Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №5 Matematika attestatsiya Matematika attestatsiya №5 InfoMaster Yanvar 24, 2022 31 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 2 ovozlar, 1 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №5 1 / 40 180 gramm suvga 70 gramm tuz aralashtrildi. Hosil bo‘lgan aralashmaninmg necha foizi suvdan iborat? A) 75 B) 70 C) 78 D) 72 2 / 40 ABCD qavariq to‘rtburchakka aylana ichki chizilgan. AB = 8, BC = 12 bo‘lsa, CD – AD ayirmani toping. A) 3 B) 4 C) 2 D) aniqlab bo‘lmaydi 3 / 40 x,y,z∈N A=6x+1=5y+1=4z+1, min(A)=? A) 61 B) 60 C) 58 D) 59 4 / 40 Uchlari A(4;5;1), B(2;3;0) va C(2;–1;–3) nuqtalarda joylashgan uchburchakning BD medianasi uzunligini toping. A) √2 B) 1 C) 2 D) √3 5 / 40 tenglamani yeching. A) 3 B) 5 C) log(3/5)2 D) log(5/3)2 6 / 40 P(-2;2) nuqtadan o'tuvchi va a(6;4)a vektorga perpendikulyar bo'lgan to'g'ri chiziq tenglamasini toping. A) 3x-2y+2=0 B) 3x+2y-2=0 C) 3x+2y+2=0 D) 3x-2y-2=0 7 / 40 Standart shaklda yozing 0,000000000000013 1,3·10-12 1,3·10-13 1,3·10-14 1,3·10-15 A) 3 B) 2 C) 1 D) 4 8 / 40 n ning qanday qiymatida tenglik to`g`ri bo’ladi? (73)=715 A) 12 B) 5 C) 1 D) 7 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 ABCD trapetsiyaning AC diagonali CD yon tomonga perpendikulyar. Agar ∠D=69° va AB = BC bo‘lsa, B burchakni toping. A) 142° B) 135° C) 138° D) 132° 11 / 40 Agar bo'lsa, ni m orqali ifodalang. A) 2/m B) (m+4)/4 C) (4-m)/4 D) (m+1)/2 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) 25 B) 20 C) 40 D) 35 13 / 40 Tengsizlik nechta butun juft yechimga ega? A) 112 B) 116 C) 110 D) 115 14 / 40 Tenglamalar sistemani yeching: A) (9; 0), (2; 7) B) (2; 3) C) (9; 0), (28; -1) D) (7; 2), (28; -1) 15 / 40 bo‘lsa, A) 2 B) 1 C) 3/2 D) 2/3 16 / 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) 42 C) 64 D) 48 17 / 40 6-(5:3)·(-3)²-(-3)³+15:(-3) hisoblang. A) 13 B) 12 C) 15 D) 14 18 / 40 sonlari geometrik progressiyaning ketma-ket hadlari bo‘ladigan barcha n larning yig‘indisini (agar bitta qiymati bo‘lsa, o‘zini) toping. A) 14 B) 35 C) 24 D) 23 19 / 40 Chizmadan foydalanib α ni toping. A) 20° B) 30° C) 40° D) 50° 20 / 40 1+sin 2x = 7(sin x + cos x) tenglamani yeching. A) π/4+πk, k∈Z B) Ø C) -π/4+πk, k∈Z D) 0 21 / 40 a+b+c=10, bo‘lsa, ni toping. A) 11 B) 4 C) 5 D) 6 22 / 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) 6 D) 12 23 / 40 Hisoblang: A) cos10° B) 1 C) cos50° D) sin10° 24 / 40 A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping. A) 20 B) 15 C) 8 D) 10 25 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 3 B) 2 C) 4 D) 1 26 / 40 kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping. A) 11 B) 5 C) 10 D) 7 27 / 40 Ifodaning qiymatini toping. A) 0,(04) B) 0,0(4) C) 0,04 D) 0,0(2) 28 / 40 tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping. A) 8 B) -2 C) 3 D) 5 29 / 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) 6 C) 3 yoki –8 D) 4 yoki –12 30 / 40 Tenglamani yeching: A) 6; -5 B) -6 C) -6; 5 D) 5 31 / 40 Kvadratlarning yuzlari yig‘indisini toping. A) 121 B) berilganlar yetarli emas C) 11 D) 22 32 / 40 Hisoblang: A) 19/20 B) 20/21 C) 9/10 D) 10/11 33 / 40 Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping. A) 8(π+2) B) 16(π-2) C) 32 D) 64 34 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 2√2 B) 3√2 C) 2√3 D) 4√3 35 / 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) 1 C) 3 D) 4 36 / 40 Asoslari 5 va 5√7 ga teng bo‘lgan trapetsiyaning yuzini teng ikkiga bo‘luvchi kesma asoslarga parallel. Shu kesma uzunligini toping. A) 8 B) 10 C) 4√7 D) 6√7 37 / 40 √3 A) 3/2 B) 5/5 C) 21/10 D) 7/3 38 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 8 B) 6 C) 7 D) 5 39 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 10 B) 12 C) 14 D) 20 40 / 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) 4√2 B) 3√2 C) 6 D) 4√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
