Matematika attestatsiya №12 Aprel 8, 2022Aprel 8, 2022 da chop etilgan InfoMaster tomonidan Matematika attestatsiya №12 ga 3 fikr bildirilgan 0% 95 12345678910111213141516171819202122232425262728293031323334353637383940 OMAD YOR BO'LSIN! Matematika fanidan attestatsiya savollari №12 28.03.2022 da Toshkentda tushgan savollar DIQQAT! Endi siz o'z bilmingizni sinab ko'rish bilan birga sertifikatga ham ega bo'lishingiz mumkin. Sertifikat olish uchun barcha ma'lumotlarni to'g'ri kiriting! e-mail manzilini to'g'ri kriting, barcha ma'lumotlar sizga yuboriladi. Testda 76% va undan yuqori natija oling va sertifikatni yuklab oling. 1 / 40 Hisoblang. A) 0 B) 2019/4036 C) 2019/2018 D) 1 2 / 40 (5;-8) nuqtaning (-4;9) nuqtaga nisbatan simmetrik bo‘lgan nuqtasini toping. A) (-13;26) B) (-13;27) C) (-13;23) D) (-14;14) 3 / 40 Hisoblang. A) 3 B) 1 C) 2 D) 0 4 / 40 Agar bo‘lsa, y = ? A) 2 B) 3 C) 8 D) 26 5 / 40 Hisoblang. A) 24-3b B) 26-3b C) 24+3b D) 14-2b 6 / 40 Hisoblang. cos²10° + cos²30° + cos²80° A) 0 B) 3/7 C) 7/3 D) 7/4 7 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) 4x+c B) 2x+c C) x+c D) c 8 / 40 Hisoblang. A) 0 B) 2 C) 1 D) 4 9 / 40 Tenglamani yeching. A) 4 B) 3 C) 1 D) 2 10 / 40 (tgx+ctgx)⁴ yoyilmaning ozod hadini toping. A) 5 B) 4 C) 6 D) 3 11 / 40 Hisoblang. A) 96 B) 32 C) 98 D) 94 12 / 40 Hisoblang. A) 0 B) -2 yoki 0 C) -2 yoki -1 D) -2 yoki 2 13 / 40 a=10,(6) , b=80,(3) , c=20,(9) bo'lsa, (a+b)c=? A) 9 B) 25 C) 16 D) 0 14 / 40 Tenglamani yeching: A) 2 B) 2,5 C) 6 D) 5,5 15 / 40 Hisoblang. A) 2 B) 0 C) -1 D) 1 16 / 40 Agar va ular orasidagi burchak 120° bo'lsa, ni toping. A) 3√7 B) 4√5 C) √7 D) 2√7 17 / 40 Tengsizlikni yeching. A) x∈(0;1000) B) x∈(10;1000) C) x∈(1;1000] D) x∈[0;1000] 18 / 40 Tenglamaning [0;π] kesmadagi ildizlari soni 3 tadan ko'pmi? A) Ha B) Yo'q 19 / 40 5-x ≤ 0,04 x ning eng kichik butun qiymatini toping. A) -1 B) 0 C) 2 D) 3 20 / 40 Tengsizlikni yeching: 3 · 9x –10 · 21x + 7 · 49 x ≤ 0 A) [-1;0) B) (-1;3] C) [-1;1] D) [-1;0] 21 / 40 Hisoblang A) 12 B) 36 C) 9 D) 6 22 / 40 Tenglama ildizlari yig'indisini toping. A) 4 B) -1 C) 0 D) 2 23 / 40 Agar cos4x = acos⁴x+bcos2²x+1 bo'lsa, a+b=? A) 0 B) 1 C) 2 D) -1 24 / 40 Hisoblang. A) 0 B) -1 C) 1 D) 2 25 / 40 Hisoblang. A) 0 B) 7/11 C) 9/11 D) 9/13 26 / 40 To'g'ri burchakli uchburchakning yuzi 96 ga, perimetri esa 48 ga teng bo'lsa, gipatenuza uzunligini toping. A) 20 B) 18 C) 21 D) 23 27 / 40 (2; 3; -5) nuqtaning OX o'qiga nisbatan simmetrik nuqtasini toping. A) (2; -3; -5) B) (2; 3; -5) C) (2; -3; 5) D) (-2; 3; 5) 28 / 40 Hisoblang. A) 4 B) 1 C) 3 D) 2 29 / 40 Ayirmasi d ga teng bo'lgan arifmetik progrssiya uchun ni toping. A) 1/2 B) 1 C) 2 D) 2/3 30 / 40 Hisoblang. A) 5/3 B) 2/3 C) 3/2 D) 3/5 31 / 40 y = x²+1-|sinx| funksiya grafigi OX o'qidan pastda yotadimi? A) Yotadi B) Yotmaydi 32 / 40 To'g'ri burchakli uchburchakning gipotenuzasiga o'tkazilgan bissiktriksasi uni 20 va 30 ga teng kesmalarga ajratadi. To'g'ri burchakli uchburchak yuzini toping. A) 7500/13 B) 7551/15 C) 6700/13 D) 15 33 / 40 Tengsizlikni yeching. A) [-7;-5) u [-1;∞) B) [-7;-3] u [-1;∞) C) [-7;3 u [-1;∞) D) [-7;-3) u [-1;∞) 34 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) x/2+c B) x+c C) 4x+c D) 2x+c 35 / 40 Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping. A) 5/16 B) 9/16 C) 5/13 D) 7/16 36 / 40 (tgx-2ctgx)⁴ yoyilmaning ozod hadini toping. A) 6 B) 24 C) 58 D) 25 37 / 40 Agar bo'lsa, ni toping. A) 0 B) 5,2 C) 3 D) 2,5 38 / 40 y=xlnx funksiyani hosilasini toping. A) 2lnx-1 B) lnx+1 C) lnx/2+1 D) 1 39 / 40 Chizmada nechta uchburchak tasvirlangan? A) 48 B) 50 C) 52 D) 54 40 / 40 Tamonlarini ayirmasi 14 bo'lgan parallelogramning o'rmas burchagidan katta diagonaliga tushgan balandlik uni 12 va 30 ga teng kesmalarga ajratadi. Tamonlar uzunligini toping. A) 30 va 24 B) 34 va 20 C) 36 va 18 D) 32 va 22 0% Testni qayta ishga tushiring Baholash mezoni - 75 foiz va undan yuqori ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, oliy malaka toifasi (bosh o’qituvchi lavozimi) saqlansin”; - 75 foizdan past ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, birinchi malaka toifasi (yetakchi o’qituvchi lavozimi)ga tushirilsin”; - 75 foiz va undan yuqori ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, oliy malaka toifasi (bosh o’qituvchi lavozimi) berilsin”; - 74 foizdan 65 foizgacha ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, birinchi malaka toifasi (yetakchi o’qituvchi lavozimi) saqlansin”; - 65 foizdan kam ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, ikkinchi malaka toifasi (katta o’qituvchi lavozimi)ga tushirilsin” - 64 foizdan 60 foizgacha ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, ikkinchi malaka toifasi (katta o’qituvchi lavozimi) saqlansin”; - 60 foizdan kam ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, mutaxassis (oliy yoki o’rta maxsus, kasb-hunar ma'lumotli o’qituvchi) lavozimiga tushirilsin” Fikr-mulohaza yuboring Author: InfoMaster Matematika attestatsiya