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 Agar aylananing o‘zaro parallel bo‘lgan uzunliklari 38, 38 va 34 ga teng vatarlari orasidagi masofalar bir xil bo‘lsa, shu masofani toping. A) 7 B) 6 C) 6,5 D) 5,5 2 / 40 Agar arifmetik progressiyada bo‘lsa, dastlabki 10 ta hadining yig‘indisini toping. A) 155 yoki 245 B) 153 yoki 317 C) 150 yoki 245 D) 160 yoki 317 3 / 40 Ushbu tenglik da ham to‘g‘rimi? A) Yo'q B) Ha 4 / 40 sonning teskarisini toping. A) 2/9 B) -9 C) 9 D) 1/9 5 / 40 (tgx-2ctgx)⁴ yoyilmaning ozod hadini toping. A) 58 B) 24 C) 25 D) 6 6 / 40 Hisoblang. A) 49/25 B) -47/25 C) 51/29 D) -47/24 7 / 40 To'g'ri burchakli uchburchakka ichki va tashqi chizilgan aylanalar raduslari 6 va 15 ga teng bo'lsa, katetlar yig'indini toping. A) 42 B) 48 C) 35 D) 28 8 / 40 Hisoblang. cos²10° + cos²30° + cos²80° A) 0 B) 7/3 C) 7/4 D) 3/7 9 / 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) 10 / 40 y=xlnx funksiyani hosilasini toping. A) lnx+1 B) lnx/2+1 C) 2lnx-1 D) 1 11 / 40 Hisoblang. A) 4 B) 2 C) 1 D) 0 12 / 40 Hisoblang A) 12 B) 9 C) 6 D) 36 13 / 40 Hisoblang. A) -2 yoki 2 B) -2 yoki -1 C) -2 yoki 0 D) 0 14 / 40 Hisoblang. A) 3 B) 5 C) 6 D) 4 15 / 40 Uchburchak tamonlari 10, 24 va 26 ga teng bo'lsa, katta burchagi necha gradus? A) 60° B) 90° C) 75° D) 80° 16 / 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) 15 B) 7551/15 C) 6700/13 D) 7500/13 17 / 40 Agar bo'lsa f(x)=? A) x²+2 B) x²-1 C) x²-2 D) 2x²-1 18 / 40 |x²-2x|≤x yengsizlikni yeching. A) (-2;0)∪[1;3] B) {0}∪[1;3) C) {0}∪[1;3] D) {0}∪(1;3] 19 / 40 (tgx+ctgx)⁴ yoyilmaning ozod hadini toping. A) 5 B) 3 C) 4 D) 6 20 / 40 Tenglamaning [0;π] kesmadagi ildizlari soni 3 tadan ko'pmi? A) Ha B) Yo'q 21 / 40 Hisoblang. A) 3 B) 1 C) 4 D) 2 22 / 40 5-x ≤ 0,04 x ning eng kichik butun qiymatini toping. A) 3 B) -1 C) 2 D) 0 23 / 40 y = x²+1-|sinx| funksiya grafigi OX o'qidan pastda yotadimi? A) Yotadi B) Yotmaydi 24 / 40 Hisoblang. A) -1 B) 0 C) 2 D) 1 25 / 40 Tenglama ildizlari yig'indisini toping. A) 0 B) 2 C) -1 D) 4 26 / 40 Hisoblang. A) 26-3b B) 24+3b C) 24-3b D) 14-2b 27 / 40 Agar bo'lsa, ni toping. A) 3 B) 0 C) 2,5 D) 5,2 28 / 40 ABCD parallelogramm va SKBCE = 18 bo'lsa, SABCD =? A) 62 B) 48 C) 54 D) 56 29 / 40 a=10,(6) , b=80,(3) , c=20,(9) bo'lsa, (a+b)c=? A) 9 B) 16 C) 0 D) 25 30 / 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) 32 va 22 B) 34 va 20 C) 30 va 24 D) 36 va 18 31 / 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) -5 B) -4 C) -3 D) -6 32 / 40 Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping. A) 7/16 B) 5/13 C) 5/16 D) 9/16 33 / 40 Chizmada nechta uchburchak tasvirlangan? A) 54 B) 48 C) 52 D) 50 34 / 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) 6 B) 4 C) 5 D) 3 35 / 40 To'g'ri burchakli uchburchakning yuzi 96 ga, perimetri esa 48 ga teng bo'lsa, gipatenuza uzunligini toping. A) 23 B) 20 C) 21 D) 18 36 / 40 Hisoblang. A) 121 B) 133 C) 135 D) 131 37 / 40 Tenglamani yeching: A) 2 B) 2,5 C) 5,5 D) 6 38 / 40 Tenglamani yeching. A) 0 B) 2 C) 5 D) 1 39 / 40 Hisoblang. A) 98 B) 32 C) 94 D) 96 40 / 40 Hisoblang. A) 3 B) 6 C) 2 D) 1 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
