Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 131 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 4 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Hisoblang. 11+192+1993+19994+199995+1999996+19999997+199999998+1999999999 A) 2222222175 B) 22222222220 C) 222220175 D) 222222222222 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) 3 B) 6 C) √37 D) √32 3 / 30 ifodaning qiymatini toping. A) 0,5 B) 0 C) -2 D) -0,5 4 / 30 Teng yonli uchburchakning tomonlari 5, 5 va 6 ga teng. Bu uchburchakning bissektiritsalari va medianalari kesishgan nuqtalar A) 1/2 B) 1/6 C) 1,2 D) 1 5 / 30 tenglamani yeching. A) 2018 B) 2017 C) 0 D) 2019 6 / 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 7 / 30 a(x+2)=2x+1 tenglama a ning qanday qiymatida yechimga ega emas? A) (-∞;2) B) (∞;∞) C) (-∞;2)v(2;∞) D) (2;∞) 8 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 3 B) 4 C) 5 D) 6 9 / 30 Tenglamaning ildizlari yig`indisini toping. A) 3 B) 6 C) 4 D) 5 10 / 30 Tengsizlik nechta butun yechimga ega? A) cheksiz ko’p B) 3 C) 1 D) 4 11 / 30 A) 2 B) 3 C) 1 D) 5 12 / 30 Tengsizlikni yeching. A) (-1/³ √2 ;0) B) 0 C) (0;+∞) D) to'g'ri javob yo'q 13 / 30 m ning qanday eng katta butun qiymatida y=2x-mx-5+m funksiyaning grafigi 1,3,4 –choraklarda yotadi? A) 1 B) 2 C) 5 D) 3 14 / 30 Agar va b-a=4 bo’lsa, a+b ni toping. A) 3 B) 3,5 C) 2 D) 4 15 / 30 Muntazam uchburchakli piramidaning yon qirrasi asos tekisligi bilan 45o li burchak tashkil etgan bo‘lsa, u holda piramidaning yon sirti yuzining uning asosi yuziga nisbatini toping. A) 4 B) 3√3 C) 2√3 D) 2√5 16 / 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) 48√3 C) 108 D) 54 17 / 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) 12 B) 3 C) 18 D) 6 18 / 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) aniqlab bo’lmaydi B) 374389 C) 412967 D) 2 19 / 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) 48√3 D) 67,5√3 20 / 30 Quyidagi 2x-y+3z=2018 va x+5y+z=2019 tekisliklarning holatini aniqlang. A) aniqlab bo’lmaydi B) o’zaro parallel C) o’zaro perpendikulyar D) ayqash 21 / 30 Ushbu funksiyaning boshlang’ich funksiyasini toping. A) ln(|x+4+|x-1|)+c B) ln(|x+4*|x-1|)+c C) ln|x+4/x-1|+c D) ln|x-1/x+4|+c 22 / 30 A) 0 B) 1 C) √6 D) 2 23 / 30 Tomoni 2 ga teng kvadratga tashqi chizilgan aylana uzunligini toping. A) 2π B) 3π C) 2π/3 D) 4π 24 / 30 (-3;4) nuqtaga absissa, ordinata o’qlariga va koordinata boshiga nisbatan simmetrik bo’lgan nuqtalarni tutashtirishdan hosil bo’lgan uchburchakning eng kata tomonini toping. A) 10 B) 24 C) 14 D) 12 25 / 30 a=sin 1; b=sin 2; c=sin 3; d=sin 4 va e=sin 5 sonlarni kamayish tartibida joylashtiring. A) b>a>c>d>e B) e>b>a>d>c C) a>b>c>d>e D) b>c>a>d>e 26 / 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) 50 B) 2 yoki 50 C) 10 yoki 50 D) 10 27 / 30 Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusidan katta bo’lgan uchburchaklar sonini toping. A) 1 B) cheksiz ko’p C) 2019 D) 0 28 / 30 Quyidagilardan qaysi biri barcha lar uchun ma’noga ega (aniqlangan)? A) (512-1/2⁻⁹)° B) 4k+1/1/8:7-1/56 C) ²ᴷ⁺⁴v2k+1/k²+1 D) ⁴ᴷ⁺³√-√2k+1 29 / 30 Musobaqada 5 ta ishtirokchidan 3 tasiga 1, 2, 3-o’rinlarni necha xil usulda berish mumkin? A) 18 B) 47 C) 120 D) 60 30 / 30 x(t)=t2+6t+5 qonuniyat bo’yicha harakatlanayotgan moddiy nuqta harakat boshlangandan necha sekund o’tgach boshlang’ich nuqtaga nisbatan 77 metr masofaga siljiydi? A) 8 B) 10 C) 7 D) 6 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
