Uy » Choraklik online testlar » Matematika choraklik » 9-sinf Matematika 4-chorak Matematika choraklik 9-sinf Matematika 4-chorak InfoMaster Aprel 20, 2021 1406 Ko'rishlar 3 izohlar SaqlashSaqlanganOlib tashlandi 2 0 OMAD YOR BO'LSIN! Tomonidan yaratilgan InfoMaster 9-sinf Matematika 4-chorak Testni Salomov Sardor tayyorladi. 1 / 25 Ifodaning qiymatini toping: (-8)2-(-5)3-(12) 2 A) 189 B) 122 C) 48 D) 45 2 / 25 sin60° nimaga teng A) √3/2 B) 1 C) 1/2 D) √2/2 3 / 25 Funksiyaning aniqlanish sohasini toping A) x≠1 B) (-∞;+∞ ) C) x≥ 1 D) x≤1 4 / 25 Agar 90°° bo’lsa, quyidagilardan qaysi biri musbat A) ctga B) cosa C) sina D) tga 5 / 25 Sonlarni taqqoslang A) b<a B) b=a+0,8 C) b=a D) a = b + 0,5 6 / 25 Agar sina=12/13 bo’lsa, ctga ni toping A) 5/12 B) 12/5 C) 5/13 D) 1 7 / 25 Ifodaning qiymatini toping sin300° A) 1/2 B) 0 C) -1/2 D) 2/√3 8 / 25 Tenglamani yeching: cos(x-π )=0 A) 3π/2 B) 3π/2+2πk C) 2πk D) πk 9 / 25 Burchakning gradus o’lchovini toping: 3π/4 A) 150° B) 35° C) 140° D) 135° 10 / 25 Ifodaning qiymatini toping: tg300° A) √3 B) √3/3 C) 0 D) -√3 11 / 25 Parallelogrammning tomonlari 7√3 va 6 sm. Agar uning o‘tmas burchagi 120° bo‘lsa, uning yuzini toping A) 63 sm² B) 61 sm² C) 56 sm² D) 73 sm² 12 / 25 Tenglamani yeching: sin(π -x)=1 A) π/2 B) 2πk C) π/2+2πk D) π/2-2πk 13 / 25 Teng yonli trapetsiyaning ikkita burchagi ayirmasi 400 ga teng. uning burchaklarini toping A) 110°; 70°; 110°; 70° B) 115° ; 75° ; 115° ; 75° C) 135°; 45°; 135°; 45° D) 140°; 40°; 140°; 40° 14 / 25 x2 + 8x+15 ≤ 0 tengsizlikning butun yechimlari yig`indisini toping A) -12 B) 12 C) -4 D) 23 15 / 25 Tengsizlikni yeching: (x + 4)(x + 9) >0 A) (-9; -4) B) (-∞; -9) U (-4; ∞) C) (4; 9) D) (-∞;4) U (9; ∞) 16 / 25 Kvadrat funksiyaning nollarini toping: y = x2 - 5x + 6 A) x₁=2, x₂ =3 B) x₁= -1, x₂= - 6 C) x₁=1, x₂=6 D) x₁= -2, x₂= -3 17 / 25 Quyidagi funksiyalardan qaysi biri kvadrat funksiya bo`ladi ? A) y = 2010x² + 41x + 9 B) y = 3x² + x³ – 8 C) y = 2x -3 D) y = 5x⁴ + 6x 18 / 25 Quyidagi tengsizliklardan qaysi biri kvadrat tengsizlik ekanini toping A) 4x -5 > 0 B) 41x² – 9x + 1 < 0 C) x⁴-4< 0 D) 2x³ + 3x²-4x +7 > 0 19 / 25 Parallelogrammning tomonlari 7√3 va 6 sm. Agar uning o‘tmas burchagi 120° bo‘lsa, uning yuzini toping A) 68 sm² B) 63 sm² C) 78 sm² D) 70 sm² 20 / 25 Tomonlari 6 sm va 7 sm bo‘lgan to‘g‘ri to‘rtburchakka tashqi chizilgan doira yuzini toping A) 21,25π sm² B) 1,25π sm² C) 21π sm² D) 21,2π sm² 21 / 25 701 soni 1, 8, 15, 22, progressiyaning nechanchi nomerli hadi? A) 100 B) 99 C) 102 D) 101 22 / 25 Geometrik progressiyada q=√3/2, b1=c-√3 bo'lsa, S ni toping? A) 3 B) 2 C) 2√3 /3 D) 2-√3 23 / 25 Arifmetik progressiyada =44, =20 bo'lsa ni toping? A) 502 B) 335 C) 334 D) 495 24 / 25 Arifmetik progressiyada = d=1,8 bo'lsa, qaysi nomerli haddan boshlab progressiyaning barcha hadlari musbat bo'ladi? A) 18 B) 15 C) 13 D) 12 25 / 25 1002, 999, 996, progressiyaning nechanchi nomerli hadidan boshlab uning hadlari manfiy sonlar bo'ladi? A) 336 B) 334 C) 335 D) 337 0% Testni qayta ishga tushiring Baholash mezoni 86%-100% 5 baho 71%-85% 4 baho 56%-70% 3 baho 55% va kamiga 2 baho Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz