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 Yig’indi quydagilardan qaysi biriga teng? A) 4506,08 B) 45,0608 C) 450,608 D) 40560,8 2 / 40 Agar va bo'lsa f(4) ning qiymatini toping. A) 12 B) 11 C) 4 D) 35 3 / 40 To’g’ri turt burchakning bo’yining perimetriga nisbatini toping. A) 5/3 B) -452/18 C) 3/18 D) 1/2 4 / 40 Hisoblang: A) 4 B) 7 C) 27 D) 9 5 / 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) 2 B) 6 C) 0 D) 4 6 / 40 tenglamaning ildizlari yig'indisini(agar ildizi bitta bo'lsa, o'zini) toping. A) 0 B) 2 C) 1 D) 3 7 / 40 Soddalshtiring. A) 6 B) 2 C) 1 D) -1 8 / 40 Hisoblang: A) 4 B) 1 C) 2 D) 3 9 / 40 DC||AB , ∠DCE=45°,∠CEA=x va ∠EAB=115° ga tengbo`lsa x ni toping ? A) 110° B) 130° C) 100° D) 120° 10 / 40 a,b,c -1 dan katta musbat sonlar uchun a³=b² va a4=c5 bo'lsa, logbc ni toping. A) 8/15 B) 1/2 C) 1/3 D) 5/6 11 / 40 ni hisoblang. A) 1 B) 0 C) 2 D) 3 12 / 40 Bo‘yi 130 cm, eni 90 cm, balandligi 60 cm bo‘lgan idishdagi suvning 234 litri olindi. Idishda qolgan suvning (sathi) balandligini toping. A) 40 B) 25 C) 20 D) 35 13 / 40 Chizmadan foydalanib α ni toping. A) 30° B) 50° C) 40° D) 20° 14 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 3 B) 4 C) 2 D) 1 15 / 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) 105 B) 100 C) 25 D) 96 16 / 40 Agar α=60°, β=70°, γ=50° bo‘lsa, tgα+ tgβ+ tgγ yig‘indi quyidagilardan qaysi biriga teng? A) √3ctg40° tg70° B) 4 √3tg50° tg70° C) 2√3tg50° tg70° D) √3tg50° tg70° 17 / 40 Hisoblang: A) -1/2 B) -√3/2 C) 1/32 D) 0 18 / 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 19 / 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) BD = BC B) DC < AD C) BC > AD D) AB = BC 20 / 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) 23 C) 14 D) 24 21 / 40 Hisoblang: A) sin10° B) cos50° C) 1 D) cos10° 22 / 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) 2 B) 4 C) 1 D) 3 23 / 40 Tenglamani yeching: A) -6 B) 5 C) 6; -5 D) -6; 5 24 / 40 6-(5:3)·(-3)²-(-3)³+15:(-3) hisoblang. A) 15 B) 12 C) 14 D) 13 25 / 40 Ifodaning qiymatini quyidagi sonlardan qaysi biriga teng. 319·25 316·28 313·28 329·28 A) 3 B) 1 C) 4 D) 2 26 / 40 1+sin 2x = 7(sin x + cos x) tenglamani yeching. A) π/4+πk, k∈Z B) -π/4+πk, k∈Z C) Ø D) 0 27 / 40 tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping. A) 8 B) 3 C) -2 D) 5 28 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 10 B) 14 C) 20 D) 12 29 / 40 sonlarini taqqolsang. A) b B) c C) a D) c 30 / 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) 4√7 D) 8 31 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 3 B) 2 C) 4 D) 1 32 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 8 B) 5 C) 6 D) 7 33 / 40 Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping. A) 8(π+2) B) 16(π-2) C) 64 D) 32 34 / 40 kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping. A) 11 B) 7 C) 5 D) 10 35 / 40 Tengsizlik nechta butun juft yechimga ega? A) 110 B) 112 C) 116 D) 115 36 / 40 x²-√11x+1=0 0 bo‘lsa, A) 11 B) 9 C) 12 D) 10 37 / 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) 6 B) 8 C) 12 D) 10 38 / 40 Kvadratlarning yuzlari yig‘indisini toping. A) 11 B) berilganlar yetarli emas C) 22 D) 121 39 / 40 Ifodaning qiymatini toping. A) 0,(04) B) 0,0(4) C) 0,0(2) D) 0,04 40 / 40 A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping. A) 15 B) 10 C) 20 D) 8 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