
1)Сумма трёх последовательных чётных натуральных чисел равна 2/3части которого равны 44. Найдите
эти числа.2) Сумма трёх последовательных нечетных натуральных чисел равна 3/11 части от 363. Найдите эти числа СРОЧНО ПОМОГИТЕ 

Ответы на вопрос





1) Let's solve the first problem step by step. We need to find three consecutive even natural numbers whose sum is equal to 2/3 of a number which is also equal to 44.
Let's assume the first even number is x. Then the next two consecutive even numbers would be x + 2 and x + 4.
According to the problem, the sum of these three numbers should be equal to 2/3 times a certain number, which is equal to 44.
So, we can set up the equation:
x + (x + 2) + (x + 4) = (2/3) * 44
Simplifying the equation:
3x + 6 = (2/3) * 44
Multiplying both sides of the equation by 3 to get rid of the fraction:
9x + 18 = 88
Subtracting 18 from both sides:
9x = 70
Dividing both sides by 9:
x = 70 / 9
So, the first even number is approximately 7.778.
The second even number is x + 2 = 7.778 + 2 = 9.778.
The third even number is x + 4 = 7.778 + 4 = 11.778.
Therefore, the three consecutive even numbers are approximately 7.778, 9.778, and 11.778.
2) Now let's solve the second problem. We need to find three consecutive odd natural numbers whose sum is equal to 3/11 of a number that is also equal to 363.
Let's assume the first odd number is y. Then the next two consecutive odd numbers would be y + 2 and y + 4.
According to the problem, the sum of these three numbers should be equal to 3/11 times a certain number, which is equal to 363.
So, we can set up the equation:
y + (y + 2) + (y + 4) = (3/11) * 363
Simplifying the equation:
3y + 6 = (3/11) * 363
Multiplying both sides of the equation by 11 to get rid of the fraction:
33y + 66 = 363
Subtracting 66 from both sides:
33y = 297
Dividing both sides by 33:
y = 297 / 33
So, the first odd number is 9.
The second odd number is y + 2 = 9 + 2 = 11.
The third odd number is y + 4 = 9 + 4 = 13.
Therefore, the three consecutive odd numbers are 9, 11, and 13.
I hope this helps! Let me know if you have any further questions.


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