Matematika attestatsiya №12 Aprel 8, 2022Aprel 8, 2022 da chop etilgan InfoMaster tomonidan Matematika attestatsiya №12 ga 1 fikr bildirilgan 0% 0 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 f(x) = log₃(x+3) bo'lsa, f-1(3) ni toping. A) 30 B) 6 C) 1 D) 24 2 / 40 Hisoblang. A) 8!/2 B) 8!9 3 / 40 ABCD parallelogramm va ABED toʻrtburchakka aylana ichki chizilgan. Agar DE=8, AB=12 boʻlsa, BEC uchburchakning perimetrini toping. A) 22 B) 25 C) 27 D) 24 4 / 40 Soddalashtiring: A) 2√b-2√a B) -2√a C) -2√b D) 2√a-2√b 5 / 40 Hisoblang. A) 133 B) 135 C) 131 D) 121 6 / 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) 36 va 18 B) 32 va 22 C) 30 va 24 D) 34 va 20 7 / 40 Hisoblang. A) 5 B) 3 C) 4 D) 6 8 / 40 Tenglamani yeching. A) 3 B) 4 C) 2 D) 1 9 / 40 Agar bo'lsa, ni toping. A) 5,2 B) 3 C) 2,5 D) 0 10 / 40 f(x) = x²-3x-2 bo'lsa. f(a+b+c)-f(abc)=? Bunda a, b, c - kvadrat funksiya koeffitsiyentlari. A) 40 B) 38 C) 42 D) 36 11 / 40 Hisoblang. A) 1 B) 2 C) 0 D) -1 12 / 40 Tenglamani yeching. A) 2 B) 5 C) 1 D) 0 13 / 40 5-x ≤ 0,04 x ning eng kichik butun qiymatini toping. A) -1 B) 0 C) 3 D) 2 14 / 40 Ayirmasi d ga teng bo'lgan arifmetik progrssiya uchun ni toping. A) 1 B) 2/3 C) 2 D) 1/2 15 / 40 To'g'ri burchakli uchburchakning yuzi 96 ga, perimetri esa 48 ga teng bo'lsa, gipatenuza uzunligini toping. A) 18 B) 23 C) 21 D) 20 16 / 40 y=xlnx funksiyani hosilasini toping. A) lnx+1 B) 2lnx-1 C) 1 D) lnx/2+1 17 / 40 Hisoblang. A) 2021 B) 2020 C) 1 D) 2019 18 / 40 Tenglamaning [0;π] kesmadagi ildizlari soni 3 tadan ko'pmi? A) Yo'q B) Ha 19 / 40 Tengsizlikni yeching: 3 · 9x –10 · 21x + 7 · 49 x ≤ 0 A) (-1;3] B) [-1;1] C) [-1;0) D) [-1;0] 20 / 40 Hisoblang. A) 1 B) -1 C) 0 D) 2 21 / 40 Hisoblang. A) 32 B) 98 C) 94 D) 96 22 / 40 Tengsizlikni yeching. A) [-7;-3] u [-1;∞) B) [-7;-3) u [-1;∞) C) [-7;-5) u [-1;∞) D) [-7;3 u [-1;∞) 23 / 40 Agar va ular orasidagi burchak 120° bo'lsa, ni toping. A) 3√7 B) 4√5 C) 2√7 D) √7 24 / 40 Tenglamani yeching: A) 6 B) 2 C) 2,5 D) 5,5 25 / 40 y = x²+1-|sinx| funksiya grafigi OX o'qidan pastda yotadimi? A) Yotmaydi B) Yotadi 26 / 40 Hisoblang. A) -47/25 B) -47/24 C) 51/29 D) 49/25 27 / 40 Hisoblang. A) 3/2 B) 5/3 C) 2/3 D) 3/5 28 / 40 Hisoblang. A) 7/11 B) 9/11 C) 9/13 D) 0 29 / 40 Agar cos4x = acos⁴x+bcos2²x+1 bo'lsa, a+b=? A) 1 B) -1 C) 0 D) 2 30 / 40 Hisoblang. A) 6 B) 1 C) 2 D) 3 31 / 40 a=10,(6) , b=80,(3) , c=20,(9) bo'lsa, (a+b)c=? A) 16 B) 25 C) 9 D) 0 32 / 40 (tgx-2ctgx)⁴ yoyilmaning ozod hadini toping. A) 6 B) 25 C) 24 D) 58 33 / 40 Hisoblang A) 9 B) 6 C) 12 D) 36 34 / 40 Uchburchak tamonlari 10, 24 va 26 ga teng bo'lsa, katta burchagi necha gradus? A) 80° B) 90° C) 75° D) 60° 35 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) x+c B) x/2+c C) 2x+c D) 4x+c 36 / 40 |x²-2x|≤x yengsizlikni yeching. A) {0}∪[1;3] B) {0}∪[1;3) C) (-2;0)∪[1;3] D) {0}∪(1;3] 37 / 40 f(x)+(x-2)·g(x)=x²+5, f(3)=10 bo'lsa, g-1=(4) ni toping. Bunda g-1(x) g(x) ga teskari funksiya. A) 3 B) 4 C) 5 D) 6 38 / 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) 7551/15 B) 15 C) 7500/13 D) 6700/13 39 / 40 Hisoblang. A) 4 B) 1 C) 3 D) 2 40 / 40 Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping. A) 7/16 B) 5/16 C) 9/16 D) 5/13 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 tomonidan Wordpress Quiz plugin Matematika attestatsiya
