matematika attestatsiya savollari yangi

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1.karim 3minutda 213ta so`zni terib , 6ta imloviy xatoga yo`l qo`ydi. karimning matn terish sifatini toping *0,0282 2.agar a(2;1;3) va b(-1;x;2) vektorlar uzunligi teng bo’lsa ,x ni toping. * 3.funksiya hosilasini toping: f(x)=5 *0 4.markazi koordinatalar boshida , radiusi 5 ga teng bolgan sfera tenglamasini yozing. *x2+y2+z2=25 5. funksiya hosilasini toping: * 6. a(2;0;-3) va b(3;4;0) nuqtalar orasidagi masofani toping. * 7.koordinatalar boshidan y=x2-4x+3 parabolaning simmetriya o’qigacha bo’lgan masofani toping. *2 8. a̅(0;-4;2) va b̅(2;2;3) vektorlarning skalyar ko’paytmasini toping. *10 9. agar b(-2;-7) nuqta y=kx2+8x +m parabolaning uchi bo’lsa, k va m ning qiymatini toping. *k=2, m=1 10.p , q mulohazalarning dizyunksiyasi to’g’ri ko’rsatilgan qatorni troping *p˅q 11.yuzi 9 sm2 bo’lgan doirani o’rab turgan aylana uzunligini toping. *6 12.konversiya bilan teng kuchli mulohazani ko’rsating: *inversiya 13.ikkita o`xshash ko`pburchakning yuzlari mos ravishda 64sm2 va 576 sm2 bo`lib , birinchisining peremetri 112 sm bo`lsa ikkinchi ko`pburchak peremetrini toping? * 336 sm 14.parabola …

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issali nuqtada o`tkazilgan urinma tenglamasini tuzing *y=35x+30 funksiyaning statsionar nuqtalarini toping *x=-1, x=2 abc uchburchak turioni aniqlang, uning yuzini toping: a(2;4;-1), b(1-1;1;2), c(5;1;2) *teng yonli , teng yonli , teng tomonli , turli tomonli , 30. fan: matematika m.a. mirzaahmedov, sh.n.ismailov, a.q.amanov qiyinchilik darajasi: 1 funksiyaning kamayish oraliqlarini toping *(-1;0) va (0;1) (-2;0) va (0;1) (-1;0) va (0;2) (-3;0) va (0;1) 31. fan: matematika m.a. mirzaahmedov, sh.n.ismailov, a.q.amanov qiyinchilik darajasi: 1 funksiyaning lokal maksimum va lokal minimumlariga uning … deyiladi *ekstremumlari o`sish oralig`i kamayish oralig`i hosilasi 32.fan: geometriya m.a. mirzaahmedov , sh.n. ismailov. qiyinchilik darajasi 3 uchlari a(4;0;1), b(5;-2;1), c(4;8;5) nuqtalarda bo’lgan uchburchakning al bissektrisasi uzunligini toping * 33. fan: matematika m.a. mirzaahmedov, sh.n.ismailov, a.q.amanov qiyinchilik darajasi: 1 koordinatalar tekisliklari fazoni nechta oktantaga bo`ladi *8 5 4 6 34. fan: matematika m.a. mirzaahmedov, sh.n.ismailov, a.q.amanov qiyinchilik darajasi: 1 fazoda …deb yo`naltirilgan kesmaga aytiladi *vektor burchak kollinear simmetriya 35. fan: matematika …

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ik darajasi: 1 ning taqribiy qiymatini hisoblang *0,484 0,56 0,8 168 40. fan: matematika m.a. mirzaahmedov, sh.n.ismailov, a.q.amanov qiyinchilik darajasi: 1 ildizning qiymatini taqribiy hisoblang *1,01 2 1,2 1,1 41. fan: matematika m.a. mirzaahmedov, sh.n.ismailov, a.q.amanov qiyinchilik darajasi: 1 funksiyaning [-4;2] oraliqdagi eng katta qiymatini toping *17 2 20 32 42.fan: algebra .m.a.mirzaaxmedov, sh.n. ismailov qiyinchilik darajasi 2 tenglamani yeching: *x=1 x=3 x=-1 x=0 43.fan: algebra .m.a.mirzaaxmedov, sh.n. ismailov qiyinchilik darajasi 2 tenglama nechta yechimga ega: =-3 *yechimga ega emas 1 2 4 44.fan: algebra .m.a.mirzaaxmedov, sh.n. ismailov qiyinchilik darajasi 3 tenglamani yeching. *x=2 x=-2 x=-1 x=0 45.fan: algebra .m.a.mirzaaxmedov, sh.n. ismailov qiyinchilik darajasi 2 shar sirt yuzi formulasi *s=4 r2 s=2 r s=4 r s= r2 46.fan: algebra .m.a.mirzaaxmedov, sh.n. ismailov qiyinchilik darajasi 2 oddiy foizlar formulasi * 47.fan: algebra .m.a.mirzaaxmedov, sh.n. ismailov qiyinchilik darajasi 2 murakkab foizlar formulasi. * 48.fan: algebra .m.a.mirzaaxmedov, sh.n. ismailovqiyinchilikdarajasi 1 tengsizlikni yeching: x+1>7-2x …

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to'g'ri chiziqlar c to'g'ri chiziqqa parallel.a va b to’g’ri chiziqlar o’zaro qanday joylashishi mumkin? *parallel perpendikular kesishadi kesishmaydi 56.fan: algebra .m.a.mirzaaxmedov, sh.n. ismailov qiyinchilik darajasi 2 agar f(x)=2x+3 bo’lsa, f(-4) ni toping. *-5 -1 3 8 57.fan: algebra .m.a.mirzaaxmedov, sh.n. ismailov qiyinchilik darajasi 2 funksiya uchun x ning qaday qiymatida g(x) mavjud emas? *x=4 x= - 4 x=2 x=-2 58.fan: algebra .m.a.mirzaaxmedov, sh.n. ismailov qiyinchilik darajasi 1 to’g’ri tasdiqni aniqlang: * fazoda to’g’ri chiziqda yotmagan nuqtadan unga parallel yagona to’g’ri chiziq o’tkazish mumkin; uchinchi to’g’ri chiziqqa parallel to’g’ri chiziqlar o’zaro kesishadi; agar ikki to’g’ri chiziq tekislikda yotsa,ular kesishadi; to’g’ri chiziqdan va unda yotmagan nuqtadan ikkita turli tekislik o’tkazish mumkin; 59.fan: geometriya m.a. mirzaahmedov , sh.n. ismailov. qiyinchilik darajasi 2 k ning qanday qiymatida b̄(k-1;1;4) vektorning uzunligi ga teng bo’ladi. *3;-1 1;3 1;2 -3;1 60.fan: matematika m.a. mirzaahmedov, sh.n.ismailov, a.q.amanov qiyinchilik darajasi: 1 funksiyaning [-4;2] oraliqdagi eng kichik qiymatini toping *-9 …

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f(g(-4)) ni toping *36 0 10 12 67.fan: geometriya m.a. mirzaahmedov , sh.n. ismailov. qiyinchilik darajasi 3 uchlari a(4;0;1), b(5;-2;1), c(4;8;5) nuqtalarda bo’lgan uchburchakning al bissektrisasi uzunligini toping * 68. fan: matematika m.a. mirzaahmedov, sh.n.ismailov, a.q.amanov qiyinchilik darajasi: 1 agar va bo`lsa ,f(f(1)) ni toping *1 10 36 0 69. fan: matematika m.a. mirzaahmedov, sh.n.ismailov, a.q.amanov qiyinchilik darajasi: 1 tgx funksiya hosilasini toping * cosx sinx tgx 70.fan: geometriya m.a. mirzaahmedov , sh.n. ismailov. qiyinchilik darajasi 2 uchta nuqta berilgan : a(1;1;1), b(-1;0;1), c(0;1;1) . shunday d(x;y;z) nuqtani topingki, a̅b̅ va c̅d̅̄ vektorlar teng bo’lsin. *d(-2;0;1) d(2;0;-1) d(0;1;1) d(1;1;1) 71. fan: algebra .m.a.mirzaaxmedov, sh.n. ismailov qiyinchilik darajasi 3 mahsulotning narxi birinchi marta 25%, ikkinchi marta yangi bahosi 20% ga oshirildi.mahsulotning oxirgi bahosi necha % kamaytirilsa, uning narxi dastlabki bahosiga teng bo’ladi? *33 33 33 33 72. fan: algebra .m.a.mirzaaxmedov, sh.n. ismailov qiyinchilik darajasi 3 a(-2;5) nuqtadan 5x-7y-4=0 to’g’ri chiziqqa parallel …

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1.karim 3minutda 213ta so`zni terib , 6ta imloviy xatoga yo`l qo`ydi. karimning matn terish sifatini toping *0,0282 2.agar a(2;1;3) va b(-1;x;2) vektorlar uzunligi teng bo’lsa ,x ni toping. * 3.funksiya hosilasini toping: f(x)=5 *0 4.markazi koordinatalar boshida , radiusi 5 ga teng bolgan sfera tenglamasini yozing. *x2+y2+z2=25 5. funksiya hosilasini toping: * 6. a(2;0;-3) va b(3;4;0) nuqtalar orasidagi masofani toping. * 7.koordinatalar boshidan y=x2-4x+3 parabolaning simmetriya o’qigacha bo’lgan masofani toping. *2 8. a̅(0;-4;2) va b̅(2;2;3) vektorlarning skalyar ko’paytmasini toping. *10 9. agar b(-2;-7) nuqta y=kx2+8x +m parabolaning uchi bo’lsa, k va m ning qiymatini toping. *k=2, m=1 10.p , q mulohazalarning dizyunksiyasi to’g’ri ko’rsatilgan qatorni troping *p˅q 11.yuzi 9 sm2 bo’lg...

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