Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 165 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping A) 10 B) 11 C) √46 D) 12 2 / 30 ABCD kvadrat ichidan olingan O nuqtadan A, B, C uchlarigacha bo’lgan masofalar mos ravishda 3, 4, 5 ga teng bo’lsa, u holda OD kesma uzunligini toping. A) 6 B) 3 C) √32 D) √37 3 / 30 Hisoblang: A) 2 B) -1 C) 3 D) 1 4 / 30 Agar va b-a=4 bo’lsa, a+b ni toping. A) 3,5 B) 2 C) 4 D) 3 5 / 30 Teng yonli trpetsiyaning asoslari 15 va 25 ga balandligi esa 15 ga teng trapetsiyaning dioganalini toping A) 30 B) 25 C) 28 D) 20 6 / 30 Radiusi 1 ga teng aylana uchta yoyga bo`lingan. Ularga mos markaziy burchaklar 1, 2 va 6 sonlariga proporsional. Yoylardan eng kattasining uzunligini toping. A) 4π/3 B) 3π/4 C) 3π/2 D) 2π/3 7 / 30 y= funktsiyaning aniqlanish sohasini toping. A) [2;∞) B) (-∞;2)v(2;∞) C) (2;∞) D) (-∞;2) 8 / 30 Tomoni 6 ga teng bo`lgan teng tomonli uchburchakga tashqi chizilgan doiraning yuzini toping. A) 7 π B) 10 π C) 6 π D) 12π 9 / 30 Anvar tub son o’yladi va o’ylagan sonini 5 ga ko’paytirib, 8 ni ayirgan edi, yana tub son hosil bo’ldi. Anvar qanday son o’ylagan? A) 412967 B) 2 C) aniqlab bo’lmaydi D) 374389 10 / 30 sistemadan x+y+z ning qiymatini toping. A) 139/41 B) 140/41 C) -139/41 D) 150/41 11 / 30 Rombning balandligi 8 ga dioganallarining ko’paytmasi 80 ga teng. Rombning perimetrini toping A) 24 B) 32 C) 20 D) 16 12 / 30 a=sin 1; b=sin 2; c=sin 3; d=sin 4 va e=sin 5 sonlarni kamayish tartibida joylashtiring. A) a>b>c>d>e B) e>b>a>d>c C) b>c>a>d>e D) b>a>c>d>e 13 / 30 A) 1 B) 2 C) 0 D) √6 14 / 30 funksiya uchun quyidagi mulohazalardan qaysi biri o’rinli? A) juft ham emas, toq ham emas funksiya B) toq funksiya C) bunday funksiya mavjud emas D) juft funksiya 15 / 30 Bankda qo`yilgan pul bir yildan kegin foydasi bilan 2600 so`m bo`ldi; Agar bank yillik 30% foyda to`lasa, boshida qancha pul qo`yilgan bo`ladi ? A) 2000 B) 2200 C) 1900 D) 2100 16 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 6 B) 3 C) 5 D) 4 17 / 30 Konsert zalining birinchi qatorida 40 ta o’rindiq bor. Har bir keyingi qatordagi o’rindiqlar soni oldingi qatordan 4 ga ko’p. Agar konsert zalida jami 40 ta qator bo’lsa, u holda shu zaldagi barcha o’rindiqlar sonini toping. A) 4680 B) 4716 C) 4720 D) 4760 18 / 30 Tengsizlikni yeching. A) 0 B) (0;+∞) C) (-1/³ √2 ;0) D) to'g'ri javob yo'q 19 / 30 Agar x,y sonlar (x+5)2+(y-12)2=142 tenglikni qanoatlantirsa, x2+y2 ifodaning eng kichik qiymatini toping. A) √2 B) √3 C) 1 D) 2 20 / 30 tenglamani yeching. A) 2017 B) 0 C) 2019 D) 2018 21 / 30 sistemada xy ning qiymatini toping. A) 64 B) 80 C) 60 D) 75 22 / 30 Agar bank qo’yilgan pulga 40% yillik bersa, qo’yilgan 4500 so’m pul bir yildan so’ng qancha bo’ladi? A) 6300 B) 6000 C) 6200 D) 6100 23 / 30 a(x+2)=2x+1 tenglama a ning qanday qiymatida yechimga ega emas? A) (∞;∞) B) (2;∞) C) (-∞;2)v(2;∞) D) (-∞;2) 24 / 30 Soddalashtiring. A) a+1 B) 2018 C) 2018a/a+1 D) 2019 25 / 30 sonning oxirgi raqamini toping. A) 2 B) 4 C) 6 D) 8 26 / 30 Qaysi javobda faqat juft funksiyalar ko’rsatilgan? A) 1,4 B) 2, 3 C) 3, 4 D) 1, 3 27 / 30 ABCD to’gri to’rtburchak ichidan olingan O nuqtadan A, B, C, D uchlarigacha bo’lgan masofalar mos ravishda 1; 2; 1,5; 1,2 ga teng bo’ladigan barcha to’g’ri to’rtburchaklar sonini toping A) cheksiz ko’p B) 0 C) 2 D) 1 28 / 30 Fazoda (1;2;3) nuqtalardan o’tuvchi to’g’ri chiziq tenglamasini tuzing. A) 2x+3y+z=0 B) x=y/3=z/2 C) -x-y+z=0 D) x-1/2=y-2/3=z-3/4 29 / 30 Markazi nuqtada bo‘lgan aylanaga va urinmalar o‘tkazilgan bo’lib, va nuqtalar urinish nuqtalari bo’lsin. Aylanadagi Q nuqtadan o‘tkazilgan uchinchi urinma va kesmalarni X va Y nuqtalarda kesib o‘tadi. Agar uchburchakka ichki chizilgan aylana markazi bo‘lsa, u holda burchakni toping. A) 90° B) 30° C) 72° D) 60° 30 / 30 O’q kesimining diagonallari o’zaro perpendikulyar bo’lgan kesik konus yasovchisi va asos tekisligi orasidagi burchak ga teng. Agar o’q kesimining diagonali ga teng bo’lsa, kesik konus asosining yuzini toping. A) πa²/4sin²α B) πa²(1+2cosα/4sin²) C) πa²(1-2sinα/4sin²) D) πa²/4cos²α 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
