Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №5 Matematika attestatsiya Matematika attestatsiya №5 InfoMaster Yanvar 24, 2022 38 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 2 ovozlar, 1 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №5 1 / 40 bo‘lsa, y = ? A) 2 B) 26 C) 8 D) 3 2 / 40 Qavariq oltiburchakning 3 ta uchidan kesib olsak qanday shakl hosil bo’ladi ? A) 14 B) 12 C) 9 D) 6 3 / 40 Agar ni hisoblang. A) 8 B) 9 C) 0 D) Cheksiz 4 / 40 Ushbu sistemaga ko'ra (ac+bd)² ni toping. A) 1 B) 4 C) 2 D) 3 5 / 40 Tengsizlikni yeching. A) [-3;2]∪{1} B) [-2;1]∪{-3} C) [-3;-2] D) (-∞;-2] 6 / 40 y=f(x) funksiya uchun tenglik o'rinli bo'lsa, f(π/4)=? A) 1/4 B) 4 C) 1/3 D) 1/5 7 / 40 Quyida berilgan tizimda yonma - yon shakillardagisonlar ustida pasdagi ( EKUB yoki EKUK topish) amali bajariladi , natija amal pastigi shakilga yoziladi Shunga ko`ra 1 , 2 va 3 amallar EKUB yoki EKUK dan qaysi biri bo`lishi kerak A) 1-EKUB 2-EKUK 3-EKUK B) 1-EKUB 2-EKUK 3-EKUB C) 1-EKUB 2-EKUB 3-EKUK D) 1-EKUB 2-EKUB 3-EKUB 8 / 40 Soddalashtiring: A) Aniqlab bo`lmaydi B) 1 C) 3 D) 2 9 / 40 Quyida berilgan shakillar ichidagi sonlar yig`indisi nolgateng bo`lishi uchun . Nechanchi shakilni olib tashlash kerak? A) 3 B) 1 C) 4 D) 2 10 / 40 Aylana tashqarisidagi nuqtadan aylanaga kesuvchi o‘tkazilgan. Berilgan nuqtadan aylanani kesgan nuqtalarigacha bo‘lgan masofalar mos ravishda 9 va 45 ga teng bo‘lsa, shu nuqtadan aylanaga o‘tkazilgan urinmaning urinish nuqtasigacha bo‘lgan masofa uzunligini toping. A) 9√5 B) 12√3 C) 8√3 D) 6√5 11 / 40 f(x)+f(3x)+f(5x)+...+f(39x)=(3x+10)4 bo'lsa, f'(0) ni toping. A) 30 B) 25 C) 0 D) 10 12 / 40 ABCD trapetsiyaning AC diagonali CD yon tomonga perpendikulyar. Agar ∠D=69° va AB = BC bo‘lsa, B burchakni toping. A) 138° B) 142° C) 135° D) 132° 13 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 14 B) 10 C) 20 D) 12 14 / 40 Hisoblang: A) 0 B) 1/32 C) -1/2 D) -√3/2 15 / 40 Tengsizlik nechta butun juft yechimga ega? A) 112 B) 110 C) 115 D) 116 16 / 40 Ifodaning qiymatini quyidagi sonlardan qaysi biriga teng. 319·25 316·28 313·28 329·28 A) 4 B) 3 C) 1 D) 2 17 / 40 Ifodaning qiymatini toping. A) 0,(04) B) 0,0(2) C) 0,04 D) 0,0(4) 18 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 3 B) 2 C) 1 D) 4 19 / 40 6-(5:3)·(-3)²-(-3)³+15:(-3) hisoblang. A) 14 B) 12 C) 15 D) 13 20 / 40 A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping. A) 20 B) 8 C) 15 D) 10 21 / 40 a+b+c=10, bo‘lsa, ni toping. A) 11 B) 4 C) 5 D) 6 22 / 40 √3 A) 3/2 B) 21/10 C) 7/3 D) 5/5 23 / 40 Tenglikdan foydalanib a ni toping. (a+b+c+d)·(a-b-c+d)=(a-b+c-d)·(a+b-c-d) A) bc/d B) cd/b C) bd/c D) 1 24 / 40 Hisoblang: A) 10/11 B) 20/21 C) 9/10 D) 19/20 25 / 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) 12 C) 6 D) 8 26 / 40 Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping. A) 64 B) 8(π+2) C) 32 D) 16(π-2) 27 / 40 Barcha ikki xonali sonlar ko‘paytmasi 4 ning qanday eng katta darajasiga bo‘linadi? A) 45 B) 44 C) 42 D) 43 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) 96 B) 100 C) 25 D) 105 29 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 1 B) 4 C) 2 D) 3 30 / 40 To‘g‘ri to‘rtburchakning 16 ga teng diagonali yon tomoni bilan 15° li burchak tashkil etadi. To‘rtburchak yuzini toping. A) 64 B) 42 C) 32 D) 48 31 / 40 sonlari geometrik progressiyaning ketma-ket hadlari bo‘ladigan barcha n larning yig‘indisini (agar bitta qiymati bo‘lsa, o‘zini) toping. A) 24 B) 23 C) 35 D) 14 32 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 2√3 B) 2√2 C) 4√3 D) 3√2 33 / 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) 1 D) 2 34 / 40 sonlarini taqqolsang. A) a B) b C) c D) c 35 / 40 Asoslari 5 va 5√7 ga teng bo‘lgan trapetsiyaning yuzini teng ikkiga bo‘luvchi kesma asoslarga parallel. Shu kesma uzunligini toping. A) 10 B) 6√7 C) 4√7 D) 8 36 / 40 Hisoblang: A) cos50° B) 1 C) cos10° D) sin10° 37 / 40 Chizmadan foydalanib α ni toping. A) 20° B) 30° C) 40° D) 50° 38 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 8 B) 6 C) 5 D) 7 39 / 40 bo‘lsa, A) 2 B) 3/2 C) 2/3 D) 1 40 / 40 Kvadratlarning yuzlari yig‘indisini toping. A) berilganlar yetarli emas B) 22 C) 121 D) 11 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
