Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №5 Matematika attestatsiya Matematika attestatsiya №5 InfoMaster Yanvar 24, 2022 55 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 2 ovozlar, 1 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №5 1 / 40 tenglamalar sistemasi yechimga ega bo‘ladigan a ning barcha qiymatlari to‘plamini toping. A) 2 B) 1 C) 3 D) 4 2 / 40 Qavariq oltiburchakning 3 ta uchidan kesib olsak qanday shakl hosil bo’ladi ? A) 9 B) 6 C) 12 D) 14 3 / 40 Jamshid bir kunda 70,5 kgdan paxta tersa, 8460 kg paxtani necha kunda terib bo’ladi ? A) 130 kun B) 100 kun C) 120 kun D) 140 kun 4 / 40 tenglamani yeching. A) 5 B) 3 C) log(3/5)2 D) log(5/3)2 5 / 40 P(x), Q(x) va R(x) ko'phatdalar berilgan. Bunda P(x) ko'phadning odoz hadi Q(x) ko'phadning ozod hadidan ikki marta katta va P(0)≠0. P(x)=Q(x)·R(x+1) bo'lsa, R(x) ko'phadning koeffitsiyentlarining yig'indisini toping. A) 3 B) 2 C) 4 D) 1 6 / 40 f(x-1)+f(x+2)=2(x²+7) ekani ma‟lum bo'lsa, f(x) ko'phadni toping. A) f(x)=x²-x+5 B) f(x)+2x²-1 C) f(x)=x²-4 D) f(x)=x²+3x+7 7 / 40 Tengsizlikni qanoatlantiradigan eng kichik ikkita butunsonning yig`indisini toping? A) 3 B) 6 C) 5 D) 4 8 / 40 AM||KL , ∠BCD=70°,∠CDE=40°, ∠DEF =45°,∠EFG=80°, ∠FGK=70°∠MBC=x ga teng bo`lsa x ni toping ? A) 30° B) 20° C) 40° D) 10° 9 / 40 Quyida berilgan holatda stol va uni atrofida qo’yiladigan stullarning holati berilgan va raqamlangan k-holatda stullarning raqamlari yig`indisi 351 taga teng bo`lsa, k ni toping ? A) 13 B) 12 C) 10 D) 11 10 / 40 Integralni hisoblang: A) 3 B) 2 C) 4 D) 1 11 / 40 Dastlabki n ta hadining yig‘indisi formula bilan aniqlanadigan arifmetik progressiyaning 6-hadini toping. A) 6 B) 8 C) 9 D) 10 12 / 40 cos75° ·cos 45° ·cos15° ni hisoblang. A) √3/8 B) √2/8 C) √3/4 D) 1/8 13 / 40 To‘g‘ri to‘rtburchakning 16 ga teng diagonali yon tomoni bilan 15° li burchak tashkil etadi. To‘rtburchak yuzini toping. A) 64 B) 48 C) 32 D) 42 14 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 4 B) 2 C) 1 D) 3 15 / 40 Tenglikdan foydalanib a ni toping. (a+b+c+d)·(a-b-c+d)=(a-b+c-d)·(a+b-c-d) A) cd/b B) bd/c C) bc/d D) 1 16 / 40 bo‘lsa, A) 2/3 B) 3/2 C) 1 D) 2 17 / 40 a+b+c=10, bo‘lsa, ni toping. A) 4 B) 5 C) 6 D) 11 18 / 40 kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping. A) 11 B) 7 C) 5 D) 10 19 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 1 B) 2 C) 4 D) 3 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) 10 C) 8 D) 6√7 21 / 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) 96 C) 25 D) 105 22 / 40 Chizmadan foydalanib α ni toping. A) 50° B) 30° C) 20° D) 40° 23 / 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) 35 C) 24 D) 14 24 / 40 A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping. A) 20 B) 8 C) 10 D) 15 25 / 40 Ifodaning qiymatini toping. A) 0,0(4) B) 0,(04) C) 0,0(2) D) 0,04 26 / 40 Kvadratlarning yuzlari yig‘indisini toping. A) 22 B) 121 C) berilganlar yetarli emas D) 11 27 / 40 Rasmdagi shakl perimetrini toping. A) 24 B) 30 C) 32 D) 28 28 / 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) DC < AD B) BD = BC C) BC > AD D) AB = BC 29 / 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) 3√2 B) 4√2 C) 4√3 D) 6 30 / 40 Ifodaning qiymatini quyidagi sonlardan qaysi biriga teng. 319·25 316·28 313·28 329·28 A) 3 B) 2 C) 4 D) 1 31 / 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) 8 C) 6 D) 10 32 / 40 tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping. A) 5 B) 3 C) 8 D) -2 33 / 40 Hisoblang: A) 0 B) -1/2 C) 1/32 D) -√3/2 34 / 40 √3 A) 21/10 B) 3/2 C) 5/5 D) 7/3 35 / 40 sonlarini taqqolsang. A) c B) c C) b D) a 36 / 40 Tengsizlik nechta butun juft yechimga ega? A) 116 B) 110 C) 115 D) 112 37 / 40 Hisoblang: A) sin10° B) cos50° C) 1 D) cos10° 38 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 20 B) 10 C) 12 D) 14 39 / 40 1+sin 2x = 7(sin x + cos x) tenglamani yeching. A) Ø B) π/4+πk, k∈Z C) 0 D) -π/4+πk, k∈Z 40 / 40 b ning qanday qiymatlarida M(2;1) nuqtadan 4x - 3y + b = 0 to‘g‘ri chiziqqacha masofa 2 ga teng bo‘ladi? A) 5 yoki –15 B) 3 yoki –8 C) 6 D) 4 yoki –12 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
