Как построить график уравнения х+у-2=0

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Объяснение:

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1) Функция убывает там, где производная отрицательнаy ' = 6x^2 - 18x - 24 = 6(x^2 - 3x - 4) = 6(x + 1)(x - 4) < 0x ∈ (-1; 4)
2) 
По теореме косинусов

AB = 10
3) Если пар-пед описан около цилиндра, то у него в основании квадрат со стороной, равной диаметру цилиндра a = 2R = 8.Высота равна высоте цилиндра H = 5.V = a^2*H = 8*8*5 = 320 куб.см.
4) Область определения логарифмаx^2 - 14x > 0x(x - 14) > 0x ∈ (-oo; 0) U (14; +oo)Основание логарифма 0 < 1/2 < 1, поэтому функция убывает.

x^2 - 14x - 32 <= 0(x + 2)(x - 16) <= 0x ∈ [-2; 16]С учетом области определенияx ∈ [-2; 0) U (14; 16]
5) 
1 уравнение возводим в квадрат
Подставляем 2 уравнение в 1 уравнение

y = 3x; подставляем в 1 уравнение
Умножаем все на 3x3x^2 - 2x - 1 = 0(x - 1)(3x + 1) = 0x1 = 1; y1 = 3x2 = -1/3; y2 = -1