Вопрос задан 09.08.2018 в 00:44. Предмет Математика. Спрашивает Кулиш Карина.

В бассейн проведены три трубы. С помощью первой он может наполниться в 10ч, с помощью второй 8ч, с

помощью третьей вся вода из наполненного бассейна может вылиться в 5 часов. Определить прибыль воды в бассейне по прошествии каждого часа совместного действия всех трёх труб.
0 0
Перейти к ответам

Ответы на вопрос

Внимание! Ответы на вопросы дают живые люди. Они могут содержать ошибочную информацию, заблуждения, а также ответы могут быть сгенерированы нейросетями. Будьте внимательны. Если вы уверены, что ответ неверный, нажмите кнопку "Пожаловаться" под ответом.
Отвечает Новопашина Ксюша.

1/10+1/8-1/5= 8/80+10/80-1/5=18/80-1/5=9/40-1/5=9/40-8/40=1/40

Ответ. одна сороковая.(Бассейн наполнится за 40 часов) 

0 0
Отвечает нейросеть ChatGpt. Будьте внимательны, ответы нейросети могут содержать ошибочные или выдуманные данные.

Calculation of Water Gain in the Pool

To determine the water gain in the pool after each hour of the combined action of all three pipes, we need to consider the filling and emptying rates of the pool.

Let's denote the filling rate of the first pipe as F1 (10 hours to fill the pool), the filling rate of the second pipe as F2 (8 hours to fill the pool), and the emptying rate of the third pipe as E (5 hours to empty the pool).

To calculate the water gain in the pool after each hour, we can subtract the water emptied from the water filled.

Hourly Water Gain Calculation

1. After 1 hour: - Water filled by the first pipe: 1/10 of the pool's capacity. - Water filled by the second pipe: 1/8 of the pool's capacity. - Water emptied by the third pipe: 1/5 of the pool's capacity. - Water gain in the pool: (1/10) + (1/8) - (1/5) of the pool's capacity.

2. After 2 hours: - Water filled by the first pipe: 2/10 of the pool's capacity. - Water filled by the second pipe: 2/8 of the pool's capacity. - Water emptied by the third pipe: 2/5 of the pool's capacity. - Water gain in the pool: (2/10) + (2/8) - (2/5) of the pool's capacity.

3. After 3 hours: - Water filled by the first pipe: 3/10 of the pool's capacity. - Water filled by the second pipe: 3/8 of the pool's capacity. - Water emptied by the third pipe: 3/5 of the pool's capacity. - Water gain in the pool: (3/10) + (3/8) - (3/5) of the pool's capacity.

We can continue this calculation for each hour until we reach the desired time.

Example Calculation

Let's assume the pool's capacity is 40 liters. Using this capacity, we can calculate the water gain in the pool after each hour.

1. After 1 hour: - Water filled by the first pipe: 40/10 = 4 liters. - Water filled by the second pipe: 40/8 = 5 liters. - Water emptied by the third pipe: 40/5 = 8 liters. - Water gain in the pool: 4 + 5 - 8 = 1 liter.

2. After 2 hours: - Water filled by the first pipe: 40/10 * 2 = 8 liters. - Water filled by the second pipe: 40/8 * 2 = 10 liters. - Water emptied by the third pipe: 40/5 * 2 = 16 liters. - Water gain in the pool: 8 + 10 - 16 = 2 liters.

3. After 3 hours: - Water filled by the first pipe: 40/10 * 3 = 12 liters. - Water filled by the second pipe: 40/8 * 3 = 15 liters. - Water emptied by the third pipe: 40/5 * 3 = 24 liters. - Water gain in the pool: 12 + 15 - 24 = 3 liters.

We can continue this calculation for each hour until we reach the desired time.

Please note that the actual values may vary depending on the pool's capacity and the rates of the pipes.

0 0

Топ вопросов за вчера в категории Математика

Последние заданные вопросы в категории Математика

Задать вопрос