Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №5 Matematika attestatsiya Matematika attestatsiya №5 InfoMaster Yanvar 24, 2022 64 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 2 ovozlar, 1 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №5 1 / 40 Kamayish tartibida joylashtiring. a)5√6 b)12 c)4√10 A) c>a>b B) a>b>c C) b>a>c D) a>c>b 2 / 40 Jaloliddin uyiga tezroq qaytish uchun qaysi yo’lni tanlash kerak. A) Birinchi yo'l B) To'rtinchi yo'l C) Ikkinchi yo'l D) Uchunchi yo'l 3 / 40 Teng yonli trapetsiya asosidagi burchakning sinusi 0,6 ga, asoslarining ayirmasi 4 ga teng bo‘lsa, trapetsiyaning yon tomonini toping. A) 3 B) 2 C) 4 D) 2,5 4 / 40 Tenglama ildizlari ayirmasining modulini toping. A) √5 B) 5 C) √5+5 D) 2√5 5 / 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) 16 B) √6+√14 C) 2√14+√6 D) 18 6 / 40 f(x-1)+f(x+2)=2(x²+7) ekani ma‟lum bo'lsa, f(x) ko'phadni toping. A) f(x)+2x²-1 B) f(x)=x²-x+5 C) f(x)=x²-4 D) f(x)=x²+3x+7 7 / 40 Eldor velosipetida 13m/s bilan 3 soniyada 15m tepalikka ko`tarilgan bo`lsa AC kesma uzunligi necha metr? A) 32 B) 40 C) 36 D) 34 8 / 40 Quyidagi berilgan qoldiqli bo`lishlardan qaysi biri to`g`ribajarilgan ? a)76 : 9 = 7 (qoldiq 13); b)20 : 7 = 2 (qoldiq 6); d)54 : 15 = 4 (qoldiq 1) A) b B) a,b C) )a,b,d D) b,d 9 / 40 Quyida berilgan tizimda yonma - yon shakillardagisonlar ustida pasdagi ( EKUB yoki EKUK topish) amali bajariladi , natija amal pastigi shakilga yoziladi Shunga ko`ra 1 , 2 va 3 amallar EKUB yoki EKUK dan qaysi biri bo`lishi kerak A) 1-EKUB 2-EKUK 3-EKUK B) 1-EKUB 2-EKUK 3-EKUB C) 1-EKUB 2-EKUB 3-EKUK D) 1-EKUB 2-EKUB 3-EKUB 10 / 40 cos75° ·cos 45° ·cos15° ni hisoblang. A) √3/4 B) 1/8 C) √3/8 D) √2/8 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) 3 C) 4 D) 1 12 / 40 Tenglamani yeching: A) 9/5 B) 5*4 C) 5/9 D) 6/5 13 / 40 Hisoblang: A) 19/20 B) 20/21 C) 9/10 D) 10/11 14 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 3 B) 2 C) 4 D) 1 15 / 40 tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping. A) 5 B) 8 C) 3 D) -2 16 / 40 kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping. A) 7 B) 10 C) 11 D) 5 17 / 40 Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping. A) 8(π+2) B) 32 C) 64 D) 16(π-2) 18 / 40 Hisoblang: A) -√3/2 B) -1/2 C) 0 D) 1/32 19 / 40 Ifodaning qiymatini toping. A) 0,0(4) B) 0,(04) C) 0,0(2) D) 0,04 20 / 40 √3 A) 3/2 B) 7/3 C) 5/5 D) 21/10 21 / 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) 6 B) 4√2 C) 3√2 D) 4√3 22 / 40 sonlarini taqqolsang. A) b B) c C) a D) c 23 / 40 Kvadratlarning yuzlari yig‘indisini toping. A) 121 B) 22 C) berilganlar yetarli emas D) 11 24 / 40 Hisoblang: A) sin10° B) cos50° C) cos10° D) 1 25 / 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) 4√7 D) 6√7 26 / 40 Barcha ikki xonali sonlar ko‘paytmasi 4 ning qanday eng katta darajasiga bo‘linadi? A) 44 B) 43 C) 42 D) 45 27 / 40 bo‘lsa, A) 3/2 B) 2/3 C) 2 D) 1 28 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 3√2 B) 4√3 C) 2√3 D) 2√2 29 / 40 A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping. A) 8 B) 15 C) 10 D) 20 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) 12 B) 10 C) 6 D) 8 31 / 40 Chizmadan foydalanib α ni toping. A) 40° B) 50° C) 20° D) 30° 32 / 40 To‘g‘ri to‘rtburchakning 16 ga teng diagonali yon tomoni bilan 15° li burchak tashkil etadi. To‘rtburchak yuzini toping. A) 48 B) 42 C) 32 D) 64 33 / 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 34 / 40 x²-√11x+1=0 0 bo‘lsa, A) 12 B) 9 C) 10 D) 11 35 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 14 B) 10 C) 20 D) 12 36 / 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) -6 D) -3 37 / 40 Tengsizlik nechta butun juft yechimga ega? A) 110 B) 115 C) 112 D) 116 38 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 7 B) 5 C) 8 D) 6 39 / 40 Tenglamani yeching: A) 6; -5 B) 5 C) -6; 5 D) -6 40 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 2 B) 4 C) 1 D) 3 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 14 Matematika fanidan attestatsiya savollari №16
