Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 158 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 1 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Rombning yuzi 96 ga, dioganallaridan biri 16 ga teng. Romb tomonini toping. A) 10 B) 11 C) 12 D) 9 2 / 30 Bekzodda 50 so’m va Sobirda 70 so’m pul bor edi. Anvar Bekzodga o’z pulining 10 foizini bergandan so’ng, Bekzod Sobirga pulining yarmini berdi. So’ng Sobir Anvarga pulining 10 foizini berdi. Anvar o’zidagi pullarini hisoblab, pullari dastlabki holdagi puli bilan teng ekanligini bildi. Anvarda qancha pul bo’lgan? A) 127 B) 375 C) 400 D) 100 3 / 30 Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusiga teng bo’lgan uchburchak qanday uchburchak deyiladi? A) o’tmas burchakli uchburchak B) to’gri burchakli uchburchak C) o’tkir burchakli uchburchak D) teng tomonli uchburchak 4 / 30 Ostki asosining yuzi 20π va ustki asosining yuzi 10π ga teng bo‘lgan kesik konus berilgan. Agar kesik konusga shar ichki chizilgan bo‘lsa, u holda kesik konus hajmining shar hajmiga nisbatini toping. A) 2√2/3 B) 3√3+1/5 C) 5√3+3/3 D) 3√2+2/4 5 / 30 A) 3 B) 1 C) 2 D) 5 6 / 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 7 / 30 integralning qiymatini toping. A) π/4 B) 0 C) π/2 D) -π/2 8 / 30 Rombning balandligi 8 ga dioganallarining ko’paytmasi 80 ga teng. Rombning perimetrini toping A) 20 B) 16 C) 24 D) 32 9 / 30 Soddalashtiring. A) 2019 B) 2018 C) 2018a/a+1 D) a+1 10 / 30 Agar sinx+cosx=a bo’lsa, ning qiymatini toping. A) 2/5 B) 1/3 C) 1/2 D) -1/2 11 / 30 a(x+2)=2x+1 tenglama a ning qanday qiymatida yechimga ega emas? A) (2;∞) B) (-∞;2)v(2;∞) C) (-∞;2) D) (∞;∞) 12 / 30 bo’lsa, ni x orqali ifodalang. A) x/25 B) 2-x C) 25/x D) 2/x 13 / 30 Tengsizlikni yeching. A) 0 B) (0;+∞) C) to'g'ri javob yo'q D) (-1/³ √2 ;0) 14 / 30 Parallelogrammning tomonlari nisbati 3:5 kabi. Agar parallelogrmmning perimetri 48 ga burchaklaridan biri 1200 ga teng bo’lsa, uning yuzini toping. A) 67,5√3 B) 67,5 C) 135√3/4 D) 48√3 15 / 30 Tenglamaning ildizlari yig`indisini toping. A) 5 B) 4 C) 6 D) 3 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) 108 C) 48√3 D) 54 17 / 30 Agar f(x)=sin2x va g(x)=cos2x bo’lsa, u holda f(g(x)) funksiyaning hosilasini toping. A) 4sin2x*cos(2cos2x) B) -4sin2x*cos(2cos2x) C) -4sin2x*cos(cos2x) D) 4sin2x*cos(cos2x) 18 / 30 Ostki asosining yuzi 32π va ustki asosining yuzi 18π ga teng bo‘lgan kesik konus berilgan. Agar kesik konusga shar ichki chizilgan bo‘lsa, u holda sharning sirtini toping. A) 72π B) 56π C) 100π D) 96π 19 / 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) 1/√3 D) 0,5 20 / 30 bo’lsa ni toping. Bunda funksiya f(x) ga teskari funksiya. A) -10 B) -12 C) -14 D) -4 21 / 30 sistemadan x+y ning qiymatini toping. A) -12 B) 6 C) 35/4 D) 12 22 / 30 Radiuslari orasidagi burchagi 36o va radiusi 5 ga teng bo`lgan sektor yoyining uzunligini toping. A) 2π B) π/2 C) π D) 2π/3 23 / 30 ifodaning qiymatini toping. A) 0 B) -2 C) -0,5 D) 0,5 24 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 5 ga, bir katetining gipotenuzadagi proyeksiyasi 1,6 ga teng. Ikkinchi katetning kvadratini toping. A) 14 B) 18 C) 16 D) 17 25 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -0,75 B) -0,5 C) -1 D) -0,25 26 / 30 Tomoni 2 ga teng kvadratga tashqi chizilgan aylana uzunligini toping. A) 2π/3 B) 4π C) 3π D) 2π 27 / 30 y= funksiyaning aniqlanish sohasini toping A) (-∞;0])v[2;∞) B) (-∞;0)v(2;∞) C) (0,2) D) (2;∞) 28 / 30 AA1A2A3A4A5A6 muntazam oltiburchakli piramidaning hajmi ga va balandligi 2 ga teng bo‘lsa, u holda AA2A6 kesim yuzini toping. A) 3 B) 2 C) √5 D) √15 29 / 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 30 / 30 Parallelogrammning yuzi 213 ga, tomonlaridan biri 7 ga va o’tkir burchagi 600 ga teng bo’lsa, ikkinchi tomonini toping A) 5 B) 6 C) 4 D) 8 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
