Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 129 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 4 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 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) 2 B) 1 C) cheksiz ko’p D) 0 2 / 30 x(t)=t2+7t-6 qonuniyat bo’yicha harakatlanayotgan moddiy nuqtaning tezligi harakat boshlangandan necha sekund o’tgach 87 m/s ga teng bo’ladi? A) 40 B) 50 C) 54 D) 36 3 / 30 Agar arctga+ arctgb + arctgc= bo’lsa, a+b+c ni toping. A) abc B) ab/c C) 1 D) ab 4 / 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) 2100 D) 1900 5 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 5 ga, bir katetining gipotenuzadagi proyeksiyasi 1,6 ga teng. Ikkinchi katetning kvadratini toping. A) 16 B) 17 C) 14 D) 18 6 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 13 ga, katetlaridan biri 52 ga teng. Gipotenuzaga tushirilgan balandlik uzunligini toping A) 4 B) 5 C) 7 D) 6 7 / 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) 2π/3 B) 4π/3 C) 3π/2 D) 3π/4 8 / 30 To‘g‘ri to‘rtburchakning eni 25% ga orttirildi, bo‘yi esa 25% ga kamaytirildi. Natijada uning yuzi qanday o‘zgardi? A) 6,25% ga kamayadi B) 6,25% ga ortadi C) o‘zgarmaydi D) 2,5% ga ortadi 9 / 30 Soddalashtiring. (0<m<7) A) 2m-7 B) 7-2m C) m D) 7 10 / 30 Arifmetik progressiyada a17=33 va a45=89. Progressiyaning birinchi hadi hamda ayirmasining o’rta geometrigini toping. A) 2 B) 4 C) 2√2 D) √2 11 / 30 Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusidan katta bo’lgan uchburchaklar sonini toping. A) cheksiz ko’p B) 2019 C) 0 D) 1 12 / 30 Markazi O nuqtada bo‘lgan aylanaga PA va PB urinmalar o‘tkazilgan bo’lib, A va B nuqtalar urinish nuqtalari bo’lsin. Aylanadagi Q nuqtadan o‘tkazilgan uchinchi urinma PA va PB kesmalarni X va Y nuqtalarda kesib o‘tadi. Agar XQ=YQ bo‘lsa, u holda PXY uchburchak qanday uchburchak bo‘ladi? A) teng yonli uchburchak B) ixtiyoriy uchburchak C) muntazam uchburchak D) to`g`ri burchakli uchburchak 13 / 30 y=cos(2sinx) funksiyaning qiymatlar sohasini toping. A) [cos2;1] B) [0;1] C) [0;cos2] D) [-1;1] 14 / 30 To’rtburchakning uchi M (0, 4) ; N(-4,0) ; P(-3;2) uchlari berilgan. Agar bo’lsa, Q uchining koordinatalarini toping. A) (-7; 1) B) (4; -3) C) (7; 1) D) (-7;-1) 15 / 30 200 kishidan iborat turistlar guruxida 140 kishi ingliz tilini, 90 kishi nemis tilini va 46 kishi ikkala tilni biladi. Ikkala tilni xam bilmaydigan turistlar necha foizni tashkil qiladi. A) 12 B) 16 C) 4 D) 8 16 / 30 Tenglamaning ildizlari yig’indisi va ko’paytmasining yig’indisini toping. |x+1|.|x-4|=5 A) -6 B) 6 C) 9 D) -3 17 / 30 To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping A) √46 B) 12 C) 10 D) 11 18 / 30 To’rtburchakli muntazam piramidaning yon qirrasidagi ikki yoqli burchak 120 ga teng. Diagonal kesimining yuzasi S ga teng bo’lsa, uning yon sirtini toping. A) 3S B) 2S C) 0,5S D) 4S 19 / 30 sistemada xy ning qiymatini toping. A) 60 B) 75 C) 64 D) 80 20 / 30 tenglamalar sistemasini yeching A) (3;–4) B) (–4;3) C) (4;–3) D) (4;3) 21 / 30 Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusiga teng bo’lgan uchburchak qanday uchburchak deyiladi? A) teng tomonli uchburchak B) o’tmas burchakli uchburchak C) to’gri burchakli uchburchak D) o’tkir burchakli uchburchak 22 / 30 “Tutgan balig‘ining og‘irligi qancha?” degan savolga baliqchi: “Baliqning dumi 3 kg, boshi uning dumi hamda tanasi yarmining og‘irligiga teng, tanasi esa boshi va dumining og‘irligiga teng”, deb javob berdi. Baliqning og‘irligini (kg) toping. A) 6 B) 12 C) 18 D) 3 23 / 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) πa²/4cos²α B) πa²(1+2cosα/4sin²) C) πa²(1-2sinα/4sin²) D) πa²/4sin²α 24 / 30 Fazoda (1;2;3) nuqtalardan o’tuvchi to’g’ri chiziq tenglamasini tuzing. A) -x-y+z=0 B) 2x+3y+z=0 C) x-1/2=y-2/3=z-3/4 D) x=y/3=z/2 25 / 30 Radiuslari orasidagi burchagi 36o va radiusi 5 ga teng bo`lgan sektor yoyining uzunligini toping. A) π/2 B) 2π/3 C) π D) 2π 26 / 30 Ifodani soddalashtiring. 2 cos55o.cos40o.sin55o+cos110o.sin40o A) 2 B) 1 C) 0 D) 0,5 27 / 30 ABC muntazam uchburchak ichidan ixtiyoriy P nuqta olinib, undan BC, CA va AB tomonlarga mos ravishda PD, PE va PF perpendikulyarlar tushirilgan bo’lsa,ni toping. A) 0,5 B) 1/√2 C) 1/√3 D) 1 28 / 30 Parallelogrammning tomonlari nisbati 3:5 kabi. Agar parallelogrmmning perimetri 48 ga burchaklaridan biri 1200 ga teng bo’lsa, uning yuzini toping. A) 135√3/4 B) 67,5 C) 67,5√3 D) 48√3 29 / 30 Ostki asosining yuzi 16π ga va ustki asosining yuzi 4π ga teng bo‘lgan kesik konus berilgan. Agar kesik konusga shar ichki chizilgan bo‘lsa, u holda sharning hajmini toping. A) 64√2π/3 B) (5√3+3)π/3 C) 8√2π/5 D) 3√2π/4 30 / 30 Bir vaqtning o’zida 9,13, . . . ,405 va 15,21, . . . ,255 ketma–ketliklarning hadlari bo’lgan sonlarning eng kattasi va eng kichigining ayirmasini toping A) 150 B) 231 C) 228 D) 147 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
