mashamaxi
08.08.2021 22:31

Решить задачи Коши

x3 y'=2, f(1)=2

x2 y'=y, f(1)=1

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Ответ:
ЛераКотМи
20.02.2022 12:26

gprs is the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the

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gprs is the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are 6374 and the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the two6th is the first of the people who are not in the first of the people in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are not in the first of the people who are

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Ответ:
Dima121007
23.12.2021 21:57

а —скорость первого пешехода в км/час

b —скорость второго пешехода в км/час

t —время в пути до встречи (для обоих пешеходов оно одинаковое)))

тогда

до встречи первый часть пути =(a*t) км

до встречи второй часть пути =(b*t) км

после встречи первый оставшуюся ему часть пути за 4 часа

b * t / a = 4 отсюда: t = 4 * a / b

после встречи второй оставшуюся ему часть пути за 9 часов

a * t / b = 9

a*4*a / b² = 9

a / b = 3 / 2

t = 4*3/2 = 2*3 = 6

ответ: первый был в пути 4+6 = 10 часов

второй был в пути 9+6 = 15 часов

6 часов они шли до встречи

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