Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №5 Matematika attestatsiya Matematika attestatsiya №5 InfoMaster Yanvar 24, 2022 39 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 2 ovozlar, 1 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №5 1 / 40 bo‘lsa, ning qiymatini toping. A) 0,7 B) 1 C) 0,5 D) 2 2 / 40 |x+1|=2|x–2| tenglamaning ildizlari yig’indisini toping. A) 5 B) 6 C) 7 D) 0 3 / 40 Jamshid bir kunda 70,5 kgdan paxta tersa, 8460 kg paxtani necha kunda terib bo’ladi ? A) 120 kun B) 130 kun C) 100 kun D) 140 kun 4 / 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) 0 B) 2 C) 4 D) 6 5 / 40 Muntazam uchburchak ichidan olingan nuqtadan uchburchak tomonlarigacha bo'lgan masofalar mos holda c(2;3;1), b(1;2;1) va a(1;2;3) vektorlarning absolut qiymatlariga teng bo'lsa, uchburchakning balandligini toping. A) √6+√14 B) 16 C) 2√14+√6 D) 18 6 / 40 Umumiy hadi xn=3+5n-n2 formula bo'yicha berilgan sonli ketma-ketlikning eng katta hadi 15 dan qanchaga kam? A) 13 B) 6 C) 5,75 D) 9,25 7 / 40 Soddalashtiring: A) 1 B) Aniqlab bo`lmaydi C) 2 D) 3 8 / 40 Quyidakilardan qaysi biri juft: 200956+200855 31210+0! 55!+877 222+333+4444 A) 4 B) 2 C) 1 D) 3 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) 11 B) 10 C) 13 D) 12 10 / 40 Agar bo‘lsa, m/n ni toping. A) 4 B) 2 C) 3 D) (1+√5)/2 11 / 40 Agar tenglamaning ildizi m/n bo‘lsa, m+ n ni toping. Bunda EKUB(m; n)=1. A) 41 B) 50 C) 32 D) 25 12 / 40 ni hisoblang. A) 3 B) 1 C) 2 D) 0 13 / 40 x²-√11x+1=0 0 bo‘lsa, A) 11 B) 10 C) 12 D) 9 14 / 40 Tengsizlik nechta butun juft yechimga ega? A) 112 B) 110 C) 116 D) 115 15 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 4 B) 1 C) 3 D) 2 16 / 40 sonlari geometrik progressiyaning ketma-ket hadlari bo‘ladigan barcha n larning yig‘indisini (agar bitta qiymati bo‘lsa, o‘zini) toping. A) 24 B) 23 C) 35 D) 14 17 / 40 tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping. A) 5 B) -2 C) 3 D) 8 18 / 40 Ifodaning qiymatini quyidagi sonlardan qaysi biriga teng. 319·25 316·28 313·28 329·28 A) 3 B) 1 C) 4 D) 2 19 / 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) 1 D) bc/d 20 / 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) -3 B) -6 C) 5 D) -4 21 / 40 a+b+c=10, bo‘lsa, ni toping. A) 4 B) 11 C) 5 D) 6 22 / 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) BC > AD B) DC < AD C) AB = BC D) BD = BC 23 / 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) 10 C) 8 D) 6 24 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 10 B) 14 C) 12 D) 20 25 / 40 Rasmdagi shakl perimetrini toping. A) 24 B) 28 C) 30 D) 32 26 / 40 1+sin 2x = 7(sin x + cos x) tenglamani yeching. A) -π/4+πk, k∈Z B) 0 C) π/4+πk, k∈Z D) Ø 27 / 40 bo‘lsa, A) 2 B) 3/2 C) 2/3 D) 1 28 / 40 sonlarini taqqolsang. A) b B) c C) a D) c 29 / 40 √3 A) 3/2 B) 21/10 C) 5/5 D) 7/3 30 / 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 31 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 7 B) 5 C) 8 D) 6 32 / 40 Asoslari 5 va 5√7 ga teng bo‘lgan trapetsiyaning yuzini teng ikkiga bo‘luvchi kesma asoslarga parallel. Shu kesma uzunligini toping. A) 10 B) 8 C) 4√7 D) 6√7 33 / 40 Agar α=60°, β=70°, γ=50° bo‘lsa, tgα+ tgβ+ tgγ yig‘indi quyidagilardan qaysi biriga teng? A) 4 √3tg50° tg70° B) √3tg50° tg70° C) √3ctg40° tg70° D) 2√3tg50° tg70° 34 / 40 Hisoblang: A) 1/32 B) -1/2 C) -√3/2 D) 0 35 / 40 To‘g‘ri to‘rtburchakning 16 ga teng diagonali yon tomoni bilan 15° li burchak tashkil etadi. To‘rtburchak yuzini toping. A) 32 B) 48 C) 64 D) 42 36 / 40 Chizmadan foydalanib α ni toping. A) 40° B) 20° C) 50° D) 30° 37 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 1 B) 4 C) 3 D) 2 38 / 40 kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping. A) 7 B) 11 C) 5 D) 10 39 / 40 Hisoblang: A) 10/11 B) 19/20 C) 9/10 D) 20/21 40 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 3√2 B) 4√3 C) 2√2 D) 2√3 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
