Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 225 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Agar va b-a=4 bo’lsa, a+b ni toping. A) 4 B) 3 C) 2 D) 3,5 2 / 30 Teng yonli uchburchakning tomonlari 5, 5 va 6 ga teng. Bu uchburchakning bissektiritsalari va medianalari kesishgan nuqtalar A) 1/2 B) 1,2 C) 1 D) 1/6 3 / 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) 2200 B) 1900 C) 2000 D) 2100 4 / 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) 1 B) 1/√2 C) 0,5 D) 1/√3 5 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -1 B) -0,5 C) -0,75 D) -0,25 6 / 30 y= funksiyaning aniqlanish sohasini toping A) (-∞;0)v(2;∞) B) (0,2) C) (-∞;0])v[2;∞) D) (2;∞) 7 / 30 Agar funksiya berilgan bo’lsa, u holda M=ning qiymatini toping. A) e⁻⁴⁹⁵⁰ B) e⁻⁵⁰⁵⁰ C) e⁴⁹⁵⁰ D) e⁵⁰⁵⁰ 8 / 30 Tenglamani yeching. |x2-11x+10|=x2-11x+10 A) (-∞;1] B) (-∞;1]v[10;∞) C) [10;∞) D) 1; 10 9 / 30 A) 0 B) 2 C) 1 D) √6 10 / 30 Tomoni 2 ga teng kvadratga tashqi chizilgan aylana uzunligini toping. A) 2π/3 B) 3π C) 4π D) 2π 11 / 30 y= funktsiyaning aniqlanish sohasini toping. A) (2;∞) B) (-∞;2)v(2;∞) C) (-∞;2) D) [2;∞) 12 / 30 Hisoblang. A) 17/34 B) 2/17 C) 15/34 D) 2/34 13 / 30 Hisoblang A) √3/2 B) 1/2 C) √3 D) 2 14 / 30 O’qishni bilmaydigan bola alifbening A,A,A, N,N, S- 6 ta harflarini ixtiyoriy ravishda terib chiqadi. Bunda ANANAS so’zining hosil bo’lish ehtimolini toping. A) 5/24 B) 1/60 C) 5/720 D) 6/720 15 / 30 Parallelogrammning yuzi 213 ga, tomonlaridan biri 7 ga va o’tkir burchagi 600 ga teng bo’lsa, ikkinchi tomonini toping A) 6 B) 4 C) 5 D) 8 16 / 30 Tengsizlikni yeching. A) (-3;2) B) (2;4) C) (-4;2)v(2;3) D) (-3;2)v(2;4) 17 / 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) (5√3+3)π/3 B) 8√2π/5 C) 3√2π/4 D) 64√2π/3 18 / 30 Yuzasi 10 ga teng bo’lgan kvadratning ketma-ket ikki uchidan o’tuvchi aylana chizilgan. Uchinchi uchidan aylanaga urunma o’tkazilgan. Urunma tomondan ikki marta katta bo’lsa, aylana radiusini toping. A) 4 B) 6 C) 10 D) 5 19 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 4 B) 3 C) 5 D) 6 20 / 30 Agar arctga+ arctgb + arctgc= bo’lsa, a+b+c ni toping. A) 1 B) ab C) ab/c D) abc 21 / 30 Perimetri 60 ga teng bo’lgan parallelogrammning tomonlari nisbati 2:3 ga, o’tkir burchagi esa 300 ga teng. Parallelogrammning yuzini toping. A) 52√3 B) 108 C) 54 D) 48√3 22 / 30 sistemadan x+y+z ning qiymatini toping. A) -139/41 B) 150/41 C) 139/41 D) 140/41 23 / 30 Uchburchakning balandligi 12 ga teng bo’lib, u asosni 5:16 nidbatda bo’ladi. Agar asosning uzunligi 21 ga teng bo’lsa, uchburchakning perimetrini toping A) 52 B) 54 C) 48 D) 108 24 / 30 Quyidagilardan qaysi biri barcha lar uchun ma’noga ega (aniqlangan)? A) ²ᴷ⁺⁴v2k+1/k²+1 B) ⁴ᴷ⁺³√-√2k+1 C) (512-1/2⁻⁹)° D) 4k+1/1/8:7-1/56 25 / 30 tenglamalar sistemasini yeching A) (4;3) B) (–4;3) C) (3;–4) D) (4;–3) 26 / 30 To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping A) √46 B) 12 C) 11 D) 10 27 / 30 Ushbu (y6+y3+1)(y3+1)(y3-1)-y6+y3+1 ifodani soddalashtish natijasida ko’phad hosil qilindi. Uning nechta hadi bor? A) 1 B) 3 C) 2 D) 4 28 / 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) 0,5S B) 2S C) 4S D) 3S 29 / 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) 36 C) 54 D) 50 30 / 30 Agar geometrik progressiyaning ketma–ket dastlabki uchta hadining yig’indisi 62 ga, ularning o’nli logarifmlari yig’indisi 3 ga teng bo’lsa, shu geometrik progressiyaning birinchi hadini toping. A) 10 yoki 50 B) 10 C) 50 D) 2 yoki 50 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz