Question

1)(-3/7 × 1/4) × (-4) 2)(-4 1/5 × 1,3) × 10/13 Плизз срочно 10 баллов

Answer 1

Ответ:

вот

Пошаговое объяснение:

вроде правильно виполнено все на фото

Answer 2

1) To solve the expression (-3/7 * 1/4) * (-4), we can follow the order of operations, which is parentheses, multiplication, and then addition or subtraction. Let's break it down step by step:

First, let's simplify the expression inside the parentheses: -3/7 * 1/4. To multiply fractions, we multiply the numerators together and the denominators together. So, (-3/7 * 1/4) becomes (-3 * 1) / (7 * 4) = -3/28.

Now, we have (-3/28) * (-4). To multiply a fraction by a whole number, we multiply the numerator by the whole number and keep the denominator the same. So, (-3/28) * (-4) becomes (-3 * -4) / 28 = 12/28.

To simplify the fraction 12/28, we can find the greatest common divisor (GCD) of the numerator and denominator, which is 4. Dividing both the numerator and denominator by 4, we get 3/7.

Therefore, the value of the expression (-3/7 * 1/4) * (-4) is 3/7.

2) Now let's solve the expression (-4 + 1/5 * 1.3) * 10/13:

First, let's simplify the expression inside the parentheses: 1/5 * 1.3. To multiply a fraction by a decimal, we multiply the numerator by the decimal. So, (1/5 * 1.3) becomes 1/5 * 1.3 = 1.3/5.

Now, we have (-4 + 1.3/5) * 10/13. To add a whole number and a decimal, we convert the whole number to a fraction with the same denominator as the decimal. So, -4 becomes -4/1. Now we have (-4/1 + 1.3/5) * 10/13.

To add fractions, we need to have a common denominator. The common denominator for 1 and 5 is 5. So, (-4/1 + 1.3/5) becomes (-20/5 + 1.3/5) = (-20 + 1.3) / 5 = -18.7/5.

Now, we have (-18.7/5) * 10/13. To multiply decimals, we multiply the numerators together and the denominators together. So, (-18.7/5 * 10/13) becomes (-18.7 * 10) / (5 * 13) = -187/65.

Therefore, the value of the expression (-4 + 1/5 * 1.3) * 10/13 is -187/65.

Please note that the above calculations are based on the given expressions and the order of operations.