Вопрос задан 27.09.2023 в 02:25. Предмет Математика. Спрашивает Динкова Виолетта.

{14-4х больше или равно 3(2-х) {14+(х+1) меньше или равно 8хмало времени!​

Смотреть ответ sabirzelena sabirzelena Ответ: Пошаговое объяснение: Решить систему уравнений. {14-4х≥3(2-х) {14+х+1≤8х {14-4х≥6-3х {14+х+1≤8х {-4х+3х≥6-14 {х-8х≤-14 {-х≥-8 {-7х≤-14 х≤8 х≥2 2≤х≤8 х∈[2; 8] Новые вопросы в Математика В числе 234* замените * цифрой так, чтобы полученное число: 1. делилось на 5, но не делилось на 10; 2. делилось на 6; 3. не делилось ни на 2, ни на 5 … ; 4. делилось на 2 и на 5. из всех чисел х, удовлетворяющих неравенству 1420&lt;х&lt;1435, выберите числа которые: 1. делятся на 3; 2. делятся на 9; 3. делятся на 9 и на 2; … 4. не делятся ни на 2, ни на 3 цена предмета была 80 долларов и стала 128 долларов. а) на сколько процентов подорожал этот предмет? б), на сколько процентов старая цена меньше новой … цены?​ помогите пожалуйста задание на фото Числа якої з наведених пар є взаємно простими? А 12 і 21 Б 245 і 50 В 14 і 51 Г 54 і 111 Допоможіть будь ласка! ^ ^ ( • - • )( \/ \/)— —​ Предыдущий Следующий Вместе мы знаем больше Этот сайт использует cookies Политика Cookies . Вы можете указать условия хранения и доступ к cookies в своем браузере. 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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 нейросеть ChatGpt. Будьте внимательны, ответы нейросети могут содержать ошибочные или выдуманные данные.

Давайте разберемся с данным решением системы неравенств:

У нас есть два неравенства:

  1. 144x3(2x)14 - 4x \geq 3(2 - x)
  2. 14+(x+1)8x14 + (x + 1) \leq 8x

Давайте пошагово решим их:

  1. Раскроем скобки в первом неравенстве:

    144x63x14 - 4x \geq 6 - 3x

  2. Теперь выразим xx на одной стороне:

    4x+3x614-4x + 3x \geq 6 - 14

    x8-x \geq -8

  3. Теперь умножим обе стороны на -1, но помним, что при умножении на отрицательное число, меняется направление неравенства:

    x8x \leq 8

Теперь переходим ко второму неравенству:

  1. Раскроем скобки во втором неравенстве:

    14+x+18x14 + x + 1 \leq 8x

    15+x8x15 + x \leq 8x

  2. Теперь выразим xx на одной стороне:

    x8x15x - 8x \geq -15

    7x15-7x \geq -15

  3. Теперь поделим обе стороны на -7, но помним, что при делении на отрицательное число, меняется направление неравенства:

    x157x \leq \frac{-15}{-7}

    x157x \leq \frac{15}{7}

Теперь у нас есть два неравенства:

  1. x8x \leq 8
  2. x157x \leq \frac{15}{7}

Чтобы найти общее решение системы, возьмем пересечение решений этих двух неравенств:

2x82 \leq x \leq 8

Итак, общее решение системы неравенств - это интервал 2x82 \leq x \leq 8, что можно записать как x[2,8]x \in [2, 8].

Теперь ответы на ваши другие вопросы:

  1. Для того чтобы найти число, которое делится на 5, но не делится на 10, можно взять число 235.

  2. Чтобы число делилось на 6, оно должно быть четным и делиться на 3. Из чисел от 2 до 8 это только 6.

  3. Чтобы число не делилось ни на 2, ни на 5, нужно выбрать число 7.

  4. Чтобы число делилось и на 2, и на 5, нужно выбрать число 10.

  5. Для чисел от 1421 до 1434:

    1. Делятся на 3: 1422, 1425, 1428, 1431, 1434.
    2. Делятся на 9: 1422, 1431.
    3. Делятся и на 9, и на 2: 1422.
    4. Не делятся ни на 2, ни на 3: 1423, 1424, 1426, 1427, 1429, 1430, 1432, 1433.
  6. Для определения процента подорожания предмета, используйте следующую формулу:

    Процент подорожания=Новая ценаСтарая ценаСтарая цена×100%\text{Процент подорожания} = \frac{\text{Новая цена} - \text{Старая цена}}{\text{Старая цена}} \times 100\%

    а) Процент подорожания=1288080×100%=4880×100%=60%\text{Процент подорожания} = \frac{128 - 80}{80} \times 100\% = \frac{48}{80} \times 100\% = 60\%

    б) Процент старой цены меньше новой=80128128×100%=48128×100%=37.5%\text{Процент старой цены меньше новой} = \frac{80 - 128}{128} \times 100\% = \frac{-48}{128} \times 100\% = -37.5\%

Если у вас есть еще какие-либо вопросы или нужна дополнительная помощь, пожалуйста, дайте знать!

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