Matematika attestatsiya №12 Aprel 8, 2022Aprel 8, 2022 da chop etilgan InfoMaster tomonidan Matematika attestatsiya №12 ga 1 fikr bildirilgan 0% 2 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 Agar cos9α = 4cosα bo'lsa, (4cos²3α-3)(4cos²α-3) ning qiymatini toping. A) 16 B) 4 C) 0 D) 8 2 / 40 ABC uchburchakning har bir tomoni chizmada ko’rsatilgandek o’z uzunligiga bir yarim barobarga teng uzunlikda davom ettirilgan. Agar SABC=12 bo’lsa, SA1B1C1 ni toping. A) 135 B) 130 C) 132 D) 147 3 / 40 a²·b·c<0, (a·b)>0 va a·c³<0 tengsizliklardan foydalanib a,b,c larning ishoralarini aniqlang: A) -, -, + B) -, +, - C) +, -, + D) -, -, - 4 / 40 Teng yonli trapetsiyaning diagionali o‘rta chizig‘ini 1 va 5 ga teng kesmalarga ajratadi. Trapetsiyaning yuzi 48 ga teng bo‘lsa, yon tomonini toping. A) 6 B) 4√5 C) 5 D) 3√2 5 / 40 1, 2, 3, 4 raqamlaridan foydalanib 400 dan kichik kechta uc xonali son yasash mumkin? A) 51 B) 48 C) 49 D) 47 6 / 40 Agar bo'lsa, ni toping. A) 3 B) 5,2 C) 2,5 D) 0 7 / 40 Hisoblang. A) 6 B) 2 C) 3 D) 1 8 / 40 Agar va ular orasidagi burchak 120° bo'lsa, ni toping. A) 2√7 B) 3√7 C) 4√5 D) √7 9 / 40 Hisoblang. A) 135 B) 133 C) 121 D) 131 10 / 40 y = x²+1-|sinx| funksiya grafigi OX o'qidan pastda yotadimi? A) Yotadi B) Yotmaydi 11 / 40 To'g'ri burchakli uchburchakning yuzi 96 ga, perimetri esa 48 ga teng bo'lsa, gipatenuza uzunligini toping. A) 23 B) 18 C) 21 D) 20 12 / 40 Hisoblang. A) -47/24 B) 49/25 C) 51/29 D) -47/25 13 / 40 (tgx-2ctgx)⁴ yoyilmaning ozod hadini toping. A) 6 B) 58 C) 25 D) 24 14 / 40 Hisoblang. A) 5/3 B) 3/5 C) 3/2 D) 2/3 15 / 40 Hisoblang. A) 5 B) 4 C) 3 D) 6 16 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) 4x+c B) c C) x+c D) 2x+c 17 / 40 Hisoblang. A) 0 B) 9/11 C) 9/13 D) 7/11 18 / 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) 19 / 40 Tenglamaning [0;π] kesmadagi ildizlari soni 3 tadan ko'pmi? A) Yo'q B) Ha 20 / 40 Tengsizlikni yeching. A) [-7;-3) u [-1;∞) B) [-7;-3] u [-1;∞) C) [-7;3 u [-1;∞) D) [-7;-5) u [-1;∞) 21 / 40 Tenglamani yeching. A) 5 B) 2 C) 1 D) 0 22 / 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) 32 va 22 C) 34 va 20 D) 36 va 18 23 / 40 Agar bo'lsa f(x)=? A) x²-2 B) 2x²-1 C) x²+2 D) x²-1 24 / 40 Hisoblang. A) 1 B) 2019 C) 2020 D) 2021 25 / 40 Arimetik progerssiyada s12=354, P=a1+a2+...+a11 , Q=a2+a4+...+a12 va P/Q=32/27 ga teng bo'lsa, d ni toping. A) -6 B) -3 C) -4 D) -5 26 / 40 Uchburchak tamonlari 10, 24 va 26 ga teng bo'lsa, katta burchagi necha gradus? A) 60° B) 80° C) 75° D) 90° 27 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) 2x+c B) 4x+c C) x/2+c D) x+c 28 / 40 Hisoblang A) 12 B) 9 C) 36 D) 6 29 / 40 Tenglamani yeching. A) 4 B) 2 C) 3 D) 1 30 / 40 To'g'ri burchakli uchburchakka ichki va tashqi chizilgan aylanalar raduslari 6 va 15 ga teng bo'lsa, katetlar yig'indini toping. A) 28 B) 42 C) 35 D) 48 31 / 40 Ayirmasi d ga teng bo'lgan arifmetik progrssiya uchun ni toping. A) 2/3 B) 1/2 C) 2 D) 1 32 / 40 Hisoblang. A) 32 B) 98 C) 96 D) 94 33 / 40 Hisoblang. cos²10° + cos²30° + cos²80° A) 7/3 B) 0 C) 3/7 D) 7/4 34 / 40 Hisoblang. A) -1 B) 2 C) 1 D) 0 35 / 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) 4 B) 3 C) 5 D) 6 36 / 40 Hisoblang. A) -1 B) 2 C) 0 D) 1 37 / 40 Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping. A) 7/16 B) 5/16 C) 9/16 D) 5/13 38 / 40 a=10,(6) , b=80,(3) , c=20,(9) bo'lsa, (a+b)c=? A) 16 B) 0 C) 25 D) 9 39 / 40 |x²-2x|≤x yengsizlikni yeching. A) {0}∪[1;3] B) {0}∪(1;3] C) (-2;0)∪[1;3] D) {0}∪[1;3) 40 / 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) 6700/13 B) 7551/15 C) 7500/13 D) 15 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