Question

ПОЖАЛУЙСТА ОЧЕНЬ СРОЧНО!!!! На завтраМожно ли чтобы:а. числа 276; 715 были членами последовательности 3,6,9,12...?б. числа 819; 826 были членами последовательности 9,18,27,36...?Ответ обоснуй ​

Answer 1

Ответ:

а) все числа кратные трем

{3,6,9,12,.......,276,279,♾} 715 не кратно 3

б) все числа кратные девяти

{9,18,27,36,......,819,♾} 826 не кратно 9

Answer 2

Can the numbers 276 and 715 be members of the sequence 3, 6, 9, 12...?

To determine if the numbers 276 and 715 can be members of the sequence 3, 6, 9, 12..., we need to check if they follow the pattern of the sequence.

The given sequence has a common difference of 3, meaning that each term is obtained by adding 3 to the previous term.

To check if 276 and 715 can be members of this sequence, we can subtract the first term (3) from each number and see if the result is divisible by the common difference (3).

Let's do the calculations:

- For 276: 276 - 3 = 273. Since 273 is not divisible by 3, 276 is not a member of the sequence 3, 6, 9, 12....

- For 715: 715 - 3 = 712. Since 712 is not divisible by 3, 715 is not a member of the sequence 3, 6, 9, 12....

Therefore, neither 276 nor 715 can be members of the sequence 3, 6, 9, 12....