Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №5 Matematika attestatsiya Matematika attestatsiya №5 InfoMaster Yanvar 24, 2022 57 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 2 ovozlar, 1 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №5 1 / 40 Bobur 14 m/s tezlik bilan 5 soat, 36 km/soat tezlik bilan 5 minut yugurgan bo’lsa , Boburning umumiy yurgan yo’li topilsin A) 670 km B) 13 km C) 255 km D) 456 km 2 / 40 ni hisoblang. A) (e+1)/2 B) (e-1)/2 C) e-1 D) e+1 3 / 40 funksiyaning eng kichik musbat davrini toping. A) 24 B) 18 C) 12 D) 26 4 / 40 cosx=0,2x tenglama nechta yechimga ega? A) 1 B) 4 C) 3 D) 2 5 / 40 Ushbu sistemaga ko'ra (ac+bd)² ni toping. A) 1 B) 2 C) 3 D) 4 6 / 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) √3 D) √2 7 / 40 Soddalashtiring: A) 3 B) 2 C) 1 D) Aniqlab bo`lmaydi 8 / 40 DC||AB , ∠DCE=45°,∠CEA=x va ∠EAB=115° ga tengbo`lsa x ni toping ? A) 120° B) 110° C) 130° D) 100° 9 / 40 Quyidagi rasmda konus va silindr zaytun moyi bilanto`ldirilmoqda . Ikkala shaklning balandliklari va taglik doiralarining radiuslari teng uzunlikda 2sm . Shunga ko`ra idishga jami necha ?m³ A) 11π B) (28π)/3 C) (29π)/3 D) (32π)/3 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) 8√3 B) 6√5 C) 12√3 D) 9√5 11 / 40 Dastlabki n ta hadining yig‘indisi formula bilan aniqlanadigan arifmetik progressiyaning 6-hadini toping. A) 9 B) 8 C) 6 D) 10 12 / 40 f(x)+f(3x)+f(5x)+...+f(39x)=(3x+10)4 bo'lsa, f'(0) ni toping. A) 10 B) 25 C) 30 D) 0 13 / 40 Ifodaning qiymatini quyidagi sonlardan qaysi biriga teng. 319·25 316·28 313·28 329·28 A) 4 B) 1 C) 3 D) 2 14 / 40 tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping. A) 3 B) 8 C) -2 D) 5 15 / 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√3 B) 6 C) 4√2 D) 3√2 16 / 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) 4 yoki –12 C) 6 D) 5 yoki –15 17 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 1 B) 3 C) 4 D) 2 18 / 40 Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping. A) 16(π-2) B) 32 C) 8(π+2) D) 64 19 / 40 Barcha ikki xonali sonlar ko‘paytmasi 4 ning qanday eng katta darajasiga bo‘linadi? A) 42 B) 43 C) 45 D) 44 20 / 40 sonlarini taqqolsang. A) a B) c C) b D) c 21 / 40 Hisoblang: A) 19/20 B) 9/10 C) 20/21 D) 10/11 22 / 40 kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping. A) 7 B) 5 C) 11 D) 10 23 / 40 Tenglamalar sistemani yeching: A) (9; 0), (28; -1) B) (7; 2), (28; -1) C) (9; 0), (2; 7) D) (2; 3) 24 / 40 Ifodaning qiymatini toping. A) 0,0(2) B) 0,04 C) 0,0(4) D) 0,(04) 25 / 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) 4 B) 2 C) 3 D) 1 26 / 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) 14 D) 35 27 / 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) 32 C) 64 D) 48 28 / 40 Chizmadan foydalanib α ni toping. A) 50° B) 40° C) 20° D) 30° 29 / 40 bo‘lsa, A) 2/3 B) 3/2 C) 2 D) 1 30 / 40 x²-√11x+1=0 0 bo‘lsa, A) 10 B) 12 C) 11 D) 9 31 / 40 Hisoblang: A) 1/32 B) -√3/2 C) 0 D) -1/2 32 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 2√3 B) 2√2 C) 4√3 D) 3√2 33 / 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 34 / 40 A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping. A) 10 B) 15 C) 8 D) 20 35 / 40 Tengsizlik nechta butun juft yechimga ega? A) 116 B) 110 C) 115 D) 112 36 / 40 √3 A) 7/3 B) 21/10 C) 3/2 D) 5/5 37 / 40 a+b+c=10, bo‘lsa, ni toping. A) 5 B) 11 C) 4 D) 6 38 / 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) 12 B) 6 C) 8 D) 10 39 / 40 Tenglamani yeching: A) 6; -5 B) 5 C) -6 D) -6; 5 40 / 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° 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