1), b(1;2;1) va a(1;2;3) vektorlarning absolut qiymatlariga teng bo'lsa, uchburchakning balandligini toping. A) 2√14+√6 B) 16 C) 18 D) √6+√14 4 / 40 y=f(x) funksiya uchun tenglik o'rinli bo'lsa, f(π/4)=? A) 1/4 B) 4 C) 1/5 D) 1/3 5 / 40 Soddalshtiring. A) 2 B) -1 C) 6 D) 1 6 / 40 Hisoblang? A) 2 B) 4 C) 0,5 D) 0,25 7 / 40 275+330 sonini 41 ga bo‘lganda qoladigan qoldiqni aniqlang. A) 1 B) 9 C) 5 D) 0 8 / 40 Agar x =17 bo‘lsa, quyidagi ifodaning qiymatini toping. A) 4 B) -√17 C) √17 D) -4 9 / 40 Agar bo‘lsa, m/n ni toping. A) 4 B) 2 C) 3 D) (1+√5)/2 10 / 40 cos75° ·cos 45° ·cos15° ni hisoblang. A) √2/8 B) √3/4 C) √3/8 D) 1/8 11 / 40 Hisoblang: A) cos50° B) cos10° C) 1 D) sin10° 12 / 40 Kvadratlarning yuzlari yig‘indisini toping. A) 121 B) 22 C) berilganlar yetarli emas D) 11 13 / 40 Hisoblang: A) 10/11 B) 19/20 C) 9/10 D) 20/21 14 / 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 15 / 40 kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping. A) 5 B) 10 C) 11 D) 7 16 / 40 b ning qanday qiymatlarida M(2;1) nuqtadan 4x - 3y + b = 0 to‘g‘ri chiziqqacha masofa 2 ga teng bo‘ladi? A) 3 yoki –8 B) 5 yoki –15 C) 4 yoki –12 D) 6 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) 96 B) 25 C) 100 D) 105 18 / 40 sonlari geometrik progressiyaning ketma-ket hadlari bo‘ladigan barcha n larning yig‘indisini (agar bitta qiymati bo‘lsa, o‘zini) toping. A) 23 B) 35 C) 24 D) 14 19 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 14 B) 12 C) 10 D) 20 20 / 40 a+b+c=10, bo‘lsa, ni toping. A) 4 B) 5 C) 6 D) 11 21 / 40 tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping. A) -2 B) 3 C) 8 D) 5 22 / 40 x²-√11x+1=0 0 bo‘lsa, A) 10 B) 11 C) 12 D) 9 23 / 40 bo‘lsa, A) 2/3 B) 2 C) 1 D) 3/2 24 / 40 sonlarini taqqolsang. A) c B) a C) b D) c 25 / 40 A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping. A) 15 B) 10 C) 8 D) 20 26 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 3√2 B) 4√3 C) 2√3 D) 2√2 27 / 40 Chizmadan foydalanib α ni toping. A) 40° B) 50° C) 20° D) 30° 28 / 40 Ifodaning qiymatini quyidagi sonlardan qaysi biriga teng. 319·25 316·28 313·28 329·28 A) 1 B) 2 C) 3 D) 4 29 / 40 √3 A) 3/2 B) 5/5 C) 21/10 D) 7/3 30 / 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) 10 B) 8 C) 6 D) 12 31 / 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) bd/c C) cd/b D) bc/d 32 / 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) -4 B) 5 C) -3 D) -6 33 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 5 B) 6 C) 8 D) 7 34 / 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) AB = BC C) BC > AD D) BD = BC 35 / 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) 4 B) 3 C) 2 D) 1 36 / 40 Hisoblang: A) 1/32 B) -1/2 C) 0 D) -√3/2 37 / 40 Rasmdagi shakl perimetrini toping. A) 30 B) 28 C) 24 D) 32 38 / 40 Tengsizlik nechta butun juft yechimga ega? A) 110 B) 112 C) 116 D) 115 39 / 40 Asoslari 5 va 5√7 ga teng bo‘lgan trapetsiyaning yuzini teng ikkiga bo‘luvchi kesma asoslarga parallel. Shu kesma uzunligini toping. A) 10 B) 4√7 C) 6√7 D) 8 40 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 4 B) 1 C) 3 D) 2 O'rtacha ball 0% 0% Testni qayta ishga tushiring Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
Istaklar ro'yxatiga qo'shildiIstaklar ro'yxatidan olib tashlandi 13 Matematika fanidan attestatsiya savollari №16