Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №6 Matematika attestatsiya Matematika attestatsiya №6 InfoMaster Yanvar 30, 2022 97 Ko'rishlar 2 izohlar SaqlashSaqlanganOlib tashlandi 0 0% 0 ovozlar, 0 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №6 1 / 40 Agar bo'lsa ifodaning qiymatini toping. A) 7 B) 0 C) 9 D) 13 2 / 40 tenglamalar sistemasi yechimga ega bo‘ladigan a ning barcha qiymatlari to‘plamini toping. A) 1 B) 2 C) 4 D) 3 3 / 40 tenglamani yeching. A) log(5/3)2 B) log(3/5)2 C) 5 D) 3 4 / 40 cosx=0,2x tenglama nechta yechimga ega? A) 1 B) 2 C) 4 D) 3 5 / 40 Chizmaga ko`ra x ning eng kichik butun qiymatinitoping ? A) 8 B) 6 C) 7 D) 9 6 / 40 Bir idishda 32/5 kg, ikkinchisida esa unga qaraganda 16/5 kg ortiq yog` bor. Ikkala idishda qancha yog` bor? A) 10 B) 48/5 C) 64/5 D) 16 7 / 40 Quyidagi sonini 9 ga bo‘lganda qoladigan qoldiqni toping. A) 0 B) 8 C) 5 D) 1 8 / 40 Integralni hisoblang: A) 3 B) 2 C) 1 D) 4 9 / 40 a,b,c -1 dan katta musbat sonlar uchun a³=b² va a4=c5 bo'lsa, logbc ni toping. A) 5/6 B) 1/3 C) 8/15 D) 1/2 10 / 40 Uchburchakning 3 va 4 ga teng bo‘lgan tomonlariga o‘tkazilgan medianalar o‘zaro perpendikulyar bo‘lsa, bu uchburchakning uchinchi tomonini toping. A) 2,5 B) √5 C) √6 D) 2,4 11 / 40 A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping. A) 20 B) 10 C) 8 D) 15 12 / 40 √3 A) 7/3 B) 21/10 C) 5/5 D) 3/2 13 / 40 Tenglamani yeching: A) -6; 5 B) 6; -5 C) -6 D) 5 14 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 1 B) 4 C) 2 D) 3 15 / 40 bo‘lsa, A) 1 B) 2 C) 2/3 D) 3/2 16 / 40 Hisoblang: A) 1 B) sin10° C) cos10° D) cos50° 17 / 40 Barcha ikki xonali sonlar ko‘paytmasi 4 ning qanday eng katta darajasiga bo‘linadi? A) 42 B) 44 C) 45 D) 43 18 / 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 19 / 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) 64 C) 42 D) 32 20 / 40 6-(5:3)·(-3)²-(-3)³+15:(-3) hisoblang. A) 13 B) 14 C) 15 D) 12 21 / 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) 5 B) -4 C) -6 D) -3 22 / 40 Tenglamalar sistemani yeching: A) (9; 0), (28; -1) B) (7; 2), (28; -1) C) (9; 0), (2; 7) D) (2; 3) 23 / 40 Rasmdagi shakl perimetrini toping. A) 28 B) 24 C) 30 D) 32 24 / 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) 25 B) 105 C) 96 D) 100 25 / 40 sonlarini taqqolsang. A) c B) a C) c D) b 26 / 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 27 / 40 Ifodaning qiymatini toping. A) 0,(04) B) 0,0(4) C) 0,0(2) D) 0,04 28 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 5 B) 8 C) 6 D) 7 29 / 40 kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping. A) 7 B) 11 C) 5 D) 10 30 / 40 Chizmadan foydalanib α ni toping. A) 20° B) 50° C) 30° D) 40° 31 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 4√3 B) 2√3 C) 2√2 D) 3√2 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) 3 yoki –8 B) 6 C) 4 yoki –12 D) 5 yoki –15 33 / 40 Hisoblang: A) 19/20 B) 9/10 C) 10/11 D) 20/21 34 / 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) 1 C) 3 D) 2 35 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 14 B) 10 C) 20 D) 12 36 / 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 37 / 40 Tengsizlik nechta butun juft yechimga ega? A) 112 B) 115 C) 116 D) 110 38 / 40 Hisoblang: A) -1/2 B) 0 C) 1/32 D) -√3/2 39 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 2 B) 1 C) 3 D) 4 40 / 40 x²-√11x+1=0 0 bo‘lsa, A) 11 B) 9 C) 12 D) 10 O'rtacha ball 0% 0% Testni qayta ishga tushiring Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
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