Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №5 Matematika attestatsiya Matematika attestatsiya №5 InfoMaster Yanvar 24, 2022 64 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 2 ovozlar, 1 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №5 1 / 40 ABCD qavariq to‘rtburchakka aylana ichki chizilgan. AB = 8, BC = 12 bo‘lsa, CD – AD ayirmani toping. A) 2 B) aniqlab bo‘lmaydi C) 3 D) 4 2 / 40 Muntazam oltiburchakning tomoni 2√6 ga teng. Shu ko‘pburchakka tengdosh bo‘lgan teng tomonli uchburchakning tomonini toping. A) 12 B) 30 C) 18 D) 24 3 / 40 Quyidagi javoblardan qaysi biri tengsizlikni yechimi bo'la oladi? A) 1 B) 4 C) 3 D) 2 4 / 40 Hisoblang: A) 7 B) 9 C) 4 D) 27 5 / 40 Uchlari A(4;5;1), B(2;3;0) va C(2;–1;–3) nuqtalarda joylashgan uchburchakning BD medianasi uzunligini toping. A) 1 B) 2 C) √2 D) √3 6 / 40 P(-2;2) nuqtadan o'tuvchi va a(6;4)a vektorga perpendikulyar bo'lgan to'g'ri chiziq tenglamasini toping. A) 3x-2y-2=0 B) 3x-2y+2=0 C) 3x+2y+2=0 D) 3x+2y-2=0 7 / 40 Eldor velosipetida 13m/s bilan 3 soniyada 15m tepalikka ko`tarilgan bo`lsa AC kesma uzunligi necha metr? A) 34 B) 36 C) 32 D) 40 8 / 40 Standart shaklda yozing 0,000000000000013 1,3·10-12 1,3·10-13 1,3·10-14 1,3·10-15 A) 1 B) 3 C) 4 D) 2 9 / 40 Quyida berilgan ko`rinishda mashinadan yuknitushurish uchun necha metr taxta kerak ? A) 14 B) 13 C) 11 D) 12 10 / 40 Tenglamalar sistemasidan foydalanib x + y + z ni toping: A) berilganlar yetarli emas B) 8 C) 6 D) 10 11 / 40 Agar f(x) funksiya (-∞;∞) da qat’iy o‘suvchi funksiya bo‘lsa, y =3 f(x)-8 funksiya uchun quyidagi mulohazalardan qaysi biri to‘g‘ri bo‘ladi? A) dastlab o‘sadi, keyin kamayadi B) qat’iy kamayuvchi C) dastlab kamayadi, keyin o‘sadi D) qat’iy o‘suvchi 12 / 40 cos75° ·cos 45° ·cos15° ni hisoblang. A) 1/8 B) √2/8 C) √3/4 D) √3/8 13 / 40 kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping. A) 5 B) 11 C) 7 D) 10 14 / 40 Chizmadan foydalanib α ni toping. A) 20° B) 50° C) 30° D) 40° 15 / 40 sonlari geometrik progressiyaning ketma-ket hadlari bo‘ladigan barcha n larning yig‘indisini (agar bitta qiymati bo‘lsa, o‘zini) toping. A) 14 B) 23 C) 24 D) 35 16 / 40 Asoslari 5 va 5√7 ga teng bo‘lgan trapetsiyaning yuzini teng ikkiga bo‘luvchi kesma asoslarga parallel. Shu kesma uzunligini toping. A) 8 B) 4√7 C) 10 D) 6√7 17 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 6 B) 7 C) 8 D) 5 18 / 40 A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping. A) 15 B) 8 C) 10 D) 20 19 / 40 a+b+c=10, bo‘lsa, ni toping. A) 5 B) 11 C) 6 D) 4 20 / 40 ABC to‘g‘ri burchakli uchburchakning og‘irlik markazi G nuqta. Bunda AB ⊥ BC, AG ⊥ GB va AG = 8 bo‘lsa, BG ni toping. A) 3√2 B) 4√2 C) 6 D) 4√3 21 / 40 Ifodaning qiymatini toping. A) 0,(04) B) 0,0(4) C) 0,04 D) 0,0(2) 22 / 40 ABC o‘tkir burchakli uchburchakning BC asosiga AD balandlik, AC yon tomoniga BE balandlik o‘tkazilgan. Bunda CE =2AE = 8, DC = 6 bo‘lsa, BD ni toping. A) 10 B) 8 C) 12 D) 6 23 / 40 b ning qanday qiymatlarida M(2;1) nuqtadan 4x - 3y + b = 0 to‘g‘ri chiziqqacha masofa 2 ga teng bo‘ladi? A) 6 B) 3 yoki –8 C) 4 yoki –12 D) 5 yoki –15 24 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 1 B) 2 C) 3 D) 4 25 / 40 sonlarini taqqolsang. A) c B) b C) c D) a 26 / 40 Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping. A) 32 B) 8(π+2) C) 16(π-2) D) 64 27 / 40 Tenglikdan foydalanib a ni toping. (a+b+c+d)·(a-b-c+d)=(a-b+c-d)·(a+b-c-d) A) cd/b B) bc/d C) bd/c D) 1 28 / 40 3x3 o‘lchamli kvadratning tugunlarida 16 ta nuqta belgilanib, ularning o‘ng tomondan eng yuqorisidagi A bilan belgilangan. Bir uchi A nuqtada, qolgan uchlari qolgan 15 ta nuqtada orasidan tanlanadigan uchburchaklarning sonini toping. A) 105 B) 96 C) 25 D) 100 29 / 40 Ifodaning qiymatini quyidagi sonlardan qaysi biriga teng. 319·25 316·28 313·28 329·28 A) 2 B) 4 C) 3 D) 1 30 / 40 Agar α=60°, β=70°, γ=50° bo‘lsa, tgα+ tgβ+ tgγ yig‘indi quyidagilardan qaysi biriga teng? A) √3tg50° tg70° B) 2√3tg50° tg70° C) 4 √3tg50° tg70° D) √3ctg40° tg70° 31 / 40 bo‘lsa, A) 2 B) 2/3 C) 1 D) 3/2 32 / 40 1+sin 2x = 7(sin x + cos x) tenglamani yeching. A) 0 B) -π/4+πk, k∈Z C) π/4+πk, k∈Z D) Ø 33 / 40 Kvadratlarning yuzlari yig‘indisini toping. A) 121 B) berilganlar yetarli emas C) 22 D) 11 34 / 40 Rasmdagi shakl perimetrini toping. A) 30 B) 32 C) 28 D) 24 35 / 40 Tengsizlik nechta butun juft yechimga ega? A) 112 B) 116 C) 115 D) 110 36 / 40 x²-√11x+1=0 0 bo‘lsa, A) 10 B) 11 C) 12 D) 9 37 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 2√2 B) 2√3 C) 3√2 D) 4√3 38 / 40 P(x)=x¹ºº ko‘phadni x³-3x+2 ga bo‘lganda qoladigan qoldiqni toping. 2¹ºº-1 (2¹ºº-1)x+2(299-1) (2¹ºº-1)x-2(299-1) 2¹ººx-3·2100 A) 3 B) 4 C) 2 D) 1 39 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 2 B) 1 C) 4 D) 3 40 / 40 To‘g‘ri to‘rtburchakning 16 ga teng diagonali yon tomoni bilan 15° li burchak tashkil etadi. To‘rtburchak yuzini toping. A) 42 B) 48 C) 32 D) 64 O'rtacha ball 0% 0% Testni qayta ishga tushiring Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
Istaklar ro'yxatiga qo'shildiIstaklar ro'yxatidan olib tashlandi 13 Matematika fanidan attestatsiya savollari №16
