Вопрос задан 19.07.2023 в 21:47. Предмет Алгебра. Спрашивает Фомина Ксюша.

{3х-у=7 {2х+3у=1 системные уравнения​

Смотреть ответ lapsinaoleksandra lapsinaoleksandra Ответ: x=2; y= -1: Объяснение:{3x-y=7 {2x+3y=1; { y= -7+3x {2x+3у=1; 2x+3(-7+3х)=1; x=2; y= -7+3×2; у=1; Новые вопросы в Алгебра ПОМОГИТЕ ПОЖАЛУЙСТА!!! алгебра 7 класс! фото прикреплю, задание не очень сложное, но я не могу решит) Шахматист играет не менее одной партии в день и не более 12 за любую неделю (неде-ля начинается в понедельник и заканчивается в воскресенье). Докажите … , что найдется несколько подряд идущих дней, за которые он сыграет ровно 20 партий. морская вода содержит 5% солей.Сколько чистой воды надо добавить в 40 литров морской воды,чтобы процент соли был 2%​ найдите числа M и N, чтобы выполнялось равенство (фото ниже) если не тяжело, то с подробным решением Найдите такие числа M и N, чтобы выполнялось равенство(фото ниже) если не тяжело, то с подробным решением ^@[email protected]^ Предыдущий Следующий Вместе мы знаем больше Этот сайт использует cookies Политика Cookies . 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Отвечает нейросеть ChatGpt. Будьте внимательны, ответы нейросети могут содержать ошибочные или выдуманные данные.

I can help you with the algebraic problem, but unfortunately, as an AI text model, I'm unable to view or process images, so I won't be able to see the photos you mentioned. However, you can describe the problems to me in detail, and I'll do my best to assist you with the solutions.

Let's start with the first problem:

Problem 1: Prove that there exist several consecutive days during which a chess player will play exactly 20 games.

Solution: To prove this, let's consider two cases:

Case 1: The chess player plays exactly 20 games on a single day. In this case, the consecutive days will be that single day itself.

Case 2: The chess player doesn't play exactly 20 games on any single day. In this case, let's assume that the chess player plays "a" games on day 1, "b" games on day 2, "c" games on day 3, and so on, until day 7 (one week). Since the player can play a maximum of 12 games in a week, we have the following inequality:

a + b + c + d + e + f + g ≤ 12

Now, let's consider the total number of games played over two weeks (14 days):

Total games in two weeks = a + b + c + d + e + f + g + a + b + c + d + e + f + g = 2(a + b + c + d + e + f + g)

Since the player can play a minimum of one game each day, the total number of games in two weeks should be at least 14 (one game each day for 14 days).

So, we have:

2(a + b + c + d + e + f + g) ≥ 14

Dividing both sides by 2:

a + b + c + d + e + f + g ≥ 7

Now, if the player plays more than 7 games in two weeks (which is greater than the minimum), it implies that there are some days where the player plays more than one game. We need to find a combination of "a," "b," "c," "d," "e," "f," and "g" that satisfies the condition above and results in exactly 20 games in consecutive days.

It's important to note that there are many possible combinations that will satisfy the condition, and you need to find one such combination to prove that the statement is true.

If you provide the other algebraic problems, I'd be happy to help you solve them. Just describe the problems in detail, and I'll provide step-by-step solutions.

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