Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №5 Matematika attestatsiya Matematika attestatsiya №5 InfoMaster Yanvar 24, 2022 73 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 2 ovozlar, 1 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №5 1 / 40 Teng yonli trapetsiya asosidagi burchakning sinusi 0,6 ga, asoslarining ayirmasi 4 ga teng bo‘lsa, trapetsiyaning yon tomonini toping. A) 2 B) 4 C) 2,5 D) 3 2 / 40 180 gramm suvga 70 gramm tuz aralashtrildi. Hosil bo‘lgan aralashmaninmg necha foizi suvdan iborat? A) 72 B) 75 C) 70 D) 78 3 / 40 Jaloliddin uyiga tezroq qaytish uchun qaysi yo’lni tanlash kerak. A) Birinchi yo'l B) Ikkinchi yo'l C) To'rtinchi yo'l D) Uchunchi yo'l 4 / 40 f(x)=26sin6x·sin7x funksiya uchun boshlang'ich funksiyasini toping. A) 13sinx+sin13x + C B) -13cosx- cos13 x +C C) 13cosx-cos13x + C D) 13sinx-sin13x + C 5 / 40 tenglamani yeching. A) 5 B) log(3/5)2 C) log(5/3)2 D) 3 6 / 40 a va b raqamlar yig'indisi 7 ga qoldiqsiz bo'linadi. Agar ko'rinishdagi uch xonali sonlarning 7 ga bo'lganda bir xil qoldiq qolsa, shu qoldiqni toping. A) 4 B) 0 C) 2 D) 6 7 / 40 Shaharlar orasidagi masofa xaritada 5 sm ga teng bo`lsava xaritaning masshtabi 1:4000000 bo`lsa , shaharlar orasidagi haqiqiy masofa qanchaga teng ? A) 20 km B) 200 km C) 2 km D) 0,2km 8 / 40 Quyida berilgan shakillar ichidagi sonlar yig`indisi nolgateng bo`lishi uchun . Nechanchi shakilni olib tashlash kerak? A) 4 B) 3 C) 2 D) 1 9 / 40 Quyidakilardan qaysi biri juft: 200956+200855 31210+0! 55!+877 222+333+4444 A) 1 B) 4 C) 2 D) 3 10 / 40 Dastlabki n ta hadining yig‘indisi formula bilan aniqlanadigan arifmetik progressiyaning 6-hadini toping. A) 6 B) 9 C) 8 D) 10 11 / 40 Tenglamani yeching: A) 5/9 B) 6/5 C) 5*4 D) 9/5 12 / 40 x>0 bo‘lsa, 128x+1/x² yig‘indining eng kichik qiymatini toping. A) 64 B) 16 C) 48 D) 32 13 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 2 B) 3 C) 1 D) 4 14 / 40 1+sin 2x = 7(sin x + cos x) tenglamani yeching. A) -π/4+πk, k∈Z B) Ø C) 0 D) π/4+πk, k∈Z 15 / 40 bo‘lsa, A) 1 B) 3/2 C) 2/3 D) 2 16 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 12 B) 10 C) 20 D) 14 17 / 40 b ning qanday qiymatlarida M(2;1) nuqtadan 4x - 3y + b = 0 to‘g‘ri chiziqqacha masofa 2 ga teng bo‘ladi? A) 4 yoki –12 B) 6 C) 5 yoki –15 D) 3 yoki –8 18 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 3√2 B) 2√2 C) 2√3 D) 4√3 19 / 40 tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping. A) 5 B) -2 C) 8 D) 3 20 / 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) 8 B) 12 C) 10 D) 6 21 / 40 sonlari geometrik progressiyaning ketma-ket hadlari bo‘ladigan barcha n larning yig‘indisini (agar bitta qiymati bo‘lsa, o‘zini) toping. A) 23 B) 24 C) 35 D) 14 22 / 40 Chizmadan foydalanib α ni toping. A) 40° B) 20° C) 30° D) 50° 23 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 6 B) 5 C) 8 D) 7 24 / 40 Asoslari 5 va 5√7 ga teng bo‘lgan trapetsiyaning yuzini teng ikkiga bo‘luvchi kesma asoslarga parallel. Shu kesma uzunligini toping. A) 6√7 B) 10 C) 8 D) 4√7 25 / 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 26 / 40 Hisoblang: A) sin10° B) cos10° C) cos50° D) 1 27 / 40 Hisoblang: A) 19/20 B) 10/11 C) 9/10 D) 20/21 28 / 40 Tenglamalar sistemani yeching: A) (7; 2), (28; -1) B) (9; 0), (28; -1) C) (2; 3) D) (9; 0), (2; 7) 29 / 40 Ifodaning qiymatini toping. A) 0,(04) B) 0,04 C) 0,0(2) D) 0,0(4) 30 / 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 31 / 40 x²-√11x+1=0 0 bo‘lsa, A) 11 B) 9 C) 12 D) 10 32 / 40 6-(5:3)·(-3)²-(-3)³+15:(-3) hisoblang. A) 15 B) 13 C) 14 D) 12 33 / 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) 1 D) bd/c 34 / 40 To‘g‘ri to‘rtburchakning 16 ga teng diagonali yon tomoni bilan 15° li burchak tashkil etadi. To‘rtburchak yuzini toping. A) 48 B) 42 C) 64 D) 32 35 / 40 Rasmdagi shakl perimetrini toping. A) 28 B) 30 C) 24 D) 32 36 / 40 Tengsizlik nechta butun juft yechimga ega? A) 115 B) 116 C) 110 D) 112 37 / 40 √3 A) 3/2 B) 21/10 C) 7/3 D) 5/5 38 / 40 a+b+c=10, bo‘lsa, ni toping. A) 11 B) 6 C) 4 D) 5 39 / 40 Tenglamani yeching: A) -6; 5 B) 5 C) 6; -5 D) -6 40 / 40 Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping. A) 16(π-2) B) 64 C) 8(π+2) D) 32 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 14 Matematika fanidan attestatsiya savollari №16
