Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №5 Matematika attestatsiya Matematika attestatsiya №5 InfoMaster Yanvar 24, 2022 68 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 2 ovozlar, 1 avg 2 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №5 1 / 40 Abdulla, Samandar va Jamshid 3 ta stulga necha xil usulda utirishi mumkin? A) 7 B) 6 C) 27 D) 9 2 / 40 Qavariq oltiburchakning 3 ta uchidan kesib olsak qanday shakl hosil bo’ladi ? A) 9 B) 14 C) 6 D) 12 3 / 40 tengsizlikni yeching. A) 4 B) 1 C) 2 D) 3 4 / 40 y=f(x) funksiya uchun tenglik o'rinli bo'lsa, f(π/4)=? A) 1/5 B) 1/4 C) 4 D) 1/3 5 / 40 cosx=0,2x tenglama nechta yechimga ega? A) 4 B) 1 C) 3 D) 2 6 / 40 Bir odam shunday vasiyat qildi: “Naqd 10 dirham pulim bor. Bir kishiga qarz ham berganman. Qarzning miqdori o'g'lim oladigan merosga teng. Ikkala o'g'lim barobar meros olishsin. Ukamga jami merosning 0,2 qismini va yana 1 dirham beringlar”. U kishining o'g'illari necha dirhamdan meros olishgan? A) 6 B) 25/3 C) 35/6 D) 8 7 / 40 Tengsizlikni qanoatlantiradigan eng kichik ikkita butunsonning yig`indisini toping? A) 4 B) 6 C) 3 D) 5 8 / 40 Soddalshtiring. A) -1 B) 1 C) 6 D) 2 9 / 40 Chizmaga ko`ra x ning eng kichik butun qiymatinitoping ? A) 8 B) 9 C) 6 D) 7 10 / 40 Raqamlari yig‘indisi 10 bo‘lgan nechta turli 3 xonali son bor? A) 0,432 B) 0,144 C) 0,288 D) 0,216 11 / 40 Radiusi 25/4 bo‘lgan sferaga balandligi 8 ga teng bo‘lgan konus ichki chizilgan. Konusning hajmini toping. A) 96π B) 144π C) 72π D) 192π 12 / 40 275+330 sonini 41 ga bo‘lganda qoladigan qoldiqni aniqlang. A) 9 B) 0 C) 5 D) 1 13 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 1 B) 3 C) 2 D) 4 14 / 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) 3 C) 1 D) 2 15 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 14 B) 12 C) 20 D) 10 16 / 40 √3 A) 3/2 B) 21/10 C) 7/3 D) 5/5 17 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 4√3 B) 2√3 C) 2√2 D) 3√2 18 / 40 sonlarini taqqolsang. A) c B) c C) b D) a 19 / 40 6-(5:3)·(-3)²-(-3)³+15:(-3) hisoblang. A) 15 B) 14 C) 13 D) 12 20 / 40 Asoslari 5 va 5√7 ga teng bo‘lgan trapetsiyaning yuzini teng ikkiga bo‘luvchi kesma asoslarga parallel. Shu kesma uzunligini toping. A) 4√7 B) 8 C) 6√7 D) 10 21 / 40 Barcha ikki xonali sonlar ko‘paytmasi 4 ning qanday eng katta darajasiga bo‘linadi? A) 42 B) 43 C) 45 D) 44 22 / 40 tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping. A) -2 B) 3 C) 5 D) 8 23 / 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) √3ctg40° tg70° D) 4 √3tg50° tg70° 24 / 40 P(x+1)=x³+3x²-2x+a+3 ko‘phadi berilgan. P(x+2) ko‘phadining koeffitsiyentlari yig‘indisini 8 ga teng bo‘lsa, a nechaga teng? A) -4 B) -3 C) -6 D) 5 25 / 40 bo‘lsa, A) 2/3 B) 1 C) 3/2 D) 2 26 / 40 Kvadratlarning yuzlari yig‘indisini toping. A) berilganlar yetarli emas B) 22 C) 11 D) 121 27 / 40 Hisoblang: A) cos10° B) sin10° C) 1 D) cos50° 28 / 40 sonlari geometrik progressiyaning ketma-ket hadlari bo‘ladigan barcha n larning yig‘indisini (agar bitta qiymati bo‘lsa, o‘zini) toping. A) 35 B) 24 C) 14 D) 23 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 a+b+c=10, bo‘lsa, ni toping. A) 11 B) 6 C) 5 D) 4 31 / 40 Tenglamalar sistemani yeching: A) (9; 0), (28; -1) B) (7; 2), (28; -1) C) (9; 0), (2; 7) D) (2; 3) 32 / 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) 4 yoki –12 C) 3 yoki –8 D) 5 yoki –15 33 / 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 34 / 40 Tengsizlik nechta butun juft yechimga ega? A) 115 B) 116 C) 112 D) 110 35 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 8 B) 5 C) 7 D) 6 36 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 3 B) 4 C) 1 D) 2 37 / 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) 1 C) cd/b D) bd/c 38 / 40 Ifodaning qiymatini toping. A) 0,0(4) B) 0,04 C) 0,0(2) D) 0,(04) 39 / 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 40 / 40 Hisoblang: A) 1/32 B) -1/2 C) 0 D) -√3/2 O'rtacha ball 34% 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