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 tengsizlik yechimi bo’la oladigan tub sonlar nechta? A) 5 B) 2 C) 4 D) 3 2 / 40 Muntazam oltiburchakning tomoni 2√6 ga teng. Shu ko‘pburchakka tengdosh bo‘lgan teng tomonli uchburchakning tomonini toping. A) 18 B) 30 C) 12 D) 24 3 / 40 funksiyaning eng kichik musbat davrini toping. A) 26 B) 24 C) 18 D) 12 4 / 40 f(x-1)+f(x+2)=2(x²+7) ekani ma‟lum bo'lsa, f(x) ko'phadni toping. A) f(x)=x²-4 B) f(x)=x²+3x+7 C) f(x)=x²-x+5 D) f(x)+2x²-1 5 / 40 Teng yonli trapetsiyaning diagonallari o'zaro perpendikulyar hamda yuzi 32 ga teng bo'lsa, uning diagonali uzunligini toping. A) 4 B) 6 C) 8 D) 2 6 / 40 Ushbu sistemaga ko'ra (ac+bd)² ni toping. A) 3 B) 2 C) 4 D) 1 7 / 40 Tengsizlikni qanoatlantiradigan eng kichik ikkita butunsonning yig`indisini toping? A) 5 B) 4 C) 6 D) 3 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) 4 C) 2 D) 3 9 / 40 n ning qanday qiymatida tenglik to`g`ri bo’ladi? (73)=715 A) 7 B) 1 C) 12 D) 5 10 / 40 Integralni hisoblang: A) 1 B) 2 C) 4 D) 3 11 / 40 Agar x =17 bo‘lsa, quyidagi ifodaning qiymatini toping. A) √17 B) -√17 C) -4 D) 4 12 / 40 cos75° ·cos 45° ·cos15° ni hisoblang. A) 1/8 B) √3/8 C) √3/4 D) √2/8 13 / 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) 4√2 B) 3√2 C) 4√3 D) 6 14 / 40 Hisoblang: A) cos50° B) sin10° C) cos10° D) 1 15 / 40 6-(5:3)·(-3)²-(-3)³+15:(-3) hisoblang. A) 15 B) 13 C) 14 D) 12 16 / 40 tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping. A) -2 B) 8 C) 5 D) 3 17 / 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) 6 C) 12 D) 8 18 / 40 A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping. A) 20 B) 15 C) 10 D) 8 19 / 40 sonlarini taqqolsang. A) b B) c C) a D) c 20 / 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) 100 B) 105 C) 96 D) 25 21 / 40 b ning qanday qiymatlarida M(2;1) nuqtadan 4x - 3y + b = 0 to‘g‘ri chiziqqacha masofa 2 ga teng bo‘ladi? A) 3 yoki –8 B) 5 yoki –15 C) 6 D) 4 yoki –12 22 / 40 Hisoblang: A) -1/2 B) 0 C) -√3/2 D) 1/32 23 / 40 Kvadratlarning yuzlari yig‘indisini toping. A) berilganlar yetarli emas B) 11 C) 22 D) 121 24 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 8 B) 7 C) 5 D) 6 25 / 40 ABC uchburchakning A burchgi 30° ga, B burchagi 75° ga teng. B uchidan AC tomonga BD kesma o‘tkazilgan. ABD burchak 45° ga teng bo‘lsa, quyidagilardan qaysi biri noto‘g‘ri? A) BC > AD B) BD = BC C) AB = BC D) DC < AD 26 / 40 Ifodaning qiymatini quyidagi sonlardan qaysi biriga teng. 319·25 316·28 313·28 329·28 A) 3 B) 1 C) 4 D) 2 27 / 40 a+b+c=10, bo‘lsa, ni toping. A) 4 B) 11 C) 5 D) 6 28 / 40 x²-√11x+1=0 0 bo‘lsa, A) 9 B) 11 C) 12 D) 10 29 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 2 B) 3 C) 4 D) 1 30 / 40 kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping. A) 11 B) 5 C) 7 D) 10 31 / 40 Chizmadan foydalanib α ni toping. A) 30° B) 40° C) 50° D) 20° 32 / 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) 14 D) 35 33 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 2√3 B) 2√2 C) 4√3 D) 3√2 34 / 40 Rasmdagi shakl perimetrini toping. A) 28 B) 24 C) 32 D) 30 35 / 40 Tenglikdan foydalanib a ni toping. (a+b+c+d)·(a-b-c+d)=(a-b+c-d)·(a+b-c-d) A) 1 B) bc/d C) cd/b D) bd/c 36 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 1 B) 2 C) 4 D) 3 37 / 40 Hisoblang: A) 10/11 B) 19/20 C) 9/10 D) 20/21 38 / 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) 32 C) 42 D) 64 39 / 40 1+sin 2x = 7(sin x + cos x) tenglamani yeching. A) 0 B) -π/4+πk, k∈Z C) Ø D) π/4+πk, k∈Z 40 / 40 bo‘lsa, A) 2 B) 2/3 C) 3/2 D) 1 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
