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Елленаа
Елленаа
04.08.2020 16:07 •  Алгебра

Приведите уравнение к виду ax^2+bx+c=0


Приведите уравнение к виду ax^2+bx+c=0

👇
Ответ:
ghcfgnyhvjt
ghcfgnyhvjt
04.08.2020

3) (2 - 3х)(5х - 3) - х(2 - х) = 3 - 12х²,

10х - 6 - 15х² + 9х - 2х + х² - 3 + 12х² = 0,

-2х² + 17х - 9 = 0,

2х² - 17х + 9 = 0,

a = 2, b = -17, c = 9;

4) (1 - 2x)(2x - 4) - 3(2 - x) = 3 - 9x²,

2x - 4 - 4x² + 8x - 6 + 3x - 3 + 9x² = 0,

5x² + 13x - 13 = 0,

a = 5, b = 13, c = -13;

5) (5 + 2x)(4x - 1) - 2(2 + 3x) = -13x²,

20x - 5 + 8x² - 2x - 4 - 6x + 13x² = 0,

21x² + 12x - 9 = 0,

7x² + 4x - 3 = 0,

a = 7, b = 4, c = -3;

6) (2 - 6x)(x - 4) - 3x(1 - x) = -22x²,

2x - 8 - 6x² + 24x - 3x + 3x² + 22x² = 0,

19x² + 23x - 8 = 0,

a = 19, b = 23, c = -8.

4,5(49 оценок)
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Ответ:
willzymustdie
willzymustdie
04.08.2020

26,

т.к. по условию в графу ответа надо писать

l / \sqrt{\pi}

Объяснение:

Из условия ни разу не ясно, что есть такое некая непонятная "его длина".

Но по всей видимости,

а) это диаметр условной окружности, которую образует Кольцевая линия.

б) это (ну, блин, грамотеи!) длина окружности, которую образует Кольцевая линия.

а) Найдем диаметр условной окружности, которую образует Кольцевая линия.

Обозначим её как d.

Площадь Центрального района S можно вычислить следующим образом:

S = \pi r^2

где r - это радиус условной окружности Кольцевой, или половина диаметра, т.е. d/2. Отсюда.

S = \pi (d/2)^2 \: = \frac{\pi d {}^{2} }{4} = \\ = d {}^{2} = \frac{ 4S}{\pi} \: \: = d = \sqrt{\frac{ 4S}{\pi}} = 2{\frac{\sqrt{S}}{\sqrt{\pi}}} \\ d = 2 \frac{ \sqrt{169} }{\sqrt{\pi} } \: = 2 \times \frac{ 13 }{\sqrt{\pi} } = 26 / \sqrt{\pi}

б) Найдем длину окружности, которую образует Кольцевая линия. Обозначим её как l.

Длина окружности равна

l = \pi d

где d - условный диаметр (см. (а)).

l = \pi \times 26 / \sqrt{\pi}

l = 26 \times ( \pi / \sqrt{\pi})

l = 26 \sqrt{\pi}

Согласно требованиям задачи в ответ записываем

l = 26 (\sqrt{\pi} / \sqrt{\pi}) = 26

т.е.

ответ: 26

4,6(50 оценок)
Ответ:
fhctybq123
fhctybq123
04.08.2020
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4,4(31 оценок)
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