Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №6 Matematika attestatsiya Matematika attestatsiya №6 InfoMaster Yanvar 30, 2022 69 Ko'rishlar 2 izohlar SaqlashSaqlanganOlib tashlandi 0 0% 0 ovozlar, 0 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №6 1 / 40 Jaloliddin uyiga tezroq qaytish uchun qaysi yo’lni tanlash kerak. A) Ikkinchi yo'l B) Birinchi yo'l C) To'rtinchi yo'l D) Uchunchi yo'l 2 / 40 tengsizlikni yeching. A) x∈(-∞;5) B) x∈(-3;1) C) x∈(-5;1) D) x∈(-∞;5)∪(-3;1) 3 / 40 funksiyaning [2; 3] kesmadagi eng katta qiymatini A) 3 B) 4,5 C) 4 D) 7,5 4 / 40 Qadam uzunligi deb biricnhi iz tovon oxiridan ikkinchi iz tovon oxirigacha bo'lgan masofaga aytiladi. Erkak kishi yurayotganda uning qadami va qadamlar soni orasidagi bog'lanish quyidagi formula bilan ifodalanadi: (n/P) = 140. Bu yerda n – bir minutdagi qadamlar soni. P – qadam uzunligi (m). Hikmat 1 minutda 70 qadam bossa, formula yordamida uning qadami uzunligini toping. A) 0,6 m yoki 60 cm B) 0,5 m yoki 50 cm C) 0,7 m yoki 70 cm D) 0,9 m yoki 90 cm 5 / 40 Eldor velosipetida 13m/s bilan 3 soniyada 15m tepalikka ko`tarilgan bo`lsa AC kesma uzunligi necha metr? A) 32 B) 36 C) 40 D) 34 6 / 40 Hisoblang: -121+(-135)-(-1)28 A) 0 B) -3 C) -1 D) -2 7 / 40 Radiusi 25/4 bo‘lgan sferaga balandligi 8 ga teng bo‘lgan konus ichki chizilgan. Konusning hajmini toping. A) 72π B) 144π C) 96π D) 192π 8 / 40 To‘g‘ri burchakli uchburchakning bir burchagi 52° ga teng bo‘lsa, to‘g‘ri burchak uchidan tushirilgan balandlik va mediana orasidagi burchakni toping. A) 17° B) 24° C) 14° D) 7° 9 / 40 Ayniyatdan foydalanib x + y + z ni toping: A) 9 B) 6 C) 12 D) 3 10 / 40 Yig‘indini toping. A) 1/4 B) 1/6 C) 1/9 D) 1/12 11 / 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) 105 D) 96 12 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 6 B) 8 C) 7 D) 5 13 / 40 sonlarini taqqolsang. A) b B) c C) a D) c 14 / 40 a+b+c=10, bo‘lsa, ni toping. A) 6 B) 11 C) 4 D) 5 15 / 40 Tenglamalar sistemani yeching: A) (2; 3) B) (9; 0), (28; -1) C) (7; 2), (28; -1) D) (9; 0), (2; 7) 16 / 40 Hisoblang: A) 9/10 B) 19/20 C) 20/21 D) 10/11 17 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 3√2 B) 4√3 C) 2√3 D) 2√2 18 / 40 Asoslari 5 va 5√7 ga teng bo‘lgan trapetsiyaning yuzini teng ikkiga bo‘luvchi kesma asoslarga parallel. Shu kesma uzunligini toping. A) 6√7 B) 4√7 C) 10 D) 8 19 / 40 Barcha ikki xonali sonlar ko‘paytmasi 4 ning qanday eng katta darajasiga bo‘linadi? A) 42 B) 44 C) 45 D) 43 20 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 3 B) 4 C) 1 D) 2 21 / 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) -3 B) -4 C) -6 D) 5 22 / 40 Rasmdagi shakl perimetrini toping. A) 30 B) 24 C) 32 D) 28 23 / 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 24 / 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) 1 D) 2 25 / 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) 42 C) 48 D) 64 26 / 40 Tenglamani yeching: A) -6 B) -6; 5 C) 6; -5 D) 5 27 / 40 √3 A) 5/5 B) 3/2 C) 7/3 D) 21/10 28 / 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) 24 C) 35 D) 14 29 / 40 bo‘lsa, A) 2/3 B) 3/2 C) 1 D) 2 30 / 40 Hisoblang: A) 1/32 B) -√3/2 C) 0 D) -1/2 31 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 20 B) 10 C) 12 D) 14 32 / 40 x²-√11x+1=0 0 bo‘lsa, A) 10 B) 12 C) 9 D) 11 33 / 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 34 / 40 Hisoblang: A) sin10° B) cos10° C) cos50° D) 1 35 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 1 B) 4 C) 2 D) 3 36 / 40 Tengsizlik nechta butun juft yechimga ega? A) 110 B) 112 C) 115 D) 116 37 / 40 Ifodaning qiymatini toping. A) 0,04 B) 0,0(4) C) 0,(04) D) 0,0(2) 38 / 40 A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping. A) 20 B) 8 C) 10 D) 15 39 / 40 1+sin 2x = 7(sin x + cos x) tenglamani yeching. A) π/4+πk, k∈Z B) Ø C) -π/4+πk, k∈Z D) 0 40 / 40 Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping. A) 16(π-2) B) 8(π+2) C) 64 D) 32 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